Difference between revisions of "Aufgaben:Exercise 2.6: Two-Way Channel"

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{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Lineare Verzerrungen
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{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Linear_Distortions
 
}}
 
}}
  
[[File:P_ID912__LZI_A_2_6.png|right|frame|Impulsantwort des Zweiwegekanals]]
+
[[File:P_ID912__LZI_A_2_6.png|right|frame|Impulse response of the two-way channel]]
Der so genannte Zweiwegekanal wird durch folgende Impulsantwort charakterisiert&nbsp; $($mit&nbsp; $T_1 < T_2)$:
+
The so-called&nbsp; "two-way channel"&nbsp; is characterised by the following impulse response&nbsp; $($with&nbsp; $T_1 < T_2)$:
 
:$$h(t) = z_1 \cdot \delta ( t - T_1) + z_2 \cdot \delta (
 
:$$h(t) = z_1 \cdot \delta ( t - T_1) + z_2 \cdot \delta (
 
t - T_2).$$
 
t - T_2).$$
  
*Bis auf wenige Kombinationen der Systemparameter&nbsp; $z_1$,&nbsp; $T_1$,&nbsp; $z_2$&nbsp; und&nbsp;  $T_2$&nbsp; wird dieser Kanal zu linearen Verzerrungen führen.
+
*Except for a few combinations of the system parameters&nbsp; $z_1$,&nbsp; $T_1$,&nbsp; $z_2$&nbsp; and&nbsp;  $T_2$,&nbsp; this channel will result in linear distortions.
* Man spricht nur dann von einem verzerrungsfreien Kanal, wenn durch ihn kein einziges Eingangssignal verzerrt wird. Das bedeutet: &nbsp; Auch bei verzerrendem Kanal kann es Sonderfälle geben, bei denen tatsächlich&nbsp; $y(t) = \alpha \cdot x(t - \tau)$&nbsp; gilt.
+
*There is a distortion-free channel at hand only if not a single input signal is distorted by it.  
 +
*This means: &nbsp; Even if the two-way channel is per se distorting,&nbsp;  there may be special cases where indeed&nbsp; $y(t) = \alpha \cdot x(t - \tau)$.
  
  
Als Testsignale werden an den Systemeingang angelegt:
+
The test signals applied to the system input are:
* ein&nbsp; [[Signal_Representation/Zeitdiskrete_Signaldarstellung#Diracpuls_im_Zeit-_und_im_Frequenzbereich|Diracpuls]]&nbsp; $x_1(t)$&nbsp; im Zeitabstand&nbsp; $T_0 = 1 \ \rm ms$, dessen Spektralfunktion&nbsp; $X_1(f)$&nbsp; ebenfalls ein Diracpuls ist, und zwar mit Abstand&nbsp; $f_0 = 1/T_0 = 1 \ \rm kHz$:
+
*a&nbsp; [[Signal_Representation/Time_Discrete_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|Dirac comb]]&nbsp; $x_1(t)$&nbsp; at a time interval of&nbsp; $T_0 = 1 \ \rm ms$,&nbsp; whose spectral function&nbsp; $X_1(f)$&nbsp; is also a Dirac comb with an interval of&nbsp; $f_0 = 1/T_0 = 1 \ \rm kHz$:
 
:$$x_1(t) = \sum_{n = - \infty}^{+\infty} \delta ( t - n \cdot T_0) ,\hspace{0.5cm} X_1(f) = T_0 \cdot \sum_{k = - \infty}^{+\infty} \delta ( f - k
 
:$$x_1(t) = \sum_{n = - \infty}^{+\infty} \delta ( t - n \cdot T_0) ,\hspace{0.5cm} X_1(f) = T_0 \cdot \sum_{k = - \infty}^{+\infty} \delta ( f - k
 
\cdot f_0) ,$$
 
\cdot f_0) ,$$
* ein Cosinussignal mit der Frequenz&nbsp; $f_2 = 250 \ \rm Hz$:
+
*a cosine signal with frequency&nbsp; $f_2 = 250 \ \rm Hz$:
 
:$$x_2(t) =  \cos(2 \pi \cdot f_2 \cdot  t) ,$$
 
:$$x_2(t) =  \cos(2 \pi \cdot f_2 \cdot  t) ,$$
  
* die Summe zweier Cosinussignale mit den Frequenzen&nbsp; $f_2 = 250 \ \rm Hz$&nbsp; und&nbsp; $f_3 = 1250 \ \rm Hz$:
+
*the sum of two cosine signals with frequencies&nbsp; $f_2 = 250 \ \rm Hz$&nbsp; and&nbsp; $f_3 = 1250 \ \rm Hz$:
 
:$$x_3(t)  = \cos(2 \pi \cdot f_2 \cdot  t) +  \cos(2 \pi \cdot f_3 \cdot  t) .$$
 
:$$x_3(t)  = \cos(2 \pi \cdot f_2 \cdot  t) +  \cos(2 \pi \cdot f_3 \cdot  t) .$$
  
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''Hinweise:''  
+
''Please note:''  
*Die Aufgabe gehört zum Kapitel&nbsp; [[Linear_and_Time_Invariant_Systems/Lineare_Verzerrungen|Lineare Verzerrungen]].
+
*The task belongs to the chapter &nbsp; [[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]].
 
   
 
   
*Um Ihnen Rechnungen zu ersparen, geben wir das Ergebnis für den Parametersatz&nbsp; $\big [z_1 = 1$,&nbsp; $T_1 = 0$,&nbsp; $z_2 =0.5$,&nbsp;  $T_2 = 1 \ \rm ms\big ]$&nbsp; an:
+
*To spare you calculations&nbsp; the result for the parameter set&nbsp; $\big [z_1 = 1$,&nbsp; $T_1 = 0$,&nbsp; $z_2 =0.5$,&nbsp;  $T_2 = 1 \ \rm ms\big ]$&nbsp; is given:
 
:$$|H(f = f_2)| = |H(f = f_3)| = \sqrt{1.25}  \approx 1.118, \; \; \; \;  b(f = f_2) = b(f = f_3) = \arctan (0.5)  \approx  0.464.$$
 
:$$|H(f = f_2)| = |H(f = f_3)| = \sqrt{1.25}  \approx 1.118, \; \; \; \;  b(f = f_2) = b(f = f_3) = \arctan (0.5)  \approx  0.464.$$
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welche der folgenden Aussagen sind zutreffend?
+
{Which of the following statements are true?
 
|type="[]"}
 
|type="[]"}
+ Der Parametersatz&nbsp;  $\big[z_1 = 1$, $T_1 = 0$, $z_2 =0 \big]$&nbsp; ist der einzig mögliche zur Beschreibung des idealen Kanals.
+
+ The parameter set&nbsp;  $\big[z_1 = 1$, $T_1 = 0$, $z_2 =0 \big]$&nbsp; is the only possible one to describe the ideal channel.
+ Jeder verzerrungsfreie Kanal wird durch die beiden Kombinationen&nbsp;  $\big[z_1 \ne 0, \; z_2 = 0 \big]$&nbsp;  bzw.&nbsp;  $\big[z_1 = 0, \; z_2 \ne 0 \big]$&nbsp; erfasst.
+
+ Any distortion-free channel is captured by the two combinations&nbsp;  $\big[z_1 \ne 0, \; z_2 = 0 \big]$&nbsp;  or&nbsp;  $\big[z_1 = 0, \; z_2 \ne 0 \big]$&nbsp;.
- Die Werte&nbsp;  $\big[z_1 \ne 0\big]$&nbsp; und&nbsp;  $\big[z_2 \ne 0\big]$&nbsp; führen zu einem verzerrungsfreien Kanal, wenn&nbsp; $T_1$&nbsp; und&nbsp; $T_2$&nbsp; bestmöglich angepasst sind.
+
- The values&nbsp;  $\big[z_1 \ne 0\big]$&nbsp; and&nbsp;  $\big[z_2 \ne 0\big]$&nbsp; result in a distortion-free channel&nbsp; if&nbsp; $T_1$&nbsp; and&nbsp; $T_2$&nbsp; are optimally adjusted.
  
  
{Es gelte&nbsp; $\big[z_1 = 1$,&nbsp; $T_1 = 0$,&nbsp; $z_2 =0.5$,&nbsp;  $T_2 = 1 \ \rm ms\big ]$.&nbsp; Berechnen Sie den Frequenzgang&nbsp; $H(f)$&nbsp; dieses Kanals. <br>Welche Werte gibt es bei Vielfachen von&nbsp; $1 \ \rm kHz$?
+
{The following holds: $\big[z_1 = 1$,&nbsp; $T_1 = 0$,&nbsp; $z_2 =0.5$,&nbsp;  $T_2 = 1 \ \rm ms\big ]$.&nbsp; Compute the frequency response&nbsp; $H(f)$&nbsp; of this channel. <br>What are the values at multiples of&nbsp; $1 \ \rm kHz$?
 
|type="{}"}
 
|type="{}"}
 
${\rm Re}\big[H(f = n \cdot 1 \ {\rm kHz})\big] \ = \ $ { 1.5 3% }
 
${\rm Re}\big[H(f = n \cdot 1 \ {\rm kHz})\big] \ = \ $ { 1.5 3% }
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{Am Eingang des Systems mit gleichen Parametern wie in Teilaufgabe&nbsp; '''(2)'''&nbsp; liegt nun der Diracpuls &nbsp;$x_1(t)$&nbsp;  an. <br>Welche Aussagen treffen für das Ausgangssignal &nbsp;$y_1(t)$&nbsp; zu?
+
{The Dirac comb&nbsp; $x_1(t)$&nbsp; is applied to the input of the system with the same parameters as in subtask&nbsp; '''(2)'''&nbsp;. <br>Which statements are true for the output signal&nbsp; $y_1(t)$&nbsp;?
 
|type="[]"}
 
|type="[]"}
+ $y_1(t)$&nbsp; ist gegenüber &nbsp;$x_1(t)$&nbsp; um eine Konstante gedämpft/verstärkt.
+
+ $y_1(t)$&nbsp; is attenuated/amplified by a constant compared to&nbsp;$x_1(t)$&nbsp;.
- $y_1(t)$&nbsp; ist gegenüber &nbsp;$x_1(t)$&nbsp; verschoben.
+
- $y_1(t)$&nbsp; is shifted with respect to&nbsp;$x_1(t)$&nbsp;.
- $y_1(t)$&nbsp; weist gegenüber &nbsp;$x_1(t)$&nbsp; Verzerrungen auf.
+
- $y_1(t)$&nbsp; exhibits distortions with respect to&nbsp;$x_1(t)$&nbsp;.
  
  
{Berechnen Sie das Signal &nbsp;$y_2(t)$&nbsp; als Systemantwort auf das Cosinussignal &nbsp;$x_2(t)$. &nbsp; Welcher Signalwert tritt zum Zeitpunkt &nbsp;$t = 0$&nbsp; auf?
+
{Compute the signal&nbsp; $y_2(t)$&nbsp; as the system response to the cosine signal&nbsp; $x_2(t)$. &nbsp; What is the signal value at time&nbsp; $t = 0$&nbsp;?
 
|type="{}"}
 
|type="{}"}
 
$y_2(t = 0) \ = \ $  { 0.996 3% }
 
$y_2(t = 0) \ = \ $  { 0.996 3% }
  
  
{Welche Aussagen treffen bezüglich der Signale &nbsp;$x_3(t)$&nbsp; und &nbsp;$y_3(t)$&nbsp; zu?
+
{Which statements are true regarding the signals&nbsp; $x_3(t)$&nbsp; and &nbsp; $y_3(t)$&nbsp;?
 
|type="[]"}
 
|type="[]"}
- $y_3(t)$&nbsp; weist gegenüber &nbsp;$x_3(t)$&nbsp; keine Verzerrungen auf.
+
- $y_3(t)$&nbsp; does not exhibit any distortions with respect to&nbsp; $x_3(t)$&nbsp;.
- $y_3(t)$&nbsp; weist gegenüber &nbsp;$x_3(t)$&nbsp; Dämpfungsverzerrungen auf.
+
- $y_3(t)$&nbsp; exhibits attenuation distortions with respect to&nbsp; $x_3(t)$&nbsp;.
+ $y_3(t)$&nbsp; weist gegenüber &nbsp;$x_3(t)$&nbsp; Phasenverzerrungen auf.
+
+ $y_3(t)$&nbsp; exhibits phase distortions with respect to&nbsp; $x_3(t)$&nbsp;.
  
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig sind die <u>Aussagen 1 und 2</u>:
+
'''(1)'''&nbsp; <u>Statements 1 and 2</u>&nbsp; are correct:
*Mit &nbsp;$z_1 = 1$, $T_1 = 0$, $z_2 =0$&nbsp; ist &nbsp;$h(t) = \delta(t)$&nbsp; und dementsprechend &nbsp;$H(f) = 1$, so dass stets &nbsp;$y(t) = x(t)$&nbsp; gelten wird.  
+
*$h(t) = \delta(t)$&nbsp; is true with&nbsp; $z_1 = 1$,&nbsp; $T_1 = 0$&nbsp; and&nbsp; $z_2 =0$&nbsp; and&nbsp; correspondingly&nbsp; $H(f) = 1$&nbsp; so that&nbsp; $y(t) = x(t)$&nbsp; will always hold.  
*Jede verzerrungsfreie Kanalimpulsantwort &nbsp;$h(t)$&nbsp; besteht aus einer einzigen Diracfunktion, zum Beispiel bei &nbsp;$t = T_1$.  
+
*Each distortion-free channel impulse response&nbsp; $h(t)$&nbsp; consists of a single Dirac function,&nbsp; for example at&nbsp; $t = T_1$.  
*Dieser Fall ist im Modell durch &nbsp;$z_2 =0$&nbsp; berücksichtigt. Damit lautet der Frequenzgang:
+
*This case is accounted for in the model by&nbsp; $z_2 =0$.&nbsp; Thus, the frequency response is:
 
:$$H(f)=  z_1\cdot {\rm e}^{-{\rm j}\cdot \hspace{0.05cm}2 \pi f T_1} \ \Rightarrow \ y(t) = z_1 \cdot x(t- T_1).$$
 
:$$H(f)=  z_1\cdot {\rm e}^{-{\rm j}\cdot \hspace{0.05cm}2 \pi f T_1} \ \Rightarrow \ y(t) = z_1 \cdot x(t- T_1).$$
*Dagegen wird der Kanal immer dann zu linearen Verzerrungen führen, wenn gleichzeitig &nbsp;$z_1$&nbsp; und &nbsp;$z_2$&nbsp; von Null verschieden sind.  
+
*In contrast, the channel will lead to linear distortions whenever&nbsp; $z_1$&nbsp; and&nbsp; $z_2$&nbsp; are simultaneously non-zero.  
  
  
  
'''(2)'''&nbsp; Die Fouriertransformation der Impulsantwort &nbsp;$h(t)$&nbsp; führt auf die Gleichung:
+
'''(2)'''&nbsp; The Fourier transform of the impulse response&nbsp; $h(t)$&nbsp; results in the equation:
 
:$$H(f) =  z_1\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_1}+ z_2\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_2}
 
:$$H(f) =  z_1\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_1}+ z_2\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_2}
 
   .$$
 
   .$$
  
*Mit &nbsp;$z_1 = 1$, $T_1 = 0$, $z_2 =0.5$&nbsp; und  &nbsp;$T_2 = 1 \ \rm ms$&nbsp; erhält man daraus:
+
*The following is obtained with&nbsp; $z_1 = 1$,&nbsp; $T_1 = 0$,&nbsp; $z_2 =0.5$&nbsp; and&nbsp; $T_2 = 1 \ \rm ms$:
 
:$$H(f) =1 + 0.5 \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_2}.$$
 
:$$H(f) =1 + 0.5 \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_2}.$$
  
*Aufgeschlüsselt nach Real&ndash; und Imaginärteil liefert dies:
+
*Broken down by real and imaginary part,&nbsp; this yields:
 
:$${\rm Re}\big[H(f)\big] = 1 + 0.5 \cdot \cos(2 \pi f \cdot 1\,{\rm  ms}) \ \Rightarrow \ \underline{{\rm Re}[H(f = f_1 =1 \ \rm kHz)] = 1.5}, $$
 
:$${\rm Re}\big[H(f)\big] = 1 + 0.5 \cdot \cos(2 \pi f \cdot 1\,{\rm  ms}) \ \Rightarrow \ \underline{{\rm Re}[H(f = f_1 =1 \ \rm kHz)] = 1.5}, $$
 
:$${\rm Im}\big[H(f)\big] = -0.5 \cdot \sin(2 \pi f \cdot 1\,{\rm  ms}) \ \Rightarrow \ \underline{{\rm Im}\big[H(f = f_1 =1 \ \rm kHz)\big] = 0}, $$
 
:$${\rm Im}\big[H(f)\big] = -0.5 \cdot \sin(2 \pi f \cdot 1\,{\rm  ms}) \ \Rightarrow \ \underline{{\rm Im}\big[H(f = f_1 =1 \ \rm kHz)\big] = 0}, $$
  
  
'''(3)'''&nbsp; Richtig ist nur die <u>erste Antwort</u>:
+
'''(3)'''&nbsp; The&nbsp; <u>first answer</u>&nbsp; is the only correct one:
*Aus&nbsp; '''(2)'''&nbsp; folgt für alle Vielfachen von &nbsp;$f_1 =1 \ \rm kHz$&nbsp; die Betragsfunktion &nbsp;$|H(f)| = 1.5$&nbsp; und die Phasenfunktion &nbsp;$b(f) \equiv 0$.  
+
*From&nbsp; '''(2)'''&nbsp; it follows the absolute value function is&nbsp; $|H(f)| = 1.5$&nbsp; and the phase function &nbsp; $b(f) \equiv 0$&nbsp; for all multiples of&nbsp; $f_1 =1 \ \rm kHz$ &nbsp; &rArr; &nbsp; $f= n \cdot f_1$.  
*Damit ist für diese diskreten Frequenzwerte auch die Phasenlaufzeit jeweils Null.
+
*Thus, the phase delay time is also zero in each case for these discrete frequency values.
*Da aber das Spektrum &nbsp;$X_1(f)$&nbsp; des Diracpulses genau bei diesen Frequenzen Spektrallinien aufweist, gilt &nbsp;$y_1(t) = 1.5 \cdot x_1(t)$.
+
*But since the spectrum&nbsp; $X_1(f)$&nbsp; of the Dirac comb has spectral lines exactly at these frequencies, &nbsp; $y_1(t) = 1.5 \cdot x_1(t)$&nbsp; holds.
 
   
 
   
  
  
'''(4)'''&nbsp; Die Betragsfunktion lautet:
+
'''(4)'''&nbsp; The absolute value function is:
 
:$$|H(f)| = \sqrt{{\rm Re}[H(f)]^2 + {\rm Im}[H(f)]^2} $$
 
:$$|H(f)| = \sqrt{{\rm Re}[H(f)]^2 + {\rm Im}[H(f)]^2} $$
 
:$$\Rightarrow \; |H(f)|  = \sqrt{1 + 0.25 \cdot \cos^2(2 \pi f \cdot T_2)+ \cos(2 \pi f \cdot T_2) +  0.25 \cdot \sin^2(2 \pi f \cdot T_2)}
 
:$$\Rightarrow \; |H(f)|  = \sqrt{1 + 0.25 \cdot \cos^2(2 \pi f \cdot T_2)+ \cos(2 \pi f \cdot T_2) +  0.25 \cdot \sin^2(2 \pi f \cdot T_2)}
 
   = \sqrt{1.25 +  \cos(2 \pi f \cdot T_2) }.$$
 
   = \sqrt{1.25 +  \cos(2 \pi f \cdot T_2) }.$$
  
*Für die Frequenz &nbsp;$f_2 =0.25 \ \rm kHz$&nbsp; erhält man somit:
+
*Thus, the following is obtained for the frequency&nbsp; $f_2 =0.25 \ \rm kHz$:
 
:$$|H(f)| = \sqrt{1.25 +  \cos(\frac{\pi}{2} ) }= \sqrt{1.25} = 1.118.$$
 
:$$|H(f)| = \sqrt{1.25 +  \cos(\frac{\pi}{2} ) }= \sqrt{1.25} = 1.118.$$
  
*Die Phasenfunktion lautet allgemein bzw. bei der Frequenz&nbsp; $f_2 =0.25 \ \rm kHz$:
+
*The phase function is generally or at the frequency&nbsp; $f_2 =0.25 \ \rm kHz$:
 
:$$b(f) = - {\rm arctan}\hspace{0.1cm}\frac{{\rm Im}[H(f)]}{{\rm
 
:$$b(f) = - {\rm arctan}\hspace{0.1cm}\frac{{\rm Im}[H(f)]}{{\rm
 
Re}[H(f)]} = - {\rm arctan}\hspace{0.1cm}\frac{-0.5 \cdot \sin(2
 
Re}[H(f)]} = - {\rm arctan}\hspace{0.1cm}\frac{-0.5 \cdot \sin(2
Line 120: Line 121:
 
arctan}\hspace{0.1cm}\frac{0.5}{1} = 0.464.$$
 
arctan}\hspace{0.1cm}\frac{0.5}{1} = 0.464.$$
  
*Damit beträgt die Phasenlaufzeit für diese Frequenz:
+
*Thus, the phase delay time for this frequency is:
 
:$$\tau_2 = \frac {b(f_2)}{2 \pi f_2}  = \frac {0.464}{2 \pi \cdot
 
:$$\tau_2 = \frac {b(f_2)}{2 \pi f_2}  = \frac {0.464}{2 \pi \cdot
0.25\,{\rm  kHz}} \approx 0.3\,{\rm  ms},$$
+
0.25\,{\rm  kHz}} \approx 0.3\,{\rm  ms}.$$
  
* Für das Ausgangssignal gilt somit:
+
*Hence, the following holds for the output signal:
 
:$$y_2(t) = 1.118 \cdot \cos(2 \pi \cdot 0.25\,{\rm  kHz}\cdot (t -
 
:$$y_2(t) = 1.118 \cdot \cos(2 \pi \cdot 0.25\,{\rm  kHz}\cdot (t -
 
0.3\,{\rm  ms})).$$
 
0.3\,{\rm  ms})).$$
  
*Der Signalwert zum Nullzeitpunkt ist somit:
+
*The signal value at zero-time is therefore:
 
:$$y_2(t=0) = 1.118 \cdot \cos(-2 \pi \cdot 0.25\,{\rm  kHz} \cdot
 
:$$y_2(t=0) = 1.118 \cdot \cos(-2 \pi \cdot 0.25\,{\rm  kHz} \cdot
 
0.3\,{\rm  ms}) \approx 1.118 \cdot  0.891 \hspace{0.15cm}\underline{= 0.996}.$$
 
0.3\,{\rm  ms}) \approx 1.118 \cdot  0.891 \hspace{0.15cm}\underline{= 0.996}.$$
  
  
'''(5)'''&nbsp; Beide Frequenzen haben den gleichen Dämpfungsfaktor &nbsp;$\alpha = 1.118$&nbsp;, daher sind keine Dämpfungsverzerrungen festzustellen.  
+
'''(5)'''&nbsp; Both frequencies have the same attenuation factor&nbsp; $\alpha = 1.118$&nbsp;. Therefore, no attenuation distortions are observed.  
  
*Mit &nbsp;$f_3 = 1.25 \ \rm kHz$&nbsp; und &nbsp;$T_2 = 1 \ \rm ms$&nbsp; ergibt sich für die Phasenfunktion:
+
*The following is obtained for the phase function with&nbsp; $f_3 = 1.25 \ \rm kHz$&nbsp; and &nbsp; $T_2 = 1 \ \rm ms$&nbsp;:
 
:$$b(f = f_3)  = - {\rm arctan}\hspace{0.1cm}\frac{-0.5 \cdot \sin(
 
:$$b(f = f_3)  = - {\rm arctan}\hspace{0.1cm}\frac{-0.5 \cdot \sin(
 
2.5 \pi)}{1+0.5 \cdot \cos(2.5 \pi)}= 0.464 = b(f = f_2),$$
 
2.5 \pi)}{1+0.5 \cdot \cos(2.5 \pi)}= 0.464 = b(f = f_2),$$
  
:also genau der gleiche Wert wie bei der Frequenz &nbsp;$f_2 = 0.25 \ \rm kHz$.  
+
:so exactly the same value as for the frequency&nbsp; $f_2 = 0.25 \ \rm kHz$.  
*Trotzdem kommt es aber nun zu Phasenverzerrungen, da für &nbsp;$f_3$&nbsp; die Phasenlaufzeit nur mehr &nbsp;$\tau  = 60 \  &micro; \rm s$&nbsp; beträgt.
+
*Despite this,&nbsp; however,&nbsp; phase distortions now occur since for&nbsp; $f_3$&nbsp; the phase delay time is only&nbsp; $\tau  = 60 \  &micro; \rm s$&nbsp; anymore.
  
*Für das Ausgangssignal kann also geschrieben werden:
+
*So,&nbsp; the following can be formulated for the output signal:
 
:$$y_3(t) = 1.118 \cdot \cos(2 \pi f_2 \cdot (t - 0.3\,{\rm  ms}) +
 
:$$y_3(t) = 1.118 \cdot \cos(2 \pi f_2 \cdot (t - 0.3\,{\rm  ms}) +
 
1.118
 
1.118
Line 149: Line 150:
 
\cos(2 \pi f_3 \cdot  t - 27^\circ).$$
 
\cos(2 \pi f_3 \cdot  t - 27^\circ).$$
  
Richtig ist demnach die <u>Antwort 3</u>:
+
Accordingly, the correct&nbsp; <u>answer is 3</u>:
*Es gibt also Phasenverzerrungen, obwohl für beide Schwingungen &nbsp;$\varphi_2 = \varphi_3= 27^\circ$&nbsp; gilt.  
+
*So, there are phase distortions although for both oscillations,&nbsp; $\varphi_2 = \varphi_3= 27^\circ$&nbsp; holds.  
*Damit keine Phasenverzerrungen auftreten, müssten
+
*To avoid phase distortions
**die Phasenlaufzeiten $\tau_2$ und $\tau_3$ gleich sein, und
+
**the phase delay times&nbsp; $\tau_2$&nbsp; and&nbsp; $\tau_3$&nbsp; would have to be equal and
**die Phasenwerte &nbsp;$\varphi_2$&nbsp; und &nbsp;$\varphi_3$&nbsp; linear mit den zugehörigen Frequenzen ansteigen.
+
**the phase values&nbsp; $\varphi_2$&nbsp; and&nbsp; $\varphi_3$&nbsp; would have to increase linearly with the corresponding frequencies.
 
{{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 10:29, 6 October 2021

Impulse response of the two-way channel

The so-called  "two-way channel"  is characterised by the following impulse response  $($with  $T_1 < T_2)$:

$$h(t) = z_1 \cdot \delta ( t - T_1) + z_2 \cdot \delta ( t - T_2).$$
  • Except for a few combinations of the system parameters  $z_1$,  $T_1$,  $z_2$  and  $T_2$,  this channel will result in linear distortions.
  • There is a distortion-free channel at hand only if not a single input signal is distorted by it.
  • This means:   Even if the two-way channel is per se distorting,  there may be special cases where indeed  $y(t) = \alpha \cdot x(t - \tau)$.


The test signals applied to the system input are:

  • Dirac comb  $x_1(t)$  at a time interval of  $T_0 = 1 \ \rm ms$,  whose spectral function  $X_1(f)$  is also a Dirac comb with an interval of  $f_0 = 1/T_0 = 1 \ \rm kHz$:
$$x_1(t) = \sum_{n = - \infty}^{+\infty} \delta ( t - n \cdot T_0) ,\hspace{0.5cm} X_1(f) = T_0 \cdot \sum_{k = - \infty}^{+\infty} \delta ( f - k \cdot f_0) ,$$
  • a cosine signal with frequency  $f_2 = 250 \ \rm Hz$:
$$x_2(t) = \cos(2 \pi \cdot f_2 \cdot t) ,$$
  • the sum of two cosine signals with frequencies  $f_2 = 250 \ \rm Hz$  and  $f_3 = 1250 \ \rm Hz$:
$$x_3(t) = \cos(2 \pi \cdot f_2 \cdot t) + \cos(2 \pi \cdot f_3 \cdot t) .$$





Please note:

  • To spare you calculations  the result for the parameter set  $\big [z_1 = 1$,  $T_1 = 0$,  $z_2 =0.5$,  $T_2 = 1 \ \rm ms\big ]$  is given:
$$|H(f = f_2)| = |H(f = f_3)| = \sqrt{1.25} \approx 1.118, \; \; \; \; b(f = f_2) = b(f = f_3) = \arctan (0.5) \approx 0.464.$$


Questions

1

Which of the following statements are true?

The parameter set  $\big[z_1 = 1$, $T_1 = 0$, $z_2 =0 \big]$  is the only possible one to describe the ideal channel.
Any distortion-free channel is captured by the two combinations  $\big[z_1 \ne 0, \; z_2 = 0 \big]$  or  $\big[z_1 = 0, \; z_2 \ne 0 \big]$ .
The values  $\big[z_1 \ne 0\big]$  and  $\big[z_2 \ne 0\big]$  result in a distortion-free channel  if  $T_1$  and  $T_2$  are optimally adjusted.

2

The following holds: $\big[z_1 = 1$,  $T_1 = 0$,  $z_2 =0.5$,  $T_2 = 1 \ \rm ms\big ]$.  Compute the frequency response  $H(f)$  of this channel.
What are the values at multiples of  $1 \ \rm kHz$?

${\rm Re}\big[H(f = n \cdot 1 \ {\rm kHz})\big] \ = \ $

${\rm Im}\big[H(f = n \cdot 1 \ {\rm kHz})\big] \ = \ $

3

The Dirac comb  $x_1(t)$  is applied to the input of the system with the same parameters as in subtask  (2) .
Which statements are true for the output signal  $y_1(t)$ ?

$y_1(t)$  is attenuated/amplified by a constant compared to $x_1(t)$ .
$y_1(t)$  is shifted with respect to $x_1(t)$ .
$y_1(t)$  exhibits distortions with respect to $x_1(t)$ .

4

Compute the signal  $y_2(t)$  as the system response to the cosine signal  $x_2(t)$.   What is the signal value at time  $t = 0$ ?

$y_2(t = 0) \ = \ $

5

Which statements are true regarding the signals  $x_3(t)$  and   $y_3(t)$ ?

$y_3(t)$  does not exhibit any distortions 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  $x_3(t)$ .


Solution

(1)  Statements 1 and 2  are correct:

  • $h(t) = \delta(t)$  is true with  $z_1 = 1$,  $T_1 = 0$  and  $z_2 =0$  and  correspondingly  $H(f) = 1$  so that  $y(t) = x(t)$  will always hold.
  • Each distortion-free channel impulse response  $h(t)$  consists of a single Dirac function,  for example at  $t = T_1$.
  • This case is accounted for in the model by  $z_2 =0$.  Thus, the frequency response is:
$$H(f)= z_1\cdot {\rm e}^{-{\rm j}\cdot \hspace{0.05cm}2 \pi f T_1} \ \Rightarrow \ y(t) = z_1 \cdot x(t- T_1).$$
  • In contrast, the channel will lead to linear distortions whenever  $z_1$  and  $z_2$  are simultaneously non-zero.


(2)  The Fourier transform of the impulse response  $h(t)$  results in the equation:

$$H(f) = z_1\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_1}+ z_2\cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_2} .$$
  • The following is obtained with  $z_1 = 1$,  $T_1 = 0$,  $z_2 =0.5$  and  $T_2 = 1 \ \rm ms$:
$$H(f) =1 + 0.5 \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}2 \pi f T_2}.$$
  • Broken down by real and imaginary part,  this yields:
$${\rm Re}\big[H(f)\big] = 1 + 0.5 \cdot \cos(2 \pi f \cdot 1\,{\rm ms}) \ \Rightarrow \ \underline{{\rm Re}[H(f = f_1 =1 \ \rm kHz)] = 1.5}, $$
$${\rm Im}\big[H(f)\big] = -0.5 \cdot \sin(2 \pi f \cdot 1\,{\rm ms}) \ \Rightarrow \ \underline{{\rm Im}\big[H(f = f_1 =1 \ \rm kHz)\big] = 0}, $$


(3)  The  first answer  is the only correct one:

  • From  (2)  it follows the absolute value function is  $|H(f)| = 1.5$  and the phase function   $b(f) \equiv 0$  for all multiples of  $f_1 =1 \ \rm kHz$   ⇒   $f= n \cdot f_1$.
  • Thus, the phase delay time is also zero in each case for these discrete frequency values.
  • But since the spectrum  $X_1(f)$  of the Dirac comb has spectral lines exactly at these frequencies,   $y_1(t) = 1.5 \cdot x_1(t)$  holds.


(4)  The absolute value function is:

$$|H(f)| = \sqrt{{\rm Re}[H(f)]^2 + {\rm Im}[H(f)]^2} $$
$$\Rightarrow \; |H(f)| = \sqrt{1 + 0.25 \cdot \cos^2(2 \pi f \cdot T_2)+ \cos(2 \pi f \cdot T_2) + 0.25 \cdot \sin^2(2 \pi f \cdot T_2)} = \sqrt{1.25 + \cos(2 \pi f \cdot T_2) }.$$
  • Thus, the following is obtained for the frequency  $f_2 =0.25 \ \rm kHz$:
$$|H(f)| = \sqrt{1.25 + \cos(\frac{\pi}{2} ) }= \sqrt{1.25} = 1.118.$$
  • The phase function is generally or at the frequency  $f_2 =0.25 \ \rm kHz$:
$$b(f) = - {\rm arctan}\hspace{0.1cm}\frac{{\rm Im}[H(f)]}{{\rm Re}[H(f)]} = - {\rm arctan}\hspace{0.1cm}\frac{-0.5 \cdot \sin(2 \pi f T_2)}{1+0.5 \cdot \cos(2 \pi f T_2)},$$
$$b(f = f_2) = - {\rm arctan}\hspace{0.1cm}\frac{-0.5 \cdot \sin( \pi/2)}{1+0.5 \cdot \cos(\pi/2)}={\rm arctan}\hspace{0.1cm}\frac{0.5}{1} = 0.464.$$
  • Thus, the phase delay time for this frequency is:
$$\tau_2 = \frac {b(f_2)}{2 \pi f_2} = \frac {0.464}{2 \pi \cdot 0.25\,{\rm kHz}} \approx 0.3\,{\rm ms}.$$
  • Hence, the following holds for the output signal:
$$y_2(t) = 1.118 \cdot \cos(2 \pi \cdot 0.25\,{\rm kHz}\cdot (t - 0.3\,{\rm ms})).$$
  • The signal value at zero-time is therefore:
$$y_2(t=0) = 1.118 \cdot \cos(-2 \pi \cdot 0.25\,{\rm kHz} \cdot 0.3\,{\rm ms}) \approx 1.118 \cdot 0.891 \hspace{0.15cm}\underline{= 0.996}.$$


(5)  Both frequencies have the same attenuation factor  $\alpha = 1.118$ . Therefore, no attenuation distortions are observed.

  • The following is obtained for the phase function with  $f_3 = 1.25 \ \rm kHz$  and   $T_2 = 1 \ \rm ms$ :
$$b(f = f_3) = - {\rm arctan}\hspace{0.1cm}\frac{-0.5 \cdot \sin( 2.5 \pi)}{1+0.5 \cdot \cos(2.5 \pi)}= 0.464 = b(f = f_2),$$
so exactly the same value as for the frequency  $f_2 = 0.25 \ \rm kHz$.
  • Despite this,  however,  phase distortions now occur since for  $f_3$  the phase delay time is only  $\tau = 60 \ µ \rm s$  anymore.
  • So,  the following can be formulated for the output signal:
$$y_3(t) = 1.118 \cdot \cos(2 \pi f_2 \cdot (t - 0.3\,{\rm ms}) + 1.118 \cdot \cos(2 \pi f_3 \cdot (t - 0.06\,{\rm ms})$$
$$\Rightarrow \; \; y_3(t) = 1.118 \cdot \cos(2 \pi f_2 \cdot t - 27^\circ) + 1.118 \cdot \cos(2 \pi f_3 \cdot t - 27^\circ).$$

Accordingly, the correct  answer is 3:

  • So, there are phase distortions although for both oscillations,  $\varphi_2 = \varphi_3= 27^\circ$  holds.
  • To avoid phase distortions
    • the phase delay times  $\tau_2$  and  $\tau_3$  would have to be equal and
    • the phase values  $\varphi_2$  and  $\varphi_3$  would have to increase linearly with the corresponding frequencies.