Aufgaben:Exercise 2.7: Two-Way Channel once more

• Return to book

Magnitude frequency response and phase function of the two-way channel

As in  Exercise 2.6,  a two-way channel is considered whose impulse response is:

$$h(t) = \delta ( t - T_1) + \delta ( t - T_2).$$

In contrast to the general representation in task 2.6, the two attenuation factors are equal here:   $z_1 = z_2 = 1$.

• For mobile communications, this corresponds to an echo at a distance of  $T_2 - T_1$  and of the same strength as the signal on the main path, for example.
• For this, the transit time  $T_1$  is assumed.

Using the transit times $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$  considered in the subtasks  (1) ... (4),  the following is obtained for the frequency response of the two-way channel, whose magnitude is depicted in the upper graph:

$$H(f) = 1 + {\rm e}^{-{\rm j}\hspace{0.04cm}2 \pi f T} = 1 + \cos(2 \pi f T) - {\rm j} \cdot \sin(2 \pi f T)$$

$$\Rightarrow \hspace{0.4cm}|H(f)| = \sqrt{2\left(1 + \cos(2 \pi f T)\right)}= 2 \cdot |\cos(\pi f T)|.$$

The bottom graph shows the phase function:

$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) = \arctan \frac{\sin(2 \pi f T)}{1 + \cos(2 \pi f T)} = \arctan \big[\tan(\pi f T)\big].$$

The following trigonometric transformation was used here:

$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$

The bottom graph shows thep hase function for $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$:

• In the frequency range  $|f| < 1/(2T)$,    $b(f)$  increases linearly:   $b(f) = \pi \cdot f \cdot T.$
• Also in further sections of the phase function, the phase always increases linearly from  $-\pi/2$  to  $+\pi/2$ .

In the questions, $y_i(t)$  denotes the signal at the output of the two-way channel if the signal $x_i(t)$  is applied to the input  $( i = 1, 2, 3, 4)$.

These signals are examined as input signals:

• a rectangular pulse  $x_1(t)$  with height  $1$  between  $t= 0$  and  $t= T$;
   $x_1(t) = 0$  holds for  $t < 0$  and  $t > T$  $($the value$0.5$ occurs at the two jump discontinuities, respectively$)$;
• a rectangular pulse  $x_2(t)$  with height  $1$  in the range of  $0 \ \text{...} \ 2T$;
• a periodic rectangular pulse  $x_3(t)$  with period  $T = T_0$:

$$x_3(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \quad \begin{array}{c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} { 0 < t < T/2,} \\ { T/2 < t < T,} \\ \end{array}$$

• a periodic rectangular pulse  $x_4(t)$  with period  $T = 2T_0$:

$$x_4(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \quad \begin{array}{c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} { 0 < t < T,} \\ { T < t < 2T.} \\ \end{array}$$

Please note:

• The task belongs to the chapter  Linear Distortions.
• For subtasks (1) to (4),  $T_1 = 0$  and  $T_2 = T = 4 \ \rm ms$ hold.
• In subtask (5), the case  $T_1 = 1 \ \rm ms$  and  $T_2 = 5 \ \rm ms$  is considered.

Questions

1

Compute the output signal  $y_1(t)$  for the input signal  $x_1(t)$. Which of the statements are true?

	$y_1(t)$  is rectangular like  $x_1(t)$ .
	$y_1(t)$  is triangular in shape.
	The absolute pulse duration is  $2T$.
	$y_1(t)$  exhibits attenuation distortions with respect to  $x_1(t)$ .
	$y_1(t)$  exhibits phase distortions with respect to  $x_1(t)$ .

2

Compute the signal  $y_2(t)$. What values are obtained at times  $t= 0.5 T$,  $t= 1.5 T$  and  $t= 2.5 T$?

$y_2(t = 0.5T) \ = \ $

$y_2(t = 1.5T) \ = \ $

$y_2(t = 2.5T) \ = \ $

3

Compute the signal  $y_3(t)$. Check which statements are true.

	$y_3(t)$  is undistorted with respect to  $x_3(t)$ .
	$y_3(t)$  exhibits attenuation distortions with respect to  $x_3(t)$ .
	$y_3(t)$  exhibits phase distortions with respect to  $x_3(t)$ .

4

Which statements are true for the output signal  $y_4(t)$ ?

	$y_4(t)$  is undistorted with respect to  $x_4(t)$ .
	$y_4(t)$  exhibits attenuation distortions with respect to  $x_4(t)$ .
	$y_4(t)$  exhibits phase distortions with respect to  $x_4(t)$ .

5

Now  $T_1 = 1 \ \rm ms$  and  $T_2 = 5 \ \rm ms$ hold. What are the changes compared to the previous results?

	The above statements regarding distortions are still valid.
	Well-founded statements are only possible after a revaluation.
	The combination  $T_1 = 1 \ \rm ms$  and  $T_2 = 5 \ \rm ms$  results in distortions for all signals.

Solution

Solution

(1)  The solution in the time domain leads faster to the final result:

$$y_1(t) = x_1(t) \star h(t) = x_1(t) \star \delta (t) + x_1(t) \star \delta (t - T) = x_1(t) + x_1(t-T).$$

• Thus,  $y_1(t)$  is a rectangular pulse of height  $1$  and width  $2T$.

• The same result – but in a more time-consuming way – is obtained by computing in the spectral domain:

$$Y_1(f) = X_1(f) \cdot H(f) = T \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot \big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$

• The complex exponential functions can be converted using the  Euler theorem  as follows:

$${\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot \big[ {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$

• Hence, the following can be formulated for the output spectrum:

$$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} , \; \; {\rm mit } \; \; Y_{11}(f) = 2T \cdot \frac {\sin(\pi f T) \cdot \cos(\pi f T)}{\pi f T} = 2T \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$

Input and output signals

Here, the relation  $\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$  is used.

• The inverse Fourier transform of  $Y_{11}$  results in a rectangle of width  $2T$, that is symmetric about  $t = 0$ .
• Due to the phase function this is shifted into the range  $0$ ... $2T$  and the result of the time domain computation is confirmed.

Despite the fact that $y_1(t)$ is rectangular just as $x_1(t)$, there are distortions:

• These are linear because of  $T_y > T_x$ . In the frequency range of interest  $($that is all frequencies for a si–shaped spectrum$)$,    $|H(f)|$  is not constant. So, there are attenuation distortions.
• In addition, there are also phase distortions since the phase does not increase linearly with  $f$ in the whole range   ⇒   The proposed solutions 1, 3, 4 and 5 are correct.

(2)  Aufgrund der bereits in  (1)  angegebenen Gleichung

$$y_2(t) = x_2(t) + x_2(t-T)$$

erhält man einen stufenförmigen Verlauf entsprechend dem unteren Diagramm der obere Grafik.

Die gesuchten Zahlenwerte sind:   $y_2(t = 0.5 T) \hspace{0.15cm}\underline{= 1}, \hspace{0.3cm} y_2(t = 1.5 T) \hspace{0.15cm}\underline{= 2}, \hspace{0.3cm}y_2(t = 2.5 T) \hspace{0.15cm}\underline{ = 1}.$

Solutions  (3)  and  (4)

(3)  Die Periodendauer  $T_0 = T$  des periodischen Signals  $x_3(t)$  ist genau so groß wie die Verzögerung auf dem zweiten Pfad. Deshalb ist  $y_3(t) = 2 \cdot x_3(t) $  und es sind keine Verzerrungen feststellbar.

Die Spektralbereichsberechnung führt zum gleichen Ergebnis.

• $X_3(f)$  ist ein Linienspektrum mit Anteilen bei den Frequenzen  $f = 0$,  $f = \pm f_0 = \pm 1/T$,  $f = \pm 3f_0$,  usw..
• Bei diesen diskreten Frequenzen gilt aber exakt:

$$|H(f)| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\tau_{\rm P}(f) = 0.$$

• Auch daraus folgt wieder  $y_3(t) = 2 \cdot x_3(t) $.
• Richtig ist somit nur der Lösungsvorschlag 1.

(4)  Aus der unteren Skizze der zweiten Grafik geht hervor, dass  $y_4(t) = 1$  gegenüber  $x_4(t)$  verzerrt ist. Dabei handelt es sich um Dämpfungsverzerrungen  ⇒  Lösungsvorschlag 2, wie die folgende Überlegung zeigt.

• Wegen  $T_0 = 2T$  weist das Signal  $x_4(t)$  die Grundfrequenz  $f_0 = 1/(2T)$ auf.
• Bei allen ungeraden Vielfachen von  $f_0$  hat somit der Frequenzgang Nullstellen.
• Die einzige verbleibende Spektrallinie von  $Y_4(f)$  liegt bei  $f = 0$, wobei gilt:

$$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f) \hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$

(5)  Der Frequenzgang lautet nun mit  $T_1 = 1 \ \rm ms$,  $T_2 = 5 \ \rm ms$  und  $T = T_2 -T_1 = 4 \ \rm ms$:

$$H(f) = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_1}+ {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_2}= \big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big]\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_1}.$$

• Der Klammerausdruck beschreibt den bereits bisher betrachteten Frequenzgang.
• Der zweite Term bewirkt eine zusätzliche Laufzeit um  $ \tau = T_1$, und es gilt für alle Signale  $(i = 1, 2, 3, 4)$:

$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$

Alle Aussagen hinsichtlich der Verzerrungen sind weiter gültig. Dies entspricht dem Lösungsvorschlag 1.