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Applets:Sampling of Analog Signals and Signal Reconstruction

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{{LntAppletLinkEnDe|sampling_en|sampling}}

== Applet Description==
 
The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  [[Modulation_Methods/Pulscodemodulation|"Puls code modulation"]]  $({\rm PCM})$  for example.   The upper graphic shows the model on which this applet is based.  Below it are the samples  $x(\nu \cdot T_{\rm A})$  of the time continuous signal  $x(t)$. The (infinite) sum over all these samples is called the sampled signal  $x_{\rm A}(t)$.

[[File:EN_Abtastung_1.png|center|frame|Top:    Underlying model for sampling and signal reconstruction Bottom:   Example for time discretization of the continuous–time signal  $x(t)$]]

*At the transmitter, the time discrete (sampled) signal  $x_{\rm A}(t)$  is obtained from the continuous–time signal  $x(t)$.  This process is called  '''sampling'''   or  '''A/D conversion'''.
*The corresponding program parameter for the transmitter is the sampling rate  $f_{\rm A}= 1/T_{\rm A}$.  The lower graphic shows the sampling distance  $T_{\rm A}$ .
*In the receiver, the discrete-time received signal  $y_{\rm A}(t)$  is used to generate the continuous-time sink signal  $y(t)$    ⇒   '''signal reconstruction'''  or  '''D/A conversion'''  corresponding to the receiver frequency response  $H_{\rm E}(f)$.

The applet does not consider the PCM blocks  "Quantization"and  "encoding/decoding".   The digital transmission channel is assumed to be ideal. 

[[File:Abtastung_2_neu.png|right|frame|Receiver frequency response  $H_{\rm E}(f)$]]

The following consequences result from this:
*In the program simplifying  $y_{\rm A}(t) = x_{\rm A}(t)$  is set.
* With suitable system parameters, the error signal   $\varepsilon(t) = y(t)-x(t)\equiv 0$  is therefore also possible.

The sampling theorem and the signal reconstruction can be better explained in the frequency domain.  Therefore all spectral functions are displayed in the program;

             $X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t)$,  $X_{\rm A}(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x_{\rm A}(t)$,  $Y(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ y(t)$,  $E(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ \varepsilon(t).$ 

Parameters for the receiver frequency response  $H_{\rm E}(f)$  are the cut–off frequency and the rolloff factor  (see lower graph):
:$$f_{\rm G} = \frac{f_2 +f_1}{2},\hspace{1cm}r = \frac{f_2 -f_1}{f_2 +f_1}.$$

''Notes:''

'''(1)'''   All signal values are normalized to  $\pm 1$.

'''(2)'''   The power calculation is done by integration over the respective period duration  $T_0$:
:$$P_x = \frac{1}{T_0} \cdot \int_0^{T_0} x^2(t)\ {\rm d}t,\hspace{0.8cm}P_\varepsilon = \frac{1}{T_0} \cdot \int_0^{T_0} \varepsilon^2(t).$$

'''(3)'''   The signal power  $P_x$  and the distortion power  $P_\varepsilon$  are also output in normalized form, which implicitly assumes the reference resistance  $R = 1\, \rm \Omega$ ;

'''(4)'''   From these the signal–distortion–distance  $10 \cdot \lg \ (P_x/P_\varepsilon)$  can be calculated.

'''(5)'''   Does the spectral function  $X(f)$  for positive frequencies consists of  $I$  Dirac delta lines with the (possibly complex) weights  $X_1$, ... , $X_I$,           so applies to the transmission power taking into account the mirror-image lines at the negative frequencies:

:$$P_x = 2 \cdot \sum_{i=1}^I |X_k|^2.$$

'''(6)'''   Correspondingly, the following applies to the distortion power if the spectral function  $E(f)$  in the range  $f>0$  has  $J$  Dirac delta lines with weights  $E_1$, ... , $E_J$:

:$$P_\varepsilon = 2 \cdot \sum_{j=1}^J |E_j|^2.$$

==Theoretical Background==

===Description of sampling in the time domain===

[[File:P_ID1120__Sig_T_5_1_S1_neu.png|center|frame|For the time discretization of the continuous-time signal  $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be  $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
*Out of the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) \equiv 0$.
*The run variable  $\nu$  be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:     $\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} $.
*In contrast, at the equidistant sampling times with the constant  $K$, the result is:

:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$

The constant depends on the type of time discretization. For the above sketch $K = 1$ is valid.
 
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.  The following pages will show that these equations, which take some getting used to, do lead to useful results if they are applied consistently.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

* By  '''sampling'''  we mean here the multiplication of the time-continuous signal  $x(t)$  by a  '''Dirac delta pulse''':

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$

*The  '''Dirac delta pulse (in the time domain)'''  consists of infinitely many Dirac delta pulses, each equally spaced  $T_{\rm A}$  and all with equal pulse weight  $T_{\rm A}$:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot\delta(t- \nu \cdot T_{\rm A})\hspace{0.05cm}.$$}}

Based on this definition, the following properties result for the sampled signal:
:$$x_{\rm A}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot\delta (t- \nu \cdot T_{\rm A})\hspace{0.05cm}.$$

*The sampled signal at the considered time  $(\nu \cdot T_{\rm A})$  ist gleich  $T_{\rm A} \cdot x(\nu \cdot T_{\rm A}) · \delta (0)$.
*Since  $\delta (t)$  at time  $t = 0$  is infinite, actually all signal values  $x_{\rm A}(\nu \cdot T_{\rm A})$  are also infinite and also the factor  $K$ introduced above.
*Two samples  $x_{\rm A}(\nu_1 \cdot T_{\rm A})$  and  $x_{\rm A}(\nu_2 \cdot T_{\rm A})$  however, differ in the same proportion as the signal values  $x(\nu_1 \cdot T_{\rm A})$  and  $x(\nu_2 \cdot T_{\rm A})$.
*The samples of  $x(t)$  appear in the pulse weights of the Dirac delta functions:
*The additional multiplication by  $T_{\rm A}$  is necessary so that  $x(t)$  and  $x_{\rm A}(t)$  have the same unit.  Note here that  $\delta (t)$  itself has the unit "1/s".

===Description of sampling in the frequency domain===

The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain|"Convolution Theorem".]] This states that multiplication in the time domain corresponds to convolution in the spectral domain:

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

If one develops the  Dirac delta pulse  $p_{\delta}(t)$   (in the time domain)   into a  [[Signal_Representation/Fourier_Series|"Fourier Series"]]  and transforms it using the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Shifting Theorem"]]  into the frequency domain, the following correspondence   ⇒   [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"proof"]] results with the distance  $f_{\rm A} = 1/T_{\rm A}$  of two adjacent dirac delta lines in the frequency domain:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot\delta(t- \nu \cdot T_{\rm A})\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f) = \sum_{\mu = - \infty }^{+\infty} \delta(f- \mu \cdot f_{\rm A} ).$$

[[File:P_ID1121__Sig_T_5_1_S3_NEU.png|right|frame|Dirac delta pulse in time and frequency domain with  $T_{\rm A} = 50\ {\rm µs}$  und  $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
The result states:
*The Dirac delta pulse  $p_{\delta}(t)$  in the time domain consists of infinitely many Dirac delta pulses, each at the same distance  $T_{\rm A}$  and all with the same pulse weight  $T_{\rm A}$.
*The Fourier transform of  $p_{\delta}(t)$  again gives a Dirac delta pulse, but now in the frequency domain   ⇒   $P_{\delta}(f)$.
*Also  $P_{\delta}(f)$  consists of infinitely many Dirac delta pulses, now in the respective spacing  $f_{\rm A} = 1/T_{\rm A}$  and all with pulse weight  $1$.
*The distances of the Dirac delta lines in time and frequency domain thus follow the  [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity Theorem"]]:   $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$

From this follows:   From the spectrum  $X(f)$  is obtained by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$ :

:$$X(f) \star \delta(f- \mu \cdot f_{\rm A})= X (f- \mu \cdot f_{\rm A})\hspace{0.05cm}.$$

Applying this result to all Dirac delta lines of the Dirac delta pulse, we finally obtain:

:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta(f- \mu \cdot f_{\rm A}) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A})\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Sampling the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  results in the spectral domain in a  '''periodic continuation'''  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$.}}

[[File:P_ID1122__Sig_T_5_1_S4_neu.png|right|frame|Spectrum of the sampled signal]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The upper graph shows  '''(schematic!)'''  the spectrum  $X(f)$  of an analog signal  $x(t)$, which contains frequencies up to  $5 \text{ kHz}$ .

Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$, i.e., at the respective spacing  $T_{\rm A}\, = {\rm 50 \, µs}$  yields the periodic spectrum  $X_{\rm A}(f)$ sketched below.
*Since the Dirac delta functions are infinitely narrow, the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high frequency components.
*Correspondingly, the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity.}}

===Signal reconstruction===

[[File:EN_Sig_T_5_1_S2_v2.png|right|frame|Joint model of "signal sampling" and "signal reconstruction"]]
Signal sampling is not an end in itself in a digital transmission system, but it must be reversed at some point  For example, consider the following system:
*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.
*At the output of an ideal transmission system, the also discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.
*The question now is how the block   '''signal reconstruction'''   has to be designed so that also  $y(t) = x(t)$  holds.

[[File:P_ID1124__Sig_T_5_1_S5b_neu.png|right|frame|Frequency domain representation of the "signal reconstruction"]]
 The solution is simple if you look at the spectral functions:  

One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass filter with the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_. E2.80.93_Transfer_function|"Frequency response"]]  $H_{\rm E}(f)$, which 

*passes the low frequencies unaltered:
:$$H_{\rm E}(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \le B_{\rm NF}\hspace{0.05cm},$$
*completely suppresses the high frequencies:
:$$H_{\rm E}(f) = 0 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \ge f_{\rm A} - B_{\rm NF}\hspace{0.05cm}.$$

Further, it can be seen from the accompanying graph:   As long as the above two conditions are satisfied,  $H_{\rm E}(f)$  can be arbitrarily shaped in the range from  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$  ,
*for example linearly descending (dashed line)
*or also rectangular.

===The Sampling Theorem===

The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  corresponding to the bandwidth  $B_{\rm NF}$  of the message signal has been chosen correctly.

From the above graph, it can be seen that the following condition must be satisfied:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

{{BlaueBox|TEXT=
$\text{Sampling theorem:}$  If an analog signal  $x(t)$  has only spectral components in the range  $\vert f \vert < B_{\rm NF}$, it can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  only if the sampling rate is sufficiently large:
:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$

Accordingly, the following must apply to the distance between two samples:

:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}

If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,
*so, for signal reconstruction of the analog signal from its samples.
*an ideal, rectangular low-pass filter with cut off frequency  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$  must be used.

{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows above the spectrum  $\pm\text{ 5 kHz}$  of an analog signal limited to  $X(f)$  below the spectrum  $X_{\rm A}(f)$  of the signal sampled at distance  $T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.
[[File:P_ID1125__Sig_T_5_1_S6_neu.png|right|frame|Sampling theorem in the frequency domain]]
Additionally drawn is the frequency response  $H_{\rm E}(f)$  of the low-pass receiving filter for signal reconstruction, whose cutoff frequency must be exactly  $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$ .

*With any other  $f_{\rm G}$ value, there would be  $Y(f) \neq X(f)$.
*For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ portions are missing.
* At  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$ due to convolution products.

If at the transmitter the sampling had been done with a sampling rate  $f_{\rm A} < 10\,\text{ kHz}$    ⇒   $T_{\rm A} >100 \ {\rm µ s}$, the analog signal  $y(t) = x(t)$  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case. }}
 
==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*All signal values are to be understood as normalized to  $\pm 1$.  Powers are normalized values, too.

{{BlaueBox|TEXT=
'''(1)'''  Source signal:  $x(t) = A \cdot \cos (2\pi \cdot f_0 \cdot t -\varphi)$  with  $f_0 = \text{4 kHz}$.   Sampling with  $f_{\rm A} = \text{10 kHz}$.  Rectanglular low-pass;  cut off frequency:  $f_{\rm G} = \text{5 kHz}$.             Interpret the shown graphics and evaluate the present signal reconstruction for all permitted parameter values of $A$  and $\varphi$. }}

* The spectrum  $X(f)$  consists of two Dirac functions at  $\pm \text{4 kHz}$, each with impulse weight  $0.5$.
* By the periodic continuation  $X_{\rm A}(f)$  has lines of equal height at  $\pm \text{4 kHz}$,  $\pm \text{6 kHz}$,  $\pm \text{14 kHz}$,  $\pm \text{16 kHz}$,  $\pm \text{24 kHz}$,  $\pm \text{26 kHz}$,  etc.
* The rectangular low-pass with the cut off frequency  $f_{\rm G} = \text{5 kHz}$  removes all lines except the two at  $\pm \text{4 kHz}$  ⇒  $Y(f) =X(f)$  ⇒  $y(t) =x(t)$  ⇒   $P_\varepsilon = 0$.
* The signal reconstruction works here perfectly  $(P_\varepsilon = 0)$  for all amplitudes $A$  and any phases $\varphi$.

{{BlaueBox|TEXT=
'''(2)'''  Continue with  $A=1$,  $f_0 = \text{4 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$.   What is the influence of the rolloff–factors  $r=0.2$,  $r=0.5$  and   $r=1$?           Specify the power values  $P_x$  and  $P_\varepsilon$ .   For which  $r$–values is  $P_\varepsilon= 0$?  Do these results also apply to other  $A$  and  $\varphi$? }}

:* With  $|X(f = \pm \text{4 kHz})|=0.5$  the signal power is  $P_x = 2\cdot 0.5^2 = 0.5$.  The distortion power  $P_\varepsilon$  depends significantly on the rolloff–factor  $r$ .
:* $P_\varepsilon$  is zero for  $r \le 0.2$.  Then the  $X_{\rm A}(f)$ line at  $f_0 = \text{4 kHz}$  is not changed by the low-pass and the unwanted  line at  $\text{6 kHz}$  is fully suppressed.
:* $r = 0.5$ :  $Y(f = \text{4 kHz}) = 0.35$,  $Y(f = \text{6 kHz}) = 0.15$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 15$  ⇒  $P_\varepsilon = 0.09$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=7.45\ \rm dB$.
:*$r = 1.0$ :  $Y(f = \text{4 kHz}) = 0.3$,  $Y(f = \text{6 kHz}) = 0.2$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 2$  ⇒  $P_\varepsilon = 0.16$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=4.95\ \rm dB$.
:* For all  $r$  the distortion power $P_\varepsilon$  is independent of  $\varphi$.   The amplitude  $A$  affects  $P_x$  and  $P_\varepsilon$  in the same way   ⇒   the quotient is independent of  $A$.

{{BlaueBox|TEXT=
'''(3)'''  Now apply  $A=1$,  $f_0 = \text{5 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$,  $r=0$  $($rectangular low–pass$)$.   Interpret the result of the signal reconstruction.}}

:*  $X(f)$  consists of two Dirac delta lines at  $\pm \text{5 kHz}$  $($weight  $0.5)$.  By periodic continuation  $X_{\rm A}(f)$  has lines at  $\pm \text{5 kHz}$,  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$,  etc.
:*  The  rectanglular low-pass  removes the lines at  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$.  The lines at  $\pm \text{5 kHz}$  are halved because of  $H_{\rm E}(\pm f_{\rm G}) = H_{\rm E}(\pm \text{5 kHz}) = 0.5$.
:*   ⇒   $\text{Weights of }X(f = \pm \text{5 kHz})$:  $0.5$   #   $\text{Weights of }X(f_{\rm A} = \pm \text{5 kHz})$:  $1. 0$;     #   $\text{Weights of }Y(f = \pm \text{5 kHz})$:  $0.5$   ⇒   $Y(f)=X(f)$.
:* So the signal reconstruction works perfectly here too  $(P_\varepsilon = 0)$.  The same is true for the phase  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

{{BlaueBox|TEXT=
'''(4)'''  The settings of  $(3)$  continue to apply except for  $\varphi=30^\circ$.  Interpret the differences from the setting  $(3)$   ⇒   $\varphi=0^\circ$.}}

:* Phase relations are lost.  The sink signal  $y(t)$  is cosine-shaped  $(\varphi_y=0^\circ)$  with by the factor  $\cos(\varphi_x)$  smaller amplitude than the source signal  $x(t)$.
:* Justification in the frequency domain:  In the periodic continuation of  $X(f)$  ⇒  $X_{\rm A}(f)$  only the real parts are to be added.  The imaginary parts cancel out.
:* The Dirac delta line of  $X(f)$  at frequency  $f_0$   ⇒   $X(f_0)$  is complex,   $Y(f_0)$  is real, and  $E(f_0)$  is imaginary   ⇒   $\varepsilon(t)$  is minus–sinusoidal   ⇒   $P_\varepsilon = 0. 125$.

{{BlueBox|TEXT=
'''(5)'''  Illustrate again the result of  $(4)$  compared to the settings  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* With this setting, the spectrum  $X_{\rm A}(f)$  also has a positive imaginary part at  $\text{5 kHz}$  and a negative imaginary part of the same magnitude at  $\text{6 kHz}$.
:* The rectangular low-pass with cutoff frequency  $\text{5.5 kHz}$  removes this second component.  Thus, with the new setting  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Any $f_0$ oscillation of arbitrary phase is error-free reconstructible from its samples if  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($any small $\mu>0)$.
:* For value–continuous spectrum with   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$no diraclines at $\pm f_0 \big ]$  the sampling rate  $f_{\rm A} = 2 \cdot f_{\rm 0}$  is sufficient in principle.

{{BlaueBox|TEXT=
'''(5)'''  Verdeutlichen Sie sich nochmals das Ergebnis von  '''(4)'''  im Vergleich zu den Einstellungen  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* Bei dieser Einstellung hat das  $X_{\rm A}(f)$–Spektrum auch einen positiven Imaginärteil bei  $\text{5 kHz}$  und einen negativen Imaginärteil gleicher Höhe bei  $\text{6 kHz}$.
:* Der Rechteck–Tiefpass mit der Grenzfrequenz  $\text{5.5 kHz}$  entfernt diesen zweiten Anteil.  Somit ist bei dieser Einstellung  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Jede  $f_0$–Schwingung beliebiger Phase ist fehlerfrei aus seinen Abtastwerten rekonstruierbar, falls  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($beliebig kleines $\mu>0)$.
:* Bei wertkontinuierlichem Spektrum mit   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$keine Diraclinien bei $\pm f_0 \big ]$ genügt grundsätzlich die Abtastrate  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

{{BlueBox|TEXT=
'''(6)'''  The settings of  $(3)$  and  $(4)$  continue to apply except for  $\varphi=90^\circ$.  Interpret the plots in the time and frequency domain.}}
:* The source signal is sampled exactly at its zero crossings   ⇒   $x_{\rm A}(t) \equiv 0$   ⇒     $y(t) \equiv 0$   ⇒   $\varepsilon(t)=-x(t)$   ⇒   $P_\varepsilon = P_x$   ⇒   $10 \cdot \lg \ (P_x/P_\varepsilon)=0\ \rm dB$.
:* Description in the frequency domain:   As in  $(4)$  the imaginary parts of  $X_{\rm A}(f)$  cancel out.  Also the real parts of  $X_{\rm A}(f)$  are zero because of the sinusoid.

{{BlueBox|TEXT=
'''(7)'''  Now consider the  $\text {Source Signal 2}$.  Let the other parameters be  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2.5 kHz}$,  $r=0$.  Interpret the results.}}
:* The source signal has spectral components up to  $\pm \text{2 kHz}$.  The signal power is $P_x = 2 \cdot \big[0.1^2 + 0.25^2+0.15^2\big]= 0.19 $. 
:* With the sampling rate  $f_{\rm A} = \text{5 kHz}$  and the receiver parameters  $f_{\rm G} = \text{2.5 kHz}$  and  $r=0$, the signal reconstruction works perfectly:  $P_\varepsilon = 0$.
:* Likewise with the trapezoidal low–pass with  $f_{\rm G} = \text{2.5 kHz}$, if for the rolloff factor holds:  $r \le 0.2$.

{{BlueBox|TEXT=
'''(8)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{1.5 kHz}$  of the rectangular low–pass filter is too small?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=-0.3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t -60^\circ)=0. 3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t +120^\circ)$  is equal to the (negated) signal component at  $\text{2 kHz}$. 
:* The distortion power is  $P_\varepsilon=2 \cdot 0.15^2= 0.045$  and the signal–to–distortion ratio  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(9)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{3.5 kHz}$  of the rectangular low–pass filter is too large?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=0.3 \cdot \cos(2\pi \cdot \text{3 kHz} \cdot t +60^\circ)$  is now equal to the  $\text{3 kHz}$  portion of the sink signal  $y(t)$  not removed by the low-pass filter.
:* Compared to the subtask  $(8)$  the frequency changes from  $\text{2 kHz}$  to  $\text{3 kHz}$  and also the phase relationship.
:* The amplitude of this  $\text{3 kHz}$ error signal is equal to the amplitude of the  $\text{2 kHz}$ portion of  $x(t)$.  Again  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(10)'''  Finally, we consider the  $\text {source signal 4}$  $($portions until  $\pm \text{4 kHz})$, as well as  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2. 5 kHz}$,  $0 \le r\le 1$.  Interpretation of results.}}
:* Up to  $r=0.2$  the signal reconstruction works perfectly  $(P_\varepsilon = 0)$.  If one increases  $r$, then  $P_\varepsilon$  increases continuously and  $10 \cdot \lg \ (P_x/P_\varepsilon)$  decreases.
:* With  $r=1$  the signal frequencies  $\text{0.5 kHz}$,  ...,  $\text{4 kHz}$  are attenuated, the more the higher the frequency is,  for example  $H_{\rm E}(f=\text{4 kHz}) = 0.6$.
:* Similarly,  $Y(f)$  also includes components at frequencies  $\text{6 kHz}$,  $\text{7 kHz}$,  $\text{8 kHz}$,  $\text{9 kHz}$  and  $\text{9.5 kHz}$ due to periodic continuation.
:* At the sampling times  $t\hspace{0.05cm}' = n \cdot T_{\rm A}$, the signals  $x(t\hspace{0.05cm}')$  and  $y(t\hspace{0.05cm}')$  agree exactly  ⇒   $\varepsilon(t\hspace{0.05cm}') = 0$.  In between, not  ⇒   small distortion power   $P_\varepsilon = 0.008$.

==Applet Manual==

[[File:Exercise_Abtast_v2.png|right|600px|frame|Screenshot]]
    '''(A)'''     Selection of one of the four given source signals,               Adjustment of amplitudes, frequencies and phases.

    '''(B)'''     Output of all set parameters of the source signal:

    '''(C)'''     Sampling & signal reconstruction parameters:
::*Sampling frequency  $f_{\rm A}$, 
::*Limit frequency of the receiving filter  $f_{\rm G}$,
::*Rolloff factor of the receiving filter  $r$,
:::⇒   Trapezoidal–corner frequencies:  $f_{1,\ 2} = f_{\rm G}\cdot (1\mp r)$

    '''(D)'''     Numerical result output:
::*Signal power  $P_{x}$
::*Distortion power  $P_{\varepsilon}$
::*Signal to distortion ratio  $10 \cdot \lg \ P_{x}/P_{\varepsilon}$

    '''(E)'''     Graphical output range for time domain:
::*Source signal  $x(t)$   ⇒   blue,
::*Sampled signal  $x_{\rm A}(t)$   ⇒   blue,
::*Reconstructed signal  $y(t)$   ⇒   green,
::*Differential signal  $\varepsilon(t)=y(t) - x(t)$   ⇒   purple

    '''(F)'''     Graphical output area for frequency domain:
::*$X(f)$   ⇒   blue,
::*$X_{\rm A}(f)$   ⇒   blue,
::*$Y(f)$   ⇒   green,
::*$E(f)=Y(f) - X(f)$   ⇒   purple

    '''(G)'''     Exercise selection

    '''(H)'''     Questions and solutions
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._Günter_Söder_(at_LNT_from_1974-2024)|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}
[[de:Applets:Sampling of Analog Signals and Signal Reconstruction]]

Applets:Zur Verdeutlichung digitaler Filter

2026-03-16T15:59:00Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{LntAppletLink|digitalFilters}}

==Programmbeschreibung==
 
Das Applet behandelt die Systemkomponenten  "Abtastung"  und  "Signalrekonstruktion", zwei Komponenten, die zum Beispiel für das Verständnis der  [[Modulation_Methods/Pulscodemodulation|Pulscodemodulation]]  $({\rm PCM})$  von großer Wichtigkeit sind.  Die obere Grafik zeigt das für dieses Applet zugrundeliegende Modell.  Darunter gezeichnet sind die Abtastwerte  $x(\nu \cdot T_{\rm A})$  des zeitkontinuierlichen Signals  $x(t)$. Die (unendliche) Summe über alle diese Abtastwerte bezeichnen wir als das abgetastete Signal  $x_{\rm A}(t)$.

*Beim Sender wird aus dem zeitkontinuierlichen Quellensignal  $x(t)$  das zeitdiskrete (abgetastete) Signal  $x_{\rm A}(t)$  gewonnen.  Man nennt diesen Vorgang  '''Abtastung'''  oder  '''A/D–Wandlung'''.
*Der entsprechende Programmparameter für den Sender ist die Abtastrate  $f_{\rm A}= 1/T_{\rm A}$. In der unteren Grafik ist der Abtastabstand  $T_{\rm A}$  eingezeichnet.
*Beim Empfänger wird aus dem zeitdiskreten Empfangssignal  $y_{\rm A}(t)$  das zeitkontinuierliche Sinkensignal  $y(t)$  erzeugt   ⇒   '''Signalrekonstruktion'''  oder  '''D/A–Wandlung'''  entsprechend dem Empfänger–Frequenzgang  $H_{\rm E}(f)$.

Das Applet berücksichtigt nicht die PCM–Blöcke  "Quantisierung",  "Codierung / Decodierung" und der Digitale Übertragungskanal ist als ideal angenommen. 

==Theoretischer Hintergrund==
 
===Allgemeines Blockschaltbild===

Jedes Signal  $x(t)$  kann an einem Rechner nur durch die Folge  $〈x_ν〉$  seiner Abtastwerte dargestellt werden, wobei  $x_ν$  für  $x(ν · T_{\rm A})$  steht.
[[File:P_ID552__Sto_T_5_2_S1_neu.png |right|frame| Blockschaltbild eines digitalen (IIR–) Filters  $M$–Ordnung]]
*Der zeitliche Abstand  $T_{\rm A}$  zwischen zwei Abtastwerten ist dabei durch das  [[Signal_Representation/Discrete-Time_Signal_Representation#Sampling_theorem|Abtasttheorem]]  nach oben begrenzt.
*Wir beschränken uns hier auf kausale Signale und Systeme, das heißt, es gilt  $x_ν \equiv 0$  für  $ν \le 0$.

*Um den Einfluss eines linearen Filters mit Frequenzgang  $H(f)$  auf das zeitdiskrete Eingangssignal  $〈x_ν〉$  zu erfassen, bietet es sich an, auch das Filter zeitdiskret zu beschreiben.  Im Zeitbereich geschieht das mit der zeitdiskreten Impulsantwort  $〈h_ν〉$.
*Rechts sehen Sie das entsprechende Blockschaltbild.  Für die Abtastwerte des Ausgangssignals  $〈y_ν〉$  gilt somit:
:$$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu } \cdot x_{\nu - \mu } + \sum\limits_{\mu = 1}^M {b_\mu } \cdot y_{\nu - \mu } .$$

Hierzu ist Folgendes zu bemerken:
*Der Index  $\nu$  bezieht sich auf Folgen, zum Beispiel   Eingang $〈x_ν〉$  und Ausgang   $〈y_ν〉$.
*Den Index  $\mu$  verwenden wir dagegen für die Kennzeichnung der  $a$– und  $b$–Filterkoeffizienten.
*Die erste Summe beschreibt die Abhängigkeit des aktuellen Ausgangs  $y_ν$  vom aktuellen Eingang  $x_ν$  und von den  $M$  vorherigen Eingangswerten  $x_{ν-1}$, ... , $x_{ν-M}.$
*Die zweite Summe kennzeichnet die Beeinflussung von  $y_ν$  durch die vorherigen Werte  $y_{ν-1}$, ... , $y_{ν-M}$  am Filterausgang.  Sie gibt den rekursiven Teil des Filters an.
*Den ganzzahligen Parameter  $M$  bezeichnet man als die ''Ordnung''  des digitalen Filters.  Im Programm ist dieser Wert auf  $M\le 2$  begrenzt.

{{BlaueBox|TEXT=
$\text{Definitionen:}$ 

'''(1)'''  Man bezeichnet die Ausgangsfolge  $〈y_ν〉$  als die  '''zeitdiskrete Impulsantwort'''  $〈h_ν〉$, wenn am Eingang die  "zeitdiskrete Diracfunktion"  anliegt:
:$$〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉 .$$
'''(2)'''  Man bezeichnet die Ausgangsfolge  $〈y_ν〉$  als die  '''zeitdiskrete Sprungantwort'''  $〈\sigma_ν〉$, wenn am Eingang die  "zeitdiskrete Sprungfunktion"  anliegt:
:$$〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉 .$$
'''(3)'''  Man bezeichnet die Ausgangsfolge  $〈y_ν〉$  als die  '''zeitdiskrete Recheckantwort'''  $〈\rho_ν^{(2, 4)}〉$, wenn am Eingang die  "zeitdiskrete Rechteckfunktion"  anliegt:
:$$〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉 .$$
:In Hochkommata angegeben sind hier der Beginn der Einsen  $(2)$  und die Stelle der letzten Eins  $(4)$.
}}

===Nichtrekursives Filter   ⇒   FIR–Filter ===

{{BlaueBox|TEXT=
[[File:P_ID553__Sto_T_5_2_S2_neu.png|right |frame| Nichtrekursives digitales Filter  $($FIR–Filter$)$  $M$–Ordnung]]
$\text{Definition:}$  Sind alle Rückführungskoeffizienten  $b_{\mu} = 0$, so spricht von einem  '''nichtrekursiven Filter'''.  Insbesondere in der englischsprachigen Literatur ist hierfür auch die Bezeichnung  '''FIR Filter'''  (''Finite Impulse Response'') gebräuchlich.

Für die Ordnung  $M$  gilt:

*Der Ausgangswert  $y_ν$  hängt nur vom aktuellen und den  $M$  vorherigen Eingangswerten ab:
:$$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu \cdot x_{\mu - \nu } } .$$
*Zeitdikrete Impulsantwort mit $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉$:
:$$〈h_\mu〉= 〈a_0,\ a_1,\ \text{...},\ a_M〉 .$$}}

{{GraueBox|TEXT=
$\text{Beispiel 1:}$  Ein Zweiwegekanal, bei dem
*das Signal auf dem Hauptpfad gegenüber dem Eingangssignal ungedämpft, aber um  $2\ \rm µ s$  verzögert ankommt, und
*in  $4\ \rm µ s$  Abstand – also absolut zur Zeit  $t = 6\ \rm µ s$  – ein Echo mit halber Amplitude nachfolgt,

kann durch ein nichtrekursives Filter entsprechend obiger Skizze nachgebildet werden, wobei folgende Parameterwerte einzustellen sind:
:$$M = 3,\quad T_{\rm A} = 2\;{\rm{µ s} },\quad a_{\rm 0} = 0,\quad a_{\rm 1} = 1, \quad a_{\rm 2} = 0, \quad a_{\rm 3} = 0.5.$$}}

{{GraueBox|TEXT=
$\text{Beispiel 2:}$  Betrachtet wird ein nichtrekursives Filter mit den Filterkoeffizienten  $a_0 = 1,\hspace{0.5cm} a_1 = 2,\hspace{0.5cm} a_2 = 1.$ 
[[File:P_ID608__Sto_Z_5_3.png|right|frame|Nichtrekursives Filter]]

'''(1)'''   Die herkömmliche Impulsantwort lautet:   $h(t) = \delta (t) + 2 \cdot \delta ( {t - T_{\rm A} } ) + \delta ( {t - 2T_{\rm A} } ).$         ⇒   Zeitdiskrete Impulsantwort:  $〈h_\mu〉= 〈1,\ 2,\ 1〉 .$

'''(2)'''   Der Frequenzgang  $H(f)$  ist die Fouriertransformierte von  $h(t)$.  Durch Anwendung des Verschiebungssatzes:
:$$H(f) = 2\big [ {1 + \cos ( {2{\rm{\pi }\cdot }f \cdot T_{\rm A} } )} \big ] \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }fT_{\rm A} }\hspace{0.5cm}\Rightarrow \hspace{0.5cm}H(f = 0) = 4.$$

'''(3)'''   Daraus folgt:  Die  '''zeitdiskrete Sprungantwort'''  $〈\sigma_ν〉$  tendiert für große  $\nu$  gegen  $4$.

'''(4)'''   Die zeitdiskrete Faltung der Eingangsfolge  $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;0,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$  mit  $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$  ergibt
:$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;2,\;1,\;0,\;1,\;2,\;1,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$

'''(5)'''   Die zeitdiskrete Faltung der Eingangsfolge  $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;1,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$  mit  $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$  ergibt
:$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;3,\;3,\;1,\;0,\;0,\;0,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$}}

===Rekursives Filter   ⇒   IIR–Filter ===

{{BlaueBox|TEXT=
[[File:P_ID607__Sto_A_5_3.png|right|frame|Rekursives Filter erster Ordnung]]
$\text{Definition:}$ 
*Ist zumindest einer der Rückführungskoeffizienten  $b_{\mu} \ne 0$, so spricht von einem  '''rekursiven Filter'''  (siehe rechte Grafik).  Insbesondere in der englischsprachigen Literatur ist hierfür auch die Bezeichnung  '''IIR Filter'''  (''Infinite Impulse Response'') gebräuchlich.  Dieses Filter wird in der Verrsuchsdurchführung ausführlich behandelt.

*Sind zusätzlich alle Vorwärtskoeffizienten identisch  $a_\mu = 0$  mit Ausnahme von  $a_0$,   so liegt ein  '''rein rekursives Filter'''  vor   (siehe linke Grafik).

[[File:P_ID554__Sto_T_5_2_S3_neu.png|left|frame| Rein rekursives Filter erster Ordnung]] }}

Im Folgenden beschränken wir uns auf den Sonderfall  "Rein rekursives Filter erster Ordnung".  Dieses Filter weist folgende Eigenschaften auf:
*Der Ausgangswert  $y_ν$  hängt (indirekt) von unendlich vielen Eingangswerten ab:
:$$y_\nu = \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot x_{\nu - \mu } .}$$
*Dies zeigt die folgende Rechung:
:$$y_\nu = a_0 \cdot x_\nu + b_1 \cdot y_{\nu - 1} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + {b_1} ^2 \cdot y_{\nu - 2} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + a_0 \cdot {b_1} ^2 \cdot x_{\nu - 2} + {b_1} ^3 \cdot y_{\nu - 3} = \text{...}. $$

*Die zeitdiskrete Impulsantwort ist definitionsgemäß der Ausgangsfolge, wenn am Eingang eine einzelne „Eins” bei  $t =0$  anliegt.
:$$h(t)= \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot \delta ( {t - \mu \cdot T_{\rm A} } )}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉= 〈\hspace{0.05cm}a_0, \ a_0\cdot {b_1}, \ a_0\cdot {b_1}^2 \ \text{...} \hspace{0.05cm}〉.$$

{{BlaueBox|TEXT=
$\text{Fazit:}$  Bei einem rekursiven Filter reicht die (zeitdiskrete) Impulsantwort schon mit  $M = 1$  bis ins Unendliche:
*Aus Stabilitätsgründen muss  $b_1 < 1$  gelten.
*Bei  $b_1 = 1$  würde sich die Impulsantwort  $h(t)$  bis ins Unendliche erstrecken und bei  $b_1 > 1$  würde  $h(t)$  sogar bis ins Unendliche anklingen.
*Bei einem solchen rekursiven Filter erster Ordnung ist jede einzelne Diraclinie genau um den Faktor  $b_1$  kleiner als die vorherige Diraclinie:
:$$h_{\mu} = h(\mu \cdot T_{\rm A}) = {b_1} \cdot h_{\mu -1}.$$}}

{{GraueBox|TEXT=
[[File:Sto_T_5_2_S3_version2.png |frame| Zeitdiskrete Impulsantwort | rechts]]
$\text{Beispiel 3:}$  Die nebenstehende Grafik zeigt die zeitdiskrete Impulsantwort  $〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉$  eines rekursiven Filters erster Ordnung mit den Parametern  $a_0 = 1$  und  $b_1 = 0.6$.
*Der Verlauf ist exponentiell abfallend und erstreckt sich bis ins Unendliche.
*Das Verhältnis der Gewichte zweier aufeinander folgender Diracs ist jeweils  $b_1 = 0.6$.
}}

===Rekursives Filter als Sinus–Generator===
[[File:P_ID622__Sto_A_5_4.png|right|frame|Vorgeschlagene Filterstruktur '''ändern auf''' $T_{\rm A}$]]
Die Grafik zeigt ein digitales Filter zweiter Ordnung, das zur Erzeugung einer zeitdiskreten Sinusfunktion auf einem digitalen Signalprozessor (DSP) geeignet ist, wenn die Eingangsfolge  $\left\langle \hspace{0.05cm} {x_\nu } \hspace{0.05cm}\right\rangle$  eine (zeitdiskrete) Diracfunktion ist:
:$$\left\langle \hspace{0.05cm}{y_\nu }\hspace{0.05cm} \right\rangle = \left\langle {\, \sin ( {\nu \cdot T_{\rm A} \cdot \omega _0 } )\, }\right\rangle .$$

Die fünf Filterkoeffizienten ergeben sich aus der  [https://de.wikipedia.org/wiki/Z-Transformation $Z$-Transformation]:
:$$Z \{ {\sin ( {\nu T{\rm A}\cdot \omega _0 } )} \} = \frac{{z \cdot \sin \left( {\omega _0 \cdot T_{\rm A}} \right)}}{{z^2 - 2 \cdot z \cdot \cos \left( {\omega _0 \cdot T_{\rm A}} \right) + 1}}.$$
Nach Umsetzung dieser Gleichung durch ein rekursives Filter zweiter Ordnung erhält man folgende Filterkoeffizienten:
:$$a_0 = 0,\quad a_1 = \sin \left( {\omega _0 \cdot T_{\rm A}} \right),\quad a_2 = 0, \quad b_1 = 2 \cdot \cos \left( {\omega _0 \cdot T_{\rm A}} \right),\quad b_2 = - 1.$$

*Auf die Filterkoeffizienten  $a_0$  und  $a_2$  kann verzichtet werden und  $b_2=-1$  hat einen festen Wert. 
*Die Kreisfrequenz  $\omega_0$  der Sinusschwingung wird also nur durch  $a_0$  und  $a_0$  festelegt.

{{GraueBox|TEXT=
$\text{Beispiel 3:}$  Es gelte  $a_1 = 0.5$,  $b_1 = \sqrt 3$,  $x_0 = 1$  und  $x_{\nu \hspace{0.05cm}\ne\hspace{0.05cm} 0} = 0$.

'''(1)'''  Dann gilt für die Ausgangswerte  $y_\nu$  zu den Zeitpunkten  $\nu \ge 0$: 
:*  $y_0 = 0;$
:*  $y_1 = 0.5$                                                                                         ⇒  die "$1$" am Eingang wirkt sich wegen  $a_0= 0$  am Ausgang erst zum Zeitpunkt  $\nu = 1$  aus;
:*  $y_2 = b_1 \cdot y_1 - y_0 = {\sqrt 3 }/{2} \approx 0.866$                             ⇒   bei  $\nu = 2$  wird auch der rekursive Teil des Filters wirksam;
:*  $y_3 = \sqrt 3 \cdot y_2 - y_1 = \sqrt 3 \cdot {\sqrt 3 }/{2} - {1}/{2} = 1$          ⇒  für  $\nu \ge 2$  ist das Filter rein rekursiv:     $y_\nu = b_1 \cdot y_{\nu - 1} - y_{\nu - 2}$;
:*  $y_4 = \sqrt 3 \cdot y_3 - y_2 = \sqrt 3 \cdot 1 - {\sqrt 3 }/{2} = {\sqrt 3 }/{2};$
:*  $y_5 = \sqrt 3 \cdot y_4 - y_3 = \sqrt 3 \cdot {\sqrt 3 }/{2} - 1 = 0.5;$
:*  $y_6 = \sqrt 3 \cdot y_5 - y_4 = \sqrt 3 \cdot {1}/{2} - {\sqrt 3 }/{2} = 0;$
:*  $y_7 = \sqrt 3 \cdot y_6 - y_5 = \sqrt 3 \cdot 0 - {1}/{2} = - 0.5.$

'''(2)'''  Durch Fortsetzung des rekursiven Algorithmuses erhält man für große  $\nu$–Werte:     $y_\nu = y_{\nu - 12} $   ⇒   $T_0/T_{\rm A}= 12.$ }}

==Versuchsdurchführung==

[[File:Exercises_binomial_fertig.png|right]]
*Wählen Sie zunächst die Nummer  '''1'''  ...  '''10'''  der zu bearbeitenden Aufgabe.
*Eine Aufgabenbeschreibung wird angezeigt. Die Parameterwerte sind angepasst.
*Lösung nach Drücken von "Musterlösung".
*Die Nummer  '''0'''  entspricht einem "Reset":  Gleiche Einstellung wie beim Programmstart.

'''Noch ersetzen'''
In den folgenden Aufgabenbeschreibungen werden folgende Kurzbezeichnungen verwendet:
*'''Rot''':     Regressionsgerade  $R_{Y \to X}$  (im Applet rot gezeichnet),
*'''Blau''':   Regressionsgerade  $R_{X \to Y}$  (im Applet blau gezeichnet).
'''bis hierher'''

{{BlaueBox|TEXT=
'''(1)'''  Die Filterkoeffizienten seien  $a_0=0.25$,  $a_1=0.5$, $a_2=0.25$,  $b_1=b_2=0$.  Um welches Filter handelt es sich?          Interpretieren Sie die Impulsantwort  $〈h_ν〉$,  die Sprungantwort  $〈\sigma_ν〉$  und  die Rechteckantwort  $〈\rho_ν^{(2, 8)}〉$  jeweils in zeitdiskreter Darstellung.}}

:*  Aufgrund der fehlenden  $b$–Koeffizienten handelt es sich um ein nichtrekursives digitales Filter   ⇒   '''FIR–Filter'''  (''Finite impulse Response'').
:*  Die Impulsantwort setzt sich aus  $M+1=3$  Diraclinien gemäß den  $a$–Koeffizienten zusammen:    $〈h_ν〉= 〈a_0, \ a_1,\ a_2〉= 〈0.25, \ 0.5,\ 0.25,\ 0, \ 0, \ 0,\text{...}〉 $.
:*  Die Sprungantwort lautet:    $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1,\text{...}〉 $.  Der Endwert ist gleich dem Gleichsignalübertragungsfaktor  $H(f=0)=a_0+a_1+a_2 = 1$.
:*  Die Verzerrungen bei Anstieg und Abfall erkennt man auch aus der Rechteckantwort  $〈\rho_ν^{(2, 8)}〉= 〈0,\ 0, 0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1, \ 1, \ 0.75, \ 0.25, \ \text{...}〉$.

{{BlaueBox|TEXT=
'''(2)'''  Wie unterscheiden sich die Ergebnisse mit  $a_2=-0.25$? }}

:*  Unter Berücksichtigung von  $H(f=0)= 0.5$  ergeben sich vergleichbare Folgen   ⇒   Sprungantwort:    $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 0.5,\ 0.5, \ 0.5, \ 0.5,\text{...}〉 $.

{{BlaueBox|TEXT=
'''(3)'''  Nun seien die Filterkoeffizienten  $a_0=1$,  $b_1=0.9$  sowie  $a_1=a_2= b_2=0$.  Um welches Filter handelt es sich?  Interpretieren Sie die Impulsantwort  $〈h_ν〉$.}}

:*  Es handelt sich um ein rekursives digitales Filter   ⇒   '''IIR–Filter'''  (''Infinite impulse Response'')  erster Ordnung.  Es ist das zeitdiskrete Analogon zum RC–Tiefpass.
:*  Ausgehend von  $h_0= 1$  gilt  $h_1= h_0 \cdot b_0= 0.9$,  $h_2= h_1 \cdot b_0= b_0^2=0.81$,  $h_3= h_2 \cdot b_0= b_0^3=0.729$,  usw.   ⇒   $〈h_ν〉$  reicht bis ins Unendliche.
:*  Impulsantwort  $h(t) = {\rm e}^{-t/T}$  mit  $T$:  Schnittpunkt $($Tangente bei  $t=0$, Abszisse$)$   ⇒   $h_\nu= h(\nu \cdot T_{\rm A}) = {\rm e}^{-\nu/(T/T_{\rm A})}$  mit  $T/T_{\rm A} = 1/(h_0-h_1)= 10$.
:*  '''Diskrepanz zu h(t) wertkontinuierlich ???''' 1.0 0.9048 0.8187 ...

{{BlaueBox|TEXT=
'''(4)'''  Die Filtereinstellung wird beibehalten.  Interpretieren Sie die Sprungantwort  $〈h_ν〉$  und  die Rechteckantwort  $〈\rho_ν^{(2, 8)}〉$.  Welcher Wert ergibt sich für  $H(f=0)$?}}

:*  Die Sprungantwort ist das Ingral über die Impulsantwort  $\sigma(t) = T \cdot (1-{\rm e}^{-t/T}) ]$   ⇒   $\sigma_\nu= 10 \cdot (1-{\rm e}^{-\nu/10})$   ⇒   $\sigma_0=1$,  $\sigma_1=1.9$,  $\sigma_2=2.71$, ...
:*  Für große $\nu$–Werte tendiert die (zeitdiskrete) Sprungantwort gegen den Gleichsignalübertragungsfaktor  $H(f=0)= 10$:  $\sigma_{40}=9.867$,  $\sigma_{50}=9.954$,  $\sigma_\infty=10$.
:* Die Rechteckantwort  $〈\rho_ν^{(2, 8)}〉$  steigt mit einer Verzögerung von $2$ in gleicher Weise an wie  $〈\sigma_ν〉$.  Im Bereich  $\nu \ge 8$  fallen die  $\rho_ν$– Werte exponentiell ab.

{{BlaueBox|TEXT=
'''(5)'''  Wir betrachten weiterhin das Filter mitnbsp; $a_0=1$,  $b_1=0.9$,  $a_1=a_2= b_2=0$.  Welche Ausgangsfolge  $〈y_ν〉$ für die Eingangsfolge  $〈x_ν〉= 〈1,\ 0.,\ -0.5〉$?         ''Hinweis'':  Die Aufgabe lässt sich ebenfalls mit diesem Programm lösen, obwohl die hier betrachtete Konstellation nicht direkt einstellbar ist.}}

:*  Man behilft sich, indem man den Koeffizienten  $a_2=-0.5$  setzt und dafür die Eingangsfolge auf   $〈x_ν〉= 〈1,\ 0.,\ 0.,\ \text{ ...}〉$   ⇒   „Diracfunktion” reduziert.
:*  Die tatsächliche Impulsantwort dieses Filters $($mit  $a_2=0)$  wurde in Aufgabe  '''(3)'''  ermittelt:   $h_0= 1$,   $h_1= 0.9$,   $h_2= 0.81$,   $h_3= 0.729$,   $h_4= 0.646$.  
:*  Die Lösung dieser Aufgabe lautet somit:   $y_0 = h_0= 1$,   $y_1= h_1= 0.9$,   $y_2 =h_2-h_0/2= 0.31$,   $y_3 =h_3-h_1/2= 0.279$,   $y_4 =h_4-h_2/2= 0.251$.  
:*  Vorsicht:  Sprungantwort und Rechteckantwort beziehen sich nun auf das fiktive Filter $($mit  $a_2=-0.5)$  und nicht auf das eigentliche Filter $($mit  $a_2=0)$.

{{BlaueBox|TEXT=
'''(6)'''  Betrachten und interpretieren Sie die Impulsanwort und die Sprungantwort für die Filterkoeffizienten  $a_0=1$,  $b_1=1$,  $a_1=a_2= b_2=0$.  }}

:*  '''Das System ist instabil''':   Eine zeitdiskrete Diracfunktion am Eingang  $($zur Zeit  $t=0)$  bewirkt im Ausgangsignal unendlich viele Diracs gleicher Höhe.
:*  Eine zeitdiskrete Sprungfunktion am Eingang bewirkt im Ausgangsignal unendlich viele Diracs mit monoton ansteigenden Gewichten (bis ins Unendliche).

{{BlaueBox|TEXT=
'''(7)'''  Betrachten und interpretieren Sie Impulsanwort und Sprungantwort für die Filterkoeffizienten  $a_0=1$,  $b_1=-1$,  $a_1=a_2= b_2=0$.  }}
:*  Im Gegensatz zur Aufgabe  '''(6)'''  sind hier die Gewichte der Impulsantwort  $〈h_ν〉$  nicht konstant gleich  $1$, sondern alternierend  $\pm 1$.  Das System ist ebenfalls instabil.
:*  Bei der Sprunganwort  $〈\sigma_ν〉$  wechseln sich dagegen die Gewichte alternierend zwischen  $0$  $($bei geradem $\nu)$  und  $1$  $($bei ungeradem $\nu)$  ab.

{{BlaueBox|TEXT=
'''(8)'''  Wir betrachten den  "Sinusgenerator":  $a_1=0.5$,  $b_1=\sqrt{3}= 1.732$,  $b_2=-1.$  Vergleichen Sie die Impulsantwort mit den berechneten Werten in  $\text{Beispiel 4}$.         Wie beinflussen die Parameter  $a_1$  und  $b_1$  die Periodendauer  $T_0/T_{\rm A}$  und die Amplitude  $A$  der Sinusfunktion? }}
:*  $〈x_ν〉=〈1, 0, 0, \text{...}〉$   ⇒   $〈y_ν〉=〈0, 0.5, 0.866, 1, 0.866, 0.5, 0, -0.5, -0.866, -1, -0.866, -0.5, 0, \text{...}〉$   ⇒   '''Sinus''',  Periode  $T_0/T_{\rm A}= 12$,  Amplitude  $1$.
:*  Die Vergrößerung/Verkleinerung von  $b_1$  führt zur größeren/kleineren Periodendauer  $T_0/T_{\rm A}$  und zur größeren/kleineren Amplitude  $A$.  Es muss  $b_1 < 2$  gelten. '''Stimmt das?'''
:*  $a_1$  beinflusst nur die Amplitude, nicht die Periodendauer.  Für  $a_1$  gibt es keine Wertebegrenzumg.  Bei negativem  $a_1$  ergibt sich die Minus–Sinusfunktion.
:*  '''Gibt es hier keine Diskrepanz zu h(t) wertkontinuierlich ???'''

{{BlaueBox|TEXT=
'''(9)'''  Die Grundeinstellung bleibt erhalten.  Mit welchen  $a_1$  und  $b_1$ ergibt sich eine Sinusfunktion mit Periodendauer  $T_0/T_{\rm A}=16$  und Amplitude  $A=1$? }}
:*  Durch Probieren erreicht man mit  $b_1= 1.8478$  tatsächlich die Periodendauer  $T_0/T_{\rm A}=16.$  Allerdings erhöht sich dadurch die Amplitude auf  $A=1.307$.
:*  Die Anpassung des Parameters   $a_1= 0.5/1.307=0.3826$  führt dann zur gewünschten Amplitude  $A=1$.
:*  Oder man kann das auch wie im Beispiel berechnen:  $b_1 = 2 \cdot \cos ( {2{\rm{\pi }}\cdot{T_{\rm A}}/{T_0 }})= 2 \cdot \cos (\pi/8)=1.8478$,     $a_1 = \sin (\pi/8)=0.3827$.

{{BlaueBox|TEXT=
'''(10)'''  Wir gehen weiter vom "Sinusgenerator" aus.  Welche Modifikationen muss man vornehmen, um damit einen "Cosinus" zu generieren?}}
:*    Mit  $a_1=0.5$,  $b_1=\sqrt{3}= 1.732$,  $b_2=-1$  sowie  $〈x_ν〉=〈1, 1, 1, \text{...}〉$  ist die Ausgangsfolge  $〈y_ν〉$  das zeitdiskrete Analogon der Sprungantwort  $\sigma(t)$.
:*  '''Hier noch auf die Diskrepanz zu sigma(t) wertkontinuierlich eingehen. Es fehlen noch einige Statements'''

==Zur Handhabung des Applets==
[[File:Handhabung_binomial.png|left|600px]]
    '''(A)'''     Vorauswahl für blauen Parametersatz

    '''(B)'''     Parametereingabe $I$ und $p$ per Slider

    '''(C)'''     Vorauswahl für roten Parametersatz

    '''(D)'''     Parametereingabe $\lambda$ per Slider

    '''(E)'''     Graphische Darstellung der Verteilungen

    '''(F)'''     Momentenausgabe für blauen Parametersatz

    '''(G)'''     Momentenausgabe für roten Parametersatz

    '''(H)'''     Variation der grafischen Darstellung

$\hspace{1.5cm}$"$+$" (Vergrößern),

$\hspace{1.5cm}$ "$-$" (Verkleinern)

$\hspace{1.5cm}$ "$\rm o$" (Zurücksetzen)

$\hspace{1.5cm}$ "$\leftarrow$" (Verschieben nach links), usw.

    '''( I )'''     Ausgabe von ${\rm Pr} (z = \mu)$ und ${\rm Pr} (z \le \mu)$

    '''(J)'''     Bereich für die Versuchsdurchführung
 
 '''Andere Möglichkeiten zur Variation der grafischen Darstellung''':
*Gedrückte Shifttaste und Scrollen: Zoomen im Koordinatensystem,
*Gedrückte Shifttaste und linke Maustaste: Verschieben des Koordinatensystems.

==Über die Autoren==
Dieses interaktive Berechnungstool wurde am [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik] der [https://www.tum.de/ Technischen Universität München] konzipiert und realisiert.
*Die erste Version wurde 2005 von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Bettina_Hirner_.28Diplomarbeit_LB_2005.29|Bettina Hirner]] im Rahmen ihrer Diplomarbeit mit "FlashMX–Actionscript" erstellt (Betreuer: [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._Günter_Söder_(at_LNT_from_1974-2024)|Günter Söder]]).
*2020 wurde das Programm von [[Andre Schulz]] (Bachelorarbeit LB, Betreuer: [[Benedikt Leible]] und [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) unter "HTML5" neu gestaltet.

==Nochmalige Aufrufmöglichkeit des Applets in neuem Fenster==

{{LntAppletLink|digitalFilters}}
[[de:Applets:Zur Verdeutlichung digitaler Filter]]

Applets:Physikalisches Signal & Analytisches Signal

2026-03-16T15:58:59Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{LntAppletLink|physAnSignal_en}}
    ''Hinweis:''   Das Applet ist für den '''CHROME'''–Browser optimiert. Bei anderen Browsern kommt es teilweise zu Darstellungsproblemen.

==Programmbeschreibung==
 
Dieses Applet zeigt den Zusammenhang zwischen dem physikalischen Bandpass–Signal $x(t)$ und dem dazugehörigen analytischen Signal $x_+(t)$. Ausgegangen wird stets von einem Bandpass–Signal $x(t)$ mit frequenzdiskretem Spektrum $X(f)$:
:$$x(t) = x_{\rm U}(t) + x_{\rm T}(t) + x_{\rm O}(t) = A_{\rm U}\cdot \cos\left(2\pi f_{\rm U}\cdot t- \varphi_{\rm U}\right)+A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t- \varphi_{\rm T}\right)+A_{\rm O}\cdot \cos\left(2\pi f_{\rm O}\cdot t- \varphi_{\rm O}\right). $$
Das physikalische Signal $x(t)$ setzt sich also aus drei [[Signal_Representation/Harmonische_Schwingung|harmonischen Schwingungen]] zusammen, einer Konstellation, die sich zum Beispiel bei der [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#AM_signals_and_spectra_with_a_harmonic_input_signal|Zweiseitenband-Amplitudenmodulation]] des Nachrichtensignals $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t- \varphi_{\rm N}\right)$ mit dem Trägersignal $x_{\rm T}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t - \varphi_{\rm T}\right)$ ergibt. Die Nomenklatur ist ebenfalls an diesen Fall angepasst:
* $x_{\rm O}(t)$ bezeichnet das "Obere Seitenband" mit der Amplitude $A_{\rm O}= A_{\rm N}/2$, der Frequenz $f_{\rm O} = f_{\rm T} + f_{\rm N}$ und der Phase $\varphi_{\rm O} = \varphi_{\rm T} + \varphi_{\rm N}$.
*Entsprechend gilt für das "Untere Seitenband" $x_{\rm U}(t)$ mit $f_{\rm U} = f_{\rm T} - f_{\rm N}$, $A_{\rm U}= A_{\rm O}$ und $\varphi_{\rm U} = -\varphi_{\rm O}$.

Das dazugehörige analytische Signal lautet:

:$$x_+(t) = x_{\rm U+}(t) + x_{\rm T+}(t) + x_{\rm O+}(t) = A_{\rm U}\cdot {\rm e}^{\hspace{0.05cm} {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm U})}\hspace{0.1cm}+ \hspace{0.1cm}A_{\rm T}\cdot {\rm e}^{\hspace{0.05cm} {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm T})}\hspace{0.1cm}+\hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{\hspace{0.05cm} {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm O})}. $$

[[File:Zeigerdiagramm_2a_version2.png|right|frame|Analytische Signal zur Zeit $t=0$]]
Im Programm dargestellt wird $x_+(t)$ als vektorielle Summe dreier Drehzeiger (alle mit positiver Drehrichtung   ⇒   entgegen dem Uhrzeigersinn) als violetter Punkt (siehe beispielhafte Grafik für den Startzeitpunkt $t=0$):

*Der (rote) Zeiger des Trägers $x_{\rm T+}(t)$ mit der Länge $A_{\rm T}$ und der Nullphasenlage $\varphi_{\rm T} = 0$ dreht mit konstanter Winkelgeschwindigkeit $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm T}$ (eine Umdrehung in der Zeit $1/f_{\rm T})$.

*Der (blaue) Zeiger des Oberen Seitenbandes $x_{\rm O+}(t)$ mit der Länge $A_{\rm O}$ und der Nullphasenlage $\varphi_{\rm O}$ dreht mit der Winkelgeschwindigkeit $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}$, also etwas schneller als $x_{\rm T+}(t)$.

*Der (grüne) Zeiger des Unteren Seitenbandes $x_{\rm U+}(t)$ mit der Länge $A_{\rm U}$ und der Nullphasenlage $\varphi_{\rm U}$ dreht mit der Winkelgeschwindigkeit $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}$, also etwas langsamer als $x_{\rm T+}(t)$.

Den zeitlichen Verlauf von $x_+(t)$ bezeichnen wir im Folgenden auch als '''Zeigerdiagramm'''. Der Zusammenhang zwischen dem physikalischen Bandpass–Signal $x(t)$ und dem dazugehörigen analytischen Signal $x_+(t)$ ist sehr einfach:

:$$x(t) = {\rm Re}\big [x_+(t)\big ].$$

''Hinweis:''   Die Grafik gilt für $\varphi_{\rm O} = +30^\circ$. Daraus folgt für den Startzeitpunkt $t=0$ der Winkel gegenüber dem Koordinatensystem:   $\phi_{\rm O} = -\varphi_{\rm O} = -30^\circ$. Ebenso folgt aus der Nullphasenlage $\varphi_{\rm U} = -30^\circ$ des unteren Seitenbandes für den in der komplexen Ebene zu berücksichtigenden Phasenwinkel:   $\phi_{\rm U} = +30^\circ$.

[[Applets:Physical_Signal_%26_Analytical_Signal|'''Englische Beschreibung''']]

==Theoretischer Hintergrund==
 
===Beschreibungsmöglichkeiten von Bandpass-Signalen===
[[File:Zeigerdiagramm_1a.png|right|frame|Bandpass–Spektrum $X(f)$ |class=fit]]
Wir betrachten hier '''Bandpass-Signale''' $x(t)$ mit der Eigenschaft, dass deren Spektren $X(f)$ nicht im Bereich um die Frequenz $f = 0$ liegen, sondern um eine Trägerfrequenz $f_{\rm T}$. Meist kann auch davon ausgegangen werden, dass die Bandbreite $B \ll f_{\rm T}$ ist.

Die Grafik zeigt ein solches Bandpass–Spektrum $X(f)$. Unter der Annahme, dass das zugehörige $x(t)$ ein physikalisches Signal und damit reell ist, ergibt sich für die Spektralfunktion $X(f)$ eine Symmetrie bezüglich der Frequenz $f = 0$. Ist $x(t)$ eine gerade Funktion   ⇒   $x(-t)=x(+t)$, so ist auch $X(f)$ reell und gerade.

Neben dem physikalischen Signal $x(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)$ verwendet man zur Beschreibung von Bandpass-Signalen gleichermaßen:
*das analytische Signal $x_+(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X_+(f)$, wie im nächsten Unterabschnitt beschrieben,
*das äquivalente Tiefpass–Signal $x_{\rm TP}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X_{\rm TP}(f)$, siehe Applet [[Applets:Physikalisches_Signal_%26_Äquivalentes_TP-Signal|Physikalisches Signal und Äquivalentes Tiefpass–Signal]].
 
===Analytisches Signal – Spektralfunktion===

Das zum physikalischen Signal $x(t)$ gehörige '''analytische Signal''' $x_+(t)$ ist diejenige Zeitfunktion, deren Spektrum folgende Eigenschaft erfüllt:
[[File:Zeigerdiagramm_3a.png|right|frame|Konstruktion der Spektralfunktion $X_+(f)$ |class=fit]]
:$$X_+(f)=\big[1+{\rm sign}(f)\big] \cdot X(f) = \left\{ {2 \cdotX(f) \; \hspace{0.2cm}\rm f\ddot{u}r\hspace{0.2cm} {\it f} > 0, \atop {\,\,\,\, \rm 0 \; \hspace{0.9cm}\rm f\ddot{u}r\hspace{0.2cm} {\it f} < 0.} }\right.$$

Die so genannte ''Signumfunktion'' ist dabei für positive Werte von $f$ gleich $+1$ und für negative $f$–Werte gleich $-1$.
*Der (beidseitige) Grenzwert liefert $\sign(0) = 0$.
*Der Index „+” soll deutlich machen, dass $X_+(f)$ nur Anteile bei positiven Frequenzen besitzt.

Aus der Grafik erkennt man die Berechnungsvorschrift für $X_+(f)$:   Das Bandpass–Spektrum $X(f)$ wird
*bei den positiven Frequenzen verdoppelt, und
*bei den negativen Frequenzen zu Null gesetzt.

Aufgrund der Unsymmetrie von $X_+(f)$ bezüglich der Frequenz $f = 0$ kann man bereits jetzt schon sagen, dass die Zeitfunktion $x_+(t)$ bis auf den trivialen Sonderfall $x_+(t)= 0 \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ X_+(f)= 0$ stets komplex ist.
 
===Analytisches Signal – Zeitverlauf===
An dieser Stelle ist es erforderlich, kurz auf eine weitere Spektraltransformation einzugehen.

{{BlaueBox|TEXT=
$\text{Definition:}$ 
Für die '''Hilberttransformierte''' $ {\rm H}\left\{x(t)\right\}$ einer Zeitfunktion $x(t)$ gilt:

:$$y(t) = {\rm H}\left\{x(t)\right\} = \frac{1}{ {\rm \pi} } \cdot\hspace{0.03cm}\int_{-\infty}^{+\infty}\frac{x(\tau)}{ {t -\tau} }\hspace{0.15cm} {\rm d}\tau.$$

Dieses bestimmte Integral ist nicht auf einfache, herkömmliche Art lösbar, sondern muss mit Hilfe des [https://de.wikipedia.org/wiki/Cauchyscher_Hauptwert Cauchy–Hauptwertsatzes] ausgewertet werden.

Entsprechend gilt im Frequenzbereich:
:$$Y(f) = {\rm -j \cdot sign}(f) \cdot X(f) \hspace{0.05cm} .$$}}

Das obige Ergebnis lässt sich mit dieser Definition wie folgt zusammenfassen:
*Man erhält aus dem physikalischen BP–Signal $x(t)$ das analytische Signal $x_+(t)$, indem man zu $x(t)$ einen Imaginärteil gemäß der Hilberttransformierten hinzufügt:

:$$x_+(t) = x(t)+{\rm j} \cdot {\rm H}\left\{x(t)\right\} .$$

*$\text{H}\{x(t)\}$ verschwindet nur für den Fall $x(t) = \rm const.$   ⇒   Gleichsignal. Bei allen anderen Signalformen ist somit das analytische Signal $x_+(t)$ komplex.

*Aus dem analytischen Signal $x_+(t)$ kann das physikalische Bandpass–Signal in einfacher Weise durch Realteilbildung ermittelt werden:
:$$x(t) = {\rm Re}\big[x_+(t)\big] .$$

{{GraueBox|TEXT=
$\text{Beispiel 1:}$  Das Prinzip der Hilbert–Transformation wird durch die nachfolgende Grafik nochmals verdeutlicht:
*Nach der linken Darstellung $\rm(A)$ kommt man vom physikalischen Signal $x(t)$ zum analytischen Signal $x_+(t)$, indem man einen Imaginärteil ${\rm j} \cdot y(t)$ hinzufügt.
*Hierbei ist $y(t) = {\rm H}\left\{x(t)\right\}$ eine reelle Zeitfunktion, die sich im Spektralbereich durch die Multiplikation des Spektrums $X(f)$ mit ${\rm - j} \cdot \sign(f)$ angeben lässt.

[[File:P_ID2729__Sig_T_4_2_S2b_neu.png|center|frame|Zur Verdeutlichung der Hilbert–Transformierten]]

Die rechte Darstellung $\rm(B)$ ist äquivalent zu $\rm(A)$. Nun gilt $x_+(t) = x(t) + z(t)$ mit der rein imaginären Funktion $z(t)$. Ein Vergleich der beiden Bilder zeigt, dass tatsächlich $z(t) = {\rm j} \cdot y(t)$ ist.}}
 
===Darstellung der harmonischen Schwingung als analytisches Signal===

Die Spektralfunktion $X(f)$ einer harmonischen Schwingung $x(t) = A \cdot \text{cos}(2\pi f_{\rm T} \cdot t - \varphi)$ besteht bekanntlich aus zwei Diracfunktionen bei den Frequenzen
* $+f_{\rm T}$ mit dem komplexen Gewicht $A/2 \cdot \text{e}^{-\text{j}\hspace{0.05cm}\varphi}$,
* $-f_{\rm T}$ mit dem komplexen Gewicht $A/2 \cdot \text{e}^{+\text{j}\hspace{0.05cm}\varphi}$.

Somit lautet das Spektrum des analytischen Signals (also ohne die Diracfunktion bei der Frequenz $f =-f_{\rm T}$, aber Verdoppelung bei $f =+f_{\rm T}$):

:$$X_+(f) = A \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\varphi}\cdot\delta (f - f_{\rm T}) .$$

Die dazugehörige Zeitfunktion erhält man durch Anwendung des [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|Verschiebungssatzes]]:

:$$x_+(t) = A \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm}( 2 \pi f_{\rm T} t\hspace{0.05cm}-\hspace{0.05cm} \varphi)}.$$

Diese Gleichung beschreibt einen mit konstanter Winkelgeschwindigkeit $\omega_{\rm T} = 2\pi f_{\rm T}$ drehenden Zeiger.

{{GraueBox|TEXT=
$\text{Beispiel 2:}$  Aus Darstellungsgründen wird das Koordinatensystem entgegen der üblichen Darstellung um $90^\circ$ gedreht (Realteil nach oben, Imaginärteil nach links).

[[File:P_ID712__Sig_T_4_2_S3.png|center|frame|Zeigerdiagramm einer harmonischen Schwingung]]

Anhand dieser Grafik sind folgende Aussagen möglich:
*Zum Startzeitpunkt $t = 0$ liegt der Zeiger der Länge $A$ (Signalamplitude) mit dem Winkel $-\varphi$ in der komplexen Ebene. Im gezeichneten Beispiel gilt $\varphi = 45^\circ$.
*Für Zeiten $t > 0$ dreht der Zeiger mit konstanter Winkelgeschwindigkeit (Kreisfrequenz) $\omega_{\rm T}$ in mathematisch positiver Richtung, das heißt entgegen dem Uhrzeigersinn.
*Die Spitze des Zeigers liegt somit stets auf einem Kreis mit Radius $A$ und benötigt für eine Umdrehung genau die Zeit $T_0$, also die Periodendauer der harmonischen Schwingung $x(t)$.
*Die Projektion des analytischen Signals $x_+(t)$ auf die reelle Achse, durch rote Punkte markiert, liefert die Augenblickswerte von $x(t)$.}}
 
===$x_+(t)$–Darstellung einer Summe aus drei harmonischen Schwingungen===

In unserem Applet setzen wir stets einen Zeigerverbund aus drei Drehzeigern voraus. Das physikalische Signal lautet:
:$$x(t) = x_{\rm U}(t) + x_{\rm T}(t) + x_{\rm O}(t) = A_{\rm U}\cdot \cos\left(2\pi f_{\rm U}\cdot t- \varphi_{\rm U}\right)+A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t- \varphi_{\rm T}\right)+A_{\rm O}\cdot \cos\left(2\pi f_{\rm O}\cdot t- \varphi_{\rm O}\right). $$
* Jede der drei harmonischen Schwingungen $x_{\rm T}(t)$, $x_{\rm U}(t)$ und $x_{\rm O}(t)$ wird durch eine Amplitude $(A)$, eine Frequenz $(f)$ und einen Phasenwert $(\varphi)$ charakterisiert.
*Die Indizes sind an das Modulationsverfahren [[Modulation_Methods/Zweiseitenband-Amplitudenmodulation|Zweiseitenband–Amplitudenmodulation]] angelehnt. "T" steht für "Träger", "U" für "Unteres Seitenband" und "O" für "Oberes Seitenband". Entsprechend gilt stets $f_{\rm U} < f_{\rm T}$ und $f_{\rm O} > f_{\rm T}$. Für die Amplituden und Phasen gibt es keine Einschränkungen.

Das dazugehörige analytische Signal lautet:
:$$x_+(t) = x_{\rm U+}(t) + x_{\rm T+}(t) + x_{\rm O+}(t) = A_{\rm U}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm U})}\hspace{0.1cm}+ \hspace{0.1cm}A_{\rm T}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm T})}\hspace{0.1cm}+\hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm O})}. $$

{{GraueBox|TEXT=
$\text{Beispiel 3:}$ 
Die hier angegebene Konstellation ergibt sich zum Beispiel bei der [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#AM_signals_and_spectra_with_a_harmonic_input_signal|Zweiseitenband-Amplitudenmodulation]] (mit Träger) des Nachrichtensignals $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t- \varphi_{\rm N}\right)$ mit dem Trägersignal $x_{\rm T}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t - \varphi_{\rm T}\right)$. Hierauf wird in der Versuchsdurchführung häufiger eingegangen.

[[File:Zeigerdiagramm_5.png|center|frame|Spektum $X_+(f)$ des analytischen Signals für verschiedene Phasenkonstellationen |class=fit]]

Bei dieser Betrachtungsweise gibt es einige Einschränkungen bezüglich der Programmparameter:
* Für die Frequenzen gelte stets $f_{\rm O} = f_{\rm T} + f_{\rm N}$ und $f_{\rm U} = f_{\rm T} - f_{\rm N}$.

*Ohne Verzerrungen sind die Amplitude der Seitenbänder $A_{\rm O}= A_{\rm U}= A_{\rm N}/2$.
*Die jeweiligen Phasenverhältnisse können der Grafik entnommen werden.}}

==Versuchsdurchführung==
[[File:Zeigerdiagramm_aufgabe_2.png|right]]
*Wählen Sie zunächst die Aufgabennummer.
*Eine Aufgabenbeschreibung wird angezeigt.
*Alle Parameterwerte sind angepasst.
*Lösung nach Drücken von "Hide solition".

Mit der Nummer "0" wird auf die gleichen Einstellung wie beim Programmstart zurückgesetzt und es wird ein Text mit weiteren Erläuterungen zum Applet ausgegeben.
 

Im Folgenden bezeichnet $\rm Grün$ das Untere Seitenband   ⇒   $\big (A_{\rm U}, f_{\rm U}, \varphi_{\rm U}\big )$,  
$\rm Rot$ den Träger   ⇒   $\big (A_{\rm T}, f_{\rm T}, \varphi_{\rm T}\big )$ und
$\rm Blau$ das Obere Seitenband   ⇒   $\big (A_{\rm O}, f_{\rm O}, \varphi_{\rm O}\big )$.

{{BlaueBox|TEXT=
'''(1)'''   Betrachten und interpretieren Sie das analytische Signal $x_+(t)$ für $\text{Rot:} \hspace{0.15cm} A_{\rm T} = 1.5\ \text{V}, \ f_{\rm T} = 50 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$. Außerdem gelte $A_{\rm U} = A_{\rm O} = 0$.

:Welche Signalwerte $x_+(t)$ ergeben sich für $t = 0$, $t = 5 \ \rm µ s$ und $t = 20 \ \rm µ s$? Wie groß sind die entsprechenden Signalwerte von $x(t)$? }}

:: Für ein Cosinussignal gilt $x_+(t= 0) = A_{\rm T} = 1.5\ \text{V}$. Danach dreht $x_+(t)$ in mathematisch positiver Richtung (eine Umdrehung pro Periodendauer $T_0 = 1/f_{\rm T}$):

:: $x_+(t= 20 \ {\rm µ s}) = x_+(t= 0) = 1.5\ \text{V}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}x(t= 20 \ {\rm µ s}) = 1.5\ \text{V,}$
:: $x_+(t= 5 \ {\rm µ s}) = {\rm j} \cdot 1.5\ \text{V}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}x(t= 5 \ {\rm µ s}) = {\rm Re}[x_+(t= 5 \ {\rm µ s})] = 0$.

{{BlaueBox|TEXT=
'''(2)'''   Wie ändern sich die Verhältnisse für $\text{Rot:} \hspace{0.15cm} A_{\rm T} = 1.0\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 90^\circ$?}}

::Das Signal $x(t)$ ist nun ein Sinussignal mit kleinerer Amplitude. Das analytische Signal startet nun wegen $\varphi_{\rm T} = 90^\circ$   ⇒   $\phi_{\rm T} = -90^\circ$ bei $x_+(t= 0) = -{\rm j} \cdot A_{\rm T}$. Danach dreht $x_+(t)$ wieder in mathematisch positiver Richtung, aber wegen $T_0 = 10 \ \rm µ s$ doppelt so schnell als bei $\rm (1)$.

{{BlaueBox|TEXT=
'''(3)'''   Nun gelte   $\text{Rot:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$,   $\text{Grün:} \hspace{0.15cm} A_{\rm U} = 0.4\ \text{V}, \ f_{\rm U} = 80 \ \text{kHz}, \ \varphi_{\rm U} = 0^\circ$,   $\text{Blau:} \hspace{0.15cm} A_{\rm O} = 0.4\ \text{V}, \ f_{\rm O} = 120 \ \text{kHz}, \ \varphi_{\rm O} = 0^\circ$.

:Betrachten und interpretieren Sie das physikalische Signal $x(t)$ das analytische Signal $x_+(t)$.}}

::Das Signal $x(t)$ ergibt sich bei der Zweiseitenband–Amplitudenmodulation '''(ZSB–AM)''' des Nachrichtensignals $A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t\right)$ mit $A_{\rm N} = 0.8\ \text{V}$, $f_{\rm N} = 20\ \text{kHz}$. Der Träger $x_{\rm T}(t)$ mit $f_{\rm T} = 100\ \text{kHz}$ ist ebenfalls cosinusförmig. Der Modulationsgrad ist $m = A_{\rm N}/A_{\rm T} = 0.8$ und die Periodendauer $T_{\rm 0} = 50\ \text{µs}$.

::Im Zeigerdiagramm dreht sich der (rote) Träger schneller als das (grüne) Untere Seitenband und langsamer als das (blaue) Obere Seitenband. Das analytische Signal $x_+(t)$ ergibt sich als die geometrische Summe der drei rotierenden Zeiger. Es scheint so, als würde der blaue Zeiger dem Träger vorauseilen und der grüne Zeiger dem Träger nachlaufen.

{{BlaueBox|TEXT=
'''(4)'''   Es gelten weiter die Einstellungen der Aufgabe '''(3)'''. Welche Signalwerte ergeben sich bei $t=0$, $t=2.5 \ \rm µ s$, $t= 5 \ \rm µ s$ und $t=10 \ \rm µ s$? }}

::Zur Zeit $t=0$ liegen alle Zeiger in Richtung der reellen Achse, so dass $x(t=0) = {\rm Re}\big [x+(t= 0)\big] = A_{\rm U} + A_{\rm T} + A_{\rm O} = 1.8\ \text{V}$ gilt.

::Bis zur Zeit $t=2.5 \ \rm µ s$ hat sich der rote Träger um $90^\circ$ gedreht, der blaue Zeiger um $108^\circ$ und der grüne um $72^\circ$. Es gilt $x(t=2.5 \ \rm µ s) = {\rm Re}\big [x_+(t= 2.5 \ \rm µ s)\big] = 0$, da nun der Zeigerverbund in Richtung der imaginären Achse zeigt. Die weiteren gesuchten Signalwerte sind $x(t=5 \ \rm µ s) = {\rm Re}\big [x_+(t= 5 \ \rm µ s)\big] = -1.647\ \text{V}$ und $x(t=10 \ \rm µ s) = {\rm Re}\big [x_+(t= 10 \ \rm µ s)\big] = 1.247\ \text{V}$.
::Für $x_+(t)$ ergibt sich ein spiralförmiger Verlauf, abwechselnd mit kleiner werdenem Radius und anschließend mit größerem Radius.

{{BlaueBox|TEXT=
'''(5)'''   Wie müssen die Phasenparameter $\varphi_{\rm T}$, $\varphi_{\rm U}$ und $\varphi_{\rm O}$ eingestellt werden, wenn sowohl der Träger $x_{\rm T}(t)$ als auch das Nachrichtensignal $x_{\rm N}(t)$ sinusförmig verlaufen?}}

::Die Parameterwahl $\varphi_{\rm T} = \varphi_{\rm U} = \varphi_{\rm O}=90^\circ$ beschreibt die Signale $x_{\rm T}(t) = A_{\rm T}\cdot \sin\left(2\pi f_{\rm T}\cdot t\right)$ und $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t\right)$. Soll zusätzlich die Nachricht $x_{\rm N}(t)$ sinusförmig verlaufen, so muss $\varphi_{\rm O}=\varphi_{\rm T} - 90^\circ = 0$ und $\varphi_{\rm U}=\varphi_{\rm T} + 90^\circ = 180^\circ$ eingestellt werden.

{{BlaueBox|TEXT=
'''(6)'''   Es gelten die Einstellungen der Aufgabe '''(3)''' mit Ausnahme von $A_{\rm T} = 0.6\ \text{V}$. Welches Modulationsverfahren wird hiermit beschrieben?

: Welche Konsequenzen ergeben sich hieraus? Was ändert sich mit $A_{\rm T} = 0$? }}

::Es handelt sich um eine '''ZSB–AM mit Träger''' mit dem Modulationsgrad $m=0.8/0.6 = 1.333$. Für $m > 1$ ist allerdings eine [[Modulation_Methods/Synchrondemodulation|Synchrondemodulation]] erforderlich. [[Modulation_Methods/Hüllkurvendemodulation|Hüllkurvendemodulation]] funktioniert nicht mehr. Ein Grund hierfür ist, dass nun die Nulldurchgänge von $x(t)$ nicht mehr im äquidistanten Abstand von $5\ \rm µ s$ auftreten   ⇒   zusätzliche Phasenmodulation.

::Mit $A_{\rm T} = 0$   ⇒   $m \to \infty$ ergibt sich eine '''ZSB–AM ohne Träger'''. Auch hierfür benötigt man unbedingt die Synchrondemodulation.

{{BlaueBox|TEXT=
'''(7)'''   Nun gelte   $\text{Rot:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$,   $\text{Grün:} \hspace{0.15cm} A_{\rm U} = 0$,   $\text{Blau:} \hspace{0.15cm} A_{\rm O} = 0.8\ \text{V}, \ f_{\rm O} = 120 \ \text{kHz}, \ \varphi_{\rm O} = 90^\circ$.

:Welche Konstellation wird hiermit beschrieben? Welche Figur ergibt sich für das äquivalente Tiefpass–Signal $x_{\rm TP}(t)$?   ⇒   "Ortskurve"? Was ändert sich mit $A_{\rm U} = 0.8\ \text{V}$ und $A_{\rm O} = 0$?}}

::In beiden Fällen handelt es sich um eine [[Modulation_Methods/Einseitenbandmodulation|Einseitenbandmodulation]] '''(ESB–AM)''' mit dem Modulationsgrad $\mu = 0.8$ (bei ESB bezeichnen wir den Modulationsgrad mit $\mu$ anstelle von $m$). Das Trägersignal ist cosinusförmig und das Nachrichtensignal sinusförmig. Das äquivalente Tiefpass–Signal $x_{\rm TP}(t)$ hat in der komplexen Ebene einen kreisförmigen Verlauf.

::Mit $A_{\rm O} = 0.8\ \text{V}$, $A_{\rm U} = 0$ handelt es sich um eine OSB–Modulation. Der grüne Zeiger fehlt und der blaue Zeiger dreht im Vergleich zum roten Träger schneller.

::Mit $A_{\rm U} = 0.8\ \text{V}$, $A_{\rm O} = 0$ handelt es sich um eine USB–Modulation. Der blaue Zeiger fehlt und der grüne Zeiger dreht im Vergleich zum roten Träger langsamer.

{{BlaueBox|TEXT=
'''(8)'''   Es gelte   $\text{Rot:} \hspace{0.05cm} A_{\rm T} = 1\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$,   $\text{Grün:} \hspace{0.05cm} A_{\rm U} = 0.4\ \text{V}, \ f_{\rm U} = 80 \ \text{kHz}, \ \varphi_{\rm U} = -90^\circ$,   $\text{Blau:} \hspace{0.05cm} A_{\rm O} = 0.2\ \text{V}, \ f_{\rm O} = 120 \ \text{kHz}, \ \varphi_{\rm O} = +90^\circ$.

:Welche Konstellation könnte hiermit beschrieben werden? Welche Figur ergibt sich für das äquivalente Tiefpass–Signal $x_{\rm TP}(t)$?   ⇒   "Ortskurve"?}}

::Es könnte eine ZSB–AM eines Sinussignals mit cosinusförmigem Träger und Modulationsgrad $m=0.8$ wie in '''(3)''' vorliegen, bei dem aber das Obere Seitenband um den Faktor $2$ gedämpft ist. Das äquivalente Tiefpass–Signal $x_{\rm TP}(t)$ hat in der komplexen Ebene einen elliptischen Verlauf.

==Zur Handhabung des Applets==
 
[[File:Zeigerdiagramm_abzug.png|right]]

* Die roten Parameter $(A_{\rm T}, \ f_{\rm T}, \ \varphi_{\rm T})$ und der rote Zeiger kennzeichnen den '''T'''räger.
* Die grünen Parameter $(A_{\rm U}, \ f_{\rm U} < f_{\rm T}, \ \varphi_{\rm U})$ kennzeichnen das '''U'''ntere Seitenband.
* Die blauen Parameter $(A_{\rm O}, \ f_{\rm O} > f_{\rm T}, \ \varphi_{\rm O})$ kennzeichnen das '''O'''bere Seitenband.
*Alle Zeiger drehen in mathematisch positiver Richtung (entgegen dem Uhrzeigersinn).

 
Bedeutung der Buchstaben in nebenstehender Grafik:

    '''(A)'''     Grafikfeld für das analytische Signal $x_{\rm +}(t)$

    '''(B)'''     Grafikfeld für das physikalische Signal $x(t)$

    '''(C)'''     Parametereingabe per Slider: Amplituden, Frequenzen, Phasenwerte

    '''(D)'''     Bedienelemente:   Start – Step – Pause/Continue – Reset

    '''(E)'''     Geschwindigkeit der Animation:   "Speed"   ⇒   Werte: 1, 2 oder 3

    '''(F)'''     "Trace"   ⇒   Ein oder Aus, Spur der komplexen Signalwerte $x_{\rm +}(t)$

    '''(G)'''     Numerikausgabe der Zeit $t$ und der Signalwerte  ${\rm Re}[x_{\rm +}(t)] = x(t)$  und  ${\rm Im}[x_{\rm +}(t)]$

    '''(H)'''     Variationsmöglichkeiten für die grafische Darstellung

$\hspace{1.5cm}$Zoom–Funktionen "$+$" (Vergrößern), "$-$" (Verkleinern) und $\rm o$ (Zurücksetzen)

$\hspace{1.5cm}$Verschieben mit "$\leftarrow$" (Ausschnitt nach links, Ordinate nach rechts), "$\uparrow$" "$\downarrow$" und "$\rightarrow$"

    '''(I)'''     Bereich für die Versuchsdurchführung:  Aufgabenauswahl und Aufgabenstellung

    '''(J)'''     Bereich für die Versuchsdurchführung:  Musterlösung
 
==Über die Autoren==
Dieses interaktive Berechnungstool wurde am [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik] der [https://www.tum.de/ Technischen Universität München] konzipiert und realisiert.
*Die erste Version wurde 2005 von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]] im Rahmen ihrer Diplomarbeit mit "FlashMX–Actionscript" erstellt (Betreuer: [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._Günter_Söder_(at_LNT_from_1974-2024)|Günter Söder]]).
*2018 wurde dieses Programm von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Xiaohan_Liu_.28Bachelorarbeit_2018.29|Xiaohan Liu]] im Rahmen ihrer Bachelorarbeit (Betreuer: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) neu gestaltet und erweitert.

Die Fertigstellung dieses Applets wurde durch  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  der Fakultät EI der TU München ermöglicht. Wir bedanken uns.

==Nochmalige Aufrufmöglichkeit des Applets in neuem Fenster==

{{LntAppletLink|physAnSignal_en}}
    ''Hinweis:''   Das Applet ist für den '''CHROME'''–Browser optimiert. Bei anderen Browsern kommt es teilweise zu Darstellungsproblemen.
[[de:Applets:Physikalisches Signal & Analytisches Signal]]

Applets:Dämpfung von Kupferkabeln

2026-03-16T15:58:59Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{LntAppletLink|attenuationCopperCables_en}}

==Programmbeschreibung==
 
Dieses Applet berechnet die Dämpfungsfunktion $a_{\rm K}(f)$ von leitungsgebundenen Übertragungsmedien (jeweils mit der Kabellänge $l$):
*Für Koaxialkabel verwendet man meist die Gleichung $a_{\rm K}(f)=(\alpha_0+\alpha_1\cdot f+\alpha_2\cdot \sqrt{f}) \cdot l$.
*Dagegen werden Zweidrahtleitungen oft in der Form $a_{\rm K}(f)=(k_1+k_2\cdot (f/{\rm MHz})^{k_3}) \cdot l$ dargestellt.
*Realisiert ist auch die Umrechnung der $(k_1, \ k_2, \ k_3)$–Darstellung in die $(\alpha_0, \ \alpha_1, \ \alpha_2)$–Form für $B = 30 \ \rm MHz$ und umgekehrt.

Außer der Dämpfungsfunktion $a_{\rm K}(f)$ können graphisch dargestellt werden:
*der zugehörige Betragsfrequenzgang $\left | H_{\rm K}(f)\right |=10^{-a_\text{K}(f)/20},$
*der Entzerrer–Frequenzgang $\left | H_{\rm E}(f)\right | = \left | H_{\rm CRO}(f) / H_{\rm K}(f)\right | $, der zu einem Nyquist–Gesamtfrequenzgang $ H_{\rm CRO}(f) $ führt,
*der entsprechende Betrags–Quadrat–Frequenzgang $\left | H_{\rm E}(f)\right |^2 $.

Das Integral über $\left | H_{\rm E}(f)\right |^2 $ ist ein Maß für die Rauschüberhöhung des ausgewählten Nyquist–Gesamtfrequenzgangs und damit auch für zu erwartende Fehlerwahrscheinlichkeit. Aus dieser wird der ''Gesamt–Wirkungsgrad''  $\eta_\text{K+E}$ für '''K'''anal und '''E'''ntzerrer berechnet, der im Applet in $\rm dB$ ausgegeben wird.

Durch Optimierung des Roll-off–Faktors $r$ des Cosinus–Roll-off–Frequenzgangs $ H_{\rm CRO}(f) $ kommt man zum ''Kanal–Wirkungsgrad''  $ \eta_\text{K}$. Dieser gibt also die Verschlechterung des Gesamtsystems aufgrund der Dämpfungsfunktion $a_{\rm K}(f)$ des Übertragungsmediums an.

==Theoretischer Hintergrund==
 
===Betragsfrequenzgang und Dämpfungsfunktion===
Es besteht folgender Zusammenhang zwischen dem Betragsfrequenzgang und der Dämpfungsfunktion:
:$$\left | H_{\rm K}(f)\right |=10^{-a_\text{K}(f)/20} = {\rm e}^{-a_\text{K, Np}(f)}.$$
*Der Index "K" soll deutlich machen, dass das betrachtete LZI–System ein '''K'''abel ist.
*Bei der ersten Berechnungsvorschrift ist die Dämpfungsfunktion $a_\text{K}(f)$ in $\rm dB$ (Dezibel) einzusetzen.
*Bei der zweiten Berechnungsvorschrift ist die Dämpfungsfunktion $a_\text{K, Np}(f)$ in $\rm Np$ (Neper) einzusetzen.
* Es gelten folgende Umrechnungen $\rm 1 \ dB = 0.05 \cdot \ln (10) \ Np= 0.1151 \ Np$ bzw. $\rm 1 \ Np = 20 \cdot \lg (e) \ dB= 8.6859 \ dB$.
* In diesem Applet werden ausschließlich die dB–Werte verwendet.

===Dämpfungsfunktion eines Koaxialkabels===
Die Dämpfungsfunktion eines Koaxialkabels der Länge $l$ wird in [Wel77]<ref name ='Wel77'>Wellhausen, H. W.: Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren. Frequenz 31, S. 23-28, 1977.</ref> wie folgt angegeben:
:$$a_{\rm K}(f)=(\alpha_0+\alpha_1\cdot f+\alpha_2\cdot \sqrt{f}) \cdot l.$$
*Beachten Sie bitte den Unterschied zwischen der Dämpfungsfunktion $a_{\rm K}(f)$ in $\rm dB$ und den "alpha"–Koeffizienten $\alpha_{\rm K}(f)=a_{\rm K}(f)/l$ mit anderen Pseudo–Einheiten.
*Die Dämpfungsfunktion $a_{\rm K}(f)$ ist direkt proportional zur Kabellänge $l$. Man bezeichnet den Quotienten $a_{\rm K}(f)/l$ als "Dämpfungsmaß" oder "kilometrische Dämpfung".
*Der frequenzunabhängige Anteil $α_0$ des Dämpfungsmaßes berücksichtigt die Ohmschen Verluste ("Leitungsverluste").
*Der frequenzproportionale Anteil $α_1 · f$ des Dämpfungsmaßes ist auf die Ableitungsverluste ("Querverluste") zurückzuführen.
*Der dominante Anteil $α_2$ geht auf den [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Frequency_response_of_a_coaxial_cable|Skineffekt]] zurück, der bewirkt, dass bei höherfrequentem Wechselstrom die Stromdichte im Leiterinneren niedriger ist als an der Oberfläche. Dadurch steigt der Widerstandsbelag einer elektrischen Leitung mit der Wurzel aus der Frequenz an.

Die Konstanten für das ''Normalkoaxialkabel'' mit 2.6 mm Innendurchmesser und 9.5 mm Außendurchmesser   ⇒  kurz '''Coax (2.6/9.5 mm)''' lauten:
:$$\alpha_0 = 0.014\, \frac{ {\rm dB} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_1 = 0.0038\, \frac{ {\rm dB} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 2.36\, \frac{ {\rm dB} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}.$$

Entsprechend gilt für das ''Kleinkoaxialkabel''   ⇒  kurz '''Coax (1.2/4.4 mm)''':
:$$\alpha_0 = 0.068\, \frac{ {\rm dB} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm}\alpha_1 = 0.0039\, \frac{ {\rm dB} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 =5.2\, \frac{ {\rm dB} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}.$$

Diese Werte können aus den geometrischen Abmessungen der Kabel berechnet werden und wurden durch Messungen am Fernmeldetechnischen Zentralamt in Darmstadt bestätigt – siehe [Wel77]<ref name ='Wel77'>Wellhausen, H. W.: Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren. Frequenz 31, S. 23-28, 1977.</ref> . Sie gelten für eine Temperatur von 20°C (293 K) und Frequenzen größer als 200 kHz.

===Dämpfungsfunktion einer Zweidrahtleitung===
Die Dämpfungsfunktion einer Zweidrahtleitung (englisch: ''Two–wired Line'') der Länge $l$ wird in [PW95]<ref name ='PW95'>Pollakowski, M.; Wellhausen, H.W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Mitteilung aus dem Forschungs- und Technologiezentrum der Deutschen Telekom AG, Darmstadt, Verlag für Wissenschaft und Leben Georg Heidecker, 1995.</ref> wie folgt angegeben:
:$$a_{\rm K}(f)=(k_1+k_2\cdot (f/{\rm MHz})^{k_3}) \cdot l.$$
Dieser Funktionsverlauf ist nicht direkt interpretierbar, sondern es handelt sich um eine phänomenologische Beschreibungsform.

Ebenfalls in [PW95]<ref name ='PW95'>Pollakowski, M.; Wellhausen, H.W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Mitteilung aus dem Forschungs- und Technologiezentrum der Deutschen Telekom AG, Darmstadt, Verlag für Wissenschaft und Leben Georg Heidecker, 1995.</ref>findet man die aus Messergebnissen ermittelten Konstanten für verschiedene Leitungsdurchmesser $d$:
* $d = 0.35 \ {\rm mm}$:   $k_1 = 7.9 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 15.1 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.62$,
* $d = 0.40 \ {\rm mm}$:   $k_1 = 5.1 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 14.3 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.59$,
* $d = 0.50 \ {\rm mm}$:   $k_1 = 4.4 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 10.8 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.60$,
* $d = 0.60 \ {\rm mm}$:   $k_1 = 3.8 \ {\rm dB/km}, \hspace{0.2cm}k_2 = \hspace{0.25cm}9.2 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.61$.

Man erkennt aus diesen Zahlenwerten:
*Dämpfungsmaß $α(f)$ und Dämpfungsfunktion $a_{\rm K}(f) = α(f) · l$ hängen signifikant vom Leitungsdurchmesser ab. Die seit 1994 verlegten Kabel mit $d = 0.35$ mm und $d = 0.5$ mm haben etwa ein um $10\%$ größeres Dämpfungsmaß als die älteren Leitungen mit $d = 0.4$bzw. $0.6$ mm.
*Dieser mit den Herstellungs– und Verlegungskosten begründete kleinere Durchmesser vermindert allerdings die Reichweite $l_{\rm max}$ der auf diesen Leitungen eingesetzten Übertragungssysteme signifikant, so dass im schlimmsten Fall teuere Zwischenregeneratoren eingesetzt werden müssen.
*Die heute üblichen Übertragungsverfahren für Kupferleitungen belegen allerdings nur ein relativ schmales Frequenzband, zum Beispiel sind dies bei [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|ISDN]] $120\ \rm kHz$ und bei [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_DSL|DSL]] ca. $1100 \ \rm kHz$. Für $f = 1 \ \rm MHz$ beträgt das Dämpfungsmaß für ein 0.4 mm–Kabel etwa $20 \ \rm dB/km$, so dass selbst bei einer Kabellänge von $l = 4 \ \rm km$ der Dämpfungswert nicht über $80 \ \rm dB$ liegt.

===Umrechnung zwischen $k$– und $\alpha$– Parametern===
Es besteht die Möglichkeit, die $k$–Parameter des Dämpfungsmaßes   ⇒   $\alpha_{\rm I} (f)$ in entsprechende $\alpha$–Parameter   ⇒   $\alpha_{\rm II} (f)$ umzurechnen:
:$$\alpha_{\rm I} (f) = k_1 + k_2 \cdot (f/f_0)^{k_3}\hspace{0.05cm}, \hspace{0.2cm}{\rm mit} \hspace{0.15cm} f_0 = 1\,{\rm MHz},$$
:$$\alpha_{\rm II} (f) = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}.$$

Als Kriterium dieser Umrechnung gehen wir davon aus, dass die quadratische Abweichung dieser beiden Funktionen innerhalb einer Bandbreite $B$ minimal ist:
:$$\int_{0}^{B} \left [ \alpha_{\rm I} (f) - \alpha_{\rm II} (f)\right ]^2 \hspace{0.1cm}{\rm d}f \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Minimum} \hspace{0.05cm} .$$
Es ist offensichtlich, dass $α_0 = k_1$ gelten wird. Die Parameter $α_1$ und $α_2$ sind von der zugrundegelegten Bandbreite $B$ abhängig und lauten:
:$$\begin{align*}\alpha_1 & = 15 \cdot (B/f_0)^{k_3 -1}\cdot \frac{k_3 -0.5}{(k_3 + 1.5)(k_3 + 2)}\cdot {k_2}/{ {f_0} }\hspace{0.05cm} ,\\ \alpha_2 & = 10 \cdot (B/f_0)^{k_3 -0.5}\cdot \frac{1-k_3}{(k_3 + 1.5)(k_3 + 2)}\cdot {k_2}/{\sqrt{f_0} }\hspace{0.05cm} .\end{align*}$$

In der Gegenrichtung lautet die Umrechnungsvorschrift für den Exponenten:

:$$k_3 = \frac{A + 0.5} {A +1}, \hspace{0.2cm}\text{Hilfsgröße: }A = \frac{2} {3} \cdot \frac{\alpha_1 \cdot \sqrt{f_0}}{\alpha_2} \cdot \sqrt{B/f_0}.$$

Mit diesem Ergebnis lässt sich $k_2$ mit jeder der oberen Gleichungen angeben.

{{GraueBox|TEXT=
$\text{Beispiel 1:}$  Im Folgenden verwenden wir die Normierunggröße $f_0 = 1 \ \rm MHz$.
*Für $k_3 = 1$ (frequenzproportionales Dämpfungsmaß) ergeben sich folgerichtig   $\alpha_0 = k_0\hspace{0.05cm} ,\hspace{0.2cm} \alpha_1 = {k_2}/{ {f_0} }\hspace{0.05cm} ,\hspace{0.2cm} \alpha_2 = 0\hspace{0.05cm} .$
*Für $k_3 = 0.5$ (entsprechend Skineffekt) erhält man folgende Koeffizienten:   $\alpha_0 = k_0\hspace{0.05cm} ,\hspace{0.2cm}\alpha_1 = 0\hspace{0.05cm} ,\hspace{0.2cm} \alpha_2 = {k_2}/{\sqrt{f_0} }\hspace{0.05cm}.$
*Für $k_3 < 0.5$ ergibt sich ein negatives $\alpha_1$. Umrechnung ist nur für $0.5 \le k_3 \le 1$ möglich.
*Für $0.5 \le k_3 \le$ ergeben sich Koeffizienten $\alpha_1 > 0$ und $\alpha_2 > 0$, die auch von $B/f_0$ abhängen.
*Aus $\alpha_1 = 0.3\, {\rm dB}/ ({\rm km \cdot MHz}) \hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 3\, {\rm dB}/ ({\rm km \cdot \sqrt{MHz} })\hspace{0.05cm},\hspace{0.2cm}B = 30 \ \rm MHz$ folgt $k_3 = 0.63$ und $k_2 = 2.9 \ \rm dB/km$.}}

===Zum Kanaleinfluss auf die binäre Nyquistentzerrung===
[[File:Applet_Kabeldaempfung_1_version3.png|right|frame|Vereinfachtes Blockschaltbild des optimalen Nyquistentzerrers|class=fit]]
Wir gehen vom skizzierten Blockschaltbild aus. Zwischen der Diracquelle und dem Entscheider liegen die Frequenzgänge für Sender  ⇒  $H_{\rm S}(f)$, Kanal  ⇒  $H_{\rm K}(f)$ und Empfänger   ⇒  $H_{\rm E}(f)$.

In diesem Applet
*vernachlässigen wir den Einfluss der Sendeimpulsform   ⇒   $H_{\rm S}(f) \equiv 1$   ⇒   diracförmiges Sendesignal $s(t)$,
*setzen ein binäres Nyquistsystem mit Cosinus–Roll-off um die Nyquistfrequenz $f_{\rm Nyq} = [f_1 + f_2]/2 =1(2T)$ voraus:
:$$H_{\rm K}(f) · H_{\rm E}(f) = H_{\rm CRO}(f).$$

[[File:Applet_Kabeldaempfung_2_version2.png|right|frame|Frequenzgang mit Cosinus–Roll-off|class=fit]]

Das bedeutet: Das [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#First_Nyquist_criterion_in_the_frequency_domain|erste Nyquistkriterium]] wird erfüllt  ⇒   zeitlich aufeinander folgende Impulse stören sich nicht gegenseitig   ⇒   es gibt keine [[Digital_Signal_Transmission/Ursachen_und_Auswirkungen_von_Impulsinterferenzen|Impulsinterferenzen]] (englisch: ''Intersymbol Interference'', ISI).

Bei weißem Rauschen wird somit die Übertragungsqualität allein durch die Rauschleistung vor dem Empfänger bestimmt:

:$$P_{\rm N} =\frac{N_0}{2} \cdot \int_{-\infty}^{+\infty} |H_{\rm E}(f)|^2 \ {\rm d}f\hspace{1cm}\text{mit}\hspace{1cm}|H_{\rm E}(f)|^2 = \frac{|H_{\rm CRO}(f)|^2}{|H_{\rm K}(f)|^2}.$$
 
Die kleinstmögliche Rauschleistung ergibt sich bei idealem Kanal   ⇒   $H_{\rm K}(f) \equiv 1$ und gleichzeitig dem Frequenzgang $H_{\rm CRO}(f)$ mit Roll-off–Faktor $r = 1$ im Bereich $|f| \le 2 \cdot f_{\rm Nyq}$ (siehe Skizze):

:$$P_\text{N, min} = P_{\rm N} \ \big [\text{optimales System: }H_{\rm K}(f) \equiv 1; \ \text{ Roll-off–Faktor } r=r_{\rm opt} =1 \big ] = N_0 \cdot 3/4 \cdot f_{\rm Nyq} .$$

{{BlaueBox|TEXT=
$\text{Definitionen:}$ 
*Als Gütekriterium für ein gegebenes System verwenden wir den '''Gesamt–Wirkungsgrad''':

:$$\eta_\text{K+E} = \frac{P_{\rm N} \ \big [\text{optimales System: }H_{\rm K}(f) \equiv 1, \ r=r_{\rm opt} =1 \big ]}{P_{\rm N} \ \big [\text{gegebenes System: Kanal }H_{\rm K}(f), \ \text{Roll-off-Faktor }r \big ]} =\left [ \frac{1}{3/4 \cdot f_{\rm Nyq} } \cdot \int_{0}^{+\infty} \vert H_{\rm E}(f) \vert^2 \ {\rm d}f \right ]^{-1}\le 1.$$

:Diese Systemgröße wird im Applet für beide Parametersätze in logarithmierter Form angegeben:   $10 \cdot \lg \ \eta_\text{K+E} \le 0 \ \rm dB$.

*Durch Variation und Optimierung des Roll-off-Faktors $r$ erhält man den '''Kanal–Wirkungsgrad''':

:$$\eta_\text{K} = \max_{0 \le r \le 1} \ \eta_\text{K+E} .$$}}

[[File:Applet_Kabeldaempfung_3_version2.png|right|frame|Betrags–Quadrat–Frequenzgang $\left \vert H_{\rm E}(f)\right \vert ^2 $|class=fit]]
{{GraueBox|TEXT=
$\text{Beispiel 2:}$ 
Die Grafik zeigt den Betrags–Quadrat–Frequenzgang $\left \vert H_{\rm E}(f)\right \vert ^2 $ mit $\left \vert H_{\rm E}(f)\right \vert = H_{\rm CRO}(f) / \left \vert H_{\rm K}(f)\right \vert$ für folgende Randbedingungen:
*Dämpfungsfunktion des Kanals:   $a_{\rm K}(f) = 1 \ {\rm dB} \cdot \sqrt{f/\ {\rm MHz} }$,
*Nyquist–Frequenz: :   $f_{\rm Nyq} = 20 \ {\rm MHz}$, Roll-off-Faktor $r = 0.5$

Daraus ergeben sich folgende Konsequenzen:
*Im Bereich bis $f_{1} = 10 \ {\rm MHz}$ ist $H_{\rm CRO}(f) = 1$   ⇒   $\left \vert H_{\rm E}(f)\right \vert ^2 = \left \vert H_{\rm K}(f)\right \vert ^{-2}$ (siehe gelbe Hinterlegung).
*Erst im Bereich von $f_{1}$ bis $f_{2} = 30 \ {\rm MHz}$ ist die Flanke von $H_{\rm CRO}(f)$ wirksam und $\left \vert H_{\rm E}(f)\right \vert ^2$ wird immer kleiner.
*Das Maximum von $\left \vert H_{\rm E}(f_{\rm max})\right \vert ^2$ bei $f_{\rm max} \approx 11.5 \ {\rm MHz}$ ist mehr als doppelt so groß wie $\left \vert H_{\rm E}(f = 0)\right \vert ^2 = 1$.
*Das Integral über $\left \vert H_{\rm E}(f)\right \vert ^2$ ist ein Maß für die wirksame Rauschleistung. Diese ist im Beispiel um den Faktor $4.6$ größer als die minimale Rauschleistung (für $a_{\rm K}(f) = 0 \ {\rm dB}$ und $r=1$)   ⇒   $10 \cdot \lg \ \eta_\text{K+E} \approx - 6.6 \ {\rm dB}.$}}

==Versuchsdurchführung==

[[File:Applet_Kabeldaempfung_6_version1.png|right]]
*Wählen Sie zunächst die Nummer '''1''' ... '''11''' der zu bearbeitenden Aufgabe.
*Der Aufgabentext wird angezeigt. Die Parameterwerte sind angepasst.
*Lösung nach Drücken von "Hide solution".
*Aufgabenstellung und Lösung in Englisch.

Die Nummer '''0''' entspricht einem "Reset":
*Gleiche Einstellung wie beim Programmstart.
*Ausgabe eines "Reset–Textes" mit weiteren Erläuterungen zum Applet.

In der folgenden Beschreibung bezeichnet '''Blue''' den linken Parametersatz (im Applet blau markiert) '''Red''' den rechten Parametersatz (im Applet rot markiert). Alle Angaben mit Hochkomma sind ohne Einheit, zum Beispiel steht ${\alpha_2}' =2$   für   $\alpha_2 =2\, {\rm dB} / ({\rm km \cdot \sqrt{MHz} })$.

{{BlaueBox|TEXT=
'''(1)'''  Setzen Sie '''Blue''' zunächst auf $\text{Coax (1.2/4.4 mm)}$ und anschließend auf $\text{Coax (2.6/9.5 mm)}$. Die Kabellänge sei jeweils $l_{\rm Blue}= 5\ \rm km$.
:Betrachten und Interpretieren Sie $a_{\rm K}(f)$ und $\vert H_{\rm K}(f) \vert$, insbesondere die Funktionswerte $a_{\rm K}(f = f_\star = 30 \ \rm MHz)$ und $\vert H_{\rm K}(f = 0) \vert$.}}

$\Rightarrow\hspace{0.3cm}\text{Näherungsweise steigt die Dämpfungsfunktion mit }\sqrt{f}\text{ und der Betragsfrequenzgang fällt ähnlich einer Exponentialfunktion};$

$\hspace{1.15cm}\text{Coax (1.2/4.4 mm): }a_{\rm K}(f = f_\star) = 143.3\text{ dB;}\hspace{0.5cm}\vert H_{\rm K}(f = 0) \vert = 0.96.$

$\hspace{1.15cm}\text{Coax (2.6/9.5 mm): }a_{\rm K}(f = f_\star) = 65.3\text{ dB;}\hspace{0.5cm}\vert H_{\rm K}(f = 0) \vert = 0.99;$

{{BlaueBox|TEXT=
'''(2)'''  Für '''Blue''' gelte $\text{Coax (2.6/9.5 mm)}$ und $l_{\rm Blue} = 5\ \rm km$. Wie wird $a_{\rm K}(f =f_\star = 30 \ \rm MHz)$ von $\alpha_0$, $\alpha_1$ und $\alpha_2$ beeinflusst?}}

$\Rightarrow\hspace{0.3cm}\text{Entscheidend ist }\alpha_2\text{ (Skineffekt). Die Beitrag von } \alpha_0\text{ ist nur ca. 0.1 dB und der von }\alpha_1 \text{ nur ca. 0.6 dB.}$

{{BlaueBox|TEXT=
'''(3)'''  Setzen Sie zusätzlich '''Red''' auf $\text{Two–wired Line (0.5 mm)}$ und $l_{\rm Red} = 1\ \rm km$. Welcher Wert ergibt sich für $a_{\rm K}(f =f_\star= 30 \ \rm MHz)$?
:Bis zu welcher Länge $l_{\rm Red}$ ist die rote Dämpfungsfunktion vergleichbar mit der blauen?}}

$\Rightarrow\hspace{0.3cm}\text{Für die rote Kurve gilt: }a_{\rm K}(f = f_\star) = 87.5 {\ \rm dB} \text{. Obige Bedingung wird erfüllt für }l_{\rm Red} = 0.7\ {\rm km} \ \Rightarrow \ a_{\rm K}(f = f_\star) = 61.3 {\ \rm dB}.$

{{BlaueBox|TEXT=
'''(4)'''  Setzen Sie '''Red''' auf ${k_1}' = 0, {k_2}' = 10, {k_3}' = 0.75, {l_{\rm red} } = 1 \ \rm km$ und variieren Sie den Parameter $0.5 \le k_3 \le 1$.
:Was erkennt man anhand von $a_{\rm K}(f)$ und $\vert H_{\rm K}(f) \vert$? }}

$\Rightarrow\hspace{0.3cm}\text{Bei festem }k_2\text {wird }a_{\rm K}(f)\text{ mit größerem }k_3\text{ immer größer und }\vert H_{\rm K}(f) \vert \text{ nimmt immer schneller ab. Mit }k_3 =1: a_{\rm K}(f)\text{ steigt linear.}$

$\hspace{1.15cm}\text{Mit }k_3 \to 0.5\text{ wird die Dämpfungsfunktion wie beim Koaxialkabel immer mehr durch den Skineffekt bestimmt.}$

{{BlaueBox|TEXT=
'''(5)'''  Setzen Sie '''Red''' auf $\text{Two–wired Line (0.5 mm)}$ und '''Blue''' auf $\text{Conversion of Red}$. Es gelte $l_{\rm Red} = l_{\rm Blue} = 1\ \rm km$.
:Betrachten und interpretieren Sie die dargestellten Funktionsverläufe für $a_{\rm K}(f)$ und $\vert H_{\rm K}(f) \vert$.}}

$\Rightarrow\hspace{0.3cm}\text{Sehr gute Approximation der Zweidrahtleitung durch den blauen Parametersatz, sowohl bezüglich }a_{\rm K}(f) \text{ als auch }\vert H_{\rm K}(f) \vert.$

$\hspace{1.15cm}\text{Die errechneten Parameterwerte nach der Konvertierung sind }{\alpha_0}' = {k_1}' = 4.4, \ {\alpha_1}' = 0.76, \ {\alpha_2}' = 11.12.$

{{BlaueBox|TEXT=
'''(6)'''  Es gelten die Einstellungen von '''(5)'''. Welche Anteile der Dämpfungsfunktion gehen auf Ohmschen Verlust, Querverluste und Skineffekt zurück? }}

$\Rightarrow\hspace{0.3cm}\text{Lösung anhand '''Blue''': }a_{\rm K}(f = f_\star= 30 \ {\rm MHz}) = 88.1\ {\rm dB}, \hspace{0.2cm}\text{ohne }\alpha_0\text{: }83.7\ {\rm dB}, \hspace{0.2cm}\text{ohne }\alpha_0 \text{ und } \alpha_1\text{: }60.9\ {\rm dB}.$

$\hspace{1.15cm}\text{Bei einer Zweidrahtleitung ist der Einfluss der Längs– und der Querverluste signifikant größer als bei einem Koaxialkabel.}$

{{BlaueBox|TEXT=
'''(7)'''  Setzen Sie '''Blue''' auf ${\alpha_0}' = {\alpha_1}' ={\alpha_2}' = 0$ und '''Red''' auf ${k_1}' = 2, {k_2}' = 0, {l_{\rm red} } = 1 \ \rm km$. Zusätzlich gelte ${f_{\rm Nyq} }' =15$ und $r= 0.5$.
:Wie groß ist jeweils der Gesamt–Wirkungsgrad $\eta_\text{K+E}$ und der Kanal–Wirkungsgrad $\eta_\text{K}$?}}

$\Rightarrow\hspace{0.3cm}\text{Es gilt }10 \cdot \lg \ \eta_\text{K+E} = -0.7\ \ {\rm dB}\text{ (Blue: ideales System) und }10 \cdot \lg \ \eta_\text{K+E} = -2.7\ \ {\rm dB}\text{ (Red: nur Gleichsignaldämpfung)}$.

$\hspace{0.95cm}\text{Der bestmögliche Rolloff–Faktor ist }r = 1.\text{ Somit ist }10 \cdot \lg \ \eta_\text{K} = 0 \ {\rm dB}\text{ (Blue) bzw. }10 \cdot \lg \ \eta_\text{K} = -2\ {\rm dB}\text{ (Red)}.$

{{BlaueBox|TEXT=
'''(8)'''  Es gilt die Einstellung von '''(7)'''. Mit welcher Sendeleistung $P_{\rm red}$ in Bezug zu $P_{\rm blue}$ erreichen beide Systeme gleiche Fehlerwahrscheinlichkeit? }}

$\Rightarrow\hspace{0.3cm}\text{Es muss gelten: }10 \cdot \lg \ P_{\rm red}/P_{\rm blue} =2 \ {\rm dB} \ \ \text{ ⇒ } \ \ P_{\rm red}/P_{\rm blue} = 10^{0.2} = 1.585.$

{{BlaueBox|TEXT=
'''(9)'''  Setzen Sie '''Blue''' auf ${\alpha_0}' = {\alpha_1}' = 0, \ {\alpha_2}' = 3, \ {l_{\rm blue} }' = 2$ und '''Red''' auf "Inactive". Zusätzlich gelte ${f_{\rm Nyq} }' =15$ und $r= 0.7$.
:Welchen Verlauf hat $\vert H_{\rm E}(f) \vert$? Wie groß ist sind Gesamt–Wirkungsgrad $\eta_\text{K+E}$ und Kanal–Wirkungsgrad $\eta_\text{K}$?}}

$\Rightarrow\hspace{0.3cm}\text{Für } f < 7.5 {\ \rm MHz}\text{ ist } \vert H_{\rm E}(f) \vert = \vert H_{\rm K}(f) \vert ^{-1}.\text{ Für }(f > 22.5 {\ \rm MHz)}\text{ ist: }\vert H_{\rm E}(f) \vert = 0.\text{ Dazwischen Einfluss der CRO–Flanke.}$

$\hspace{0.95cm}\text{Der bestmögliche Rolloff–Faktor }r = 0.7\text{ ist bereits eingestellt: }\Rightarrow \ 10 \cdot \lg \ \eta_\text{K+E} = 10 \cdot \lg \ \eta_\text{K} \approx - 18.1 \ {\rm dB}.$

{{BlaueBox|TEXT=
'''(10)'''  Setzen Sie '''Blue''' auf ${\alpha_0}' = {\alpha_1}' = 0, \ {\alpha_2}' = 3, \ {l_{\rm blue} }' = 8$ sowie '''Red''' auf "Inactive". Zusätzlich gelte ${f_{\rm Nyq} }' =15$ und $r= 0.7$.
:Welchen Wert hat $\vert H_{\rm E}(f = 0) \vert$? Was ist der Maximalwert von $\vert H_{\rm E}(f) \vert$? Wie groß ist ist der Kanal–Wirkungsgrad $\eta_\text{K}$?}}

$\Rightarrow\hspace{0.3cm}\text{Es gilt }\vert H_{\rm E}(f = 0) \vert = \vert H_{\rm E}(f = 0) \vert ^{-1}= 1 \text{ und das Maximum von } \vert H_{\rm E}(f) \vert \text{ ist ca. }37500\text{ für }r=0.7 \Rightarrow 10 \cdot \lg \ \eta_\text{K+E} \approx -89.2 \ {\rm dB},$

$\hspace{0.95cm}\text{weil das Intergral über }\vert H_{\rm E}(f) \vert^2\text{sehr groß ist. Nach Optimierung von }r=0.17 \text{ erhält man }10 \cdot \lg \ \eta_\text{K} \approx -82.6 \ {\rm dB}.$

{{BlaueBox|TEXT=
'''(11)'''  Es gelten die Einstellungen von '''(10) und $r= 0.17$. Variieren Sie die Kabellänge bis $l_{\rm blue} =10 \ \rm km$.
:Wie ändert sich der Maximalwert von $\vert H_{\rm E}(f) \vert$, der Kanal–Wirkungsgrad $\eta_\text{K}$ und der optimale Roll–off–Faktor $r_{\rm opt}$?}}

$\Rightarrow\hspace{0.3cm}\text{Der Maximalwert von } \vert H_{\rm E}(f) \vert \text{wird immer größer und }10 \cdot \lg \ \eta_\text{K}\text{ immer kleiner.}$

$\hspace{0.95cm}\text{Bei 10 km Länge ist } 10 \cdot \lg \ \eta_\text{K} \approx -104.9 \ {\rm dB} \text{ und } r_{\rm opt}=0.14\text{. Für }f_\star \approx 14.5\ {\rm MHz} \text{ ist } \vert H_{\rm E}(f = f_\star) = 352000 \cdot \approx \vert H_{\rm E}(f =0)\vert$.

==Zur Handhabung des Applets==
[[File:Applet_Kabeldaempfung_5_version2.png|left|600px]]
    '''(A)'''     Vorauswahl für blauen Parametersatz

    '''(B)'''     Eingabe der $\alpha$–Parameter per Slider

    '''(C)'''     Vorauswahl für roten Parametersatz

    '''(D)'''     Eingabe der $k$–Parameter per Slider

    '''(E)'''     Eingabe der Parameter $f_{\rm Nyq}$ und $r$

    '''(F)'''     Auswahl für die graphische Darstellung

    '''(G)'''     Darstellung $a_\text{K}(f)$, $|H_\text{K}(f)|$, $|H_\text{E}(f)|$, ...

    '''(H)'''     Skalierungsfaktor $H_0$ für $|H_\text{E}(f)|$, $|H_\text{E}(f)|^2$

    '''(I)'''     Auswahl der Frequenz $f_\star$ für Numerikausgabe

    '''(J)'''     Numerikausgabe für blauen Parametersatz

    '''(K)'''     Numerikausgabe für roten Parametersatz

    '''(L)'''     Ausgabe Systemwirkungsgrad $\eta_\text{K+E}$ in dB

    '''(M)'''     Store & Recall von Einstellungen

    '''(N)'''     Bereich für die Versuchsdurchführung

    '''(O)'''     Variation der grafischen Darstellung:$\hspace{0.5cm}$"$+$" (Vergrößern),

$\hspace{0.5cm}$ "$-$" (Verkleinern)

$\hspace{0.5cm}$ "$\rm o$" (Zurücksetzen)

$\hspace{0.5cm}$ "$\leftarrow$" (Verschieben nach links), usw.

'''Andere Möglichkeiten zur Variation der grafischen Darstellung''':
*Gedrückte Shifttaste und Scrollen: Zoomen im Koordinatensystem,
*Gedrückte Shifttaste und linke Maustaste: Verschieben des Koordinatensystems.

==Über die Autoren==
Dieses interaktive Berechnungstool wurde am [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik] der [https://www.tum.de/ Technischen Universität München] konzipiert und realisiert.
*Die erste Version wurde 2009 von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Sebastian_Seitz_.28Diplomarbeit_LB_2009.29|Sebastian Seitz]] im Rahmen seiner Diplomarbeit erstellt (Betreuer: [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._Günter_Söder_(at_LNT_from_1974-2024)|Günter Söder]] und [[Biographies_and_Bibliographies/LNTwww_members_from_LÜT#Dr.-Ing._Bernhard_Göbel_(at_LÜT_from_2004-2010)|Bernhard Göbel]]).
*2018 wurde das Programm von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Jimmy_He_.28Bachelorarbeit_2018.29|Jimmy He]] (Bachelorarbeit, Betreuer: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) auf "HTML5" umgesetzt und neu gestaltet.

{{LntAppletLink|attenuationCopperCables_en}}
[[de:Applets:Dämpfung von Kupferkabeln]]

Applets:Physikalisches Signal & Äquivalentes TP-Signal

2026-03-16T15:58:58Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{LntAppletLink|physAnLPSignal_en}}

==Programmbeschreibung==
 
Dieses Applet zeigt den Zusammenhang zwischen dem physikalischen Bandpass–Signal $x(t)$ und dem dazugehörigen äquivalenten Tiefpass–Signal $x_{\rm TP}(t)$. Ausgegangen wird stets von einem Bandpass–Signal $x(t)$ mit frequenzdiskretem Spektrum $X(f)$:
:$$x(t) = x_{\rm T}(t) + x_{\rm O}(t) + x_{\rm U}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t- \varphi_{\rm T}\right)+A_{\rm O}\cdot \cos\left(2\pi f_{\rm O}\cdot t- \varphi_{\rm O}\right)+ A_{\rm U}\cdot \cos\left(2\pi f_{\rm U}\cdot t- \varphi_{\rm U}\right). $$
Das physikalische Signal $x(t)$ setzt sich also aus drei [[Signal_Representation/Harmonische_Schwingung|harmonischen Schwingungen]] zusammen, einer Konstellation, die sich zum Beispiel bei der [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#AM_signals_and_spectra_with_a_harmonic_input_signal|Zweiseitenband-Amplitudenmodulation]] des Nachrichtensignals $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t- \varphi_{\rm N}\right)$ mit dem Trägersignal $x_{\rm T}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t - \varphi_{\rm T}\right)$ ergibt. Die Nomenklatur ist ebenfalls an diesen Fall angepasst:
* $x_{\rm O}(t)$ bezeichnet das "Obere Seitenband" mit der Amplitude $A_{\rm O}= A_{\rm N}/2$, der Frequenz $f_{\rm O} = f_{\rm T} + f_{\rm N}$ und der Phase $\varphi_{\rm O} = \varphi_{\rm T} + \varphi_{\rm N}$.
*Entsprechend gilt für das "Untere Seitenband" $x_{\rm U}(t)$ mit $f_{\rm U} = f_{\rm T} - f_{\rm N}$, $A_{\rm U}= A_{\rm O}$ und $\varphi_{\rm U} = -\varphi_{\rm O}$.

Das dazugehörige äquivalente Tiefpass–Signal lautet mit $f_{\rm O}\hspace{0.01cm}' = f_{\rm O}- f_{\rm T} > 0$,   $f_{\rm U}\hspace{0.01cm}' = f_{\rm U}- f_{\rm T} < 0$  und  $f_{\rm T}\hspace{0.01cm}' = 0$:

:$$x_{\rm TP}(t) = x_\text{TP, T}(t) + x_\text{TP, O}(t) + x_\text{TP, U}(t) = A_{\rm T}\cdot {\rm e}^{-{\rm j} \varphi_{\rm T} } \hspace{0.1cm}+ \hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{-{\rm j} \varphi_{\rm O} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm}+ \hspace{0.1cm}A_{\rm U}\cdot {\rm e}^{-{\rm j} \varphi_{\rm U} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t} . $$

[[File:Ortskurve_1.png|right|frame|Äquivalentes TP–Signal zur Zeit $t=0$ bei cosinusförmigem Träger   ⇒   $\varphi_{\rm T} = 0$]]
Im Programm dargestellt wird $x_{\rm TP}(t)$ als vektorielle Summe dreier Drehzeiger als violetter Punkt (siehe beispielhafte Grafik für den Startzeitpunkt $t=0$ und cosinusförmigem Träger):

*Der (rote) Zeiger des Trägers $x_\text{TP, T}(t)$ mit der Länge $A_{\rm T}$ und der Nullphasenlage $\varphi_{\rm T} = 0$ liegt in der komplexen Ebene fest. Es gilt also für alle Zeiten $t$:   $x_{\rm TP}(t)= A_{\rm T}\cdot {\rm e}^{-{\rm j} \varphi_{\rm T} }$.

*Der (blaue) Zeiger des Oberen Seitenbandes $x_\text{TP, O}(t)$ mit der Länge $A_{\rm O}$ und der Nullphasenlage $\varphi_{\rm O}$ dreht mit der Winkelgeschwindigkeit $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.01cm}'$ in mathematisch positiver Richtung (eine Umdrehung in der Zeit $1/f_{\rm O}\hspace{0.01cm}')$.

*Der (grüne) Zeiger des Unteren Seitenbandes $x_{\rm U+}(t)$ mit der Länge $A_{\rm U}$ und der Nullphasenlage $\varphi_{\rm U}$ dreht mit der Winkelgeschwindigkeit $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.01cm}'$, wegen $f_{\rm U}\hspace{0.01cm}'<0$ im Uhrzeigersinn (mathematisch negative Richtung).

*Mit $f_{\rm U}\hspace{0.01cm}' = -f_{\rm O}\hspace{0.01cm}'$ drehen der blaue und der grüne Zeiger gleich schnell, aber in unterschiedlichen Richtungen. Gilt zudem $A_{\rm O} = A_{\rm U}$ und $\varphi_{\rm U} = -\varphi_{\rm O}$, so bewegt sich $x_{\rm TP}(t)$ auf einer Geraden mit einer Neigung von $\varphi_{\rm T}$.

''Hinweis:''   Die Grafik gilt für $\varphi_{\rm O} = +30^\circ$. Daraus folgt für den Startzeitpunkt $t=0$ der Winkel des blauen Zeigers (OSB) gegenüber dem Koordinatensystem:   $\phi_{\rm O} = -\varphi_{\rm O} = -30^\circ$. Ebenso folgt aus der Nullphasennlage $\varphi_{\rm U} = -30^\circ$ des unteren Seitenbandes (USB, grüner Zeiger) für den in der komplexen Ebene zu berücksichtigenden Phasenwinkel:   $\phi_{\rm U} = +30^\circ$.

Den zeitlichen Verlauf von $x_{\rm TP}(t)$ bezeichnen wir im Folgenden auch als '''Ortskurve'''. Der Zusammenhang zwischen $x_{\rm TP}(t)$ und dem physikalischen Bandpass–Signal $x(t)$ wird im Abschnitt [[???]] angegeben. Der Zusammenhang zwischen $x_{\rm TP}(t)$ und dem dazugehörigen analytischen Signal $x_+(t)$ lautet:

:$$x_{\rm TP}(t) = x_{\rm +}(t)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \cdot f_{\rm T}\cdot \hspace{0.05cm}t},$$
:$$x_{\rm +}(t) = x_{\rm TP}(t)\cdot {\rm e}^{+{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \cdot f_{\rm T}\cdot \hspace{0.05cm}t}.$$

[[Applets:Physical_Signal_%26_Equivalent_Low-pass_Signal|'''Englische Beschreibung''']]

==Theoretischer Hintergrund==
 
===Beschreibungsmöglichkeiten von Bandpass-Signalen===
[[File:Zeigerdiagramm_1a.png|right|frame|Bandpass–Spektrum $X(f)$ |class=fit]]
Wir betrachten hier '''Bandpass-Signale''' $x(t)$ mit der Eigenschaft, dass deren Spektren $X(f)$ nicht im Bereich um die Frequenz $f = 0$ liegen, sondern um eine Trägerfrequenz $f_{\rm T}$. Meist kann auch davon ausgegangen werden, dass die Bandbreite $B \ll f_{\rm T}$ ist.

Die Grafik zeigt ein solches Bandpass–Spektrum $X(f)$. Unter der Annahme, dass das zugehörige $x(t)$ ein physikalisches Signal und damit reell ist, ergibt sich für die Spektralfunktion $X(f)$ eine Symmetrie bezüglich der Frequenz $f = 0$. Ist $x(t)$ eine gerade Funktion   ⇒   $x(-t)=x(+t)$, so ist auch $X(f)$ reell und gerade.

Neben dem physikalischen Signal $x(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)$ verwendet man zur Beschreibung von Bandpass-Signalen gleichermaßen:
*das analytische Signal $x_+(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X_+(f)$, siehe Applet [[Applets:Physikalisches_Signal_%26_Analytisches_Signal|Physikalisches Signal & Analytisches Signal]],
*das äquivalente Tiefpass–Signal $x_{\rm TP}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X_{\rm TP}(f)$, wie im nächsten Unterabschnitt beschrieben.
 
===Spektralfunktionen des analytischen und des äquivalenten TP–Signals===

Das zum physikalischen Signal $x(t)$ gehörige '''analytische Signal''' $x_+(t)$ ist diejenige Zeitfunktion, deren Spektrum folgende Eigenschaft erfüllt:
[[File:Ortskurve_2.png|right|frame|Spektralfunktionen $X_+(f)$ und $X_{\rm TP}(f)$ |class=fit]]
:$$X_+(f)=\big[1+{\rm sign}(f)\big] \cdot X(f) = \left\{ {2 \cdotX(f) \; \hspace{0.2cm}\rm f\ddot{u}r\hspace{0.2cm} {\it f} > 0, \atop {\,\,\,\, \rm 0 \; \hspace{0.9cm}\rm f\ddot{u}r\hspace{0.2cm} {\it f} < 0.} }\right.$$

Die so genannte ''Signumfunktion'' ist dabei für positive Werte von $f$ gleich $+1$ und für negative $f$–Werte gleich $-1$.
*Der (beidseitige) Grenzwert liefert $\sign(0) = 0$.
*Der Index „+” soll deutlich machen, dass $X_+(f)$ nur Anteile bei positiven Frequenzen besitzt.

Aus der Grafik erkennt man die Berechnungsvorschrift für $X_+(f)$: Das tatsächliche BP–Spektrum $X(f)$ wird
*bei den positiven Frequenzen verdoppelt, und
*bei den negativen Frequenzen zu Null gesetzt.

Aufgrund der Unsymmetrie von $X_+(f)$ bezüglich der Frequenz $f = 0$ kann man bereits jetzt schon sagen, dass die Zeitfunktion $x_+(t)$ bis auf einen trivialen Sonderfall $x_+(t)= 0 \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ \ X_+(f)= 0$ stets komplex ist.

Zum Spektrum $X_{\rm TP}(f)$ des äquivalenten TP–Signals kommt man, indem man $X_+(f)$ um die Trägerfrequenz $f_{\rm T}$ nach links verschiebt:
:$$X_{\rm TP}(f)= X_+(f+f_{\rm T}).$$

Im Zeitbereich entspricht diese Operation der Multiplkation von $x_{\rm +}(t)$ mit der komplexen Exponentialfunktion mit negativem Exponenten:
:$$x_{\rm TP}(t) = x_{\rm +}(t)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \cdot f_{\rm T}\cdot \hspace{0.05cm}t}.$$

Man erkennt, dass $x_{\rm TP}(t)$ im Allgemeinen komplexwertig ist. Ist aber $X_+(f)$ symmetrisch um die Trägerfrequenz $f_{\rm T}$, so ist $X_{\rm TP}(f)$ symmetrisch um die Frequenz $f=0$ und es ergibt sich dementsprechend eine reelle Zeitfunktion $x_{\rm TP}(t)$.

===$x_{\rm TP}(t)$–Darstellung einer Summe aus drei harmonischen Schwingungen===

In unserem Applet setzen wir stets einen Zeigerverbund aus drei Drehzeigern voraus. Das physikalische Signal lautet:
:$$x(t) = x_{\rm U}(t) + x_{\rm T}(t) + x_{\rm O}(t) = A_{\rm U}\cdot \cos\left(2\pi f_{\rm U}\cdot t- \varphi_{\rm U}\right)+A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t- \varphi_{\rm T}\right)+A_{\rm O}\cdot \cos\left(2\pi f_{\rm O}\cdot t- \varphi_{\rm O}\right). $$
* Jede der drei harmonischen Schwingungen harmonischen Schwingungen $x_{\rm T}(t)$, $x_{\rm U}(t)$ und $x_{\rm O}(t)$ wird durch eine Amplitude $(A)$, eine Frequenz $(f)$ und einen Phasenwert $(\varphi)$ charakterisiert.
*Die Indizes sind an das Modulationsverfahren [[Modulation_Methods/Zweiseitenband-Amplitudenmodulation|Zweiseitenband–Amplitudenmodulation]] angelehnt. "T" steht für "Träger", "U" für "Unteres Seitenband" und "O" für "Oberes Seitenband". Entsprechend gilt stets $f_{\rm U} < f_{\rm T}$ und $f_{\rm O} > f_{\rm T}$. Für die Amplituden und Phasen gibt es keine Einschränkungen.

Das dazugehörige äquivalente Tiefpass–Signal lautet mit $f_{\rm O}\hspace{0.01cm}' = f_{\rm O}- f_{\rm T} > 0$,   $f_{\rm U}\hspace{0.01cm}' = f_{\rm U}- f_{\rm T} < 0$  und  $f_{\rm T}\hspace{0.01cm}' = 0$:

:$$x_{\rm TP}(t) = x_\text{TP, T}(t) + x_\text{TP, O}(t) + x_\text{TP, U}(t) = A_{\rm T}\cdot {\rm e}^{-{\rm j} \varphi_{\rm T} } \hspace{0.1cm}+ \hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{-{\rm j} \varphi_{\rm O} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm}+ \hspace{0.1cm}A_{\rm U}\cdot {\rm e}^{-{\rm j} \varphi_{\rm U} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t} . $$

{{GraueBox|TEXT=
$\text{Beispiel 1:}$ 
Die hier angegebene Konstellation ergibt sich zum Beispiel bei der [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#AM_signals_and_spectra_with_a_harmonic_input_signal|Zweiseitenband-Amplitudenmodulation]] des Nachrichtensignals $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t- \varphi_{\rm N}\right)$ mit dem Trägersignal $x_{\rm T}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t - \varphi_{\rm T}\right)$. Hierauf wird in der Versuchsdurchführung häufiger eingegangen.

[[File:Ortskurve_5.png|center|frame|Spektum $X_{\rm TP}(f)$ des äquivalenten TP–Signals für verschiedene Phasenkonstellationen |class=fit]]

Bei dieser Betrachtungsweise gibt es einige Einschränkungen bezüglich der Programmparameter:
* Für die Frequenzen gelte stets $f\hspace{0.05cm}'_{\rm O} = f_{\rm N}$ und $f\hspace{0.05cm}'_{\rm U} = -f_{\rm N}$.
*Ohne Verzerrungen sind die Amplitude der Seitenbänder $A_{\rm O}= A_{\rm U}= A_{\rm N}/2$.
*Die jeweiligen Phasenverhältnisse können der Grafik entnommen werden.

}}

===Darstellung des äquivalenten TP–Signals nach Betrag und Phase===

Das im Allgemeinen komplexwertige äquivalenten TP–Signal
:$$x_{\rm TP}(t) = a(t) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \phi(t) }$$
kann entsprechend der hier angegebenen Gleichung in eine Betragsfunktion $a(t)$ und eine Phasenfunktion $\phi(t)$ aufgespalten werden, wobei gilt:
:$$a(t) = \vert x_{\rm TP}(t)\vert = \sqrt{ {\rm Re}^2\big [x_{\rm TP}(t)\big ] + {\rm Im}^2\big [x_{\rm TP}(t)\big ] }\hspace{0.05cm},$$
:$$\phi(t) = \text{arc }x_{\rm TP}(t) = \arctan \frac{{\rm Im}\big [x_{\rm TP}(t)\big ]}{{\rm Re}\big [x_{\rm TP}(t)\big ]}.$$

Der Grund dafür, dass man ein Bandpass–Signal $x(t)$ meist durch das äquivalente TP–Signal $x_{\rm TP}(t)$ beschreibt ist, dass die Funktionen $a(t)$ und $\phi(t)$ in beiden Darstellungen interpretierbar sind:
*Der Betrag $a(t)$ des äquivalentes TP–Signals $x_{\rm TP}(t)$ gibt die (zeitabhängige) Hüllkurve von $x(t)$ an.
*Die Phase $\phi(t)$ von $x_{\rm TP}(t)$ kennzeichnet die Lage der Nulldurchgänge von $x(t)$, wobei gilt:
:–   Bei $\phi(t)>0$ ist der Nulldurchgang früher als seine Solllage   ⇒   das Signal ist hier vorlaufend.
:–  Bei $\phi(t)<0$ ist der Nulldurchgang später als seine Solllage   ⇒   das Signal ist hier nachlaufend.

{{GraueBox|TEXT=
$\text{Beispiel 2:}$ 
Die Grafik soll diesen Zusammenhang verdeutlichen, wobei $A_{\rm U} > A_{\rm O}$ vorausgesetzt ist   ⇒   der grüne Zeiger (für das untere Seitenband) ist länger als der blaue Zeiger (oberes Seitenband). Es handelt sich um eine Momentaufnahme zum Zeitpunkt $t_0$:

[[File:Ortskurve_3_neu.png|center|frame|Bandpass–Spektrum $X(f)$ |class=fit]]

*Bei diesen Systemparametern liegt die Spitze des Zeigerverbundes $x_{\rm TP}(t)$ – also die geometrisch Summe aus rotem, blauem und grünem Zeiger – auf einer Ellipse.
* In der linken Grafik schwarz eingezeichnet ist der Betrag $a(t_0) = \vert x_{\rm TP}(t_0) \vert$ und in brauner Farbe angedeutet ist der Phasenwert $\phi(t_0) = \text{arc }x_{\rm TP}(t_0) > 0.$
*In der rechten Grafik gibt der Betrag $a(t_0) = \vert x_{\rm TP}(t_0) \vert$ des äquivalenten TP–Signals die Hüllkurve des physikalischen Signals $x(t)$ an.
* Bei $\phi(t) \equiv 0$ würden alle Nulldurchgänge von $x(t)$ in äquidistenten Abständen auftreten. Wegen $\phi(t_0) > 0$ ist zum Zeitpunkt $t_0$ das Signal vorlaufend, das heißt: Die Nulldurchgänge kommen früher, als es das Raster vorgibt. }}

==Versuchsdurchführung==
[[File:Zeigerdiagramm_aufgabe_2.png|right]]
*Wählen Sie zunächst die Aufgabennummer.
*Eine Aufgabenbeschreibung wird angezeigt.
*Parameterwerte sind angepasst.
*Lösung nach Drücken von "Hide solition".

Mit der Nummer "0" wird auf die gleichen Einstellung wie beim Programmstart zurückgesetzt und es wird ein Text mit weiteren Erläuterungen zum Applet ausgegeben.

Im Folgenden bezeichnet $\rm Grün$ das Untere Seitenband   ⇒   $\big (A_{\rm U}, f_{\rm U}, \varphi_{\rm U}\big )$,  
$\rm Rot$ den Träger   ⇒   $\big (A_{\rm T}, f_{\rm T}, \varphi_{\rm T}\big )$ und
$\rm Blau$ das Obere Seitenband   ⇒   $\big (A_{\rm O}, f_{\rm O}, \varphi_{\rm O}\big )$.

{{BlaueBox|TEXT=
'''(1)'''   Es gelte   $\text{Rot:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, f_{\rm T} = 100 \ \text{kHz}, \varphi_{\rm T} = 0^\circ$,   $\text{Grün:} \hspace{0.15cm} A_{\rm U} = 0.4 \text{V}, \ f_{\rm U} = 80 \ \text{kHz}, \varphi_{\rm U} = -90^\circ$,   $\text{Blau:} \hspace{0.15cm} A_{\rm O} = 0.4\ \text{V}, f_{\rm O} = 120 \ \text{kHz}, \varphi_{\rm O} = 90^\circ$.

:Betrachten und interpretieren Sie das äquivalente TP–Signal $x_{\rm TP}(t)$ und das physikalische Signal $x(t)$. Welche Periodendauer $T_0$ erkennt man?}}

:: Das äquivalente TP–Signal $x_{\rm TP}(t)$ nimmt ausgehend von $x_{\rm TP}(t=0)=1\ \text{V}$ auf der reellen Achse Werte zwischen $0.2\ \text{V}$ und $1.8\ \text{V}$ an   ⇒   Phase $\phi(t) \equiv 0$.  Der Betrag $|x_{\rm TP}(t)|$ gibt die Hüllkurve $a(t)$ des physikalischen Signals $x(t)$ an. Es gilt mit $A_{\rm N} = 0.8\ \text{V}$ und $f_{\rm N} = 20\ \text{kHz}$:   $a(t) = A_{\rm T}+ A_{\rm N} \cdot \sin(2\pi\cdot f_{\rm N} \cdot t)$.  Sowohl $x_{\rm TP}(t)$ als auch $x(t)$ sind periodisch mit der Periodendauer $T_0 = 1/f_{\rm N} = 50\ \rm µ s$.

{{BlaueBox|TEXT=
'''(2)'''   Wie ändern sich die Verhältnisse gegenüber '''(1)''' mit $f_{\rm U} = 99 \ \text{kHz}$ und $f_{\rm O} = 101 \ \text{kHz}$ ? Wie könnte $x(t)$ entstanden sein?}}

:: Für die Hüllkurve $a(t)$ des Signals $x(t)$ gilt weiterhin $a(t) = A_{\rm T}+ A_{\rm N} \cdot \sin(2\pi\cdot f_{\rm N} \cdot t)$, aber nun mit $f_{\rm N} = 1\ \text{kHz}$. Auch wenn es nicht zu erkennen ist:  $x_{\rm TP}(t)$ und $x(t)$ sind weiterhin periodisch:   $T_0 = 1\ \rm ms$. Beispiel: Zweiseitenband–Amplitudenmodulation '''(ZSB–AM)''' eines Sinussignals mit Cosinus–Träger.

{{BlaueBox|TEXT=
'''(3)'''   Welche Einstellungen müssen gegenüber '''(2)''' geändert werden, um zur ZSB–AM eines Cosinussignals mit Sinus–Träger zu gelangen. Was ändert sich gegenüber '''(2)'''?}}

::Die Trägerphase muss auf $\varphi_{\rm T} = 90^\circ$ geändert werden   ⇒   Sinus–Träger. Ebenso muss $\varphi_{\rm O} =\varphi_{\rm U} =\varphi_{\rm T} = 90^\circ$ eingestellt werden   ⇒   cosinusförmige Nachricht  Die Ortskurve liegt nun auf der imaginären Achse  ⇒   $\phi(t) \equiv -90^\circ$. Zu Beginn gilt $x_{\rm TP}(t=0)= - {\rm j} \cdot 1.8 \ \text{V}$.

{{BlaueBox|TEXT=
'''(4)'''   Nun gelte   $\text{Rot:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$,   $\text{Grün:} \hspace{0.15cm} A_{\rm U} = 0.4 \text{V}, \ f_{\rm U} = 80 \ \text{kHz}, \ \varphi_{\rm U} = 0^\circ$,   $\text{Blau:} \hspace{0.15cm} A_{\rm O} = 0.4\ \text{V}, \ f_{\rm O} = 120 \ \text{kHz}, \ \varphi_{\rm O} = 0^\circ$.

:Welche Eigenschaften weist dieses System "ZSB–AM, wobei Nachrichtensignal und Träger jeweils cosinusförmig" auf? Wie groß ist der Modulationsgrad $m$?}}

:: Das äquivalente TP–Signal $x_{\rm TP}(t)$ nimmt ausgehend von $x_{\rm TP}(t=0)=1.8\ \text{V}$ auf der reellen Achse Werte zwischen $0.2\ \text{V}$ und $1.8\ \text{V}$ an   ⇒   Phase $\phi(t) \equiv 0$.  Bis auf den Startzustand $x_{\rm TP}(t=0)$ gleiches Verhalten wie bei der Einstellung '''(1)'''. Der Modulationsgrad ist jeweils $m = 0.8$.

{{BlaueBox|TEXT=
'''(5)'''   Es gelten weiter die Parameter gemäß '''(4)''' mit Ausnahme von $A_{\rm T}= 0.6 \text{V}$. Wie groß ist nun der Modulationsgrad $m$? Welche Konsequenzen hat das?}}

:: Es liegt nun eine ZSB–AM mit Modulationsgrad $m = 1.333$ vor. Bei $m > 1$ ist die einfachere [[Modulation_Methods/Hüllkurvendemodulation|Hüllkurvendemodulation]] nicht anwendbar, da nun die Phasenfunktion $\phi(t) \in \{ 0, \ \pm 180^\circ\}$ nicht mehr konstant ist und die Hüllkurve $a(t)$ nicht mehr mit dem Nachrichtensignal übereinstimmt. Vielmehr muss die aufwändigere [[Modulation_Methods/Synchrondemodulation|Synchrondemodulation]] verwendet werden. Bei Hüllkurvendemodulation käme es zu nichtlinearen Verzerrungen.

{{BlaueBox|TEXT=
'''(6)'''   Es gelten weiter die Parameter gemäß '''(4)''' bzw. '''(5)''' mit Ausnahme von $A_{\rm T}= 0$ an   ⇒   $m \to \infty$. Welches Modulationsverfahren wird so beschrieben?}}

::Es handelt sich um eine '''ZSB–AM ohne Träger''' und es ist eine eine Synchrondemodulation erforderlich. Das äquivalente TP–Signal $x_{\rm TP}(t)$ liegt zwar auf der reellen Achse, aber nicht nur in der rechten Halbebene. Damit gilt auch hier für die Phasenfunktion $\phi(t) \in \{ 0, \ \pm 180^\circ\}$, wodurch Hüllkurvendemodulation nicht anwendbar ist.

{{BlaueBox|TEXT=
'''(7)'''   Nun gelte   $\text{Rot:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, f_{\rm T} = 100 \ \text{kHz}, \varphi_{\rm T} = 0^\circ$,   $\text{Grün:} \hspace{0.15cm} A_{\rm U} = 0, \ f_{\rm U} = 80 \ \text{kHz}, \varphi_{\rm U} = -90^\circ$,   $\text{Blau:} \hspace{0.15cm} A_{\rm O} = 0.8\ \text{V}, f_{\rm O} = 120 \ \text{kHz}, \varphi_{\rm O} = 90^\circ$.

:Welches Konstellation wird hiermit beschrieben? Welche Eigenschaften dieses Verfahrens erkennt man aus der Grafik?}}

::Es handelt es sich um eine [[Modulation_Methods/Einseitenbandmodulation|Einseitenbandmodulation]] '''(ESB–AM)''', genauer gesagt um eine '''OSB–AM''': Der rote Träger liegt fest, der grüne Zeiger fehlt und der blaue Zeiger (OSB) dreht entgegen dem Uhrzeigersinn. Der Modulationsgrad ist $\mu = 0.8$ (bei ESB bezeichnen wir den Modulationsgrad mit $\mu$ anstelle von $m$). Das Trägersignal ist cosinusförmig und das Nachrichtensignal sinusförmig. Die Ortskurve ist ein Kreis. $x_{\rm TP}(t)$ bewegt sich darauf in mathematisch positiver Richtung. Wegen $\phi(t) \ne \text{const.}$ ist auch hier die Hüllkurvendemodulation nicht anwendbar:  Dies erkennt man daran, dass die Hüllkurve $a(t)$ nicht cosinusförmig ist. Vielmehr ist die untere Halbwelle spitzer als die obere   ⇒   starke lineare Verzerrungen.

{{BlaueBox|TEXT=
'''(8)'''   Es gelten weiter die Parameter gemäß '''(7)''' mit Ausnahme von $A_{\rm O}= 0$ und $A_{\rm U}= 0.8 \text{V}$. Welche Unterschiede ergeben sich gegenüber '''(7)'''?}}

::Nun handelt es sich um eine '''USB–AM''': Der rote Träger liegt fest, der blaue Zeiger fehlt und der grüne Zeiger (USB) dreht im Uhrzeigersinn. Alle anderen Aussagen von '''(7)''' treffen auch hier zu.

{{BlaueBox|TEXT=
'''(9)'''   Es gelten weiter die Parameter gemäß '''(7)''' mit Ausnahme von $A_{\rm O} = 0.2 \text{ V} \ne A_{\rm U} = 0.4 \text{ V} $. Welche Unterschiede ergeben sich gegenüber '''(7)'''?}}

::Die Ortskurve $x_{\rm TP}(t)$ ist nun keine horizontale Gerade, sondern eine Ellipse mit dem Realteil zwischen $0.4 \text{ V}$ und $1.6 \text{ V}$ sowie dem Imaginärteil im Bereich $\pm 0.2 \text{ V}$. Wegen $\phi(t) \ne \text{const.}$ würde auch hier die Hüllkurvendemodulation zu nichtlinearen Verzerrungen führen Die hier simulierte Konstellation beschreibt die Situation von '''(4)''', nämlich eine ZSB–AM mit Modulationsgrad $m = 0.8$, wobei das obere Seitenband aufgrund der Kanaldämpfung auf $50\%$ reduziert wird.

==Zur Handhabung des Applets==
 
[[File:Ortskurve_abzug3.png|right]]
* Die roten Parameter $(A_{\rm T}, \ f_{\rm T}, \ \varphi_{\rm T})$ und der rote Zeiger kennzeichnen den '''T'''räger.
* Die grünen Parameter $(A_{\rm U}, \ f_{\rm U} < f_{\rm T}, \ \varphi_{\rm U})$ kennzeichnen das '''U'''ntere Seitenband.
* Die blauen Parameter $(A_{\rm O}, \ f_{\rm O} > f_{\rm T}, \ \varphi_{\rm O})$ kennzeichnen das '''O'''bere Seitenband.
* Der rote Zeiger dreht nicht.
* Der grüne Zeiger dreht in mathematisch negativer Richtung (im Uhrzeigersinn).
* Der blaue Zeiger dreht entgegen dem Uhrzeigersinn.

Bedeutung der Buchstaben in nebenstehender Grafik:

    '''(A)'''     Grafikfeld für das äquivalente TP–Signal $x_{\rm TP}(t)$

    '''(B)'''     Grafikfeld für das physikalische Signal $x(t)$

    '''(C)'''     Parametereingabe per Slider:   Amplituden, Frequenzen, Phasenwerte

    '''(D)'''     Bedienelemente:   Start – Step – Pause/Continue – Reset

    '''(E)'''     Geschwindigkeit der Animation:   "Speed"   ⇒   Werte: 1, 2 oder 3

    '''(F)'''     "Trace"   ⇒   Ein oder Aus, Spur des äquivalenten TP–Signals   $x_{\rm TP}(t)$

    '''(G)'''     Numerikausgabe:   Zeit $t$, Signalwerte  ${\rm Re}[x_{\rm TP}(t)]$  und  ${\rm Im}[x_{\rm TP}(t)]$,

$\text{}\hspace{4.2cm}$   Hüllkurve $a(t) = |x_{\rm TP}(t)|$  und  Phase $\phi(t) = {\rm arc} \ x_{\rm TP}(t)$

    '''(H)'''     Variationsmöglichkeiten für die grafische Darstellung

$\hspace{1.5cm}$Zoom–Funktionen "$+$" (Vergrößern), "$-$" (Verkleinern) und $\rm o$ (Zurücksetzen)

$\hspace{1.5cm}$Verschieben mit "$\leftarrow$" (Ausschnitt nach links, Ordinate nach rechts), "$\uparrow$" "$\downarrow$" "$\rightarrow$"

    '''(I)'''     Bereich für die Versuchsdurchführung:  Aufgabenauswahl und Aufgabenstellung

    '''(J)'''     Bereich für die Versuchsdurchführung:  Musterlösung
 

==Über die Autoren==
Dieses interaktive Berechnungstool wurde am [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik] der [https://www.tum.de/ Technischen Universität München] konzipiert und realisiert.
*Die erste Version wurde 2005 von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]] im Rahmen ihrer Diplomarbeit mit "FlashMX–Actionscript" erstellt (Betreuer: [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._Günter_Söder_(at_LNT_from_1974-2024)|Günter Söder]]).
*2018 wurde dieses Programm von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Xiaohan_Liu_.28Bachelorarbeit_2018.29|Xiaohan Liu]] im Rahmen ihrer Bachelorarbeit (Betreuer: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) neu gestaltet und erweitert.

==Nochmalige Aufrufmöglichkeit des Applets in neuem Fenster==

{{LntAppletLink|physAnLPSignal_en}}
[[de:Applets:Physikalisches Signal & Äquivalentes TP-Signal]]

Applets:Physical Signal & Equivalent Lowpass Signal

2026-03-16T15:58:57Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{LntAppletLink|physAnSignal_en}}         [https://lntwww.lnt.ei.tum.de/Applets:Physikalisches_Signal_%26_%C3%84quivalentes_TP-Signal '''English Applet with German WIKI description''']
==Applet Description==
 
This applet shows the relationship between the physical band-pass signal $x(t)$ and the associated equivalent low-pass signal $x_{\rm TP}(t)$. It is assumed that the band-pass signal $x(t)$ has a frequency-discrete spectrum $X(f)$:
:$$x(t) = x_{\rm T}(t) + x_{\rm O}(t)+ x_{\rm U}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t- \varphi_{\rm T}\right)+A_{\rm O}\cdot \cos\left(2\pi f_{\rm O}\cdot t- \varphi_{\rm O}\right) + A_{\rm U}\cdot \cos\left(2\pi f_{\rm U}\cdot t- \varphi_{\rm U}\right). $$
The physical signal $x(t)$ is thus composed of three harmonic oscillations, a constellation that can be found, for example, in the ''Double-sideband Amplitude Modulation''
*of the message signal $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t- \varphi_{\rm N}\right)$   ⇒   in German:   '''N'''achrichtensignal
*with the carrier signal $x_{\rm T}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t - \varphi_{\rm T}\right)$   ⇒   in German:   '''T'''rägersignal.

The nomenclature is also adapted to this case:
* $x_{\rm O}(t)$ denotes the "upper sideband"   (in German:   '''O'''beres Seitenband) with the amplitude $A_{\rm O}= A_{\rm N}/2$, the frequency $f_{\rm O} = f_{\rm T} + f_{\rm N}$ and the phase $\varphi_{\rm O} = \varphi_{\rm T} + \varphi_{\rm N}$.
*Similarly, for the "lower sideband"   (in German:   '''U'''nteres Seitenband) $x_{\rm U}(t)$ with $f_{\rm U} = f_{\rm T} - f_{\rm N}$, $A_{\rm U}= A_{\rm O}$ and $\varphi_{\rm U} = -\varphi_{\rm O}$.

The associated equivalent low-pass signal is $f_{\rm O}\hspace{0.01cm}' = f_{\rm O}- f_{\rm T} > 0$,   $f_{\rm U}\hspace{0.01cm}' = f_{\rm U}- f_{\rm T} < 0$  and  $f_{\rm T}\hspace{0.01cm}' = 0$:

:$$x_{\rm TP}(t) = x_\text{TP, T}(t) + x_\text{TP, O}(t) + x_\text{TP, U}(t) = A_{\rm T}\cdot {\rm e}^{-{\rm j} \varphi_{\rm T} } \hspace{0.1cm}+ \hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{-{\rm j} \varphi_{\rm O} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm}+ \hspace{0.1cm}A_{\rm U}\cdot {\rm e}^{-{\rm j} \varphi_{\rm U} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t} . $$

[[File:Ortskurve_1.png|right|frame|Equivalent low-pass signal currently $t=0$ for cosinusoidal carrier   ⇒   $\varphi_{\rm T} = 0$]]
The program shows $x_{\rm TP}(t)$ as the vectorial sum of three rotation pointers as a violet dot (see figure for start time $t=0$ and cosinusoidal carrier):

*The (red) pointer of the carrier $x_\text{TP, T}(t)$ with the length $A_{\rm T}$ and the zero phase position $\varphi_{\rm T}=0$ is fixed in the complex plane. So it applies to all times $t$:   $x_{\rm TP}(t)= A_{\rm T}\cdot {\rm e}^{-{\rm j} \varphi_{\rm T} }$.

*The (blue) pointer of the upper sideband $x_\text{TP, O}(t)$ with the length $A_{\rm O}$ and the zero phase position $\varphi_{\rm O}$ rotates at the angular velocity $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.01cm}'$ in mathematically positive direction (one revolution in time $1/f_{\rm O}\hspace{0.01cm}')$.

*The (green) pointer of the lower sideband $x_{\rm U+}(t)$ with the length $A_{\rm U}$ and the zero phase position $\varphi_{\rm U}$ rotates at the angular velocity $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.01cm}'$, because of $f_{\rm U}\hspace{0.01cm}'<0$ counterclockwise.

*With $f_{\rm U}\hspace{0.01cm}' = -f_{\rm O}\hspace{0.01cm}'$ the blue and green pointers will spin at the same speed but in different directions. Also, if $A_{\rm O} = A_{\rm U}$ and $\varphi_{\rm U} = -\varphi_{\rm O}$, then $x_{\rm TP}(t)$ moves on a straight line with a incline of $\varphi_{\rm T}$.

''Note:''   In the figure $\varphi_{\rm O} = +30^\circ$. From this follows for the start time $t=0$ the angle of the upper sideband (OSB, blue pointer) with respect to the coordinate system:   $\phi_{\rm O} = -\varphi_{\rm O} = -30^\circ$. Likewise, the zero phase position $\varphi_{\rm U} = -30^\circ$ of the lower sideband (USB, green pointer) follows for the phase angle to be considered in the complex plane:   $\phi_{\rm U} = +30^\circ$.

The temporal process of $x_{\rm TP}(t)$ is also referred to below as "locus". The relationship between $x_{\rm TP}(t)$ and the physical band-pass signal $x(t)$ is given in the section and the associated analytic signal is $x_+(t)$ :

:$$x_{\rm TP}(t) = x_{\rm +}(t)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \cdot f_{\rm T}\cdot \hspace{0.05cm}t},$$
:$$x_{\rm +}(t) = x_{\rm TP}(t)\cdot {\rm e}^{+{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \cdot f_{\rm T}\cdot \hspace{0.05cm}t}.$$

==Theoretical Background==
 
===Description of Band-pass Signals===
[[File:Zeigerdiagramm_1a.png|right|frame|band-pass spectrum $X(f)$ |class=fit]]
We consider '''band-pass signals''' $x(t)$ with the property that their spectra $X(f)$ are not in the range around the frequency $f=0$, but around a carrier frequency $f_{\rm T}$. In most cases it can also be assumed that the bandwidth is $B \ll f_{\rm T}$.

The figure shows such a band-pass spectrum $X(f)$. Assuming that the associated $x(t)$ is a physical signal and thus real, the spectral function $X(f)$ has a symmetry with respect to the frequency $f = 0$, if $x(t)$ is an even function   ⇒   $x(-t)=x(t)$, $X(f)$ is real and even.

Beside the physical signal $x(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)$, one can also use the following descriptions of band-pass signals:
*the analytic signal $x_+(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X_+(f)$, see applet [[Applets:Physical_Signal_%26_Analytic_Signal|"Physical Signal & Analytic Signal"]],
*the equivalent low-pass signal $x_{\rm TP}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X_{\rm TP}(f)$, see next section
 

===Spectral Functions of the Analytic and the Equivalent Low-pass Signal===

The '''analytic signal''' $x_+(t)$ belonging to the physical signal $x(t)$ is the time function whose spectrum fulfills the following property:
[[File:EN_Sig_T_4_2_S1c.png|right|frame|spectral functions $X(f)$, $X_+(f)$ and $X_{\rm TP}(f)$|class=fit]]
:$$X_+(f)=\big[1+{\rm sign}(f)\big] \cdot X(f) = \left\{ {2 \cdotX(f) \; \hspace{0.2cm}\rm for\hspace{0.2cm} {\it f} > 0, \atop {\,\,\,\, \rm 0 \; \hspace{0.9cm}\rm for\hspace{0.2cm} {\it f} < 0.} }\right.$$

The ''Signum function'' is for positive values of $f$ equal to $+1$ and for negative $f$ values equal to $-1$.
*The (double-sided) limit returns $\sign(0) = 0$.
*The index „+” should make it clear that $X_+(f)$ only has parts at positive frequencies.

From the graph you can see the calculation rule for $X_+(f)$:   The actual band-pass spectrum $X(f)$ becomes
*doubled at the positive frequencies, and
*set to zero at the negative frequencies.

Due to the asymmetry of $X_+(f)$ with respect to the frequency $f = 0$, it can already be said that the time function $x_+(t)$ except for a trivial special case $x_+(t)= 0 \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ \ X_+(f)= 0$ is always complex.

The spectrum $X_{\rm TP}(f)$ of the equivalent low-pass signal is obtained by shifting $X_+(f)$ to the left by the carrier frequency $f_{\rm T}$:
:$$X_{\rm TP}(f)= X_+(f+f_{\rm T}).$$

In the time domain this operation corresponds to the multiplication of $x_{\rm +}(t)$ with the complex exponential function with negative exponent:
:$$x_{\rm TP}(t) = x_{\rm +}(t)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \cdot f_{\rm T}\cdot \hspace{0.05cm}t}.$$

It can be seen that $x_{\rm TP}(t)$ is generally complex. But if $X_+(f)$ is symmetric about the carrier frequency $f_{\rm T}$, $X_{\rm TP}(f)$ is symmetric about the frequency $f=0$ and there is accordingly a real time function $x_{\rm TP}(t)$.
 

===$x_{\rm TP}(t)$ Representation of a Sum of Three Harmonic Oscillations===

In our applet, we always assume a set of three rotating pointers. The physical signal is:
:$$x(t) = x_{\rm T}(t) + x_{\rm O}(t) + x_{\rm U}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t- \varphi_{\rm T}\right)+A_{\rm O}\cdot \cos\left(2\pi f_{\rm O}\cdot t- \varphi_{\rm O}\right) + A_{\rm U}\cdot \cos\left(2\pi f_{\rm U}\cdot t- \varphi_{\rm U}\right). $$
* Each of the three harmonic oscillations $x_{\rm T}(t)$, $x_{\rm U}(t)$ and $x_{\rm O}(t)$ is represented by an amplitude $(A)$, a frequency $(f)$ and a phase value $(\varphi)$.
*The indices are based on the modulation method [[Modulation_Methods/Zweiseitenband-Amplitudenmodulation|"double-sideband amplitude modulation"]]. "T" stands for "carrier", "U" for "lower sideband" and "O" for "upper Sideband". Similarly, $f_{\rm U} < f_{\rm T}$ and $f_{\rm O} > f_{\rm T}$. There are no restrictions for the amplitudes and phases.

The associated equivalent low-pass signal is with $f_{\rm O}\hspace{0.01cm}' = f_{\rm O}- f_{\rm T} > 0$,   $f_{\rm U}\hspace{0.01cm}' = f_{\rm U}- f_{\rm T} < 0$  and  $f_{\rm T}\hspace{0.01cm}' = 0$:

:$$x_{\rm TP}(t) = x_\text{TP, T}(t) + x_\text{TP, O}(t) + x_\text{TP, U}(t) = A_{\rm T}\cdot {\rm e}^{-{\rm j} \varphi_{\rm T} } \hspace{0.1cm}+ \hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{-{\rm j} \varphi_{\rm O} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm}+ \hspace{0.1cm}A_{\rm U}\cdot {\rm e}^{-{\rm j} \varphi_{\rm U} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t} . $$

{{GraueBox|TEXT=
$\text{Example 1:}$ 
The constellation given here results, for example, in the [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#AM_signals_and_spectra_with_a_harmonic_input_signal|"double-sideband amplitude modulation"]] of the message signal $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t- \varphi_{\rm N}\right)$ with the carrier signal $x_{\rm T}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t - \varphi_{\rm T}\right)$. This is discussed frequently in the exercises.

[[File:Ortskurve_5.png|center|frame|Spectrum $X_{\rm TP}(f)$ of the equivalent low–pass signal for different phase constellations |class=fit]]

There are some limitations to the program parameters in this approach:
* The frequencies are always $f\hspace{0.05cm}'_{\rm O} = f_{\rm N}$ and $f\hspace{0.05cm}'_{\rm U} = -f_{\rm N}$.
*Without distortion, the amplitude of the sidebands is $A_{\rm O}= A_{\rm U}= A_{\rm N}/2$.
*The respective phase relationships can be seen in the following graphic.

}}
 

===Representation of the Equivalent Low-pass Signal by Magnitude and Phase===

The generally complex equivalent low-pass signal
:$$x_{\rm TP}(t) = a(t) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \phi(t) }$$
can be split into a magnitude function $a(t)$ and a phase function $\phi(t)$ according to the equation given here, where:
:$$a(t) = \vert x_{\rm TP}(t)\vert = \sqrt{ {\rm Re}^2\big [x_{\rm TP}(t)\big ] + {\rm Im}^2\big [x_{\rm TP}(t)\big ] }\hspace{0.05cm},$$
:$$\phi(t) = \text{arc }x_{\rm TP}(t) = \arctan \frac{{\rm Im}\big [x_{\rm TP}(t)\big ]}{{\rm Re}\big [x_{\rm TP}(t)\big ]}.$$

The reason for this is that a band-pass signal $x(t)$ is usually described by the equivalent low-pass signal $x_{\rm TP}(t)$ that the functions $a(t)$ and $\phi(t)$ are interpretable in both representations:
*The magnitude $a(t)$ of the equivalent low-pass signal $x_{\rm TP}(t)$ indicates the (time-dependent) envelope of $x(t)$.
*The phase $\phi(t)$ of $x_{\rm TP}(t)$ denotes the location of the zero crossings of $x(t)$, where:
:–   For $\phi(t)>0$ the zero crossing is earlier than its nominal position   ⇒   the signal is leading here.
:–  When $\phi(t)<0$, the zero crossing is later than its target position   ⇒   the signal is trailing here.

{{GraueBox|TEXT=
$\text{Example 2:}$ 
The graph is intended to illustrate this relationship, assuming $A_{\rm U} > A_{\rm O}$   ⇒   the green pointer (for the lower sideband) is longer than the blue pointer (upper sideband). This is a snapshot at time $t_0$:

[[File:EN_Sig_T_4_2_ortskurve.png|center|frame|band-pass spectrum $X(f)$|class=fit]]

*For these system parameters, the top of the pointer cluster $x_{\rm TP}(t)$ – that is, the geometric sum of red, blue and green pointers – on an ellipse.
* The magnitude $a(t_0) = \vert x_{\rm TP}(t_0) \vert$ is drawn in black in the left-hand diagram and the phase value $\phi(t_0) = \text{arc }x_{\rm TP}(t_0) > 0$ is indicated in brown color.
*In the graph on the right, the magnitude $a(t_0) = \vert x_{\rm TP}(t_0) \vert$ of the equivalent low-pass signal indicates the envelope of the physical signal $x(t)$.
* At $\phi(t) \equiv 0$, all zero crossings of $x(t)$ would occur at equidistant intervals. Because of $\phi(t_0) > 0$, the signal is leading at the time $t_0$, that is: the zero crossings come earlier than the grid dictates. }}

==Exercises==
[[File:Exercises_verzerrungen.png|right]]
*First select the task number.
*A task description is displayed.
*Parameter values are adjusted.
*Solution after pressing "Hide solition".

The number "0" will reset the program and output a text with the further explanation of the applet.

In the following, $\rm Green$ denotes the lower sideband   ⇒   $\big (A_{\rm U}, f_{\rm U}, \varphi_{\rm U}\big )$,  
$\rm Red$ the carrier   ⇒   $\big (A_{\rm T}, f_{\rm T}, \varphi_{\rm T}\big )$ and
$\rm Blue$ the upper sideband   ⇒   $\big (A_{\rm O}, f_{\rm O}, \varphi_{\rm O}\big )$.

{{BlaueBox|TEXT=
'''(1)'''   Let   $\text{Red:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, f_{\rm T} = 100 \ \text{kHz}, \varphi_{\rm T} = 0^\circ$,   $\text{Green:} \hspace{0.15cm} A_{\rm U} = 0.4 \text{V}, \ f_{\rm U} = 80 \ \text{kHz}, \varphi_{\rm U} = -90^\circ$,   $\text{Blue:} \hspace{0.15cm} A_{\rm O} = 0.4\ \text{V}, f_{\rm O} = 120 \ \text{kHz}, \varphi_{\rm O} = 90^\circ$.

:Consider and interpret the equivalent low-pass signal $x_{\rm TP}(t)$ and the physical signal $x(t)$. Which period $T_0$ recognizable?}}

:: The equivalent low-pass signal $x_{\rm TP}(t)$ takes from $x_{\rm TP}(t=0)=1\ \text{V}$ on the real axis values between $0.2\ \text{V}$ and $1.8\ \text{V}$   ⇒   phase $\phi(t) \equiv 0$.  The magnitude $|x_{\rm TP}(t)|$ indicates the envelope $a(t)$ of the physical signal $x(t)$. It holds $A_{\rm N} = 0.8\ \text{V}$ and $f_{\rm N} = 20\ \text{kHz}$:   $a(t) = A_{\rm T}+ A_{\rm N} \cdot \sin(2\pi\cdot f_{\rm N} \cdot t)$.  Both $x_{\rm TP}(t)$ and $x(t)$ are periodically with the period $T_0 = 1/f_{\rm N} = 50\ \rm µ s$.

{{BlaueBox|TEXT=
'''(2)'''   How do the ratios change to '''(1)''' with $f_{\rm U} = 99 \ \text{kHz}$ and $f_{\rm O} = 101 \ \text{kHz}$ ? How could $x(t)$ have arisen?}}

:: For the envelope $a(t)$ of the signal $x(t)$ we still have $a(t) = A_{\rm T}+ A_{\rm N} \cdot \sin(2\pi\cdot f_{\rm N} \cdot t)$, but now $f_{\rm N} = 1\ \text{kHz}$. Even though it can not be recognized:  $x_{\rm TP}(t)$ and $x(t)$ are still periodically:   $T_0 = 1\ \rm ms$. Example: Double-sideband Amplitude modulation '''(DSB–AM)''' of a sine signal with cosine carrier.

{{BlaueBox|TEXT=
'''(3)'''   Which settings have to be changed from '''(2)''' in order to arrive at the DSB–AM of a cosine signal with sine carrier. What changes over '''(2)'''?}}

::The carrier phase must be changed to $\varphi_{\rm T} = 90^\circ$   ⇒   sine carrier. Similarly, $\varphi_{\rm O} =\varphi_{\rm U} =\varphi_{\rm T} = 90^\circ$ must be set   ⇒   cosinusoidal message  The locus now lies on the imaginary axis  ⇒   $\phi(t) \equiv -90^\circ$. At the beginning $x_{\rm TP}(t=0)= - {\rm j} \cdot 1.8 \ \text{V}$.

{{BlaueBox|TEXT=
'''(4)'''   Now let   $\text{Red:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$,   $\text{Green:} \hspace{0.15cm} A_{\rm U} = 0.4 \text{V}, \ f_{\rm U} = 80 \ \text{kHz}, \ \varphi_{\rm U} = 0^\circ$,   $\text{Blue:} \hspace{0.15cm} A_{\rm O} = 0.4\ \text{V}, \ f_{\rm O} = 120 \ \text{kHz}, \ \varphi_{\rm O} = 0^\circ$.

:What are the characteristics of this system "DSB–AM, where the message signal and carrier are respectively cosinusoidal"? What is the degree of modulation $m$?}}

:: The equivalent low-pass signal $x_{\rm TP}(t)$ takes from $x_{\rm TP}(t=0)=1.8\ \text{V}$ on the real axis values between $0.2\ \text{V}$ and $1.8\ \text{V}$   ⇒   phase $\phi(t) \equiv 0$.  Except for the start state $x_{\rm TP}(t=0)$ same behavior as at the setting '''(1)'''. The degree of modulation is $m = 0.8$.

{{BlaueBox|TEXT=
'''(5)'''   The parameters are still valid according to '''(4)''' with the exception of $A_{\rm T}= 0.6 \text{V}$. What is the degree of modulation $m$? What are the consequences?}}

:: There is now a DSB–AM with modulation degree $m = 1.333$. For $m > 1$, the simpler ''Envelope Demodulation'' is not applicable, since the phase function $\phi(t) \in \{ 0, \ \pm 180^\circ\}$ is no more constant and the envelope $a(t)$ no more matches the message signal. Rather, the complex ''Synchronous Demodulation'' must be used. Envelope detection would produce nonlinear distortions.

{{BlaueBox|TEXT=
'''(6)'''   The parameters are still valid according to '''(4)''' or '''(5)''' with the exception from $A_{\rm T}= 0$ on   ⇒   $m \to \infty$. Which modulation method is described in this way?}}

::It is a '''DSB–AM without carrier''' and a synchronous demodulation is required. The equivalent low-pass signal $x_{\rm TP}(t)$ is on the real axis, but not only in the right half-plane. Thus, the phase function $\phi(t) \in \{ 0, \ \pm 180^\circ\}$, also applies here, which means that ''Envelope Demodulation'' is not applicable.

{{BlaueBox|TEXT=
'''(7)'''   Now let   $\text{Red:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, f_{\rm T} = 100 \ \text{kHz}, \varphi_{\rm T} = 0^\circ$,   $\text{Green:} \hspace{0.15cm} A_{\rm U} = 0, \ f_{\rm U} = 80 \ \text{kHz}, \varphi_{\rm U} = -90^\circ$,   $\text{Blue:} \hspace{0.15cm} A_{\rm O} = 0.8\ \text{V}, f_{\rm O} = 120 \ \text{kHz}, \varphi_{\rm O} = 90^\circ$.

:Which constellation is described here? Which characteristics of this procedure can be recognized from the graphic?}}

::It is a [[Modulation_Methods/Einseitenbandmodulation|"single-sideband modulation"]] '''(SSB–AM)''', more specifically an '''OSB–AM''': the red carrier is fixed, the green pointer missing and the blue pointer (OSB) turns counterclockwise. The degree of modulation is $\mu = 0.8$ (in the case of SSB we denote the degree of modulation with $\mu$ instead of $m$). The carrier signal is cosinusoidal and the message signal sinusoidal. The locus is a circle. $x_{\rm TP}(t)$ moves in a mathematically positive direction. Because of $\phi(t) \ne \text{const.}$ the envelope demodulation is not applicable here:  This can be seen by the fact that the envelope $a(t)$ is not cosinusoidal. Rather, the lower half-wave is sharper than the upper one   ⇒   strong linear distortions.

{{BlaueBox|TEXT=
'''(8)'''   The parameters are still valid according to '''(7)''' with the exception of $A_{\rm O}= 0$ and $A_{\rm U}= 0.8 \text{ V}$. What differences arise opposite '''(7)'''?}}

::Now it is a '''LSB–AM''': The red carrier is fixed, the blue pointer is missing and the green pointer (LSB) rotates clockwise. All other statements of '''(7)''' apply here as well.

{{BlaueBox|TEXT=
'''(9)'''   The parameters according to '''(7)''' are still valid with the exception of $A_{\rm O} = 0.2 \text{ V} \ne A_{\rm U} = 0.4 \text{ V} $. What are the differences from '''(7)'''?}}

::The locus $x_{\rm TP}(t)$ is not a horizontal straight line, but an ellipse with the real part between $0.4 \text{ V}$ and $1.6 \text{ V}$ and the imaginary part in the range $\pm 0.2 \text{ V}$. Because of $\phi(t) \ne \text{const.}$ , Envelope demodulation would lead to non-linear distortions here too. The constellation simulated here describes the situation of '''(4)''', namely a DSB–AM with modulation degree $m = 0.8$, where the upper sideband is reduced to $50\%$ due to channel attenuation.

==Applet Manual==

[[File:Ortskurve_abzug3.png|right|frame|Screenshot]]

    '''(A)'''     Plot of the equivalent low-pass signal $x_{\rm TP}(t)$

    '''(B)'''     Plot of the physical signal $x(t)$

    '''(C)'''     Parameter input via slider:   amplitudes, frequencies, phase values

    '''(D)'''     Control elements:   Start – Step – Pause/Continue – Reset

    '''(E)'''     Speed of animation:   "Speed"   ⇒   Values: 1, 2 oder 3

    '''(F)'''     "Trace"   ⇒   On or Off, trace of equivalent low-pass signal   $x_{\rm TP}(t)$

    '''(G)'''     Numerical output:   time $t$, the signal values  ${\rm Re}[x_{\rm TP}(t)]$  and  ${\rm Im}[x_{\rm TP}(t)]$,

$\text{}\hspace{4.2cm}$   envelope $a(t) = |x_{\rm TP}(t)|$  and  phase $\phi(t) = {\rm arc} \ x_{\rm TP}(t)$

    '''(H)'''     Variations for the graphical representation

$\hspace{1.5cm}$Zoom–Functions "$+$" (Enlarge), "$-$" (Decrease) and $\rm o$ (Reset to default)

$\hspace{1.5cm}$Move with "$\leftarrow$" (Section to the left, ordinate to the right), "$\uparrow$" "$\downarrow$" "$\rightarrow$"

    '''(I)'''     Experiment section:   Task selection and task

    '''(J)'''     Experiment section:  solution
 

In all applets top right:    Changeable graphical interface design   ⇒   '''Theme''':
* Dark:   black background  (recommended by the authors).
* Bright:   white background  (recommended for beamers and printouts)
* Deuteranopia:   for users with pronounced green–visual impairment
* Protanopia:   for users with pronounced red–visual impairment
 
Note:
*Red parameters  $(A_{\rm T}, \ f_{\rm T}, \ \varphi_{\rm T})$  and the red pointer mark the "Carrier"  (German:  $\rm T$räger).  The red pointer does not turn.
* Green parameters  $(A_{\rm U}, \ f_{\rm U} < f_{\rm T}, \ \varphi_{\rm U})$  mark the "Lower sideband"  (German:  $\rm U$nteres Seitenband).  The green pointer rotates in a mathematically negative direction.
* Blue parameters  $(A_{\rm O}, \ f_{\rm O} > f_{\rm T}, \ \varphi_{\rm O})$  mark the "Upper sideband"  (German:  $\rm O$beres Seitenband).  The blue pointer turns counterclockwise.

==About the Authors==
This interactive calculation was designed and realized at the  [https://www.ei.tum.de/en/lnt/home//startseite Institute for Communications Engineering]  of the  [https://www.tum.de/ Technical University of Munich] .
*The original version was created in 2005 by  [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]]  as part of her Diploma thesis using "FlashMX–Actionscript"  (Supervisor:  [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._Günter_Söder_(at_LNT_from_1974-2024)|Günter Söder]]).
*In 2018 this Applet was redesigned and updated to "HTML5" by  [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Xiaohan_Liu_.28Bachelorarbeit_2018.29|Xiaohan Liu]]  as part of her Bachelor's thesis (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]).

==Once again: Open Applet in new Tab==

{{LntAppletLink|physAnSignal_en}}         [https://lntwww.lnt.ei.tum.de/Applets:Physikalisches_Signal_%26_%C3%84quivalentes_TP-Signal '''English Applet with German WIKI description''']
[[de:Applets:Physical Signal & Equivalent Lowpass Signal]]

Applets:Lineare Verzerrungen periodischer Signale

2026-03-16T15:58:57Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{LntAppletLink|linDistortions_en}}

==Programmbeschreibung==
 
Dieses Applet veranschaulicht die Auswirkungen von linearen Verzerrungen (Dämpfungsverzerrungen und Phasenverzerrungen) anhand
[[File:Modell_version2.png|right|frame|Bedeutung der verwendeten Signale]]
*des Eingangssignals $x(t)$   ⇒   Leistung $P_x$:
:$$x(t) = x_1(t) + x_2(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right), $$
*des Ausgangssignals $y(t)$   ⇒   Leistung $P_y$:
:$$y(t) = \alpha_1 \cdot x_1(t-\tau_1) + \alpha_2 \cdot x_2(t-\tau_2),$$
*des Matching–Ausgangssignals $z(t)$   ⇒   Leistung $P_z$:
:$$z(t) = k_{\rm M} \cdot y(t-\tau_{\rm M}) + \alpha_2 \cdot x_2(t-\tau_2),$$
*des Differenzsignals   $\varepsilon(t) = z(t) - x(t)$   ⇒   Leistung $P_\varepsilon$.

Als nächster Block im obigen Modell folgt das "Matching": Dabei wird das Ausgangssignal $y(t)$ mit für alle Frequenzen einheitlichen Größen $k_{\rm M}$ und $\tau_{\rm M}$ in Amplitude bzw. Phase angepasst. Dies ist also keine frequenzabhängige Entzerrung. Anhand des Signals $z(t)$ kann unterschieden werden
*zwischen einer Dämpfungsverzerrung und einer frequenzunabhängigen Dämpfung, sowie
*zwischen einer Phasenverzerrung und einer für alle Frequenzen gleichen Laufzeit.

Als Maß für die Stärke der linearen Verzerrungen wird die Verzerrungsleistung (englisch: ''Distortion Power'') $P_{\rm D}$ verwendet. Für diese gilt:
:$$P_{\rm D} = \min_{k_{\rm M}, \ \tau_{\rm M}} P_\varepsilon.$$

[[Applets:Linear_Distortions_of_Periodic_Signals|'''Englische Beschreibung''']]

==Theoretischer Hintergrund==
 
Unter '''Verzerrungen''' (englisch: ''Distortions'') versteht man allgemein die unerwünschte deterministische Veränderungen eines Nachrichtensignals durch ein Übertragungssystem. Sie sind bei vielen Nachrichtensystemen neben den stochastischen Störungen (Rauschen, Nebensprechen, etc.) eine entscheidende Begrenzung für die Übertragungsqualität und die Übertragungsrate.

Ebenso wie man die "Stärke" von Rauschen durch
*die Rauschleistung (englisch: ''Noise Power'') $P_{\rm N}$ und
*das Signal–zu–Rauschleistungsverhältnis (englisch: ''Signal–to–Noise Ratio'', SNR) $\rho_{\rm N}$

bewertet, verwendet man zur Quantifizierung der Verzerrungen

*die Verzerrungsleistung (englisch: ''Distortion Power'') $P_{\rm D}$ und
*das Signal–zu–Verzerrungsleistungsverhältnis (englisch: ''Signal–to–Distortion Ratio'', SDR)
:$$\rho_{\rm D}=\frac{\rm Signalleistung}{\rm Verzerrungsleistung} = \frac{P_x}{P_{\rm D} }.$$

=== Lineare und nichtlineare Verzerrungen ===
 
Man unterscheidet zwischen linearen und nichtlinearen Verzerrungen:
*'''Nichtlineare Verzerrungen''' gibt es, wenn zu allen Zeiten $t$ zwischen dem Signalwert $x = x(t)$ am Eingang und dem Ausgangssignalwert $y = y(t)$ der nichtlineare Zusammenhang $y = g(x) \ne {\rm const.} \cdot x$ besteht, wobei $y = g(x)$ die nichtlineare Kennlinie des Systems bezeichnet. Legt man an den Eingang ein Cosinussignal der Freuenz $f_0$ an, so beinhaltet das Ausgangssignal neben $f_0$ auch Vielfache hiervon   ⇒   so genannte ''Oberwellen''. Durch nichtlineare Verzerrungen entstehen also neue Frequenzen.

[[File:LZI_T_2_2_S3_vers2.png|center|frame|Zur Verdeutlichung nichtlinearer Verzerrungen |class=fit]]

[[File:P_ID899__LZI_T_2_3_S1_neu.png|right |frame| Beschreibung eines linearen Systems|class=fit]]
*'''Lineare Verzerrungen''' entstehen dann, wenn der Übertragungskanal durch einen Frequenzgang $H(f) \ne \rm const.$ charakterisiert wird. Dann werden unterschiedliche Frequenzen unterschiedlich gedämpft und unterschiedlich verzögert. Charakteristisch hierfür ist, dass zwar Frequenzen verschwinden können (zum Beispiel durch einen Tiefpass, einen Hochpass oder einen Bandpass), dass aber keine neuen Frequenzen entstehen.

In diesem Applet werden nur lineare Verzerrungen betrachtet.

=== Beschreibungsformen für den Frequenzgang ===
 
Der im Allgemeinen komplexe Frequenzgang kann auch wie folgt dargestellt werden:
:$$H(f) = |H(f)| \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot\hspace{0.05cm} b(f)} = {\rm e}^{-a(f)}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} b(f)}.$$

Daraus ergeben sich folgende Beschreibungsgrößen:
*Der Betrag $|H(f)|$ wird als '''Amplitudengang''' und in logarithmierter Form als '''Dämpfungsverlauf''' bezeichnet:
:$$a(f) = - \ln |H(f)|\hspace{0.2cm}{\rm in \hspace{0.1cm}Neper\hspace{0.1cm}(Np) } = - 20 \cdot \lg |H(f)|\hspace{0.2cm}{\rm in\hspace{0.1cm}Dezibel \hspace{0.1cm}(dB) }.$$
*Der '''Phasengang''' $b(f)$ gibt den negativen frequenzabhängigen Winkel von $H(f)$ in der komplexen Ebene an, bezogen auf die reelle Achse:
:$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) \hspace{0.2cm}{\rm in\hspace{0.1cm}Radian \hspace{0.1cm}(rad)}.$$

=== Tiefpass N–ter Ordnung ===
 
[[File:Tiefpass_version2.png|right|frame|Dämpfungsverlauf und Phasenverlauf eines Tiefpasses N–ter Ordnung]]
Der Frequenzgang eines realisierbaren Tiefpasses N–Ordnung lautet:
:$$H(f) = \left [\frac{1}{1 + {\rm j}\cdot f/f_0 }\right ]^N\hspace{0.05cm}.$$
Ein einfacher RC–Tiefpass hat diesen Verlauf mit $N=1$. Damit erhält man
*den Dämpfungsverlauf:
:$$a(f) =N/2 \cdot \ln [1+( f/f_0)^2] \hspace{0.05cm},$$
*den Phasenverlauf:
:$$b(f) =N \cdot \arctan( f/f_0) \hspace{0.05cm},$$
*den Dämpfungsfaktor für die Frequenz $f=f_i$:
:$$\alpha_i =|H(f = f_i)| = [1+( f/f_0)^2]^{-N/2}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)= \alpha_i \cdot A_i\cdot \cos(2\pi f_i t)\hspace{0.05cm},$$
*die Phasenlaufzeit für die Frequenz $f=f_i$:
:$$\tau_i =\frac{b(f_i)}{2 \pi f_i} = \frac{N \cdot \arctan( f_i/f_0)}{2 \pi f_i}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)=A_i\cdot \cos(2\pi f_i (t- \tau_i))\hspace{0.05cm}.$$

=== Hochpass N–ter Ordnung ===
 
[[File:Hochpass_version2.png|right|frame|Dämpfungsverlauf und Phasenverlauf eines Hochpasses N–ter Ordnung]]
Der Frequenzgang eines realisierbaren Hochpasses N–Ordnung lautet:
:$$H(f) = \left [\frac{ {\rm j}\cdot f/f_0 }{1 + {\rm j}\cdot f/f_0 }\right ]^N\hspace{0.05cm}.$$
Ein einfacher LC–Tiefpass hat diesen Verlauf mit $N=1$. Damit erhält man
*den Dämpfungsverlauf:
:$$a(f) =N/2 \cdot \ln [1+( f_0/f)^2] \hspace{0.05cm},$$
*den Phasenverlauf:
:$$b(f) =-N \cdot \arctan( f_0/f) \hspace{0.05cm},$$
*den Dämpfungsfaktor für die Frequenz $f=f_i$:
:$$\alpha_i =|H(f = f_i)| = [1+( f_0/f)^2]^{-N/2}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)= \alpha_i \cdot A_i\cdot \cos(2\pi f_i t)\hspace{0.05cm},$$
*die Phasenlaufzeit für die Frequenz $f=f_i$:
:$$\tau_i =\frac{b(f_i)}{2 \pi f_i} = \frac{-N \cdot \arctan( f_0/f_i)}{2 \pi f_i}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)=A_i\cdot \cos(2 \pi f_i (t- \tau_i))\hspace{0.05cm}.$$

[[File:Verzerrungen_HP_TP_1_englisch.png|right|frame|Phasenfunktion $b(f)$ von Tiefpass und Hochpass]]
{{GraueBox|TEXT=
$\text{Beispiel:}$ 
Die Grafik zeigt jeweils für die Grenzfrequenz $f_0 = 1\ \rm kHz$ und die Ordnung $N=1$ die Phasenfunktion $b(f)$
* eines Tiefpasses (englisch: ''low–pass'') als grüne Kurve, und
* eines Hochpasses (englisch: ''high–pass'') als violette Kurve.

Das Eingangssignal sei jeweils sinusförmig mit der Frequenz $f_{\rm S} = 1.25\ {\rm kHz}$, wobei dieses Signal erst zum Zeitpunkt $t=0$ eingeschaltet wird:
:$$x(t) = \left\{ \begin{array}{l} \hspace{0.75cm}0 \\ \sin(2\pi \cdot f_{\rm S} \cdot t ) \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r} } \\ {\rm{f\ddot{u}r} } \\ \end{array}\begin{array} \ t < 0, \\ t>0. \\ \end{array}$$

In der linken (blau umrandeten) Grafik ist dieses Signal $x(t)$ dargestellt. Der Zeitpunkt $t = T_0 = 0.8\ {\rm ms}$ der ersten Nullstelle ist durch eine gestrichelte Linie markiert. Die beiden anderen Grafiken zeigen die Ausgangssignale $y_{\rm TP}(t)$ und $y_{\rm HP}(t)$ von Tiefpass und Hochpass, wobei in beiden Fällen die Amplitudenänderungen ausgeglichen wurden.

[[File:Verzerrungen_HP_TP_2_version2.png|center|frame|Eingangssignal $x(t)$ sowie Ausgangssignale $y_{\rm TP}(t)$ und $y_{\rm HP}(t)$]]

*Die erste Nullstelle des Signals $y_{\rm TP}(t)$ nach dem Tiefpass kommt um $\tau_{\rm TP} = 0.9/(2\pi) \cdot T_0 \approx 0.115 \ {\rm ms}$ später als die erste Nullstelle von $x(t)$   ⇒   markiert mit grünem Pfeil, wobei $b_{\rm TP}(f/f_{\rm S} )= 0.9 \ {\rm rad}$ berücksichtigt wurde.
* Dagegen ist die Laufzeit des Hochpasses negativ: $\tau_{\rm HP} = -0.67/(2\pi) \cdot T_0 \approx 0.085 \ {\rm ms}$ und die erste Nullstelle von $y_{\rm HP}(t)$ kommt deshalb vor der weißen Markierung.
*Nach diesem Einschwingvorgang kommen in beiden Fällen die Nulldurchgänge wieder im Raster der Periodendauer $T_0 = 0.8 \ {\rm ms}.$

''Anmerkung:'' Die gezeigten Signalverläufe wurden mit dem intereaktiven Applet [[Applets:Kausale_Systeme_-_Laplacetransformation|Kausale Systeme – Laplacetransformation]] erstellt. }}

=== Dämpfungsverzerrungen und Phasenverzerrungen ===
 
[[File:P_ID900__LZI_T_2_3_S2_neu.png|frame| Voraussetzung für einen nichtverzerrenden Kanal|right|class=fit]]
Die nebenstehende Grafik zeigt
*den geraden Dämpfungsverlauf $a(f)$   ⇒   $a(-f) = a(f)$, und
*den ungeraden Phasenverlauf $b(f)$   ⇒   $b(-f) = -b(- f)$

eines verzerrungsfreien Systems. Man erkennt:
*Bei einem verzerrungsfreien Systems muss in einem Bereich von $f_{\rm U}$ bis $f_{\rm O}$ um die Trägerfrequenz $f_{\rm T}$, in dem das Signal $x(t)$ Anteile besitzt, die Dämpfungsfunktion $a(f)$ konstant sein.
*Aus dem angegebenen konstanten Dämpfungswert $6 \ \rm dB$ folgt für den Amplitudengang $|H(f)| = 0.5$   ⇒   die Signalwerte aller Frequenzen werden somit durch das System halbiert   ⇒   keine Dämpfungsverzerrungen.
*Zusätzlich muss bei einem solchen Systems der Phasenverlauf $b(f)$ zwischen $f_{\rm U}$ und $f_{\rm O}$ linear mit der Frequenz ansteigen. Dies hat zur Folge, dass alle Frequenzanteile um die gleiche Phasenlaufzeit $τ$ verzögert werden   ⇒   keine Phasenverzerrungen.
*Die Verzögerung $τ$ liegt durch die Steigung von $b(f)$ fest. Mit $b(f) = 0$ würde sich ein laufzeitfreies System ergeben   ⇒   $τ = 0$.

Die folgende Zusammenfassung berücksichtigt, dass in diesem Applet das Einganssignal stets die Summe zweier harmonischer Schwingungen ist:
:$$x(t) = x_1(t) + x_2(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right).$$
Damit wird der Kanaleinfluss durch die Dämpfungsfaktoren $\alpha_1$ und $\alpha_2$ sowie die Phasenlaufzeiten $\tau_1$ und $\tau_2$ vollständig beschrieben:
:$$y(t) = \alpha_1 \cdot x_1(t-\tau_1) + \alpha_2 \cdot x_2(t-\tau_2).$$

{{BlaueBox|TEXT=
$\text{Fazit:}$ 
*Ein Signal $y(t)$ ist gegenüber dem Eingang $x(t)$ nur dann unverzerrt, wenn $\alpha_1 = \alpha_2= \alpha$   und   $\tau_1 = \tau_2= \tau$ gilt   ⇒   $y(t) = \alpha \cdot x(t-\tau)$.
* Dämpfungsverzerrungen ergeben sich, falls $\alpha_1 \ne \alpha_2$ ist . Ist $\alpha_1 \ne \alpha_2$ und $\tau_1 = \tau_2$, so liegen ausschließlich Dämpfungsverzerrungen vor.
* Phasenverzerrungen gibt es für $\tau_1 \ne \tau_2$. Ist $\tau_1 \ne \tau_2$ und $\alpha_1 = \alpha_2$, so liegen ausschließlich Phasenverzerrungen vor. }}

==Versuchsdurchführung==
[[File:Exercises_verzerrungen.png|right]]
*Wählen Sie zunächst die Aufgabennummer.
*Eine Aufgabenbeschreibung wird angezeigt.
*Parameterwerte sind angepasst.
*Lösung nach Drücken von "Hide solition".

Die Nummer "0" entspricht einem "Reset":
*Gleiche Einstellung wie beim Programmstart.
*Ausgabe eines "Reset–Textes" mit weiteren Erläuterungen zum Applet.

{{BlaueBox|TEXT=
'''(1)'''   Für das Eingangssignal $x(t)$ gelte $A_1 = 0.8\ {\rm V}, \ A_2 = 0.6\ {\rm V}, \ f_1 = 0.5\ {\rm kHz}, \ f_2 = 1.5\ {\rm kHz}, \ \varphi_1 = 90^\circ, \ \varphi_2 = 30^\circ$.
:Wie groß ist die Periodendauer $T_0$? Welche Leistung $P_x$ weist dieses Signal auf? Wo kann man diesen Wert im Programm ablesen? }}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}T_0 = \big [\hspace{-0.1cm}\text{ größter gemeinsamer Teiler }(0.5 \ {\rm kHz}, \ 1.5 \ {\rm kHz})\big ]^{-1}\hspace{0.15cm}\underline{ = 2.0 \ {\rm ms}};$

$\hspace{1.85cm} P_x = A_1^2/2 + A_2^2/2 \hspace{0.15cm}\underline{= 0.5 \ {\rm V^2}} = P_\varepsilon\text{, wenn }\hspace{0.15cm}\underline{k_{\rm M} = 0} \ \Rightarrow \ z(t) \equiv 0$.

{{BlaueBox|TEXT=
'''(2)'''   Variieren Sie bei sonst gleicher Einstellung wie unter '''(1)''' die Phase $\varphi_2$ im gesamten möglichen Bereich $\pm 180^\circ$. Wie ändern sich $T_0$ und $P_x$?}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{Keine Veränderungen:}\hspace{0.2cm}\hspace{0.15cm}\underline{ T_0 = 2.0 \ {\rm ms}; \hspace{0.2cm} P_x = 0.5 \ {\rm V^2}}$.

{{BlaueBox|TEXT=
'''(3)'''   Variieren Sie bei sonst gleicher Einstellung wie unter '''(1)''' die Frequenz $f_2$ im Bereich $0 \le f_2 \le 5\ {\rm kHz}$. Wie ändert sich die Signalleistung $P_x$?}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{Keine Veränderungen, falls }f_2 \ne 0\text{ und } f_2 \ne f_1\text{:}\hspace{0.3cm} \hspace{0.15cm}\underline{P_x = 0.5 \ {\rm V^2}}\text{.} \hspace{0.2cm} T_0 \text{ ändert sich, falls }f_2\text{ kein Vielfaches von }f_1$.

$\hspace{1.85cm}\text{Falls }f_2 = 0\text{:}\hspace{0.2cm} P_x = A_1^2/2 + A_2^2\hspace{0.15cm}\underline{ = 0.68 \ {\rm V^2}}$. $\hspace{3cm}\text{Allgemeine Formel noch überprüfen}$

$\hspace{1.85cm}\text{Falls }f_2 = f_1\text{:}\hspace{0.2cm} P_x = [A_1 \cdot \cos(\varphi_1) + A_2 \cdot \cos(\varphi_2)]^2/2 + [A_1\sin \cdot (\varphi_1) + A_2 \cdot \sin(\varphi_2)]^2/2 \text{. Mit } \varphi_1 = 90^\circ, \ \varphi_2 = 30^\circ\text{:}\hspace{0.3cm}\hspace{0.15cm}\underline{ P_x = 0.74 \ {\rm V^2}}\text{.} $

{{BlaueBox|TEXT=
'''(4)'''   Ausgehend vom bisherigen Eingangssignal $x(t)$ gelte für den Kanal: $\alpha_1 = \alpha_2 = 0.5, \ \tau_1 = \tau_2 = 0.5\ {\rm ms}$. Zudem sei $k_{\rm M} = 1 \text{ und } \tau_{\rm M} = 0$ .
:Gibt es lineare Verzerrungen? Wie groß ist die Empfangsleistung $P_y$ und die Leistung $P_\varepsilon$ des Differenzsignals $\varepsilon(t) = z(t) - x(t)$? }}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{ y(t) = 0.5 \cdot x(t- 1\ {\rm ms})}\text{ ist unverzerrt, nur gedämpft und verzögert.}$

$\hspace{1.85cm}\text{Empfangsleistung:}\hspace{0.2cm} P_y = (A_1/2)^2/2 + (A_2/2)^2/2\hspace{0.15cm}\underline{ = 0.125 \ {\rm V^2}}\text{. } P_\varepsilon \text{ ist deutlich größer:} \hspace{0.1cm} \hspace{0.15cm}\underline{P_\varepsilon = 0.625 \ {\rm V^2}}.$

{{BlaueBox|TEXT=
'''(5)'''   Variieren Sie bei sonst gleicher Einstellung wie unter '''(4)''' die Matchingparameter $k_{\rm M} \text{ und } \tau_{\rm M}$. Wie groß ist die Verzerrungsleistung $P_{\rm D}$?}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D}\text{ ist gleich der Leistung }P_\varepsilon \text{ des Differenzsignals bei bestmöglicher Anpassung:} \hspace{0.2cm}k_{\rm M} = 2 \text{ und } \tau_{\rm M}=T_0 - 0.5\ {\rm ms} = 1.5\ {\rm ms}$

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm}z(t) = x(t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\varepsilon(t) = 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}P_{\rm D}\hspace{0.15cm}\underline{ = P_\varepsilon = 0} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\text{weder Dämpfungs- noch Phasenverzerrungen.}$

{{BlaueBox|TEXT=
'''(6)'''   Für den Kanal gelte nun $\alpha_1 = 0.5, \hspace{0.15cm}\underline{\alpha_2 = 0.2}, \ \tau_1 = \tau_2 = 0.5\ {\rm ms}$. Wie groß sind nun die Verzerrungsleistung $P_{\rm D}$ und das Signal–zu–Verzerrungsverhäldnis $(\rm SDR)$ $\rho_{\rm D}$?}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{ bei bestmöglicher Anpassung:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 2.24} \text{ und } \hspace{0.15cm}\underline{\tau_{\rm M} = 1.5\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.059 \ {\rm V^2}}$.

$\hspace{1.85cm}\text{Nur Dämpfungsverzerrungen.} \hspace{0.3cm}\text{Signal-zu-Verzerrung-Leistungsverhältnis}\ \hspace{0.15cm}\underline{\rho_{\rm D} = P_x/P_\varepsilon \approx 8.5}$.

{{BlaueBox|TEXT=
'''(7)'''   Für den Kanal gelte nun $\alpha_1 = \alpha_2 = 0.5, \ \tau_1 \hspace{0.15cm}\underline{= 2\ {\rm ms} }, \ \tau_2 = 0.5\ {\rm ms}$. Wie groß sind nun $P_{\rm D}$ und $\rho_{\rm D}$?}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{ bei bestmöglicher Anpassung:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.82} \text{ und } \tau_{\rm M}\hspace{0.15cm}\underline{ = 0.15\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.072 \ {\rm V^2}}$.

$\hspace{1.85cm}\text{Nur Phasenverzerrungen.} \hspace{0.3cm}\text{Signal-zu-Verzerrung-Leistungsverhältnis}\ \hspace{0.15cm}\underline{\rho_{\rm D} = P_x/P_\varepsilon \approx 7}$.

{{BlaueBox|TEXT=
'''(8)'''   Die Kanalparameter seien nun $\hspace{0.15cm}\underline{\alpha_1 = 0.5} , \hspace{0.15cm}\underline{\alpha_2 = 0.2} , \ \hspace{0.15cm}\underline{\tau_1= 0.5\ {\rm ms} }, \ \hspace{0.15cm}\underline{\tau_2 = 0.3\ {\rm ms} }$. Gibt es Dämpfungs– und/oder Phasenverzerrungen?
:Wie kann man $y(t)$ annähern? ''Hinweis:'' $\cos(3x) = 4 \cdot \cos^3(x) - 3\cdot \cos(x).$}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{Es gibt sowohlDämpfungs– als auch Phasenverzerrungen, weil }\alpha_1 \ne \alpha_2\text{ und }\tau_1 \ne \tau_2$.

$\hspace{1.85cm}\text{Es gilt }y(t) = y_1(t) + y_2(t)\ \Rightarrow \ y_1(t) = A_1 \cdot \alpha_1 \cdot \sin[2\pi f_1\ (t- 0.5\ \rm ms)] = -0.4 \ {\rm V} \cdot \cos(2\pi f_1 t)$

$\hspace{1.85cm} y_2(t) = \alpha_2 \cdot x_2(t- \tau_2) \text{ mit }x_2(t) = A_2 \cdot \cos[2\pi f_2\ (t- 30^\circ)] \approx A_2 \cdot \cos[2\pi f_2\ (t- 1/36 \ \rm ms)]$

$\hspace{1.85cm} \Rightarrow \ y_2(t) = 0.12 \ {\rm V} \cdot \cos[2\pi f_2\ (t- 0.328 \ {\rm ms})] \approx -0.12 \ { \rm V} \cdot \cos[2\pi f_2t] $.

$\hspace{1.85cm} \Rightarrow \ y(t) = y_1(t) + y_2(t) \approx -0.4 \ {\rm V} \cdot [\cos(2\pi \cdot f_1\cdot t) + 1/3 \cdot \cos(2\pi \cdot 3 f_1 \cdot t) = -0.533 \ {\rm V} \cdot \cos^3(2\pi f_1 t)$.

{{BlaueBox|TEXT=
'''(9)'''   Es gelten weiter die Parameter von '''(8)'''. Wie groß ist die Verzerrungsleistung $P_{\rm D}$ and das Signal-zu-Verzerrungsleistungsverhältnis $\rho_{\rm D}$?}}

$\hspace{1.0cm}\text{Bestmögliche Anpassung:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.96} \text{, } \hspace{0.15cm}\underline{\tau_{\rm M} = 1.65\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.156 \ {\rm V^2} },\hspace{0.1cm}\hspace{0.15cm}\underline{\rho_{\rm D} = 0.500/0.15 \approx 3.3}$.

{{BlaueBox|TEXT=
'''(10)'''  Nun gelte $A_2 = 0$ sowie $A_1 = 1\ {\rm V}, \ f_1 = 1\ {\rm kHz}, \varphi_1 = 0^\circ$. Der Kanal sei ein Tiefpass erster Ordnung $(f_0 = 1\ {\rm kHz})$.
:Gibt es Dämpfungs– und/oder Phasenverzerrungen? Wie groß sind die Kanalkoeffizienten $\alpha_1$ and $\tau_1$?}}

$\hspace{1.0cm}\text{Bei nur einer Frequenz gibt es weder Dämpfungs– noch Phasenverzerrungen.}$
$\hspace{1.0cm}\text{Dämpfungsfaktor für }f_1=f_0\text{ und }N=1\text{: }\alpha_1 =|H(f = f_1)| = [1+( f_1/f_0)^2]^{-N/2} = 2^{-1/2}= 1/\sqrt{2}\hspace{0.15cm}\underline{=0.707},$
$\hspace{1.0cm}\text{Phasenlaufzeit für}f_1=f_0\text{ und }N=1\text{: }\tau_1 = N \cdot \arctan( f_1/f_0)/(2 \pi f_1)=\arctan( 1)/(2 \pi f_1) =1/(8f_1) \hspace{0.15cm}\underline{=0.125 \ \rm ms}.$

{{BlaueBox|TEXT=
'''(11)'''   Wie ändern sich die Kanalparameter durch einen Tiefpass zweiter Ordnung gegenüber einem Tiefpass erster Ordnung $(f_0 = 1\ {\rm kHz})$?}}

$\hspace{1.0cm}\text{Es gilt }\hspace{0.15cm}\alpha_1 = 0.707^2 = 0.5$ und $\tau_1 = 2 \cdot 0.125 = 0.25 \ {\rm ms}$.

$\hspace{1.0cm}\text{Das Signal }y(t)\text{ ist nur halb so groß wie }x(t)\text{ und läuft diesem nach: Aus dem Cosinusverlauf wird die Sinusfunktion}$.

{{BlaueBox|TEXT=
'''(12)'''   Welche Unterschiede ergeben sich bei einem Hochpass zweiter Ordnung gegenüber einem Tiefpass zweiter Ordnung $(f_0 = 1\ {\rm kHz})$. }}

$\hspace{1.0cm}\text{Wegen }f_1 = f_0\text{ ergibt sich der gleiche Dämpfungsfaktor }\alpha_1 = 0.5\text{ und es gilt }\tau_1 = -0.25 \ {\rm ms}\text{ Das heißt:}$.

$\hspace{1.0cm}\text{Das Signal }y(t)\text{ ist halb so groß wie }x(t)\text{ und läuft diesem vor: Aus dem Cosinusverlauf wird die Minus–Sinusfunktion}$.

{{BlaueBox|TEXT=
'''(13)'''   Welche Unterschiede erkennen Sie am Signalverlauf $y(t)$ zwischen dem Tiefpass zweiter Ordnung und dem Hochpass zweiter Ordnung $(f_0 = 1\ {\rm kHz})$, wenn Sie vom Eingangssignal gemäß'''(1)''' ausgehen und Sie die Frequenz $f_2$ kontinuierlich bis auf $10 \ \rm kHz$ erhöhen. }}

$\hspace{1.0cm}\text{Nach dem Tiefpass wird der zweite Anteil mehr und mehr unterdrückt. Für }f_2 = 10 \ \rm kHz\text{ gilt: }y_{\rm LP}(t) \approx 0.8 \cdot x_1(t-0.3 \ \rm ms).$

$\hspace{1.0cm}\text{Nach dem Hochpass überwiegt dagegen der zweite Anteil. Für }f_2 = 10 \ \rm kHz\text{ gilt: }y_{\rm HP}(t) \approx 0.2 \cdot x_1(t+0.7 \ {\rm ms)} + x_2(t).$

==Zur Handhabung des Applets==
[[File:Handhabung_verzerrungen.png|center]]
 
    '''(A)'''     Parametereingabe für das Eingangssignal $x(t)$ per Slider: Amplituden, Frequenzen, Phasenwerte

    '''(B)'''     Vorauswahl für die Kanalparameter: per Slider, Tiefpass oder Hochpass

    '''(C)'''     Eingabe der Kanalparameter per Slider: Dämpfungsfaktoren und Phasenlaufzeiten

    '''(D)'''     Eingabe der Kanalparameter für Hoch– und Tiefpass: Ordnung $n$, Grenzfrequenz $f_0$

    '''(E)'''     Eingabe der Matching–Parameter $k_{\rm M}$ und $\varphi_{\rm M}$

    '''(F)'''     Auswahl der darzustellenden Signale: $x(t)$, $y(t)$, $z(t)$, $\varepsilon(t)$, $\varepsilon^2(t)$

    '''(G)'''     Graphische Darstellung der Signale

    '''(H)'''     Eingabe der Zeit $t_*$ für die Numerikausgabe

    '''( I )'''     Numerikausgabe der Signalwerte $x(t_*)$, $y(t_*)$, $z(t_*)$ und $\varepsilon(t_*)$

    '''(J)'''     Numerikausgabe des Hauptergebnisses $P_\varepsilon$

    '''(K)'''     Abspeichern und Zurückholen von Parametersätzen

    '''(L)'''     Bereich für die Versuchsdurchführung: Aufgabenauswahl, Aufgabenstellung und Musterlösung

    '''(M)'''     Variationsmöglichkeiten für die grafische Darstellung

$\hspace{1.5cm}$Zoom–Funktionen "$+$" (Vergrößern), "$-$" (Verkleinern) und $\rm o$ (Zurücksetzen)

$\hspace{1.5cm}$Verschieben mit "$\leftarrow$" (Ausschnitt nach links, Ordinate nach rechts), "$\uparrow$" "$\downarrow$" und "$\rightarrow$"

$\hspace{1.5cm}$'''Andere Möglichkeiten''':

$\hspace{1.5cm}$Gedrückte Shifttaste und Scrollen: Zoomen im Koordinatensystem,

$\hspace{1.5cm}$Gedrückte Shifttaste und linke Maustaste: Verschieben des Koordinatensystems.

==Über die Autoren==
Dieses interaktive Berechnungstool wurde am [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik] der [https://www.tum.de/ Technischen Universität München] konzipiert und realisiert.
*Die erste Version wurde 2005 von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Bettina_Hirner_.28Diplomarbeit_LB_2005.29|Bettina Hirner]] im Rahmen ihrer Diplomarbeit mit "FlashMX–Actionscript" erstellt (Betreuer: [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._Günter_Söder_(at_LNT_from_1974-2024)|Günter Söder]]).
*2018 wurde dieses Programm von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Jimmy_He_.28Bachelorarbeit_2018.29|Jimmy He]] im Rahmen seiner Bachelorarbeit (Betreuer: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) neu gestaltet und erweitert.

==Nochmalige Aufrufmöglichkeit des Applets in neuem Fenster==

{{LntAppletLink|linDistortions_en}}
[[de:Applets:Lineare Verzerrungen periodischer Signale]]

Applets:Frequenzgang und Impulsantwort

2026-03-16T15:58:56Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{LntAppletLinkEn|frequImpResp_en}}

==Programmbeschreibung==
 

Dargestellt werden reelle und symmetrische Tiefpässe $H(f)$ und die dazugehörigen Impulsantworten $h(t)$, nämlich
*Gauß–Tiefpass (englisch: ''Gaussian low–pass''),
*Rechteck–Tiefpass (englisch: ''Rectangular low–pass''),
*Dreieck–Tiefpass (englisch: ''Triangular low–pass''),
*Trapez–Tiefpass (englisch: ''Trapezoidal low–pass''),
*Cosinus–Rolloff–Tiefpass (englisch: ''Cosine-rolloff low–pass''),
*Cosinus-Quadrat-Tiefpass (englisch: ''Cosine-rolloff -squared Low–pass'').

Das aufzurufende Applet verwendet die englischen Begriffe im Gegensatz zu dieser deutschen Beschreibung.

Die englische Beschreibung finden Sie unter [[englische Version: Frequency & Pulse response]] (ist derzeit noch nicht realisiert).

Weiter ist zu beachten:
* Die Funktionen $H(f)$ bzw. $h(t)$ werden für bis zu zwei Parametersätzen in jeweils einem Diagramm dargestellt.
* Die orangenfarbenen ("roten") Kurven und Zahlenangaben gelten für den linken Parametersatz, die blauen für den rechten Parametersatz.
* Die Abszissen $t$ (Zeit) und $f$ (Frequenz) sowie die Ordinaten $H(f)$ und $h(t)$ sind jeweils normiert.

==Theoretischer Hintergrund==
 
===Frequenzgang $H(f)$ und Impulsantwort $h(t)$===
*Der [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#.C3.9Cbertragungsfunktion_-_Frequenzgang|Frequenzgang]] (oder auch die ''Übertragungsfunktion'') eines linearen zeitinvarianten Übertragungssystems $H(f)$ gibt das Verhältnis zwischen dem Ausgangsspektrum $Y(f)$ und dem dem Eingangsspektrum $X(f)$ an:
:$$H(f) = \frac{Y(f)}{X(f)}.$$
*Ist das Übertragungsverhalten bei tiefen Frequenzen besser als bei höheren, so spricht man von einem '''Tiefpass''' (englisch: ''Low-pass'').
*Die Eigenschaften von $H(f)$ werden im Zeitbereich durch die [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|Impulsantwort]] $h(t)$ ausgedrückt. Entsprechend dem [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|zweiten Fourierintegral]] gilt:
:$$h(t)={\rm IFT} [H(f)] = \int_{-\infty}^{+\infty}H(f)\cdot {\rm e}^{+{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}f\hspace{1cm}{\rm IFT}\hspace{-0.1cm}: \rm Inverse \ Fouriertransformation.$$
*Die Gegenrichtung wird durch das [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|erste Fourierintegral]] beschrieben:
:$$H(f)={\rm FT} [h(t)] = \int_{-\infty}^{+\infty}h(t)\cdot {\rm e}^{-{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}t\hspace{1cm}\rm FT\hspace{-0.1cm}: \ Fouriertransformation.$$
*In allen Beispielen verwenden wir reelle und gerade Funktionen. Somit gilt:
:$$h(t)=\int_{-\infty}^{+\infty}H(f)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}f \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ H(f)=\int_{-\infty}^{+\infty}h(t)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}t .$$
*Bei einem Vierpol   ⇒   $X(f)$ und $Y(f)$ haben gleiche Einheiten]   ist $Y(f)$ dimensionslos. Die Einheit der Impulsantwort ist $\rm 1/s$. Es gilt zwar $\rm 1/s = 1 \ Hz$, aber die Einheit "Hertz" ist in diesem Zusammenhang unüblich.
*Der Zusammenhang zwischen diesem Modul "Frequenzgang & Impulsantwort" und dem ähnlich aufgebauten Applet [[Applets:Impulse_und_Spektren]] basiert auf dem [[Signal_Representation/The_Fourier_Transform_Theorems#Duality_Theorem|Vertauschungssatz]].
*Alle Zeiten sind auf eine Normierungszeit $T$ normiert und alle Frequenzen auf $1/T \Rightarrow$ die Impulsantwortwerte $h(t)$ müssen noch durch die Normierungszeit $T$ dividiert werden.

{{GraueBox|TEXT=
$\text{Beispiel:}$  Stellt man einen Rechteck–Tiefpass mit Höhe $K_1 = 1$ und äquivalenter Bandbreite $\Delta f_1 = 1$ ein, so ist der Frequenzgang $H_1(f)$ im Bereich $-1 < f < 1$ gleich $1$ und außerhalb dieses Bereichs gleich $0$. Die Impulsantwort $h_1(t)$ verläuft si–förmig mit $h_1(t= 0) = 1$ und der ersten Nullstelle bei $t=1$.

Mit dieser Einstellung soll nun ein Rechteck–Tiefpass mit $K = 1.5$ und $\Delta f = 2 \ \rm kHz$ nachgebildet werden, wobei wir die Normierungszeit $T= 1 \ \rm ms$. Dann liegt die erste Nullstelle bei $t=0.5\ \rm ms$ und das Impulsantwortmaximum ist dann $h(t= 0) = 3 \cdot 10^3 \ \rm 1/s$.}}

===Gauß–Tiefpass   $\Rightarrow$   Gaussian Low–pass ===

*Der Gauß–Tiefpass lautet mit der Höhe $K$ und der (äquivalenten) Bandbreite $\Delta f$:
:$$H(f)=K\cdot {\rm e}^{-\pi\cdot(f/\Delta f)^2}.$$
*Die äquivalente Bandbreite $\Delta f$ ergibt sich aus dem flächengleichen Rechteck.
*Der Wert bei $f = \Delta f/2$ ist um den Faktor $0.456$ kleiner als der Wert bei $f=0$.
*Für die Impulsantwort erhält man gemäß der Fourierrücktransformation:
:$$h(t)=K\cdot \Delta f \cdot {\rm e}^{-\pi(t\cdot \Delta f)^2} .$$
*Je kleiner $\Delta f$ ist, um so breiter und niedriger ist die Impulsantwort   ⇒   [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|Reziprozitätsgesetz von Bandbreite und Impulsdauer]].
*Sowohl $H(f)$ als auch $h(t)$ sind zu keinem $f$- bzw. $t$-Wert exakt gleich Null.
*Für praktische Anwendungen kann der Gaußimpuls jedoch in Zeit und Frequenz als begrenzt angenommen werden. Zum Beispiel ist $h(t)$ bereits bei $t=1.5 \cdot \Delta t$ auf weniger als $0.1\% $ des Maximums abgefallen.

===Idealer (rechteckförmiger) Tiefpass   $\Rightarrow$   Rectangular Low–pass ===
*Der Rechteck–Tiefpass lautet mit der Höhe $K$ und der (äquivalenten) Bandbreite $\Delta f$:

:$$H(f) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K /2 \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} f\hspace{0.05cm} \right| < \Delta f/2,} \\ {\left| \hspace{0.05cm}f\hspace{0.05cm} \right| = \Delta f/2,} \\ {\left|\hspace{0.05cm} f \hspace{0.05cm} \right| > \Delta f/2.} \\ \end{array}$$

*Der $\pm \Delta f/2$–Wert liegt mittig zwischen links- und rechtsseitigem Grenzwert.
*Für die Impulsantwort $h(t)$ erhält man entsprechend den Gesetzmäßigkeiten der Fourierrücktransformation (2. Fourierintegral):
:$$h(t)=K\cdot \Delta f \cdot {\rm si}(\pi\cdot \Delta f \cdot t) \quad \text{mit} \ {\rm si}(x)={\sin(x)}/{x}.$$
*Der $h(t)$–Wert bei $t=0$ ist gleich der Rechteckfläche des Frequenzgangs.
*Die Impulsantwort besitzt Nullstellen in äquidistanten Abständen $1/\Delta f$.
*Das Integral über die Impulsantwort $h(t)$ ist gleich dem Frequenzgang $H(f)$ bei der Frequenz $f=0$, also gleich $K$.

===Dreieck–Tiefpass $\Rightarrow$ Triangular Low–pass===

*Der Dreieck–Tiefpass lautet mit der Höhe $K$ und der (äquivalenten) Bandbreite $\Delta f$:

:$$H(f) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot \Big(1-\frac{|f|}{\Delta f}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} f\hspace{0.05cm} \right| < \Delta f,} \\ {\left| \hspace{0.05cm}f\hspace{0.05cm} \right| \ge \Delta f.} \\ \end{array}$$

*Die absolute physikalische Bandbreite $B$   ⇒   nur positive Frequenzen]   ist ebenfalls gleich $\Delta f$, also so groß wie beim Rechteck–Tiefpass.
*Für die Impulsantwort $h(t)$ erhält man gemäß der Fouriertransformation:
:$$h(t)=K\cdot \Delta f \cdot {\rm si}^2(\pi\cdot \Delta f \cdot t) \quad \text{mit} \ {\rm si}(x)={\sin(x)}/{x}.$$
*$H(f)$ kann man als Faltung zweier Rechteckfunktionen (jeweils mit Breite $\Delta f$) darstellen.
*Daraus folgt: $h(t)$ beinhaltet anstelle der ${\rm si}$-Funktion die ${\rm si}^2$-Funktion.
*$h(t)$ weist somit ebenfalls Nullstellen im äquidistanten Abständen $1/\Delta f$ auf.
*Der asymptotische Abfall von $h(t)$ erfolgt hier mit $1/t^2$, während zum Vergleich beim Rechteck–Tiefpass $h(t)$ mit $1/t$ abfällt.

===Trapez–Tiefpass   $\Rightarrow$   Trapezoidal Low–pass ===
Der Trapez–Tiefpass lautet mit der Höhe $K$ und den Eckfrequenzen $f_1$ und $f_2$:
:$$H(f) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \frac{f_2-|f|}{f_2-f_1} \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} f\hspace{0.05cm} \right| \le f_1,} \\ {f_1\le \left| \hspace{0.05cm}f\hspace{0.05cm} \right| \le f_2,} \\ {\left|\hspace{0.05cm} f \hspace{0.05cm} \right| \ge f_2.} \\ \end{array}$$

*Für die äquivalente Bandbreite (flächengleiches Rechteck) gilt: $\Delta f = f_1+f_2$.
*Der Rolloff-Faktor (im Frequenzbereich) kennzeichnet die Flankensteilheit:
:$$r=\frac{f_2-f_1}{f_2+f_1}.$$
*Der Sonderfall $r=0$ entspricht dem Rechteck–Tiefpass und der Sonderfall $r=1$ dem Dreieck–Tiefpass.
*Für die Impulsantwort erhält man gemäß der Fourierrücktransformation:
:$$h(t)=K\cdot \Delta f \cdot {\rm si}(\pi\cdot \Delta f \cdot t)\cdot {\rm si}(\pi \cdot r \cdot \Delta f \cdot t) \quad \text{mit} \ {\rm si}(x)={\sin(x)}/{x}.$$
*Der asymptotische Abfall von $h(t)$ liegt zwischen $1/t$ (für Rechteck–Tiefpass oder $r=0$) und $1/t^2$ (für Dreieck–Tiefpass oder $r=1$).

===Cosinus-Rolloff-Tiefpass   $\Rightarrow$   Cosine-rolloff Low–pass ===
Der Cosinus–Rolloff–Tiefpass lautet mit der Höhe $K$ und den Eckfrequenzen $f_1$ und $f_2$:

:$$H(f) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \cos^2\Big(\frac{|f|-f_1}{f_2-f_1}\cdot \frac{\pi}{2}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} f\hspace{0.05cm} \right| \le f_1,} \\ {f_1\le \left| \hspace{0.05cm}f\hspace{0.05cm} \right| \le f_2,} \\ {\left|\hspace{0.05cm} f \hspace{0.05cm} \right| \ge f_2.} \\ \end{array}$$

*Für die äquivalente Bandbreite (flächengleiches Rechteck) gilt: $\Delta f = f_1+f_2$.
*Der Rolloff-Faktor (im Frequenzbereich) kennzeichnet die Flankensteilheit:
:$$r=\frac{f_2-f_1}{f_2+f_1}.$$
*Der Sonderfall $r=0$ entspricht dem Rechteck–Tiefpass der Sonderfall $r=1$ dem Cosinus-Quadrat-Tiefpass.
*Für die Impulsantwort erhält man gemäß der Fourierrücktransformation:
:$$h(t)=K\cdot \Delta f \cdot \frac{\cos(\pi \cdot r\cdot \Delta f \cdot t)}{1-(2\cdot r\cdot \Delta f \cdot t)^2} \cdot {\rm si}(\pi \cdot \Delta f \cdot t).$$
*Je größer der Rolloff-Faktor $r$ ist, desto schneller nimmt $h(t)$ asymptotisch mit $t$ ab.

===Cosinus-Quadrat-Tiefpass   $\Rightarrow$   Cosine-rolloff -squared Low–pass ===
*Dies ist ein Sonderfall des Cosinus–Rolloff–Tiefpasses und ergibt sich aus diesem für $r=1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}f_1=0, f_2= \Delta f$:

:$$H(f) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot \cos^2\Big(\frac{|f|\cdot \pi}{2\cdot \Delta f}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} f\hspace{0.05cm} \right| < \Delta f,} \\ {\left| \hspace{0.05cm}f\hspace{0.05cm} \right| \ge \Delta f.} \\ \end{array}$$

*Für die Impulsantwort erhält man gemäß der Fourierrücktransformation:
:$$h(t)=K\cdot \Delta f \cdot {\pi}/{4}\cdot \big [{\rm si}(\pi(\Delta f\cdot t +0.5))+{\rm si}(\pi(\Delta f\cdot t -0.5))\big ]\cdot {\rm si}(\pi \cdot \Delta f \cdot t).$$
*Wegen der letzten ${\rm si}$-Funktion ist $h(t)=0$ für alle Vielfachen von $T=1/\Delta f$   ⇒   Die äquidistanten Nulldurchgänge des Cosinus–Rolloff–Tiefpasses bleiben erhalten.
*Aufgrund des Klammerausdrucks weist $h(t)$ nun weitere Nulldurchgänge bei $t=\pm1.5 T$, $\pm2.5 T$, $\pm3.5 T$, ... auf.
*Für $t=\pm T/2$ hat die Impulsanwort den Wert $K\cdot \Delta f/2$.
*Der asymptotische Abfall von $h(t)$ verläuft in diesem Sonderfall mit $1/t^3$.

==Vorschlag für die Versuchsdurchführung==
 
"Rot" bezieht sich stets auf den ersten Parametersatz   ⇒   $H_1(f) \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\ h_1(t)$ und "Blau" auf den zweiten   ⇒   $H_2(f) \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\ h_2(t)$.

{{BlaueBox|TEXT=
'''(1)'''   Vergleichen Sie den '''roten Gauß–Tiefpass''' $(K_1 = 1, \Delta f_1 = 1)$ mit dem '''blauen Rechteck–Tiefpass''' $(K_2 = 1, \Delta f_2 = 1)$   ⇒   Voreinstellung  ]   und beantworten Sie folgende Fragen: 
*Welche Signale $y(t)$ treten am Ausgang der Tiefpässe auf, wenn am Eingang das Signal $x(t) = 2 \cdot \cos (2\pi f_0 t -\varphi_0)$ mit $f_0 = 0.5$ anliegt?
*Welche Unterschiede ergeben sich bei beiden Tiefpässen mit $f_0 = 0.5 \pm f_\varepsilon$ und $f_\varepsilon \ne 0, \ f_\varepsilon \to 0$?}}

*In beiden Fällen gilt $y(t) = A \cdot \cos (2\pi f_0 t -\varphi_0)$ mit $A = 2 \cdot H(f = f_0) \ \Rightarrow \ A_1 = 2 \cdot 0.456 = 0.912, A_2 = 2 \cdot 0.5 =1.000$. Die Phase $\varphi_0$ bleibt erhalten.

*Beim Gauß–Tiefpass gilt weiterhin $ A_1 = 0.912$. Beim Rechteck–Tiefpass ist $A_2 = 0$ für $f_0 = 0.5000\text{...}001$ und $A_2 = 2$ für $f_0 = 0.4999\text{...}999$.

{{BlaueBox|TEXT=
'''(2)'''   Lassen Sie die Einstellungen unverändert. Welcher Tiefpass kann das [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#First_Nyquist_criterion_in_the_time_domain|erste Nyquistkriterium]] oder das [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#Second_Nyquist_criterion|zweite Nyquistkriterium]] erfüllen, wenn $H(f)$ den Gesamtfrequenzgang von Sender, Kanal und Empfangsfilter bezeichnet?}}

*Um das erste Nyquistkriterium zu erfüllen, muss die Impulsantwort $h(t)$ äquidistante Nulldurchgänge bei Vielfachen der (normierten) Zeit $t = 1, 2$, ... aufweisen. Die Impulsantwort $h(t) = {\rm si}(\pi \cdot \Delta f \cdot t)$ des Rechteck–Tiefpasses erfüllt dieses Kriterium mit $\Delta f = 1$. Dagegen ist beim Gauß–Tiefpass das erste Nyquistkriterium nie erfüllt und es kommt immer zu [[Digital_Signal_Transmission/Ursachen_und_Auswirkungen_von_Impulsinterferenzen|Impulsinterferenzen]].
*Das zweite Nyquistkriterium erfüllt der Rechteck–Tiefpass ebenso nicht wie der Gauß–Tiefpass.

{{BlaueBox|TEXT=
'''(3)'''   Vergleichen Sie den '''roten Rechteck–Tiefpass''' $(K_1 = 0.5, \Delta f_1 = 2)$ mit dem '''blauen Rechteck–Tiefpass''' $(K_2 = 1, \Delta f_2 = 1)$ und variieren Sie anschließend $\Delta f_1$ zwischen $2$ und $0.5$. }}

*Bei der Einstellung $\Delta f_1 = 2$ liegen die Nullstellen der Impulsantwort bei Vielfachen von $0.5$. Die Impulsantwort $h_1(t)$ klingt also doppelt so schnell ab als die Impulsantwort $h_2(t)$ des schmalbandigeren Tiefpasses $H_2(f)$.
*Mit dieser Einstellung gilt $h_1(t = 0) = h_2(t = 0)$, da die Rechteckflächen von $H_1(f)$ und $H_2(f)$ gleich sind.
*Verringert man man $\Delta f_1$, so wird die Impulsantwort $h_1(t)$ immer breiter und niedriger. Mit $\Delta f_1 = 0.5$ ist $h_1(t)$ doppelt so breit wie $h_2(t)$, gleichzeitig aber um den Faktor $4$ niedriger.

{{BlaueBox|TEXT=
'''(4)'''   Vergleichen Sie den '''roten Trapez–Tiefpass''' $(K_1 = 1, \Delta f_1 = 1, r_1 = 0.5)$ mit dem '''blauen Rechteck–Tiefpass''' $(K_2 = 1, \Delta f_2 = 1)$ und variieren Sie anschließend $r_1$ zwischen $0$ und $1$. }}

*Bei der Einstellung $r_1 = 0.5$ sind die Unterschwinger in der Impulsantwort $h(t)$ beim Trapez–Tiefpass aufgrund des flacheren Flankenabfalls geringer als beim Rechteck–Tiefpass.
*Je kleiner der Roll–off–Faktor $r_1$ wird, desto größer werden die Unterschwinger. Bei $r_1= 0$ ist der Trapez–Tiefpass identisch mit dem Rechteck–Tiefpass   ⇒   $h(t)= {\rm si}(\pi \cdot t)$.
*Erhöht man dagegen den Roll–off–Faktor $r_1$, so größer werden die Unterschwinger kleiner. Bei $r_1= 1$ ist der Trapez–Tiefpass identisch mit dem Dreieck–Tiefpass   ⇒   $h(t)= {\rm si}^2(\pi \cdot t)$.

{{BlaueBox|TEXT=
'''(5)'''   Vergleichen Sie den '''roten Trapez–Tiefpass''' $(K_1 = 1, \Delta f_1 = 1, r_1 = 0.5)$ mit dem '''blauen Cosinus-Rolloff-Tiefpass''' $(K_2 = 1,\Delta f_2 = 1, r_2 = 0.5)$. Variieren Sie $r_2$ zwischen $0$ und $1$. Interpretieren Sie die Impulsantwort $h_2(t)$ für $r_2 = 0.75$. Welcher Tiefpass erfüllt das [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#First_Nyquist_criterion_in_the_time_domain|erste Nyquistkriterium]] ?}}

*Bei gleichem Rolloff-Faktor $r_1 = r_2= 0.5$ verläuft der Flankenabfall des Cosinus-Rolloff-Tiefpasses $H_2(f)$ um die Frequenz $f = 0.5$ steiler als der Flankenabfall des Trapez–Tiefpasses $H_1(f)$.
*Der Vergleich der zugehörigen Impulsantworten bei gleichem Rolloff-Faktor $r= 0.5$ zeigt, dass $h_2(t)$ für $t > 1$ betragsmäßig größere Anteile besitzt als $h_1(t)$.
*Mit $r_1 = 0.5$ und $r_2 = 0.75$ gilt $H_1(f) \approx H_2(f)$ und damit auch $h_1(t) \approx h_2(t)$.
*Beide Frequenzgänge $H_1(f)$ und $H_2(f)$ erfüllen das erste Nyquistkriterium, da die Funktionen bei $\Delta f = 1$ punktsymmetrisch um den Punkt $f = f_{\rm Nyq} = 1/2, \ H(f_{\rm Nyq}) = K/2$ sind.
*Wegen $\Delta f = 1$ besitzen sowohl $h_1(t)$ als auch $h_2(t)$ Nulldurchgänge bei $\pm 1$, $\pm 2$, ...   ⇒   die vertikale Augenöffnung ist in beiden Fällen maximal.

{{BlaueBox|TEXT=
'''(6)'''   Vergleichen Sie den '''roten Cosinus–Quadrat–Tiefpass''' $(K_1 = 1, \Delta f_1 = 1)$ mit dem '''blauen Cosinus-Rolloff-Tiefpass''' $(K_2 = 1,\Delta f_2 = 1, r_2 = 0.5)$. Variieren Sie $r_2$ zwischen $0$ und $1$. Interpretieren Sie die Ergebnisse. Welcher Tiefpass erfüllt das [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#Second_Nyquist_criterion|zweite Nyquistkriterium]]?}}

*Der Cosinus–Quadrat–Tiefpass $H_1(f)$ ist ein Sonderfall Cosinus–Rolloff–Tiefpasses $H_2(f)$ mit Rolloff-Faktor $r_2 =1$. Das erste Nyquistkriterium wird auch mit $r_2 \ne 1$ erfüllt.
*Soll das zweite Nyquistkriterium erfüllt sein, so muss die Impulsantwort weitere Nulldurchgänge bei $t=\pm 1.5$, $\pm 2.5$, $\pm 3.5$, ... aufweisen (nicht jedoch bei $t = \pm 0.5$).
*Für den Cosinus–Quadrat–Tiefpass $H_1(f)$ gilt also $h_1(t=\pm 1) = h_1(t=\pm 1.5) = h_1(t=\pm 2)= h_1(t=\pm 2.5) = \text{...} =0$. Dagegen ist $h_1(t=\pm 0.5) = 0.5$. Der asymptotische Abfall von $h(t)$ verläuft in diesem Sonderfall mit $1/t^3$.
*Kein anderer Tiefpass als der Cosinus–Quadrat–Tiefpass erfüllt das erste und zweite Nyquistkriterium gleichzeitig. Demzufolge ist sowohl die vertikale als auch die horizontale Augenöffnung maximal.

==Zur Handhabung des Programms==
[[File:Frequenzgang_fertig_version1.png|left]]
    '''(A)'''     Bereich der graphischen Darstellung für $H(f)$

    '''(B)'''     Bereich der graphischen Darstellung für $h(t)$

    '''(C)'''     Variationsmöglichkeit für die graphischen Darstellungen

    '''(D)'''     Parametereingabe per Slider                       links (rot): "Low–pass 1",         rechts (blau): "Low–pass 2"

    '''(E)'''     Parameter entsprechend der Voreinstellung   ⇒   "Reset"

    '''(F)'''     Einstellung von $t_*$ und $f_*$ für Numerikausgabe

    '''(G)'''     Numerikausgabe von $H(f_*)$ und $h(t_*)$                       links (rot): "Low–pass 1",         rechts (blau): "Low–pass 2"

'''Details zum obigen Punkt (C)'''

    '''(*)'''   Zoom–Funktionen "$+$" (Vergrößern), "$-$" (Verkleinern)                      und $\rm o$ (Zurücksetzen)

    '''(*)'''   Verschiebe–Funktionen "$\leftarrow$" (Bildausschnitt nach links,                      Ordinate nach rechts) sowie "$\uparrow$" "$\downarrow$" "$\rightarrow$"

'''Andere Möglichkeiten''':

*Bei gedrückter Shifttaste und Scrollen kann im Koordinatensystem gezoomt werden.
*Bei gedrückter Shifttaste und gedrückter linker Maustaste kann das Koordinatensystem verschoben werden.
 

==Über die Autoren==
Dieses interaktive Berechnungstool wurde am [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik] der [https://www.tum.de/ Technischen Universität München] konzipiert und realisiert.
*Die erste Version wurde 2005 von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]] im Rahmen ihrer Diplomarbeit mit "FlashMX–Actionscript" erstellt (Betreuer: [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._Günter_Söder_(at_LNT_from_1974-2024)|Günter Söder]] und [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Klaus_Eichin_.28at_LNT_from_1972-2011.29|Klaus Eichin]]).
*2017 wurde "Impulse & Spektren" von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#David_Jobst_.28Ingenieurspraxis_Math_2017.29|David Jobst]] im Rahmen seiner Ingenieurspraxis (Betreuer: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) auf "HTML5" umgesetzt und neu gestaltet.

==Nochmalige Aufrufmöglichkeit des Applets in neuem Fenster==
{{LntAppletLinkEn|frequImpResp_en}}
[[de:Applets:Frequenzgang und Impulsantwort]]

Applets:Periodendauer periodischer Signale

2026-03-16T15:58:56Z

Maintenance script: Fix interlanguage link: resolve redirect chain

Wir bieten hier zwei Applets zur gleichen Thematik mit unterschiedlichem Layout an:

{{LntAppletLink|signalPeriod_en|Applet-Variante 1 in neuem Tab öffnen}}     {{LntAppletLink|signalPeriodS_en|Applet-Variante 2 in neuem Tab öffnen}}

==Programmbeschreibung==
 
Dieses Applet zeichnet den Verlauf und berechnet die Periodendauer $T_0$ der periodischen Funktion
:$$x(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right).$$

Bitte beachten Sie:
*Die Phasen $\varphi_i$ sind hier im Bogenmaß einzusetzen. Umrechnung aus dem Eingabewert:   $\varphi_i \text{[im Bogenmaß]} =\varphi_i \text{[in Grad]}/360 \cdot 2\pi$.
*Ausgegeben werden auch der Maximalwert $x_{\rm max}$ und ein Signalwert $x(t_*)$ zu einer vorgebbaren Zeit $t_*$.
*Das aufzurufende Applet verwendet die englischen Begriffe im Gegensatz zu dieser deutschen Beschreibung.

Die englische Beschreibung finden Sie unter [[Period Duration of Periodic Signals]] (derzeit noch nicht realisiert) .

==Theoretischer Hintergrund==
 
*Ein ''periodisches Signal'' $x(t)$ liegt genau dann vor, wenn dieses nicht konstant ist und für alle beliebigen Werte von $t$ und alle ganzzahligen Werte von $i$ mit einem geeigneten $T_{0}$ gilt:   $x(t+i\cdot T_{0}) = x(t)$. Man bezeichnet $T_0$ als die '''Periodendauer''' und $f_0 = 1/T_0$ als die '''Grundfrequenz'''.

*Bei einer harmonischen Schwingung $x_1(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)$ gilt $f_0 = f_1$ und $T_0 = 1/f_1$, unabhängig von der Phase $\varphi_1$ und der Amplitude $A_1 \ne 0$.

{{BlaueBox|TEXT=
$\text{Berechnungsvorschrift:}$  Setzt sich das periodisches Signal $x(t)$ wie in diesem Applet aus zwei Anteilen $x_1(t)$ und $x_2(t)$ zusammen, dann gilt mit $A_1 \ne 0$, $f_1 \ne 0$, $A_2 \ne 0$, $f_2 \ne 0$ für Grundfrequenz und Periodendauer:

:$$f_0 = {\rm ggT}(f_1, \ f_2) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}T_0 = 1/f_0,$$
wobei "ggT" den ''größten gemeinsamen Teiler'' bezeichnet.}}

{{GraueBox|TEXT=
$\text{Beispiele:}$   Im Folgenden bezeichnen $f_0'$, $f_1'$ und $f_2'$ die auf $1\ \rm kHz$ normierten Signalfrequenzen:

'''(a)'''   $f_1' = 1.0$,   $f_2' = 3.0$   ⇒   $f_0' = {\rm ggt}(1.0, \ 3.0) = 1.0$   ⇒   $T_0 = 1.0\ \rm ms$;

'''(b)'''   $f_1' = 1.0$,   $f_2' = 3.5$   ⇒   $f_0' = {\rm ggt}(1.0, \ 3.5)= 0.5$   ⇒   $T_0 = 2.0\ \rm ms$;

'''(c)'''   $f_1' = 1.0$,   $f_2' = 2.5$   ⇒   $f_0' = {\rm ggt}(1.0, \ 2.5) = 0.5$   ⇒   $T_0 = 2.0\ \rm ms$;

'''(d)'''   $f_1' = 0.9$,   $f_2' = 2.5$   ⇒   $f_0' = {\rm ggt}(0.9, \ 2.5) = 0.1$   ⇒   $T_0 = 10.0 \ \rm ms$;

'''(e)'''   $f_2' = \sqrt{2} \cdot f_1' $   ⇒   $f_0' = {\rm ggt}(f_1', \ f_2') \to 0$   ⇒   $T_0 \to \infty$  ⇒   Das Signal $x(t)$ ist nicht periodisch.}}

$\text{Anmerkung:}$  Die Periodendauer könnte auch als ''kleinstes gemeinsame Vielfache'' (kgV) entsprechend $T_0 = {\rm kgV}(T_1, \ T_2)$ ermittelt werden:

'''(c)'''   $T_1 = 1.0\ \rm ms$,   $T_2 = 0.4\ \rm kHz$   ⇒   $T_0 = {\rm kgV}(1.0, \ 0.4) \ \rm ms = 2.0\ \rm ms$

Bei allen anderen Parameterwerten würde es aber zu numerischen Problemen kommen, zum Beispiel

'''(a)'''   $T_1 = 1.0\ \rm ms$ und $T_2 = 0.333\text{...} \ \rm ms$ besitzen aufgrund der begrenzten Darstellung reeller Zahlen kein kleinstes gemeinsames Vielfaches.

==Vorschlag für die Versuchsdurchführung==
 
Im Folgenden bezeichnen $A_1'$ und $A_2'$ die auf $1\ \rm V$ normierten Signalamplituden und $f_0'$, $f_1'$ und $f_2'$ die auf $1\ \rm kHz$ normierte Frequenzen:

{{BlaueBox|TEXT=
'''(1)'''   nach Voreinstellung:     $A_1' = 1.0, \ A_2' = 0.5, \ f_1' = 2.0, \ A_2' = 2.5, \ \varphi_1 = 0^\circ \ \varphi_2 = 90^\circ\text{:}$}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$Die Periodendauer ist $T_0 = 2.0 \ \rm ms$   wegen   ${\rm ggt}(2.0, 2.5) = 0.5$.

{{BlaueBox|TEXT=
'''(2)'''   Variieren Sie bei der bestehenden Einstellung $\varphi_1$ und $\varphi_2$ im gesamten möglichen Bereich $\pm 180^\circ\text{:}$}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$Die Periodendauer $T_0 = 2.0 \ \rm ms$ bleibt erhalten.

{{BlaueBox|TEXT=
'''(3)'''   Wählen Sie die Voreinstellung   ⇒   "Recall Parameters" und variieren Sie $A_1'$ im gesamten möglichen Bereich $0 \le A_1' \le 1\text{:}$}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$Die Periodendauer $T_0 = 2.0 \ \rm ms$ bleibt erhalten mit Ausnahme von $A_1' =0$. In diesem Fall ist $T_0 = 0.4 \ \rm ms$.

{{BlaueBox|TEXT=
'''(4)'''   Wählen Sie die Voreinstellung   ⇒   "Recall Parameters" und ändern Sie $f_2' = 0.2\text{:}$}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$Die Periodendauer ist $T_0 = 5.0 \ \rm ms$   wegen   ${\rm ggt}(2.0, 0.2) = 0.2$.

{{BlaueBox|TEXT=
'''(5)'''   Wählen Sie die Voreinstellung   ⇒   "Recall Parameters" und ändern Sie $f_1' = 0.2$. Speichern Sie diese Einstellung mit "Store Parameters":}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$Die Periodendauer ist $T_0 = 10.0 \ \rm ms$   wegen   ${\rm ggt}(0.2, 2.5) = 0.1$.

{{BlaueBox|TEXT=
'''(6)'''   Wählen Sie die letzte Einstellung   ⇒   "Recall Parameters" und ändern Sie $f_2' = 0.6$. Speichern Sie diese Einstellung mit "Store Parameters":}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$Die Periodendauer ist $T_0 = 5.0 \ \rm ms$   wegen   ${\rm ggt}(0.2,0.6) = 0.2$.

{{BlaueBox|TEXT=
'''(7)'''   Wie groß ist bei gleicher Einstellung der maximale Signalwert $x_{\rm max}\text{?}$}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$Der maximale Signalwert ist $x_{\rm max} =x(t_* + i \cdot T_0) = 1.39 \ \rm V$ mit $t_* = 0.3 \ \rm ms$ und $T_0 = 5.0 \ \rm ms$
{{BlaueBox|TEXT=
'''(8)'''   Wählen Sie die letzte Einstellung   ⇒   "Recall Parameters" und ändern Sie $\varphi_2 = 0^\circ \hspace{0.1cm}\Rightarrow\hspace{0.1cm}$ Summe zweier Cosinusschwingungen:}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$Der maximale Signalwert ist nun mit $x_{\rm max} =x(t_* + i \cdot T_0) = 1.5 \ \rm V$, also gleich $A_1 + A_2$   ⇒   $t_* = 0$, $T_0 = 5.0 \ \rm ms$.

{{BlaueBox|TEXT=
'''(9)'''   Wählen Sie die vorletzte Einstellung   ⇒   "Recall Parameters" und ändern Sie $\varphi_1 = 90^\circ \hspace{0.1cm}\Rightarrow\hspace{0.1cm}$ Summe zweier Sinusschwingungen:}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}$Der maximale Signalwert ist nun mit $x_{\rm max} = 1.08 \ \rm V$, also ungleich $A_1 + A_2$  ⇒   $t_* = 0.6 \ \rm ms$, $T_0 = 5.0 \ \rm ms$.

==Zur Handhabung der Applet-Variante 1==
[[File:Periodendauer_fertig_version1.png|left]]
    '''(A)'''     Parametereingabe per Slider

    '''(B)'''     Bereich der graphischen Darstellung

    '''(C)'''     Variationsmöglichkeit für die graphische Darstellung

    '''(D)'''     Abspeichern und Zurückholen von Parametersätzen

    '''(E)'''     Numerikausgabe des Hauptergebnisses $T_0$; graphische Verdeutlichung durch rote Linie

    '''(F)'''     Ausgabe von $x_{\rm max}$ und der Signalwerte $x(t_*) = x(t_* + T_0)= x(t_* + 2T_0)$

    '''(G)'''     Darstellung der Signalwerte $x(t_*) = x(t_* + T_0)= x(t_* + 2T_0)$ durch grüne Punkte

    '''(H)'''     Einstellung der Zeit $t_*$ für die Signalwerte $x(t_*) = x(t_* + T_0)= x(t_* + 2T_0)$

'''Details zum obigen Punkt (C)'''

    '''(*)'''   Zoom–Funktionen "$+$" (Vergrößern), "$-$" (Verkleinern) und $\rm o$ (Zurücksetzen)

    '''(*)'''   Verschieben mit "$\leftarrow$" (Ausschnitt nach links, Ordinate nach rechts), "$\uparrow$" "$\downarrow$" und "$\rightarrow$"

'''Andere Möglichkeiten''':

    '''(*)'''   Gedrückte Shifttaste und Scrollen: Zoomen im Koordinatensystem,

    '''(*)'''   Gedrückte Shifttaste und linke Maustaste: Verschieben des Koordinatensystems.
 

==Zur Handhabung der Applet-Variante 2==
[[File:Periodendauer_SB_version2.png|left]]
    '''(A)'''     Parametereingabe

    '''(B)'''     Bereich der graphischen Darstellung

    '''(C)'''     Größe der graphischen Darstellung

    '''(D)'''     Speichern/Zurückholen von Eingaben

    '''(E)'''     Numerikausgabe des Hauptergebnisses $T_0$;                       in Grafik:     blaue Linien im Abstand $T_0$

    '''(F)'''     Eingabe $t_\star$,   Ausgabe von $x(t_*)$ und $x_{\rm max}$
 

==Über die Autoren==
Dieses interaktive Berechnungstool wurde am [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik] der [https://www.tum.de/ Technischen Universität München] konzipiert und realisiert.
*Die erste Version wurde 2004 von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]] im Rahmen ihrer Diplomarbeit mit "FlashMX–Actionscript" erstellt (Betreuer: [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._Günter_Söder_(at_LNT_from_1974-2024)|Günter Söder]] ).
*2017 wurde dieses Programm von [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#David_Jobst_.28Ingenieurspraxis_Math_2017.29|David Jobst]] im Rahmen seiner Ingenieurspraxis (Betreuer: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) auf "HTML5" umgesetzt und neu gestaltet   ⇒   Applet-Variante 1.
*Parallel dazu erarbeitete [[Biographies_and_Bibliographies/Students_involved_in_LNTwww#Bastian_Siebenwirth_.28Bachelorarbeit_LB_2017.29|Bastian Siebenwirth]] im Rahmen seiner Bachelorarbeit (Betreuer: [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._Günter_Söder_(at_LNT_from_1974-2024)|Günter Söder]]) die HTML5-Variante 2.

==Nochmalige Aufrufmöglichkeit der Applets in neuem Fenster==
Wir bieten hier zwei Applets zur gleichen Thematik mit unterschiedlichem Layout an:

{{LntAppletLink|signalPeriod_en|Applet-Variante 1 in neuem Tab öffnen}}     {{LntAppletLink|signalPeriodS_en|Applet-Variante 2 in neuem Tab öffnen}}
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Applets:Principle of QAM

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Aufgaben:Exercise 2.8: COST Delay Models

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{{quiz-Header|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}

[[File:Mob_A_2_8_version2.png|right|frame|COST delay models]]
On the right, four delay power density spectra are plotted logarithmically as a function of the delay time  $\tau$ 
:$$10 \cdot {\rm lg}\hspace{0.15cm} ({{\it \Phi}_{\rm V}(\tau)}/{\it \Phi}_{\rm 0}) \hspace{0.05cm}.$$

Here the abbreviation  $\phi_0 = \phi_{\rm V}(\tau = 0)$  is used.  These are the so-called COST delay models.

The upper sketch contains the two profiles  ${\rm RA}$  ("Rural Area") and  ${\rm TU}$  ("Typical Urban").  Both of these are exponential:
:$${{\it \Phi}_{\rm V}(\tau)}/{\it \Phi}_{\rm 0} = {\rm e}^{ -\tau / \tau_0} \hspace{0.05cm}.$$

The value of the parameter  $\tau_0$  (time constant of the auto-correlation function) should be determined from the graphic in subtask  '''(1)'''.  Note the specified values of   $\tau_{-30}$ for  ${\it \Phi}_{\rm V}(\tau_{-30})=-30 \ \rm dB$:
:$${\rm RA}\text{:}\hspace{0.15cm}\tau_{-30} = 0.75\,{\rm µ s} \hspace{0.05cm},\hspace{0.2cm}{\rm TU}\text{:}\hspace{0.15cm}\tau_{-30} = 6.9\,{\rm µ s} \hspace{0.05cm}. $$

The lower graph applies to less favourable conditions in
* urban areas ("Bad Urban",  ${\rm BU}$):
:$${{\it \Phi}_{\rm V}(\tau)}/{{\it \Phi}_{\rm 0}}= \left\{ \begin{array}{c} {\rm e}^{ -\tau / \tau_0} \\0.5 \cdot {\rm e}^{ (5\,{\rm \mu s}-\tau) / \tau_0} \end{array} \right.\quad\begin{array}{*{1}c} \hspace{-0.55cm} {\rm if}\hspace{0.15cm}0 < \tau < 5\,{\rm µ s}\hspace{0.05cm},\hspace{0.15cm}\tau_0 = 1\,{\rm µ s} \hspace{0.05cm},\\ \hspace{-0.15cm} {\,\, \,\, \rm if}\hspace{0.15cm}5\,{\rm µ s} < \tau < 10\,{\rm µ s}\hspace{0.05cm},\hspace{0.15cm}\tau_0 = 1\,{\rm µ s} \hspace{0.05cm}, \end{array}$$

* in rural areas ("Hilly Terrain",  ${\rm HT}$):
:$${{\it \Phi}_{\rm V}(\tau)}/{{\it \Phi}_{\rm 0}}= \left\{ \begin{array}{c} {\rm e}^{ -\tau / \tau_0} \\{0.04 \cdot \rm e}^{ (15\,{\rm \mu s}-\tau) / \tau_0} \end{array} \right.\quad\begin{array}{*{1}c} \hspace{-0.55cm} {\rm if}\hspace{0.15cm}0 < \tau < 2\,{\rm µ s}\hspace{0.05cm},\hspace{0.15cm}\tau_0 = 0.286\,{\rm µ s} \hspace{0.05cm},\\ \hspace{-0.35cm} {\rm if}\hspace{0.15cm}15\,{\rm µ s} < \tau < 20\,{\rm µ s}\hspace{0.05cm},\hspace{0.15cm}\tau_0 = 1\,{\rm µ s} \hspace{0.05cm}. \end{array}$$

For the models  ${\rm RA}$,  ${\rm TU}$  and  ${\rm BU}$  the following parameters are to be determined:
* The  delay spread  $T_{\rm V}$  is the standard deviation of the delay  $\tau$. If the delay power-spectral density  ${\it \Phi}_{\rm V}(\tau)$  has an exponential course as with the profiles  ${\rm RA}$  and  ${\rm TU}$, then  $T_{\rm V} = \tau_0$, see  [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth|Exercise 2.7]].

* The coherence bandwidth  $B_{\rm K}$  is the value of  $\Delta f$ at which the magnitude of the frequency correlation function  $\varphi_{\rm F}(\Delta f)$  has dropped to half its value for the first time. With exponential  ${\it \Phi}_{\rm V}(\tau)$  as with  ${\rm RA}$  and  ${\rm TU}$  the product is  $T_{\rm V} \cdot B_{\rm K} \approx 0.276$, see  [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth|Exercise 2.7]].

''Notes:''
*This exercise belongs to the chapter [[Mobile_Communications/The_GWSSUS_Channel_Model|The GWSSUS Channel Model]].
* The following integrals are given:
:$$\frac{1}{\tau_0} \cdot \int_{0}^{\infty}\hspace{-0.15cm} {\rm e}^{ -\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau = 1\hspace{0.05cm},\hspace{0.6cm}\frac{1}{\tau_0} \cdot \int_{0}^{\infty}\hspace{-0.15cm} {\tau} \cdot{\rm e}^{ -\tau / \tau_0}\hspace{0.15cm}{\rm d} \tau = \tau_0\hspace{0.05cm},\hspace{0.6cm}\frac{1}{\tau_0} \cdot \int_{0}^{\infty} \hspace{-0.15cm}{\tau^2} \cdot{\rm e}^{ -\tau / \tau_0}\hspace{0.15cm}{\rm d} \tau = 2\tau_0^2\hspace{0.05cm}.$$

===Questionnaire===
<quiz display=simple>
{Specify the parameter  $\tau_0$  of the delay power-spectral density for the profiles  ${\rm RA}$  and  ${\rm TU}$ .
|type="{}"}
${\rm RA} \text{:} \ \hspace{0.4cm} \tau_0 \ = \ ${ 0.109 3% } $\ \rm µ s$
${\rm TU} \text{:} \ \hspace{0.4cm} \tau_0 \ = \ ${ 1 3% } $\ \rm µ s$

{How large is the delay spread  $T_{\rm V}$  of these channels?
|type="{}"}
${\rm RA} \text{:} \ \hspace{0.4cm} T_{\rm V} \ = \ ${ 0.109 3% } $\ \rm µ s$
${\rm TU} \text{:} \ \hspace{0.4cm} T_{\rm V} \ = \ ${ 1 3% } $\ \rm µ s$

{What is the coherence bandwidth  $B_{\rm K}$  of these channels?
|type="{}"}
${\rm RA} \text{:} \ \hspace{0.4cm} B_{\rm K} \ = \ ${ 2500 3% } $\ \ \rm kHz$
${\rm TU} \text{:} \ \hspace{0.4cm} B_{\rm K} \ = \ ${ 276 3% } $\ \ \rm kHz$

{For which channel does frequency selectivity play a greater role?
|type="()"}
- Rural Area  $({\rm RA})$.
+ Typical urban  $({\rm TU})$.

{How large is the (normalized) power density for  "Bad Urban"  $({\rm BU})$   with   $\tau = 5.001 \ \rm µ s$  and with   $\tau = 4.999 \ \rm µ s$?
|type="{}"}
${\it \Phi}_{\rm V}(\tau = 5.001 \ \rm µ s) \ = \ ${ 0.5 3% } $\ \cdot {\it \Phi}_0$
${\it \Phi}_{\rm V}(\tau = 4.999 \ \rm µ s) \ = \ ${ 0.00674 3% } $\ \cdot {\it \Phi}_0$

{We consider  ${\rm BU}$ again. Let $P_1$ be the power of the signal between $0$  and  $5 \ \rm µ s$, and let $P_2$ be the remaining signal power. What percentage of the total signal power comes from the interval
$0< t < 5 \ \rm µ s$?
|type="{}"}
$P_1/(P_1 + P_2) \ = \ ${ 66.7 3% } $\ \rm \%$

{Calculate the delay spread  $T_{\rm V}$  of the profile  ${\rm BU}$. ''Note'':  The average delay is  $m_{\rm V} = E[\tau] = 2.667 \ \rm µ s$.
|type="{}"}
$T_{\rm V} \ = \ ${ 2.56 3% } $\ \rm µ s$
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The following property can be seen from the graph:
:$$10 \cdot {\rm lg}\hspace{0.1cm} (\frac{{\it \Phi}_{\rm V}(\tau_{\rm -30})}{{\it \Phi}_0}) =10 \cdot {\rm lg}\hspace{0.1cm}\left [{\rm exp}[ -\frac{\tau_{\rm -30}}{ \tau_{\rm 0}}]\right ] \stackrel {!}{=} -30\,{\rm dB}$$
:$$\Rightarrow \hspace{0.3cm} {\rm lg}\hspace{0.1cm}\left [{\rm exp}[ -\frac{\tau_{\rm -30}}{ \tau_{\rm 0}}]\right ] = -3\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\rm ln}\hspace{0.1cm}\left [{\rm exp}[ -\frac{\tau_{\rm -30}}{ \tau_{\rm 0}}]\right ] = -3 \cdot{\rm ln}\hspace{0.1cm}(10)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \tau_{\rm 0} = \frac{\tau_{\rm -30}}{ 3 \cdot {\rm ln}\hspace{0.1cm}(10)}\approx \frac{\tau_{\rm -30}}{ 6.9}\hspace{0.05cm}.$$

Here  $\tau_{-30}$  denotes the delay that leads to the logarithmic ordinate value  $-30 \ \rm dB$.  Thus one obtains
* for "Rural Area"  $\rm (RA)$  with  $\tau_{–30} = 0.75 \ \rm µ s$:
:$$\tau_{\rm 0} = \frac{0.75\,{\rm µ s}}{ 6.9} \hspace{0.1cm}\underline {\approx 0.109\,{\rm µ s}}\hspace{0.05cm},$$
* for urban and suburban areas  ⇒   "Typical Urban" $\rm (TU)$  with  $\tau_{–30} = 6.9 \ \rm µ s$:
:$$\tau_{\rm 0} = \frac{6.9\,{\rm µ s}}{ 6.9} \hspace{0.1cm}\underline {\approx 1\,{\rm µ s}}\hspace{0.05cm},$$

'''(2)'''  In  [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth|Exercise 2.7]], it was shown that the delay spread is  $T_{\rm V} =\tau_0$  when the delay power-spectral density decreases exponentially according to  ${\rm e}^{-\tau/\tau_0}$.  Thus the following applies:
* for "Rural Area":  $\hspace{0.4cm} T_{\rm V} \ \underline {= 0.109 \ \rm µ s}$,
* for "Typical Urban":  $\hspace{0.4cm} T_{\rm V} \ \underline {= 1 \ \rm µ s}$.

'''(3)'''  In Exercise 2.7 it was also shown that for the coherence bandwidth  $B_{\rm K} \approx 0.276/\tau_0$  applies.  It follows:
*$B_{\rm K} \ \underline {\approx 2500 \ \rm kHz}$ ("Rural Area"),
*$B_{\rm K} \ \underline {\approx 276 \ \ \rm kHz}$ ("Typical Urban")

'''(4)''' The second solution is correct:
*Frequency selectivity of the mobile radio channel is present if the signal bandwidth  $B_{\rm S}$  is larger than the coherence bandwidth  $B_{\rm K}$  (or at least of the same order of magnitude).
*The smaller  $B_{\rm K}$  is, the more often this happens.

'''(5)'''  According to the given equation, we have  ${\it \Phi}_{\rm V}(\tau = 5.001 \ \rm µ s)/{\it \Phi}_0 \hspace{0.15cm}\underline{\approx0.5}$.
*On the other hand, for slightly smaller  $\tau$  $($for example $\tau = 4.999 \ \rm µ s)$  we have approximately
:$${{\it \Phi}_{\rm V}(\tau = 4.999\,{\rm \mu s})}/{{\it \Phi}_{\rm 0}} = {\rm e}^{ -{4.999\,{\rm µ s}}/{ 1\,{\rm \mu s}}}\approx {\rm e}^{-5} \hspace{0.1cm}\underline {= 0.00674 }\hspace{0.05cm}.$$

'''(6)'''  The power  $P_1$  of all signal components with delays between  $0$  and  $5 \ µ\rm   s$  is:
:$$P_1 = {\it \Phi}_{\rm 0} \cdot \int_{0}^{5\,{\rm \mu s}} {\rm exp}[ -{\tau}/{ \tau_0}] \hspace{0.15cm}{\rm d} \tau \hspace{0.15cm} \approx \hspace{0.15cm}{\it \Phi}_{\rm 0} \cdot \int_{0}^{\infty} {\rm e}^{ -{\tau}/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau= {\it \Phi}_{\rm 0} \cdot \tau_0 \hspace{0.05cm}.$$

*The power outside  $[0\;µ \mathrm{s}, 5\;µ \mathrm{s}]$  is
:$$P_2 = \frac{{\it \Phi}_{\rm 0}}{2} \cdot \int_{5\,{\rm µ s}}^{\infty} {\rm exp}[ \frac{5\,{\rm µ s} -\tau}{ \tau_0}] \hspace{0.15cm}{\rm d} \tau \hspace{0.15cm} \approx \hspace{0.15cm}\frac{{\it \Phi}_{\rm 0}}{2} \cdot \int_{0}^{\infty} {\rm exp}[ -{\tau}/{ \tau_0}] \hspace{0.15cm}{\rm d} \tau= \frac{{\it \Phi}_{\rm 0} \cdot \tau_0}{2} \hspace{0.05cm}. $$

[[File:P_ID2184__Mob_A_2_8f.png|right|frame|Delay power density of the COST profiles  ${\rm BU}$  and  ${\rm HT}$]]

*Correspondingly, the percentage of power between  $0$  and  $5 \ µ\rm   s$  is
:$$\frac{P_1}{P_1+ P_2} = \frac{2}{3} \hspace{0.15cm}\underline {\approx 66.7\%}\hspace{0.05cm}.$$

The figure shows  ${\it \Phi}_{\rm V}(\tau)$  in linear scale.  The areas  $P_1$  and  $P_2$  are labeled.
*The left graph is for  ${\rm BU}$, the right graph is for  ${\rm HT}$.
*For the latter, the power percentage of all later echoes  $($later than  $15 \ \rm µ s)$  is only about  $12\%$.
 
'''(7)'''  The area of the entire power-spectral density gives  $P = 1.5 \cdot \phi_0 \cdot \tau_0$.
[[File:EN_Mob_A_2_8.png|right|frame|Delay PDF of profile  ${\rm BU}$ ]]

*Normalizing  ${\it \Phi}_{\rm V}(\tau)$  to this value yields the probability density function  $f_{\rm V}(\tau)$, as shown in the graph on the right (left diagram).

*With  $\tau_0 = 1 \ \ \rm µ s$  and  $\tau_5 = 5 \ \ \rm µ s$, the mean is:
:$$m_{\rm V}= \int_{0}^{\infty} f_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau$$
:$$\Rightarrow \hspace{0.3cm}m_{\rm V}= \frac{2}{3\tau_0} \cdot \int_{0}^{\tau_5} \tau \cdot {\rm e}^{ - {\tau}/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau \ + $$
:$$ \hspace{1.7cm}+\ \frac{1}{3\tau_0} \cdot \int_{\tau_5}^{\infty} \tau \cdot {\rm e}^{ (\tau_5 -\tau)/\tau_0}\hspace{0.15cm}{\rm d} \tau \hspace{0.05cm}. $$

*The first integral is equal to  $2\tau_0/3$  according to the provided expression.

*With the substitution  $\tau' = \tau \, -\tau_5$  you finally obtain using the integral solutions given above:
:$$m_{\rm V} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{2\tau_0}{3} + \frac{1}{3\tau_0} \cdot \int_{0}^{\infty} (\tau_5 + \tau') \cdot{\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' = \frac{2\tau_0}{3} +\frac{\tau_5}{3\tau_0} \cdot \int_{0}^{\infty} \cdot{\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' +\frac{1}{3\tau_0} \cdot \int_{0}^{\infty} \tau' \cdot \cdot{\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' $$
:$$\Rightarrow \hspace{0.3cm}m_{\rm V}= \frac{2\tau_0}{3} + \frac{\tau_5}{3}+ \frac{\tau_0}{3} = \tau_0 + \frac{\tau_5}{3}\hspace{0.15cm}\underline {\approx 2.667\,{\rm µ s}}\hspace{0.05cm}. $$

*The variance  $\sigma_{\rm V}^2$  is equal to the second moment (mean of the square) of the zero-mean random variable  $\theta = \tau \, –m_{\rm V}$, whose PDF is shown in the right graph
*From this  $T_{\rm V} = \sigma_{\rm V}$  can be specified.

A second possibility is to first calculate the mean square value of the random variable  $\tau$  and from this the variance  $\sigma_{\rm V}^2$  using Steiner's theorem.
*With the substitutions and approximations already described above, one obtains
:$$m_{\rm V2} \hspace{-0.1cm} \ \approx \ \hspace{-0.1cm} \frac{2}{3\tau_0} \cdot \int_{0}^{\infty} \tau^2 \cdot {\rm e}^{ - {\tau}/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau + \frac{1}{3\tau_0} \cdot \int_{0}^{\infty} (\tau_5 + \tau')^2 \cdot {\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' $$
:$$\Rightarrow \hspace{0.3cm}m_{\rm V2} = \frac{2}{3} \cdot \int_{0}^{\infty} \frac{\tau^2}{\tau_0} \cdot {\rm e}^{ - {\tau}/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau + \frac{\tau_5^2}{3} \cdot \int_{0}^{\infty} \frac{1}{\tau_0} \cdot {\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' +\frac{2\tau_5}{3} \cdot \int_{0}^{\infty} \frac{\tau '}{\tau_0} \cdot {\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' + \frac{1}{3} \cdot \int_{0}^{\infty} \frac{{\tau '}^2}{\tau_0} \cdot {\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau '\hspace{0.05cm}. $$

*With the integrals given above, we have
:$$m_{\rm V2} \approx \frac{2}{3} \cdot 2 \tau_0^2 + \frac{\tau_5^2}{3} \cdot 1 + \frac{2\tau_5}{3} \cdot \tau_0 +\frac{1}{3} \cdot 2 \tau_0^2 = 2 \tau_0^2 + \frac{\tau_5^2}{3} + \frac{2 \cdot \tau_0 \cdot \tau_5}{3} $$
:$$\Rightarrow \hspace{0.3cm} \sigma_{\rm V}^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} m_{\rm V2} - m_{\rm V}^2 = 2 \tau_0^2 + \frac{\tau_5^2}{3} + \frac{2 \cdot \tau_0 \cdot \tau_5}{3}- (\tau_0 + \frac{\tau_5}{3})^2 =\tau_0^2 + \frac{2\tau_5^2}{9} = (1\,{\rm µ s})^2 + \frac{2\cdot (5\,{\rm µ s})^2}{9} = 6.55\,({\rm µ s})^2$$
:$$\Rightarrow \hspace{0.3cm} T_{\rm V} = \sigma_{\rm V} \hspace{0.15cm}\underline {\approx 2.56\,{\rm µ s}}\hspace{0.05cm}.$$

The above graph shows the parameters  $T_{\rm V}$ and $\sigma_{\rm V}$.
{{ML-Fuß}}

[[Category:Mobile Communications: Exercises|^2.3 The GWSSUS Channel Model^]]
[[de:Aufgaben:Exercise 2.8: COST Delay Models]]

Aufgaben:Exercise 3.3Z: Convolution and D-Transformation

2026-03-16T15:58:49Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Algebraic_and_Polynomial_Description}}

[[File:P_ID2628__KC_Z_3_3.png|right|frame|Predefined filter structure]]

In this exercise, we use a simple example to describe
* the finite  »impulse response«  of a filter:
:$$\underline{g} = \left (g_0, g_1, \hspace{0.05cm}\text{...}\hspace{0.1cm}, g_l, \hspace{0.05cm}\text{...}\hspace{0.1cm}, g_m \right )\hspace{0.05cm},\hspace{0.2cm}g_l \in {\rm GF(2) } = \{ 0, 1 \}\hspace{0.05cm}, $$

* the  »input sequence«  of the filter:
:$$\underline{u} = \left (u_0, u_1, \hspace{0.05cm}\text{...}\hspace{0.1cm}, u_i, \hspace{0.05cm}\text{...}\hspace{0.1cm} \right )\hspace{0.05cm},\hspace{0.2cm}u_i \in {\rm GF(2) } = \{ 0, 1 \}\hspace{0.05cm}, $$
* the  »output sequence«  of the filter:
:$$\underline{x} = \left (x_0, x_1, \hspace{0.05cm}\text{...}\hspace{0.1cm}, x_i, \hspace{0.05cm}\text{...}\hspace{0.1cm} \right )\hspace{0.05cm},\hspace{0.2cm}x_i \in {\rm GF(2) } = \{ 0, 1 \}\hspace{0.05cm}. $$

We have adapted the nomenclature for this  $($digital$)$  filter description to the book  "Introduction to Channel Coding".  In other  $\rm LNTww$ books often 
#  $\underline{x}$  denotes the filter input, 
#  $\underline{y}$  the filter output,  and
#  the impulse response is called  $\underline{h}$.

In general,  for the output sequence corresponding to the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_time_domain| $\text{convolution}$]] :
:$$\underline{x} = \underline{u}* \underline{g} = \left (x_0, x_1, \hspace{0.05cm}\text{...}\hspace{0.1cm}, x_i, \hspace{0.05cm}\text{...}\hspace{0.1cm} \right )\hspace{0.05cm},\hspace{0.1cm} {\rm with} \hspace{0.2cm} x_i = \sum_{l = 0}^{m} g_l \cdot u_{i-l}\hspace{0.05cm}.$$

*We now represent the time functions   $\underline{g}, \ \underline{u}$   and   $\underline{x}$  by polynomials in a dummy variable  $D$  and call these the  D–transforms:
:$$\underline{g} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}{G}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \sum_{l = 0}^{m} g_l \cdot D\hspace{0.03cm}^l = g_0 + g_1 \cdot D + g_2 \cdot D^2 + \text{...} + g_m \cdot D\hspace{0.03cm}^m\hspace{0.05cm},$$
:$$\underline{u} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}{U}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \sum_{i = 0}^{\infty} u_i \cdot D\hspace{0.03cm}^i = u_0 + u_1 \cdot D + u_2 \cdot D^2 + \text{...} \hspace{0.05cm},$$
:$$\underline{x} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}{X}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \sum_{i = 0}^{\infty} x_i \cdot D\hspace{0.03cm}^i = x_0 + x_1 \cdot D + x_2 \cdot D^2 + \text{...} \hspace{0.05cm}.$$

*Thus the  $($more complicated$)$  convolution becomes a multiplication:
:$$\underline{x} = \underline{u}* \underline{g} \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}{X}(D) = U(D) \cdot G(D) \hspace{0.05cm}.$$

*Formally,  this relationship can be demonstrated as follows:
:$${X}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \sum_{i = 0}^{\infty} x_i \cdot D\hspace{0.03cm}^i = \sum_{i = 0}^{\infty} \sum_{l = 0}^{m}\hspace{0.1cm}g_l \cdot u_{i-l} \cdot D\hspace{0.03cm}^{i} = \sum_{l = 0}^{m} \hspace{0.1cm} g_l \cdot \sum_{j = -l}^{\infty} \hspace{0.1cm}u_{j} \cdot D\hspace{0.03cm}^{j+l} = \sum_{l = 0}^{m} \hspace{0.1cm} g_l \cdot D\hspace{0.03cm}^l \hspace{0.1cm} \cdot \hspace{0.1cm} \sum_{j = 0}^{\infty} \hspace{0.1cm}u_{j} \cdot D\hspace{0.03cm}^{j}$$
:$$\Rightarrow \hspace{0.3cm}{X}(D) = U(D) \cdot G(D)\hspace{0.05cm}.$$

*Here it was considered that all   $u_j$   for  $j < 0$  do not exist and can be set to zero.

*Both procedures for computing the initial sequence  $\underline{x}$  shall be demonstrated for the digital filter outlined above,  viz.
#   using the convolution,
#   by means of the D–transformation.

Hints:
* The exercise belongs to the chapter  [[Channel_Coding/Algebraic_and_Polynomial_Description| "Algebraic and Polynomial Description"]].

* Refer in particular to the  [[Channel_Coding/Algebraic_and_Polynomial_Description#GF.282.29_description_forms_of_a_digital_filter|"GF(2) Description Forms of a Digital Filter"]]  section.

* In the solution,  consider the following identity for calculations in  $\rm GF(2)$:
:$$1 + D + D^2 + D^3 + \hspace{0.05cm}\text{...}\hspace{0.1cm}= \frac{1}{1+D} \hspace{0.05cm}.$$

===Questions===
<quiz display=simple>
{What are the present filter coefficients?
|type="{}"}
$g_0 \ = \ ${ 1 }
$g_1 \ = \ ${ 1 }
$g_2 \ = \ ${ 0. }

{The sequence  $\underline{u} = (1, \, 0, \, 0, \, 1)$  let be finite.  What is the output sequence?
|type="()"}
- $\underline{x} = (1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
- $\underline{x} = (1, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
+ $\underline{x} = (1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
- $\underline{x} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$   ⇒   "sequence of ones".

{The sequence  $\underline{u} = (1, \, 1, \, 1)$  let be finite.  What is the output sequence?
|type="()"}
- $\underline{x} = (1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
+ $\underline{x} = (1, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
- $\underline{x} = (1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
- $\underline{x} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$   ⇒   "sequence of ones".

{What is the output sequence for  $\underline{u} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm}.)$   ⇒   "sequence of ones"?
|type="()"}
+ $\underline{x} = (1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
- $\underline{x} = (1, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
- $\underline{x} = (1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 0, \,\text{...}\hspace{0.05cm})$.
- $\underline{x} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$   ⇒   "sequence of ones".

{For which vector  $\underline{u}$  does the sequence  $\underline{x} = (1, \, 1, \, 1, \, 1, \ \text{...}\hspace{0.05cm})$  occur at the output?
|type="()"}
- $\underline{u} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$   ⇒   "sequence of ones"
+ $\underline{u} = (1, \, 0, \, 1, \, 0, \, 1, \, 0, \,\text{...}\hspace{0.05cm})$   ⇒   alternating sequence, starting with  "$1$".
- $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$   ⇒   alternating sequence, starting with  "$0$".
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The only two filter coefficients different from  $0$  are  $g_0 \ \underline{= 1}$  and  $g_1 \ \underline{= 1}$.
*From this follows  $g_2 \ \underline{= 0}$  and for the D–transform of the impulse response:
:$$\underline{g} = (1\hspace{0.05cm},\hspace{0.05cm} 1) \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}{G}(D) = 1+ D \hspace{0.05cm}.$$

'''(2)'''  The impulse response of the considered filter is  $\underline{g} = (1, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
*For the output sequence,  therefore,  we obtain the convolution product
:$$\underline{x} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \underline{u}* \underline{g} =(1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} ...\hspace{0.05cm}) *(1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} ... \hspace{0.05cm}) =(1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} ...\hspace{0.05cm})\hspace{0.05cm}.$$

*The same result is obtained using the D–transforms  $U(D) = 1 + D^3$  and  $G(D) = 1 + D$:
:$${X}(D) = U(D) \cdot G(D) = ( 1+D^3) \cdot (1+D) = 1 +D + D^3 +D^4\hspace{0.05cm}.$$

*The inverse transformation leads again to the result  $\underline{x} = (1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$   ⇒   Proposed solution 3.

'''(3)'''  Here we immediately use the path over the D–transforms:
:$${X}(D) = ( 1+D+D^2) \cdot (1+D) = 1 +D + D +D^2 +D^2 +D^3 = 1+ D^3\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \underline{x} =(1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} \text{...}\hspace{0.05cm})\hspace{0.05cm}.$$

*The result corresponds to the  proposed solution 2.  The following calculation is to illustrate the path in the time domain:
:$$(1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm}) *(1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} \text{...} \hspace{0.05cm})\hspace{-0.15cm} \ = \ \hspace{-0.15cm}(1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm},$$
:$$(0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} ...\hspace{0.05cm}) *(1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} \text{...} \hspace{0.05cm})\hspace{-0.15cm} \ = \ \hspace{-0.15cm}(0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm},$$
:$$(0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) *(1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} ... \hspace{0.05cm})\hspace{-0.15cm} \ = \ \hspace{-0.15cm}(0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm}.$$

*Since the convolution is a linear operation,  in the Galois field  ${\rm GF}(2)$  results from summation:
:$$(1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm}) *(1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} \text{...} \hspace{0.05cm})\hspace{0.08cm}=\hspace{0.08cm}(1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm}.$$

*If the convolution had been performed not in  ${\rm GF}(2)$  but for real numbers,  we would have obtained the result  $\underline{x} = (1, \, 2, \, 2, \, 1, \, 0, \, 0, \, \text{...})$.

'''(4)'''  The sample solution to the subtask  '''(3)'''  already suggests that the  proposed solution 1  is correct here.
*The way over the D–transforms confirms this result:
:$$\underline{u} = (1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm}) \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}{U}(D)= 1+ D + D^2+ D^3 + \text{...}\hspace{0.15cm}.$$

*Using the equation valid for calculations in  ${\rm GF}(2)$
:$$1 + D + D^2 + D^3 + \hspace{0.05cm}\text{...} \hspace{0.1cm}= \frac{1}{1+D}$$

:one further obtains:
:$${X}(D) = U(D) \cdot G(D) = \frac{1}{1+D} \cdot (1+D) = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\underline{x} = (1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...} \hspace{0.05cm})\hspace{0.05cm}.$$

'''(5)'''  The path over the D–transforms leads to  solution 2.
*For this alternating sequence  $\underline{u}$,  starting with  $1$,  one obtains:
:$${X}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 \cdot (1+D) + D^2 \cdot (1+D) + D^4 \cdot (1+D) + \text{...} = 1 + D + D^2 + D^3 + D^4 + D^5 +\hspace{0.05cm} ...\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\underline{x} = (1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} \text{...} \hspace{0.05cm})\hspace{0.05cm}.$$

*This result can also be read by direct application of the convolution as in subtask  '''(2)'''.

*With  $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 1, \, \text{...})$  you get  $\underline{x} = (0, \, 1, \, 1, \, 1, \, 1, \, 1, \,\text{...})$.

*This differs from the  "sequence of ones"  only in the first bit.  It is then  $x_1 = 0$  instead of  $x_1 = 1$.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^3.2 Polynomial Description^]]
[[de:Aufgaben:Aufgabe 3.3Z: Faltung und D–Transformation]]

Aufgaben:Exercise 4.14: 8-PSK and 16-PSK

2026-03-16T15:58:48Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}

[[File:P_ID2067__Dig_A_4_14.png|right|frame|Signal space constellations of the 8-PSK and 16-PSK]]
Now a signal set  $\{s_i(t)\}$  is considered,  which is limited to the time domain  $0 ≤ t ≤ T$.  The index  $i$  runs through the values  $0, \ \text{...} \ , M-1$:
:$$s_i(t) = A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) \hspace{0.05cm}.$$

*This is a  "phase modulation"  with  $M$  signal shapes. 

*This modulation process is also called  "$\rm M–PSK$".  $M$  is usually a power of two.

The graphic shows the signal space constellation for  $M = 8$  (top)  and  $M = 16$  (bottom).  All signal space points have equal energy  $||\boldsymbol{s}_i||^2 = E_{\rm S}$  ("average symbol energy").

The exact calculation of the symbol error probability is difficult for  $M ≠ 2$.  However,  the so-called  "Union Bound"  can always be given as an upper bound for the symbol error probability  $(p_{\rm UB} ≥ p_{\rm S})$:
:$$ p_{\rm UB} = 2 \cdot {\rm Q} \left ( \frac{ d/2}{ \sigma_n}\right ) = 2 \cdot {\rm Q} \left (\sqrt{ \frac{ d^2}{ 2 N_0}}\right ) \hspace{0.05cm}.$$

The following quantities are used here:
* $d$  is the distance between two neighboring points,  for example between  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$.

*If the decision boundary is exactly centered perpendicular to the line connecting  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$,  then  $d/2$  is the distance of  $\boldsymbol{s}_0$  or  $\boldsymbol{s}_1$  from this decision boundary.

* The variance of the AWGN noise is  $\sigma_n^2 = N_0/2$.

* The factor of  $2$  in the above limit takes into account that for  $M > 2$  each signal space point can be falsified in two directions,  e.g. for the  "8–PSK"  the symbol  $\boldsymbol{s}_0$  into the symbol  $\boldsymbol{s}_1$  or into the symbol  $\boldsymbol{s}_7$.

* ${\rm Q}(x)$  is the complementary Gaussian error function for which the following approximation holds:
:$${\rm Q}(x) \approx \frac{1}{\sqrt{2\pi} \cdot x} \cdot {\rm e}^{-x^2/2}\hspace{0.05cm}.$$

The last subtask deals with the bit error probability.  For this,  the following bound was given in the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "theory section"]]  under the assumption of a  "Gray code": 
:$$p_{\rm B} \le \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )\hspace{0.05cm}.$$

However,  this equation is only applicable for  $M > 4$.  In contrast,  the exact solution
* for  $M = 2$   from the identity with the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Binary_phase_shift_keying_.28BPSK.29| "BPSK"]], and

* for  $M = 4$   from the fact that the  "4–PSK"  is identical with the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"4–QAM"]]:
:$$p_{\rm B} ={\rm Q} \left ( \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right )\hspace{0.05cm}.$$

Notes:
* The exercise belongs to the topic of the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].

*Reference is made in particular to the section  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Multi-level_phase.E2.80.93shift_keying_.28M.E2.80.93PSK.29| "Multi-level Phase Shift Keying"]].

* The assignment of the  $8$  or  $16$  symbols to binary sequences of length  $3$  or  $4$ according to the Gray coding can be taken from the graphic (red labeling).

* When solving the exercise, you can use the following equations:
:$$\cos(\alpha + \beta) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \cos(\alpha ) \cdot \cos( \beta) - \sin(\alpha ) \cdot \sin( \beta)\hspace{0.05cm}, \hspace{0.25cm}1 - \cos(2\alpha ) = \sin^2(\alpha )\hspace{0.05cm},$$
:$$ \int_{0}^{T} \cos^2 ( 2\pi f_{\rm T}t) \,{\rm d} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5\hspace{0.05cm},\hspace{0.15cm}{\rm falls}\hspace{0.15cm} f_{\rm T} \gg 1/T \hspace{0.05cm}.$$

===Questions===
<quiz display=simple>
{What are the basis functions in the band-pass representation?  Let $\varphi_1(t)$  and  $\varphi_2(t)$  each be limited to the range  $0 ≤ t ≤ T$. 
|type="[]"}
- $\varphi_1(t) = \cos {(2\pi f_{\rm T}t)}$,
+ $\varphi_1(t) = \sqrt {2/T} \cdot \cos {(2\pi f_{\rm T}t)}$,
- $\varphi_2(t) = E_{\rm S} \cdot \sin {(2\pi f_{\rm T}t)}$,
+ $\varphi_2(t) = \, –\sqrt {2/T} \cdot \sin {(2\pi f_{\rm T}t)}$.

{What are the in-phase and quadrature components of the signal space point  $\boldsymbol{s}_i$?  Which statements are true?
|type="[]"}
+ $s_{\rm I \it i} = \sqrt {E_{\rm S}} \cdot \cos {(2\pi \cdot i/M)}$,
- $s_{\rm I \it i} = \sqrt {E_{\rm S}} \cdot \sin {(2\pi \cdot i/M)}$,
- $s_{\rm Q \it i} = \sqrt {E_{\rm S}} \cdot \cos {(2\pi \cdot i/M)}$,
+ $s_{\rm Q \it i} = \sqrt {E_{\rm S}} \cdot \sin {(2\pi \cdot i/M)}$.

{What is the distance  $d$  between two adjacent signal space points?  What are the values for  $M = 8$  and  $M = 16$, respectively?
|type="{}"}
$M = 8 \text{:} \hspace{0.45cm} d \ = \ $ { 0.765 3% } $\ \cdot \sqrt {E_{\rm S}}$
$M = 16 \text{:} \hspace{0.2cm} d \ = \ $ { 0.39 3% } $\ \cdot \sqrt {E_{\rm S}}$

{What is the value of the  "Union Bound"  $(p_{\rm UB})$  with  $E_{\rm S}/N_0 = 50$.
|type="{}"}
$M = 8 \text{:} \hspace{0.45cm} p_{\rm UB}$ = { 0.014 3% } $\ \%$
$M = 16 \text{:} \hspace{0.2cm} p_{\rm UB}$ = { 6 3% } $\ \%$

{Does the statement   "$p_{\rm UB}$  approximates  $p_{\rm S}$  more closely the larger  $M$   is"  hold?
|type="()"}
+ YES.
- NO.

{Which statements are true regarding the bit error probability  $p_{\rm B}$?
|type="[]"}
+ $p_{\rm B}$  is smallest for  $M = 2$  and  $M = 4$. 
- $p_{\rm B}$  is smallest for  $M = 8$. 
- $p_{\rm B}$  is smallest for  $M = 16$. 
+ $p_{\rm B}$  is not the main reason for using higher level PSK.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Solutions 2 and 4  are correct:
*The signal set  $\{s_i(t)\}$  can be represented in the band-pass domain with the given trigonometric transformation as follows:
:$$s_i(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos \left ( 2\pi f_{\rm T}t + { 2\pi }/{ M} \cdot i \right ) = A \cdot \cos \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \cos \left ( 2\pi f_{\rm T}t \right )-A \cdot \sin \left ( { 2\pi }/{ M} \cdot i \right ) \cdot \sin \left ( 2\pi f_{\rm T}t \right )\hspace{0.05cm}.$$

*The respective first terms in this difference lead to the signal space point  $\boldsymbol{s}_i$,  the respective second terms to the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.

*Here it is to be noted that these must be energy normalized in each case:
:$$\int_{0}^{T} \varphi_1 ( t)^2 \,{\rm d} t = \int_{0}^{T} \left [K \cdot \cos( 2\pi f_{\rm T}t)\right ]^2 \,{\rm d} t = 1\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \frac{K^2 \cdot T}{2} = 2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K = \sqrt{2/T } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\varphi_1 ( t) = \sqrt{2/T } \cdot \cos( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$
*Using the same arithmetic,  we arrive at the second basis function:
:$$\varphi_2 ( t) = -\sqrt{2/T } \cdot \sin( 2\pi f_{\rm T}t)\hspace{0.05cm}.$$

'''(2)'''  Solutions 1 and 4  are correct:
*Using the basis functions just calculated,  the signal set  $s_i(t)$  can be represented as follows  $($again, limited to the range  $0 ≤ t ≤ T)$:
:$$s_i(t) = A \cdot \sqrt{T/2 } \cdot \cos ( 2\pi \cdot i /M ) \cdot \varphi_1 ( t)+A \cdot \sqrt{T/2 } \cdot \sin ( 2\pi \cdot i /M ) \cdot \varphi_2 ( t)\hspace{0.05cm}.$$

*With the energy  $E = 0.5 \cdot A^2 \cdot T$,  which is the same for all  $M$  points and which is also the average signal energy per symbol  $(E_{\rm S})$,  the above equation is
:$$s_i(t) = s_{{\rm I}i} \cdot \varphi_1 ( t)+s_{{\rm Q}i} \cdot \varphi_2 ( t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm} s_{{\rm I}i} = \sqrt{E_{\rm S} }\cdot \cos ( 2\pi \cdot i /M )\hspace{0.05cm},\hspace{0.2cm}s_{{\rm Q}i} = \sqrt{E_{\rm S} }\cdot \sin ( 2\pi \cdot i /M )\hspace{0.05cm}. $$

*The signal space points sketched in the front graph for  $M = 8$  resp.  $M = 16$,  can be represented in exactly this way.

'''(3)'''  Since the distance from one to the nearest point is the same for all  $i$,  we can calculate  $d$, for example,  from the signal space points  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$. 
*In doing  so,  consider the sketch below.  According to the  "Pythagorean theorem":
[[File:P_ID2070__Dig_A_4_14c.png|right|frame|Distance calculation at the  "8-PSK"]]

:$$d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} \sin^2(\frac{2\pi}{M}) + [1- \cos(\frac{2\pi}{M})]^2 $$
:$$\Rightarrow \hspace{0.3cm} d_{\rm norm}^2 \hspace{-0.01cm}=\hspace{-0.01cm} 2 \cdot [1- \cos( \frac{2\pi}{M} )] = 4 \cdot \sin^2(\pi/M )\hspace{0.05cm} $$
:$$\Rightarrow \hspace{0.3cm} d_{\rm norm}= 2 \cdot \sin({\pi}/{M }) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} d = 2\sqrt {E_{\rm S}} \cdot \sin{\pi}/{M })\hspace{0.05cm}.$$

*This results in the following numerical values:
:$$M = 8\hspace{-0.09cm}: \hspace{0.2cm}d \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \underline{0.765} \cdot \sqrt {E_{\rm S} } \hspace{0.05cm},$$
:$$M = 16\hspace{-0.09cm}: \hspace{0.2cm} d \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \underline{0.390} \cdot \sqrt {E_{\rm S} } \hspace{0.05cm}.$$

'''(4)'''  Using the result from subtask  '''(3)''',  we obtain with the equation given in front:
:$$p_{\rm S} \le p_{\rm UB} = 2 \cdot {\rm Q} \left [ \sin ({ \pi}/{ M}) \cdot \sqrt{ { 2E_{\rm S}}/{ N_0}} \right ] \hspace{0.05cm}.$$

*With  $E_{\rm S}/N_0 = 50$  ⇒  $(2E_{\rm S}/N_0)^{\rm 0.5} = 10$   it follows:
:$$M = 8\hspace{-0.09cm}: \hspace{0.2cm}p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2 \cdot {\rm Q} ( 10 \cdot 0.383 )= 2 \cdot {\rm Q} ( 3.83 ) \approx \underline{0.014 \%} \hspace{0.05cm},$$
:$$ M = 16\hspace{-0.09cm}: \hspace{0.2cm} p_{\rm UB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2 \cdot {\rm Q} ( 10 \cdot 0.195 )= 2 \cdot {\rm Q} ( 1.95 ) \approx \underline{6 \%} \hspace{0.05cm}.$$

*Thus,  the symbol error probability becomes larger and larger as  $M$  increases,  assuming constant  $E_{\rm S}/N_0$,  as in this case.

*The most favorable value would result for  $M = 2$  (the factor  $2$  of the Union Bound is then not necessary)  as follows
:$$M = 2\hspace{-0.09cm}: \hspace{0.2cm}p_{\rm UB} = p_{\rm S} = {\rm Q} ( 10 ) \approx { 10^{-23}}.$$

*Thus,  multi-level PSK would not make sense if there were not other reasons for its use,  which will be discussed in subtask  '''(6)'''.

'''(5)'''  Correct is YES.  The graph shows the constellation for  "8–PSK"  and  "16–PSK".
[[File:P_ID2071__Dig_A_4_14e.png|right|frame|For the interpretation of the Union Bound at the  "$\rm M–PSK$"]]

*Each valid under the assumption that  $\boldsymbol{s}_0$  was transmitted.

*The contour lines of the AWGN PDF are then circles around  $\boldsymbol{s}_0$.

The following statements are valid:
# The actual error probability  $p_{\rm S}$  is composed of the components  $\rm A$,  $\rm B$  and  $\rm C$.
# On the other hand,  the  "Union Bound"  results from  $\rm A+B$  $($falsification into the symbol  $\boldsymbol{s}_1)$  plus  $\rm C+B$  $($falsification into the symbol $\boldsymbol{s}_{M-1})$.
# Thus,  $p_{\rm S} = p_{\rm UB}-B$  always holds,  but the larger  $M$  is, the smaller the fraction  $\rm B$  is.

'''(6)'''  For verification we assume  $p_{\rm B} = 10^{\rm –4}$.

*From this follows for  $M = 2$  and  $M = 4$:
:$$p_{\rm B} ={\rm Q} \left ( \sqrt{ { {2E_{\rm B}}}/{ N_0} }\right ) = 10^{-4}$$
:$$\Rightarrow \hspace{0.3cm}\sqrt{ { {2E_{\rm B}}}/{ N_0} } \approx 3.72\hspace{0.05cm}.$$

*On the other hand,  with the given equation for  $M = 8$:
:$$p_{\rm B} \hspace{-0.1cm} \ \le \ \hspace{-0.1cm} \frac{2}{{\rm log_2} \hspace{0.05cm}(M)} \cdot {\rm Q} \left ( \sqrt{{\rm log_2} \hspace{0.05cm}(M)} \cdot \sin ({ \pi}/{ M}) \cdot \sqrt{ { {2E_{\rm S}}}/{ N_0} }\right )= {2}/{3} \cdot {\rm Q} \left ( \sqrt{3} \cdot 0.383 \cdot 3.72 \right ) \approx {2}/{3} \cdot {\rm Q} \left ( 2.46 \right ) \approx 0.46 \%\hspace{0.05cm}.$$

*Accordingly,  for  $M = 16$ ⇒ ${\rm log}_2 (M) = 4$  is obtained:
:$$p_{\rm B} = {2}/{4} \cdot {\rm Q} \left ( \sqrt{4} \cdot 0.195 \cdot 3.72 \right ) \approx {1}/{2} \cdot {\rm Q} \left ( 1.45 \right ) \approx 4.8 \%\hspace{0.05cm}.$$

Thus,  The statements 1 and 4  are correct:
*Thus,  the main advantage of a higher-level PSK is not the lower bit error rate,  but the lower demand on the very expensive resource  "bandwidth".

*It should also be noted that the results are completely different when a  (highly)  distorting channel is present,  as is common in  "wireline transmission technology".
{{ML-Fuß}}

[[Category:Digital Signal Transmission: Exercises|^4.4 Coherent Demodulation^]]
[[de:Aufgaben:Aufgabe 4.14: 8-PSK und 16-PSK]]

Aufgaben:Exercise 3.6: Transient Behavior

2026-03-16T15:58:48Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1784__LZI_A_3_6.png|right|frame|Cosine and sine waves,  each causal]]
In this exercise,  we consider a cosine signal  $c(t)$  with amplitude  $1$  and period  $T = 1 \ \rm µ s$,  which is defined for all times  $t$  $($in the range  $ \pm \infty)$ :
:$$c(t) = \cos(2\pi \cdot {t}/{T})\hspace{0.05cm} .$$
The causal  $($German:  "kausal"   ⇒   subscript  "K"$)$  cosine signal  $c_{\rm K}(t)$  (red curve)  starts only at the turn-on instant  $t = 0$:
:$$c_{\rm K}(t)= \left\{ \begin{array}{c} c(t) \\0 \end{array} \right.\begin{array}{c} {\rm{for}} \\ {\rm{for}}\end{array}\begin{array}{*{20}c}{ t \ge 0\hspace{0.05cm},} \\{ t < 0\hspace{0.05cm}.}\end{array}$$Only the Fourier spectrum can be specified for the bilaterally unbounded signal  $c(t)$::$$C(f) = {1}/{ 2} \cdot \delta (f - f_0) + {1}/{ 2} \cdot \delta (f + f_0)\quad {\rm with} \quad f_0 = {1}/{ T}= 1\,\,{\rm MHz.}$$On the contrary,  for the causal cosine signal  $c_{\rm K}(t)$  the Laplace transform can also be specified::$$C_{\rm L}(p) =\frac {p} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$Accordingly,  the following holds for the Laplace transform of the causal sine function  $s_{\rm K}(t)$::$$S_{\rm L}(p) =\frac {2 \pi/T} { (p-{\rm j} \cdot 2 \pi/T)\cdot (p+{\rm j} \cdot 2 \pi/T)}\hspace{0.05cm} .$$The bilaterally unbounded sine function is denoted by  $s(t)$  and is shown as a blue–dotted curve in the below diagram.The signals  $c(t)$,  $c_{\rm K}(t)$,  $s(t)$  and  $s_{\rm K}(t)$  are applied to the input of a low-pass filter of first-order with the following transfer function  (or impulse response)::$$H_{\rm L}(p) =\frac {2 /T} { p + 2 /T} \quad\bullet\!\!-\!\!\!-^{\hspace{-0.25cm}\rm L}\!\!\!-\!\!\circ\quadh(t) = {2}/{T} \cdot {\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2\hspace{0.03cm}t/T}.$$
*The corresponding output signals are denoted by  $y_{\rm C}(t)$,  $y_{\rm CK}(t)$,  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$ .
*These signals are to be computed and correlated to each other in this exercise.

Please note:
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
*The computations for subtask  '''(6)'''  are bulky.
*For computing the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$,  for example the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]  can be used.

===Questions===

<quiz display=simple>
{Compute the frequency response  $H(f)$  by magnitude and phase using  $H_{\rm L}(p)$.   What values are obtained for frequency  $ f = f_0 = 1/T = 1 \ \rm MHz$?
|type="{}"}
$|H(f = f_0)| \ = \ $ { 0.303 3% }
$a(f = f_0)\hspace{0.2cm} = \ $ { 1.194 3% } $\ \rm Np$
${\rm arc} \ H(f = f_0)\ = \ $ { -74--70 } $\ \rm deg$
$b(f = f_0)\hspace{0.24cm} = \ $ { 72 3% } $\ \rm deg$

{Compute the signal  $y_{\rm C}(t)$  at the filter output if the cosine signal  $c(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm C}(t = 0) \ = \ $ { 0.092 3% }

{Compute the output signal  $y_{\rm S}(t)$  if the sine signal  $s(t)$  is applied to the filter input.  What value is obtained for  $t = 0$?
|type="{}"}
$y_{\rm S}(t = 0) \ = \ $ { -0.295--0.283 }

{Determine the influence length  $T_h$  of the filter impulse response,  that is the time at which  $h(t)$  has decayed to  $1\%$  of the maximum value.  Normalization to  $T$.
|type="{}"}
$T_h/T \ = \ $ { 2.3 3% }

{Which statements are true for the signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$ ?
|type="[]"}
+ The following holds:  $y_{\rm CK}(t) \equiv 0$  and  $y_{\rm SK}(t) \equiv 0$  for  $t < 0$.
+ The signal  $y_{\rm CK}(t)$  is approximately equal to  $y_{\rm C}(t)$ for  $t > T_h$ .
- The causal signal  $y_{\rm SK}(t)$  is approximately equal to $y_{\rm S}(t)$ for  $t < T_h$ .

{Using the residue theorem compute the signal  $y_{\rm CK}(t)$  after the filter if  $c_{\rm K}(t)$  is applied to the input.  What signal value occurs at time  $t = T/5$ ?
|type="{}"}
$y_\text{CK}(t = T/5) \ = \ $ { 0.24 3% }

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Replacing the parameter  $T$  by  $1/f_0$  and   $p$   by  ${\rm j} \cdot 2 \pi f$ in  $H_{\rm L}(p)$  the following is obtained for the frequency response in general or for  $f_0 = 1 \ \rm MHz$:
:$$H(f) = \frac {2 /T} { {\rm j} \cdot 2 \pi f + 2 /T}=\frac {f_0} { {\rm j} \cdot \pi f + f_0}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H(f= f_0) = \frac {1} { 1 + {\rm j} \cdot \pi }$$
:$$\Rightarrow \hspace{0.3cm} |H(f= f_0)| = \frac {1} { \sqrt{1 + \pi^2}} \hspace{0.15cm}\underline{= 0.303}\hspace{0.05cm},\hspace{0.2cm}a(f= f_0)= - {\rm ln}\,\, |H(f= f_0)| \hspace{0.15cm}\underline{\approx 1.194\,\,{\rm Np}}$$
:$$\Rightarrow \hspace{0.3cm} {\rm arc}\,H(f= f_0)= - {\rm arctan}\,(\pi) \hspace{0.15cm}\underline{\approx -72^\circ} \hspace{0.05cm}, \hspace{0.2cm}b(f= f_0)=-{\rm arc}\,H(f= f_0) \hspace{0.15cm}\underline{\approx +72^\circ} \hspace{0.05cm}.$$

'''(2)'''  The output signal  $y_{\rm C}(t)$  is attenuated by the factor  $0.303$  and delayed by  $\tau \approx 72/360 \cdot T = T/5$  compared to the input signal  $c(t)$ .

*Thus, this signal can also be described as follows:
:$$y_{\rm C}(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2}}= 0.303 \cdot \cos(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm C}(t=0) = \frac {1} { {1 + \pi^2}} \hspace{0.15cm}\underline{\approx 0.092} \hspace{0.05cm}.$$
*This signal is shown dotted in blue in the left graph for solution  '''(5)'''.

'''(3)'''  The signal  $y_{\rm S}(t)$  is also smaller by the attenuation factor  $0.303$  and delayed by the time period  $\tau = T/5$  compared to  $s(t)$ .

*It can be described as follows:
:$$y_{\rm S}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})} { {1 + \pi^2}}= 0.303 \cdot \sin(2\pi \cdot \frac{t-T/5}{T}) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} y_{\rm S}(t=0) = -\frac {\pi} { {1 + \pi^2}} \hspace{0.15cm}\underline{\approx -0.289} \hspace{0.05cm}.$$
*This signal is sketched dotted in blue in the right graph for solution  '''(5)''' .

'''(4)'''  At  $T_h$  the impulse response  $h(t)$  should have decayed to  $1\%$  of the maximum value.  Thus,  the following holds:
:$${\rm e}^{\hspace{0.05cm}-\hspace{0.03cm}2\hspace{0.03cm}T_{ h}/T} = 0.01 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {T_{ h}}/{T} =\frac{1}{2}\cdot {\rm ln}\,\, \frac{1}{0.01} \hspace{0.15cm}\underline{\approx 2.3}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}h(t=0) ={2}/{T}\hspace{0.05cm}, \hspace{0.2cm}h(t=T_{ h}) ={0.02}/{T}\hspace{0.05cm}.$$

[[File:P_ID1785__LZI_A_3_6_e.png|right|frame|Transient behavior of a causal cosine and a causal sine signal]]
 '''(5)'''  The statements 1 and 2 are correct:
*The causal signals  $y_{\rm CK}(t)$  and  $y_{\rm SK}(t)$  must be identical to zero for  $t < 0$ .
*However,  since the impulse response  $h(t)$  of the considered filter (nearly) vanishes for  $t > T_h$,  it does not matter whether the unbounded cosine signal  $c(t)$  or the causal signal  $c_{\rm K}(t)$  is applied to the input after the transient effect is over.
*The same is true for the sinusoidal signals:   For  $t >T_{ h}$,    $y_{\rm SK}(t)=y_{\rm S}(t)$ holds.

The graph shows the output signals  $y_{\rm C}(t)$  and  $y_{\rm CK}(t)$  for cosine-shaped input on the left and the signals  $y_{\rm S}(t)$  and  $y_{\rm SK}(t)$  for sinusoidal input on the right.  '''Note the transit time'''  of  $T/5$  $($corresponding to the phase  $72^\circ)$  '''in both cases'''.
*For  $y_{\rm CK}(t)$  the first wave peaks are smaller than  $1$  to achieve the correct phase position of  $y_{\rm C}(t)$ .
*In contrast,  for  $y_{\rm SK}(t)$  the first wave peaks are greater than  $1$  to achieve the correct phase position of  $y_{\rm S}(t)$ .

'''(6)'''  Considering 
$p_{\rm x1}= {\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}p_{\rm x2}= -{\rm j} \cdot {2\pi}/{T} , \hspace{0.1cm}
p_{\rm x3}= -{2}/{T}
\hspace{0.05cm}$ 
the following can be written for the Laplace transform of the signal  $y_{\rm CK}(t)$ :
:$$Y_{\rm L}(p) =\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})(p-p_{{\rm x}3})}\hspace{0.05cm}.$$
The time function  $y_{\rm CK}(t)$  is thus composed of three parts according to the residue theorem:

* Considering  $p_{\rm x2}= -p_{\rm x1}$  the first part is
:$$y_1(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}2})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm}t}\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=\frac {-p_{{\rm x}3}\cdotp_{{\rm x}1}} { (p_{{\rm x}1}-p_{{\rm x}2})(p_{{\rm x}1}-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot\hspace{0.03cm}t}= \frac {-p_{{\rm x}3}/2} { p_{{\rm x}1}-p_{{\rm x}3}}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}1}\cdot\hspace{0.03cm}t}\hspace{0.05cm} .$$* Similarly,  the following is obtained for the second part::$$y_2(t)= {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}\hspace{0.05cm}\{Y_{\rm L}(p)\cdot {\rm e}^{\hspace{0.03cm}p t}\}=\frac {-p_{{\rm x}3}\cdot p} { (p-p_{{\rm x}1})(p-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm}t}\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}=\frac {-p_{{\rm x}3}\cdotp_{{\rm x}2}} { (p_{{\rm x}2}-p_{{\rm x}1})(p_{{\rm x}2}-p_{{\rm x}3})}\cdot {\rm e}^{\hspace{0.03cm}p_{{\rm x}2}\cdot\hspace{0.03cm}t}= \frac {p_{{\rm x}3}/2} { p_{{\rm x}1}+p_{{\rm x}3}}\cdot {\rm e}^{-p_{{\rm x}1}\cdot\hspace{0.03cm}t}\hspace{0.05cm} .$$* Combining both parts and considering the numerical values of  $p_{\rm x1}$  and  $p_{\rm x3}$::$$y_{1\hspace{0.03cm}+2}(t)= \frac {1/T} { 2/T + {\rm j} \cdot 2\pi /T}\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi \hspace{0.03cm} t/T}+\frac {1/T} { 2/T - {\rm j} \cdot 2\pi /T}\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi\hspace{0.03cm}t/T}= \frac {1/2} { 1 + {\rm j} \cdot \pi }\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi\hspace{0.03cm}t/T}+\frac {1/2} { 1 - {\rm j} \cdot \pi }\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi\hspace{0.03cm}t/T}$$:$$\Rightarrow \hspace{0.3cm} y_{1\hspace{0.03cm}+2}(t)\hspace{0.25cm} = \frac {1/2 \cdot (1 - {\rm j} \cdot \pi)} { 1 + \pi^2 }\cdot {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi\hspace{0.03cm}t/T}+\frac {1/2 \cdot (1 + {\rm j} \cdot \pi)} { 1 + \pi^2 }\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi\hspace{0.03cm}t/T}\hspace{0.05cm} $$* Using Euler's theorem this can also be expressed as follows::$$y_{1\hspace{-0.03cm}+2} (t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})} { {1 + \pi^2}}= y_{\rm C}(t)\hspace{0.05cm}.$$:It can be seen that  $y_{1\hspace{0.03cm}+2}(t)$  is equal to the signal  $y_{\rm C}(t)$  computed in subtask  '''(2)''' .* Finally,  the following is obtained for the last residual::$$y_3(t)=\frac {-p_{{\rm x}3}^2} { (p_{{\rm x}3}-p_{{\rm x}1}) (p_{{\rm x}3}-p_{{\rm x}2})} \cdot {\rm e}^{\hspace{0.05cm}p_{{\rm x}3}\cdot\hspace{0.03cm}t}= \frac {-(2/T)^2 \cdot {\rm e}^{\hspace{0.05cm}-2\hspace{0.03cm}t/T}} { (-2/T-{\rm j} \cdot 2\pi /T) (-2/T+{\rm j} \cdot 2\pi /T)} =\frac {- {\rm e}^{\hspace{0.05cm}-2\hspace{0.03cm}t/T} } { (1+{\rm j} \cdot \pi ) (1-{\rm j} \cdot\pi)} =\frac {- {\rm e}^{\hspace{0.05cm}-2\hspace{0.03cm}t/T}} { 1+\pi^2}\hspace{0.05cm} .$$* Thus,  the output signal is as follows for a causal cosine signal applied to the input::$$y_{\rm CK}(t) = y_1(t)+y_2(t)+y_3(t) = \frac { \cos(2\pi {t}/{T}) + \pi \cdot \sin(2\pi {t}/{T})-{\rm e}^{\hspace{0.05cm}-2\hspace{0.03cm}t/T}} { {1 + \pi^2}}$$:$$\Rightarrow \hspace{0.3cm}y_{\rm CK}(t = {T}/{5}) = \frac { \cos(72^\circ) + \pi \cdot \sin(72^\circ)-{\rm e}^{\hspace{0.05cm}-0.4}} { {1 + \pi^2}} \hspace{0.15cm}\underline{ \approx 0.24} < 0.303\hspace{0.05cm} .$$:For comparison:   The signal  $y_{\rm C}(t)$  has the value  $0.303$   at this time.* In contrast to this,  the following arises as a result in general and specifically at the time of the first maximum at  $t = 0.45 \cdot T$  for the causal sinusoidal signal applied to the input::$$y_{\rm SK}(t) = \frac { -\pi \cdot \cos(2\pi {t}/{T}) + \sin(2\pi {t}/{T})+\pi \cdot {\rm e}^{\hspace{0.05cm}-2\hspace{0.03cm}t/T}} { {1 + \pi^2}}$$:$$\Rightarrow \hspace{0.3cm} y_{\rm SK}(t = 0.45 \cdot T) = \frac { -\pi \cdot \cos(162^\circ) + \sin(162^\circ)+\pi \cdot{\rm e}^{\hspace{0.05cm}-0.9}} { {1 + \pi^2}} \approx 0.42 > 0.303\hspace{0.05cm} .$$

{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]
[[de:Aufgaben:Aufgabe 3.6: Einschwingverhalten]]

Aufgaben:Exercise 4.1Z: Transmission Behavior of Short Cables

2026-03-16T15:58:47Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Some_Results_from_Line_Transmission_Theory
}}

[[File:EN_LZI_Z_4_1.png|right|frame|Short line section]]
We assume a homogeneous and reflection-free terminated line of length  $l$  so that the following applies to the spectral function at the output:
:$$U_2(f) = U_1(f) \cdot {\rm e}^{-\hspace{0.02cm}\gamma(f) \hspace{0.05cm} \cdot \hspace{0.05cm}l} \hspace{0.05cm}.$$
Here  $\gamma(f)$  describes the  '''complex propagation function'''  of an extremely short line of infinitesimal length  $dx$,  which can be represented with the parameters  $R\hspace{0.05cm}'$,  $L\hspace{0.05cm}'$,  $G\hspace{0.08cm}'$ and  $C\hspace{0.08cm}'$ (see diagram) as follows:
:$$\gamma(f) = \sqrt{(R\hspace{0.05cm}' + {\rm j} \cdot 2\pi f \cdot L\hspace{0.05cm}') \cdot (G\hspace{0.08cm}' + {\rm j} \cdot 2\pi f \cdot C\hspace{0.08cm}')} =\alpha (f) + {\rm j} \cdot \beta (f)\hspace{0.05cm}.$$
The real part of  $\gamma(f)$  results in
*The real part of  $\gamma(f)$  results in the attenuation function $\alpha(f)$  (per unit length).
*The imaginary part of  $\gamma(f)$  results in the phase function  $\beta(f)$ (per unit length).

After some calculation one can write for these sizes:
:$$\alpha(f) = \sqrt{{1}/{2}\cdot \left (R\hspace{0.05cm}' \cdot G\hspace{0.08cm}' - \omega^2 \cdot L\hspace{0.05cm}' \cdot C\hspace{0.08cm}'\right)+{1}/{2}\cdot \sqrt{(R\hspace{0.05cm}'\hspace{0.05cm}^2 + \omega^2 \cdot L\hspace{0.05cm}'\hspace{0.05cm}^2) \cdot (G\hspace{0.08cm}'\hspace{0.05cm}^2 + \omega^2 \cdot C\hspace{0.08cm}'\hspace{0.05cm}^2)}}\hspace{0.1cm}\bigg |_{\hspace{0.05cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pif},$$
:$$\beta(f) = \sqrt{{1}/{2}\cdot \left (-R\hspace{0.05cm}' \cdot G\hspace{0.08cm}' + \omega^2 \cdot L\hspace{0.05cm}' C\hspace{0.08cm}'\right)+{1}/{2}\cdot \sqrt{(R\hspace{0.05cm}'\hspace{0.05cm}^2 + \omega^2 \cdot L\hspace{0.05cm}'\hspace{0.05cm}^2) \cdot (G\hspace{0.08cm}'\hspace{0.05cm}^2 + \omega^2 \cdot C\hspace{0.08cm}'\hspace{0.05cm}^2)}}\hspace{0.1cm}\bigg |_{\hspace{0.05cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f}.$$

*For the attenuation function  $a(f)$  the pseudo unit  "Neper"  (Np)  has to be added additionally and for the phase function  $b(f)$  "Radian"  (rad).  
*Since the primary line parameters are each related to the line length,  $\alpha(f)$  and  $\beta(f)$  have the units  "Np/km"  and  "rad/km",  respectively.

Another important descriptive quantity besides  $\gamma(f)$  is the  '''wave impedance'''  $Z_{\rm W}(f)$,  which gives the relationship between voltage and current of the two running waves at each location.  It holds:
:$$Z_{\rm W}(f) = \sqrt{\frac {R\hspace{0.05cm}' + {\rm j} \cdot \omega L\hspace{0.05cm}'}{G\hspace{0.08cm}' + {\rm j} \cdot \omega C\hspace{0.08cm}'}}\hspace{0.1cm}\bigg |_{\hspace{0.05cm} \omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f}.$$

Notes:
*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Results_from_Line_Transmission_Theory|Some Results from Line Transmission Theory]].

*Use the following values for the numerical calculations:
:$$R\hspace{0.05cm}' = 100\,\,{\rm \Omega}/{ {\rm km} }\hspace{0.05cm},\hspace{0.3cm}G\hspace{0.08cm}' = 1\,\,{\rm µ S}/{ {\rm km}}\hspace{0.05cm},\hspace{0.3cm}2\pi L\hspace{0.03cm}' = 2\,\,{\rm mH}/{ {\rm km}} \hspace{0.05cm},\hspace{0.3cm}2\pi C\hspace{0.08cm}' = 200\,\,{\rm nF}/{ {\rm km}}\hspace{0.05cm}.$$

===Questions===

<quiz display=simple>
{Specify  $\alpha(f)$,  $\beta(f)$ and  $Z_{\rm W}(f)$  for frequency  $f = 0$  ("direct current").
|type="{}"}
$\alpha(f =0) \ =$ { 0.01 3% } $\ \rm Np/km$
$\beta(f = 0) \ =$ { 0. } $\ \rm rad/km$
$Z_{\rm W}(f = 0) \ =$ { 10000 3% } $\ \rm \Omega$

{Calculate the attenuation function  $\alpha(f)$  (per unit length)  for  $f = 100\ \rm kHz$.
|type="{}"}
$\alpha(f = 100\ \rm kHz) \ = \ $ { 0.486 3% } $\ \rm Np/km$

{Give the approximations of  $Z_{\rm W}(f)$  and  $\alpha(f)$,  valid for  $f → \infty$ .
|type="{}"}
$ Z_{\rm W}(f → \infty) \ = \ $ { 100 3% } $\ \rm \Omega$
$\alpha(f → \infty) \ = \ $ { 0.5 3% } $\ \rm Np/km$

{Use  $\omega L\hspace{0.03cm}' \ll R\hspace{0.05cm}'$  and  $\omega C\hspace{0.08cm}' \gg G\hspace{0.08cm}'$  to derive an  $\alpha(f)$  approximation for  (not too)  small frequencies. What is the attenuation function per unit length for  $ f = 1 \ \rm kHz$  and  $ f = 4 \ \rm kHz$?
|type="{}"}
$\alpha(f = 1\ \rm kHz) \ = \ $ { 0.1 3% } $\ \rm Np/km$
$\alpha(f = 4\ \rm kHz) \ = \ $ { 0.2 3% } $\ \rm Np/km$

{For the same frequency range,  give a suitable approximation for the wave impedance  $Z_{\rm W}(f)$ . What value results for  $ f = 1 \ \rm kHz$?
|type="{}"}
${\rm Re}\{Z_{\rm W}(f = 1\ \rm kHz)\} \ = \ $ { 500 3% } $\ \rm \Omega$
${\rm Im}\{Z_{\rm W}(f = 1\ \rm kHz)\} \ = \ $ { -515--485 } $\ \rm \Omega$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  If you insert the frequency  $f = 0$  into the given equations, we obtain
:$$\alpha(f = 0) = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{{1}/{2}\hspace{-0.03cm}\cdot \hspace{-0.03cm} R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm} G\hspace{0.03cm}'+ {1}/{2}\cdot R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm}G\hspace{0.03cm}'} = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{ R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm} G\hspace{0.03cm}'} = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{ 100\,{\rm \Omega/km} \hspace{-0.03cm}\cdot \hspace{-0.03cm} 10^{-6}\,{\rm (\Omega \hspace{-0.03cm}\cdot \hspace{-0.03cm} km})^{-1}}\hspace{0.15cm}\underline{= 0.01\,{\rm Np}/{ {\rm km}}}\hspace{0.05cm},$$
:$$\beta(f = 0) = [1\,{\rm rad}] \cdot \sqrt{-{1}/{2}\cdot R\hspace{0.03cm}' \cdot G\hspace{0.03cm}'+ {1}/{2}\cdot R\hspace{0.03cm}' \cdotG\hspace{0.03cm}'} \hspace{0.15cm}\underline{= 0 }\hspace{0.05cm},$$
:$$Z_{\rm W}(f = 0) = \sqrt{\frac {R\hspace{0.03cm}'}{G\hspace{0.03cm}'}} = \sqrt{\frac {100\,{\rm \Omega/km}}{{\rm 10^{-6}/(\Omega \cdot km})}}\hspace{0.15cm}\underline{= 10\, {\rm k \Omega}}\hspace{0.05cm}.$$

The DC signal attenuation becomes relevant,
*if the useful signal is to be transmitted in the baseband and has a DC component,  or
*if the network termination at the participant must be supplied with power from the local exchange  ("remote power supply").

'''(2)'''  With  $f = 10^{5} \ \rm Hz$  and the specified values,  the following holds:
:$$f \cdot 2\pi L' = 10^5\,\frac{1}{ {\rm s}} \cdot 2 \cdot10^{-3}\,\frac{\rm \Omega \cdot s}{ {\rm km}}= 200 \,\frac{\rm\Omega}{ {\rm km}} \hspace{0.05cm},\hspace{1.05cm}f \cdot 2\pi C' = 10^5\,\frac{1}{ {\rm s}} \cdot 2 \cdot10^{-7}\,\frac{\rm s}{ {\rm \Omega \cdot km}}= 0.02\,\frac{\rm 1 }{ {\rm \Omega \cdot km}} \hspace{0.05cm}.$$This results in the following for the attenuation function in "Np/km"::$$\alpha(f = 100\,{\rm kHz})= \sqrt{ {1}/{2}\cdot \left (100 \cdot 10^{-6} - 200 \cdot 0.02 \right)+{1}/{2} \cdot \sqrt{(100^2 + 200^2) \cdot (10^{-12} + 0.02^2)}} $$:$$ \Rightarrow \; \; \alpha(f = 100\,{\rm kHz}) \approx \sqrt{{1}/{2}\cdot \left (10^{-4} - 4 \right)+{1}/{2}\cdot \sqrt{5 \cdot 10^{4} \cdot 4 \cdot 10^{-4}}} \approx \sqrt {-2 + \frac{\sqrt{20}}{2}} \hspace{0.15cm}\underline{\approx 0.486 \ {\rm Np/km}} \hspace{0.05cm}.$$'''(3)'''  The limit for  $f → \infty$  results if one neglects the second terms in the numerator  $R\hspace{0.03cm}'$  and in the denominator  $G\hspace{0.08cm}'$ ::$$\lim_{f \rightarrow \infty} \hspace{0.1cm} Z_{\rm W}(f)= \lim_{\omega \rightarrow \infty} \hspace{0.1cm} \sqrt{\frac {R\hspace{0.03cm}' + {\rm j} \cdot \omega L'}{G' + {\rm j} \cdot \omega C\hspace{0.03cm}'}}=\sqrt{\frac {2 \pi L\hspace{0.03cm}' }{2 \pi C\hspace{0.03cm}'}}=\sqrt{\frac {2 \cdot 10^{-3}\,{\rm \Omega \cdot s} }{2 \cdot 10^{-73}\,{\rm s/\Omega} }} \hspace{0.15cm}\underline{= 100\,{\rm \Omega }}\hspace{0.05cm}.$$*The approximation for the attenuation function is more difficult to derive.  Starting from:$$\alpha(\omega) = \sqrt{ {1}/{2}\cdot \left (R\hspace{0.03cm}' G\hspace{0.03cm}' - \omega^2 \cdot L\hspace{0.03cm}' C\hspace{0.03cm}'\right)+{1}/{2}\sqrt{(R\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2) \cdot (G\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot C\hspace{0.03cm}\hspace{0.03cm}'^2)}}$$:then also the following applies::$$2 \cdot \alpha^2(\omega) = R\hspace{0.03cm}' G\hspace{0.03cm}' + \omega^2 \cdot L'C'\cdot\left [-1 +\sqrt{(1 + \frac{R\hspace{0.03cm}'^2}{ \omega^2 \cdot L\hspace{0.03cm}'^2}) \cdot (1 + \frac{G\hspace{0.03cm}'^2}{ \omega^2 \cdot C\hspace{0.03cm}'^2})} \hspace{0.1cm}\right]$$:$$\Rightarrow \; \; 2 \cdot \alpha^2(\omega) \approx R\hspace{0.03cm}' G\hspace{0.03cm}' + \omega^2 \cdot L'C\hspace{0.03cm}'\cdot\left [-1 +\sqrt{1 + \frac{R'^2}{ \omega^2 \cdot L'^2}+ \frac{G\hspace{0.03cm}'^2}{ \omega^2 \cdot C\hspace{0.03cm}'^2}} \hspace{0.1cm}\right].$$*Using the approximation  $\sqrt{1 + x}\approx 1+x/2$  valid for small  $x$,  one arrives at the intermediate result for (infinitely) large frequencies::$$2 \cdot \alpha^2(\omega \rightarrow \infty) = R\hspace{0.03cm}' G\hspace{0.05cm}' + \omega^2 \cdot L'C\hspace{0.05cm}'\cdot\left [ -1 +1 + {1}/{2} \cdot \left ( \frac{R\hspace{0.03cm}'\hspace{0.03cm}^2}{ \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2}+ \frac{G\hspace{0.03cm}'\hspace{0.03cm}^2}{ \omega^2 \cdot C\hspace{0.03cm}'\hspace{0.03cm}^2}\right) \hspace{0.1cm}\right] $$:$$\Rightarrow \hspace{0.3cm} 2 \cdot \alpha^2(\omega \rightarrow \infty) = \frac{2 \cdot R\hspace{0.03cm}' G\hspace{0.03cm}' C\hspace{0.03cm}' L'+ R\hspace{0.03cm}'\hspace{0.03cm}^2 C\hspace{0.03cm}'\hspace{0.03cm}^2+G\hspace{0.03cm}'\hspace{0.03cm}^2 L\hspace{0.03cm}'\hspace{0.03cm}^2}{2 \cdot C\hspace{0.03cm}' L\hspace{0.03cm}'}= \frac{(R\hspace{0.03cm}' C\hspace{0.03cm}' + G\hspace{0.03cm}' L\hspace{0.03cm}')^2}{2 \cdot C\hspace{0.03cm}' L\hspace{0.03cm}' }$$:$$\Rightarrow \hspace{0.3cm} \alpha(\omega \rightarrow \infty) ={1}/{2}\cdot \frac{R\hspace{0.03cm}' C\hspace{0.03cm}' + G\hspace{0.03cm}' L\hspace{0.03cm}'}{\sqrt{ C\hspace{0.03cm}' L\hspace{0.03cm}' }}={1}/{2}\cdot \left [R\hspace{0.03cm}' \cdot \sqrt{\frac{C\hspace{0.03cm}'}{L\hspace{0.03cm}'}}+G\hspace{0.03cm}' \cdot \sqrt{\frac{L\hspace{0.03cm}'}{C\hspace{0.03cm}'}}\right]\hspace{0.05cm}.$$*With the numerical values inserted,  we get:$$\alpha(f \rightarrow \infty) = \alpha(\omega \rightarrow \infty)= {0.5\,{\rm Np/km}}\cdot \left [100 \cdot \sqrt{\frac{2 \cdot 10^{-7}}{2 \cdot10^{-3}}}+10^{-6} \cdot\sqrt{\frac{2 \cdot10^{-3}}{2 \cdot10^{-7}}}\right]\hspace{0.15cm}\underline{\approx 0.5 \, {\rm Np}/{\rm km}}\hspace{0.05cm}.$$'''(4)'''  For small frequencies,  $\omega L\hspace{0.03cm}' \ll R\hspace{0.03cm}'$  and  $ \omega C\hspace{0.03cm}' \gg G\hspace{0.03cm}'$ apply.*Neglecting the  $\omega^2$–part, one obtains::$$\alpha(f) = \sqrt{\frac {1}{2}\cdot \left (R\hspace{0.03cm}' G\hspace{0.03cm}' - \omega^2 \cdot L\hspace{0.03cm}' C\hspace{0.03cm}'\right)+\frac {1}{2}\sqrt{(R\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2) \cdot (G\hspace{0.03cm}'\hspace{0.03cm}^2+ \omega^2 \cdot C\hspace{0.03cm}'\hspace{0.03cm}^2)}}\hspace{0.1cm}\bigg |_{\hspace{0.05cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pif}$$:$$ \Rightarrow \hspace{0.3cm} \alpha(f) \approx \sqrt{\frac {R\hspace{0.03cm}' G\hspace{0.03cm}'}{2}+\frac {R\hspace{0.03cm}' \cdot \omega C\hspace{0.03cm}'}{2}}\hspace{0.1cm}\bigg |_{\hspace{0.03cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pif} \approx \sqrt{{1}/{2} \cdot f \cdot R\hspace{0.03cm}' \cdot 2 \pi C\hspace{0.03cm}'}\hspace{0.05cm}.$$*Here it is considered that the first part can be neglected according to subtask  '''(1)'''  except for the frequency  $f = 0$ .*For the frequency  $f = 1 \ \rm kHz$  we get the approximation:$$\alpha(f = 1\,{\rm kHz}) = \sqrt{{1}/{2} \cdot 10^{3}\,{\rm Hz} \cdot 100\,\frac{\rm \Omega }{ {\rm km}} \cdot 2 \cdot 10^{-7}\,\frac{\rm s }{ {\rm \Omega \cdot km}}}\hspace{0.15cm}\underline{= 0.1\,{\rm Np }/{ {\rm km}}}\hspace{0.05cm}.$$*For frequency  $f = 4 \ \rm kHz$  the attenuation function per unit length is twice as large::$$\alpha(f = 4\,{\rm kHz}) \hspace{0.15cm}\underline{ = 0.2\,{\rm Np }/{ {\rm km}}}\hspace{0.05cm}.$$'''(5)'''  The wave impedance at low frequencies is approximated by::$$Z_{\rm W}(f) = \sqrt{\frac {R\hspace{0.03cm}' + {\rm j} \cdot f \cdot 2 \pi L\hspace{0.03cm}'}{G\hspace{0.03cm}' + {\rm j} \cdot f \cdot 2 \pi C\hspace{0.03cm}'}}\approx \sqrt\frac{1 }{ {\rm j}} \cdot \sqrt{\frac {R\hspace{0.03cm}' }{ f \cdot 2 \piC\hspace{0.03cm}'}}= (1 - {\rm j})\cdot \sqrt{\frac {R\hspace{0.03cm}' }{ 2 \cdot f \cdot 2 \piC\hspace{0.03cm}'}}\hspace{0.05cm}.$$*With the specified line fittings we obtain::$${\rm Re}\{Z_{\rm W}(f= 1\,{\rm kHz})\} = \sqrt{\frac {100\,{\rm \Omega/km }}{ 2 \cdot 10^{3}\,{\rm Hz} \cdot 2 \cdot 10^{-7}\,{\rm s/(\Omega \cdot km) }}} \hspace{0.15cm}\underline{= 500\,{\rm\Omega}}\hspace{0.05cm},$$:$$ {\rm Im}\{Z_{\rm W}(f= 1\,{\rm kHz})\} = -{\rm Re}\{Z_{\rm W}(f= 1\,{\rm kHz})\}\hspace{0.15cm}\underline{= -500\,{\rm\Omega}}\hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Linear and Time-Invariant Systems: Exercises|^4.1 Results of Line Transmission Theory^]]
[[de:Aufgaben:Aufgabe 4.1Z: Übertragungsmaß]]

Aufgaben:Exercise 4.5: On the Extrinsic L-values again

2026-03-16T15:58:47Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}

[[File:P_ID3026__KC_A_4_5_v2.png|right|frame|Table for first  $L_{\rm E}(i)$  approach]]
We assume as in the  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_log_likelihood_ratios|"theory section"]]  the  "single parity–check code"   $\rm SPC \, (3, \, 2, \, 2)$. 

The possible code words are  $\underline{x} \hspace{-0.01cm}\in \hspace{-0.01cm}
\{ \underline{x}_0,\hspace{0.05cm}
\underline{x}_1,\hspace{0.05cm}
\underline{x}_2,\hspace{0.05cm}
\underline{x}_3\}$  with
:$$\underline{x}_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (0\hspace{-0.03cm},\hspace{0.05cm}0\hspace{-0.03cm},\hspace{0.05cm}0)\hspace{0.35cm}{\rm resp. } \hspace{0.35cm}\underline{x}_0 \hspace{-0.05cm}=\hspace{-0.05cm} (+1\hspace{-0.03cm},\hspace{-0.05cm}+1\hspace{-0.03cm},\hspace{-0.05cm}+1)\hspace{0.05cm},$$
:$$\underline{x}_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (0\hspace{-0.03cm},\hspace{0.05cm}1\hspace{-0.03cm},\hspace{0.05cm}1)\hspace{0.35cm}{\rm resp. } \hspace{0.35cm}\underline{x}_1 \hspace{-0.05cm}=\hspace{-0.05cm} (+1\hspace{-0.03cm},\hspace{-0.05cm}-1\hspace{-0.03cm},\hspace{-0.05cm}-1)\hspace{0.05cm},$$
:$$\underline{x}_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (1\hspace{-0.03cm},\hspace{0.05cm}0\hspace{-0.03cm},\hspace{0.05cm}1)\hspace{0.35cm}{\rm resp. } \hspace{0.35cm}\underline{x}_2 \hspace{-0.05cm}=\hspace{-0.05cm} (-1\hspace{-0.03cm},\hspace{-0.05cm}+1\hspace{-0.03cm},\hspace{-0.05cm}-1)\hspace{0.05cm},$$
:$$\underline{x}_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (1\hspace{-0.03cm},\hspace{0.05cm}1\hspace{-0.03cm},\hspace{0.05cm}0)\hspace{0.35cm}{\rm resp. } \hspace{0.35cm}\underline{x}_3 \hspace{-0.05cm}=\hspace{-0.05cm} (-1\hspace{-0.03cm},\hspace{-0.05cm}-1\hspace{-0.03cm},\hspace{-0.05cm}+1)\hspace{0.05cm}.$$

In the exercise we mostly use the second (bipolar) representation of the code symbols:  
:$$x_i ∈ \{+1, -1\}.$$

Note:
#It is not that the  $\rm SPC \, (3, \, 2, \, 2)$  would be of much practical interest,  since,  for example,  in  "hard decision"  because of  $d_{\rm min} = 2$  only one error can be detected and none can be corrected.
#However,  the code is well suited for demonstration purposes because of the manageable effort involved.
#With  "iterative symbol-wise decoding"  one can also correct one error.
#In the present code,  the extrinsic  $L$–values  $\underline{L}_{\rm E} = \big (L_{\rm E}(1), \ L_{\rm E}(2), \ L_{\rm E}(3)\big )$  must be calculated according to the following equation:
:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} \right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} \hspace{0.05cm}\right ]}.$$

:Here  $\underline{x}^{(-1)}$  denotes all symbols except  $x_i$  and is thus a vector of length  $n - 1 = 2$.

⇒   As the  »'''first $L_{\rm E}(i)$ approach'''«  we refer to the approach corresponding to the equations
:$$L_{\rm E}(1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm tanh}^{-1} \left [{\rm tanh}(L_2/2) \cdot {\rm tanh}(L_3/2) \right ] \hspace{0.05cm},$$
:$$L_{\rm E}(2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm tanh}^{-1} \left [{\rm tanh}(L_1/2) \cdot {\rm tanh}(L_3/2) \right ] \hspace{0.05cm},$$
:$$L_{\rm E}(3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm tanh}^{-1} \left [{\rm tanh}(L_1/2) \cdot {\rm tanh}(L_2/2) \right ] \hspace{0.05cm}.$$

'''(1)'''  This  $L_{\rm E}(i)$  approach underlies the results table above  $($red entries$)$,  assuming the following a-posteriori $L$–values:
:$$\underline {L}_{\rm APP} = (+1.0\hspace{0.05cm},\hspace{0.05cm}+0.4\hspace{0.05cm},\hspace{0.05cm}-1.0) \hspace{0.5cm}\Rightarrow \hspace{0.5cm}L_1 = +1.0\hspace{0.05cm},\hspace{0.15cm}L_2 = +0.4\hspace{0.05cm},\hspace{0.15cm}L_3 = -1.0\hspace{0.05cm}.$$

'''(2)'''  The extrinsic  $L$–values for the zeroth iteration result in  $($derivation in  [[Aufgaben:Exercise_4.5Z:_Tangent_Hyperbolic_and_Inverse|$\text{Exercise 4.5Z})$]]:
:$$L_{\rm E}(1) = -0.1829, \ L_{\rm E}(2) = -0.4337, \ L_{\rm E}(3) = +0.1829.$$

'''(3)'''  The a-posteriori  $L$–values at the beginning of the first iteration are thus
:$$\underline{L_{\rm APP} }^{(I=1)} = \underline{L_{\rm APP} }^{(I=0)} + \underline{L}_{\hspace{0.02cm}\rm E}^{(I=0)} =(+0.8171\hspace{0.05cm},\hspace{0.05cm}-0.0337\hspace{0.05cm},\hspace{0.05cm}-0.8171)\hspace{0.05cm} . $$

'''(4)'''  From this,  the new extrinsic  $L$–values for the iteration loop  $I = 1$  are as follows:
:$$L_{\rm E}(1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm tanh}^{-1} \big [{\rm tanh}(-0.0337/2) \cdot {\rm tanh}(-0.8171/2) \big ] = 0.0130 = -L_{\rm E}(3)\hspace{0.05cm},$$
:$$L_{\rm E}(2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm tanh}^{-1} \big [{\rm tanh}(+0.8171/2) \cdot {\rm tanh}(-0.8171/2) \big ] = - 0.3023\hspace{0.05cm}.$$

Further,  one can see from the above table:
* A hard decision according to the signs before the first iteration  $(I = 0)$ fails,  since  $(+1, +1, -1)$  is not a valid  $\rm SPC \, (3, \, 2, \, 2)$  code word.

* But already after  $I = 1$  iterations,  a hard decision yields a valid code word,  namely  $\underline{x}_2 = (+1, -1, -1)$.

*Also in later graphs,  the rows with correct hard decisions for the first time are highlighted in blue.

* Hard decisions after further iterations  $(I ≥ 2)$  each lead to the same code word  $\underline{x}_2$. This statement is not only valid for this example, but in general.

Besides,  in this exercise we consider a  »'''second $L_{\rm E}(i)$ approach'''«,  which is given here for the example of the first symbol  $(i = 1)$:
:$${\rm sign} \big[L_{\rm E}(1)\big] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm sign} \big[L_{\rm E}(2)\big] \cdot {\rm sign} \big[L_{\rm E}(3)\big]\hspace{0.05cm},$$
:$$|L_{\rm E}(1)| \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Min} \left ( |L_{\rm E}(2)|\hspace{0.05cm}, \hspace{0.05cm}|L_{\rm E}(3)| \right ) \hspace{0.05cm}.$$

This second approach is based on the assumption that the reliability of  $L_{\rm E}(i)$  is essentially determined by the most unreliable neighbor symbol.  The better  $($larger$)$  the input log likelihood ratio is completely disregarded.

Let us consider two examples for this:

'''(1)'''  For  $L_2 = 1.0$  and  $L_3 = 5.0$  we get
* after the first approach:   $L_{\rm E}(1) =2 \cdot {\rm tanh}^{-1} \big [{\rm tanh}(0.5) \cdot {\rm tanh}(2.5) \big ] =2 \cdot {\rm tanh}^{-1}(0.4559) = 0.984\hspace{0.05cm},$

* according to the second approach:   $|L_{\rm E}(1)| \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Min} \big ( 1.0\hspace{0.05cm}, \hspace{0.05cm}5.0 \big ) = 1.000 \hspace{0.05cm}.$

'''(2)'''  On the other hand one obtains for  $L_2 = L_3 = 1.0$
* according to the first approach:   $L_{\rm E}(1) =2 \cdot {\rm tanh}^{-1} \big [{\rm tanh}(0.5) \cdot {\rm tanh}(0.5) \big ] =2 \cdot {\rm tanh}^{-1}(0.2135) = 0.433\hspace{0.05cm},$

* according to the second approach:   $|L_{\rm E}(1)| \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Min} \big ( 1.0\hspace{0.05cm}, \hspace{0.05cm}1.0 \big ) = 1.000 \hspace{0.05cm}.$

One can see the clear discrepancy between the two approaches.  The second approach  $($approximation$)$  is clearly more positive than the first  $($correct$)$  approach. However,  it is actually only important that the iterations lead to the desired decoding result.

Hints:
*The exercise belongs to the chapter  [[Channel_Coding/Soft-in_Soft-Out_Decoder|"Soft–in Soft–out Decoder"]].

*Referred to in particular  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_log_likelihood_ratios|"Calculation of extrinsic log likelihood ratios"]].

* Only the  '''second solution approach'''  is treated here.

* For the first solution approach we refer to  [[Aufgaben:Exercise_4.5Z:_Tangent_Hyperbolic_and_Inverse|$\text{Exercise 4.5Z}$]] .

===Questions===
<quiz display=simple>
{It holds  $\underline{L} = (+1.0, +0.4, -1.0)$.  Determine the extrinsic  $L$–values according to the  '''second  $L_{\rm E}(i)$–approach'''  without previous iteration  $\underline{(I = 0)}$.
|type="{}"}
$L_{\rm E}(1) \ = \ ${ -0.412--0.388 }
$L_{\rm E}(2) \ = \ ${ -1.03--0.97 }
$L_{\rm E}(3) \ = \ ${ 0.4 3% }

{What are the a-posteriori $L$–values  $L_i = L_{\rm APP} (i)$  for the first iteration  $\underline{(I = 1)}$?
|type="{}"}
$L_1 \ = \ ${ 0.6 3% }
$L_2 \ = \ ${ -0.618--0.582 }
$L_3 \ = \ ${ -0.618--0.582 }

{Which of the following statements are true for  $\underline{L} = (+1.0, +0.4, -1.0)$?
|type="[]"}
+ Hard decision  after  $I = 1$  leads to the code word  $\underline{x}_1 = (+1, -1, -1)$.
+ This does not change after further iterations.
- Further iterations do not increase the reliability for  $\underline{x}_1$ .

{Which of the following statements are true for  $\underline{L} = (+0.6, +1.0, -0.4)$?
|type="[]"}
+ The iterative decoding leads to the result  $\underline{x}_0 = (+1, +1, +1)$.
- The iterative decoding leads to the result  $\underline{x}_2 = (-1, +1, -1)$.
+ Hard decision also returns this result for  $I \ge 1$.

{Which of the following statements are true for  $\underline{L} = (+0.6, +1.0, -0.8)$?
|type="[]"}
- The iterative decoding leads to the result  $\underline{x}_0 = (+1, +1, +1)$.
+ The iterative decoding leads to the result  $\underline{x}_2 = (-1, +1, -1)$.
+ Hard decision also returns this result for  $I \ge 1$.

{Which of the following statements are true for  $\underline{L} = (+0.6, +1.0, -0.6)$?
|type="[]"}
- Iterative decoding leads to the result  $\underline{x}_0 = (+1, +1, +1)$.
- The iterative decoding leads to the result  $\underline{x}_2 = (-1, +1, -1)$.
+ The iterative decoding does not lead to the result here.
</quiz>

===Solution===
{{ML-Kopf}}
[[File:P_ID3027__KC_A_4_5a_v2.png|right|frame|Results for  $\underline{L}=(+1.0, +0.4, –1.0)$]]
'''(1)'''  According to the second  $L_{\rm E}(i)$  approach holds:
:$${\rm sign} [L_{\rm E}(1)] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm sign} [L_{\rm E}(2)] \cdot {\rm sign} [L_{\rm E}(3)] = -1 \hspace{0.05cm},$$
:$$|L_{\rm E}(1)| \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Min} \left ( |L_{\rm E}(2)|\hspace{0.05cm}, \hspace{0.05cm}|L_{\rm E}(3)| \right ) = {\rm Min} \left ( 0.4\hspace{0.05cm}, \hspace{0.05cm}1.0 \right ) = 0.4$$
:$$\Rightarrow \hspace{0.3cm}L_{\rm E}(1) \hspace{0.15cm} \underline{-0.4}\hspace{0.05cm}.$$

*In the same way you get:
:$$L_{\rm E}(2) \hspace{0.15cm} \underline{-1.0}\hspace{0.05cm}, $$
:$$L_{\rm E}(3) \hspace{0.15cm} \underline{+0.4}\hspace{0.05cm}.$$

'''(2)'''  The a-posteriori  $L$–values at the beginning of the first iteration  $(I = 1)$  are the sum
*of the previous  $L$–values  $($for  $I = 0$) 
*and the extrinsic values calculated in subtask  '''(1)''':
:$$L_1 = L_{\rm APP}(1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}1.0 + (-0.4)\hspace{0.15cm} \underline{=+0.6}\hspace{0.05cm},$$
:$$L_2 = L_{\rm APP}(2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.4 + (-1.0)\hspace{0.15cm} \underline{=-0.6}\hspace{0.05cm},$$
:$$L_3 = L_{\rm APP}(3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (-1.0) + 0.4\hspace{0.15cm} \underline{=-0.6}\hspace{0.05cm}.$$

'''(3)'''  As can be seen from the above table,  the  solutions 1 and 2  are correct in contrast to answer 3:
*With each new iteration,  the magnitudes of  $L(1), \ L(2)$ and $L(3)$  become significantly larger.

[[File:P_ID3030__KC_A_4_5d_v2.png|right|frame|Results for  $\underline{L}=(+0.6, +1.0, –0.4)$]]
 
'''(4)'''  As can be seen from the adjacent table,
  the  answers 1 and 3  are correct:
*So the decision is made for the code word $\underline{x}_0 = (+1, +1, +1)$.

*From  $I = 1$  this would also be the decision of  "hard decision".
 
[[File:P_ID3028__KC_A_4_5e_v2.png|right|frame|Results for  $\underline{L}=(+0.6, +1.0, –0.8)$]]
'''(5)'''  Correct are the  answers 2 and 3:
*Because of  $|L(3)| > |L(1)|$  the following is valid for $I /ge 1$:   $L_1 < 0 \hspace{0.05cm},\hspace{0.2cm}
L_2 > 0 \hspace{0.05cm},\hspace{0.2cm}
L_3 < 0 \hspace{0.05cm}.$

*From this iteration loop,  hard decision returns the code word  $\underline{x}_2 = (-1, +1, -1)$.
 
[[File: P_ID3029__KC_A_4_5f_v1.png|right|frame|Results for  $\underline{L}=(+0.6, +1.0, –0.6)$]]
'''(6)'''  Correct is the  proposed solution 3:
*The adjacent table shows that under the condition  $|L(1)| = |L(3)|$,  starting from the iteration loop  $I = 1$,  all extrinsic  $L$–values are zero.

*Thus,  the a-posteriori  $L$– values remain constantly equal to  $\underline{L} = (0., +0.4, 0.)$  even for  $I > 1$,  which cannot be assigned to any code word.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^4.1 Soft–in Soft–out Decoder^]]
[[de:Aufgaben:Aufgabe 4.5: Nochmals zu den extrinsischen L–Werten]]

Aufgaben:Exercise 4.5: Irrelevance Theorem

2026-03-16T15:58:46Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver}}

[[File:EN_Dig_A_4_5.png|right|frame|Considered optimal system with  "detector"  and  "decision"]]
The communication system given by the graph is to be investigated.  The binary message  $m ∈ \{m_0, m_1\}$  with equal occurrence probabilities
:$${\rm Pr} (m_0 ) = {\rm Pr} (m_1 ) = 0.5$$

is represented by the two signals
:$$s_0 = \sqrt{E_s} \hspace{0.05cm},\hspace{0.2cm}s_1 = -\sqrt{E_s}$$

where the assignments  $m_0 ⇔ s_0$  and  $m_1 ⇔ s_1$  are one-to-one.

The detector  $($highlighted in green in the figure$)$  provides two decision values
:$$r_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} s + n_1\hspace{0.05cm},$$
:$$r_2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} n_1 + n_2\hspace{0.05cm},$$

from which the decision forms the estimated values  $\mu ∈ \{m_0,\ m_1\}$  for the transmitted message  $m$.  The decision includes
*two weighting factors  $K_1$  and  $K_2$,

*a summation point,  and

*a threshold decision with the threshold at  $0$.

Three evaluations are considered in this exercises:
# Decision based on  $r_1$  $(K_1 ≠ 0,\ K_2 = 0)$,
# decision based on  $r_2$  $(K_1 = 0,\ K_2 ≠ 0)$,
# joint evaluation of  $r_1$  und  $r_2$  $(K_1 ≠ 0,\ K_2 ≠ 0)$.

Notes:
* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver|"Structure of the Optimal Receiver"]]  of this book.

* In particular,  [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver#The_irrelevance_theorem|"the irrelevance theorem"]]  is referred to here,  but besides that also the  [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver#Optimal_receiver_for_the_AWGN_channel|"Optimal receiver for the AWGN channel"]].

* For more information on topics relevant to this exercise, see the following links:
** [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver#Fundamental_approach_to_optimal_receiver_design|"Decision rules for MAP and ML receivers"]],
** [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver#Implementation_aspects|"Realization as correlation receiver or matched filter receiver"]],
** [[Digital_Signal_Transmission/Structure_of_the_Optimal_Receiver#Probability_density_function_of_the_received_values|"Conditional Gaussian probability density functions"]].

* For the error probability of a system  $r = s + n$  $($because of  $N = 1$  here  $s,\ n,\ r$  are scalars$)$  is valid:
::$$p_{\rm S} = {\rm Pr} ({\rm symbol\ \ error} ) = {\rm Q} \left ( \sqrt{{2 E_s}/{N_0}}\right ) \hspace{0.05cm},$$

:where a binary message signal  $s ∈ \{s_0,\ s_1\}$  with  $s_0 = \sqrt{E_s} \hspace{0.05cm},\hspace{0.2cm}s_1 = -\sqrt{E_s}$  is assumed.
*Let the two noise sources  $n_1$  and  $n_2$  be independent of each other and also independent of the transmitted signal  $s ∈ \{s_0,\ s_1\}$.

*$n_1$  and  $n_2$  can each be modeled by AWGN noise sources  $($white,  Gaussian distributed,  mean-free,  variance  $\sigma^2 = N_0/2)$.  

*For numerical calculations, use the values
:$$E_s = 8 \cdot 10^{-6}\,{\rm Ws}\hspace{0.05cm},\hspace{0.2cm}N_0 = 10^{-6}\,{\rm W/Hz} \hspace{0.05cm}.$$

*The  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"complementary Gaussian error function"]]  gives the following results:
:$${\rm Q}(0) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5\hspace{0.05cm},\hspace{1.35cm}{\rm Q}(2^{0.5}) = 0.786 \cdot 10^{-1}\hspace{0.05cm},\hspace{1.1cm}{\rm Q}(2) = 0.227 \cdot 10^{-1}\hspace{0.05cm},$$
:$${\rm Q}(2 \cdot 2^{0.5}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.234 \cdot 10^{-2}\hspace{0.05cm},\hspace{0.2cm}{\rm Q}(4) = 0.317 \cdot 10^{-4}\hspace{0.05cm},\hspace{0.2cm}{\rm Q}(4 \cdot 2^{0.5}) = 0.771 \cdot 10^{-8}\hspace{0.05cm}.$$

===Questions===
<quiz display=simple>
{What statements apply here regarding the receiver?
|type="()"}
- The ML receiver is better than the MAP receiver.
- The MAP receiver is better than the ML receiver.
+ Both receivers deliver the same result.

{What is the error probability with  $K_2 = 0$?
|type="{}"}
${\rm Pr(symbol\hspace{0.15cm} error)}\ = \ $ { 0.00317 3% } $\ \%$

{What is the error probability with  $K_1 = 0$?
|type="{}"}
${\rm Pr(symbol\hspace{0.15cm}error)}\ = \ $ { 50 3% } $\ \%$

{Can an improvement be achieved by using  $r_1$  and  $r_2$? 
|type="()"}
+ Yes.
- No.

{What are the equations for the estimated value  $(\mu)$  for AWGN noise?
|type="[]"}
- $\mu = {\rm arg \ min} \, \big[(\rho_1 + \rho_2) \cdot s_i \big]$,
+ $\mu = {\rm arg \ min} \, \big[(\rho_2 \, - 2 \rho_1) \cdot s_i \big]$,
+ $\mu = {\rm arg \ max} \, \big[(\rho_1 - \rho_2/2) \cdot s_i \big]$.

{How can this rule be implemented exactly with the given decision  (threshold at zero)?  Let  $K_1 = 1$.
|type="{}"}
$K_2 \ = \ $ { -0.515--0.485 }

{What is the  (minimum)  error probability with the realization according to subtask  '''(6)'''?
|type="{}"}
${\rm Minimum \ \big[Pr(symbol\hspace{0.15cm}error)\big]} \ = \ $ { 0.771 3% } $\ \cdot 10^{\rm -8}$
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The  last alternative solution  is correct:
*In general,  the MAP receiver leads to a smaller error probability.
*However,  if the occurrence probabilities  ${\rm Pr}(m = m_0) = {\rm Pr}(m = m_1) = 0.5$  are equal,  both receivers yield the same result.

'''(2)'''  With  $K_2 = 0$  and  $K_1 = 1$  the result is
:$$r = r_1 = s + n_1\hspace{0.05cm}.$$

*With bipolar  (antipodal)  transmitted signal and AWGN noise,  the error probability of the optimal receiver  (whether implemented as a correlation or matched filter receiver)  is equal to
:$$p_{\rm S} = {\rm Pr} ({\rm symbol\hspace{0.15cm} error} ) = {\rm Q} \left ( \sqrt{ E_s /{\sigma}}\right )= {\rm Q} \left ( \sqrt{2 E_s /N_0}\right ) \hspace{0.05cm}.$$

*With   $E_s = 8 \cdot 10^{\rm –6} \ \rm Ws$ and $N_0 = 10^{\rm –6} \ \rm W/Hz$,  we further obtain:
:$$p_{\rm S} = {\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Q} \left ( \sqrt{\frac{2 \cdot 8 \cdot 10^{-6}\,{\rm Ws}}{10^{-6}\,{\rm W/Hz} }}\right ) = {\rm Q} (4) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.00317 \%}\hspace{0.05cm}.$$

*This result is independent of  $K_1$,  since amplification or attenuation changes the useful power in the same way as the noise power.

'''(3)'''  With  $K_1 = 0$  and  $K_2 = 1$,  the decision variable is:
:$$r = r_2 = n_1 + n_2\hspace{0.05cm}.$$

*This contains no information about the useful signal,  only noise,  and it holds independently of  $K_2$:
:$$p_{\rm S} = {\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Q} (0) \hspace{0.15cm}\underline {= 50\%} \hspace{0.05cm}.$$

'''(4)'''  Because of  ${\rm Pr}(m = m_0) = {\rm Pr}(m = m_1)$,  the decision rule of the optimal receiver  (whether realized as MAP or as ML)  is:
:$$\hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm arg} \max_i \hspace{0.1cm} [ p_{m \hspace{0.05cm}|\hspace{0.05cm}r_1, \hspace{0.05cm}r_2 } \hspace{0.05cm} (m_i \hspace{0.05cm}|\hspace{0.05cm}\rho_1, \hspace{0.05cm}\rho_2 ) ] = {\rm arg} \max_i \hspace{0.1cm} [ p_{r_1, \hspace{0.05cm}r_2 \hspace{0.05cm}|\hspace{0.05cm} m} \hspace{0.05cm} ( \rho_1, \hspace{0.05cm}\rho_2 \hspace{0.05cm}|\hspace{0.05cm} m_i ) \cdot {\rm Pr} (m = m_i)] = {\rm arg} \max_i \hspace{0.1cm} [ p_{r_1, \hspace{0.05cm}r_2 \hspace{0.05cm}|\hspace{0.05cm} s} \hspace{0.05cm} ( \rho_1, \hspace{0.05cm}\rho_2 \hspace{0.05cm}|\hspace{0.05cm} s_i ) ]\hspace{0.05cm}.$$

*This composite probability density can be rewritten as follows:
:$$\hat{m} ={\rm arg} \max_i \hspace{0.1cm} [ p_{r_1 \hspace{0.05cm}|\hspace{0.05cm} s} \hspace{0.05cm} ( \rho_1 \hspace{0.05cm}|\hspace{0.05cm} s_i ) \cdot p_{r_2 \hspace{0.05cm}|\hspace{0.05cm} r_1, \hspace{0.05cm}s} \hspace{0.05cm} ( \rho_2 \hspace{0.05cm}|\hspace{0.05cm} \rho_1, \hspace{0.05cm}s_i ) ]\hspace{0.05cm}.$$

*Now,  since the second multiplicand also depends on the message  ($s_i$),  $r_2$  should definitely be included in the decision process.  Thus,  the correct answer is:  YES.

'''(5)'''  For AWGN noise with variance  $\sigma^2$,  the two composite densities introduced in  '''(4)'''  together with their product  $P$  give:
:$$p_{r_1\hspace{0.05cm}|\hspace{0.05cm} s}(\rho_1\hspace{0.05cm}|\hspace{0.05cm}s_i) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2\pi} \cdot \sigma}\cdot {\rm exp} \left [ - \frac{(\rho_1 - s_i)^2}{2 \sigma^2}\right ]\hspace{0.05cm},\hspace{1cm}p_{r_2\hspace{0.05cm}|\hspace{0.05cm}r_1,\hspace{0.05cm} s}(\rho_2\hspace{0.05cm}|\hspace{0.05cm}\rho_1, \hspace{0.05cm}s_i) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2\pi} \cdot \sigma}\cdot {\rm exp} \left [ - \frac{(\rho_2 - (\rho_1 - s_i))^2}{2 \sigma^2}\right ]\hspace{0.05cm}$$
:$$ \Rightarrow \hspace{0.3cm}P \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{{2\pi} \cdot \sigma^2}\cdot {\rm exp} \left [ - \frac{1}{2 \sigma^2} \cdot \left \{ (\rho_1 - s_i)^2+ (\rho_2 - (\rho_1 - s_i))^2\right \}\right ]\hspace{0.05cm}.$$

*We are looking for the argument that maximizes this product  $P$,  which at the same time means that the expression in the curly brackets should take the smallest possible value:
:$$\mu = \hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm arg} \max_i \hspace{0.1cm} P = {\rm arg} \min _i \hspace{0.1cm} \left \{ (\rho_1 - s_i)^2+ (\rho_2 - (\rho_1 - s_i))^2\right \} $$
:$$\Rightarrow \hspace{0.3cm} \mu = \hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm arg} \min _i \hspace{0.1cm}\left \{ \rho_1^2 - 2\rho_1 s_i + s_i^2+ \rho_2^2 - 2\rho_1 \rho_2 + 2\rho_2 s_i+ \rho_1^2 - 2\rho_1 s_i + s_i^2\right \}\hspace{0.05cm}.$$

*Here  $\mu$  denotes the estimated value of the message.  In this minimization,  all terms that do not depend on the message  $s_i$  can now be omitted.  Likewise,  the terms  $s_i^2$  are disregarded,  since  $s_0^2 = s_1^2$  holds.  Thus,  the much simpler decision rule is obtained:
:$$\mu = {\rm arg} \min _i \hspace{0.1cm}\left \{ - 4\rho_1 s_i + 2\rho_2 s_i \right \}={\rm arg} \min _i \hspace{0.1cm}\left \{ (\rho_2 - 2\rho_1) \cdot s_i \right \}\hspace{0.05cm}.$$

*So,  correct is already the proposed solution 2.  But after multiplication by  $–1/2$,  we also get the last mentioned decision rule:
:$$\mu = {\rm arg} \max_i \hspace{0.1cm}\left \{ (\rho_1 - \rho_2/2) \cdot s_i \right \}\hspace{0.05cm}.$$

*Thus,  the solutions 2 and 3  are correct.

'''(6)'''  Setting $K_1 = 1$  and  $\underline {K_2 = \, -0.5}$,  the optimal decision rule with realization  $\rho = \rho_1 \, – \rho_2/2$ is:
:$$\mu =\left\{ \begin{array}{c} m_0 \\m_1 \end{array} \right.\quad\begin{array}{*{1}c} {\rm f{or}} \hspace{0.15cm} \rho > 0 \hspace{0.05cm},\\ {\rm f{or}} \hspace{0.15cm} \rho < 0 \hspace{0.05cm}.\\ \end{array}$$

*Since  $\rho = 0$  only occurs with probability  $0$,  it does not matter in the sense of probability theory whether one assigns the message  $\mu = m_0$  or  $\mu = m_1$  to this event  "$\rho = 0$".

'''(7)'''  With  $K_2 = \, -0.5$  one obtains for the input value of the decision:
:$$r = r_1 - r_2/2 = s + n_1 - (n_1 + n_2)/2 = s + n \hspace{0.3cm}\Rightarrow \hspace{0.3cm} n = \frac{n_1 - n_2}{2}\hspace{0.05cm}.$$

*The variance of this random variable is
:$$\sigma_n^2 = {1}/{4} \cdot \left [ \sigma^2 + \sigma^2 \right ] = {\sigma^2}/{2}= {N_0}/{4}\hspace{0.05cm}.$$

*From this,  the error probability is analogous to subtask  '''(2)''':
:$${\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Q} \left ( \sqrt{\frac{E_s}{N_0/4}}\right ) ={\rm Q} \left ( 4 \cdot \sqrt{2}\right ) = {1}/{2} \cdot {\rm erfc}(4) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.771 \cdot 10^{-8}}\hspace{0.05cm}.$$

*Thus,  by taking  $r_2$  into account,  the error probability can be lowered from  $0.317 \cdot 10^{\rm –4}$  to the much smaller value of  $0.771 \cdot 10^{-8}$,  although the decision component  $r_2$  contains only noise.  However,  this noise  $r_2$  allows an estimate of the noise component  $n_1$  of  $r_1$.

*Halving the transmit energy from  $8 \cdot 10^{\rm –6} \ \rm Ws$  to  $4 \cdot 10^{\rm –6} \ \rm Ws$,  we still get the error probability  $0.317 \cdot 10^{\rm –4}$  here,  as calculated in subtask  '''(2)'''.  When evaluating  $r_1$  alone,  on the other hand,  the error probability would be  $0.234 \cdot 10^{\rm –2}$.
{{ML-Fuß}}

[[Category:Digital Signal Transmission: Exercises|^4.2 Structure of the Optimal Receiver^]]
[[de:Aufgaben:Aufgabe 4.5: Theorem der Irrelevanz]]

Aufgaben:Exercise 4.09: Decision Regions at Laplace

2026-03-16T15:58:46Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Digital_Signal_Transmission/Approximation_of_the_Error_Probability}}

[[File:P_ID2044__Dig_A_4_9.png|right|frame|Three Laplace decision regions]]
We consider a transmission system based on the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.  The two equally probable transmitted signals are given by the signal points
:$$\boldsymbol{ s }_0 = (-\sqrt{E}, \hspace{0.1cm}-\sqrt{E})\hspace{0.05cm}, \hspace{0.2cm}\boldsymbol{ s }_1 = (+\sqrt{E}, \hspace{0.1cm}+\sqrt{E})\hspace{0.05cm}.$$

In the following we normalize the energy parameter to  $E = 1$  for simplification and thus obtain
:$$\boldsymbol{ s }_0 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (-1, \hspace{0.1cm}-1) \hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_0\hspace{0.05cm}, $$
:$$ \boldsymbol{ s }_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (+1, \hspace{0.1cm}+1)\hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_1\hspace{0.05cm}.$$

The messages  $m_0$  and  $m_1$  are uniquely assigned to the signals  $\boldsymbol{s}_0$  and  $\boldsymbol{s}_1$  defined in this way.

Let the noise components  $n_1(t)$  and  $n_2(t)$  be independent of each other and each be Laplace distributed with parameter  $a = 1$:
:$$p_{n_1} (\eta_1) = {1}/{2} \cdot {\rm e}^{- | \eta_1|} \hspace{0.05cm}, \hspace{0.2cm}p_{n_2} (\eta_2) = {1}/{2} \cdot {\rm e}^{- | \eta_2|} \hspace{0.05cm} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \boldsymbol{ p }_{\boldsymbol{ n }} (\eta_1, \eta_2) = {1}/{4} \cdot {\rm e}^{- | \eta_1|- | \eta_2|} \hspace{0.05cm}. $$

The properties of such a Laplace noise will be discussed in detail in  [[Aufgaben:Exercise_4.09Z:_Laplace_Distributed_Noise|"Exercise 4.9Z"]]. 

The received signal  $\boldsymbol{r}$  is composed additively of the transmitted signal  $\boldsymbol{s}$  and the  noise component  $\boldsymbol{n}$:
:$$\boldsymbol{ r } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \boldsymbol{ s } + \boldsymbol{ n }\hspace{0.05cm}, \hspace{0.45cm}\boldsymbol{ r } = ( r_1, r_2)\hspace{0.05cm},\hspace{0.45cm}\boldsymbol{ s } \hspace{-0.1cm} \ = \ \hspace{-0.1cm} ( s_1, s_2)\hspace{0.05cm}, \hspace{0.8cm}\boldsymbol{ n } = ( n_1, n_2)\hspace{0.05cm}. $$

The corresponding realizations are denoted as follows:
:$$\boldsymbol{ s }\text{:} \hspace{0.4cm} (s_{01},s_{02}){\hspace{0.15cm}\rm and \hspace{0.15cm}} (s_{11},s_{12})\hspace{0.05cm},\hspace{0.8cm} \boldsymbol{ r }\text{:} \hspace{0.4cm} (\rho_{1},\rho_{2})\hspace{0.05cm}, \hspace{0.8cm}\boldsymbol{ n }\text{:} \hspace{0.4cm} (\eta_{1},\eta_{2})\hspace{0.05cm}.$$

The decision rule of the MAP and ML receivers  $($both are identical due to the same symbol probabilities$)$  are:

⇒   Decide for the symbol  $m_0$, if   $p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} |m_0 ) >
p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } (\rho_{1},\rho_{2} |m_1 ) \hspace{0.05cm}.$

⇒   With the further conditions for the  "decision for  $m_0$"  can also be written:
:$${1}/{4} \cdot {\rm exp}\left [- | \rho_1 +1|- | \rho_2 +1| \hspace{0.1cm} \right ] >{1}/{4} \cdot {\rm exp}\left [- | \rho_1 -1|- | \rho_2 -1| \hspace{0.1cm} \right ] $$
:$$\Rightarrow \hspace{0.3cm} | \rho_1 +1|+ | \rho_2 +1| <| \rho_1 -1|+ | \rho_2 -1|$$
:$$\Rightarrow \hspace{0.3cm} L (\rho_1, \rho_2) = | \rho_1 +1|+ | \rho_2 +1| -| \rho_1 -1|- | \rho_2 -1| < 0 \hspace{0.05cm}.$$

⇒   This function  $L(\rho_1, \rho_2)$  is frequently referred to in the questions.

The graph shows three different decision regions  $(I_0, \ I_1)$.
:*For AWGN noise,  only the upper variant  $\rm A$  would be optimal.

:*Also for the considered Laplace noise, variant  $\rm A$  leads to the smallest possible error probability, see  [[Aufgaben:Exercise_4.09Z:_Laplace_Distributed_Noise|"Exercise 4.9Z"]]:
::$$p_{\rm min} = {\rm Pr}({\cal{E}} \hspace{0.05cm}|\hspace{0.05cm} {\rm optimal\hspace{0.15cm} receiver}) = {\rm e}^{-2} \approx 13.5\,\%\hspace{0.05cm}.$$
:*It is to be examined whether variant  $\rm B$  or variant  $\rm C$  is also optimal, i.e. whether their error probabilities are also as small as possible equal to  $p_{\rm min}$. 

Note:   The exercise belongs to the chapter  [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]].

===Questions===
<quiz display=simple>
{Which of the decision rules are correct?  Decide for  $m_0$,  if
|type="[]"}
+ $p_{\it r\hspace{0.03cm}|\hspace{0.03cm}m}(\rho_1, \ \rho_2\hspace{0.03cm}|\hspace{0.03cm}m_0) > p_{\it r\hspace{0.03cm}|\hspace{0.03cm}m}(\rho_1, \ \rho_2\hspace{0.03cm}|\hspace{0.03cm}m_1)$,
+ $L(\rho_1, \ \rho_2) = |\rho_1+1| \, -|\rho_1 \, –1| + |\rho_2+1| \, -|\rho_2 \, –1| < 0$,
- $L(\rho_1, \ \rho_2) = \rho_1 + \rho_2 ≥ 0$.

{How can the expression  $|x+1| \ -|x \ -1|$  be transformed?
|type="[]"}
+ For  $x ≥ +1$:    $|x + 1| \, -|x -1| = 2$.
+ For  $x ≤ \, -1$:    $|x+1| \,-|x \, -1| = \, -2$.
+ For  $-1 ≤ x ≤ +1$:    $|x+1| \, -|x \, -1| = 2x$.

{What is the decision rule in the range  $-1 ≤ \rho_1 ≤ +1$,  $-1 ≤ \rho_2 ≤ +1$?
|type="()"}
+ Decision for  $m_0$,  if  $\rho_1 + \rho_2 < 0$.
- Decision for  $m_1$,  if  $\rho_1 + \rho_2 < 0$.

{What is the decision rule in the range  $\rho_1 > +1$?
|type="()"}
- Decision for  $m_0$  in the whole range.
+ Decision for  $m_1$  in the whole range.
- Decision for  $m_0$  only if  $\rho_1 + \rho_2 < 0$.

{What is the decision rule in the range  $\rho_1 < \, -1$?
|type="()"}
+ Decision for  $m_0$  in the whole range.
- Decision for  $m_1$  in the whole range.
- Decision for  $m_0$  only if  $\rho_1 + \rho_2 < 0$.

{What is the decision rule in the range  $\rho_2 > +1$?
|type="()"}
- Decision for  $m_0$  in the whole range.
+ Decision for  $m_1$  in the whole range.
- Decision for  $m_0$  only if  $\rho_1 + \rho_2 < 0$.

{What is the decision rule in the range  $\rho_2 < -1$?
|type="()"}
+ Decision for  $m_0$  in the whole range.
- Decision for  $m_1$  in the whole range.
- Decision for  $m_0$  only if  $\rho_1 + \rho_2 < 0$.

{Which of the following statements are true?
|type="[]"}
+ Variant  $\rm A$  leads to the minimum error probability.
+ Variant  $\rm B$  leads to the minimum error probability.
- Variant  $\rm C$  leads to the minimum error probability.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Solutions 1 and 2  are correct:
*The joint probability densities under conditions  $m_0$  and  $m_1$,  respectively,  are:
:$$p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} |m_0 )\hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{4} \cdot {\rm exp}\left [- | \rho_1 +1|- | \rho_2 +1| \hspace{0.05cm} \right ]\hspace{0.05cm},$$
:$$p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} |m_1 )\hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{4} \cdot {\rm exp}\left [- | \rho_1 -1|- | \rho_2 -1| \hspace{0.05cm} \right ]\hspace{0.05cm}.$$
*For equally probable symbols   ⇒   ${\rm Pr}(m_0) = {\rm Pr}(m_1) = 0.5$,  the MAP decision rule is:   Decide for symbol  $m_0$    ⇔   signal  $s_0$,  if
:$$p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } ( \rho_{1},\rho_{2} |m_0 ) >p_{\boldsymbol{ r} \hspace{0.05cm}|\hspace{0.05cm}m } (\rho_{1},\rho_{2} |m_1 ) \hspace{0.05cm}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {1}/{4} \cdot {\rm exp}\left [- | \rho_1 +1|- | \rho_2 +1| \hspace{0.05cm} \right ] >{1}/{4} \cdot {\rm exp}\left [- | \rho_1 -1|- | \rho_2 -1|\hspace{0.05cm} \right ] $$
:$$\Rightarrow \hspace{0.3cm} | \rho_1 +1|+ | \rho_2 +1| <| \rho_1 -1|+ | \rho_2 -1|\hspace{0.3cm}\Rightarrow \hspace{0.3cm} L (\rho_1, \rho_2) = | \rho_1 +1|-| \rho_1 -1|+ | \rho_2 +1| - | \rho_2 -1| < 0 \hspace{0.05cm}.$$

'''(2)'''  All statements are true:
*For $x ≥ 1$:
:$$| x +1|- | x -1| = x +1 -x +1 =2 \hspace{0.05cm}.$$

*Similarly,  for $x ≤ \, –1$,  e.g.  $x = \, –3$:
:$$| x +1|- | x -1| = | -3 +1|- | -3 -1| = 2-4 = -2 \hspace{0.05cm}.$$

*On the other hand,  in the middle range $–1 ≤ x ≤ +1$:
:$$| x+1|- | x -1| = x +1 -1 +x =2x \hspace{0.05cm}.$$

'''(3)'''  Solution 1  is correct:
*The result of subtask  '''(1)'''  was:  Decide for the symbol  $m_0$,  if
:$$L (\rho_1, \rho_2) = | \rho_1 +1| -| \rho_1 -1|+ | \rho_2 +1| - | \rho_2 -1| < 0 \hspace{0.05cm}.$$
*In the considered  (inner)  range  $-1 ≤ \rho_1 ≤ +1$,  $-1 ≤ \rho_2 ≤ +1$  holds with the result of subtask  '''(2)''':
:$$| \rho_1+1| - | \rho_1 -1| = 2\rho_1 \hspace{0.05cm}, \hspace{0.2cm} | \rho_2+1| - | \rho_2 -1| = 2\rho_2 \hspace{0.05cm}.$$

*If we insert this result above,  we have to decide for  $m_0$  exactly if
:$$L (\rho_1, \rho_2) = 2 \cdot ( \rho_1+\rho_2) < 0\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \rho_1+\rho_2 < 0\hspace{0.05cm}.$$

'''(4)'''  Correct is here  solution 2:
*For  $\rho_1 > 1$   ⇒   $|\rho_1+1| \, -|\rho_1 \, -1| = 2$,  while for  $D_2 = |\rho_2+1| \,-|\rho_2 \, -1|$   ⇒   all values between  $-2$  and  $+2$  are possible.

*Thus,  the decision variable is  $L(\rho_1, \rho_2) = 2 + D_2 ≥ 0$.  In this case,  the rule leads to an  $m_1$  decision.

'''(5)'''  Solution 1  is correct:
*After similar calculation as in subtask  '''(3)''',  one arrives at the result:
:$$L (\rho_1, \rho_2) = -2 + D_2 \le 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm decision\hspace{0.15cm} on\hspace{0.15cm}} m_0\hspace{0.05cm}.$$

'''(6)'''  Solution 2  is correct:  Decision on  $m_1$.
*Similar to subtask  '''(4)''',  the following holds here:
:$$D_1 = | \rho_1 +1| - | \rho_1 -1| \in \{-2, ... \hspace{0.05cm} , +2 \}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}L (\rho_1, \rho_2) = 2 + D_1 \ge 0\hspace{0.05cm}.$$

'''(7)'''  Solution 1  is correct:  Decision on  $m_0$.
*After similar reasoning as in the last subtask,  we arrive at the result:
:$$L (\rho_1, \rho_2) = -2 + D_1 \le 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm decision\hspace{0.15cm} on\hspace{0.15cm}} m_0\hspace{0.05cm}.$$

[[File:P_ID2050__Dig_A_4_9h.png|right|frame|Summary of the results]]
'''(8)'''  The results of subtasks  '''(3)'''  to  '''(7)'''  are summarized in the graph:

* Subarea  $T_0$:   Decision on  $m_0$  or  $m_1$  according to task  '''(3)'''.
* Subarea  $T_1$:   Decision on  $m_1$  according to task  '''(4)'''.
* Subarea  $T_2$:   Decision on  $m_0$  according to task  '''(5)'''.
* Subarea  $T_3$:   Decision on  $m_1$  according to task  '''(6)'''.
* Subarea  $T_4$:   Decision on $m_0$  according to task  '''(7)'''.
* Subarea  $T_5$:   Decision on  $m_0$  according to task  '''(5)''', and on  $m_1$  according to task  '''(6)''' ⇒   For Laplace noise,  it does not matter whether one assigns  $T_5$  to region  $I_0$  or  $I_1$.
* Subarea  $T_6$:   Again,  based on the results of task  '''(6)'''  and  '''(7)''',  one can assign this region to region  $I_0$  or region  $I_1$.

It can be seen:
#For subarea  $T_0$,  ... ,  $T_4$  there is a fixed assignment to the decision regions  $I_0$  (red)  and  $I_1$  (blue).
#In contrast,  the two yellow regions  $T_5$  and  $T_6$  can be assigned to both,  $I_0$  and  $I_1$  without loss of optimality.

Comparing this graph with variants  $\rm A$,  $\rm B$  and  $\rm C$  on the specification page,  we see that  suggestions 1 and 2  are correct:
# Variants  $\rm A$  and  $\rm B$ are equally good.  Both are optimal.  The error probability in both cases is  $p_{\rm min} = {\rm e}^{\rm -2}$.
# Variant  $\rm C$  is not optimal;  with respect to the subareas  $T_1$  and  $T_2$  there are mismatches.  The error probability is therefore greater than  $p_{\rm min}$.
{{ML-Fuß}}

[[Category:Digital Signal Transmission: Exercises|^4.3 BER Approximation^]]
[[de:Aufgaben:Aufgabe 4.09: Entscheidungsregionen bei Laplace]]

Aufgaben:Exercise 1.13: Binary Erasure Channel Decoding

2026-03-16T15:58:45Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Decoding_of_Linear_Block_Codes}}

[[File:EN_KC_A_1_13.png|right|frame|Decoding at the BEC]]

We assume here the model in section  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel|"Decoding at the Binary Erasure Channel"]]  (BEC configuration highlighted in green):

*Each information word  $\underline{u}$  is encoded blockwise and yields the code word  $\underline{x}$.  Let the block code be linear and completely given by its parity-check matrix  $\boldsymbol{\rm H}$.

*During transmission  $n_{\rm E}$  bits of the code word are erased   ⇒  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_.E2.80.93_BEC|"Binary Erasure Channel"]]  $\rm (BEC)$.  The code word  $\underline{x}$  thus becomes the received word  $\underline{y}$.

*If the number  $n_{\rm E}$  of erasures is less than the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"minimum distance"]]  $d_{\rm min}$  of the code,  it is possible to reconstruct from  $\underline{y}$  the code word  $\underline{z} = \underline{x}$  without error,  and thus the correct information word  $\underline{v} = \underline{u}$  is also obtained.

For exercise description,  let us now consider the Hamming code word  $\underline{x} = (0, 1, 0, 1, 1, 0, 0)$  and the received word  $\underline{y} = (0, 1, {\rm E} , {\rm E}, 1, 0, 0).$

*The third and fourth bit were thus erased by the channel.  The  "code word finder"  thus has the task to determine the vector  $z_{\rm E} = (z_{3}, z_{4})$  with  $z_{3}, \ z_{4} \in \{0, 1\}$  to be determined.  This is done according to the equation

:$${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T}= { \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T}\hspace{0.05cm}.$$

*In the present example:

:$$\underline{z}_{\rm K} = (0, 1, 1, 0, 0)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm K} = \begin{pmatrix} 1 &1 &1 &0 &0\\ 0 &1 &0 &1 &0\\ 1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0\\ 1 &1\\ 0 &1 \end{pmatrix} \hspace{0.05cm}.$$

*This equation gives two equations for the bits to be determined,  whose solution leads to the result  $z_{3} = 0$  and  $z_{4} = 1$.

Hints:
* This exercise belongs to the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes|"Decoding of Linear Block Codes"]].

* The algorithm for mapping the received word  $\underline{y}$  to the correct code word  $\underline{z} = \underline{x}$   is described in detail in the  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Decoding_at_the_Binary_Erasure_Channel|"theory section"]] .

* We would like to remind again that in BEC decoding we use the first decoder block  $\underline{y} → \underline{z}$  as  "code word finder"  since wrong decisions are excluded here.  Each received word is decoded correctly,  or it may not be decoded at all.

*In the BSC model,  on the other hand,  decoding errors cannot be avoided.  Accordingly,  we refer to the corresponding block there as  "code word estimator".

===Questions===
<quiz display=simple>
{  $\underline{y} = (1, {\rm E}, 0, 1, 0, 0, {\rm E})$  was received.  Which sequence does the code word finder decide to use?
|type="()"}
- $\underline{z} = (1, 0, 0, 1, 0, 0, 0),$
+ $\underline{z} = (1, 1, 0, 1, 0, 0, 1),$
- $\underline{z} = (1, 0, 0, 1, 0, 0, 1).$

{What are the consequences of the red entries for  $\boldsymbol{\rm H}_{\rm K}$  and  $z_{\rm K}$  (see graphic in the information section)?
|type="[]"}
+ The erasure vector is  $\underline{z}_{\rm E} = (z_{5}, z_{6}, z_{7}).$
+ The received word is  $\underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E}).$
- $\boldsymbol{\rm H}_{\rm E}$  is a  $2 \times 3$  matrix.
+ $\boldsymbol{\rm H}_{\rm E}$  is a  $3 \times 3$  matrix.

{Now apply  $\underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E}).$  Which code word is selected?
|type="()"}
- $\underline{z} = (1, 1, 0, 1, 1, 1, 0),$
+ $\underline{z} = (1, 1, 0, 1, 0, 0, 1),$
- $\underline{z} = (1, 1, 0, 0, 0, 1, 0).$
- For the present  $\underline{y}$   no unique decoding is possible.

{What are the consequences of the green entries for  $\boldsymbol{\rm H}_{\rm K}$  and  $z_{\rm K}$  (see graphic in the information section)?
|type="[]"}
+ The received word is  $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E}).$
- $\boldsymbol{\rm H}_{\rm K}$  differs from subtask  '''(2)'''  in the last row.
+ $\boldsymbol{\rm H}_{\rm K}$  differs from subtask  '''(2)'''  in the last column.

{Now apply  $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E}).$  Which code word is selected?
|type="()"}
- $\underline{z} = (1, 1, 0, 1, 1, 1, 0),$
- $\underline{z} = (1, 1, 0, 1, 0, 0, 1),$
- $\underline{z} = (1, 1, 0, 0, 0, 1, 0).$
+ For the present  $\underline{y}$   no unique decoding is possible.

{Which statements result for the correction capability at the BEC?  $n_{\rm E}$  indicates in the following the number of erasures.
|type="[]"}
+ For  $n_{\rm E} < d_{\rm min}$  unique decoding is always possible.
- For  $n_{\rm E} = d_{\rm min}$  unique decoding is always possible.
+ For  $n_{\rm E} = d_{\rm min}$  sometimes a unique decoding is possible.
+ For  $n_{\rm E} > d_{\rm min}$  unique decoding is never possible.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The received vector is  $\underline{y} = (1, {\rm E}, 0, 1, 0, 0, {\rm E})$.  Thus,  the code symbols at positions 2 and 7 were erased.  Based on the given parity-check matrix

:$${ \boldsymbol{\rm H}} = \begin{pmatrix} 1 &1 &1 &0 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &1 &0 &1 &0 &0 &1 \end{pmatrix}$$

of the Hamming code is obtained for vector and matrix

* with respect to all  "correctly transmitted code symbols"  $($index  "$\rm K$"  from German  "korrekt"$)$  which are known to the code word finder:

:$$\underline{z}_{\rm K} = (1, 0, 1, 0, 0)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm K} = \begin{pmatrix} 1 &1 &0 &1 &0\\ 0 &1 &1 &0 &1\\ 1 &0 &1 &0 &0 \end{pmatrix} \hspace{0.05cm},$$

*with respect to the two  "erased code symbols"  $z_{2}$  and  $z_{7}$  $($index  $\rm E)$  to be determined:

:$$\underline{z}_{\rm E} = (z_2, z_7)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0\\ 1 &0\\ 1 &1 \end{pmatrix} \hspace{0.05cm}.$$

The equation of determination is thus:
:$${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T}= { \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \begin{pmatrix} 1 &0\\ 1 &0\\ 1 &1 \end{pmatrix} \cdot \begin{pmatrix} z_2 \\ z_7 \end{pmatrix} = \begin{pmatrix} 1 &1 &0 &1 &0\\ 0 &1 &1 &0 &1\\ 1 &0 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm}.$$

This results in three equations for the two unknowns  $z_{2}$  and  $z_{7}$:

:$${\rm (a)}\ z_{2} = 1,$$
:$${\rm (b)}\ z_{2} = 1,$$
:$${\rm (c)}\ z_{2} + z_{7} = 0 \ \Rightarrow \ z_{7}= 1.$$

Thus,  the code word finder returns $\underline{z} = (1, 1, 0, 1, 0, 0, 1)$   ⇒   Suggested solution 2.

'''(2)'''  Looking at the given matrix  $\boldsymbol{\rm H}_{\rm K}$,  we see that it coincides with the first four columns of the parity-check matrix  $\boldsymbol{\rm H}$.

*The erasures thus affect the last three bits of the received word   ⇒   $\underline{z}_{\rm E} = (z_{5}, z_{6}, z_{7})$   ⇒   $\underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E})$.
*The erasure matrix reads:

:$${ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$

Correct are therefore the  statements 1, 2 and 4.

'''(3)'''  One obtains after some matrix multiplications:
:$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &0\\ 0 &1 &1 &1\\ 1 &1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \hspace{0.05cm},\hspace{1cm}{ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} \hspace{0.05cm}.$$

*By equating it follows  $z_{5} = 0, \ z_{6} = 0, \ z_{7} = 1$   ⇒   Suggested solution 2.

'''(4)'''  The matrix comparison shows that the first three columns of  $\boldsymbol{\rm H}$   and   $\boldsymbol{\rm H}_{\rm K}$   are identical.
* The fourth column of  $\boldsymbol{\rm H}_{\rm K}$  is equal to the fifth column of the parity-check matrix  $\boldsymbol{\rm H}$.

*From this follows for the vector  $z_{\rm E} = (z_{4}, z_{6}, z_{7})$  and further for the received vector  $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E})$   ⇒   Suggested solutions 1 and 3.

'''(5)'''  Analogous to subtask  '''(3)''',  we obtain:

:$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &1\\ 0 &1 &1 &0\\ 1 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm},\hspace{1cm}{ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 0 &0 &0\\ 1 &1 &0\\ 1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_4 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} 0 \\ z_4 + z_6 \\ z_4 + z_7 \end{pmatrix} \hspace{0.05cm}.$$

*If we now set the two column vectors equal,  we obtain only two equations for the three unknowns   ⇒   Suggested solution 4.  Or in other words:  

*If the number of extinctions of the BEC channel is larger than the rank of the matrix $\boldsymbol{\rm H}_{\rm E}$,  then no unique solution of the resulting system of equations results.

'''(6)'''  To solve this exercise,  we again refer to the systematic Hamming code  $(7, 4, 3)$  according to the given parity-check equation and code table.  Note: 
[[File:P_ID2540__KC_A_1_13f.png|right|frame|Code table of the systematic  $(7, 4, 3)$  Hamming code ]]
*The information bits are shown in black and the parity bits in red.  The minimum distance of this code is  $d_{\rm min} = 3$.

*Further we assume that always the code word  $\underline{x} = (1, 1, 0, 1, 0, 0, 1)$  with yellow background was sent.  Then holds:

*If the number  $n_{\rm E}$  of erasures is smaller than  $d_{\rm min} = 3$,  decoding by the method described here is always possible   ⇒   see e.g. subtask  '''(1)'''  with  $n_{\rm E}= 2$.

*For  $n_{\rm E} = d_{\rm min} = 3$  decoding is sometimes possible,  see subtask  '''(3)'''.  In the code table,  there is only one code word that could match   $\underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E})$   namely the code word highlighted in yellow   $\underline{x} = (1, 1, 0, 1, 0, 0, 1)$ .

*In contrast   $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E})$  according to subtask  '''(4)'''  could not be decoded. In the code table, besides  $(1, 1, 0, 1, 0, 0, 1)$  with  $(1, 1, 0, 0, 0, 1, 0)$  one can recognize another code word  (highlighted in green),  which becomes the received word  $\underline{y}$  due to the  $n_{\rm E} = 3$  erasures concerning bit 4, 6 and 7.

*This case,  when the  $n_{\rm E} = d_{\rm min}$  extinctions affect exactly the  $d_{\rm min}$  different bits of two code words, leads to a matrix  $\mathbf{H}_{\rm E}$  with rank smaller  $d_{\rm min}$.

*If  $n_{\rm E} > d_{\rm min}$, the number  $n - n_{\rm E}$  of bits not erased is smaller than the number  $k$  of information bits.  In this case,  of course,  the code word cannot be decoded.

That is:   Applicable are the  statements 1, 3 and 4.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^1.5 Linear Block Code Decoding^]]
[[de:Aufgaben:Aufgabe 1.13: Decodierung beim binären Auslöschungskanal (BEC)]]

Aufgaben:Exercise 4.3: Iterative Decoding at the BSC

2026-03-16T15:58:44Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}

[[File:EN_KC_A_4_3.png|right|frame|BSC model and  (above) possible received values]]
We consider in this exercise two codes:
* the  "single parity–check code"   ⇒   [[Channel_Coding/Examples_of_Binary_Block_Codes#Single_Parity-check_Codes| $\text{SPC (3, 2, 2)}$]]:
:$$\underline{x} = \big (\hspace{0.05cm}(0, 0, 0), \hspace{0.1cm}(0, 1, 1), \hspace{0.1cm}(1, 0, 1), \hspace{0.1cm}(1, 1, 0) \hspace{0.05cm} \big )\hspace{0.05cm}, $$
* the repetition code   ⇒   [[Channel_Coding/Examples_of_Binary_Block_Codes#Repetition_Codes| $\text{RC (3, 1, 3)}$]]:
:$$\underline{x} = \big (\hspace{0.05cm}(0, 0, 0), \hspace{0.1cm}(1, 1, 1) \hspace{0.05cm} \big )\hspace{0.05cm}.$$

The channel is described at bit level by the  [[Digital_Signal_Transmission/Binary_Symmetric_Channel|$\text{BSC model}$]].  According to the graphic,  the following applies:
:$${\rm Pr}(y_i \ne x_i) =\varepsilon = 0.269\hspace{0.05cm},$$
:$${\rm Pr}(y_i = x_i) =1-\varepsilon = 0.731\hspace{0.05cm}.$$

Here,  $\varepsilon$  denotes the falsification probability of the BSC model.

Except for the last subtask,  the following received value is always assumed:
:$$\underline{y} = (0, 1, 0) =\underline{y}_2\hspace{0.05cm}. $$

The chosen indexing of all possible received vectors can be taken from the graphic.
*The most considered vector  $\underline{y}_2$  is highlighted in red.

*Only for subtask  '''(6)'''  applies:
:$$\underline{y} = (1, 1, 0) =\underline{y}_6\hspace{0.05cm}. $$

For decoding purposes, the exercise will examine:
* the  [[Channel_Coding/Decoding_of_Linear_Block_Codes#Generalization_of_syndrome_coding|"syndrome decoding"]],  which follows the concept  "hard decision maximum likelihood detection"  $\rm (HD-ML)$  for the codes under consideration. $($because soft values are not available at the BSC$)$;

* the bit-wise  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Bit-wise_soft-in_soft-out_decoding|"Soft–in Soft–out Decoding"]]  $\rm (SISO)$  according to this section.

Hints:
* This exercise refers to the chapter  [[Channel_Coding/Soft-in_Soft-Out_Decoder| "Soft–in Soft–out Decoder"]].

* Reference is made in particular to the sections
:*[[Channel_Coding/Soft-in_Soft-Out_Decoder#Bit-wise_soft-in_soft-out_decoding|"Bit-wise Soft–in Soft–out Decoding"]], 
:*[[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Symmetric_Channel_.E2.80.93_BSC|"Binary Symmetric Channel"]]
* The code word selected by the decoder is denoted in the questions by  $\underline{z}$.

===Questions===
<quiz display=simple>
{Which statements are valid for decoding the  $\text{SPC (3, 2, 2)}$?
|type="()"}
- The hard decision syndrome decoding yields the result  $\underline{z} = (0, \, 1, \, 0)$.
- The hard decision syndrome decoding returns the result  $\underline{z} = (0, \, 0, \, 0)$.
+ The hard decision syndrome decoding fails here.

{Which statements are valid for the  $\text{ RC (3, 1, 3)}$?
|type="()"}
- The hard decision syndrome decoding returns the result  $\underline{z} = (0, \, 1, \, 0)$.
+ The hard decision syndrome decoding returns the result  $\underline{z} = (0, \, 0, \, 0)$.
- The hard decision syndrome decoding fails here.

{How certain is this decision if we define reliability  $($"security"$)$  $S$  as the quotient of the probabilities for a correct or an incorrect decision? Set the falsification probability of the BSC model to  $\varepsilon = 26.9\%$.
|type="{}"}
$S \ = \ ${ 2.717 3% }
$\hspace{0.75cm} \ln {(S)} \ = \ ${ 1 3% }

{What are the intrinsic log likelihood ratios for iterative bit-wise decoding of the  $\text{RC (3, 1)}$  received word  $\underline{y}_2 = (0, \, 1, \, 0)$?
|type="{}"}
$L_{\rm K}(1) \ = \ ${ 1 3% }
$L_{\rm K}(2) \ = \ ${ -1.03--0.97 }
$L_{\rm K}(3) \ = \ ${ 1 3% }

{Which statements are true for decoding the received word  $\underline{y}_2 = (0, \, 1, \, 0)$?  Continue to assume the  $\text{RC (3, 1, 3)}$.
|type="[]"}
+ From the first iteration all signs of  $L_{\rm APP}(i)$  are positive.
+ Already after the second iteration  ${\rm Pr}(\underline{x}_0\hspace{0.05cm} |\hspace{0.05cm} \underline{y}_2)$  is greater than  $99\%$.
+ With each iteration the absolute values  $|L_{\rm APP}(i)|$  become larger.

{Which statements are true for decoding the received word  $\underline{y}_6 = (1, 1, 0)$,  when  $\underline{x}_0 = (0, 0, 0)$  was sent?
|type="[]"}
- The iterative decoder decides correctly.
+ The iterative decoder decides wrong.
+ The  "reliability"  for  $\underline{y}_6 \Rightarrow \underline{x}_0$  increases with increasing  $I$.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Correct is the  proposed solution 3:
*The received word  $\underline{y}_2 = (0, 1, 0)$  is not a valid code word of the single parity–check code  $\text{SPC (3, 2)}$.  Thus,  the first statement is false.

*In addition,  since the  $\text{SPC (3, 2)}$  has only the minimum distance  $d_{\rm min} = 2$,  no error can be corrected.

'''(2)'''  Correct is the  proposed solution 2:
*The possible code words at the  $\text{RP (3, 1)}$  are  $\underline{x}_0 = (0, 0, 0)$  and  $\underline{x}_1 = (1, 1, 1)$.

*The minimum distance of this code is  $d_{\rm min} = 3$,  so  $t = (d_{\rm min} \, - 1)/2 = 1$  error can be corrected.

*In addition to  $\underline{y}_0 = (0, 0, 0)$,  the received words  $\underline{y}_1 = (0, 0, 1), \ \underline{y}_2 = (0, 1, 0)$,  and  $\underline{y}_4 = (1, 0, 0)$  are also assigned to the decoding result  $\underline{x}_0 = (0, 0, 0)$.

'''(3)'''  According to the BSC model,  the conditional probability that  $\underline{y}_2 = (0, 1, 0)$  is received,  given that  $\underline{x}_0 = (0, 0, 0)$  was sent:
:$${\rm Pr}(\underline{y} = \underline{y}_2 \hspace{0.1cm}| \hspace{0.1cm}\underline{x} = \underline{x}_0 ) = (1-\varepsilon)^2 \cdot \varepsilon\hspace{0.05cm}.$$

*The first term  $(1 \, –\varepsilon)^2$  indicates the probability that the first and third bit were transmitted correctly.  $\varepsilon$ describes the falsification probability for the second bit.

*Correspondingly,  for the second possible code word  $\underline{x}_1 = (1, 1, 1)$:
:$${\rm Pr}(\underline{y} = \underline{y}_2 \hspace{0.1cm}| \hspace{0.1cm}\underline{x} = \underline{x}_1 ) = \varepsilon^2 \cdot (1-\varepsilon) \hspace{0.05cm}.$$

*According to Bayes' theorem,  the inference probabilities are then:
:$${\rm Pr}(\hspace{0.1cm}\underline{x} = \underline{x}_0 \hspace{0.1cm}| \hspace{0.1cm}\underline{y} = \underline{y}_2 ) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Pr}(\underline{y} = \underline{y}_2 \hspace{0.1cm}| \hspace{0.1cm}\underline{x} = \underline{x}_0 ) \cdot \frac{{\rm Pr}(\underline{x} = \underline{x}_0)}{{\rm Pr}(\underline{y} = \underline{y}_2)} \hspace{0.05cm},$$
:$${\rm Pr}(\hspace{0.1cm}\underline{x} = \underline{x}_1 \hspace{0.1cm}| \hspace{0.1cm}\underline{y} = \underline{y}_2 ) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Pr}(\underline{y} = \underline{y}_2 \hspace{0.1cm}| \hspace{0.1cm}\underline{x} = \underline{x}_1 ) \cdot \frac{{\rm Pr}(\underline{x} = \underline{x}_1)}{{\rm Pr}(\underline{y} = \underline{y}_2)} $$
:$$\Rightarrow \hspace{0.3cm} S = \frac{{\rm Pr(correct \hspace{0.15cm}decision)}}{{\rm Pr(wrong \hspace{0.15cm}decision) }} = \frac{(1-\varepsilon)^2 \cdot \varepsilon}{\varepsilon^2 \cdot (1-\varepsilon)}= \frac{(1-\varepsilon)}{\varepsilon}\hspace{0.05cm}.$$

*With  $\varepsilon = 0.269$  we get the following numerical values:
:$$S = {0.731}/{0.269}\hspace{0.15cm}\underline {= 2.717}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}{\rm ln}\hspace{0.15cm}(S)\hspace{0.15cm} \underline {= 1}\hspace{0.05cm}.$$

[[File:P_ID2988__KC_A_4_3e_v1.png|rightr|frame| Iterative decoding of  $(+1, -1, +1)$]]
'''(4)'''  The sign of the channel log likelihood ratios  $L_{\rm K}(i)$  is
*positive if  $y_i = 0$,  and

*negative for  $y_i = 1$.

The absolute value   $|L_{\rm K}(i)|$  indicates the reliability of  $y_i$. 
*In the BSC model,  $|L_{\rm K}(i)| = \ln {(1 \, – \varepsilon)/\varepsilon} = 1$  for all  $i$.  Thus:
:$$\underline {L_{\rm K}}(1)\hspace{0.15cm} \underline {= +1}\hspace{0.05cm},\hspace{0.5cm}\underline {L_{\rm K}}(2)\hspace{0.15cm} \underline {= -1}\hspace{0.05cm},\hspace{0.5cm}\underline {L_{\rm K}}(3)\hspace{0.15cm} \underline {= +1}\hspace{0.05cm}.$$

'''(5)'''  The adjacent table illustrates the iterative bit-wise decoding starting from  $\underline{y}_2 = (0, \, 1, \, 0)$.

These results can be interpreted as follows:
* The preassignment  $($iteration  $I = 0)$  happens according to  $\underline{L}_{\rm APP} = \underline{L}_{\rm K}$.  A hard decision  ⇒  "$\sign \ {\underline{L}_{\rm APP}(i)}$"  would lead to the decoding result  $(0, \, 1, \, 0)$.

*The reliability of this obviously incorrect result is given as  $|{\it \Sigma}_{\rm APP}| = 1$.  This value agrees with  $\ln (S)$  calculated in subtask  '''(3)'''.

* After the first iteration  $(I = 1)$  all a-posteriori log likelihood ratios are  $L_{\rm APP}(i) = +1$.  A hard decision here would yield the  $($expected$)$  correct result  $\underline{x}_{\rm APP} = (0, \, 0, \, 0)$. 

*The probability that this outcome is correct is quantified by  $|{\it \Sigma}_{\rm APP}| = 3$:
:$${\rm ln}\hspace{0.25cm}\frac{{\rm Pr}(\hspace{0.1cm}\underline{x} = \underline{x}_0 \hspace{0.1cm}| \hspace{0.1cm}\underline{y}=\underline{y}_2)}{1-{\rm Pr}(\hspace{0.1cm}\underline{x} = \underline{x}_0 \hspace{0.1cm}| \hspace{0.1cm}\underline{y}=\underline{y}_2)} = 3\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\frac{{\rm Pr}(\hspace{0.1cm}\underline{x} = \underline{x}_0 \hspace{0.1cm}| \hspace{0.1cm}\underline{y}=\underline{y}_2)}{1-{\rm Pr}(\hspace{0.1cm}\underline{x} = \underline{x}_0 \hspace{0.1cm}| \hspace{0.1cm}\underline{y}=\underline{y}_2)} = {\rm e}^3 \approx 20$$
:$$\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Pr}(\hspace{0.1cm}\underline{x} = \underline{x}_0 \hspace{0.1cm}| \hspace{0.1cm}\underline{y}=\underline{y}_2) = {20}/{21}{\approx 95.39\%}\hspace{0.05cm}.$$
* The second iteration confirms the decoding result of the first iteration.  The reliability is even quantified here with  $9$.  This value can be interpreted as follows:
:$$\frac{{\rm Pr}(\hspace{0.1cm}\underline{x} = \underline{x}_0 \hspace{0.1cm}| \hspace{0.1cm}\underline{y}=\underline{y}_2)}{1-{\rm Pr}(\hspace{0.1cm}\underline{x} = \underline{x}_0 \hspace{0.1cm}| \hspace{0.1cm}\underline{y}=\underline{y}_2)} = {\rm e}^9\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Pr}(\hspace{0.1cm}\underline{x} = \underline{x}_0 \hspace{0.1cm}| \hspace{0.1cm}\underline{y}=\underline{y}_2) = {{\rm e}^9}/{({\rm e}^9+1)}\approx 99.99\% \hspace{0.05cm}.$$
*With each further iteration the reliability value and thus the probability  ${\rm Pr}(\underline{x}_0 | \underline{y}_2)$  increases drastically   ⇒   All proposed solutions  are correct.

[[File:P_ID2991__KC_A_4_3f_v1.png|right|frame|Iterative decoding of  $(–1, –1, +1)$]]
'''(6)'''  Correct are  the proposed solutions 2 and 3:
*For the received vector  $\underline{y}_6 = (1, \, 1, \, 0)$,  the second table applies.

*The decoder now decides for the sequence  $\underline{x}_1 = (1, \, 1, \, 1)$.

*The case  »$\underline{y}_6 = (1, \, 1, \, 0)$  received under the condition  $\underline{x}_1 = (1, \, 1, \, 1)$  sent«  would correspond exactly to the constellation  »$\underline{y}_2 = (0, \, 1, \, 0)$  received and  $\underline{x}_0 = (0, \, 0, \, 0)$ sent«  considered in the last subtask.

*But since  $\underline{x}_0 = (0, \, 0, \, 0)$  was sent,  there are now two bit errors with the following consequence:
:* The iterative decoder decides incorrectly.

:* With each further iteration the wrong decision is declared as more reliable.

{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^4.1 Soft–in Soft–out Decoder^]]
[[de:Aufgaben:Aufgabe 4.3: Iterative Decodierung beim BSC]]

Aufgaben:Exercise 3.5: Recursive Filters for GF(2)

2026-03-16T15:58:44Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Algebraic_and_Polynomial_Description}}

[[File:P_ID2647__KC_A_3_5.png|right|frame|General recursive filter  (above) and considered realization  (below)]]
The upper of the two circuits on the right shows a second order recursive filter in general form.  With
:$$A(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} a_0 + a_1 \cdot D + a_2 \cdot D^2 \hspace{0.05cm},$$
:$$B(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 + b_1 \cdot D + b_2 \cdot D^2 $$

one obtains for the transfer function
:$$G(D) = \frac{A(D)}{B(D)} = \frac{a_0 + a_1 \cdot D + a_2 \cdot D^2}{1 + b_1 \cdot D + b_2 \cdot D^2} \hspace{0.05cm}.$$

* It should be noted that all arithmetic operations refer to  ${\rm GF(2)}$. 

* Thus,  the filter coefficients   $a_0, \ a_1, \ a_2, \ b_1, \ b_2$   are also binary $(0$  or  $1)$.

The bottom graph shows the filter specific to the exercise at hand:
#   A filter coefficient results in   $a_i = 1$   if the connection is through  $(0 ≤ i ≤ 2)$.
#   Otherwise,  $a_i = 0$.   The same system applies to the coefficients  $b_1$  and  $b_2$.

In the subtasks  '''(1)''', ... , '''(3)'''  you are to determine the respective output sequence  $\underline{x}$  for different input sequences
* $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$,

* $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$,

* $\underline{u} = (1, \, 1, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$

using the given circuit.  It should be taken into account:
* If the input sequence  $\underline{u}$  consists of a  "$1$"  followed by all zeros,  this specific output sequence  $\underline{x}$  is the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|"impulse response"]]  $\underline{g}$,  and it holds:
:$$\underline{g} \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.15cm}{G}(D)\hspace{0.05cm}. $$
* Otherwise,  the output sequence results as the  [[Signal_Representation/The_Convolution_Theorem_and_Operation|"convolution product"]]  between the input sequence and the impulse response:
:$$\underline{x} = \underline{u} * \underline{g} \hspace{0.05cm}.$$
* The convolution operation can be bypassed with the  [[Channel_Coding/Algebraic_and_Polynomial_Description#GF.282.29_description_forms_of_a_digital_filter|"D-transform"]] .

Hints:
* This exercise belongs to the chapter  [[Channel_Coding/Algebraic_and_Polynomial_Description| "Algebraic and Polynomial Description"]].

* Reference is made in particular to the section  [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function| "Filter structure with fractional–rational transfer function"]]

===Questions===
<quiz display=simple>
{What statements hold for the impulse response  $\underline{g}$  of the recursive filter?
|type="[]"}
- It holds  $\underline{g} = (0, \, 1, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
+ It holds  $\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
+ The impulse response  $\underline{g}$  is infinitely extended.

{Let now  $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, 1)$.  Which statements are true?
|type="[]"}
+ The output sequence is:  $\underline{x} = (0, \, 1, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
- The output sequence is:  $\underline{x} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
+ The output sequence  $\underline{x}$  extends to infinity.

{Now apply  $\underline{u} = (1, \, 1, \, 1)$.  Which statements are true?
|type="[]"}
+ The output sequence  $\underline{x}$  starts with  $(1, \, 0, \, 1)$.
- The output sequence  $\underline{x}$  begins with  $(1, \, 1, \, 1)$.
- The output sequence  $\underline{x}$  extends to infinity.

{What statements are valid for the transfer function  $G(D)$?
|type="[]"}
+ It holds  $G(D) = (1 + D^2)/(1 + D + D^2)$.
- It holds  $G(D) = (1 + D + D^2)/(1 + D^2)$.
+ It holds  $G(D) = 1 + D + D^2 + D^4 + D^5 + D^7 + D^8 + \hspace{0.05cm} \text{...}\hspace{0.05cm} $ .
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Correct are the  proposed solutions 2 and 3:
[[File:P_ID2643__KC_A_3_5a.png|right|frame|For calculation of the impulse response  $\underline{g}$]]
[[File:P_ID2644__KC_A_3_5b.png|right|frame|For calculation of the output sequence $\underline{x}$]]

*The impulse response  $\underline{g}$  is equal to the sequence  $\underline{x}$  for the input sequence  $\underline{u} = (1, \, 0, \, 0, \, \text{...})$. 

*Based on the filter structure,  $w_0 = w_{-1} = 0$  and the equations
:$$w_i \hspace{-0.2cm} \ = \ \hspace{-0.2cm} u_i + w_{i-1} + w_{i-2} \hspace{0.05cm},$$
:$$x_i \hspace{-0.2cm} \ = \ \hspace{-0.2cm} w_i + w_{i-2} $$

:the result is  $\underline{g} = \underline{x} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...})$   corresponding to  proposed solution 2,  as shown in the adjacent calculation.

*But additionally the  proposed solution 3  is correct,  because one recognizes from this calculation scheme further following periodicities of the impulse response  $\underline{g}$   $($up to infinity$)$  because of in each case same register assignment:

:$$g_3 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} g_6 = g_9 = \hspace{0.05cm}\text{...} \hspace{0.05cm}= 1 \hspace{0.05cm},$$
:$$g_4 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} g_7 = g_{10} =\hspace{0.05cm} \text{...} \hspace{0.05cm}= 0 \hspace{0.05cm},$$
:$$g_5 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} g_8 = g_{11} =\hspace{0.05cm} \text{...} \hspace{0.05cm}= 1 \hspace{0.05cm}.$$

 
'''(2)''' After similar calculations as in subtask  '''(1)'''  one recognizes the correctness of  solutions 1 and 3:
*The initial sequence  $\underline{x}$  also extends to infinity.  Periodicities show up again.

*The same result is obtained by adding the impulse responses   $\underline{g} = (1, \, 0, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...} \hspace{0.05cm})$   in the Galois field  ${\rm GF(2)}$  shifted by one,  three,  six,  and seven positions  $($to the right,  respectively$)$:
:$$\underline{x} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm} + \hspace{0.05cm} (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm} + \hspace{0.05cm} (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm} + \hspace{0.05cm} (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) $$
:$$\Rightarrow \hspace{0.3cm}\underline{x} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm}. $$

*Due to the linearity of the system under consideration,  this is allowed.

'''(3)'''  Here we choose the way over the D–transforms:
:$$\underline{u}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quadU(D) = 1+ D + D^2 \hspace{0.05cm}.$$

*With the transfer function   $G(D) = (1 + D^2)/(1 + D + D^2)$   one thus obtains for the D–transform of the output sequence:
:$$X(D) = {U(D)} \cdot G(D) = {1+D+D^2} \cdot \frac{1+D^2}{1+D+D^2} = 1+D^2 \hspace{0.05cm}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\underline{x} = (1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.01cm} \text{...}\hspace{0.05cm} \hspace{0.05cm} ) \hspace{0.05cm}.$$

*Only the  proposed solution 1  is correct here:  Despite infinitely long impulse response  $\underline{g}$,  for this input sequence  $\underline{u}$  the output sequence  $\underline{x}$  is limited to three bits.

*The same result is again obtained by adding shifted impulse responses:
:$$\underline{x} = (1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) + (0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) + (0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) = (1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm}. $$

'''(4)'''  Correct are the  proposed solutions 1 and 3:
*On the data sheet,  the general transfer function of a second–order recursive filter is given as follows.
[[File:P_ID2645__KC_A_3_5e.png|right|frame|$\rm GF(2)$  polynomial division  $(1 + D^2)/(1 + D + D^2)$]]
:$$G(D) = \frac{a_0 + a_1 \cdot D + a_2 \cdot D^2}{1 + b_1 \cdot D + b_2 \cdot D^2} \hspace{0.05cm}.$$

*The filter considered here is determined by the coefficients  $a_0 = a_2 = b_1 = b_2 = 1$  and  $a_1 = 0$.  Thus one obtains the result according to the  solution suggestion 1:
:$$G(D) = \frac{1 + D^2}{1 + D + D^2} \hspace{0.05cm}. $$
*At the same time,  $G(D)$  is also the D–transformed of the impulse response:
:$$\underline{g}= (1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}0 ,\hspace{0.05cm} \text{ ...}\hspace{0.05cm}) \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.15cm}{G}(D)$$
:$$\Rightarrow\hspace{0.3cm}{G}(D)= 1 + D + D^2 + D^4+ D^5 +\text{...}\hspace{0.1cm}. $$
*This means:  The  proposed solution 3  is also correct.

*The same result would have been obtained by dividing the two polynomials  $1 + D^2$  and  $1 + D + D^2$,  as the calculation opposite shows.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^3.2 Polynomial Description^]]
[[de:Aufgaben:Aufgabe 3.5: Rekursive Filter für GF(2)]]

Aufgaben:Exercise 3.5Z: Kullback-Leibler Distance again

2026-03-16T15:58:43Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Information_Theory/Some_Preliminary_Remarks_on_Two-Dimensional_Random_Variables
}}

[[File:P_ID2762__Inf_Z_3_4.png|right|frame|Determined probability mass functions]]
The probability mass function is:
:$$P_X(X) = \big[\hspace{0.03cm}0.25\hspace{0.03cm}, \hspace{0.15cm} 0.25\hspace{0.15cm},\hspace{0.15cm} 0.25 \hspace{0.03cm}, \hspace{0.15cm} 0.25\hspace{0.03cm}\big]\hspace{0.05cm}.$$
The random variable  $X$  is thus characterised by
* the symbol set size  $M=4$,
* equal probabilities  $P_X(1) = P_X(2) = P_X(3) = P_X(4) = 1/4$ .

The random variable  $Y$  is always an approximation for  $X$:
*It was obtained by simulation from a uniform distribution, whereby only  $N$  random numbers were evaluated in each case.  This means:  
*$P_Y(1)$, ... , $P_Y(4)$  are not probabilities in the conventional sense.  Rather, they describe  [[Theory_of_Stochastic_Signals/Wahrscheinlichkeit_und_relative_H%C3%A4ufigkeit#Bernoullisches_Gesetz_der_gro.C3.9Fen_Zahlen| relative frequencies]].

The result of the sixth test series  (with  $N=1000)$  is thus summarised by the following probability function:

:$$P_Y(X) = \big [\hspace{0.05cm}0.225\hspace{0.15cm}, \hspace{0.05cm} 0.253\hspace{0.05cm},\hspace{0.15cm} 0.250 \hspace{0.05cm}, \hspace{0.15cm} 0.272\hspace{0.05cm}\big]\hspace{0.05cm}.$$
This notation already takes into account that the random variables  $X$  and  $Y$  are based on the same alphabet  $X = \{1,\ 2,\ 3,\ 4\}$.

With these preconditions, the  '''informational divergence'''  between the two probability functions  $P_X(.)$  and  $P_Y(.)$ :

:$$D(P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) = {\rm E}_X \hspace{-0.1cm}\left [ {\rm log}_2 \hspace{0.1cm} \frac{P_X(X)}{P_Y(X)}\right ] \hspace{0.2cm}=\hspace{0.2cm} \sum_{\mu = 1}^{M} P_X(\mu) \cdot {\rm log}_2 \hspace{0.1cm} \frac{P_X(\mu)}{P_Y(\mu)} \hspace{0.05cm}.$$

One calls  $D( P_X\hspace{0.05cm} || \hspace{0.05cm}P_Y)$  the (first)  '''Kullback-Leibler distance'''.
*This is a measure of the similarity between the two probability mass functions  $P_X(.)$  and  $P_Y(.)$.
*The expected value formation occurs here with regard to the  (actually equally distributed)  random variable  $X$.  This is indicated by the nomenclature  ${\rm E}_X\big[.\big]$.

A second form of Kullback-Leibler distance results from the formation of expected values with respect to the random variable  $Y$   ⇒   ${\rm E}_Y\big [.\big ]$:

:$$D(P_Y \hspace{0.05cm}|| \hspace{0.05cm} P_X) = {\rm E}_Y \hspace{-0.1cm} \left [ {\rm log}_2 \hspace{0.1cm} \frac{P_Y(X)}{P_X(X)}\right ] \hspace{0.2cm}=\hspace{0.2cm} \sum_{\mu = 1}^M P_Y(\mu) \cdot {\rm log}_2 \hspace{0.1cm} \frac{P_Y(\mu)}{P_X(\mu)} \hspace{0.05cm}.$$

Hints:
*The exercise belongs to the chapter  [[Information_Theory/Some_Preliminary_Remarks_on_Two-Dimensional_Random_Variables|Some preliminary remarks on two-dimensional random variables]].
*In particular, reference is made to the page  [[Information_Theory/Some_Preliminary_Remarks_on_Two-Dimensional_Random_Variables#Informational_divergence_-_Kullback-Leibler_distance|Relative entropy – Kullback-Leibler distance]].
*The entropy  $H(Y)$  and the Kullback-Leibler distance  $D( P_X \hspace{0.05cm}|| \hspace{0.05cm}P_Y)$  in the above graph are to be understood in  "bit".
* The fields marked with  "???"  in the graph are to be completed by you in this task.

===Questions===

<quiz display=simple>

{What is the entropy of the random variable  $X$ ?
|type="{}"}
$H(X)\ = \ $ { 2 1% } $\ \rm bit$

{What are the entropies of the random variables  $Y$  $($approximations for  $X)$?
|type="{}"}
$N=10^3\text{:} \hspace{0.5cm} H(Y) \ = \ $ { 1.9968 1% } $\ \rm bit$
$N=10^2\text{:} \hspace{0.5cm} H(Y) \ = \ $ { 1.941 1% } $\ \rm bit$
$N=10^1\text{:} \hspace{0.5cm} H(Y) \ = \ $ { 1.6855 1% } $\ \rm bit$

{Calculate the following Kullback-Leibler distances.
|type="{}"}
$N=10^3\text{:} \hspace{0.5cm} D( P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) \ = \ $ { 0.00328 1% } $\ \rm bit$
$N=10^2\text{:} \hspace{0.5cm} D( P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) \ = \ $ { 0.0442 1% } $\ \rm bit$
$N=10^1\text{:} \hspace{0.5cm} D( P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) \ = \ $ { 0.345 1% } $\ \rm bit$

{Does  $D(P_Y\hspace{0.05cm}|| \hspace{0.05cm} P_X)$  give exactly the same result in each case?
|type="()"}
- Yes.
+ No.

{Which statements are true for the Kullback-Leibler distances with  $N = 4$?
|type="[]"}
-   $D(P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) = 0$  is true.
-   $D(P_X\hspace{0.05cm}|| \hspace{0.05cm} P_Y) = 0.5 \ \rm bit$  is true.
+   $D(P_X\hspace{0.05cm}|| \hspace{0.05cm} P_Y)$  is infinitely large.
-   $D(P_Y\hspace{0.05cm}|| \hspace{0.05cm} P_X) = 0$  holds.
+   $D(P_Y\hspace{0.05cm}|| \hspace{0.05cm} P_X) = 0.5 \ \rm bit$  holds.
-   $D(P_Y\hspace{0.05cm}|| \hspace{0.05cm} P_X)$  is infinitely large.

{Do both  $H(Y)$  and  $D(P_X\hspace{0.05cm}|| \hspace{0.05cm} P_Y)$  change monotonically with  $N$?
|type="()"}
- Yes,
+ No.

</quiz>

===Solution===
{{ML-Kopf}}

'''(1)'''  With equal probabilities, and with  $M = 4$:
:$$H(X) = {\rm log}_2 \hspace{0.1cm} M\hspace{0.15cm} \underline {= 2\,{\rm (bit)}} \hspace{0.05cm}.$$

'''(2)'''  The probabilities for the empirically determined random variables  $Y$  generally  (not always!)  deviate from the uniform distribution the more the parameter  $N$  is smaller.  One obtains for
* $N = 1000 \ \ \Rightarrow \ \ P_Y(Y) = \big [0.225, \ 0.253, \ 0.250, \ 0.272 \big ]$:
:$$H(Y) =0.225 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.225} +0.253 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.253} +0.250 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.250} +0.272 \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{0.272}\hspace{0.15cm} \underline {= 1.9968\ {\rm (bit)}} \hspace{0.05cm},$$
* $N = 100 \ \ \Rightarrow \ \ P_Y(Y) = \big[0.24, \ 0.16, \ 0.30, \ 0.30\big]$:
:$$H(Y) = \hspace{0.05cm}\text{...} \hspace{0.15cm} \underline {= 1.9410\ {\rm (bit)}} \hspace{0.05cm},$$
* $N = 10 \ \ \Rightarrow \ \ P_Y(Y) = \big[0.5, \ 0.1, \ 0.3, \ 0.1 \big]$:
:$$H(Y) = \hspace{0.05cm}\text{...} \hspace{0.15cm} \underline {= 1.6855\ {\rm (bit)}} \hspace{0.05cm}.$$

'''(3)'''  The equation for the Kullback-Leibler distance we are looking for is:

:$$D(P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) = \sum_{\mu = 1}^{4} P_X(\mu) \cdot {\rm log}_2 \hspace{0.1cm} \frac{P_X(\mu)}{P_Y(\mu)}= \frac{1/4}{{\rm lg} \hspace{0.1cm}(2)} \cdot\left [ {\rm lg} \hspace{0.1cm} \frac{0.25}{P_Y(1)} + \frac{0.25}{P_Y(2)} + \frac{0.25}{P_Y(3)} + \frac{0.25}{P_Y(4)}\right ] $$
:$$\Rightarrow \hspace{0.3cm} D(P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) = \frac{1}{4 \cdot {\rm lg} \hspace{0.1cm}(2)} \cdot\left [ {\rm lg} \hspace{0.1cm} \frac{0.25^4}{P_Y(1) \cdot P_Y(2)\cdot P_Y(3)\cdot P_Y(4)}\right ] \hspace{0.05cm}.$$

The logarithm to the base  $ 2$  ⇒   $\log_2(.)$  was replaced by the logarithm to the base  $ 10$   ⇒   $\lg(.)$ for easy use of the calculator.

The following numerical results are obtained:
* for $N=1000$:
:$$D(P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) = \frac{1}{4 \cdot {\rm lg} \hspace{0.1cm}(2)} \cdot\left [ {\rm lg} \hspace{0.1cm} \frac{0.25^4}{0.225 \cdot 0.253\cdot 0.250\cdot 0.272}\right ] \hspace{0.15cm} \underline {= 0.00328 \,{\rm (bit)}} \hspace{0.05cm},$$
* for $N=100$:
:$$D(P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) = \frac{1}{4 \cdot {\rm lg} \hspace{0.1cm}(2)} \cdot\left [ {\rm lg} \hspace{0.1cm} \frac{0.25^4}{0.24 \cdot 0.16\cdot 0.30\cdot 0.30}\right ] \hspace{0.15cm} \underline {= 0.0442 \,{\rm (bit)}} \hspace{0.05cm},$$
* for $N=10$:
:$$D(P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) = \frac{1}{4 \cdot {\rm lg} \hspace{0.1cm}(2)} \cdot\left [ {\rm lg} \hspace{0.1cm} \frac{0.25^4}{0.5 \cdot 0.1\cdot 0.3\cdot 0.1}\right ] \hspace{0.15cm} \underline {= 0.345 \,{\rm (bit)}} \hspace{0.05cm}.$$

'''(4)'''  Correct is  '''No''',  as will be shown by the example  $N = 100$ :
:$$D(P_Y \hspace{0.05cm}|| \hspace{0.05cm} P_X) = \sum_{\mu = 1}^M P_Y(\mu) \cdot {\rm log}_2 \hspace{0.1cm} \frac{P_Y(\mu)}{P_X(\mu)} = 0.24\cdot {\rm log}_2 \hspace{0.1cm} \frac{0.24}{0.25} + 0.16\cdot {\rm log}_2 \hspace{0.1cm} \frac{0.16}{0.25} +2 \cdot 0.30\cdot {\rm log}_2 \hspace{0.1cm} \frac{0.30}{0.25} = 0.0407\ {\rm (bit)}\hspace{0.05cm}.$$

*In subtask  '''(3)'''  we got  $D(P_X\hspace{0.05cm}|| \hspace{0.05cm} P_Y) = 0.0442$  instead.
*This also means:   The designation „distance” is somewhat misleading.
*According to this, one would actually expect  $D(P_Y\hspace{0.05cm}|| \hspace{0.05cm} P_X) = D(P_X\hspace{0.05cm}|| \hspace{0.05cm} P_Y)$ .

[[File:P_ID2763__Inf_Z_3_4e.png|right|frame|Probability function, entropy and Kullback-Leibler distance]]
'''(5)'''  With  $P_Y(X) = \big [0, \ 0.25, \ 0.5, \ 0.25 \big ]$  one obtains:
:$$D(P_X \hspace{0.05cm}|| \hspace{0.05cm} P_Y) = 0.25\cdot {\rm log}_2 \hspace{0.1cm} \frac{0.25}{0} + 2 \cdot 0.25\cdot {\rm log}_2 \hspace{0.1cm} \frac{0.25}{0.25}+0.25\cdot {\rm log}_2 \hspace{0.1cm} \frac{0.25}{0.50}\hspace{0.05cm}.$$

*Because of the first term, the value of  $D(P_X\hspace{0.05cm}|| \hspace{0.05cm}P_Y)$  is infinitely large.
*For the second Kullback-Leibler distance holds:
:$$D(P_Y \hspace{0.05cm}|| \hspace{0.05cm} P_X) = 0\cdot {\rm log}_2 \hspace{0.1cm} \frac{0}{0.25} + 2 \cdot 0.25\cdot {\rm log}_2 \hspace{0.1cm} \frac{0.25}{0.25}+0.50\cdot {\rm log}_2 \hspace{0.1cm} \frac{0.5}{0.25}\hspace{0.05cm}.$$

*After looking at the limits, one can see that the first term yields the result  $0$ .  The second term also yields zero, and one obtains as the final result:
:$$D(P_Y \hspace{0.05cm}|| \hspace{0.05cm} P_X) = 0.50\cdot {\rm log}_2 \hspace{0.1cm} (2) \hspace{0.15cm} \underline {= 0.5\,{\rm (bit)}} \hspace{0.05cm}.$$

  Statements 3 and 5 are therefore correct:
*From this extreme example it is clear that  $D(P_Y\hspace{0.05cm}|| \hspace{0.05cm} P_X)$  is always different from  $D(P_X\hspace{0.05cm}|| \hspace{0.05cm} P_Y)$ .
*Only for the special case  $P_Y \equiv P_X$  are both Kullback-Leibler distances equal, namely zero.
*The adjacent table shows the complete result of this task.

'''(6)'''  Correct is again  '''No'''.   Although the tendency is clear:   The larger  $N$  is,
* the more  $H(Y)$  approaches in principle the final value  $H(X) = 2 \ \rm bit$,
* the smaller the distances  $D(P_X\hspace{0.05cm}|| \hspace{0.05cm} P_Y)$  and  $D(P_Y\hspace{0.05cm}|| \hspace{0.05cm} P_X)$ become.

However, one can also see from the table that there are exceptions:
* The entropy  $H(Y)$  is smaller for  $N = 1000$  than for  $N = 400$.
* The distance  $D(P_X\hspace{0.05cm}|| \hspace{0.05cm}P_Y)$  is greater for  $N = 1000$  than for  $N = 400$.
* The reason for this is that the experiment documented here with  $N = 400$  was more likely to lead to a uniform distribution than the experiment with  $N = 1000$.
*If, on the other hand, one were to start a very (infinitely) large number of experiments with  $N = 400$  and  $N = 1000$  and average over all of them, the actually expected monotonic course would actually result.

{{ML-Fuß}}

[[Category:Information Theory: Exercises|^3.1 General Information on 2D Random Variables^]]
[[de:Aufgaben:Aufgabe 3.5Z: Nochmals Kullback-Leibler-Distanz]]

Aufgaben:Exercise 2.11: Reed-Solomon Decoding according to "Erasures"

2026-03-16T15:58:43Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Reed-Solomon_Decoding_for_the_Erasure_Channel}}

[[File:EN_KC_Z_2_5_neu.png|right|frame|${\rm GF}(2^3)$, represented as powers, polynomials, coefficient vectors]]
We consider here an encoding and decoding system corresponding to the  [[Channel_Coding/Reed-Solomon_Decoding_for_the_Erasure_Channel#Block_diagram_and_requirements_for_RS_fault_detection| "graph in the theory section for this chapter"]]. 

* The Reed–Solomon code is given by the generator matrix  $\mathbf{G}$  and the parity-check matrix  $\mathbf{H}$  where all elements are from the Galois field  $\rm GF(2^3) \ \backslash \ \{0\}$:
:$${ \boldsymbol{\rm G}} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}\begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 & 1\\1 & \alpha^1 & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6\\1 & \alpha^2 & \alpha^4 & \alpha^6 & \alpha^1 & \alpha^{3} & \alpha^{5}\\1 & \alpha^3 & \alpha^6 & \alpha^2 & \alpha^{5} & \alpha^{1} & \alpha^{4}\end{pmatrix} \hspace{0.05cm},$$:$${ \boldsymbol{\rm H}} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}\begin{pmatrix}1 & \alpha^1 & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6\\1 & \alpha^2 & \alpha^4 & \alpha^6 & \alpha^1 & \alpha^{3} & \alpha^{5}\\1 & \alpha^3 & \alpha^6 & \alpha^2 & \alpha^{5} & \alpha^{1} & \alpha^{4}\end{pmatrix} \hspace{0.05cm}.$$

* All code symbols   $c_i ∈ \{0, \, 1, \, \alpha, \, \alpha^2, \, \alpha^3, \, \alpha^4, \, \alpha^5, \, \alpha^6\}$   are represented by  $m = 3$  bits and transmitted via the erasure channel  $(m–\rm BEC)$.  A code symbol is already marked as an erasure  $\rm E$  if one of the three associated bits is uncertain.

* The  "code word finder"  $\rm (CWF)$  has the task of generating the regenerated code word  $\underline{z}$  from the partially erased received word  $\underline{y}$.  It must be ensured that the result  $\underline{z}$  is indeed a valid Reed–Solomon code word.

* If the received word  $\underline{y}$  contains too many erasures,  the decoder outputs a message of the type  "symbol cannot be decoded".  So no attempt is made to estimate the code word.  If  $\underline{z}$  is output,  this is also correct:  $\underline{z} = \underline{c}$.

* The sought information value  $\underline{v} = \underline{u}$  results from the inverse encoder function   $\underline{v} = {\rm enc}^{-1}(\underline{z})$.  With the generator matrix  $\mathbf{G}$  this can be realized as follows:
:$$\underline{c} = {\rm enc}(\underline{u}) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \underline{u} \cdot {\boldsymbol{\rm G}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{z} = {\rm enc}(\underline{v})= \underline{v} \cdot {\boldsymbol{\rm G}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \underline{v} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm enc}^{-1}(\underline{z}) = \underline{z} \cdot {\boldsymbol{\rm G}}^{\rm T}\hspace{0.05cm}.$$

Hints:
* This exercise refers to the chapter  [[Channel_Coding/Reed-Solomon_Decoding_for_the_Erasure_Channel| "Reed–Solomon Decoding for the Erasure Channel"]].

* Regarding the code word finder,  we refer in particular to the pages 
::[[Channel_Coding/Reed-Solomon_Decoding_for_the_Erasure_Channel#Decoding_procedure_using_the_RSC_.287.2C_3.2C_5.298_as_an_example| "Decoding procedure ..."]],  and 
::[[Channel_Coding/Reed-Solomon_Decoding_for_the_Erasure_Channel#Solution_of_the_matrix_equations_using_the_example_of_the_RSC_.287.2C_3.2C_5.298| "Solution of the matrix equations ..."]].

* All calculations are to be performed in  $\rm GF(2^3)$.  The upper graph describes their  $q = 8$  elements in power, polynomial and coefficient vector representation.

===Questions===
<quiz display=simple>
{Specify the code parameters of the present Reed–Solomon code.
|type="{}"}
$n \ = \ ${ 7 }
$k \ = \ ${ 4 }
$d_{\rm min} \ = \ ${ 4 }

{Can the received vector   $\underline{y} = (0, \, 0, \, 0,\, 0, \, 0, \, 0, \, {\rm E})$   be decoded??
|type="()"}
+ YES.
- NO.

{Can the received vector   $\underline{y} = (\rm E, \, E, \, 1, \, 1, \, 1, \, 1, \, 1)$   be decoded?
|type="()"}
+ YES.
- NO.

{What is the result of decoding   "$\underline{y} = (\rm E, \, E, \, E, \, 0, \, 1, \, \alpha, \, 0)$"?
|type="()"}
- $z_0 = \alpha, \ z_1 = \alpha^3, \ z_2 = 0$.
+ $z_0 = \alpha, \ z_1 = \alpha^3, \ z_2 = \alpha^3$.
- $z_0 = 1, \ z_1 = 0, \ z_2 = \alpha^3$.
- The decoding does not lead to any result.

{What is the result of decoding   "$\underline{y} = (\rm E, \, E, \, E, \, 0, \, 1, \, \alpha, \, E)$"?
|type="()"}
- $z_0 = \alpha, \ z_1 = \alpha^3, \ z_2 = 0, \ z_6 = 1$.
- $z_0 = \alpha, \ z_1 = \alpha^3, \ z_2 = \alpha^3, \ z_6 = 1$.
- $z_0 = 1, \ z_1 = 0, \ z_2 = \alpha^3, \ z_6 = 1$.
+ The decoding does not lead to any result.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The number of columns of the parity-check matrix  $\mathbf{H}$  indicates the code length: $n \ \underline{= 7}$.
*The same result is obtained if we assume the order  $q = 8$  of the Galois field.  For the Reed–Solomon codes  $n = q - 1$  is valid.

*The number of rows of the parity-check matrix is equal to  $n - k = 3 \ \Rightarrow \ k \ \underline{= 4}$.

*Of all Reed–Solomon codes,  the  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes#Singleton_bound_and_minimum_distance| "Singleton bound"]] is satisfied   ⇒   $d_{\rm min} = n - k + 1 \ \underline{= 4}$.

*Thus,  it is the Reed–Solomon code  $(7, \, 4, \, 4)_8$.

'''(2)'''  Decoding is certainly possible as long as the number  $e$  of erasures is smaller than the minimum distance  $d_{\rm min}$.  This condition is fulfilled here   ⇒   YES.
*Since the null word is allowed in all Reed–Solomon codes and every other code word contains at least four symbols  "$\ne 0$",  it is already certain without calculation that the null word was sent.

*The formal calculation confirms this result:
:$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline {z}_{\rm K}^{\rm T} =\begin{pmatrix}0\\0\\0\end{pmatrix} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\begin{pmatrix}\alpha^6\\\alpha^{5}\\\alpha^{4}\end{pmatrix} \cdot z_6 =\begin{pmatrix}0\\0\\0\end{pmatrix} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_6 = 0\hspace{0.05cm}. $$'''(3)'''  Again  $e = 2$  is smaller than  $d_{\rm min} = 4$   ⇒   YES.*Since  $(1, \, 1, \, 1, \, 1, \, 1, \, 1, \, 1)$  is also a valid code word,  we expect  $z_0 = 1$  und  $z_1 = 1$  in the formal verification.:$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline {z}_{\rm K}^{\rm T} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}\begin{pmatrix}\alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6\\\alpha^4 & \alpha^6 & \alpha^1 & \alpha^{3} & \alpha^{5}\\\alpha^6 & \alpha^2 & \alpha^{5} & \alpha^{1} & \alpha^{4}\end{pmatrix} \cdot\begin{pmatrix}1\\1\\1\\1\\1\end{pmatrix}=\begin{pmatrix}\alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6\\\alpha^4 + \alpha^6 + \alpha^1 + \alpha^{3} + \alpha^{5}\\\alpha^6 + \alpha^2 + \alpha^{5} + \alpha^{1} + \alpha^{4}\end{pmatrix}$$:$$\Rightarrow \hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm K} \cdot \underline {z}_{\rm K}^{\rm T} \hspace{-0.15cm} \ = \\begin{pmatrix}(100) + (011) + (110) + (111) + (101)\\(110) + (101) + (010) + (011) + (111)\\(101) + (100) + (111) + (010) + (110)\end{pmatrix} =\begin{pmatrix}(011)\\(101)\\(010)\end{pmatrix} =\begin{pmatrix}\alpha^3\\\alpha^6\\\alpha^1\end{pmatrix}\hspace{0.05cm}. $$*In this calculation,  we varied between the polynomial representation and the coefficient representation on the data side.  Thus the system of equations reads::$$\begin{pmatrix}(001) + (010) \\(001) + (100)\\(001) + (011)\end{pmatrix} \cdot\begin{pmatrix}z_0\\z_1\end{pmatrix} =\begin{pmatrix}(011)\\(101)\\(010)\end{pmatrix} \hspace{0.25cm} \Rightarrow \hspace{0.25cm}\begin{pmatrix}(001) + (010) \\(001) + (100)\\(000) + (111)\end{pmatrix} \cdot\begin{pmatrix}z_0\\z_1\end{pmatrix} =\begin{pmatrix}(011)\\(101)\\(111)\end{pmatrix}\hspace{0.05cm}.$$*The second form is obtained by substituting the third row from the modulo-2 sum of rows 2 and 3.*From the last row now follows  $z_1 = 1$  and the rows 1 and 2 are then::$$(1)\hspace{0.3cm}z_0 + (010) \cdot 1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (011)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}z_0 = (001) = 1\hspace{0.05cm},$$:$$(2)\hspace{0.3cm}z_0 + (100) \cdot 1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (101)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}z_0 = (001) = 1\hspace{0.05cm}. $$*Both equations lead to the same result  $z_0 = 1, \ z_1 = 1$.  The decoding is successful.'''(4)'''  The decoding happens on the following steps::$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline {z}_{\rm K}^{\rm T} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}\begin{pmatrix}\alpha^3 & \alpha^4 & \alpha^5 & \alpha^6\\\alpha^6 & \alpha^1 & \alpha^{3} & \alpha^{5}\\\alpha^2 & \alpha^{5} & \alpha^{1} & \alpha^{4}\end{pmatrix} \cdot\begin{pmatrix}0\\1\\\alpha\\0\end{pmatrix}=\begin{pmatrix}\alpha^4 + \alpha^6\\\alpha^1 + \alpha^{4}\\\alpha^5 + \alpha^2\end{pmatrix}=\begin{pmatrix}(110) + (101)\\(010) + (110)\\(111) + (100)\end{pmatrix} =\begin{pmatrix}(011)\\(100)\\(011)\end{pmatrix}\hspace{0.05cm},$$:$${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline {z}_{\rm E}^{\rm T} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}\begin{pmatrix}1 & \alpha^1 & \alpha^2\\1 & \alpha^2 & \alpha^4\\1 & \alpha^3 & \alpha^6\end{pmatrix} \cdot\begin{pmatrix}z_0\\z_1\\z_2\end{pmatrix}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\begin{pmatrix}(001) &(010) &(100)\\(001) &(100) &(110)\\(001) &(011) &(101)\end{pmatrix} \cdot\begin{pmatrix}z_0\\z_1\\z_2\end{pmatrix}=\begin{pmatrix}(011)\\(100)\\(011)\end{pmatrix}\hspace{0.05cm}. $$*We now replace row 2 with the modulo-2 sum of rows 1 and 2, and row 3 with the modulo-2 sum of rows 1 and 3::$$\begin{pmatrix}(001) &(010) &(100)\\(000) &(110) &(010)\\(000) &(001) &(001)\end{pmatrix} \cdot\begin{pmatrix}z_0\\z_1\\z_2\end{pmatrix}=\begin{pmatrix}(011)\\(111)\\(000)\end{pmatrix}\hspace{0.05cm}.$$

*From the last row follows  $z_1 + z_2 = 0 \ \Rightarrow \ z_2 = z_1$.  Substituting into the second row of this matrix equation we get:
:$$\big[(110) + (010)\big] \cdot z_1 = (100) \cdot z_1 = (111) \hspace{0.2cm} \Rightarrow \hspace{0.2cm}\alpha^2 \cdot z_1 = \alpha^5\hspace{0.2cm}\Rightarrow \hspace{0.2cm}z_1 \hspace{0.1cm}\underline{= \alpha^3}\hspace{0.05cm},\hspace{0.2cm}z_2 \hspace{0.1cm}\underline{= \alpha^3}\hspace{0.05cm}. $$

*With this result follows from the first matrix row:
:$$z_0 + \big[(010) + (100)\big] \cdot z_1 = z_0 + (110) \cdot z_1 = (011) $$
:$$\Rightarrow \hspace{0.2cm} z_0 + \alpha^4 \cdot \alpha^3 = z_0 + 1 = \alpha^3\hspace{0.2cm} \Rightarrow \hspace{0.2cm}z_0 = \alpha^3 + 1 = ( \alpha + 1) +1\hspace{0.15cm} \underline{= \alpha}\hspace{0.05cm}.$$

*The correct solution is therefore  proposed solution 2.

'''(5)'''  Correct is the  proposed solution 4.   Reasons:
* Four information symbols cannot be obtained from the three known symbols  $0, \, 1, \, \alpha$.

*The  $\mathbf{H}$  matrix of this  $(7, \, 4, \, 4)_8$  code has exactly  $n - k = 3$  rows.

*This also means that you have only three equations.  But you would need four equations for the unknowns  $z_0, \ z_1, \ z_2$  and  $z_6$.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^2.4 Reed-Solomon Decoding at the BEC^]]
[[de:Aufgaben:Aufgabe 2.11: RS–Decodierung nach „Erasures”]]

Aufgaben:Exercise 4.4Z: Supplement to Exercise 4.4

2026-03-16T15:58:42Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}

[[File:P_ID2994__KC_Z_4_4_v3.png |right|frame|Hamming weights  $w_{\rm H}(\underline{x})$, sequence probabilities  ${\rm Pr}(\underline{x})$ ]]
The information theorist  [https://en.wikipedia.org/wiki/Robert_G._Gallager $\text{Robert G. Gallager}$]  dealt already in 1963 with the following problem:
* Given a random vector  $\underline{x} = (x_1, \, x_2, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm} , x_n)$  with  $n$  binary elements  $x_i ∈ \{0, \, 1\}$.

* Known are all probabilities  $p_i = {\rm Pr}(x_i = 1)$  and  $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$  with index  $i = 1, \hspace{-0.04cm}\text{ ...} \hspace{0.08cm} ,\ n$.

* What we are looking for is the probability that the number of  "ones"  in this vector is even.

* Or expressed using the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"Hamming weight"]]:   What is the probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$?

The graph illustrates the task for the example  $n = 4$  and  $p_1 = 0.2$,  $p_2 = 0.9$,  $p_3 = 0.3$  and  $p_4 = 0.6$.
* For the row highlighted in green   ⇒   $\underline{x} = (1, \, 0, \, 0, \, 1)$  holds  $w_{\rm H}(\underline{x}) = 2$  and
:$${\rm Pr}(\underline{x}) = p_1 \cdot q_2 \cdot q_3 \cdot p_4 = 0.0084.$$
* Blue font means  "$w_{\rm H}(\underline{x})$  is even".  Red font stands for  "$w_{\rm H}(\underline{x})$  is odd."

*The probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$  is the sum of the blue numbers in the last column.

*The sum of the red numbers gives  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ odd}] = 1 - {\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$.

Gallager solved the problem in an analytical way:
:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \right ]\hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 + \pi]\hspace{0.05cm},$$
:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right ] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 - \pi]\hspace{0.05cm}.$$

Here the following auxiliary variable is used:
:$$\pi = \prod\limits_{i =1}^{n} \hspace{0.25cm}(1-2p_i)\hspace{0.05cm}.$$

One applies the equation,  for example,  to calculate the extrinsic  L–values of a  "single parity–check code".

Indeed,  as already pointed out in  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| $\text{Exercise 4.4}$]]  the extrinsic log likelihood ratio with Hamming–weight  $w_{\rm H}$  is the truncated sequence  $\underline{x}^{(-i)}$:
:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}\hspace{0.05cm}.$$

Here it is taken into account that for  $L_{\rm E}(i)$  one may refer only to the other symbols  $(j ≠ i)$ :
:$$\underline{x}^{(-i)} = \big ( \hspace{0.03cm}x_1, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm} , \hspace{0.03cm} x_{i-1}, \hspace{0.43cm} x_{i+1}, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm} , x_{n} \hspace{0.03cm} \big )\hspace{0.03cm}. $$

Hints:
*This exercise belongs to the chapter  [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder|"Soft–in Soft–out Decoder"]].

*Reference is made in particular to the section  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_log_likelihood_ratios|"For calculating the extrinsic log likelihood ratios"]].

*The exercise is intended as a supplement to  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC| $\text{Exercise 4.4}$]] .

===Questions===
<quiz display=simple>
{We consider the vector  $\underline{x} = (x_1, \, x_2) \ \Rightarrow \ n = 2$  with  $x_i ∈ \{0, \, 1\}$  and  $p_1 = 0.2, \ p_2 = 0.9$. What is the probability that   $\underline{x}$   contains an even number of  "ones" ?
|type="{}"}
${\rm Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \ = \ ${ 0.26 3% }

{Compute the same probability for  $\underline{x} = (x_1, \, x_2, \, x_3) \ \Rightarrow \ n = 3$  and  $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3$.
|type="{}"}
${\rm Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \ = \ ${ 0.404 3% }

{Now let be  $n = 4$  and  $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6$.  Calculate the following quantities according to Gallager's equation:
|type="{}"}
${\rm Pr(blue) = Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \hspace{0.33cm} = \ ${ 0.5192 3% }
${\rm Pr(red) = Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ odd}\big ] \hspace{0.1cm} = \ ${ 0.4808 3% }
$\text{Quotient }Q = {\rm Pr(blau)/Pr(rot)} \hspace{0.4cm} = \ ${ 1.0799 3% }

{What is the extrinsic log likelihood ratio for the symbol  $i = 5$  at  $\text{SPC (5, 4, 2)}$  with  $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6, \ p_5 = 0.9$?
|type="{}"}
$L_{\rm E}(i = 5) \ = \ ${ 0.077 3% }

{How does  $L_{\rm E}(i = 5)$  change if we assume  $p_5 = 0.1$  instead?
|type="()"}
- $L_{\rm E}(i = 5)$  becomes larger.
- $L_{\rm E}(i = 5)$  becomes smaller.
+ $L_{\rm E}(i = 5)$  is not changed compared to subtask  '''(4)'''.
</quiz>

===Solution===
{{ML-Kopf}}
[[File:P_ID2996__KC_Z_4_4a_v1.png|frame|right|Derivation  "$w_{\rm H}$ is even"  for code length  $n = 2$.]]
'''(1)'''  According to the adjacent table applies:
:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.10cm}{\rm is \hspace{0.10cm} even}\right ] ={\rm Pr} \left [w_{\rm H} = 0 \right] + {\rm Pr} \left [w_{\rm H} = 2 \right]\hspace{0.05cm}. $$

*With the probabilities
:$$p_1 = {\rm Pr} (x_1 = 1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.2\hspace{0.05cm},\hspace{0.3cm}q_1 = {\rm Pr} (x_1 = 0) = 0.8\hspace{0.05cm},$$
:$$p_2 = {\rm Pr} (x_2 = 1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.9\hspace{0.05cm},\hspace{0.3cm}q_2 = {\rm Pr} (x_2 = 0) = 0.1$$

:one obtains:
:$${\rm Pr} \left [w_{\rm H}(\underline{x}) = 0\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm Pr} \left [(x_1 = 0)\cap (x_2 = 0) \right] = q_1 \cdot q_2 = 0.8 \cdot 0.1= 0.08 \hspace{0.05cm},$$
:$${\rm Pr} \left [w_{\rm H}(\underline{x}) = 2\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm Pr} \left [(x_1 = 1)\cap (x_2 = 1) \right] = p_1 \cdot p_2 = 0.2 \cdot 0.9 = 0.18$$
:$$\Rightarrow \hspace{0.3cm} {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] = 0.8 + 0.18 \hspace{0.15cm} \underline{= 0.26}\hspace{0.05cm}.$$

*The Gallager's equation provides for the same set of parameters:
:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \prod\limits_{i =1}^{2} \hspace{0.25cm}(1-2\cdot p_i) = 0.5 + 0.5 \cdot (1 - 2 \cdot 0.2)\cdot (1 - 2 \cdot 0.9) = 0.26\hspace{0.05cm}.$$

*The equation given by Gallager 1963 was hereby verified for  $n = 2$.

'''(2)'''  In the second table,  the four combinations with an even number of  "ones"  are marked in blue. 
[[File:P_ID2997__KC_Z_4_4b_v1.png|right|frame|Derivation "$w_{\rm H}$ is even"  for code length  $n = 3$.]]

*The occurrence probabilities of each combination are given in the last column.  Thus,  the result is:
:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] = 0.056 + 0.216 + 0.006 + 0.126 \hspace{0.15cm} \underline{= 0.404}\hspace{0.05cm}.$$

*The red rows provide the complementary event:
:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right] = 0.024 + 0.504 + 0.014 + 0.054= 0.596\hspace{0.05cm}.$$

*Gallager's equation again gives the exact same result,  although it should be noted that this equation is valid for all  $n$  and all arbitrary probabilities:
:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \prod\limits_{i =1}^{3} \hspace{0.25cm}(1-2\cdot p_i) $$
:$$\Rightarrow\hspace{0.3cm}{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot (+0.6) \cdot (-0.8) \cdot (+0.4) = 0.404\hspace{0.05cm}.$$

'''(3)'''  According to the specification page applies:
:$$\pi = \prod\limits_{i =1}^{4} \hspace{0.25cm}(1-2\cdot p_i) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (1 - 2 \cdot 0.2) \cdot (1 - 2 \cdot 0.9) \cdot (1 - 2 \cdot 0.3) \cdot (1 - 2 \cdot 0.6) $$
:$$\Rightarrow\hspace{0.3cm}\pi = \prod\limits_{i =1}^{4} \hspace{0.25cm}(1-2\cdot p_i) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}(+0.6) \cdot (-0.8) \cdot (+0.4) \cdot (-0.2) = 0.0384\hspace{0.05cm}.$$

*From this can be calculated:
:$${\rm Pr}({\rm blue}) = {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \pi = 0.5 + 0.5 \cdot 0.0384\hspace{0.15cm} \underline{= 0.5192}\hspace{0.05cm},$$
:$${\rm Pr}({\rm red}) = {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 - 0.5 \cdot \pi = 0.5 - 0.5 \cdot 0.0384\hspace{0.15cm} \underline{= 0.4808}\hspace{0.05cm}. $$

*If you add up the blue and red probabilities on the information page,V you get exactly the values calculated here.  For the quotient we get:
:$$Q = \frac{{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right]} { {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right]} = \frac{0.5192}{0.4808}\hspace{0.15cm} \underline{= 1.0799}\hspace{0.05cm}. $$

'''(4)'''  For the  "single parity–check code",  the extrinsic log likelihood ratio with respect to the  $i^{th}$  bit was specified as follows:
:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}\hspace{0.05cm},$$

*Or:
:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{1+\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}{1-\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}\hspace{0.05cm}.$$

*At  $\text{SPC (5, 4, 2}$)   ⇒   $n = 5$,  this product for  $i = 5$  results from the following four factors:
:$$\pi = \prod\limits_{j = 1, \hspace{0.05cm}2, \hspace{0.05cm}3, \hspace{0.05cm}4} \hspace{0.05cm}(1-2\cdot p_j) =(1-2\cdot p_1) \cdot (1-2\cdot p_2) \cdot (1-2\cdot p_3) \cdot (1-2\cdot p_4)\hspace{0.05cm}.$$

The comparison with subtask  '''(3)'''  shows that  $L_{\rm E}(i = 5) = \ln {Q} = \ln {(1.0799)} \ \underline{\approx 0.077}$.

'''(5)'''  Correct is the  proposed solution 3  because the result for  $L_{\rm E}(i = 5)$  is independent of  $p_5$.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^4.1 Soft–in Soft–out Decoder^]]
[[de:Aufgaben:Aufgabe 4.4Z: Ergänzung zur Aufgabe 4.4]]

Aufgaben:Exercise 4.3: PDF Comparison with Regard to Differential Entropy

2026-03-16T15:58:42Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Information_Theory/Differential_Entropy
}}

[[File:EN_Inf_A_4_3_v2.png|right|frame|$h(X)$  for four probability density functions]]
The adjacent table shows the comparison result with respect to the differential entropy  $h(X)$  for
* the  [[Theory_of_Stochastic_Signals/Gleichverteilte_Zufallsgrößen|uniform distribution]]   ⇒   $f_X(x) = f_1(x)$:
:$$f_1(x) = \left\{ \begin{array}{c} 1/(2A) \\ 0 \\ \end{array} \right. \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \hspace{0.1cm} |x| \le A \\ {\rm else} \\ \end{array},$$
* the  [[Aufgaben:Exercise_3.1Z:_Triangular_PDF|triangular distribution]]   ⇒   $f_X(x) = f_2(x)$:
:$$f_2(x) = \left\{ \begin{array}{c} 1/A \cdot \big [1 - |x|/A \big ] \\ 0 \\ \end{array} \right. \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \hspace{0.1cm} |x| \le A \\ {\rm else} \\ \end{array},$$
* the  [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#Two-sided_exponential_distribution_-_Laplace_distribution|Laplace distribution]]   ⇒   $f_X(x) = f_3(x)$:
:$$f_3(x) = \lambda/2 \cdot {\rm e}^{-\lambda \hspace{0.05cm} \cdot \hspace{0.05cm}|x|}\hspace{0.05cm}.$$

The values for the  [[Theory_of_Stochastic_Signals/Gaußverteilte_Zufallsgrößen|Gaussian distribution]]   ⇒   $f_X(x) = f_4(x)$  with
:$$f_4(x) = \frac{1}{\sqrt{2\pi \sigma^2}} \cdot {\rm e}^{- \hspace{0.05cm}{x ^2}/{(2 \sigma^2})}$$
are not yet entered here.  These are to be determined in subtasks  '''(1)'''  to  '''(3)''' .

Each probability density function  $\rm (PDF)$  considered here is
* symmetric about  $x = 0$    ⇒   $f_X(-x) = f_X(x)$
* and thus zero mean   ⇒  $m_1 = 0$.

In all cases considered here, the differential entropy can be represented as follows:
*Under the constraint  $|X| ≤ A$   ⇒    [[Information_Theory/Differential_Entropy#Proof:_Maximum_differential_entropy_with_peak_constraint|peak constraint]]  $($German:  "Spitzenwertbegrenzung"  or  "Amplitudenbegrenzung"   ⇒   Identifier:  $\rm A)$:
:$$h(X) = {\rm log} \hspace{0.1cm} ({\it \Gamma}_{\hspace{-0.01cm}\rm A} \cdot A)\hspace{0.05cm},$$
*Under the constraint  ${\rm E}\big [|X – m_1|^2 \big ] ≤ σ^2$   ⇒   [[Information_Theory/Differential_Entropy#Proof:_Maximum_differential_entropy_with_power_constraint|power constraint]]  $($German:  "Leistungsbegrenzung"   ⇒   Identifier:  $\rm L)$:
:$$h(X) = {1}/{2} \cdot {\rm log} \hspace{0.1cm} ({\it \Gamma}_{\hspace{-0.01cm}\rm L} \cdot \sigma^2)\hspace{0.05cm}.$$
The larger the respective parameter  ${\it \Gamma}_{\hspace{-0.01cm}\rm A}$  or  ${\it \Gamma}_{\hspace{-0.01cm}\rm L}$  is, the more favorable is the present PDF in terms of differential entropy for the agreed constraint.

Hints:
*The exercise belongs to the chapter  [[Information_Theory/Differentielle_Entropie|Differential Entropy]].
*Useful hints for solving this task can be found in particular on the pages
::[[Information_Theory/Differential_Entropy#Differential_entropy_of_some_peak-constrained_random_variables|Differential entropy of some peak-constrained random variables]], 
::[[Information_Theory/Differential_Entropy#Differential_entropy_of_some_power-constrained_random_variables|Differential entropy of some power-constrained random variables]].

===Questions===

<quiz display=simple>
{Which equation is valid for the logarithm of the Gaussian PDF?
|type="[]"}
+ It holds:   $\ln \big[f_X(x) \big] = \ln (A) - x^2/(2 \sigma^2)$   with   $A = f_X(x=0)$.
- Es It holds:   $\ln \big [f_X(x) \big] = A - \ln (x^2/(2 \sigma^2)$   with   $A = f_X(x=0)$.

{Which equation holds for the differential entropy of the Gaussian PDF?
|type="[]"}
+ It holds:   $h(X)= 1/2 \cdot \ln (2\pi\hspace{0.05cm}{\rm e}\hspace{0.01cm}\cdot\hspace{0.01cm}\sigma^2)$  with the pseudo-unit  "nat".
+ It holds:   $h(X)= 1/2 \cdot \log_2 (2\pi\hspace{0.05cm}{\rm e}\hspace{0.01cm}\cdot\hspace{0.01cm}\sigma^2)$  with the pseudo-unit  "bit".

{Complete the missing entry for the Gaussian PDF in the above table.
|type="{}"}
${\it \Gamma}_{\rm L} \ = \ $ { 17.08 3% }

{What values are obtained for the Gaussian PDF with the DC component  $m_1 = \sigma = 1$?
|type="{}"}
$P/\sigma^2 \ = \ $ { 2 3% }
$h(X) \ = \ $ { 2.047 3% } $\ \rm bit$

{Which of the statements are true for the differential entropy  $h(X)$  considering the  "power constraint"  ${\rm E}\big[|X – m_1|^2\big] ≤ σ^2$?
|type="[]"}
+ The Gaussian PDF   ⇒   $f_4(x)$  leads to the maximum  $h(X)$.
- The uniform PDF   ⇒   $f_1(x)$  leads to the maximum  $h(X)$.
- The triangular PDF   ⇒   $f_2(x)$  is very unfavorable because it is peak-constrained.
+ The triangular PDF   ⇒   $f_2(x)$  is more favorable than the Laplace PDF   ⇒   $f_3(x)$.

{Which of the statements are true for  "peak constraint"  to the range  $|X| ≤ A$.  The maximum differential entropy  $h(X)$  is obtained for
|type="[]"}
- a Gaussian PDF   ⇒   $f_4(x)$  followed by a constraint   ⇒  $|X| ≤ A$,
+ the uniform PDF   ⇒   $f_1(x)$,
- the triangular PDF   ⇒   $f_2(x)$.

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  We assume the zero mean Gaussian PDF:
:$$f_X(x) = f_4(x) =A \cdot {\rm exp} [- \hspace{0.05cm}\frac{x ^2}{2 \sigma^2}]\hspace{0.5cm}{\rm with}\hspace{0.5cm}A = \frac{1}{\sqrt{2\pi \sigma^2}}\hspace{0.05cm}.$$
*Logarithmizing this function, the result is proposed solution 1:
:$${\rm ln}\hspace{0.1cm} \big [f_X(x) \big ] = {\rm ln}\hspace{0.1cm}(A) +{\rm ln}\hspace{0.1cm}\left [{\rm exp} (- \hspace{0.05cm}\frac{x ^2}{2 \sigma^2}) \right ]= {\rm ln}\hspace{0.1cm}(A) - \frac{x ^2}{2 \sigma^2}\hspace{0.05cm}.$$

'''(2)'''  Both proposed solutions are correct:
*Using the result from  '''(1)'''  we obtain for the differential entropy in  "nat":
:$$h_{\rm nat}(X)= -\hspace{-0.1cm} \int_{-\infty}^{+\infty} \hspace{-0.15cm} f_X(x) \cdot {\rm ln} \hspace{0.1cm} [f_X(x)] \hspace{0.1cm}{\rm d}x =- {\rm ln}\hspace{0.1cm}(A) \cdot\int_{-\infty}^{+\infty} \hspace{-0.15cm} f_X(x) \hspace{0.1cm}{\rm d}x+ \frac{1}{2 \sigma^2} \cdot \int_{-\infty}^{+\infty} \hspace{-0.15cm} x^2 \cdot f_X(x) \hspace{0.1cm}{\rm d}x = - {\rm ln}\hspace{0.1cm}(A) + {1}/{2}\hspace{0.05cm}.$$
*Here it is taken into account that the first integral is equal to  $1$   (PDF area).
*The second integral also gives the variance  $\sigma^2$  (if, as here, the equal part  $m_1 = 0$ ).
*Substituting the abbreviation variable  $A$, we obtain:
:$$h_{\rm nat}(X) \hspace{-0.15cm} = \hspace{-0.15cm} - {\rm ln}\hspace{0.05cm}\left (\frac{1}{\sqrt{2\pi \sigma^2}} \right ) + {1}/{2} = {1}/{2}\cdot {\rm ln}\hspace{0.05cm}\left ({2\pi \sigma^2} \right ) + {1}/{2} \cdot {\rm ln}\hspace{0.05cm}\left ( {\rm e} \right ) = {1}/{2} \cdot {\rm ln}\hspace{0.05cm}\left ({{2\pi {\rm e} \cdot \sigma^2}} \right )\hspace{0.05cm}.$$
*If the differential entropy  $h(X)$  is not to be given in  "nat"  but in  "bit",  choose base  $2$  for the logarithm:
:$$h_{\rm bit}(X) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ({{2\pi {\rm e} \cdot \sigma^2}} \right )\hspace{0.05cm}.$$

'''(3)'''  Thus, according to the implicit definition  $h(X) = {1}/{2} \cdot {\rm log} \hspace{0.1cm} ({\it \Gamma}_{\hspace{-0.01cm}\rm L} \cdot \sigma^2)$ , the parameter is:
:$${\it \Gamma}_{\rm L} = 2\pi {\rm e} \hspace{0.15cm}\underline{\approx 17.08}\hspace{0.05cm}.$$

'''(4)'''  We now consider a Gaussian probability density function with mean  $m_1$:
:$$f_X(x) = \frac{1}{\sqrt{2\pi \sigma^2}} \cdot {\rm exp}\left [- \hspace{0.05cm}\frac{(x -m_1)^2}{2 \sigma^2} \right ]\hspace{0.05cm}.$$
* The second moment  $m_2 = {\rm E}\big [X ^2 \big ]$  can also be called the power  $P$,  while for the variance holds  (this is also the second central moment):
:$$\sigma^2 = {\rm E}\big [|X – m_1|^2 \big ] = \mu_2.$$
*According to Steiner's theorem,  $P = m_2 = m_1^2 + \sigma^2$.  Thus, assuming  $m_1 = \sigma = 1$   ⇒   $\underline{P/\sigma^2 = 2}$.

*Due to the DC component, the power is indeed doubled.  However, this does not change anything in the differential entropy.  Thus, it is still valid:
:$$h(X) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ({{2\pi {\rm e} \cdot \sigma^2}} \right )= {1}/{2} \cdot {\rm log}_2\hspace{0.05cm} (17.08)\hspace{0.15cm}\underline{\approx 2.047\,{\rm bit}}\hspace{0.05cm}.$$

[[File:EN_Inf_A_4_3_M_v2L.png|right|frame|Completed results table for  $h(X)$]]
'''(5)'''  Correct are the proposed solutions  '''(1)'''  and  '''(4)'''.  The numerical values of the characteristics  ${\it \Gamma}_{\rm L}$  and  ${\it \Gamma}_{\rm A}$  are also entered in the completed table on the right.

A probability density function  $f_X(x)$  is always particularly favorable under power constraints if the value  ${\it \Gamma}_{\rm L}$  (right column)  is as large as possible.  Then the differential entropy  $h(X)$  is also large.

The numerical results can be interpreted as follows:
* As is proved in the theory part, the Gaussian PDF  $f_4(x)$  leads here to the largest possible  ${\it \Gamma}_{\rm L} ≈ 17.08$   ⇒   the proposed solution 1 is correct (the value in the last column is marked in red).
* For the uniform PDF  $f_1(x)$  the parameter ${\it \Gamma}_{\rm L} = 12$  is the smallest in the whole table   ⇒   the proposed solution 2 is wrong.
* The triangular PDF  $f_2(x)$  with  ${\it \Gamma}_{\rm L} = 16.31$  is more favorable than the uniform PDF   ⇒   the proposed solution 3 is wrong.
*The triangular PDF  $f_2(x)$  is also better than the Laplace PDF  $f_2(x) \ \ ({\it \Gamma}_{\rm L} = 14.78)$   ⇒   the proposed solution 4 is correct.

'''(6)'''  Correct is the proposed solution  '''(2)'''.  A PDF  $f_X(x)$  is favorable in terms of differential entropy  $h(X)$ under the peak constraint   ⇒   $|X| ≤ A$,  if the weighting factor  ${\it \Gamma}_{\rm A}$  (middle column)  is as large as possible:
* As shown in the theory section, the uniform distribution   $f_1(x)$  leads here to the largest possible  ${\it \Gamma}_{\rm A}= 2$   ⇒   the proposed solution 2 is correct  (the value in the middle column is marked in red).
* The triangular PDF  $f_2(x)$,  which is also peak-constrained, is characterized by a somewhat smaller  ${\it \Gamma}_{\rm A}= 1.649$    ⇒   dthe proposed solution 3 is incorrect.
* The Gaussian PDF  $f_4(x)$  is infinitely extended.  A peak constraint on  $|X| ≤ A$  leads here to Dirac functions in the PDF   ⇒   $h(X) \to - \infty$, see sample solution to Exercise 4.2Z, subtask '''(4)'''.
* The same would be true for the Laplace PDF  $f_3(x)$ .

{{ML-Fuß}}

[[Category:Information Theory: Exercises|^4.1 Differential Entropy^]]
[[de:Aufgaben:Aufgabe 4.3: WDF–Vergleich bezüglich differentieller Entropie]]

Aufgaben:Exercise 5.4: Walsh Functions (PCCF, PACF)

2026-03-16T15:58:41Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Modulation_Methods/Spreading_Sequences_for_CDMA
}}

[[File:P_ID1889__Mod_A_5_4.png|right|frame|Hadamard matrix  ${\mathbf{H}_{8}}$]]
The so-called  "Walsh functions",  which can be constructed by means of the Hadamard matrix,  are often used for band spreading and band compression.  Starting from the matrix
:$${\mathbf{H}_{2}} = \left[ \begin{array}{ccc} +1 & +1 \\ +1 & -1 \end{array} \right] $$
the further Hadamard matrices  $ {\mathbf{H}_{4}}$,  $ {\mathbf{H}_{8}}$,  etc. can be derived by the following recursion:
:$$ {\mathbf{H}_{2J}} = \left[ \begin{array}{ccc} \mathbf{H}_J & \mathbf{H}_J \\ \mathbf{H}_J & -\mathbf{H}_J \end{array} \right] \hspace{0.05cm}.$$
The diagram shows the matrix  $ {\mathbf{H}_{8}}$  for the spreading factor  $J = 8$.  From this we can derive the spreading sequences
:$$ \langle w_\nu^{(1)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm}{+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm}{+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},$$
:$$ \langle w_\nu^{(2)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm}{+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm}{-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},$$
:$$...$$
:$$\langle w_\nu^{(7)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm}{-\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm}{+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm}$$
for seven CDMA subscribers.  The spreading sequence  $ \langle w_\nu^{(0)}\rangle$  corresponding to the first row in the Hadamard matrix is usually not assigned because it does not spread.

The questions mostly refer to the spreading factor  $J = 4$.  Thus,  correspondingly,  a maximum of three CDMA subscribers can be supplied with the spreading sequences  $ \langle w_\nu^{(1)}\rangle$,  $ \langle w_\nu^{(2)}\rangle$  and  $ \langle w_\nu^{(3)}\rangle$,  which result from the second, third and fourth rows of the matrix $ {\mathbf{H}_{4}}$.

Regarding the correlation functions, the following nomenclature shall apply in this exercise:
* The  [[Modulation_Methods/Spreading_Sequences_for_CDMA#Periodic_ACF_and_CCF|periodic cross-correlation function]]  $\rm (PCCF)$  between the sequences  $ \langle w_\nu^{(i)}\rangle$  and  $ \langle w_\nu^{(j)}\rangle$  is denoted by  $φ_{ij}(λ)$.   Here:
:$${\it \varphi}_{ij}(\lambda) = {\rm E}\left [ w_{\nu}^{(i)} \cdot w_{\nu+ \lambda}^{(j)} \right ] \hspace{0.05cm}.$$
* If  $φ_{ij} \equiv 0$  $($that is:  $φ_{ij}(λ) = 0$  for all values of  $λ)$,  the CDMA subscribers do not interfere with each other,  even if they have different propagation times.
* If at least  $φ_{ij}({\it λ} = 0) = 0$  applies,  then no interference occurs,  at least in synchronous CDMA operation  $($no or equal propagation times of all subscribers$).$ 
* The  [[Modulation_Methods/Spreading_Sequences_for_CDMA#Periodic_ACF_and_CCF|periodic auto-correlation function]]  $\rm (PACF)$  of the Walsh function  $ \langle w_\nu^{(i)}\rangle$  is denoted by  $φ_{ii}(λ)$,  and it holds:
:$${\it \varphi}_{ii}(\lambda) = {\rm E}\left [ w_{\nu}^{(i)} \cdot w_{\nu+ \lambda}^{(i)} \right ] \hspace{0.05cm}.$$

Notes:
*The exercise belongs to the chapter  [[Modulation_Methods/Spreizfolgen_für_CDMA|Spreading Sequences for CDMA]].
*Reference is made in particular to the section  [[Modulation_Methods/Spreading_Sequences_for_CDMA#Walsh_functions|Walsh functions]]  in the theory part.
* We would also like to draw your attention to the interactive applet  [[Applets:Generation_of_Walsh_functions|Generation of Walsh functions]]. 
*The abscissa is normalized to the chip duration  $T_c$.  This means that  $λ = 1$  actually describes a shift by the delay time  $τ = T_c$. 

===Questions===

<quiz display=simple>
{What are the spreading sequences for  $J = 4$?
|type="[]"}
+ $ \langle w_\nu^{(1)}\rangle = +\hspace{-0.05cm}1 -\hspace{-0.15cm}1 +\hspace{-0.15cm}1 -\hspace{-0.15cm}1$,
+ $ \langle w_\nu^{(2)}\rangle = +\hspace{-0.05cm}1 +\hspace{-0.15cm}1 -\hspace{-0.15cm}1 -\hspace{-0.15cm}1$,
+ $ \langle w_\nu^{(3)}\rangle = +\hspace{-0.05cm}1 -\hspace{-0.15cm}1 -\hspace{-0.15cm}1 +\hspace{-0.15cm}1$.

{Which statements are true regarding the PCCF values  $φ_{ij}(λ = 0)$?
|type="[]"}
+ For $J = 4$,   $φ_{12}(λ = 0) = 0$.
+ For $J = 4$,   $φ_{13}(λ = 0) = 0$.
+ For $J = 4$,   $φ_{23}(λ = 0) = 0$.
- For $J = 8$,   $φ_{ij}(λ = 0) ≠ 0$  may well hold  $(i ≠ j)$.
+ In synchronous CDMA,  the subscribers do not interfere with each other.

{Which statements are true for the PCCF values with  $λ ≠ 0$?
|type="[]"}
+ For all values of  $λ$,  the PCCF is  $φ_{12}(λ) = 0$.
+ For all values of  $λ$,  the PCCF is  $φ_{13}(λ) = 0$.
- For all values of  $λ$,  the PCCF is  $φ_{23}(λ) = 0$.
- In asynchronous CDMA,  the subscribers do not interfere with each other.

{Which statements are true for the PACF curves?
|type="[]"}
+ All   $φ_{ii}(λ)$  curves are periodic.
+  $φ_{11}(λ = 0) = +\hspace{-0.05cm}1$  and  $φ_{11}(λ = 1) = -\hspace{-0.05cm}1$  hold.
-  $φ_{22}(λ) = φ_{11}(λ)$  holds.
+  $φ_{33}(λ) = φ_{22}(λ)$  holds.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  All solutions  are correct:
*The matrix  $ {\mathbf{H}_{4}}$  is the upper left submatrix of  $ {\mathbf{H}_{8}}$.
*The spreading sequences result from the rows 2,  3  and 4  of  $ {\mathbf{H}_{4}}$,  and agree with the given sequences.

'''(2)'''  Solutions 1, 2 and 3  are correct:
*According to the equations in the data section,  the following holds:
:$${\it \varphi}_{12}(\lambda = 0) = 1/4 \cdot \left [ (+1) \cdot (+1) + (-1) \cdot (+1) + (+1) \cdot (-1) + (-1) \cdot (-1) \right ] = 0\hspace{0.05cm},$$
:$${\it \varphi}_{13}(\lambda = 0) = 1/4\cdot \left [ (+1) \cdot (+1) + (-1) \cdot (-1) + (+1) \cdot (-1) + (-1) \cdot (+1) \right ] = 0\hspace{0.05cm},$$
:$${\it \varphi}_{23}(\lambda = 0) =1/4 \cdot \left [ (+1) \cdot (+1) + (+1) \cdot (-1) + (-1) \cdot (-1) + (-1) \cdot (+1) \right ] = 0\hspace{0.05cm}.$$
*Also,  for larger values of  $J$,  for  $i ≠ j$  the PCCF value is always  $φ_{ij}(λ = 0)= 0$.
*It follows:   In synchronous CDMA,  the subscribers do not interfere with each other.

'''(3)'''  Solutions 1 and 2  are correct:
*For all values of  $λ$,  the PCCF is  $φ_{12}(λ) = 0$,  as shown by the following lines:
:$$\langle w_\nu^{(1)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},$$ $$\langle w_\nu^{(2)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm}, \hspace{0.3cm}{\rm product\hspace{0.1cm} with \hspace{0.1cm}}\langle w_\nu^{(1)}\rangle: {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.05cm},$$
:$$\langle w_{\nu+1}^{(2)}\rangle = {-\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm}, \hspace{0.3cm}{\rm product\hspace{0.1cm} with \hspace{0.1cm}}\langle w_\nu^{(1)}\rangle: {-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.05cm},$$
:$$\langle w_{\nu+2}^{(2)}\rangle = {-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.05cm},\hspace{0.3cm}{\rm product\hspace{0.1cm} with \hspace{0.1cm}}\langle w_\nu^{(1)}\rangle: {-\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},$$
:$$\langle w_{\nu+3}^{(2)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.05cm},\hspace{0.3cm}{\rm product\hspace{0.1cm} with \hspace{0.1cm}}\langle w_\nu^{(1)}\rangle: {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} \hspace{0.05cm},$$
:$$\langle w_{\nu+4}^{(2)}\rangle = {+\hspace{-0.05cm}1}\hspace{0.15cm} {+\hspace{-0.05cm}1} \hspace{0.15cm} {-\hspace{-0.05cm}1}\hspace{0.15cm} {-\hspace{-0.05cm}1} = \langle w_\nu^{(2)}\rangle \hspace{0.05cm}.$$
[[File:P_ID1890__Mod_A_5_4c.png|right|frame|Some PCCF and PACF curves]]
*The same is true for the PCCF  $φ_{13}(λ)$.
*In contrast,  for the PCCF between the sequences  $ \langle w_\nu^{(2)}\rangle$  and  $ \langle w_\nu^{(3)}\rangle$  we obtain:

:$${\it \varphi}_{23}(\lambda ) = \left\{ \begin{array}{c}0 \\+1\\ -1 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ {\rm{for}} \\ {\rm{for}} \\ \end{array} \begin{array}{*{20}c} \lambda = 0, \pm 2, \pm 4,\pm 6, ... \hspace{0.05cm}, \\ \hspace{0.14cm} \lambda = ... \hspace{0.05cm} , -3, +1, +5, ... \hspace{0.05cm}, \\ \hspace{0.14cm} \lambda = ... \hspace{0.05cm} , -5, -1, +3, ... \hspace{0.05cm}. \\ \end{array}$$
*This means:   If the signal from subscriber  $3$  is delayed by one spreading chip with respect to subscriber  $2$  or vice versa,  the subscribers can no longer be separated and there is a significant increase in the error probability.
*In the diagram,  the PCCF curves are drawn in dashed lines  (violet and red).

'''(4)'''  Statements 1,  2  and 4  are correct:
* Since the Walsh function no.  $1$  is periodic with  $T_0 = 2T_c$,  the PACF is also periodic with  $λ = 2$.
*The second statement is correct,  as shown by the following calculation  (green curve):
:$${\it \varphi}_{11}(\lambda = 0) = 1/4 \cdot \big [ (+1) \cdot (+1) + (-1) \cdot (-1) + (+1) \cdot (+1) + (-1) \cdot (-1) \big ] = +1\hspace{0.05cm},$$
:$${\it \varphi}_{11}(\lambda = 1) = 1/4 \cdot \big [ (+1) \cdot (-1) + (+1) \cdot (-1) + (+1) \cdot (-1) + (+1) \cdot (-1) \big ] = -1\hspace{0.05cm}.$$
*Since the two Walsh functions no.  $2$  and  $3$  differ only by a shift around  $T_c$  and a phase in the PACF has no effect in principle,  in fact,  according to the last statement,  $φ_{33}(λ) = φ_{22}(λ)$.  These two PACF functions are plotted in blue.
*In contrast,  $φ_{22}(λ)$  differs from  $φ_{11}(λ)$  by a different periodicity:   $φ_{22}(λ) = φ_{33}(λ)$  is twice as wide as  $φ_{11}(λ)$.

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^5.3 Spread Sequences for CDMA^]]
[[de:Aufgaben:Aufgabe 5.4: Walsh–Funktionen (PKKF, PAKF)]]

Aufgaben:Exercise 2.15: Block Error Probability with AWGN

2026-03-16T15:58:40Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Error_Probability_and_Areas_of_Application}}

[[File:P_ID2571__KC_A_2_15neu.png|right|frame|Incomplete table of results]]
Using the example of  $\rm RSC \, (7, \, 3, \, 5)_8$  with parameters
* $n = 7$  $($number of code symbols$)$,

* $k =3$  $($number of information symbols$)$,

* $t = 2$  $($correction capability$)$.

the calculation of the block error probability in  [[Channel_Coding/Error_Probability_and_Application_Areas#Block_error_probability_for_RSC_and_BDD|"Bounded Distance Decoding"]]  $\rm (BDD)$  shall be shown.  The corresponding equation is:
:$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) =\sum_{f = t + 1}^{n} {n \choose f} \cdot {\varepsilon_{\rm S}}^f \cdot (1 - \varepsilon_{\rm S})^{n-f} \hspace{0.05cm}.$$

⇒   The calculation is performed for the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"AWGN channel"
]]  characterized by the parameter  $E_{\rm B}/N_0$:

*The quotient  $E_{\rm B}/{N_0}$  can be expressed by the relation
:$$\varepsilon = {\rm Q} \big (\sqrt{{2 \cdot R \cdot E_{\rm B}}/{N_0}} \big ) $$
:into the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Symmetric_Channel_.E2.80. 93_BSC|"BSC model"]]  where  $R$  denotes the code rate  $($here:  $R = 3/7)$  and  ${\rm Q}(x)$  indicates the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"complementary Gaussian error integral"]].

*But since in the considered code the symbols come from  $\rm GF(2^3)$,  the BSC model with parameter  $\varepsilon$  must also still be adapted to the task.

*For the falsification probability of the  [[Channel_Coding/Error_Probability_and_Application_Areas#Block_error_probability_for_RSC_and_BDD|"''m''– BSC"  model]]  applies: 
:$$\varepsilon_{\rm S} = 1 - (1 - \varepsilon)^m\hspace{0.05cm}.$$
:Here it is to be set  $m = 3$  $($three bits per code symbol$)$.

⇒   For some  $E_{\rm B}/N_0$ values the results are entered in the table above.  The two rows with yellow background are briefly explained here:
* For  $10 \cdot \lg {E_{\rm B}/N_0} = 4 \ \rm dB$  we get  $\varepsilon \approx {\rm Q}(1.47) \approx 0.071$  and  $\varepsilon_{\rm S} \approx 0.2$.   The block error probability here can most easily be calculated using the complement:
:$${\rm Pr(block\:error)} = 1 - \left [ {7 \choose 0} \cdot 0.8^7 + {7 \choose 1} \cdot 0.2 \cdot 0.8^6 + {7 \choose 2} \cdot 0.2^2 \cdot 0.8^5\right ]\approx 0.148 \hspace{0.05cm}.$$

* For  $10 \cdot \lg {E_{\rm B}/N_0} = 12 \ \rm dB$  one gets  $\varepsilon \approx 1.2 \cdot 10^{-4}$  and  $\varepsilon_{\rm S} \approx 3.5 \cdot 10^{-4}$.  With this very small falsification probability,  the  $f = 3$  term dominates,  and we obtain:
:$${\rm Pr(block\:error)} \approx {7 \choose 3} \cdot (3.5 \cdot 10^{-4})^3 \cdot (1- 3.5 \cdot 10^{-4})^4\approx 1.63 \cdot 10^{-9} \hspace{0.05cm}.$$

⇒   You are to calculate the block error probabilities for the rows highlighted in red   $(10 \cdot \lg {E_{\rm B}/N_0} = 5 \ \rm dB, \ 8 \rm dB$,  $10 \ \rm dB)$.
:*The rows with blue background show some results of  [[Aufgaben:Exercise_2.15Z:_Block_Error_Probability_once_more|"Exercise 2.15Z"]].  There  ${\rm Pr}(\underline{v} ≠ \underline{u})$  is calculated for  $\varepsilon_{\rm S} = 10\%,  \ 1\%$  $0.1\%$.
:*In subtasks '''(4)''' and '''(5)''' you are to establish the relationship between the quantity  $\varepsilon_{\rm S}$  and the AWGN parameter  $E_{\rm B}/N_0$  thus completing the above table.

Hints:
* The exercise belongs to the chapter  [[Channel_Coding/Error_Probability_and_Areas_of_Application| "Error Probability and Application Areas"]].

* We refer you here to the two interactive HTML5/JavaScript applets 
:*[[Applets:Complementary_Gaussian_Error_Functions|"Complementary Gaussian error functions"]],  and 
:*[[Applets:Binomial_and_Poisson_Distribution_(Applet)|"Binomial and Poisson Distribution"]].

===Questions===
<quiz display=simple>
{What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0} \hspace{0.15cm}\underline{= 5 \ \rm dB}$?
|type="{}"}
${\rm Pr(block\:error)} \ = \ ${ 6.66 3% } $\ \cdot 10^{-2}$

{What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0} \hspace{0.15cm}\underline{= 8 \ \rm dB}$?
|type="{}"}
${\rm Pr(block\:error)} \ = \ ${ 8.63 3% } $\ \cdot 10^{-4}$

{What is the block error probability for  $10 \cdot \lg {E_{\rm B}/N_0}\hspace{0.15cm}\underline{ = 10 \ \rm dB}$?
|type="{}"}
${\rm Pr(block\:error)} \ = \ ${ 4.3 3% } $\ \cdot 10^{-6}$

{How is  $\varepsilon_{\rm S} = 0.1$  related to  $10 \cdot \lg {E_{\rm B}/N_0}$ ?   Note:  Use the given applet to calculate  ${\rm Q}(x)$.
|type="{}"}
$\varepsilon_{\rm S} = 10^{-1} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ ${ 5.87 3% } $\ \rm dB$

{Find also the  $E_{\rm B}/N_0$ values  $($in  $\rm dB)$  for  $\varepsilon_{\rm S} = 0.01$  and  $\varepsilon_{\rm S} = 0.001$. Complete the table.
|type="{}"}
$\varepsilon_{\rm S} = 10^{-2} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ $ { 9.32 3% } $\ \rm dB$
$\varepsilon_{\rm S} = 10^{-3} \text{:} \hspace{0.4cm} 10 \cdot \lg {E_{\rm B}/N_0} \ = \ $ { 11.3 3% } $\ \rm dB$
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  From the table on the information page,  the BSC parameter  $\varepsilon = 0.0505$  can be read.
*This gives  $\varepsilon_{\rm S}$  for the symbol error probability with  $m = 3$:
:$$1 - \varepsilon_{\rm S} = (1 - 0.0505)^3 \approx 0.856\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\varepsilon_{\rm S} \approx 0.144\hspace{0.05cm}.$$

*The fastest way to calculate the block error probability is here to use the formula
:$${\rm Pr(block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 - {\rm Pr}(f=0) - {\rm Pr}(f=1) - {\rm Pr}(f=2) = 1 - 1 \cdot 0.856^7 -7 \cdot 0.144^1 \cdot 0.856^6 - 21 \cdot 0.144^2 \cdot 0.856^5$$
:$$\Rightarrow \hspace{0.3cm} {\rm Pr(block\:error)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Pr}(\underline{v} \ne \underline{u}) =1 - 0.3368 - 0.3965 - 0.2001 \hspace{0.15cm} \underline{=0.0666}\hspace{0.05cm}.$$

'''(2)'''  Following the same calculation procedure as in subtask  '''(1)''',  we obtain with  $\varepsilon_{\rm S} \approx 0.03 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.97$:
:$${\rm Pr(block\:error)}\hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 \hspace{-0.05cm}-\hspace{-0.05cm} 1 \cdot 0.97^7 \hspace{-0.05cm}-\hspace{-0.05cm}7 \cdot 0.03^1 \cdot 0.97^6 \hspace{-0.05cm}-\hspace{-0.05cm} 21 \cdot 0.03^2 \cdot 0.97^5 =1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.8080 \hspace{-0.05cm}-\hspace{-0.05cm} 0.1749\hspace{-0.05cm}-\hspace{-0.05cm} 0.0162= 1 \hspace{-0.05cm}-\hspace{-0.05cm} 0.9991 = 9 \cdot 10^{-4}\hspace{0.05cm}.$$

*You can see that here the difference between two numbers of almost the same size must be formed,  so that the result could be affected by an error.

*Therefore we still calculate the following quantities:
:$${\rm Pr}(f=3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{7 \choose 3} \cdot \varepsilon_{\rm S}^3 \cdot (1 - \varepsilon_{\rm S})^4 = 35 \cdot 0.03^3 \cdot 0.97^4 = 8.366 \cdot 10^{-4}\hspace{0.05cm},$$
:$${\rm Pr}(f=4) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{7 \choose 4} \cdot \varepsilon_{\rm S}^4 \cdot (1 - \varepsilon_{\rm S})^3 = 35 \cdot 0.03^4 \cdot 0.97^3 = 0.259 \cdot 10^{-4}\hspace{0.05cm},$$
:$${\rm Pr}(f=5) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{7 \choose 5} \cdot \varepsilon_{\rm S}^5 \cdot (1 - \varepsilon_{\rm S})^2 = 21 \cdot 0.03^5 \cdot 0.97^2 = 0.005 \cdot 10^{-4}$$
:$$\Rightarrow \hspace{0.3cm} {\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) + {\rm Pr}(f=4) + {\rm Pr}(f=5) \hspace{0.15cm} \underline{=8.63 \cdot 10^{-4}} \hspace{0.05cm}.$$

*The terms for  $f = 6$  and  $f = 7$  can be omitted here.  They do not provide a relevant contribution.

'''(3)'''  Here  $\varepsilon_{\rm S} = 0.005 \ \Rightarrow \ 1 - \varepsilon_{\rm S} = 0.995$  is already given in the table.
*The  (by far)  dominant term in the calculation of the block error probability is  ${\rm Pr}(f = 3)$:
:$${\rm Pr(block\:error)} = {\rm Pr}(\underline{v} \ne \underline{u}) \approx {\rm Pr}(f=3) = {7 \choose 3} \cdot 0.005^3 \cdot 0.995^4\hspace{0.15cm} \underline{\approx 4.3 \cdot 10^{-6}} \hspace{0.05cm}.$$

'''(4)'''  For the BSC parameter  $\varepsilon$  holds with  $\varepsilon_{\rm S} = 0.1$:
:$$\varepsilon = 1 -(1 - \varepsilon_{\rm S})^{1/3} = 1 - 0.9^{1/3} \approx 0.0345\hspace{0.05cm}.$$

*The relation between  $\varepsilon$  and  $E_{\rm B}/N_0$  is:
:$$\varepsilon = {\rm Q}(x)\hspace{0.05cm}, \hspace{0.5cm} x = \sqrt{2 \cdot R \cdot E_{\rm B}/N_0}\hspace{0.05cm}.$$

*The inverse  $x = {\rm Q}^{-1}(0.0345)$  is obtained with the applet  [[Applets:QFunction|"Complementary Gaussian Error Functions"]]  to  $x = 1.82$.  This further gives:
:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{1.82^2}{2R \cdot 3/7} \approx 3.864\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0)\hspace{0.15cm} \underline{\approx 5.87 \,\, {\rm dB}} \hspace{0.05cm}. $$

'''(5)'''  After the same calculation one obtains
* for  $\varepsilon_{\rm S} = 10^{-2} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-2} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 2.71$
:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{2.71^2}{2R \cdot 3/7} \approx 8.568\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0)\hspace{0.15cm} \underline{\approx 9.32 \,\, {\rm dB}} \hspace{0.05cm}, $$

* for $\varepsilon_{\rm S} = 10^{-3} \ \Rightarrow \ \varepsilon \approx 0.33 \cdot 10^{-3} \ \Rightarrow \ x = {\rm Q}^{-1}(\varepsilon) = 3.4$:
:$$E_{\rm B}/N_0 = \frac{x^2}{2R} = \frac{3.4^2}{2R \cdot 3/7} \approx 13.487\hspace{0.3cm} \Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.15cm}(E_{\rm B}/N_0)\hspace{0.15cm} \underline{\approx 11.3 \,\, {\rm dB}} \hspace{0.05cm}. $$

[[File:EN_KC_A_2_15e.ng.png|right|frame|Results for $\rm RSC \, (7, \, 3, \, 5)_8$ decoding]]

The graph shows
*the course of the block error probability as function of $10 \cdot \lg {E_{\rm B}/N_0}$,

* and the completely filled result table.

One can see the clearly less favorable  (asymptotic)  behavior of the  (green)  code $\rm RSC \, (7, \, 5, \, 3)_8$ compared to the  (red)  comparison code $\rm RSC \, (255, \, 223, \, 33)_8$:

# For abscissa values smaller than  $10 \ \rm dB$  the result is even worse than without coding.
# It should be pointed out again that the  $\rm RSC \, (7, \, 3, \, 5)_8$  has only little practical meaning.
# It was chosen for this exercise only to be able to demonstrate with reasonable effort the calculation of the block error probability for  "Bounded Distance Decoding"  $\rm (BDD)$.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^2.6 Block Error Probability of RS Codes^]]
[[de:Aufgaben:Aufgabe 2.15: RS-Blockfehlerwahrscheinlichkeit bei AWGN]]

Aufgaben:Exercise 1.14: Bhattacharyya Bound for BEC

2026-03-16T15:58:40Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Limits_for_Block_Error_Probability}}

[[File:EN_KC_A_1_14.png|right|frame|Possible received vectors for the  $(5, 2)$ code and BEC ]]

In this exercise,  we consider the systematic  $(5, 2)$  code
*with the  $2×5$ generator matrix

:$${ \boldsymbol{\rm G}}_{(5, 2)} = \begin{pmatrix} 1 &0 &1 &1 &0\\ 0 &1 &0 &1 &1 \end{pmatrix} \hspace{0.05cm},$$

*the  $3 × 5$ parity-check matrix

:$${ \boldsymbol{\rm H}}_{(5, 2)} = \begin{pmatrix} 1 &0 &1 &0 &0\\ 1 &1 &0 &1 &0\\ 0 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm},$$

*and the  $2^k = 4$  code words

:$$\underline{x}_0 \hspace{-0.15cm}\ = \ \hspace{-0.15cm} (0, 0, 0, 0, 0) \hspace{0.05cm},\hspace{0.2cm}\underline{x}_1 = (0, 1, 0, 1, 1)\hspace{0.05cm},\hspace{0.2cm}\underline{x}_2 \hspace{-0.15cm}\ = \ \hspace{-0.15cm} (1, 0, 1, 1, 0) \hspace{0.05cm},\hspace{0.2cm}\underline{x}_3 = (1, 1, 1, 0, 1)\hspace{0.05cm}.$$

At the output of the digital channel defined by the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_.E2.80.93_BEC|"BEC model"]]  ("Binary Erasure Channel")  with the erasure probability  $\lambda = 0.001$,  the received vector

:$$\underline{y} = (y_1, \hspace{0.05cm}y_2, \hspace{0.05cm}y_3, \hspace{0.05cm}y_4, \hspace{0.05cm}y_5)$$

occurs,  where for  $i = 1, \ \text{...} \ , 5$  holds:  
:$$y_{i} \in \{0, 1, \rm E\}.$$

The BEC channel is characterized by the fact that
*falsifications  $(0 → 1,\ 1 → 0)$  are excluded,

*but erasures  $(0 → \rm E,\ 1 → E)$  may occur.

The graph explicitly shows all possible received vectors   $\underline{y}$   with three or more erasures $\rm (E)$  assuming that the all-zero vector   $(0, 0, 0, 0, 0)$  was sent.
*For less than three erausures,  for the considered  $(5, 2)$  code,  the  "code word finder"  always returns the correct decision:   $\underline{z} = \underline{x}$.

*On the other hand,  if there are three or more erasures,  wrong decisions may occur.  In this case,  the following applies to the block error probability:

:$$ {\rm Pr(block\:error)}= {\rm Pr} (\underline{z} \ne \underline{x}) = {\rm Pr}\left \{ \hspace{0.1cm} [\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}1}] \hspace{0.05cm}\cup\hspace{0.05cm}[\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}2}] \hspace{0.05cm}\cup \hspace{0.05cm}[\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}3}] \hspace{0.1cm}\right \} \hspace{0.05cm}.$$

Please note:

*The event  $[\underline{x}_{0} → \underline{x}_{1}]$  does not necessarily say that at the received vector under consideration  $\underline{y}$  is actually decided for the code word  $\underline{x}_{1}$,  but only that the decision for  $x_{1}$  would be more reasonable than the decision for  $\underline{x}_{0}$  due to statistics.

*But it could also be decided for  $\underline{x}_{2}$  or  $\underline{x}_{3}$  if the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Criteria_.C2.BBMaximum-a-posteriori.C2.AB_and_.C2.BBMaximum-Likelihood.C2.AB|"maximum-likelihood criterion"]]  is in favor.

Computing the block error probability is difficult because the events  $[\underline{x}_{0} → \underline{x}_{1}]$ ,  $[\underline{x}_{0} → \underline{x}_{2}]$  and  $[\underline{x}_{0} → \underline{x}_{3}]$  are not necessarily  [[Theory_of_Stochastic_Signals/Set_Theory_Basics#Disjoint_sets|"disjoint"]] . An upper bound is provided by the  [[Channel_Coding/Bounds_for_Block_Error_Probability#Union_Bound_of_the_block_error_probability|"Union Bound"]]:

:$${\rm Pr(Union \hspace{0.15cm}Bound)} = {\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm}\underline{x}_{\hspace{0.02cm}1}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}2}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}3}] \ge {\rm Pr(block\hspace{0.15cm}error)} \hspace{0.05cm}.$$

Another bound was given by Bhattacharyya:

:$${\rm Pr(Bhattacharyya)} = W(\beta)-1 \ge {\rm Pr(Union \hspace{0.15cm}Bound)} \ge {\rm Pr(block\:error)} \hspace{0.05cm},$$

where for the binary erasure channel,  the Bhattacharyya parameter  $\beta = \lambda$  and the  [[Channel_Coding/Bounds_for_Block_Error_Probability#Distance_spectrum_of_a_linear_code|"weight enumerator function"]]  $W(X)$  where the pseudo-variable  $X$  is to be replaced here by the Bhattacharyya parameter  $\lambda$.

*The  "Bhattacharyya Bound"  is more or less far above the  "Union Bound"  depending on the channel.

*Its importance lies in the fact that the bound can be specified in the same way for different channels.

Hints:  The exercise belongs to the chapter  [[Channel_Coding/Limits_for_Block_Error_Probability|"Bounds on block error probability"]].

===Questions===
<quiz display=simple>
{What is the pairwise error probability between the code words   $\underline{x}_{0} = (0, 0, 0, 0, 0)$   and   $\underline{x}_{1} = (0, 1, 0, 1, 1)$?
|type="{}"}
${\rm Pr}[\underline{x}_{0} → \underline{x}_{1}] \ = \ $ { 0.5 3% }$\ \cdot 10^{-3} $

{Which statements are true regarding  ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$  with index  $i = 1, \ \text{...} \ , 3$?   $d_{{\rm H},\hspace{0.05cm}i}$  denotes here the Hamming distance between  $x_{0}$  and  $x_{i}$.
|type="()"}

- It holds  ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}] \ = \ \lambda ^{d_{{\rm H},\hspace{0.05cm}i}} \ · \ (1 – \lambda)^{n \hspace{0.05cm}– \hspace{0.05cm}d_{{\rm H},\hspace{0.05cm}i}}$.
+ It holds  ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}] \ = \ 1/2 · \lambda ^{d_{{\rm H},\hspace{0.05cm}i}}.$
- ${\rm Pr}[\underline{x}_{0} → \underline{x}_{i}]$  is the falsification probability from  $x_{0}$  to  $x_{i}$.

{What are the following probabilities?
|type="{}"}
$\ {\rm Pr}[\underline{x}_{0} → \underline{x}_{2}] \ = \ $ { 0.5 3% } $\ \cdot 10^{-3} $
$\ {\rm Pr}[\underline{x}_{0} → \underline{x}_{3}] \ = \ $ { 0.05 3% } $\ \cdot 10^{-3} $

{Specify the  "Union Bound"  for the block error probability.
|type="{}"}
$\ {\rm Pr(Union\ Bound)} \ = \ ${ 1.05 3% } $\ \cdot 10^{-3} $

{What is the  "Bhattacharyya Bound"  in the present case?
|type="{}"}
$\ {\rm Pr(Bhattacharyya)} \ = \ ${ 2.1 3% } $\ \cdot 10^{-3} $

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The code words   $\underline{x}_{0}$   and   $\underline{x}_{1}$  differ in bit $2, \ 4$  and $5$.  If only one of these three binary values is transmitted correctly,  the entire code word is thus uniquely determined.  No information about the code word is obtained for the following received vectors  (see table in the information section):
* $\underline{y} = (0, {\rm E}, 0, {\rm E}, {\rm E})$  with probability  $\lambda^3 \ · \ (1 – \lambda)^2$,
* $\underline{y} = (0, {\rm E}, {\rm E}, {\rm E}, {\rm E})$  with probability  $\lambda^4 \ · \ (1 – \lambda)$,
* $\underline{y} = ({\rm E}, {\rm E}, 0, {\rm E}, {\rm E})$  with probability $\lambda^4 \ · \ (1 – \lambda)$,
* $\underline{y} = ({\rm E}, {\rm E}, {\rm E}, {\rm E}, {\rm E})$ with probability  $\lambda^5$.

The probability that due to the specific received vector  $\underline{y}$,  the code word  $\underline{x}_{1}$  is just as likely as  $\underline{x}_{0}$,  is given by

:$$\ {\rm Pr}\ [\underline{x}_0 \hspace{0.12cm}{\rm and}\hspace{0.12cm} \underline{x}_1 \hspace{0.15cm}{\rm are \hspace{0.15cm}equally \hspace{0.15cm}probable}] = \lambda^3 \cdot (1- \lambda)^2 + 2 \cdot \lambda^4 \cdot (1- \lambda) + \lambda^5 =\lambda^3 \cdot \left [ (1- \lambda)^2 + 2 \cdot \lambda \cdot (1- \lambda) + \lambda^2 \right ] = \lambda^3 \hspace{0.05cm}.$$

In this case,  one decides at random for  $\underline{x}_{0}$  (would be correct)  or for  $\underline{x}_{1}$  (unfortunately wrong),  with equal probability.  From this follows:

:$${\rm Pr} [\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}1}] = 1/2 \cdot \lambda^3 \hspace{0.15cm} \underline{= 0.5 \cdot 10^{-3}} \hspace{0.05cm}.$$

'''(2)'''  According to subtask  '''(1)''',  only answer 2  is correct:
*${\rm Pr}[\underline{x}_{0} → \underline{x}_{1}]$  does not state that with this probability the code word  $\underline{x}_{0}$  actually transitions to the incorrect code word  $\underline{x}_{1}$,  but only that it could transition to  $\underline{x}_{1}$  with this probability.

* ${\rm Pr}[\underline{x}_{0} → \underline{x}_{1}]$  also includes constellations where the decision is actually for  $\underline{x}_{2}$  or  $\underline{x}_{3}$.

'''(3)'''  Because of  $d_{\rm H}(\underline{x}_{0}, \underline{x}_{2}) = 3$  and  $d_{\rm H}(\underline{x}_{0}, \underline{x}_{3}) = 4$  it follows that

:$${\rm Pr} [\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}2}] = 1/2 \cdot \lambda^3 \hspace{0.15cm} \underline{= 0.5 \cdot 10^{-3}} \hspace{0.05cm},$$
:$$ {\rm Pr} [\underline{x}_{\hspace{0.02cm}0} \mapsto \underline{x}_{\hspace{0.02cm}3}] = 1/2 \cdot \lambda^4 \hspace{0.15cm} \underline{= 0.05 \cdot 10^{-3}} \hspace{0.05cm}.$$

'''(4)'''  The block error probability is never larger  (with a certain probability rather smaller)  than the so-called  "Union Bound":

:$${\rm Pr(Union \hspace{0.15cm}Bound)} \hspace{-0.15cm}\ = \ \hspace{-0.15cm} {\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm}\underline{x}_{\hspace{0.02cm}1}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}2}] +{\rm Pr}[\underline{x}_{\hspace{0.02cm}0} \hspace{-0.02cm}\mapsto \hspace{-0.02cm} \underline{x}_{\hspace{0.02cm}3}] = 2 \cdot \lambda^3/2 + \lambda^4/2 = 0.001 + 0.00005 \hspace{0.15cm} \underline{= 1.05 \cdot 10^{-3}} \hspace{0.05cm}.$$

'''(5)'''  Generally speaking:
:$${\rm Pr(block\:error) ≤ {\rm Pr(Union \hspace{0.15cm}Bound)} \le Pr(Bhattacharyya)} = W(\beta) - 1.$$

*For the distance spectrum or the enumerator weight function,  we obtain in the present case:

:$$W_0 = 1 \hspace{0.05cm}, \hspace{0.2cm} W_3 = 2 \hspace{0.05cm}, \hspace{0.2cm}W_4 = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} W(X) = 1+ 2 \cdot X^{3} +X^{4} \hspace{0.05cm}.$$

*For the BEC channel,  $\beta = \lambda$  also applies.  From this follows as a final result for  $\lambda = 0.001$:

:$${\rm Pr(Bhattacharyya)} = 2 \cdot \lambda^3 + \lambda^4 \hspace{0.15cm} \underline{= 2.1 \cdot 10^{-3}} \hspace{0.05cm}.$$

*Note that with the BEC model,  the  "Bhattacharyya Bound"  is always twice as large as the  "Union Bound",  which is itself an upper bound on the block error probability.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^1.6 Error Probability Bounds^]]
[[de:Aufgaben:Aufgabe 1.14: Bhattacharyya–Schranke für BEC]]

Aufgaben:Exercise 4.10: Turbo Encoder for UMTS and LTE

2026-03-16T15:58:39Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/The_Basics_of_Turbo_Codes}}

[[File:EN_KC_A_4_10_v1.png|right|frame|Turbo encoder for UMTS and LTE]]
The mobile communications standards  [[Mobile_Communications/Characteristics_of_UMTS|$\rm UMTS$]]  and  [[Mobile_Communications/General_Information_on_the_LTE_Mobile_Communications_Standard|$\rm LTE$]]  each use a turbo code that is largely identical to the encoder described in the  [[Channel_Coding/The_Basics_of_Turbo_Codes|"The Basics of Turbo Codes"]]  chapter.
* The  $1/n$ convolutional code is systematic, meaning that the encoded sequence  $\underline{x}$  includes the information sequence  $\underline{u}$  as a component.

* The parity-check sequences  $\underline{p}_1$  and  $\underline{p}_2$  are based on the same transfer function:
:$$G_1(D) = G_2(D) = G(D).$$
* $\underline{p}_1$  and  $\underline{p}_2$  however,  use different input sequences  $\underline{u}$  and  $\underline{u}_{\pi}$,  respectively.  Here,  ${\rm \Pi}$  marks the interleaver,  for UMTS and LTE mostly a  $S$–random interleaver.

[[File:P_ID3052__KC_A_4_10b_v2.png|left|frame|Given filter structure]]
 The main difference compared to the description in the theory part results from a different transfer function  $G(D)$  given by the recursive filter structure drawn on the left.
 

Hints:
* The exercise belongs to the chapter  [[Channel_Coding/The_Basics_of_Turbo_Codes| "Basics of Turbo Codes"]].

* Knowledge is expected about
** the  [[Channel_Coding/Algebraic_and_Polynomial_Description|"Algebraic and Polynomial Description of Convolutional Codes"]],
** the  [[Channel_Coding/Code_Description_with_State_and_Trellis_Diagram|"Code Description with State and Trellis Diagram"]].

* For further guidance on how to do this, see  [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes|$\text{Exercise 4.8}$]]  and  [[Aufgaben:Exercise_4.09:_Recursive_Systematic_Convolutional_Codes|$\text{Exercise 4.9}$]].

* The information sequence  $\underline{u}$  is partially specified by its  $D$–transform for easier description in the subtasks.  For example:
:$$\underline{u}= (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}\hspace{0.05cm} \text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quadU(D) = D+ D^2\hspace{0.05cm},$$
:$$\underline{u}= (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}\hspace{0.05cm} \text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quadU(D) = D+ D^8\hspace{0.05cm}.$$

===Questions===
<quiz display=simple>
{What are the characteristics of the considered turbo code  $($memory  $m$,  influence length  $\nu$,  rate $R)$?
|type="{}"}
$ m \hspace{0.2cm} = \ ${ 3 3% }
$ \nu \hspace{0.3cm} = \ ${ 9 3% }
$R \hspace{0.2cm} = \ ${ 0.333 3% }

{What are the  (identical)  transfer functions  $G_1(D) = G_2(D) = G(D)$?
|type="()"}
+   $G(D) = (1 + D + D^3)/(1 + D^2 + D^3)$.
-   $G(D) = (1 + D^2 + D^3)/(1 + D + D^3)$.

{What is the impulse response  $\underline{g}$?
|type="[]"}
-  $\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \hspace{0.05cm} \text{...}\hspace{0.05cm})$
+  $\underline{g} = (1, \, 1, \, 1, \, 1, \, 0, \, 0, \, 1, \, 0, \, 1, \, 1, \, 1, \, 0, \, 0, \, 1, \, 0, \hspace{0.05cm} \text{...}\hspace{0.05cm})$.
+  $\underline{g}$  continues to infinity.

{Are there periodic components within the impulse response  $\underline{g}$ ?
|type="()"}
+ Yes,  with period  $P = 7$.
- Yes,  with period  $P = 8$.
- No.

{Let $U(D) = D + D^2$. Which statements are true?
|type="[]"}
+ The initial sequence  $\underline{p}$  contains a periodic component.
+ The period  $P$  is unchanged from  $\underline{g}$ .
+ The Hamming weight of the input sequence is  $w_{\rm H}(\underline{u}) = 2$.
- The Hamming weight of the output sequence is  $w_{\rm H}(\underline{p}) = 6$.

{Which statements are true for  $U(D) = D + D^8$ ?
|type="[]"}
- The initial sequence  $\underline{p}$  contains a periodic component.
- The period  $P$  is unchanged from  $\underline{g}$ .
+ The Hamming weight of the input sequence is  $w_{\rm H}(\underline{u}) = 2$.
+ The Hamming weight of the output sequence  is $w_{\rm H}(\underline{p}) = 6$.
</quiz>

===Solution===
{{ML-Kopf}}
[[File:EN_KC_A_4_10c_v3_neu2.png|right|frame|Polynomial division for subtask '''(3)''': $G(D) = (1 + D + D^3) \ / \ (1 + D^2 + D^3)$]]
'''(1)'''  The code parameters are  $k = 1$  and  $n = 3$   ⇒   code rate  $\underline{R = 1/3}$.
*The memory is  $\underline{m = 3}$.

*The influence lengths result in  $\nu = 1, \ \nu_2 = 4$  and  $\nu_3 = 4$   ⇒   Total influence length  $\underline{\nu = 9}$.

'''(2)'''  As the comparison of the  "recursive filter"  on the data page with the  [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function|"filter structure"]]  in the theory section for fractional–rational  $G(D)$,  the  proposed solution 1 is correct.

'''(3)'''  Correct are the  proposed solutions 2 and 3:

The upper graph illustrates the polynomial division  $(1 + D + D^3) \ / \ (1 + D^2 + D^3)$.  For explanation:
*Cancelled is the representation with the remainder  $D^8 + D^9 = D^7 \cdot (D + D^2)$.  Thus also holds:
:$$(D^8 + D^9) \hspace{0.05cm} /\hspace{0.05cm} (1+ D^2+ D^3 ) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} D^7 \cdot (D+ D^2+ D^3 + D^6) + {\rm remainder_2}$$
*After summarizing:
:$$G(D) = 1 + D + D^2 + D^3 + D^6 + D^8+ D^9+ D^{10} + D^{13} + \hspace{0.05cm}\text{...}\hspace{0.05cm} \hspace{0.05cm}. $$

* The $D$–inverse transform gives the proposed solution 2:
:$$\underline{g}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm})\hspace{0.05cm}. $$

* The impulse response continues to infinity   ⇒   Solution proposal 3 is also correct.

[[File:P_ID3061__KC_A_4_10d_v2.png|right|frame|State transition diagram and impulse response]]
'''(4)'''  The impulse response can be expressed as follows:
:$$\underline{g}= \Big (\hspace{0.03cm}1\hspace{0.03cm},\big [ \hspace{0.03cm} 1\hspace{0.03cm},\hspace{0.03cm} 1\hspace{0.03cm},\hspace{0.03cm} 1\hspace{0.03cm},\hspace{0.03cm} 0\hspace{0.03cm},\hspace{0.03cm} 0\hspace{0.03cm},\hspace{0.03cm} 1\hspace{0.03cm},\hspace{0.03cm} 0\hspace{0.03cm} \big ]_{\rm per}\Big ) \hspace{0.15cm}\Rightarrow \hspace{0.15cm} \underline{P = 7}\hspace{0.05cm}. $$

*In the state transition diagram  $($right$)$,  the impulse response  $\underline{g}$  is highlighted in yellow. 

*The impulse response results as the parity-check sequence   $\underline{p}$   for the information sequence   $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, \text{...})$.

* The transitions in the diagram are labeled  "$u_i\hspace{0.05cm}|\hspace{0.05cm}\underline{x}_i$",  which is equivalent to  "$u_i\hspace{0.05cm}|\hspace{0.05cm}u_i \hspace{0.05cm}p_i$".

*The parity-check sequence   $\underline{p} \ (=$ impulse response  $\underline{g})$   thus results from the respective second coder output symbol.

* The impulse response  $\underline{g}$  is represented by the following states:
:$$S_0 → [S_1 → S_2 → S_5 → S_3 → S_7 → S_6 → S_4 ] → [S_1 → \ ... \ → S_4] → \ \text{...} $$

'''(5)'''  The following graph shows the solution using the generator matrix  $\mathbf{G}$.  It holds  $\underline{u} = (0, \, 1, \, 1, \, 0, \, 0, \, \text{...} )$.

[[File:EN_KC_A_4_10e_v1.png|right|frame|$\underline{p} = (0, \, 1, \, 1, \, \text{...} ) \cdot \mathbf{G}$]]

[[File:EN_KC_A_4_10f_v1.png|right|frame|$\underline{p} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{...}) \cdot \mathbf{G}$]]

It can be seen that  solutions 1, 2 and 3  are correct:
* The present parity-check sequence  $\underline{p}$  has the same period  $P = 7$  as the impulse response  $\underline{g}$.

* The Hamming weight of the  $($limited$)$  input sequence is actually  $w_{\rm H}(\underline{u}) = 2$.

* Proposition 4 is incorrect.  Rather,  for the semi–infinite output sequence:   $w_{\rm H}(\underline{p}) → \infty$.

*In the transition diagram,  the states
:$$S_0 → S_0 → S_1 → S_3 → S_7 → S_6 → S_4 → S_1$$
: are passed through first. 

*This is followed  $($infinitely often$)$ by the periodic portion
:$$S_1 → S_2 → S_5 → S_3 → S_7 → S_6 → S_4 → S_1.$$

'''(6)'''  The last graph shows the solution for 
:$$U(D) = D + D^8$$
:$$ \Rightarrow \hspace{0.3cm}\underline{u} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{...}).$$
*Correct are the  proposed solutions 3 and 4:

*The input sequence  $\underline{u}$  contains two  "ones"  and the output sequence  $\underline{p}$  six  "ones".

*From position 10  the output sequence is  $\underline{p} \equiv\underline{0}$   ⇒   proposals 1 and 2 therefore do not apply.
 
Further notes:
# For turbo codes,  especially those input sequences  $\underline{u}$  whose  $D$–transforms are representable as  $U(D) = f(D) \cdot [1 + D^{P}]$  are extremely unfavorable.
#They cause the error floor as seen on the  [[Channel_Coding/The_Basics_of_Turbo_Codes#Performance_of_the_turbo_codes|"Performance of Turbo Codes"]] page in the theory section.
#$P$  indicates here the period of the impulse response  $\underline{g}$.  In our example  $f(D) = D$  and  $P = 7$.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^4.3 About the Turbo Codes^]]
[[de:Aufgaben:Aufgabe 4.10: Turbocoder für UMTS und LTE]]

Aufgaben:Exercise 1.16: Block Error Probability Bounds for AWGN

2026-03-16T15:58:39Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Limits_for_Block_Error_Probability}}

[[File:P_ID2414__KC_A_1_15.png|right|frame|Function  ${\rm Q}(x)$  and approximations; it holds:  ${\rm Q_u}(x)\le{\rm Q}(x)\le{\rm Q_o}(x)$ ]]

We assume the following constellation:
*A linear block code with code rate  $R = k/n$  and distance spectrum  $\{W_i\}, \ i = 1, \ \text{...} \ , n$,

*an AWGN channel characterized by  $E_{\rm B}/N_{0}$   ⇒   convertible to noise power  $\sigma^2$,

*a receiver based on  "soft decision"  as well as the  "maximum likelihood criterion".

Under the assumption valid for the entire exercise that always the zero-word  $\underline{x}_{1} = (0, 0, \text{... } \ , 0)$  is sent, the  [[Channel_Coding/Bounds_for_Block_Error_Probability#Union_Bound_of_the_block_error_probability|"pairwise error probability"]]  with a different code word  $\underline{x}_{l} (l = 2,\ \text{...} \ , 2^k)$:

:$$ {\rm Pr}[\hspace{0.05cm}\underline{x}_{\hspace{0.02cm}1} \hspace{-0.02cm}\mapsto \hspace{-0.02cm}\underline{x}_{\hspace{0.02cm}l}\hspace{0.05cm}] = {\rm Q}\left ( \sqrt{w_{\rm H}(\underline{x}_{\hspace{0.02cm}l})/\sigma^2} \right ) \hspace{0.05cm}.$$

The derivation of this relation can be found in  [Liv10].  Used in this equation are:
*the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"complementary Gaussian error function"]]  ${\rm Q}(x)$,

*the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"Hamming weight"]]  $w_{\rm H}(\underline{x}_{l})$  of the code word  $\underline{x}_{l}$,

*the  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#System_optimization_with_power_limitation|"AWGN noise power"]]  $\sigma^2 = (2 \cdot R \cdot E_{\rm B}/N_{0})^{-1}.$

This allows various bounds to be specified for the block error probability:

*the so called  [[Channel_Coding/Bounds_for_Block_Error_Probability#Union_Bound_of_the_block_error_probability|"Union Bound"]]  $\rm (UB)$:

:$$p_1 = \sum_{l = 2}^{2^k}\hspace{0.05cm}{\rm Pr}[\hspace{0.05cm}\underline{x}_{\hspace{0.02cm}1} \hspace{-0.02cm}\mapsto \hspace{-0.02cm}\underline{x}_{\hspace{0.02cm}l}\hspace{0.05cm}] = \sum_{l \hspace{0.05cm}= \hspace{0.05cm}2}^{2^k}\hspace{0.05cm}{\rm Q}\left ( \sqrt{w_{\rm H}(\underline{x}_{\hspace{0.02cm}l})/\sigma^2} \right ) \hspace{0.05cm},$$

*the so called  [[Channel_Coding/Bounds_for_Block_Error_Probability#Bounds_for_the_(7,_4,_3)_Hamming_code_at_the_AWGN_channel|"Truncated Union Bound"]]  $\rm (TUB)$:

:$$p_2 = W_{d_{\rm min}} \cdot {\rm Q}\left ( \sqrt{d_{\rm min}/\sigma^2} \right ) \hspace{0.05cm},$$

*the  [[Channel_Coding/Bounds_for_Block_Error_Probability#The_upper_bound_according_to_Bhattacharyya|"Bhattacharyya Bound"]]:

:$$p_3 = W(\beta) - 1\hspace{0.05cm},\hspace{0.2cm} {\rm with}\hspace{0.15cm} \beta = {\rm e}^{ - 1/(2\sigma^2) } \hspace{0.05cm}.$$

:In this case,  replace the distance spectrum  $\{W_i\}$  with the weight enumerator function:

:$$\left \{ \hspace{0.05cm} W_i \hspace{0.05cm} \right \} \hspace{0.3cm} \Leftrightarrow \hspace{0.3cm} W(X) = \sum_{i=0 }^{n} W_i \cdot X^{i} = W_0 + W_1 \cdot X + W_2 \cdot X^{2} + ... \hspace{0.05cm} + W_n \cdot X^{n}\hspace{0.05cm}.$$

In the transition from the  "Union Bound"  $p_{1}$  to the more imprecise bound  $p_{3}$  among others
*the function  ${\rm Q}(x)$  is replaced by the  [https://en.wikipedia.org/wiki/Chernoff_bound "Chernoff-Rubin bound"]  ${\rm Q}_{\rm CR}(x)$.

*Both functions are shown in the above graph  (red and green curve, resp.).

In the  [[Aufgaben:Exercise_1.16Z:_Bounds_for_the_Gaussian_Error_Function|"Exercise 1.16Z"]]  the relationship between these functions is evaluated numerically and referenced to the bounds  ${\rm Q}_{\rm o}(x)$ and ${\rm Q}_{\rm u}(x)$  which are also drawn in the above graph.

Hints:
* This exercise belongs to the chapter  [[Channel_Coding/Bounds_for_Block_Error_Probability|"Bounds for block error probability"]].

* The above cited reference  "[Liv10]"  refers to the lecture manuscript "Liva, G.:  Channel Coding.  Chair of Communications Engineering, TU Munich and DLR Oberpfaffenhofen, 2010."

* Further we refer to the interactive HTML5/JavaScript applet  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen| "Complementary Gaussian error functions"]].

===Questions===

<quiz display=simple>
{Which equation applies to the  "Union Bound"?
|type="[]"}
- $p_{1} = \sum_{l\hspace{0.05cm}=\hspace{0.05cm}2}^{2^k} W_{l} · {\rm Q}\big[(l/\sigma^2)^{0.5}\big],$
+ $p_{1} = \sum_{i\hspace{0.05cm}=\hspace{0.05cm}1}^{n} W_{i} · {\rm Q}\big[(i/\sigma^2)^{0.5}\big].$

{Specify the Union Bound for the  $(8, 4, 4)$  code and various  $\sigma$.
|type="{}"}
$\sigma = 1.0 \text{:} \hspace{0.4cm} p_{1} \ = \ $ { 32.15 3% } $\ \%$
$\sigma = 0.5 \text{:} \hspace{0.4cm} p_{1} \ = \ $ { 0.0444 3% } $\ \%$

{Given the same boundary conditions, what does the  "Truncated Union Bound"  provide?
|type="{}"}
$\sigma = 1.0 \text{:} \hspace{0.4cm} p_{2} \ = \ $ { 31.92 3% } $\ \%$
$\sigma = 0.5 \text{:} \hspace{0.4cm} p_{2} \ = \ $ { 0.044 3% } $\ \%$

{Which statement is always true  (for all constellations)?
|type="[]"}
+ The block error probability is never greater than  $p_{1}$.
- The block error probability is never greater than  $p_{2}$.

{How do you get from  $p_{1}$  to the  "Bhattacharyya Bound"  $p_{3}$? 
|type="[]"}
+ Replace the error function  ${\rm Q}(x)$  with the function  ${\rm Q}_{\rm CR}(x)$.
- Set the Bhattacharyya parameter  $\beta = 1/\sigma$.
+ Instead of  $\{W_i\}$  uses the weight enumerator function  $W(X)$.

{Specify the Bhattacharyya Bound for  $\sigma = 1$  and  $\sigma = 0.5$ .
|type="{}"}
$\sigma = 1.0 \text{:} \hspace{0.4cm} p_{3} \ = \ $ { 191.3 3% } $\ \%$
$\sigma = 0.5 \text{:} \hspace{0.4cm} p_{3} \ = \ $ { 0.47 3% } $\ \%$
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The correct solution is suggestion 2:

*The distance spectrum  $\{W_i\}$  is defined for  $i = 0, \ \text{...} \ , \ n$:

#$W_{1}$  indicates how often the Hamming weight  $w_{\rm H}(\underline{x}_{i}) = 1$  occurs.
#$W_{n}$  indicates how often the Hamming weight  $w_{\rm H}(\underline{x}_{i}) = n$  occurs.

*With that,  the  "Union Bound"  is:

:$$p_1 = {\rm Pr(Union \hspace{0.15cm}Bound)}= \sum_{i = 1}^{n}\hspace{0.05cm}W_i \cdot {\rm Q}\left ( \sqrt{i/\sigma^2} \right ) \hspace{0.05cm}.$$

'''(2)'''  The distance spectrum of the  $(8, 4, 4)$  code was given as  $W_{0} = 1 , \ W_{4} = 14, \ W_{8} = 1$. 
*Thus,  one obtains for $\sigma = 1$:
:$$p_1 = W_4 \cdot {\rm Q}\left ( 2 \right ) + W_8 \cdot {\rm Q}\left ( 2 \cdot \sqrt{2} \right )= 14 \cdot 2.28 \cdot 10^{-2}+ 1 \cdot 0.23 \cdot 10^{-2} \hspace{0.15cm}\underline{\approx 32.15\%}\hspace{0.05cm},$$

*For  $\sigma = 0.5$:
:$$p_1 = 14 \cdot {\rm Q}\left ( 4 \right ) + {\rm Q}\left ( 4 \cdot \sqrt{2} \right )= 14 \cdot 3.17 \cdot 10^{-5}+ 1.1 \cdot 10^{-8} \hspace{0.15cm}\underline{\approx 0.0444 \%}\hspace{0.05cm}.$$

'''(3)'''  With the minimum distance  $d_{\rm min} = 4$  we get:

:$$\sigma = 1.0\text{:} \hspace{0.4cm} p_2 \hspace{-0.15cm}\ = \ \hspace{-0.15cm} W_4 \cdot {\rm Q}\left ( 2 \right ) \hspace{0.15cm}\underline{= 31.92\%}\hspace{0.05cm},$$
:$$\sigma = 0.5\text{:} \hspace{0.4cm} p_2 \hspace{-0.15cm}\ = \ \hspace{-0.15cm}W_4 \cdot {\rm Q}\left ( 4 \right ) \approx p_1 \hspace{0.15cm}\underline{ = 0.0444 \%}\hspace{0.05cm}.$$

'''(4)'''  The correct solution is  suggestion 1:
*The  "Union Bound"  - denoted here by  $p_{1}$ - is an upper bound on the block error probability in all cases.

*For the bound  $p_{2}$  ("Truncated Union Bound")  this is not always true.

*For example,  in the  $(7, 4, 3)$  Hamming code   ⇒   $W_{3} = W_{4} = 7, \ W_{7} = 1$  is obtained with standard deviation  $\sigma = 1$:

:$$p_2 \hspace{-0.15cm}\ = \ \hspace{-0.15cm} 7 \cdot {\rm Q}\left ( \sqrt{3} \right ) = 7 \cdot 4.18 \cdot 10^{-2} \approx 0.293\hspace{0.05cm},$$
:$$p_1 \hspace{-0.15cm}\ = \ \hspace{-0.15cm} p_2 + 7 \cdot {\rm Q}\left ( \sqrt{4} \right )+ 1 \cdot {\rm Q}\left ( \sqrt{7} \right ) \approx 0.455 \hspace{0.05cm}.$$

*The actual block error probability is likely to be between  $p_{2} = 29.3\%$  and  $p_{1} = 45.5\%$  (but this has not been verified). That is,   $p_{2}$ is not an upper bound.

'''(5)'''  Correct are  suggested solutions 1 and 3,  as the following calculation for the  $(8, 4, 4)$  code shows:

*It holds  ${\rm Q}(x) ≤ {\rm Q_{CR}}(x) = {\rm e}^{-x^2/2}$.  Thus,  for the Union Bound

:$$p_1 = W_4 \cdot {\rm Q}\left ( \sqrt{4/\sigma^2} \right ) +W_8 \cdot {\rm Q}\left ( \sqrt{8/\sigma^2} \right )$$

:another upper bound can be specified:

:$$p_1 \le W_4 \cdot {\rm e}^{ - {4}/(2 \sigma^2) } +W_8 \cdot {\rm e}^{ - {8}/(2 \sigma^2) } \hspace{0.05cm}.$$

*With  $\beta = {\rm e}^{-1/(2\sigma^2)}$  can be written for this also  (so the given  $\beta = 1/\sigma$  is wrong):

:$$p_1 \le W_4 \cdot \beta^4 + W_8 \cdot \beta^8 \hspace{0.05cm}.$$

*The weight function of the  $(8, 4, 4)$  code is:

:$$W(X) = 1 + W_4 \cdot X^4 + W_8 \cdot X^8 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} W(\beta) - 1 = W_4 \cdot \beta^4 + W_8 \cdot \beta^8\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_3 = W(\beta) - 1 \ge p_1\hspace{0.05cm}.$$

'''(6)'''  With  $\sigma = 1$,  the Bhattacharyya parameter is  $\beta = {\rm e}^{-0.5} = 0.6065$,  and thus one obtains for the Bhattacharyya Bound:

:$$p_3 = 14 \cdot \beta^4 + \beta^8 = 14 \cdot 0.135 + 0.018= 1.913 \hspace{0.15cm}\underline{= 191.3%}\hspace{0.05cm}.$$

*Considering that  $p_{3}$  is a bound for a probability,  $p_{3} = 1.913$  is only a trivial bound.

*For  $\sigma = 0.5$,  on the other hand,  $\beta = {\rm e}^{-2} \approx 0.135.$  Then holds:

:$$p_3 = 14 \cdot \beta^4 + \beta^8 = 14 \cdot 3.35 \cdot 10^{-4} + 1.1 \cdot 10^{-7} \hspace{0.15cm}\underline{= 0.47 \%}\hspace{0.05cm}.$$

A comparison with subtask  '''(2)'''  shows that in the present example the Bhattacharyya Bound  $p_{3}$  is above the  "Union Bound"  $p_{1}$  by a factor 
:$$(0.47 - 10^{-2})/(0.044 - 10^{-2}) > 10.$$

*The reason for this large deviation is the Chernoff-Rubin bound,  which is well above the  ${\rm Q}$ function.

*In  [[Aufgaben:Exercise_1.16Z:_Bounds_for_the_Gaussian_Error_Function|"Exercise 1.16Z"]],  the deviation between  ${\rm Q}_{\rm CR}$  and  ${\rm Q}(x)$  is also calculated quantitatively:

:$${{\rm Q_{CR}}( x )}/{{\rm Q}( x )} \approx 2.5 \cdot x \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {{\rm Q_{CR}}( x = 4 )}/{{\rm Q}( x = 4)} \approx 10 \hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^1.6 Error Probability Bounds^]]
[[de:Aufgaben:Aufgabe 1.16: Fehlerwahrscheinlichkeitsschranken für AWGN]]

Aufgaben:Exercise 2.8: Huffman Application for a Markov Source

2026-03-16T15:58:38Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Information_Theory/Entropy_Coding_According_to_Huffman
}}

[[File:EN_Inf_A_2_8.png|right|frame|Binary symmetric Markov source]]
We consider the symmetric Markov source according to the graph, which is completely given by the single parameter
:$$q = {\rm Pr}(\boldsymbol{\rm X}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm X}) ={\rm Pr}(\boldsymbol{\rm Y}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm Y}).$$

*The given source symbol sequences apply to the conditional probabilities  $q = 0.2$  and  $q = 0.8$, respectively.
*In subtask  '''(1)'''  it has to be clarified which symbol sequence – the red or the blue one – was generated with  $q = 0.2$  and which with  $q = 0.8$ .

The properties of Markov sources are described in detail in the chapter  [[Information_Theory/Discrete_Sources_with_Memory|Discrete Sources with Memory]].  Due to the symmetry assumed here with regard to the binary symbols  $\rm X$  and  $\rm Y$,  some serious simplifications result, as is derived in  [[Aufgaben:1.5Z_Symmetrische_Markovquelle|Exercise 1.5Z]]:
* The symbols  $\rm X$  and  $\rm Y$  are equally probable, that is,  $p_{\rm X} = p_{\rm Y} = 0.5$ holds.  Thus the first entropy approximation is  $H_1 = 1\,\,{\rm bit/source\hspace{0.15cm}symbol}\hspace{0.05cm}. $
* The entropy of the Markov source for  $q = 0.2$  as well as for  $q = 0.8$  results in
:$$H = q \cdot {\rm log_2}\hspace{0.15cm}\frac{1}{q} + (1-q) \cdot {\rm log_2}\hspace{0.15cm}\frac{1}{1-q}= 0.722\,\,{\rm bit/source\hspace{0.15cm}symbol}\hspace{0.05cm}.$$
* For Markov sources, all entropy approximations  $H_k$  with order  $k \ge 2$  are determined by  $H_1$  and  $H = H_{k \to \infty}$:
:$$H_k = {1}/{k}\cdot \big [ H_1 + H \big ] \hspace{0.05cm}.$$
*The following numerical values again apply equally to  $q = 0.2$  and  $q = 0.8$ :
:$$H_2 = {1}/{2}\cdot \big [ H_1 + H \big ] = 0.861\,\,{\rm bit/source\hspace{0.15cm}symbol}\hspace{0.05cm},$$
:$$H_3 = {1}/{3} \cdot \big [ H_1 + 2H \big ] = 0.815\,\,{\rm bit/source\hspace{0.15cm}symbol}\hspace{0.05cm}.$$

In this exercise, the Huffman algorithm is to be applied to  $k$–tuples, where we restrict ourselves to  $k = 2$  and  $k = 3$.

Hints:
*The exercise belongs to the chapter  [[Information_Theory/Entropiecodierung_nach_Huffman|Entropy Coding according to Huffman]].
*In particular, reference is made to the page  [[Information_Theory/Entropy_Coding_According_to_Huffman#Application_of_Huffman_coding_to_.7F.27.22.60UNIQ-MathJax168-QINU.60.22.27.7F.E2.80.93tuples|Application of Huffman coding to  $k$-tuples]].
*Useful information can also be found in the specification sheets for  [[Aufgaben:Exercise_2.7:_Huffman_Application_for_Binary_Two-Tuples|Exercise 2.7]]  and  [[Aufgaben:Exercise_2.7Z:_Huffman_Coding_for_Two-Tuples_of_a_Ternary_Source|Exercise 2.7Z]].
*To check your results, please refer to the (German language) SWF module  [[Applets:Huffman_Shannon_Fano|Coding according to Huffman and Shannon/Fano]].

===Questions===

<quiz display=simple>
{Which of the example sequences given at the front is true for  $q = 0.8$?
|type="()"}
- The red source symbol sequence  '''1''',
+ the blue source symbol sequence  '''2'''.

{Which of the following statements are true?
|type="[]"}
- The direct application of Huffman is also useful here.
+ Huffman makes sense when forming two-tuples  $(k = 2)$.
+ Huffman makes sense when forming tuples of three  $(k = 3)$.

{What are the probabilities of two-tuples  $(k = 2)$  for  $\underline{q = 0.8}$?
|type="{}"}
$p_{\rm A} = \rm Pr(XX)\ = \ $ { 0.4 3% }
$p_{\rm B} = \rm Pr(XY)\ = \ $ { 0.1 3% }
$p_{\rm C} = \rm Pr(YX)\ = \ $ { 0.1 3% }
$p_{\rm D} = \rm Pr(YY)\ = \ $ { 0.4 3% }

{Find the Huffman code for  $\underline{k = 2}$.  What is the average code word length in this case?
|type="{}"}
$L_{\rm M} \ = \ $ { 0.9 3% } $\ \rm bit/source\hspace{0.15cm}symbol$

{What is the bound on the average code word length when two-tuples are formed  $(k = 2)$? Interpretation.
|type="()"}
- $L_{\rm M} \ge H_1 = 1.000$  $\ \rm bit/source\hspace{0.15cm}symbol$
+ $L_{\rm M} \ge H_2 \approx 0.861$  $\ \rm bit/source\hspace{0.15cm}symbol$
- $L_{\rm M} \ge H_3 \approx 0.815$  $\ \rm bit/source\hspace{0.15cm}symbol$
- $L_{\rm M} \ge H_{k \to \infty} \approx 0.722$  $\ \rm bit/source\hspace{0.15cm}symbol$
- $L_{\rm M} \ge 0.5$  $\ \rm bit/source\hspace{0.15cm}symbol$

{Calculate the probabilities of the three-tuple  $(k = 3)$  for  $\underline{q = 0.8}$?
|type="{}"}
$p_{\rm A} = \rm Pr(XXX)\ = \ $ { 0.32 3% }
$p_{\rm B} = \rm Pr(XXY)\ = \ $ { 0.08 3% }
$p_{\rm C} = \rm Pr(XYX)\ = \ $ { 0.02 3% }
$p_{\rm D} = \rm Pr(XYY)\ = \ $ { 0.08 3% }
$p_{\rm E} = \rm Pr(YXX)\ = \ $ { 0.08 3% }
$p_{\rm F} = \rm Pr(YXY)\ = \ $ { 0.02 3% }
$p_{\rm G} = \rm Pr(YYX)\ = \ $ { 0.08 3% }
$p_{\rm H} = \rm Pr(YYY)\ = \ $ { 0.32 3% }

{Find the Huffman code for $\underline{k = 3}$.  What is the average code word length in this case?
|type="{}"}
$L_{\rm M} \ = \ $ { 0.84 3% } $\ \rm bit/source\hspace{0.15cm}symbol$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Correct is the solution suggestion 2:
*In the blue source symbol sequence  '''2'''  one recognizes much less symbol changes than in the red sequence.
*The symbol sequence  '''2'''  was generated with the parameter  $q = {\rm Pr}(\boldsymbol{\rm X}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm X}) =
{\rm Pr}(\boldsymbol{\rm Y}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm Y}) = 0.8$  and the red symbol sequence  '''1'''  with  $q = 0.2$.

'''(2)''' Answers 2 and 3 are correct:
*Since here the source symbols  $\rm X$  and  $\rm Y$  were assumed to be equally probable, the direct application of Huffman makes no sense.
*In contrast, one can use the inner statistical depenndecies of the Markov source for data compression if one forms  $k$–tuples   $(k ≥ 2)$.
*The larger  $k$  is, the more the average code word length  $L_{\rm M}$  approaches the entropy  $H$.

'''(3)'''  The symbol probabilities are  $p_{\rm X} = p_{\rm Y} = 0.5$, which gives us for the two-tuples: 
:$$p_{\rm A} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XX}) = p_{\rm X} \cdot {\rm Pr}(\boldsymbol{\rm X}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm X}) = 0.5 \cdot q = 0.5 \cdot 0.8 \hspace{0.15cm}\underline{ = 0.4} \hspace{0.05cm},$$
:$$p_{\rm B} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XY}) = p_{\rm X} \cdot {\rm Pr}(\boldsymbol{\rm Y}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm X}) = 0.5 \cdot (1-q)= 0.5 \cdot 0.2 \hspace{0.15cm}\underline{ = 0.1} \hspace{0.05cm},$$
:$$p_{\rm C} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm YX}) = p_{\rm Y} \cdot {\rm Pr}(\boldsymbol{\rm X}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm Y}) = 0.5 \cdot (1-q)= 0.5 \cdot 0.2 \hspace{0.15cm}\underline{ = 0.1} \hspace{0.05cm},$$
:$$p_{\rm D} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm YY}) = p_{\rm Y} \cdot {\rm Pr}(\boldsymbol{\rm Y}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm Y}) = 0.5 \cdot q = 0.5 \cdot 0.8\hspace{0.15cm}\underline{ = 0.4} \hspace{0.05cm}.$$

[[File:P_ID2462__Inf_A_2_8d.png|right|frame|For Huffman coding for $k = 2$]]
'''(4)'''  Opposite screen capture of the (earlier) SWF applet  [[Applets:Huffman_Shannon_Fano|Coding according to Huffman and Shannon/Fano]]  shows the construction of the Huffman code for  $k = 2$  with the probabilities just calculated.
*Thus, the average code word length is:
:$$L_{\rm M}\hspace{0.01cm}' = 0.4 \cdot 1 + 0.4 \cdot 2 + (0.1 + 0.1) \cdot 3 = 1.8\,\,\text { bit/two-tuple}$$
:$$\Rightarrow\hspace{0.3cm}L_{\rm M} = {L_{\rm M}\hspace{0.01cm}'}/{2}\hspace{0.15cm}\underline{ = 0.9\,\text{ bit/source symbol}}\hspace{0.05cm}.$$

'''(5)'''  Correct is the suggested solution 2:
*According to the source coding theorem  $L_{\rm M} ≥ H$ holds.
*However, if we apply Huffman coding and disregard ties between non-adjacent symbols  $(k = 2)$, the lower bound of the code word length is not  $H = 0.722$, but  $H_2 = 0.861$  (the addition  "bit/source symbol"  is omitted for the rest of the task).
*The result of subtask  '''(4)'''  was  $L_{\rm M} = 0.9.$
*If an asymmetrical Markov chain were present and in such a way that for the probabilities  $p_{\rm A}$, ... , $p_{\rm D}$  the values  $50\%$,  $25\%$  and twice  $12.5\%$  would result, then one would come to the average code word length  $L_{\rm M} = 0.875$.
*How the exact parameters of this asymmetrical Markov source look, however, is not known even to the task creator (G. Söder).
*Nor how the value  $0.875$  could be reduced to  $0.861$.  In any case, the Huffman algorithm is unsuitable for this.

'''(6)'''  With  $q = 0.8$  and  $1 - q = 0.2$  we get:
:$$p_{\rm A} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XXX}) = 0.5 \cdot q^2 \hspace{0.15cm}\underline{ = 0.32} = p_{\rm H} = {\rm Pr}(\boldsymbol{\rm YYY})\hspace{0.05cm},$$
:$$p_{\rm B} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XXY}) = 0.5 \cdot q \cdot (1-q) \hspace{0.15cm}\underline{ = 0.08}= p_{\rm G} = {\rm Pr}(\boldsymbol{\rm YYX}) \hspace{0.05cm},$$
:$$p_{\rm C} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XYX}) = 0.5 \cdot (1-q)^2\hspace{0.15cm}\underline{ = 0.02} = p_{\rm F}= {\rm Pr}(\boldsymbol{\rm YXY}) \hspace{0.05cm},$$
:$$p_{\rm D} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XYY}) = 0.5 \cdot (1-q) \cdot q \hspace{0.15cm}\underline{ = 0.08} = p_{\rm E} = {\rm Pr}(\boldsymbol{\rm YXX})\hspace{0.05cm}.$$

[[File:P_ID2463__Inf_A_2_8g.png|right|frame|On the Huffman coding for  $k = 3$]]
'''(7)'''  The screen capture of the of the (earlier) SWF applet  [[Applets:Huffman_Shannon_Fano|Coding according to Huffman and Shannon/Fano]]  coding illustrates the constellation of the Huffman code for  $k = 3$. 

This gives us for the average code word length:
:$$L_{\rm M}\hspace{0.01cm}' = 0.64 \cdot 2 + 0.24 \cdot 3 + 0.04 \cdot 5 = 2.52\,\,{\rm bit/three tupel}$$
:$$\Rightarrow\hspace{0.3cm}L_{\rm M} = {L_{\rm M}\hspace{0.01cm}'}/{3}\hspace{0.15cm}\underline{ = 0.84\,{\rm bit/source\:symbol}}\hspace{0.05cm}.$$

*One can see the improvement over subtask  '''(4)'''.
*The bound  $k = 2$  valid for  $H_2 = 0.861$  is now undercut by the average code word length  $L_{\rm M}$.
*The new bound for  $k = 3$  is  $H_3 = 0.815$.
*However, to reach the source entropy  $H = 0.722$   (or better:  to come close to this final value up to an  $ε$ ), one would have to form infinitely long tuples  $(k → ∞)$.

{{ML-Fuß}}

[[Category:Information Theory: Exercises|^2.3 Entropy Coding according to Huffman^]]
[[de:Aufgaben:Aufgabe 2.8: Huffman-Anwendung bei einer Markovquelle]]

Aufgaben:Exercise 2.10: Reed-Solomon Error Detection

2026-03-16T15:58:37Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes}}

[[File: P_ID2524__KC_A_2_10.png|right|frame|Distance spectra of two different Reed-Solomon codes]]
For a linear block code,  up to   $e = d_{\rm min} - 1$   errors can be detected.  For all Reed–Solomon codes the minimum distance is
:$$d_{\rm min} = n-k+1 \hspace{0.05cm}.$$

One must distinguish the following cases:
* If no more than  $e = d_{\rm min} - 1= n - k$  symbol errors occur,  the block is always detected as faulty.

* Error detection can still work even if more than  $n - k$  symbol errors occur.  This is when the received word is not a valid code word of the Reed–Solomon code:
:$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_{\hspace{0.05cm}n -1} \}\hspace{0.05cm}. $$
* However,  if the falsified received word  $(\underline{y} ≠ \underline{c})$  is a valid code word  $(\underline{y\in C_{\rm RS}})$,  the decoding process will leave the falsified block undetected.

We define as  '''block error probability''':
:$${\rm Pr}({\rm block\hspace{0.15cm}error}) = {\rm Pr}(\underline {y} \ne \underline {c})\hspace{0.05cm}.$$

In this exercise,  this probability is to be determined for the following codes:
* Reed–Solomon code  $(7, \, 3, \, 5)_8 \ \Rightarrow \ d_{\rm min} = 5$,

* Reed–Solomon code  $(7, \, 5, \, 3)_8 \ \Rightarrow \ d_{\rm min} = 3$.

Furthermore,  let it hold:
* Each symbol is falsified into another symbol with probability  $\varepsilon_{\rm S} = 0.1$  and correctly transferred with probability  $1 - \varepsilon_{\rm S} = 0.9$.
* For the distance spectrum of a Reed–Solomon code of length  $n$  holds with  $d = d_{\rm min}$:
:$$W_i = {n \choose i} \cdot \sum_{j = 0}^{i-d}\hspace{0.15cm}(-1)^j \cdot {i \choose j} \cdot \big [\hspace{0.03cm}q^{i\hspace{0.03cm}-\hspace{0.03cm}j\hspace{0.03cm}-\hspace{0.03cm}d\hspace{0.03cm}+\hspace{0.03cm}1}-1 \hspace{0.03cm} \big ]\hspace{0.05cm}.$$

Besides,  two bounds for the block error probability will be considered and evaluated:
* If alone the minimum distance is known,  one can derive an  "upper bound"  from it.  The weight enumerating factors  $W_i$  are to be chosen in such a way,  that certainly (that is:   with all constellations)  applies:
:$${\rm Pr}({\rm upper\hspace{0.15cm} bound}) \ge {\rm Pr}({\rm block\hspace{0.15cm}error})\hspace{0.05cm}. $$
* A  "lower bound"  additionally requires knowledge of the weight function  $W_i$  for  $i = d_{\rm min}$.  Thus,  the following condition can be satisfied:
:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) \le {\rm Pr}({\rm block\hspace{0.15cm}error})\hspace{0.05cm}.$$

Hints:
*The exercise belongs to the chapter  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes|"Definition and Properties of Reed-Solomon Codes"]].

*Reference is made in particular to the page  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes#Singleton_bound_and_minimum_distance|"Singleton Bound and Minimum Distance"]].

* The weights  $W_i$  marked in red in the above graph are to be calculated.

===Questions===
<quiz display=simple>
{Calculate the distance spectrum for the  $\rm RSC \, (7, \, 3, \, 5)_8$.
|type="{}"}
$W_3 \ = \ ${ 0. }
$W_4 \ = \ ${ 0. }
$W_5 \ = \ ${ 147 }
$W_6 \ = \ ${ 147 }
$W_7 \ = \ ${ 217 }

{What is the missing weight of the  $\rm RSC \, (7, \, 5, \, 3)_8$  in the graph?
|type="{}"}
$W_3 \ = \ ${ 245 }

{What is the probability that an incorrect block remains undetected? Let the falsification probability of each symbol be  $\varepsilon = 0.1$.
|type="{}"}
$\rm RSC \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} Pr(block error) \ = \ ${ 0.263 3% } $\ \cdot 10^{-5}$
$\rm RSC \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} Pr(block error) \ = \ ${ 0.942 3% } $\ \cdot 10^{-5}$

{Calculate and evaluate for both codes the  "upper bound"  proposed in the specification:  $p_{\rm upper} = \rm Pr(upper \ bound)$.
|type="{}"}
${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm upper} \ = \ ${ 0.81 3% } $\ \cdot 10^{-5}$
${\rm RSC} \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} p_{\rm upper} \ = \ ${ 65.6 3% } $\ \cdot 10^{-5}$

{Calculate and evaluate for both codes the  "lower bound"  proposed in the specification:  $p_{\rm lower} = \rm Pr(lower \ bound)$.
|type="{}"}
${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm lower} \ = \ ${ 0.233 3% } $\ \cdot 10^{-5}$
${\rm RSC} \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} p_{\rm lower} \ = \ ${ 0.494 3% } $\ \cdot 10^{-5}$
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The equation describing the weights  $W_i$  is  $($the minimum distance  $d_{\rm min}$  is abbreviated here as  $d)$:
:$$W_i = {n \choose i} \cdot \sum_{j = 0}^{i-d}\hspace{0.15cm}(-1)^j \cdot {i \choose j} \cdot \big [\hspace{0.03cm}q^{i\hspace{0.03cm}-\hspace{0.03cm}j\hspace{0.03cm}-\hspace{0.03cm}d\hspace{0.03cm}+\hspace{0.03cm}1}-1 \hspace{0.03cm} \big ]\hspace{0.05cm}.$$

*Because of the minimum distance  $d_{\min} = 5$,  $W_3 \ \underline{= 0}$  and  $W_4 \ \underline{= 0}$.  The other weights result in
:$$W_5 = {7 \choose 5} \cdot (8^1 - 1) = \frac {7 \cdot 6 \cdot 5 \cdot4 \cdot 3}{1 \cdot 2 \cdot 3 \cdot4 \cdot 5} \cdot 7 = 21 \cdot 7\hspace{0.15cm}\underline {= 147}\hspace{0.05cm},$$
:$$W_6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {7 \choose 6} \cdot \sum_{j = 0}^{1}\hspace{0.15cm}(-1)^j \cdot {6 \choose j} \cdot \big (\hspace{0.03cm}8^{2\hspace{0.03cm}-\hspace{0.03cm}j} -1 \big )=7 \cdot \left [ (8^2 - 1) - 6 \cdot (8^1 - 1)\right ] = 7 \cdot (63-42)\hspace{0.15cm}\underline {= 147}\hspace{0.05cm},$$
:$$W_7 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {7 \choose 7} \cdot \sum_{j = 0}^{2}\hspace{0.15cm}(-1)^j \cdot {7 \choose j} \cdot \big (\hspace{0.03cm}8^{3\hspace{0.03cm}-\hspace{0.03cm}j} -1 \big )= (8^3 - 1) - 7 \cdot (8^2 - 1) +21 \cdot (8^1 - 1) = 511 - 7 \cdot 63 + 21 \cdot 7\hspace{0.15cm}\underline {= 217}\hspace{0.05cm},$$
:$$\Rightarrow \hspace{0.3cm}W_0 + W_5 + W_6 + W_7 = 1 + 147 + 147 + 217 = 512 = 8^3 = q^k\hspace{0.05cm}.$$

'''(2)'''  Analogous to subtask  '''(1)''',  we obtain:
:$$W_3 = {7 \choose 3} \cdot (8^1 - 1) = \frac {7 \cdot 6 \cdot 5 }{1 \cdot 2 \cdot 3 } \cdot 7 = 35 \cdot 7\hspace{0.15cm}\underline {= 245}\hspace{0.05cm}.$$

'''(3)'''  The falsification probability of a symbol is given by  $\varepsilon_{\rm S} = 0.1$.

:Then,  for the probability that in a code word with  $n = 7$  symbols
* exactly three symbols are falsified:
:$$p_3 = 0.1^3 \cdot 0.9^4 = 0.6561 \cdot 10^{-3}\hspace{0.05cm},$$
* exactly four symbols are falsified:
:$$p_4 = 0.1^4 \cdot 0.9^3 = 0.729 \cdot 10^{-4}\hspace{0.05cm},$$
* exactly five symbols are falsified:
:$$p_5 = 0.1^5 \cdot 0.9^2 = 0.81 \cdot 10^{-5}\hspace{0.05cm},$$
* exactly six symbols are falsified:
:$$p_6 = 0.1^6 \cdot 0.9 = 0.9 \cdot 10^{-6}\hspace{0.05cm},$$
* all $n = 7$ symbols are falsified:
:$$p_7 = 0.1^7 = 10^{-7}\hspace{0.05cm}.$$

⇒   At the  $\rm RSC \, (7, \, 3, \, 5)_8$  the zero word can be falsified by symbol errors into one of  $q^k - 1 = 8^3 - 1 = 511$  other code words.  Thus,  using the weight enumerating functions of subtask  '''(1)''',  we obtain:
:$${\rm Pr}({\rm block\hspace{0.15cm}error}) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac {W_5 \cdot p_5 + W_6 \cdot p_6 + W_7 \cdot p_7}{511} =\frac {147 \cdot 0.81 \cdot 10^{-5} + 147 \cdot 0.9 \cdot 10^{-6} + 217 \cdot 10^{-7}}{511}\hspace{0.15cm}\underline {\approx 0.263 \cdot 10^{-5}} \hspace{0.05cm}.$$

⇒   At the  $\rm RSC \, (7, \, 5, \, 3)_8$  we have to average over  $8^5 - 1 = 32767$  falsification probabilities because of  $k = 5$.  With  $W_3 = 245$  from subtask  '''(2)'''  and the weights 
:$$W_4 = 1224, \ W_5 = 5586, \ W_6 = 12838, \ W_7 = 12873$$
according to the specification sheet,  we obtain for this:
:$${\rm Pr}({\rm block\hspace{0.15cm}error}) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac {W_3 \cdot p_3 + W_4 \cdot p_4 + W_5 \cdot p_5 + W_6 \cdot p_6 + W_7 \cdot p_7}{32767} =\frac {245 \cdot 0.656 \cdot 10^{-3} + \hspace{0.15cm}... \hspace{0.15cm}+ 12873 \cdot 10^{-7}}{32767}\hspace{0.15cm}\underline {\approx 0.942 \cdot 10^{-5}} \hspace{0.05cm}. $$

'''(4)'''  Let only  $d_{\rm min}$  $($further abbreviated as  $d)$  be known and thus also  $p_d = \varepsilon_{\rm S}^d \cdot (1 - \varepsilon_{\rm S})^{n-d}$. 
*This is also the upper bound we are looking for:
:* ${\rm RSC} \, (7, \, 3, \, 5)_8 \text{:} \hspace{0.33cm} {\rm Pr(upper \ bound)} = p_5 = \underline{0.81 \cdot 10^{-5}}$,

:* ${\rm RSC} \, (7, \, 5, \, 3)_8 \text{:} \hspace{0.33cm} {\rm Pr(upper \ bound)} = p_3 = \underline{65.6 \cdot 10^{-5}}$.

*Since the weight  $W_d$  was assumed to be unknown,  it is set to the maximum possible value  $(W_5 = 511$  or  $W_3 = 32767)$  so that the prefactors in the equations for subtask  '''(3)'''  disappear.  This is the only way to ensure an upper bound.

*In both cases,  the upper bound is significantly higher than the results of subtask  '''(3)''':
:* $\rm RSC \, (7, \, 3, \, 5)_8 \text{:} \hspace{0.33cm} 0.810 \cdot 10^{-5}$  instead of  $0.263 \cdot 10^{-5}$  $($faktor approx.  $3)$,

:* $\rm RSC \, (7, \, 5, \, 3)_8 \text{:} \hspace{0.33cm} 65.6 \cdot 10^{-5}$  instead of  $0.942 \cdot 10^{-5}$  $($faktor approx.  $70)$.

'''(5)'''  Using the abbreviation  $d = d_{\rm min}$,  one obtains for the lower bound:
:$${\rm Pr}({\rm lower\hspace{0.15cm} bound})= \frac{W_d \cdot p_d}{q^k -1}\hspace{0.05cm}. $$
* For the  $\rm RSC \, (7, \, 3, \, 5)_8$,  because of  $W_d = W_5$ and  $p_d = p_5$,  the lower bound is about  $11\%$  below the actual value $(0.263 \cdot 10^{-5})$:
:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) = \frac{147 \cdot 0.81 \cdot 10^{-5}}{511}\hspace{0.15cm}\underline {\approx 0.233 \cdot 10^{-5}}$$
* For the  $\rm RSC \, (7, \, 5, \, 3)_8$  holds with  $W_d = W_3$  and  $p_d = p_3$,  the lower bound here deviates from the actual value  $(0.942 \cdot 10^{-5})$  more strongly,  because in this code the contributions of the higher weights  $(W_4, \ W_5, \ W_6, \ W_7)$  are more relevant in relation to  $W_3$:
:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) = \frac{245 \cdot 0.656 \cdot 10^{-3}}{32767}\hspace{0.15cm}\underline {\approx 0.494 \cdot 10^{-5}}\hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^2.3 Reed–Solomon Codes^]]
[[de:Aufgaben:Aufgabe 2.10: Fehlererkennung bei Reed–Solomon]]

Aufgaben:Exercise 4.14: ACF and CCF for Square Wave Signals

2026-03-16T15:58:37Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Cross-Correlation_Function_and_Cross_Power_Density
}}

[[File:P_ID436__Sto_A_4_14.png|right|framed|ACF and CCF for rectangular signals]]
We consider a periodic square wave signal  $p(t)$  corresponding to the top sketch with the two possible amplitude values  $0 \hspace{0.05cm} \rm V$  and  $1 \hspace{0.05cm} \rm V$  and the rectangle duration  $T$.  Thus, the period duration is  $T_0 = 2T$.

Below this is drawn the random signal  $z(t)$:
*This is  $z(t)=0 \hspace{0.05cm} \rm V$  between  $(2i-1)\cdot T$  and  $2i \cdot T$  respectively   (highlighted in red in the figure).
*In the intervals drawn in blue between  $2i \cdot T$  and  $(2i+1) \cdot T$  the signal value is two-point distributed  $(\pm 1 \hspace{0.05cm} \rm V)$.

The probability that in the intervals shown in blue  $z(t)=+1 \hspace{0.05cm} \rm V$  holds is generally equal  $p$  and independent of the previously selected values.

The lowest signal in the adjacent graph can be constructed from the first two. It holds:
:$$s(t) = {1}/{2} \cdot \big[p(t) + z(t)\big].$$

*In the time intervals drawn in red between  $(2i-1) \cdot T$  and  $2i \cdot T$  $(i$  integer$)$  holds  $s(t)=0 \hspace{0.05cm} \rm V$,  since here both  $p(t)$  and  $z(t)$  are equal to zero.
*In the intervening intervals,  the amplitude value is two-point distributed between  $0 \hspace{0.05cm} \rm V$  and  $1 \hspace{0.05cm} \rm V$,  where the value  $1 \hspace{0.05cm} \rm V$  occurs again with probability  $p$.
*Or in other words:   The signals  $z(t)$  and  $s(t)$  are equivalent pattern signals of the identical random process with bipolar  $(-1 \hspace{0.05cm} \rm V, \ +1 \hspace{0.05cm} \rm V)$  resp. unipolar  $(0 \hspace{0.05cm} \rm V, \ 1 \hspace{0.05cm} \rm V)$  signal representation.

Hints:
*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Cross-Correlation_Function_and_Cross_Power_Density|Cross-Correlation Function and Cross Power-Spectral Density]].
*Refer is also made to the chapter  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function|Auto-Correlation Function]].
*Sketch the sought correlation functions in each case in the range from  $-7T$  to  $+7T$.

===Questions===

<quiz display=simple>
{Calculate the ACF  $\varphi_z(\tau)$  and sketch it for  $p = 0.25$.  What values result for  $\tau = 0$,  $\tau = 3T$  and  $\tau = 6T$?
|type="{}"}
$\varphi_z(\tau= 0) \ = \ $ { 0.5 3% } $\ \rm V^2$
$\varphi_z(\tau= 3T) \ = \ $ { 0. } $\ \rm V^2$
$\varphi_z(\tau= 6T) \ = \ $ { 0.125 3% } $\ \rm V^2$

{Now,  using the result from  '''(1)'''  calculate the ACF  $\varphi_p(\tau)$.  What values result for  $\tau = 0$,  $\tau = 3T$  and  $\tau = 6T$?
|type="{}"}
$\varphi_p(\tau= 0) \ = \ $ { 0.5 3% } $\ \rm V^2$
$\varphi_p(\tau= 3T) \ = \ $ { 0. } $\ \rm V^2$
$\varphi_p(\tau= 6T) \ = \ $ { 0.5 3% } $\ \rm V^2$

{It holds again  $p = 0.25$.  Calculate the cross-correlation function  $\varphi_{pz}(\tau)$ for  $\tau = 0$,  $\tau = 3T$  and  $\tau = 6T$ ?
|type="{}"}
$\varphi_{pz}(\tau= 0) \ = \ $ { -0.26--0.24 } $\ \rm V^2$
$\varphi_{pz}(\tau= 3T) \ = \ $ { 0. } $\ \rm V^2$
$\varphi_{pz}(\tau= 6T) \ = \ $ { -0.26--0.24 } $\ \rm V^2$

{What ACF  $\varphi_c(\tau)$  results in general for the sum  $c(t) = a(t) + b(t)$ ?
|type="()"}
- $\varphi_c(\tau) = \varphi_a(\tau) + \varphi_b(\tau)$.
+ $\varphi_c(\tau) = \varphi_a(\tau) + \varphi_{ab}(\tau) + \varphi_{ba}(\tau) + \varphi_b(\tau)$.
- $\varphi_c(\tau) = \varphi_a(\tau) \star \varphi_b(\tau)$.

{Calculate the ACF  $\varphi_s(\tau)$,  taking into account the result of  '''(4)'''.   What values result with  $p = 0.25$  for  $\tau = 0$,  $\tau = 3T$  and  $\tau = 6T$ ?
|type="{}"}
$\varphi_s(\tau= 0) \ = \ $ { 0.125 3% } $\ \rm V^2$
$\varphi_s(\tau= 3T) \ = \ $ { 0. } $\ \rm V^2$
$\varphi_s(\tau= 6T) \ = \ $ { -0.03175--0.03075 } $\ \rm V^2$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The ACF value at  $\tau = 0$  gives the average power:
:$$\varphi_z ( \tau = 0) = {1}/{2} \cdot (1 {\rm V})^2 \hspace{0.15cm}\underline{= 0.5 {\rm V}^2}.$$

*For  $\tau = \pm T$,  $\underline{\tau = \pm 3T}$, ...   results  $\varphi_z ( \tau)\hspace{0.15cm}\underline{ = 0}$.
*For intermediate values  $\tau = \pm 2T$,  $\tau = \pm 4T$,  $\underline{\tau = \pm 6T}$, ...   applies:
:$$\varphi_z ( \tau) = \frac {1 {\rm V}^2}{2} \left(p \hspace{0.02cm} \cdot \hspace{0.02cm}p \hspace{0.2cm} + \hspace{0.2cm}p \hspace{0.02cm}\cdot \hspace{0.02cm}(p-1) \hspace{0.2cm}+\hspace{0.2cm} (p-1)\hspace{0.02cm} \cdot \hspace{0.02cm}p \hspace{0.2cm}+\hspace{0.2cm} (p-1)\hspace{0.02cm} \cdot \hspace{0.02cm}(p-1)\right) = \hspace{0.1cm}\text{...} \hspace{0.1cm}= 0.5\, {\rm V}^2 \cdot (1-2p)^2 .$$

*Here  $p$  stands for  $p \cdot (+1)$  and  $(p-1)$  for $(1-p) \cdot (-1)$, i.e. probability times normalized amplitude value,  respectively.
[[File:P_ID437__Sto_A_4_14_a.png|frame|right|Auto-correlation functions and cross-correlation function]]
*With  $p = 0.25$  one gets  $\varphi_z ( \tau = \pm 6 T) \hspace{0.15cm}\underline{=0.125 \rm V^2}$.

The blue curve shows  $\varphi_z(\tau)$  for  $p = 0.25$  in the range of  $-7T \le \tau \le +7T$:
*Because of the rectangular signal waveform,  a sum of triangular functions is obtained.
*For  $p = 0.5$  the outer  (smaller)  triangles would disappear.
 
'''(2)'''  The ACF  $\varphi_p(\tau)$  of the unipolar periodic signal  $p(t)$  is in the generalized representation of  '''(1)'''   ⇒   ACF  $\varphi_z(\tau)$ as a special case for  $p = 1$  .
*Now one obtains a periodic ACF  (see red curve in the above sketch)  with
:$$\varphi_p ( \tau = 0) = \varphi_p ( \tau = \pm 2 T) = \varphi_p ( \tau = \pm 4 T) = \hspace{0.1cm} \text{...} \hspace{0.1cm}\hspace{0.15cm}\underline{= 0.5 {\rm V}^2},$$
:$$\varphi_p ( \tau = \pm T) = \varphi_p ( \tau = \pm 3T) = \hspace{0.1cm} ... \hspace{0.1cm}\hspace{0.15cm}\underline{= 0}.$$

'''(3)'''  For the cross-correlation function results for  $\tau = \pm T$,  $\underline{\tau = \pm 3T}$, ...   always the value zero.
*In contrast,  the CCF values for  $\tau = \pm 2T$,  $\tau = \pm 2T$, ...   identical to those for  $\tau = 0$:
:$$\varphi_{pz} ( \tau = 0) = \varphi_{pz} ( \tau = \pm 2 T) = \varphi_{pz} ( \tau = \pm 4 T) = \hspace{0.1cm} \text{...} \hspace{0.1cm}= \frac {1 {\rm V}^2}{2} \left( p - (1-p)\right) = \frac {2p -1}{2}\, {\rm V}^2 .$$

*You get the following results with  $p = 0.25$  (see green curve in above sketch):
:$$\varphi_{pz} ( \tau = 0)\hspace{0.15cm}\underline{= -0.25 {\rm V}^2},\hspace{0.5cm}\varphi_{pz} ( \tau = 3T)\hspace{0.15cm}\underline{= 0},\hspace{0.5cm}\varphi_{pz} ( \tau = 6T)\hspace{0.15cm}\underline{= -0.25 {\rm V}^2}.$$

*With  $p = 1$  on the other hand  $z(t) \equiv p(t)$  would hold and so of course  $\varphi_{pz}(\tau) \equiv \varphi_{p}(\tau) \equiv \varphi_{z}(\tau)$.
*For the special case  $p = 0.5$  there would be no correlation between  $p(t)$  and  $z(t)$  and thus  $\varphi_{pz}(\tau) \equiv 0$.

'''(4)'''  Substituting  $c(t) = a(t) + b(t)$  into the general ACF definition yields:
:$$\varphi_c ( \tau ) = \overline{c(t)\hspace{0.02cm} \cdot \hspace{0.02cm} c(t + \tau)} = \overline{a(t)\hspace{0.02cm} \cdot \hspace{0.02cm} a(t + \tau)} \hspace{0.1cm}+\hspace{0.1cm}\overline{a(t)\hspace{0.02cm} \cdot \hspace{0.02cm} b(t + \tau)} +\overline{b(t)\hspace{0.02cm} \cdot \hspace{0.02cm} a(t + \tau)} \hspace{0.1cm}+\hspace{0.1cm}\overline{b(t)\hspace{0.02cm} \cdot \hspace{0.02cm} b(t + \tau)}. $$
:$$\Rightarrow \hspace{0.5cm} \varphi_c ( \tau ) = \varphi_{a} ( \tau ) \hspace{0.1cm} + \hspace{0.1cm} \varphi_{ab} ( \tau ) \hspace{0.1cm} + \hspace{0.1cm} \varphi_{ba} ( \tau ) \hspace{0.1cm} + \hspace{0.1cm}\varphi_{a} ( \tau ). $$

*The correct solution is thus the  proposed solution 2.
*The proposed solution 1 is true only if  $a(t)$  and  $b(t)$  are uncorrelated.
*The last proposition:  The convolution operation is always false.
*A similar equation would result only if we consider the PDF  $f_c(c)$  of the sum  $c(t) = a(t) + b(t)$  and  $a(t)$  and  $b(t)$  are statistically independent:  
:$$f_c (c) = f_a (a) \star f_b (b) .$$

'''(5)'''  Using the result from  '''(4)'''  and taking into account the factor  $1/2$  we get:
:$$\varphi_s ( \tau ) = {1}/{4} \cdot \big[ \varphi_{p} ( \tau ) \hspace{0.1cm} + \hspace{0.1cm} \varphi_{z} ( \tau ) \hspace{0.1cm} + \hspace{0.1cm} 2 \cdot \varphi_{pz} ( \tau ) \big] . $$

*This already takes into account that the CCF between  $p(t)$  and  $z(t)$  is an even function,  so that  $\varphi_{pz}(\tau) = \varphi_{zp}(\tau)$  holds.
*For  $\tau = 0$  one therefore obtains in general with the above results:
:$$\varphi_s( \tau = 0) = {1}/{4} \cdot \left( 0.5 {\rm V}^2 +0.5 {\rm V}^2 + 2 \cdot \frac{2p-1}{2} {\rm V}^2\right) .$$
*With  $p = 0.25$  we get  $\varphi_{pz} ( \tau = 0 ) = 0.125\rm V^2$.  This result is plausible.  On average,  only in every eighth interval  $s(t)=1 \hspace{0.05cm} \rm V$;  otherwise  $s(t)=0 \hspace{0.05cm} \rm V$.

*For even multiples of  $T$  holds:
:$$ \varphi_s ( \tau = \pm 2 T) = \varphi_s ( \tau = \pm 4 T) = \hspace{0.1cm} \text{ ...} \hspace{0.1cm} = \frac {0.5 {\rm V}^2}{4} \left( (1-2p)^2 +1 + 2 \cdot (2p -1)\right) = 0.5 \, {\rm V}^2 \hspace{0.02cm} \cdot \hspace{0.02cm} p^2.$$

*With  $p = 0.5$  we obtain for this the value  $0.03125 \hspace{0.1cm}{\rm V}^2$.  All ACF values at odd multiples of  $T$  are zero again.
*This gives the outlined ACF–curve.

[[File:P_ID441__Sto_A_4_14_e.png|framed|right|ACF of a unipolar rectangular signal]]
Thus, the numerical values we are looking for are:
:$$\varphi_{s} ( \tau = 0)\hspace{0.15cm}\underline{= 0.125 {\rm V}^2},$$
:$$\varphi_{s} ( \tau = 3T)\hspace{0.15cm}\underline{= 0},$$
:$$\varphi_{s} ( \tau = 6T)\hspace{0.15cm}\underline{= -0.03125 {\rm V}^2}.$$

A comparison with the sketch for subtask  '''(1)'''  shows that the binary signal  $s(t)$  has the same ACF as the ternary signal  $z(t)$ except for the factor  $1/4$.
{{ML-Fuß}}

[[Category:Theory of Stochastic Signals: Exercises|^4.6 CCF and Cross Power-Spectral Density^]]
[[de:Aufgaben:Aufgabe 4.14: AKF und KKF bei Rechtecksignalen]]

Aufgaben:Exercise 4.2: AM/PM Oscillations

2026-03-16T15:58:36Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces}}

[[File:P_ID1997__Dig_A_4_2.png|right|frame|Two possible AM/PM oscillations]]
We consider the signal set   $\{s_i(t)\}$   with the indexing variable  $i = 1, \ \text{...} \, M$.  All signals  $s_i(t)$  can be represented in the same way:
:$$s_i(t) =\left\{ \begin{array}{c} A_i \cdot \cos(2\pi f_{\rm T}t + \phi_i) \\0 \end{array} \right.\quad\begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm},\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

The signal duration  $T$  is an integer multiple of  $1/f_{\rm T}$,  where  $f_{\rm T}$  is the signal frequency  ("carrier frequency").

*For the sketch,  the duration of the energy-limited signals is  $T = 4/f_{\rm T}$,  i.e. exactly four oscillations are recognized within  $T$  in each case.

*The individual signals  $s_i(t)$  differ in amplitude  $(A_i)$  and/or phase  $(\phi_i)$.

For the two signals  (shown in the graph)  holds:
:$$s_1(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos(2\pi f_{\rm T}t ) \hspace{0.05cm},$$
:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2A \cdot \cos(2\pi f_{\rm T}t + \pi/4) \hspace{0.05cm}. $$

If we first restrict ourselves to these two signals  $s_1(t)$  and  $s_2(t)$,  they can be completely described by the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.  These are orthonormal to each other,  that is,  taking into account the time constraint on  $T$  holds:
:$$\int_{0}^{T}\varphi_1^2(t) \, {\rm d} t = \int_{0}^{T}\varphi_2^2(t) \, {\rm d} t = 1 \hspace{0.05cm},$$
:$$ \int_{0}^{T}\varphi_1(t) \cdot \varphi_2(t)\, {\rm d} t = 0\hspace{0.05cm}.$$

With these basis functions,  the two signals can be represented as follows:
:$$s_1(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_{11} \cdot \varphi_1(t) \hspace{0.05cm},$$
:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_{21} \cdot \varphi_1(t) + s_{22} \cdot \varphi_2(t) \hspace{0.05cm}. $$

In subtask  '''(7)'''  we want to check whether all signals  $s_i(t)$  according to the above definition   $($with arbitrary amplitude  $A_i$  and arbitrary phase  $\phi_i)$   can be described by the following equation:
:$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}. $$

The basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$  are to be found here by the   [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#The_Gram-Schmidt_process|"Gram–Schmidt process"]],   which was described in detail in the theory section.  The required equations are summarized here again:
:$$\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||}\hspace{0.4cm}{\rm with}\hspace{0.4cm}s_{11} = ||s_1(t)|| = \sqrt{\int_{0}^{T}s_1^2(t) \, {\rm d} t}\hspace{0.05cm},\hspace{0.4cm}s_{21} = \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_1(t) \hspace{-0.1cm} > \hspace{0.1cm} =\int_{0}^{T}s_2(t) \cdot \varphi_1(t)\, {\rm d} t\hspace{0.05cm},$$
:$$\theta_2(t) = s_2(t) - s_{21} \cdot \varphi_1(t)\hspace{0.05cm}, \hspace{0.2cm}\varphi_2(t) = \frac{\theta_2(t)}{||\theta_2(t)||}\hspace{0.05cm}.$$

Notes:
*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces|"Signals, Basis Functions and Vector Spaces"]].

* For abbreviation,  use the energy  $E = 1/2 \cdot A^2 \cdot T$.

* Furthermore,  the following trigonometric relation is given:  
:$$\cos(\alpha \pm \beta) = \cos(\alpha )\cdot \cos(\beta) \mp \sin(\alpha )\cdot \sin(\beta)\hspace{0.05cm}.$$

===Questions===
<quiz display=simple>
{What is the energy and the  "2–norm"  of the signal  $s_1(t)$,  expressed by  $E$?
|type="{}"}
$E_1\ = \ $ { 1 3% } $\ \cdot E$
$||s_1(t)|| \ = \ $ { 1 3% } $\ \cdot \sqrt{E}$

{What is the Gram–Schmidt basis function  $\varphi_1(t)$? 
|type="[]"}
- $\varphi_1(t) = \sqrt{E} \cdot {\rm cos}(2\pi f_{\rm T}t)$,
- $\varphi_1(t) = \cos(2\pi f_{\rm T}t)$,
+ $\varphi_1(t) = \sqrt{2/T} \cdot {\rm cos}(2\pi f_{\rm T}t)$.

{What is the relationship between  $s_1(t)$  and  $\varphi_1(t)$?
|type="[]"}
+ $s_1(t) = \sqrt{E} \cdot \varphi_1(t)$,
- $s_1(t) = A \cdot \varphi_1(t)$,
- $s_1(t) = \sqrt{2/T} \cdot \varphi_1(t)$.

{What is the  "inner product"  $s_{\rm 21} = 〈 s_2(t) \cdot \varphi_1(t)〉$?
|type="{}"}
$s_{\rm 21} \ = \ $ { 1.414 3% } $\ \cdot \sqrt{E}$

{What is the auxiliary function  $\theta_2(t)$?
|type="[]"}
- $\theta_2(t) = +\sqrt{2} \cdot A \cdot {\rm sin}(2\pi f_{\rm T}t)$,
+ $\theta_2(t) =-\sqrt{2} \cdot A \cdot {\rm sin}(2\pi f_{\rm T}t)$,
- $\theta_2(t) = \sqrt{2/T} \cdot {\rm sin}(2\pi f_{\rm T}t)$.

{Give the coefficients of   $s_2(t) = s_{\rm 21} \cdot \varphi_1(t) + s_{\rm 22} \cdot \varphi_2(t)$. 
|type="{}"}
$s_{\rm 21}\ = \ $ { 1.414 3% } $\ \cdot \sqrt{E}$
$s_{\rm 22}\ = \ $ { 1.414 3% } $\ \cdot \sqrt{E}$

{Which of the statements are generally true for the basis functions of the signal set  $\{s_i(t)\}$  with  $i = 1, \ \text{ ...} \ , M$,  if  $M \gg 2$?
|type="[]"}
- The number of basis functions is always  $N = M$.
+ The number of basis functions is always  $N = 2$.
+ Possible basis functions are cosine and (minus) sine.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  The energy can be calculated using the following equation:
:$$E_{1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}\int_{0}^{T}A^2 \cdot \cos^2(2\pi f_{\rm T}t )\, {\rm d} t = \frac{A^2 \cdot T}{2}\hspace{0.05cm}+\hspace{0.05cm}\frac{A^2 }{2}\int_{0}^{T} \cos(4\pi f_{\rm T}t )\, {\rm d} t = \frac{A^2 \cdot T}{2} \hspace{0.05cm}\underline{= 1 \cdot E}\hspace{0.05cm}. $$

*Here it is considered that  $T$  is an even multiple of  $1/f_{\rm T}$,  so the second integral vanishes.

*Further:
:$$||s_1(t)|| = \sqrt{E_1} = \sqrt{E} = \hspace{0.1cm}\hspace{0.15cm}\underline{1 \cdot\sqrt{E}}\hspace{0.05cm}.$$

'''(2)'''  Solution 3  is correct:  The basis function  $\varphi_1(t)$  is equal in form to  $s_1(t)$,  where holds:
:$$\varphi_1(t) = \frac{s_1(t)}{||s_1(t)||}= \frac{A \cdot \cos(2\pi f_{\rm T}t )}{\sqrt{E}}= \frac{A \cdot \cos(2\pi f_{\rm T}t )}{\sqrt{1/2 \cdot A^2 \cdot T}} = \sqrt{{2}/{T}}\cdot \cos(2\pi f_{\rm T}t )\hspace{0.05cm}.$$

'''(3)'''  Solution 1  is correct  since according to the equation given in  '''(2)''':
:$$s_1(t) = ||s_1(t)|| \cdot \varphi_1(t) = \sqrt{E} \cdot \varphi_1(t)\hspace{0.05cm}.$$

'''(4)'''  Using the signal  $s_2(t)$  according to the given information,  the basis function  $\varphi_1(t)$  according to subtask  '''(2)'''  and the given trigonometric relation we get:
:$$s_{21} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_1(t) \hspace{-0.1cm} > \hspace{0.1cm} =\int_{0}^{T}2A \cdot \cos(2\pi f_{\rm T}t + {\pi}/{4}) \cdot \sqrt{{2}/{T}}\cdot \cos(2\pi f_{\rm T}t )\, {\rm d} t = $$
:$$\Rightarrow \hspace{0.3cm}s_{21} = \sqrt{\frac{8A^2}{T}}\cdot \int_{0}^{T}\cos({\pi}/{4})\cdot \cos^2(2\pi f_{\rm T}t )\, {\rm d} t \hspace{0.1cm}-\sqrt{\frac{8A^2}{T}}\cdot \int_{0}^{T}\sin({\pi}/{4})\cdot \sin(2\pi f_{\rm T}t )\cdot \cos(2\pi f_{\rm T}t )\, {\rm d} t\hspace{0.05cm}. $$

*The second component yields the value  $0$  (orthogonality).  The first component yields:
:$$s_{21} = \sqrt{\frac{8A^2}{T}}\cdot \frac{1}{\sqrt{2}}\cdot \frac{T}{2} = \sqrt{A^2 \cdot T} = \sqrt{2E} \hspace{0.1cm}\hspace{0.15cm}\underline { = 1.414 \cdot \sqrt{E}}\hspace{0.05cm}.$$

'''(5)'''  According to the Gram–Schmidt process,  we obtain
:$$\theta_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_2(t) - s_{21} \cdot \varphi_1(t)\hspace{0.05cm} =2A \cdot \cos(2\pi f_{\rm T}t + {\pi}/{4}) - \sqrt{A^2 \cdot T}\cdot \sqrt{{2}/{T}}\cdot \cos(2\pi f_{\rm T}t ) $$
:$$\Rightarrow \hspace{0.3cm}\theta_2(t) =2A \cdot \cos({\pi}/{4})\cdot \cos(2\pi f_{\rm T}t )\hspace{0.1cm} - \hspace{0.1cm}2A \cdot \sin({\pi}/{4})\cdot \sin(2\pi f_{\rm T}t )\hspace{0.1cm} - \sqrt{2} \cdot A \cdot \cos(2\pi f_{\rm T}t )\hspace{0.05cm}. $$

*With  $\cos {(\pi/4)} = \sin (\pi/4) =\sqrt{0.5}$  it follows:
:$$\theta_2(t) = - \sqrt{2} \cdot A \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}.$$

*Therefore,  solution 2 is correct.

'''(6)'''  Analogous to subtask  '''(2)''',  the orthonormal basis function  $\varphi_2(t)$  is given by
:$$\varphi_2(t) = \frac{\theta_2(t)}{||\theta_2(t)||} = - \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}.$$

*Thus,  the signal  $s_2(t)$  can be represented by  $s_{21}$  according to subtask  '''(4)'''  as follows:
:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_{21} \cdot \varphi_1(t) + s_{22} \cdot \varphi_2(t) \hspace{0.05cm}, \hspace{0.2cm}s_{21} = \underline{ = 1.414 \cdot \sqrt {E}}\hspace{0.05cm},$$
:$$s_{22}\hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{\theta_2(t)}{\varphi_2(t)} = \frac{-\sqrt{2} \cdot A \cdot \sin(2\pi f_{\rm T}t )}{-\sqrt{2/T}\cdot \sin(2\pi f_{\rm T}t )} = \sqrt{2} \cdot \sqrt{1/2 \cdot A^2 \cdot T}\hspace{0.05cm} \underline{ = 1.414 \cdot \sqrt {E}}\hspace{0.05cm}.$$

'''(7)'''  We consider very many energy-limited signals  $(M \gg 2)$  of the following form:
:$$s_i(t)=\left\{ \begin{array}{c} A_i \cdot \cos(2\pi f_{\rm T}t + \phi_i) \\0 \end{array} \right.\quad\begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm},\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

The indexing variable can take the values  $i = 1, 2, \ \text{...} \ , M$.  Then holds:
* All  $M$  signals can be completely described by only  $N = 2$  basis functions:
:$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}. $$

* If one proceeds according to the Gram–Schmidt process,  one obtains for the two basis functions
:$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t + \phi_1)\hspace{0.05cm},\hspace{0.5cm}\varphi_2(t) = \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t + \phi_1 \pm {\pi}/{2})\hspace{0.05cm}.$$

* The sign in the argument of the second cosine function  $(± \pi/2)$  is not unique.  Rather,  the sign of  $s_{i 2}$  also depends on whether the plus sign or the minus sign was used for  $\varphi_2(t)$.

* However,  possible basis functions that then lead to other coefficients are also:
:$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t )\hspace{0.05cm},\hspace{0.5cm}\varphi_2(t) \pm \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}.$$

⇒   So the  solutions 2 and 3  are correct.
{{ML-Fuß}}

[[Category:Digital Signal Transmission: Exercises|^4.1 Basis Functions & Vector Spaces^]]
[[de:Aufgaben:Aufgabe 4.2: AM/PM-Schwingungen]]

Aufgaben:Exercise 3.4: Optimization of the Cutoff Frequency

2026-03-16T15:58:36Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission
}}

[[File:EN_Dig_A_3_4.png|right|frame|System sizes for binary and quaternary systems and for different cutoff frequencies]]
We compare a redundancy-free binary system  $(M = 2)$  and a redundancy-free quaternary system  $(M = 4)$  in terms of their worst-case S/N ratios:
:$$\rho_{\rm U} = \frac{\big[\ddot{o}(T_{\rm D})/2 \big]^2}{ \sigma_d^2} \hspace{0.05cm}.$$

$\ddot{o}(T_{\rm D})$  is the vertical eye opening and  $\sigma_d^2$  indicates the detection noise power.  The same boundary conditions apply to both system configurations  (similar to  [[Aufgaben:Exercise_3.4Z:_Eye_Opening_and_Level_Number|Exercise 3.4Z]]):
* The rectangular basic transmission pulse  $g_s(t)$  in NRZ format has the height  $s_0 = 1 \, {\rm V}$.

* The (equivalent) bit rate is  $R_{\rm B} = 100 \, {\rm Mbit/s}$ in both cases.

* The channel consists of a coaxial cable with the characteristic cable attenuation  $a_* = 80 \, {\rm dB}\Rightarrow 9.2 \, {\rm Np}$.

* Let the receiver filter be a Gaussian low-pass filter with cutoff frequency  $f_{\rm G}$,  which is to be optimized:
:$$H_{\rm G}(f) = {\rm e}^{{- \pi \cdot f^2}/{(2f_{\rm G})^2}}\hspace{0.05cm}.$$
* AWGN noise with  (one-sided)  power-spectral density  $N_0$  is present at the channel output.

* The decision thresholds are optimally selected and the detection time  $T_{\rm D} = 0$  is also best possible.

In contrast to  [[Aufgaben:Exercise_3.4Z:_Eye_Opening_and_Level_Number|Exercise 3.4Z]]  $($fixed cutoff frequency  $f_{\rm G} = 30 \, {\rm MHz}),$  the cutoff frequency of the low-pass filter is variable here.  This is to be determined in such a way that the worst-case S/N ratio  $\rho_{\rm U}$  is maximized and thus the worst-case error probability  $p_{\rm U}$  is minimized.

The table shows
*the  (normalized)  half eye opening,  and
*the  (normalized)  detection noise rms value

for the binary system  $(M = 2)$  and the quaternary system  $(M = 4)$  as well as for different  (normalized)  cutoff frequencies.  The normalization is based on the bit rate  $R_{\rm B}$.

To be noted:
* The table is valid for  $E_{\rm B}/N_0 = 5 \cdot 10^8$  and  $a_* = 80 \, {\rm dB}\Rightarrow 9.2 \, {\rm Np}$.

* Taking into account the ideal channel equalizer,  the normalized noise power results in
:$$\frac{ \sigma_d^2}{N_{\rm 0} \cdot R_{\rm B}} = \frac{ 1}{R_{\rm B}} \cdot \int_{0}^{\infty}{\rm exp}\left [2 \cdot 9.2\cdot \sqrt{2 \cdot f/R_{\rm B}} - 2\pi \cdot \frac{(f/R_{\rm B})^2}{(2 f_{\rm G}/R_{\rm B})^2} \right ]{\rm d} \hspace{0.05cm} f\hspace{0.05cm}.$$
* As will be derived in  [[Aufgaben:Exercise_3.4Z:_Eye_Opening_and_Level_Number|Exercise 3.4Z]],  the following holds for the  (normalized)  half eye opening:
:$$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = \frac{1}{ M-1}\cdot \big [1- 2 \cdot M \cdot {\rm Q} \left(\sqrt{2\pi} \cdot {\rm log_2}\hspace{0.1cm}(M) \cdot {f_{\rm G}}/{R_{\rm B}}\right)\big]\hspace{0.05cm}.$$
* Thus,  for the worst case S/N ratio,  the last term can be interpreted as  "energy per bit related to noise power density"  for the NRZ rectangular pulse considered here:
:$$\rho_{\rm U} = \left [\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} \right ]^2 \cdot\frac{N_{\rm 0} \cdot R_{\rm B}}{ \sigma_d^2} \cdot \frac{ s_0^2}{N_{\rm 0} \cdot R_{\rm B}}\hspace{0.05cm}.$$

Notes:
*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission|"Intersymbol Interference for Multi-Level Transmission"]].

*In the table,  $\sigma_d/s_0$  is given,  which means that the second and third terms of the above equation are combined here.

*By dividing the first column element  ("normalized half eye opening")  by the second element  $(\sigma_d/s_0)$  and squaring the quotient,  the result  $\rho_{\rm U}$  is obtained very easily.

===Questions===
<quiz display=simple>
{The table is not complete with respect to  $\sigma_d$.  Determine the following values:
|type="{}"}
$M = 2, f_{\rm G} = 0.33\text{:}\hspace{0.4cm} \sigma_d/s_0 \ = \ $ { 0.047 3% }
$M = 4, f_{\rm G} = 0.28\text{:}\hspace{0.4cm} \sigma_d/s_0 \ = \ $ { 0.021 3% }

{Determine the optimal cutoff frequency and the achievable worst-case signal-to-noise ratio for the binary system.
|type="{}"}
$M = 2\text{:}\hspace{0.9cm} f_{\rm G, \ opt}/R_{\rm B}\ = \ $ { 0.33 3% }
$M = 2\text{:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U, \ max}\ = \ $ { 11.85 3% } ${\ \rm dB}$

{Find the optimal cutoff frequency and the achievable worst-case signal-to-noise ratio for the quaternary system.
|type="{}"}
$M = 4\text{:}\hspace{0.97cm} f_{\rm G, \ opt}/R_{\rm B}\ = \ ${ 0.28 3% }
$M = 4\text{:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U,\ max}\ = \ $ { 15.21 3% } ${\ \rm dB}$

{Evaluate the results of subtasks  '''(2)'''  and  '''(3)'''  using the following statements.  Which of them are true?
|type="[]"}
+ If the normalized cutoff frequency  $f_{\rm G}/R_{\rm B} ≥ 0.35$,  the binary system is superior to the quaternary system.
- If the normalized cutoff frequency  $f_{\rm G}/R_{\rm B} = 0.33$,  the binary system is superior to the quaternary system.
+ The main reason for the superiority of the quaternary system over the binary system  (optimized in each case)  is the lower symbol rate.
+ From the present values,  it can be concluded that the (optimal) quaternary system is also better for  $a_* = 100 \, {\rm dB}$. 
- From the present values,  it can be concluded that the (optimal) quaternary system is also better for  $a_* = 40 \, {\rm dB}$. 
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Normalizing the cutoff frequency  $f_{\rm G}$  to the bit rate  $R_{\rm B}$  $($and not to the symbol rate $1/T)$,  the given noise rms values hold independently of the level number.  Thus one obtains:
:$$M = 2, \hspace{0.1cm}f_{\rm G}/R_{\rm B} = 0.33\text{:} \hspace{0.2cm}\sigma_d/s_0 \ \hspace{0.15cm}\underline { = 0.047}\hspace{0.05cm},$$
:$$M = 4, \hspace{0.1cm}f_{\rm G}/R_{\rm B} = 0.28\text{:} \hspace{0.2cm}\sigma_d/s_0 \ \hspace{0.15cm}\underline { = 0.021}\hspace{0.05cm}.$$

'''(2)'''  The optimum cutoff frequency is given when the quotient of the (half) eye opening and the noise rms value is maximum.
*The optimum results for the binary system for  $f_{\rm G}/R_{\rm B} \underline {= 0.33}$:
:$$\rho_{\rm U,\hspace{0.05cm} max} = \frac{0.184^2}{ 0.047^2} = 15.32\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U,\hspace{0.05cm} max}\hspace{0.15cm}\underline { = 11.85\,{\rm dB}} \hspace{0.05cm}.$$

*In contrast,  for the adjacent cutoff frequency values:
:$$f_{\rm G}/R_{\rm B} = 0.32\text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.155^2}{ 0.040^2} = 15.02\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 11.76\,{\rm dB} \hspace{0.05cm},$$
:$$f_{\rm G}/R_{\rm B} = 0.34 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.212^2}{ 0.055^2} =14.86\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 11.72\,{\rm dB} \hspace{0.05cm}.$$
*From this,  one can see the optimum,  albeit a flat one.

'''(3)'''  For $M = 4$, the following results are obtained:
:$$f_{\rm G}/R_{\rm B} = 0.27 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.097^2}{ 0.017^2} =32.56 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 15.13\,{\rm dB} \hspace{0.05cm},$$
:$$f_{\rm G}/R_{\rm B} = 0.28 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.121^2}{ 0.021^2} =33.20 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {= 15.21\,{\rm dB}} \hspace{0.05cm},$$
:$$f_{\rm G}/R_{\rm B} = 0.29 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.139^2}{ 0.025^2} =30.91 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 14.90\,{\rm dB} \hspace{0.05cm}.$$

*Thus,  the optimum cutoff frequency is  $f_{\rm G}/R_{\rm B} \underline {= 0.28}$  for the quaternary system.
*The signal-to-noise ratio is then more than  $3 \, {\rm dB}$  larger than for the binary system with optimized cutoff frequency.

'''(4)'''  Statements 1, 3 and 4  are correct:
*The correctness of the first statement is confirmed by the table.  For  $f_{\rm G}/R_{\rm B} ≥ 0.35$,  the binary system has a larger eye opening than the quaternary system.  Moreover,  by normalizing all frequencies to the bit rate,  the noise rms value is independent of the level number  $M$,  so that the optimization can be restricted to the eye opening.

*For  $f_{\rm G}/R_{\rm B} < 0.35$,  on the other hand,  the quaternary system is better,  so also for  $f_{\rm G}/R_{\rm B} = 0.33$.  Although this cutoff frequency is optimal for  $M = 2$  and not for  $M = 4$,  with $f_{\rm G}/R_{\rm B} = 0.33$  the quaternary system is better than the binary system by about  $0.85 \, {\rm dB}$.

*The third statement is true.  Due to the lower (more precisely:  half) symbol rate,  for  $M = 4$  the eye is still open even with  $f_{\rm G}/R_{\rm B} = 0.23$,  while for a binary system there is already an (almost) closed eye for  $f_{\rm G}/R_{\rm B} = 0.27$.

*With larger characteristic cable attenuation,  the tendency goes to smaller and smaller cutoff frequency in order to keep the increase of the noise as small as possible.  But if already at  $a_* = 80 \, {\rm dB}$  the quaternary system  (optimized with respect to cutoff frequency)  is better,  the same is true for  $100 \, {\rm dB}$.  The gain is larger  than $15.21 - 11.85 \approx 3.4 \, {\rm dB}$.  These values were obtained in questions  '''(2)'''  and  '''(3)'''.

*In contrast,  no statement is possible for the characteristic cable attenuation  $a_* = 40 \, {\rm dB}$  based on the available numerical material.  A system simulation provided the following results for this  $($for $E_{\rm B}/N_0 = 50 \, {\rm dB})$:
:$$M =2\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 15.43\,{\rm dB} \hspace{0.2cm}{\rm with}\hspace{0.2cm}f_{\rm G}/R_{\rm B} \approx 0.4 \hspace{0.05cm},$$
:$$M =4\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 14.65\,{\rm dB} \hspace{0.2cm}{\rm with}\hspace{0.2cm}f_{\rm G}/R_{\rm B} \approx 0.32 \hspace{0.05cm}.$$
{{ML-Fuß}}

[[Category:Digital Signal Transmission: Exercises|^3.4 Eye Opening with Multilevel Systems^]]
[[de:Aufgaben:Aufgabe 3.4: Grenzfrequenzoptimierung]]

Aufgaben:Exercise 3.10Z: Maximum Likelihood Decoding of Convolutional Codes

2026-03-16T15:58:35Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Decoding_of_Convolutional_Codes}}

[[File:EN_KC_Z_3_10.png|right|frame|Overall system model,  given for this exercise]]
The Viterbi algorithm represents the best known realization form for the maximum likelihood decoding of a convolutional code.  We assume here the following model:
* The information sequence  $\underline{u}$  is converted into the code sequence  $\underline{x}$  by a convolutional code.

*It is valid  $u_i ∈ \{0, \, 1\}$.  In contrast,  the code symbols are represented bipolar   ⇒   $x_i ∈ \{–1, \, +1\}$.

* Let the channel be given by the models  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Symmetric_Channel_.E2.80.93_BSC|$\text{BSC}$]]    ⇒   received values  $y_i ∈ \{–1, \, +1\}$  or  [[Channel_Coding/Channel_Models_and_Decision_Structures#AWGN_channel_at_binary_input| $\text{AWGN}$]]   ⇒ $y_i$  real-valued.

* Given a received sequence  $\underline{y}$  the Viterbi algorithm decides for the sequence  $\underline{z}$  according to
:$$\underline{z} = {\rm arg} \max_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.03cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} {\rm Pr}( \underline{x}_{\hspace{0.03cm}i} |\hspace{0.05cm} \underline{y} ) \hspace{0.05cm}.$$

*This corresponds to the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Criteria_.C2.BBMaximum-a-posteriori.C2.AB_and_.C2.BBMaximum-Likelihood.C2.AB| "Maximum a posteriori"]]  $\rm (MAP)$  criterion.  If all information sequences  $\underline{u}$  are equally likely,  this transitions to the somewhat simpler  [[Channel_Coding/Channel_Models_and_Decision_Structures#Criteria_.C2.BBMaximum-a-posteriori.C2.AB_and_.C2.BBMaximum-Likelihood.C2.AB| "Maximum likelihood criterion"]]  $\rm (ML)$:
:$$\underline{z} = {\rm arg} \max_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} {\rm Pr}( \underline{y} \hspace{0.05cm}|\hspace{0.05cm} \underline{x}_{\hspace{0.03cm}i} ) \hspace{0.05cm}.$$

*As a further result,  the Viterbi algorithm additionally outputs the sequence  $\underline{v}$  as an estimate for the information sequence  $\underline{u}$.

In this exercise,  you should determinethe relationship between the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding| $\text{Hamming distance}$]]  $d_{\rm H}(\underline{x}, \, \underline{y})$  and the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Maximum-likelihood_decision_at_the_AWGN_channel|$\text{Euclidean distance}$]]
:$$d_{\rm E}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y}) =\sqrt{\sum_{i=1}^{L} \hspace{0.2cm}(x_i - y_i)^2}\hspace{0.05cm}.$$

Then,  the above maximum likelihood criterion is to be formulated with
* the Hamming distance  $d_{\rm H}(\underline{x}, \, \underline{y})$,

* the Euclidean distance  $d_{\rm E}(\underline{x}, \, \underline{y})$,  and

* the  [[Channel_Coding/Decoding_of_Convolutional_Codes#Relationship_between_Hamming_distance_and_correlation|$\text{correlation value}$]]  $〈 x \cdot y 〉$.

Hints:
* This exercise refers to the chapter  [[Channel_Coding/Decoding_of_Convolutional_Codes| "Decoding of Convolutional Codes"]].

* Reference is made in particular to the section  [[Channel_Coding/Decoding_of_Convolutional_Codes#Viterbi_algorithm_based_on_correlation_and_metrics|"Viterbi algorithm – based on correlation and metrics"]].

* For simplicity,  "tilde"  and  "apostrophe"  are omitted.

* For more information on this topic,  see the following sections in this book:
:* [[Channel_Coding/Channel_Models_and_Decision_Structures#Criteria_.C2.BBMaximum-a-posteriori.C2.AB_and_.C2.BBMaximum-Likelihood.C2.AB| "MAP and ML criterion"]],

:* [[Channel_Coding/Channel_Models_and_Decision_Structures#Maximum-likelihood_decision_at_the_BSC_channel| "ML decision at the BSC channel"]],

:* [[Channel_Coding/Channel_Models_and_Decision_Structures#Maximum-likelihood_decision_at_the_AWGN_channel| "ML decision at the AWGN channel"]],

:* [[Channel_Coding/Decoding_of_Linear_Block_Codes#Block_diagram_and_requirements| "Decoding linear block codes"]].

===Questions===
<quiz display=simple>
{How are   $d_{\rm H}(\underline{x}, \, \underline{y})$   and   $d_{\rm E}(\underline{x}, \, \underline{y})$   related in the BSC model?
|type="()"}
-   $d_{\rm H}(\underline{x}, \, \underline{y}) = d_{\rm E}(\underline{x}, \, \underline{y})$  is valid.
-   $d_{\rm H}(\underline{x}, \, \underline{y}) = d_{\rm E}^2(\underline{x}, \, \underline{y})$  is valid.
+   $d_{\rm H}(\underline{x}, \, \underline{y}) = d_{\rm E}^2(\underline{x}, \, \underline{y})/4$  is valid.

{Which of the equations describe the maximum likelihood decoding in the BSC model?  The minimization/maximization refers alwaysto all  $\underline{x} ∈\mathcal{ C}$.
|type="[]"}
+ $\underline{z} = \arg \min {d_{\rm H}(\underline{x}, \, \underline{y})}$,
+ $\underline{z} = \arg \min {d_{\rm E}(\underline{x}, \, \underline{y})}$,
+ $\underline{z} = \arg \min {d_{\rm E}^2(\underline{x}, \, \underline{y})}$,

{Which equation describes the maximum likelihood decision in the BSC model?
|type="()"}
- $\underline{z} = \arg \min 〈 \underline{x} \cdot \underline{y} 〉$,
+ $\underline{z} = \arg \max 〈 \underline{x} \cdot \underline{y} 〉$.

{What equations apply to the maximum likelihood decision in the AWGN model?
|type="[]"}
- $\underline{z} = \arg \min {d_{\rm H}(\underline{x}, \, \underline{y})}$,
+ $\underline{z} = \arg \min {d_{\rm E}(\underline{x}, \, \underline{y})}$,
+ $\underline{z} = \arg \max 〈 \underline{x} \cdot \underline{y} 〉$.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Correct is the  proposed solution 3:
*Let the two binary sequences be   $\underline{x}$   and   $\underline{y}$   with   $x_i ∈ \{-1, \, +1\}, \ y_i ∈ \{-1, \, +1\}$.  Let the sequence length be  $L$  in each case.

*The Hamming distance   $d_{\rm H}(\underline{x}, \, \underline{y})$   gives the number of bits in which  $\underline{x}$  and  $\underline{y}$  differ,  for which thus  $x_i \, - y_i = ±2$   ⇒   $ (x_i \, - y_i)^2 = 4$  holds.

*Equal symbols  $(x_i = y_i)$  do not contribute to the Hamming distance and give  $(x_i \, – y_i)^2 = 0$.  According to the  solution 3,  we can therefore write:
:$$ d_{\rm H}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y}) =\frac{1}{4} \cdot \sum_{i=1}^{L} \hspace{0.2cm}(x_i - y_i)^2= \frac{1}{4} \cdot d_{\rm E}^2(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y})\hspace{0.05cm}.$$

'''(2)'''  All proposed solutions  are correct:
*In the BSC model,  it is common practice to select the code word  $\underline{x}$  with the smallest Hamming distance  $d_{\rm H}(\underline{x}, \, \underline{y})$  for the given received vector  $\underline{y}$:
:$$\underline{z} = {\rm arg} \min_{\underline{x} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm}d_{\rm H}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y})\hspace{0.05cm}.$$
*But according to the subtask  '''(1)'''  also applies:
:$$\underline{z} = {\rm arg} \min_{\underline{x} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm}d_{\rm E}^{\hspace{0.15cm}2}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y})/4\hspace{0.2cm}\Rightarrow \hspace{0.2cm} \underline{z} = {\rm arg} \min_{\underline{x} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm}d_{\rm E}^{\hspace{0.15cm}2}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y})\hspace{0.2cm}\Rightarrow \hspace{0.2cm} \underline{z} = {\rm arg} \min_{\underline{x} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm}d_{\rm E}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y})\hspace{0.05cm}.$$

*The factor  $1/4$  does not matter for the minimization.  Since  $d_{\rm E}(\underline{x}, \, \underline{y}) ≥ 0$,  it does not matter whether the minimization is done with respect to   $d_{\rm E}(\underline{x}, \, \underline{y})$   or   $d_{\rm E}^2(\underline{x}, \, \underline{y})$.

'''(3)'''  Correct is the  proposed solution 2:
*The square of the Euclidean distance can be expressed as follows:
:$$d_{\rm E}^{\hspace{0.15cm}2}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y}) =\sum_{i=1}^{L} \hspace{0.2cm}(x_i - y_i)^2 =\hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} x_i^{\hspace{0.15cm}2} \hspace{0.1cm}+ \hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} y_i^{\hspace{0.15cm}2}\hspace{0.1cm}-2 \cdot \sum_{i=1}^{L} \hspace{0.1cm} x_i \cdot y_i\hspace{0.05cm}.$$

*The first two summands are each equal to  $L$  and need not be considered for minimization.

*For the last expression in this equation,  $–2 \cdot 〈 \underline{x}, \, \underline{y} 〉$  can be written.

*Due to the negative sign,  minimization becomes maximization   ⇒   answer 2.

'''(4)'''  Correct are the  proposed solutions 2 and 3:
*For the AWGN channel,  unlike the BSC,  no Hamming distance can be specified.

*Based on the equation
:$$d_{\rm E}^{\hspace{0.15cm}2}(\underline{x} \hspace{0.05cm}, \hspace{0.1cm}\underline{y}) =\hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} x_i^{\hspace{0.15cm}2} \hspace{0.1cm}+ \hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} y_i^{\hspace{0.15cm}2}\hspace{0.1cm}-2 \cdot \sum_{i=1}^{L} \hspace{0.1cm} x_i \cdot y_i$$

:the same statements apply for the first and last summands as for the BSC model – see subtask  '''(3)'''.

*For the middle summand,  $y_i = x_i + n_i$  and  $x_i ∈ \{–1, \, +1\}$  hold:
:$$\sum_{i=1}^{L} \hspace{0.1cm} y_i^{\hspace{0.15cm}2} =\hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} x_i^{\hspace{0.15cm}2} \hspace{0.1cm}+ \hspace{0.1cm}\sum_{i=1}^{L} \hspace{0.1cm} n_i^{\hspace{0.15cm}2}\hspace{0.1cm}+2 \cdot \sum_{i=1}^{L} \hspace{0.1cm} x_i \cdot n_i \hspace{0.05cm}.$$

*The first summand gives again  $L$,  the second is proportional to the noise power,  and the last term vanishes since  $\underline{x}$  and  $\underline{n}$  are uncorrelated.

*So for minimizing  $d_{\rm E}(\underline{x}, \, \underline{y})$,  the sum over  $y_i^2$  need not be considered since there is no relation to the code sequences  $\underline{x}$.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^3.4 Decoding of Convolutional Codes^]]
[[de:Aufgaben:Aufgabe 3.10Z: ML–Decodierung von Faltungscodes]]

Aufgaben:Exercise 4.12: Root-Nyquist Systems

2026-03-16T15:58:35Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Modulationsverfahren/Quadratur–Amplitudenmodulation
}}

[[File:P_ID1722__Mod_A_4_11.png|right|frame|Spectra of transmission pulse (above) and detection pulse (below)]]
In  "quadrature amplitude modulation"  $\rm (QAM)$  systems,  the  "root-Nyquist variant"  is often chosen  (which gets its name from the spectral range)  instead of a rectangular basic transmission pulse.  The reason for this is the significantly smaller bandwidth.

*In this case,  the basic detection pulse  $g_d(t)$  satisfies the  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#First_Nyquist_criterion_in_the_time_domain|first Nyquist criterion]],  since  $G_d(f)$  is point-symmetric about the so-called  "Nyquist frequency"  $f_{\rm Nyq} = 1/T$ .
*$G_d(f)$  is a  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Raised-cosine_low-pass_filter|raised-cosine spectrum]],  where the rolloff factor  $r$  can take values from $0$  to  $1$  (including these limits).

Furthermore,  the following holds for the Nyquist frequency response:
* When  $|f| < f_1 = f_{\rm Nyq} · (1 – r)$   ⇒   $G_d(f)$  is constant and equal to  $g_0 · T$.
* At frequencies greater than  $f_2 = f_{\rm Nyq} · (1 + r)$   ⇒   $G_d(f)$  has no components.
* In between,  the slope is cosine.

The optimization of digital communication systems requires that the receiver frequency response  $H_{\rm E}(f)$  should be of the same shape as the transmission spectrum $G_s(f)$ .

To obtain dimensionally correct spectral functions for this task and the graph,  it is assumed that
:$$G_s(f) = \sqrt{g_0 \cdot T \cdot G_d(f)},$$
:$$ H_{\rm E}(f) = \frac{1}{g_0 \cdot T}\cdot G_s(f)\hspace{0.05cm}.$$

The top graph shows the transmission spectrum  $G_s(f)$  for the rolloff factors
*$r = 0$   (green dotted rectangle),
*$r = 0.5$   (blue solid curve),
*$r = 1$   (red dashed curve).

Below,  the spectrum  $G_d(f)$  of the basic detection pulse before the decider is shown in the same colors.
*The associated pulse   $g_d(t)$  is a [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#First_Nyquist_criterion_in_the_time_domain|Nyquist pulse]]  for all valid rolloff factors   $(0 ≤ r ≤ 1)$  as opposed to the basic transmission pulse   $g_s(t)$.
*For this,  the following equation is given in the literature - for example in  '''[Kam04]''' :
:$$g_s(t) = g_0 \cdot \frac{4 r t/T \cdot \cos \left [\pi \cdot (1+r) \cdot t/T \right ]+ \sin \left [\pi \cdot (1-r) \cdot t/T \right ]}{\left [1- (4 r t/T)^2 \right ] \cdot \pi \cdot t/T}\hspace{0.05cm}.$$

Hints:
*This exercise belongs to the chapter  [[Modulation_Methods/Quadrature_Amplitude_Modulation|"Quadrature Amplitude Modulation"]].
*Particular reference is made to the page   [[Modulation_Methods/Quadrature_Amplitude_Modulation#Nyquist_and_Root-Nyquist_QAM_systems|"Nyquist and Root-Nyquist systems"]]  in this chapter.
*Further useful informations can be found in the chapter  [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems|Properties of Nyquist Systems]]  in the book  "Digital Signal Transmission".
* '''[Kam04]'''  refers to the textbook  "Kammeyer, K.D.:  Nachrichtenübertragung.  Stuttgart: B.G. Teubner, 4. Auflage, 2004".
*Energies are to be specified in  $\rm V^2s$;  they thus refer to the reference resistance  $R = 1 \ \rm \Omega$.

===Questions===

<quiz display=simple>
{What is the basic transmission pulse  $g_s(t)$  for the rolloff factor  $r = 0$?  What is the signal value at time  $t = 0$?
|type="{}"}
$g_s(t = 0) \ = \ $ { 1 3% } $\ \cdot g_0$

{What is the basic transmission pulse  $g_s(t)$  for the rolloff factor $r = 1$?  What is the signal value at time  $t = 0$?
|type="{}"}
$g_s(t = 0) \ = \ $ { 1.273 3% } $\ \cdot g_0$

{Let  $r = 1$.  At what times does  $g_s(t)$  cross the axis?
|type="()"}
- At all multiples of the symbol duration  $T$.
- At  $t = ±0.25 T, \ ±0.75 T, \ ±1.25 T, \ ±1.75 T$, ...
+ At  $t = ±0.75 T, \ ±1.25 T,\ ±1.75 T$, ...

{What is the basic transmission pulse  $g_s(t)$  for the rolloff factor  $r = 0.5$?  What is the signal value at time  $t = 0$?
|type="{}"}
$g_s(t = 0) \ = \ $ { 1.137 3% } $\ \cdot g_0$

{Which statements are valid for the pulse amplitude, independent of  $r$ ?  Solve using the frequency domain.
|type="()"}
- The pulse amplitude can take any value in the range   $0 ≤ g_s(t = 0) ≤ g_0$  .
- The pulse amplitude can take any value in the range   $g_0 ≤ g_s(t = 0) ≤ 2 g_0$  .
+ The pulse amplitude can take any value in the range   $g_0 ≤ g_s(t = 0) ≤ 4 g_0/π$  .

{What is the energy  $E_{g_s}$  of the basic transmission pulse  $g_s(t)$  when  $r = 0$  and  $r = 1$?
|type="{}"}
$r = 0\text{:} \ \ \ \ E_{g_s} \ = \ $ { 1 3% } $\ \cdot g_0^2 \cdot T$
$r = 1\text{:} \ \ \ \ E_{g_s} \ = \ $ { 1 3% } $\ \cdot g_0^2 \cdot T$

</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  If we substitute  $r = 0$  into the given equation,  the first terms in the numerator and denominator disappear and we get:
: $$g_s(t) = g_0 \cdot \frac{\sin \left (\pi \cdot t/T \right )}{\pi \cdot t/T} = g_0 \cdot {\rm sinc} \left ( {t}/{T} \right )\hspace{0.05cm}.$$
*At time  $t = 0$,  ${\rm sinc} \left ( {t}/{T} \right ) =g_0$:  
:$$ g_s(t) \hspace{0.15cm}\underline { = 1.0 } \cdot g_0 \hspace{0.05cm}.$$

'''(2)'''  When  $r = 1$,  the given equation simplies as follows:
:$$g_s(t) = \frac{4 \cdot g_0}{\pi} \cdot \frac{ \cos \left (2 \pi \cdot t/T \right )}{\left [1- (4 t/T)^2 \right ] }\hspace{0.3cm}\Rightarrow \hspace{0.3cm} g_s(t = 0) = \frac{4 \cdot g_0}{\pi} \hspace{0.15cm}\underline {= 1.273 }\cdot g_0 \hspace{0.05cm}.$$

'''(3)'''  The  last answer  is correct:
*Zero intercepts are only possible for  $r = 1$  if the cosine function in the numerator is zero,  that is,  for all integer values of   $k$:
:$$2 \pi \cdot t/T = {\pi}/{2} + k \cdot \pi \hspace{0.3cm}\Rightarrow \hspace{0.3cm} t = \pm 0.25T, \hspace{0.15cm} \pm 0.75T, \hspace{0.15cm}\pm 1.25T, \hspace{0.15cm} ...$$
*However,  only the last answer is correct,  since the zero values at   $±0.25T$  are cancelled by the zero in the denominator.
*Applying de l'Hospital's rule yields   $g_s(t = ± 0.25T) = g_0$.

'''(4)'''  With  $r = 0.5$  and the shortcut  $x = t/T$,  one gets:
:$$g_s(x) = \frac{g_0}{\pi} \cdot \frac{2 \cdot x \cdot \cos \left (1.5\pi \cdot x \right )+ \sin \left (0.5\pi \cdot x \right )}{\left (1- 4 \cdot x^2 \right ) \cdot x}\hspace{0.05cm}.$$

*For the calculation at time  $t = 0$,  de l'Hospital's rule must be applied.
*The derivatives of the numerator and denominator give:
:$$Z'(x) = 2 \cdot \cos \left (1.5\pi \cdot x \right ) - 3 \pi \cdot x \cdot \sin \left (1.5\pi \cdot x \right ) + 0.5 \pi \cdot \cos \left (0.5\pi \cdot x \right ),$$
[[File:P_ID1723__Mod_A_4_11b.png|right|frame|Basic transmission pulse  (root-Nyquist)  and basic detection pulse  (Nyquist)]]

:$$N'(x) = \left (1- 4 \cdot x^2 \right ) - 8 \cdot x^2 \hspace{0.05cm}.$$
*The two boundary transitions for  $x → 0$  yield:
:$$\lim_{x \rightarrow 0} Z'(x) = 2 +{\pi }/{2},\hspace{0.2cm} \lim_{x \rightarrow 0} N'(x) = 1 \hspace{0.05cm}.$$
*Thus, for the signal amplitude at time   $t = 0$:
:$$g_s(t=0) = \frac{g_0}{\pi} \cdot \left ( 2 +{\pi }/{2} \right ) = {g_0} \cdot \left ( 0.5 + {2}/{\pi } \right )\hspace{0.15cm}\underline {= 1.137} \cdot g_0 \hspace{0.05cm}.$$

Here,  the graph illustrates the results calculated again:
*$g_d(t)$  is a Nyquist pulse,  meaning that it has zero crossings at least at all multiples of the symbol duration  $T$  (and possibly others depending on the rolloff factor).
*On the other hand,  the pulse  $g_s(t)$  does not satisfy the Nyquist criterion.  Moreover,  from this plot one can once again see that for   $r ≠ 0$  the pulse amplitude $g_s(t = 0)$  is always larger than $g_0$.

'''(5)'''  The  last answer  is correct  $($the first answer is ruled out from the results in questions  '''(2)'''  and  '''(4)''' $)$.  The validity of the lower bound   $g_0$  and the upper bound   $4g_0/π$  can be proved as follows:
* The pulse amplitude  $g_s(t = 0)$  is generally equal to the area under the spectral function  $G_s(f)$.
* The smallest area is obtained for  $r = 0$.  Here,   $G_s(f) = g_0 · T$  is in the range  $|f| < ±1/(2T)$.  Thus, the area is equal to  $g_0$.
* The largest area is obtained for  $r = 1$. Here,   $G_s(f)$  extends to the range   $±1/T$  and has a cosine shape.
*The result  $g_s(t = 0) = 4g_0/π$  was already calculated in question   '''(3)''' .  Though it still holds that:
:$$g_s(t=0) = 2 \cdot {g_0} \cdot \int_{ 0 }^{1/T} {\cos\left(\frac{\pi }{2}\cdot f \cdot T \right)}\hspace{0.1cm} {\rm d}f = \frac{4 g_0}{\pi} \cdot \int_{ 0 }^{\pi/2} {\cos\left(x \right)}\hspace{0.1cm} {\rm d}x = {4 g_0}/{\pi} \cdot \big[\sin(\pi/2) - \sin(0) \big] = {4 g_0}/{\pi}\hspace{0.05cm}.$$

'''(6)'''  The energy of the basic transmission pulse   $g_s(t)$  can be found in the time or frequency domain according to Parseval's theorem:
:$$E_{g_s} = \int_{ -\infty }^{+\infty} {[g_s(t)]^2}\hspace{0.1cm} {\rm d}t = \int_{ -\infty }^{+\infty} {|G_s(f)|^2}\hspace{0.1cm} {\rm d}f \hspace{0.05cm}.$$
*From the equations and graph on the exercise page,  we can see that   $|G_s(f)|^2$  has the same shape as   $G_d(f)$,  but the height is now   $(g_0 · T)^2$  instead of   $g_0 · T$:
:$$E_{g_s} = \int_{ -\infty }^{+\infty} {|G_s(f)|^2}\hspace{0.1cm} {\rm d}f = \frac{g_0^2 \cdot T^2}{g_0 \cdot T} \cdot \int_{ -\infty }^{+\infty} {G_d(f)}\hspace{0.1cm} {\rm d}f \hspace{0.05cm}.$$
*Due to the Nyquist form of  $G_d(f)$,  it holds independently of   $r$:
:$$\int_{ -\infty }^{+\infty} {G_d(f)}\hspace{0.1cm} {\rm d}f = g_0 \hspace{0.05cm}.$$
*Thus,  the pulse energy is also independent of  $r$,  so it is also valid for   $r = 0$  and  $r = 1$.  In  both cases,  $E_ {g_s}\hspace{0.15cm}\underline { = 1.0} · g_0^2 · T.$

{{ML-Fuß}}

[[Category:Modulation Methods: Exercises|^4.3 Quadrature Amplitude Modulation^]]
[[de:Aufgaben:Aufgabe 4.12: Wurzel–Nyquist–Systeme]]

Aufgaben:Exercise 2.6: GF(P power m). Which P, which m?

2026-03-16T15:58:34Z

Maintenance script: Fix interlanguage link: resolve redirect chain

{{quiz-Header|Buchseite=Channel_Coding/Extension_Field}}

[[File:P_ID2510__KC_A_2_6_neu.png|right|frame|Underlying tables for addition and multiplication]]
A Galois field  ${\rm GF}(q)$  with  $q = P^m$  elements defined by the adjacent tables is to be analyzed
*for addition  $($marked with "$+$"$)$,  and

*for multiplication  $($marked with "$\hspace{0.05cm}\cdot\hspace{0.05cm}$)".

This Galois field   ${\rm GF}(q) = \{\hspace{0.1cm}z_0,\hspace{0.1cm} z_1,\hspace{0.05cm} \text{...} , \hspace{0.1cm}z_{q-1}\}$   satisfies all the requirements for a finite field listed in the chapter  [[Channel_Coding/Some_Basics_of_Algebra|"Some Basics of Algebra"]] . Thus,  commutative, associative and distributive laws are also satisfied.

Furthermore there is
* a neutral element with respect to addition   ⇒   $N_{\rm A}$:
:$$\exists \hspace{0.15cm} z_j \in {\rm GF}(q)\text{:}\hspace{0.25cm}z_i + z_j = z_i \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_j = N_{\rm A} \hspace{0.25cm}{\rm (zero\hspace{0.15cm}element)}\hspace{0.05cm},$$
* a neutral element with respect to multiplication   ⇒   $N_{\rm M}$:
:$$\exists \hspace{0.15cm} z_j \in {\rm GF}(q)\text{:}\hspace{0.25cm}z_i \cdot z_j = z_i \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_j = N_{\rm M} \hspace{0.25cm}{\rm (identity\hspace{0.15cm}element)}\hspace{0.05cm},$$
* for all elements  $z_i$  an additive inverse   ⇒   ${\rm Inv_A}(z_i)$:
:$$\forall \hspace{0.15cm} z_i \in {\rm GF}(q)\hspace{0.15cm} \exists \hspace{0.15cm} {\rm Inv_A}(z_i) \in {\rm GF}(q)\text{:}$$
::$$z_i + {\rm Inv_A}(z_i) = N_{\rm A} = {\rm "\hspace{-0.15cm}0\hspace{-0.1cm}"}\hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm short}\text{:}\hspace{0.15cm}{\rm Inv_A}(z_i) = - z_i \hspace{0.05cm}, $$
* for all elements  $z_i$  except the zero element a multiplicative inverse  ⇒  ${\rm Inv_M}(z_i)$:
:$$\forall \hspace{0.15cm} z_i \in {\rm GF}(q),\hspace{0.15cm} z_i \ne N_{\rm A} \hspace{0.15cm} \exists \hspace{0.15cm} {\rm Inv_M}(z_i) \in {\rm GF}(q)\text{:}$$
::$$z_i \cdot {\rm Inv_M}(z_i) = N_{\rm M} = {\rm "\hspace{-0.15cm}1\hspace{-0.1cm}"}\hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm short}\text{:}\hspace{0.15cm}{\rm Inv_M}(z_i) = z_i^{-1}\hspace{0.05cm}. $$

Hints:
* This exercise belongs to the chapter  [[Channel_Coding/Extension_Field| "Extension Field"]].

* In the tables,  the elements   $z_0, \hspace{0.05cm} \text{...} \hspace{0.1cm} , \ z_8$   are called  "coefficient vectors".

* For example,  "$2 \hspace{0.03cm}1$"  stands for  "$2 \cdot \alpha + 1$".

===Questions===
<quiz display=simple>
{Specify the parameters of the Galois field considered here.
|type="{}"}
$P \ = \ ${ 3 }
$m \ = \ ${ 2 }
$q \ = \ ${ 9 }

{What is the neutral element of addition?
|type="()"}
+ The neutral element of addition is  $N_{\rm A} = \,$ "$0\hspace{0.03cm}0$",
- The neutral element of addition is  $N_{\rm A} = \,$ "$0\hspace{0.03cm}1$".

{What is the neutral element of multiplication?
|type="()"}
- The neutral element of multiplication is  $N_{\rm M} = \,$ "$0\hspace{0.03cm}0$",
+ The neutral element of multiplication is  $N_{\rm M} = \,$ "$0\hspace{0.03cm}1$".

{What statements are true regarding additive inverses?
|type="[]"}
+ It holds  ${\rm Inv_A} ($"$0\hspace{0.03cm}2$") $\, = \, $ "$0\hspace{0.03cm}1$",
+ It holds  ${\rm Inv_A} ($"$1\hspace{0.03cm}1$") $\, = \, $ "$2\hspace{0.03cm}2$",
- It holds  ${\rm Inv_A} ($"$2\hspace{0.03cm}2$") $\, = \, $ "$0\hspace{0.03cm}0$".

{Which of the following statements are true about the multiplication?
|type="()"}
- The multiplication is defined here modulo  $p(\alpha) = \alpha^2 + 2$.
+ The multiplication is defined here modulo  $p(\alpha) = \alpha^2 + 2\alpha + 2$.

{What statements are true regarding multiplicative inverses?
|type="[]"}
- There is a multiplicative inverse for all elements  $z_i ∈ {\rm GF}(P^m)$  .
+ It holds  ${\rm Inv_M} ($"$1\hspace{0.03cm}2$") $\, = \, $"$1\hspace{0.03cm}0$".
- It holds  ${\rm Inv_M} ($"$2\hspace{0.03cm}1$") $\, = \, $ "$1\hspace{0.03cm}2$".

{Does  $($"$2\hspace{0.03cm}0$" $\, + \,$ "$1\hspace{0.03cm}2$"$)$ $\, \cdot\, $ "$1\hspace{0.03cm}2$" $\, = \, $"$2\hspace{0.03cm}0$" $\, \cdot\, $ "$1\hspace{0.03cm}2$" $\, + \, $"$1\hspace{0.03cm}2$" $\, \cdot\, $ "$1\hspace{0.03cm}2$" hold?
|type="()"}
+ Yes.
- No.
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''  Each element consists of two ternaries   ⇒   $\underline{P = 3}, \ \underline{m = 2}$.  There are  $q = P^m = 3^8 = \underline{9 \rm \hspace{0.2cm}elements}$.

'''(2)'''  Correct is the  proposed solution 1:
*The neutral element of the addition   $(N_{\rm A})$  satisfies for all  $z_i ∈ {\rm GF}(P^m)$  the condition  $z_i + N_{\rm A} = z_i$.
*From the addition table it can be read that  "$0\hspace{0.03cm}0$"  satisfies this condition.

'''(3)'''  Correct is the  proposed solution 2:
*The neutral element of the multiplication   $(N_{\rm M})$  must always satisfy the condition  $z_i \cdot N_{\rm M} = z_i$.

*From the multiplication table,  $N_{\rm M} = \, "\hspace{-0.15cm}0\hspace{0.03cm}1\hspace{-0.1cm}"$.

*In polynomial notation,  this corresponds to  $k_1 = 0$  and  $k_0 = 1$:
:$$k_1 \cdot \alpha + k_0 = 1 \hspace{0.05cm}.$$

'''(4)'''  With the polynomial representation,  the following calculations result:
:$${\rm Inv_A}("\hspace{-0.05cm}0\hspace{0.03cm}2") \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(2) = (-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = 1 \hspace{0.25cm}\Rightarrow \hspace{0.25cm}{\rm vector}\hspace{0.15cm}"\hspace{-0.1cm}0\hspace{0.03cm}1\hspace{-0.1cm}"\hspace{0.05cm},$$
:$${\rm Inv_A}("\hspace{-0.05cm}1\hspace{0.03cm}1")\hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(\alpha + 1) = \big[(-\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] +\big[(-1) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] =2\alpha + 2 \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}\hspace{-0.05cm}2\hspace{0.03cm}2\hspace{-0.1cm}"\hspace{0.05cm},$$
:$${\rm Inv_A}("\hspace{-0.05cm}2\hspace{0.03cm}2")\hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(2\alpha + 2) = \big[(-2\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] +\big[(-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] =\alpha + 1 \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}\hspace{-0.05cm}1\hspace{0.03cm}1\hspace{-0.1cm}"\hspace{0.05cm}.$$

Consequently,  only the  first two proposed solutions  are correct.

However,  the exercise can also be solved without calculation using the addition table alone:
*For example,  you can find the inverse of  "$2\hspace{0.03cm}2$"  by looking for the column with the entry  "$0\hspace{0.03cm}0$"  in the last row.

*You find the column labeled  "$1\hspace{0.03cm}1$"  and thus  ${\rm Inv_A}("\hspace{-0.15cm}2\hspace{0.03cm}2\hspace{-0.15cm}") = \, "\hspace{-0.15cm}1\hspace{0.03cm}1\hspace{-0.15cm}"$.

'''(5)'''  Multiplying  $\alpha$  $($vector "$1\hspace{0.03cm}0$"$)$  by itself gives  $\alpha^2$.
* If the first proposed solution were valid,  the condition  $\alpha^2 + 2 = 0$  and thus  $\alpha^2 = (-2) \, {\rm mod} \, 3 = 1$,  thus yielding the vector  "$0\hspace{0.03cm}1$".

* Assuming the second proposed solution,  it follows from the condition  $\alpha^2 + 2\alpha + 2 = 0$   in polynomial notation:
:$$\alpha^2 = \big [(-2\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + \big[(-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big ] = \alpha + 1 $$
:and thus the coefficient vector  "$1\hspace{0.03cm}1$".

*In the multiplication table,  in row 4, column 4,  we find exactly the entry  "$1\hspace{0.03cm}1$"   ⇒   So,  the correct answert is the  proposed solution 2.

'''(6)'''  The multiplicative inverse to  "$1\hspace{0.03cm}2$"  can be found in row 6 of the multiplication table as the column with the entry  "$0\hspace{0.03cm}1$"  
*So the  proposed solution 2  is correct in contrast to proposal 3.  Namely,  ${\rm Inv_M}("\hspace{-0.15cm}21\hspace{-0.15cm}") = \, "\hspace{-0.15cm}2\hspace{0.03cm}0\hspace{-0.1cm}"$ holds.

*We check these results considering  $\alpha^2 + 2\alpha + 2 = 0$  by multiplications:
:$$"\hspace{-0.15cm}1\hspace{0.03cm}2\hspace{-0.1cm}" \hspace{0.05cm}\cdot \hspace{0.05cm}"\hspace{-0.15cm}1\hspace{0.03cm}0\hspace{-0.1cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm} (\alpha + 2) \cdot \alpha = \alpha^2 + 2\alpha = (-2\alpha-2) + 2\alpha = -2 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = 1 \hspace{0.15cm} \Rightarrow \hspace{0.15cm} {\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}0\hspace{0.03cm}1\hspace{-0.15cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm}{\rm multiplicative \hspace{0.15cm}inverse}\hspace{0.05cm}.$$
:$$"\hspace{-0.15cm}2\hspace{0.03cm}1\hspace{-0.1cm}" \hspace{0.05cm}\cdot \hspace{0.05cm}"\hspace{-0.15cm}1\hspace{0.03cm}2\hspace{-0.1cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm} (2\alpha + 1) \cdot (\alpha + 2) = 2 \alpha^2 + \alpha + 4\alpha + 2 = 2 \alpha^2 + 5\alpha + 2 = 2 \cdot (-2\alpha - 2) + 5\alpha + 2 = (\alpha - 2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = \alpha +1 $$
:$$\hspace{2.725cm} \Rightarrow \ {\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}1\hspace{0.03cm}1\hspace{-0.15cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm}{\rm no\hspace{0.15cm}multiplicative \hspace{0.15cm}inverse}\hspace{0.05cm}.$$

*The solution suggestion 1 is therefore not correct,  because there is no multiplicative inverse for  "$00$".

'''(7)'''  The two expressions agree   ⇒   YES, as the following calculations show:
:$$("\hspace{-0.15cm}20\hspace{-0.1cm}" + "\hspace{-0.15cm}12\hspace{-0.1cm}") \ \cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}02\hspace{-0.1cm}"\cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}21\hspace{-0.1cm}"\hspace{0.05cm},$$
:$$"\hspace{-0.15cm}20\hspace{-0.1cm}" \cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" + "\hspace{-0.15cm}12\hspace{-0.1cm}" \cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}02\hspace{-0.1cm}" + "\hspace{-0.15cm}22\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}21\hspace{-0.1cm}"\hspace{0.05cm}.$$

This means:   The distributive law has been proved at least on a single example.
{{ML-Fuß}}

[[Category:Channel Coding: Exercises|^2.2 Extension Field^]]
[[de:Aufgaben:Aufgabe 2.6: GF(P hoch m). Welches P, welches m?]]