# Sampling of Analog Signals and Signal Reconstruction

## Applet Description

The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  "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  $x_{\rm A}(t)$.

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.

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

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  "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.

$\text{Definitions:}$

• 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  "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  "Fourier Series"  and transforms it using the  "Shifting Theorem"  into the frequency domain, the following correspondence   ⇒   "proof" results with the distance  $f_{\rm A} = 1/T_{\rm A}$  of two adjacent dirac delta lines in the frequency domain:

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

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

$\text{Conclusion:}$  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  $f_{\rm A} = 1/T_{\rm A}$.

Spectrum of the sampled signal

$\text{Example 1:}$  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

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.
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  "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:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

$\text{Sampling theorem:}$  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.

$\text{Example 2:}$  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  $T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.

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.

(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 $\varphi$.

(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  $A$.

(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  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

(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   ⇒   $P_\varepsilon = 0. 125$.

(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.

(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  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

(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.

(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:  $r \le 0.2$.

(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  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

(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  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

(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

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