View source for Applets:Sampling of Analog Signals and Signal Reconstruction

{{LntAppletLinkEnDe|sampling_en|sampling}}

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

[[File:EN_Abtastung_1.png|center|frame|Top: &nbsp;&nbsp; Underlying model for sampling and signal reconstruction<br>Bottom: &nbsp; Example for time discretization of the continuous&ndash;time signal&nbsp; $x(t)$]]

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


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

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

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


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

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

Parameters for the receiver frequency response&nbsp; $H_{\rm E}(f)$&nbsp; are the cut&ndash;off frequency and the rolloff factor&nbsp; (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)''' &nbsp; All signal values are normalized to&nbsp; $\pm 1$. 

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

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

'''(4)''' &nbsp; From these the <u>signal&ndash;distortion&ndash;distance</u>&nbsp; $10 \cdot \lg \ (P_x/P_\varepsilon)$&nbsp; can be calculated.
 
'''(5)''' &nbsp; Does the spectral function&nbsp; $X(f)$&nbsp; for positive frequencies consists of&nbsp; $I$&nbsp; Dirac delta lines with the (possibly complex) weights&nbsp; $X_1$, ... , $X_I$,<br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 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)''' &nbsp; Correspondingly, the following applies to the distortion power if the spectral function&nbsp; $E(f)$&nbsp; in the range&nbsp; $f>0$&nbsp; has&nbsp; $J$&nbsp; Dirac delta lines with weights&nbsp; $E_1$, ... , $E_J$: 

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

==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&nbsp; $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be&nbsp; $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals&nbsp; $T_{\rm A}$&nbsp; be&nbsp; $x_{\rm A}(t)$.
*Out of the sampling time points&nbsp; $\nu \cdot T_{\rm A}$&nbsp; always holds&nbsp; $x_{\rm A}(t) \equiv 0$.
*The run variable&nbsp; $\nu$&nbsp; be an&nbsp; [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:  &nbsp; &nbsp; $\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&nbsp; $K$, 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.
<br><br>
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.&nbsp; 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=
$\text{Definitions:}$&nbsp;

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

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

*The&nbsp; '''Dirac delta pulse (in the time domain)'''&nbsp; consists of infinitely many Dirac delta pulses, each equally spaced&nbsp; $T_{\rm A}$&nbsp; and all with equal pulse weight&nbsp; $T_{\rm A}$:
 
:$$p_{\delta}(t) =  \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
 \delta(t- \nu \cdot T_{\rm A}
 )\hspace{0.05cm}.$$}}
 

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


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

The spectrum of the sampled signal&nbsp; $x_{\rm A}(t)$&nbsp; is obtained by applying the&nbsp; [[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&nbsp; Dirac delta pulse&nbsp; $p_{\delta}(t)$ &nbsp; (in the time domain) &nbsp; into a&nbsp; [[Signal_Representation/Fourier_Series|"Fourier Series"]]&nbsp; and transforms it using the&nbsp; [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Shifting Theorem"]]&nbsp; into the frequency domain, the following correspondence &nbsp; &rArr; &nbsp; [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"proof"]] results with the distance&nbsp; $f_{\rm A} = 1/T_{\rm A}$&nbsp; 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&nbsp; $T_{\rm A} = 50\ {\rm &micro;s}$&nbsp; und&nbsp; $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
The result states:
*The Dirac delta pulse&nbsp; $p_{\delta}(t)$&nbsp; in the time domain consists of infinitely many Dirac delta pulses, each at the same distance&nbsp; $T_{\rm A}$&nbsp; and all with the same pulse weight&nbsp; $T_{\rm A}$.
*The Fourier transform of&nbsp; $p_{\delta}(t)$&nbsp; again gives a Dirac delta pulse, but now in the frequency domain &nbsp; ⇒ &nbsp; $P_{\delta}(f)$.
*Also&nbsp; $P_{\delta}(f)$&nbsp; consists of infinitely many Dirac delta pulses, now in the respective spacing&nbsp; $f_{\rm A} = 1/T_{\rm A}$&nbsp; and all with pulse weight&nbsp; $1$.
*The distances of the Dirac delta lines in time and frequency domain thus follow the&nbsp; [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity Theorem"]]: &nbsp; $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$


From this follows: &nbsp; From the spectrum&nbsp; $X(f)$&nbsp; is obtained by convolution with the Dirac delta line shifted by&nbsp; $\mu \cdot f_{\rm A}$&nbsp;:
 
:$$X(f) \star \delta
 (f- \mu \cdot f_{\rm A}
 )= X (f- \mu \cdot f_{\rm A}
 )\hspace{0.05cm}.$$

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


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

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


===Signal reconstruction===

[[File:P_ID1123__Sig_T_5_1_S5a_neu.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&nbsp; For example, consider the following system: 
*The analog signal&nbsp; $x(t)$&nbsp; with bandwidth&nbsp; $B_{\rm NF}$&nbsp; is sampled as described above. 
*At the output of an ideal transmission system, the also discrete-time signal&nbsp; $y_{\rm A}(t) = x_{\rm A}(t)$&nbsp; is present. 
*The question now is how the block &nbsp; '''signal reconstruction''' &nbsp; has to be designed so that also&nbsp; $y(t) = x(t)$&nbsp; holds.

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

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

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


===The Sampling Theorem===

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

From the above graph, it can be seen that the following condition must be satisfied: &nbsp; $f_{\rm A} - B_{\rm   NF} > B_{\rm   NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot  B_{\rm   NF}\hspace{0.05cm}.$
 
{{BlaueBox|TEXT=
$\text{Sampling theorem:}$&nbsp; If an analog signal&nbsp; $x(t)$&nbsp; has only spectral components in the range&nbsp; $\vert f \vert < B_{\rm NF}$, it can be completely reconstructed from its sampled signal&nbsp; $x_{\rm A}(t)$&nbsp; 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 &nbsp; ⇒ &nbsp; $T_{\rm A} = 1/(2B_{\rm NF})$&nbsp; 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&nbsp; $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$&nbsp; must be used.


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


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


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

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


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

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


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

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

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

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


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

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


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

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


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


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


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

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


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

 


==Applet Manual==
<br>
[[File:Anleitung_Auge.png|right|600px]]
&nbsp; &nbsp; '''(A)''' &nbsp; &nbsp; Auswahl: &nbsp; Codierung <br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(binär,&nbsp; quaternär,&nbsp; AMI&ndash;Code,&nbsp; Duobinärcode) 

&nbsp; &nbsp; '''(B)''' &nbsp; &nbsp; Auswahl: &nbsp; Detektionsgrundimpuls<br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (nach Gauß&ndash;TP,&nbsp; CRO&ndash;Nyquist,&nbsp; nach Spalt&ndash;TP}

&nbsp; &nbsp; '''(C)''' &nbsp; &nbsp; Prametereingabe zu&nbsp; '''(B)'''<br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(Grenzfrequenz,&nbsp; Rolloff&ndash;Faktor,&nbsp; Rechteckdauer)  

&nbsp; &nbsp; '''(D)''' &nbsp; &nbsp; Steuerung der Augendiagrammdarstellung<br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(Start,&nbsp; Pause/Weiter,&nbsp; Einzelschritt,&nbsp; Gesamt,&nbsp; Reset)

&nbsp; &nbsp; '''(E)''' &nbsp; &nbsp; Geschwindigkeit der Augendiagrammdarstellung

&nbsp; &nbsp; '''(F)''' &nbsp; &nbsp; Darstellung:&nbsp; Detektionsgrundimpuls &nbsp;$g_d(t)$ 

&nbsp; &nbsp; '''(G)''' &nbsp; &nbsp; Darstellung:&nbsp; Detektionsnutzsignal &nbsp;$d_{\rm S}(t - \nu \cdot T)$

&nbsp; &nbsp; '''(H)''' &nbsp; &nbsp; Darstellung:&nbsp; Augendiagramm im Bereich &nbsp;$\pm T$

&nbsp; &nbsp; '''( I )''' &nbsp; &nbsp; Numerikausgabe:&nbsp; $ö_{\rm norm}$&nbsp; (normierte Augenöffnung)  

&nbsp; &nbsp; '''(J)''' &nbsp; &nbsp; Prametereingabe &nbsp;$10 \cdot \lg \ E_{\rm B}/N_0$&nbsp; für&nbsp; '''(K)'''

&nbsp; &nbsp; '''(K)''' &nbsp; &nbsp; Numerikausgabe:&nbsp; $\sigma_{\rm norm}$&nbsp; (normierter Rauscheffektivwert)

&nbsp; &nbsp; '''(L)''' &nbsp; &nbsp; Numerikausgabe:&nbsp; $p_{\rm U}$&nbsp; (ungünstigste Fehlerwahrscheinlichkeit)

&nbsp; &nbsp; '''(M)''' &nbsp; &nbsp; Bereich für die Versuchsdurchführung: &nbsp;  Aufgabenauswahl

&nbsp; &nbsp; '''(N)''' &nbsp; &nbsp; Bereich für die Versuchsdurchführung: &nbsp;  Aufgabenstellung

&nbsp; &nbsp; '''(O)''' &nbsp; &nbsp; Bereich für die Versuchsdurchführung: &nbsp;  Musterlösung einblenden
<br clear=all>
==About the Authors==
This interactive calculation tool was designed and implemented at the&nbsp; [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]&nbsp; at the&nbsp; [https://www.tum.de/en Technical University of Munich]. 
*The first version was created in 2008 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]&nbsp; as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]). 

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


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



==Once again:&nbsp; Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}