Difference between revisions of "Aufgaben:Exercise 2.7: Two-Way Channel once more"

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{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Lineare Verzerrungen
+
{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Linear_Distortions
 
}}
 
}}
  
[[File:P_ID914__LZI_A_2_7.png|right|frame|Betragsfrequenzgang und Phasenfunktion des Zweiwegekanals]]
+
[[File:P_ID914__LZI_A_2_7.png|right|frame|Magnitude frequency response and phase function of the two-way channel]]
Wie in [[Aufgaben:2.6_Zweiwegekanal|Aufgabe 2.6]] wird ein Zweiwegekanal betrachtet, für dessen Impulsantwort gelte:
+
As in  [[Aufgaben:Exercise_2.6:_Two-Way_Channel|Exercise 2.6]],  a two-way channel is considered whose impulse response is:
 
:$$h(t) = \delta ( t - T_1) + \delta ( t - T_2).$$
 
:$$h(t) = \delta ( t - T_1) + \delta ( t - T_2).$$
  
Entgegen der allgemeinen Darstellung in Aufgabe 2.6 sind hier die beiden Dämpfungsfaktoren gleich:   $z_1 = z_2 = 1$.  
+
In contrast to the general representation in exercise 2.6, the two attenuation factors are equal here:   $z_1 = z_2 = 1$.  
*Dies entspricht zum Beispiel beim Mobilfunk einem Echo im Abstand $T_2 - T_1$ in gleicher Stärke wie das Signal auf dem Hauptpfad.  
+
For mobile communications, this corresponds for example to an echo at a distance of  $T_2 - T_1$  and of the same strength as the signal on the main path.  For this, the transit time  $T_1$  is assumed.
*Für dieses wird die Laufzeit $T_1$ vorausgesetzt.
 
  
  
Mit den in den Teilaufgaben '''(1)''' ... '''(4)'''  betrachteten Laufzeiten  $T_1 = 0$  und  $T_2 = T = 4 \ \rm ms$  erhält man für den Frequenzgang des Zweiwegekanals, dessen Betrag in der oberen Grafik dargestellt ist:
+
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 +
 
:$$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)$$
 
\cos(2 \pi f T) - {\rm j} \cdot \sin(2 \pi f T)$$
Line 18: Line 17:
 
T)\right)}= 2 \cdot |\cos(\pi f T)|.$$
 
T)\right)}= 2 \cdot |\cos(\pi f T)|.$$
  
Die untere Grafik zeigt die Phasenfunktion:
+
The bottom graph shows the phase function  for  $T_1 = 0$  and   $T_2 = T = 4 \ \rm ms$:
 
:$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) = \arctan \frac{\sin(2 \pi f
 
:$$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].$$
 
T)}{1 + \cos(2 \pi f T)} = \arctan \big[\tan(\pi f T)\big].$$
  
Hierbei wurde folgende trigonometrische Umformung benutzt:
+
*In the frequency region&nbsp; $|f| < 1/(2T)$,&nbsp; &nbsp; $b(f)$&nbsp; increases linearly: &nbsp; $b(f) = \pi \cdot f \cdot T.$
 +
*Also in further sections of the phase function,&nbsp; the phase always increases linearly from&nbsp; $-\pi/2$&nbsp; to&nbsp; $+\pi/2$&nbsp;.
 +
*The following trigonometric transformation was used here:
 
:$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$
 
:$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$
  
Die untere Grafik zeigt diePhasenfunktion für &nbsp;$T_1 = 0$&nbsp; und &nbsp;$T_2 = T = 4 \ \rm ms$:
 
*Im Frequenzbereich $|f| < 1/(2T)$ steigt $b(f)$ linear an: &nbsp; $b(f) = \pi \cdot f \cdot T.$
 
*Auch in den weiteren Abschnitten der Phasenfunktion nimmt die Phase stets von $-\pi/2$ bis $+\pi/2$ linear zu.
 
  
 +
In the questions,&nbsp; $y_i(t)$&nbsp; denotes the signal at the output of the two-way channel if the signal&nbsp; $x_i(t)$&nbsp; is applied to the input&nbsp; $(i = 1, 2, 3, 4)$.
  
Im Fragenkatalog bezeichnet &nbsp;$y_i(t)$&nbsp; das Signal am Ausgang des Zweiwegekanals, wenn am Eingang das Signal &nbsp;$x_i(t)$&nbsp; anliegt ($ i = 1, 2, 3, 4$).
+
These signals are examined as input signals:
 
+
*a rectangular pulse &nbsp;$x_1(t)$&nbsp; with height&nbsp; $1$&nbsp; between &nbsp;$t= 0$&nbsp; and &nbsp;$t= T$;&nbsp; $x_1(t) = 0$&nbsp; holds for &nbsp;$t < 0$&nbsp; and &nbsp;$t > T$&nbsp; $($the value $0.5$ occurs at the two jump discontinuities$)$;
Als Eingangssignale werden untersucht:
+
*a rectangular pulse &nbsp;$x_2(t)$&nbsp; with height&nbsp; $1$&nbsp; in the range of&nbsp; $0 \ \text{...} \ 2T$;
*ein Rechteckimpuls &nbsp;$x_1(t)$&nbsp; mit der Höhe $1$ zwischen &nbsp;$t= 0$&nbsp; und &nbsp;$t= T$; für &nbsp;$t < 0$&nbsp; und &nbsp;$t > T$&nbsp; ist &nbsp;$x_1(t) = 0$ (an den beiden Sprungstellen tritt jeweils der Wert $0.5$ auf);
+
*a periodic rectangular signal &nbsp;$x_3(t)$&nbsp; with period duration &nbsp;$T = T_0$:
*ein Rechteckimpuls &nbsp;$x_2(t)$&nbsp; mit Höhe $1$ im Bereich von $0 ...  2T$;
 
*ein periodisches Rechtecksignal &nbsp;$x_3(t)$&nbsp; mit der Periodendauer &nbsp;$T = T_0$:
 
 
:$$x_3(t) = \left\{ \begin{array}{c} 1 \\
 
:$$x_3(t) = \left\{ \begin{array}{c} 1 \\
 
  0 \\  \end{array} \right.\quad \quad
 
  0 \\  \end{array} \right.\quad \quad
\begin{array}{c}  {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}}
+
\begin{array}{c}  {\rm{for}}  \\ {\rm{for}}
 
\\    \end{array}\begin{array}{*{20}c}
 
\\    \end{array}\begin{array}{*{20}c}
 
{ 0 < t  < T/2,}  \\
 
{ 0 < t  < T/2,}  \\
 
{ T/2 < t  < T,}  \\
 
{ T/2 < t  < T,}  \\
 
\end{array}$$
 
\end{array}$$
*ein periodisches Rechtecksignal &nbsp;$x_4(t)$&nbsp; mit der Periodendauer &nbsp;$T = 2T_0$:
+
*a periodic rectangular signal &nbsp;$x_4(t)$&nbsp; with period duration &nbsp;$T = 2T_0$:
 
:$$x_4(t) = \left\{ \begin{array}{c} 1 \\
 
:$$x_4(t) = \left\{ \begin{array}{c} 1 \\
 
  0 \\  \end{array} \right.\quad \quad
 
  0 \\  \end{array} \right.\quad \quad
\begin{array}{c}   {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}}
+
\begin{array}{c} {\rm{for}}  \\ {\rm{for}}
 
\\    \end{array}\begin{array}{*{20}c}
 
\\    \end{array}\begin{array}{*{20}c}
 
{ 0 < t  < T,}  \\
 
{ 0 < t  < T,}  \\
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+
Please note:  
''Hinweise:''
+
*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]].
*Die Aufgabe gehört zum Kapitel&nbsp; [[Lineare_zeitinvariante_Systeme/Lineare_Verzerrungen|Lineare Verzerrungen]].
+
*For subtasks&nbsp; '''(1)''' to '''(4)''', &nbsp;$T_1 = 0$&nbsp; and &nbsp;$T_2 = T = 4 \ \rm ms$&nbsp; hold.
*Für die Teilaufgaben '''(1)''' bis '''(4)''' gelte &nbsp;$T_1 = 0$&nbsp; und &nbsp;$T_2 = T = 4 \ \rm ms$.
+
*In subtask&nbsp; '''(5)''',&nbsp; the case &nbsp;$T_1 = 1 \ \rm ms$&nbsp; and &nbsp;$T_2  = 5 \ \rm ms$&nbsp; is considered.  
*In Teilaufgabe '''(5)''' wird der Fall &nbsp;$T_1 = 1 \ \rm ms$&nbsp; und &nbsp;$T_2  = 5 \ \rm ms$&nbsp; betrachtet.  
 
 
   
 
   
  
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Berechnen Sie das Ausgangssignal $y_1(t)$ für das Eingangssignal $x_1(t)$. Welche der Aussagen sind zutreffend?
+
{Compute the output signal&nbsp; $y_1(t)$&nbsp; for the input signal&nbsp; $x_1(t)$.&nbsp; Which of the statements are true?
 
|type="[]"}
 
|type="[]"}
+ $y_1(t)$ ist wie $x_1(t)$ rechteckförmig.
+
+ $y_1(t)$&nbsp; is rectangular like&nbsp; $x_1(t)$.
- $y_1(t)$ ist dreieckförmig.
+
- $y_1(t)$&nbsp; is triangular in shape.
+ Die absolute Impulsdauer ist $2T$.
+
+ The absolute pulse duration is&nbsp; $2T$.
+ $y_1(t)$ weist gegenüber $x_1(t)$ Dämpfungsverzerrungen auf.
+
+ $y_1(t)$&nbsp; exhibits attenuation distortions with respect to&nbsp; $x_1(t)$.
+ $y_1(t)$ weist gegenüber $x_1(t)$ Phasenverzerrungen auf.
+
+ $y_1(t)$&nbsp; exhibits phase distortions with respect to&nbsp; $x_1(t)$.
  
  
{Berechnen Sie das Signal $y_2(t)$. Welche Werte ergeben sich zu den Zeitpunkten $t= 0.5 T$, $t= 1.5 T$ und $t= 2.5 T$?
+
{Compute the signal&nbsp; $y_2(t)$.&nbsp; What values are obtained at times&nbsp; $t= 0.5 T$,&nbsp; $t= 1.5 T$&nbsp; and&nbsp; $t= 2.5 T$?
 
|type="{}"}
 
|type="{}"}
 
$y_2(t = 0.5T) \ = \ $  { 1 3% }
 
$y_2(t = 0.5T) \ = \ $  { 1 3% }
Line 83: Line 79:
  
  
{Berechnen Sie das Signal $y_3(t)$. Überprüfen Sie, welche Aussagen zutreffen.
+
{Compute the signal&nbsp; $y_3(t)$.&nbsp; Check which statements are true.
 
|type="[]"}
 
|type="[]"}
+ $y_3(t)$ ist gegenüber $x_3(t)$ unverzerrt.
+
+ $y_3(t)$&nbsp; is undistorted with respect to&nbsp; $x_3(t)$.
- $y_3(t)$ weist gegenüber $x_3(t)$ Dämpfungsverzerrungen auf.
+
- $y_3(t)$&nbsp; exhibits attenuation distortions with respect to&nbsp; $x_3(t)$.
- $y_3(t)$ weist gegenüber $x_3(t)$ Phasenverzerrungen auf.
+
- $y_3(t)$&nbsp; exhibits phase distortions with respect to&nbsp; $x_3(t)$.
  
  
{Welche Aussagen treffen für das Ausgangssignal <i>y</i><sub>4</sub>(<i>t</i>)  
+
{Which statements are true for the output signal&nbsp; $y_4(t)$?
zu?
 
 
|type="[]"}
 
|type="[]"}
- $y_4(t)$ ist gegenüber $x_4(t)$ unverzerrt.
+
- $y_4(t)$&nbsp; is undistorted with respect to&nbsp; $x_4(t)$.
+ $y_4(t)$ weist gegenüber $x_4(t)$ Dämpfungsverzerrungen auf.
+
+ $y_4(t)$&nbsp; exhibits attenuation distortions with respect to&nbsp; $x_4(t)$.
- $y_4(t)$ weist gegenüber $x_4(t)$ Phasenverzerrungen auf.
+
- $y_4(t)$&nbsp; exhibits phase distortions with respect to&nbsp; $x_4(t)$.
  
  
{Es gelte nun &nbsp;$T_1 = 1 \ \rm ms$&nbsp; und &nbsp;$T_2  = 5 \ \rm ms$. Welche Veränderungen ergeben sich gegenüber den bisherigen Ergebnissen?
+
{Now &nbsp;$T_1 = 1 \ \rm ms$&nbsp; and &nbsp;$T_2  = 5 \ \rm ms$ hold.&nbsp; What are the changes compared to the previous results?
 
|type="[]"}
 
|type="[]"}
+ Die obigen Aussagen hinsichtlich Verzerrungen sind weiterhin gültig.
+
+ The above statements regarding distortions are still valid.
- Fundierte Aussagen sind erst nach einer Neuberechnung möglich.
+
- Well-founded statements are only possible after a revaluation.
- Die Kombination &nbsp;$T_1 = 1 \ \rm ms$&nbsp; und &nbsp;$T_2  = 5 \ \rm ms$&nbsp; führt bei allen Signalen zu Verzerrungen.
+
- The combination &nbsp;$T_1 = 1 \ \rm ms$&nbsp; and &nbsp;$T_2  = 5 \ \rm ms$&nbsp; results in distortions for all signals.
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die Lösung im Zeitbereich führt schneller zum Endergebnis:
+
'''(1)'''&nbsp; 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)
 
:$$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).$$
 
+ x_1(t) \star \delta (t - T) = x_1(t) + x_1(t-T).$$
  
Somit ist $y_1(t)$ ein Rechteckimpuls der Höhe $1$ und der Breite $2T$.
+
*Thus,&nbsp; $y_1(t)$&nbsp; is a rectangular pulse of height&nbsp; $1$&nbsp; and width&nbsp; $2T$.
  
Zum gleichen Ergebnis &ndash; aber zeitaufwändiger &ndash; kommt man durch die Berechnung im Spektralbereich:
+
*The same result &ndash; but in a more time-consuming way &ndash; is obtained by computing in the frequency 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
 
:$$Y_1(f) = X_1(f) \cdot H(f) = T  \cdot \frac {\sin(\pi f T)}{\pi f T}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \cdot
  \left[  1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \right].$$
+
  \big[  1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big].$$
  
Die komplexen Exponentialfunktionen können mit dem [[Signaldarstellung/Zum_Rechnen_mit_komplexen_Zahlen#Darstellung_nach_Betrag_und_Phase|Satz von Euler]] wie folgt umgewandelt werden:
+
*The complex exponential functions can be converted using the&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_magnitude_and_phase|Euler theorem]]&nbsp; as follows:
 
:$${\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T}
 
:$${\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T}
  \left[  1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \right] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}
+
  \big[  1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}
  \cdot \left[  {\rm e}^{{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \pi f T} \right] = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \cdot 2 \cos(\pi f T) .$$
+
  \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) .$$
  
Somit kann für das Ausgangsspektrum geschrieben werden:
+
*Hence,&nbsp; 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}
 
:$$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}
 
  , \; \; {\rm mit }  \; \;
 
  , \; \; {\rm mit }  \; \;
Line 131: Line 126:
 
f T} = 2T  \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$
 
f T} = 2T  \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$
  
[[File:P_ID925__LZI_A_2_7_a.png|right|frame|Eingangs– und Ausgangssignale]]
+
[[File:P_ID925__LZI_A_2_7_a.png|right|frame|Solutions&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)''']]
Hierbei ist die $t = 0$ symmetrischen Rechteck der Breite $2T$. Durch die Phasenfunktion wird dieser in den Bereich $0$ ... $2T$ verschoben und das Ergebnis der Zeitbereichsberechnung bestätigt.
+
Here,&nbsp; the relation &nbsp;$\sin(\alpha) \cdot \cos(\alpha) = \sin(2\alpha)/2$&nbsp; is used.  
  
Trotz der Tatsache, dass $y_1(t)$ ebenso wie $x_1(t)$ rechteckförmig ist, liegen hier Verzerrungen vor:
+
*The inverse Fourier transform of&nbsp; $Y_{11}$&nbsp; results in a rectangle of width&nbsp; $2T$,&nbsp; that is symmetric about&nbsp;  $t = 0$&nbsp;.  
*Wegen $T_y > T_x$ sind diese linear. Im interessierenden Frequenzbereich (das sind bei einem si&ndash;förmigem Spektrum alle Frequenzen) ist $|H(f)|$ nicht konstant. Also gibt es Dämpfungsverzerrungen.
+
*Due to the phase function this is shifted into the range&nbsp; $0$ ... $2T$&nbsp; and the result of the time domain computation is confirmed.
*Da zudem die Phase nicht im gesamten Bereich linear mit $f$ ansteigt, gibt es auch Phasenverzerrungen &nbsp;&rArr;&nbsp; Richtig sind die <u> Lösungsvorschläge 1, 3, 4 und 5</u>.
 
  
  
'''(2)'''&nbsp; Aufgrund der bereits in (1) angegebenen Gleichung
+
Despite the fact that&nbsp; $y_1(t)$&nbsp; is rectangular just as&nbsp; $x_1(t)$,&nbsp; there are distortions:
 +
*These are linear because of&nbsp; $T_y > T_x$.&nbsp; In the frequency range of interest&nbsp; $($that is all frequencies for a sinc&ndash;shaped spectrum$)$,&nbsp; &nbsp; $|H(f)|$&nbsp; is not constant.&nbsp; So, there are attenuation distortions.
 +
*In addition, there are also phase distortions since the phase does not increase linearly with&nbsp; $f$&nbsp; in the whole range &nbsp; &rArr; &nbsp; The&nbsp; <u>proposed solutions 1, 3, 4 and 5</u>&nbsp; are correct.
 +
 
 +
 
 +
 
 +
'''(2)'''&nbsp; Due to the equation
 
:$$y_2(t)  = x_2(t) + x_2(t-T)$$
 
:$$y_2(t)  = x_2(t) + x_2(t-T)$$
  
erhält man einen stufenförmigen Verlauf entsprechend zweiten Grafik. Die gesuchten Werte sind:
+
already specified in&nbsp; '''(1)'''&nbsp; a step-like curve shape corresponding to the lower diagram of the upper graph is obtained.  
:$$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}.$$
 
  
 +
The numerical values that are looked for are: &nbsp; $y_2(t = 0.5 T)  \hspace{0.15cm}\underline{= 1}, \hspace{0.3cm} y_2(t = 1.5 T)  \hspace{0.15cm}\underline{= 2},
 +
\hspace{0.3cm}y_2(t = 2.5 T)  \hspace{0.15cm}\underline{ = 1}.$
  
[[File:P_ID927__LZI_A_2_7_c.png|right|frame|Lösungen zu Aufgabe (3) und (4)]]
 
'''(3)'''&nbsp; 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 $f  = 0$,  
+
[[File:P_ID927__LZI_A_2_7_c.png|right|frame|Solutions&nbsp; '''(3)'''&nbsp; and&nbsp; '''(4)''']]
$f  =  \pm f_0 = \pm 1/T$, $f  =  \pm 3f_0$, usw.. Bei diesen diskreten Frequenzen gilt aber exakt:
+
'''(3)'''&nbsp; The period&nbsp; $T_0 = T$&nbsp; of the periodic signal &nbsp;$x_3(t)$&nbsp; is exactly as large as the delay on the second path. Therefore, &nbsp;$y_3(t) = 2 \cdot x_3(t) $&nbsp; holds and no distortions are observable.
 +
 
 +
The frequency domain computation leads to the same result.
 +
*$X_3(f)$&nbsp; is a line spectrum with components at frequencies&nbsp; $f  = 0$,&nbsp; $f  =  \pm f_0 = \pm 1/T$,&nbsp; $f  =  \pm 3f_0$,&nbsp; etc..  
 +
*However,&nbsp; the following holds at these discrete frequencies:
 
:$$|H(f)| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow
 
:$$|H(f)| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow
 
  \hspace{0.3cm}\tau_{\rm P}(f) = 0.$$
 
  \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 <u>Lösungsvorschlag 1</u>.
+
*From this it follows again that&nbsp; $y_3(t) = 2 \cdot x_3(t) $.  
 +
*Thus,&nbsp; only&nbsp; <u>proposed solution 1</u>&nbsp; is correct.
 +
 
 +
 
  
  
'''(4)'''&nbsp; 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 &nbsp;&#8658;&nbsp; <u>Lösungsvorschlag 2</u>, wie die folgende Überlegung zeigt.  
+
'''(4)'''&nbsp; It can be seen that &nbsp;$y_4(t) = 1$&nbsp; is distorted with respect to &nbsp;$x_4(t)$&nbsp; from the lower sketch of the second graph. These are attenuation distortions &nbsp;&#8658;&nbsp; <u>Proposed solution 2</u>&nbsp; is correct as the following consideration shows.  
*Wegen $T_0 = 2T$ weist das Signal $x_4(t)$ die Grundfrequenz $f_0 = 1/(2T)$ auf.  
+
*Due to&nbsp; $T_0 = 2T$&nbsp; the signal &nbsp;$x_4(t)$&nbsp; has the basic frequency &nbsp;$f_0 = 1/(2T)$.  
*Bei allen ungeraden Vielfachen von $f_0$ hat somit der Frequenzgang Nullstellen.  
+
*The frequency response thus has zeros for all odd multiples of &nbsp;$f_0$.  
*Die einzige verbleibende Spektrallinie von  $Y_4(f)$ liegt bei $f = 0$, wobei gilt:
+
*The only remaining spectral line of &nbsp;$Y_4(f)$&nbsp; is at &nbsp;$f = 0$&nbsp; where the following holds:
 
:$$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f)
 
:$$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.$$
 
\hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$
  
  
'''(5)'''&nbsp; Der Frequenzgang lautet nun mit $T_1 = 1 \ \rm ms$, $T_2  = 5 \ \rm ms$ und $T = T_2 -T_1 = 4 \ \rm ms$:
+
 
 +
'''(5)'''&nbsp; Now the frequency response is with &nbsp;$T_1 = 1 \ \rm ms$, &nbsp;$T_2  = 5 \ \rm ms$&nbsp; and &nbsp;$T = T_2 -T_1 = 4 \ \rm ms$:
 
:$$H(f) =  {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f
 
:$$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_1}+ {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f
  T_2}=  \left[  1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}
+
  T_2}=  \big[  1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T}
  \right]\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f
+
  \big]\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f
 
  T_1}.$$
 
  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$):
+
*The expression in parentheses describes the frequency response already considered so far.  
$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$
+
*The second term causes an additional runtime around &nbsp;$ \tau = T_1$&nbsp; and the following holds for all signals&nbsp; $(i = 1, 2, 3, 4)$:
 +
:$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$
  
Alle Aussagen hinsichtlich der Verzerrungen sind weiter gültig. Dies entspricht dem <u>Lösungsvorschlag 1</u>.
+
All statements regarding the distortions are still valid.&nbsp; This corresponds to&nbsp; <u>proposed solution 1</u>.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Lineare zeitinvariante Systeme|^2.3 Lineare Verzerrungen^]]
+
[[Category:Linear and Time-Invariant Systems: Exercises|^2.3 Linear Distortions^]]

Latest revision as of 16:11, 6 October 2021

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 exercise 2.6, the two attenuation factors are equal here:   $z_1 = z_2 = 1$. For mobile communications, this corresponds for example to an echo at a distance of  $T_2 - T_1$  and of the same strength as the signal on the main path.  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 for  $T_1 = 0$  and   $T_2 = T = 4 \ \rm ms$:

$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) = \arctan \frac{\sin(2 \pi f T)}{1 + \cos(2 \pi f T)} = \arctan \big[\tan(\pi f T)\big].$$
  • In the frequency region  $|f| < 1/(2T)$,    $b(f)$  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  $+\pi/2$ .
  • The following trigonometric transformation was used here:
$$ \frac{\sin(2 \alpha)}{1 + \cos(2 \alpha)} = \tan(\alpha).$$


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$)$;
  • a rectangular pulse  $x_2(t)$  with height  $1$  in the range of  $0 \ \text{...} \ 2T$;
  • a periodic rectangular signal  $x_3(t)$  with period duration  $T = T_0$:
$$x_3(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \quad \begin{array}{c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} { 0 < t < T/2,} \\ { T/2 < t < T,} \\ \end{array}$$
  • a periodic rectangular signal  $x_4(t)$  with period duration  $T = 2T_0$:
$$x_4(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \quad \begin{array}{c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} { 0 < t < T,} \\ { T < t < 2T.} \\ \end{array}$$



Please note:

  • The exercise 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

(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 frequency 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:
$$Y_1(f) = Y_{11}(f) \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} , \; \; {\rm mit } \; \; Y_{11}(f) = 2T \cdot \frac {\sin(\pi f T) \cdot \cos(\pi f T)}{\pi f T} = 2T \cdot \frac {\sin(2\pi f T) }{2\pi f T}.$$
Solutions  (1)  and  (2)

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 sinc–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)  Due to the equation

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

already specified in  (1)  a step-like curve shape corresponding to the lower diagram of the upper graph is obtained.

The numerical values that are looked for are:   $y_2(t = 0.5 T) \hspace{0.15cm}\underline{= 1}, \hspace{0.3cm} y_2(t = 1.5 T) \hspace{0.15cm}\underline{= 2}, \hspace{0.3cm}y_2(t = 2.5 T) \hspace{0.15cm}\underline{ = 1}.$


Solutions  (3)  and  (4)

(3)  The period  $T_0 = T$  of the periodic signal  $x_3(t)$  is exactly as large as the delay on the second path. Therefore,  $y_3(t) = 2 \cdot x_3(t) $  holds and no distortions are observable.

The frequency domain computation leads to the same result.

  • $X_3(f)$  is a line spectrum with components at frequencies  $f = 0$,  $f = \pm f_0 = \pm 1/T$,  $f = \pm 3f_0$,  etc..
  • However,  the following holds at these discrete frequencies:
$$|H(f)| = 2, \hspace{0.3cm} b(f) = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\tau_{\rm P}(f) = 0.$$
  • From this it follows again that  $y_3(t) = 2 \cdot x_3(t) $.
  • Thus,  only  proposed solution 1  is correct.



(4)  It can be seen that  $y_4(t) = 1$  is distorted with respect to  $x_4(t)$  from the lower sketch of the second graph. These are attenuation distortions  ⇒  Proposed solution 2  is correct as the following consideration shows.

  • Due to  $T_0 = 2T$  the signal  $x_4(t)$  has the basic frequency  $f_0 = 1/(2T)$.
  • The frequency response thus has zeros for all odd multiples of  $f_0$.
  • The only remaining spectral line of  $Y_4(f)$  is at  $f = 0$  where the following holds:
$$Y_4(f) = 2 \cdot 0.5 \cdot \delta (f) = 1 \cdot \delta (f) \hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_4(t) = 1.$$


(5)  Now the frequency response is with  $T_1 = 1 \ \rm ms$,  $T_2 = 5 \ \rm ms$  and  $T = T_2 -T_1 = 4 \ \rm ms$:

$$H(f) = {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_1}+ {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_2}= \big[ 1 + {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T} \big]\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_1}.$$
  • The expression in parentheses describes the frequency response already considered so far.
  • The second term causes an additional runtime around  $ \tau = T_1$  and the following holds for all signals  $(i = 1, 2, 3, 4)$:
$$y_i^{\rm (5)}(t) = y_i(t-T_1).$$

All statements regarding the distortions are still valid.  This corresponds to  proposed solution 1.