Difference between revisions of "Aufgaben:Exercise 2.1Z: Distortion and Equalisation"

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{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Klassifizierung der Verzerrungen
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{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Classification_of_the_Distortions
 
}}
 
}}
  
[[File:P_ID880__LZI_Z_2_1.png|right|frame|Drei kontinuierliche Spektralfunktionen]]
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[[File:P_ID880__LZI_Z_2_1.png|right|frame|Three continuous spectral functions]]
Die Grafik zeigt drei kontinuierliche Spektralfunktionen:
+
Three continuous spectral functions are depicted in the graph:
  
* ein cos<sup>2</sup>&ndash;Spektrum, das nur Anteile im Bereich&nbsp; $|f| < 1 \ \rm kHz$&nbsp; besitzt, wobei gilt:
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* a cos<sup>2</sup>&ndash;spectrum  that has components only in the range&nbsp; $|f| < 1 \ \rm kHz$&nbsp; where the following holds:
 
:$$A(f)  = 10^{\rm -3} \  {\rm V}/{\rm Hz} \cdot \cos^2(\frac{|f|}{1 \, \rm kHz} \cdot \frac{\pi}{ 2}  ) ,$$
 
:$$A(f)  = 10^{\rm -3} \  {\rm V}/{\rm Hz} \cdot \cos^2(\frac{|f|}{1 \, \rm kHz} \cdot \frac{\pi}{ 2}  ) ,$$
  
* ein Dreieckspektrum, ebenfalls begrenzt auf den Frequenzbereich&nbsp; $|f| < 1 \ \rm kHz$:
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* a triangular spectrum which is also limited to the frequency range&nbsp; $|f| < 1 \ \rm kHz$:
 
:$$B(f)  = 10^{\rm -3}  \ {\rm V}/{\rm Hz}  \cdot \left(1-\frac{|f|}{1 \, \rm kHz} \right),$$
 
:$$B(f)  = 10^{\rm -3}  \ {\rm V}/{\rm Hz}  \cdot \left(1-\frac{|f|}{1 \, \rm kHz} \right),$$
  
* ein so genanntes Gaußspektrum:
+
* a so-called Gaussian spectrum:
 
:$$C(f)  = 10^{\rm -3} \ {\rm V}/{\rm Hz} \cdot {\rm e}^{-\pi \hspace{0.05cm}\cdot \hspace{0.05cm} (f/{1 \, \rm kHz})^2}  .$$
 
:$$C(f)  = 10^{\rm -3} \ {\rm V}/{\rm Hz} \cdot {\rm e}^{-\pi \hspace{0.05cm}\cdot \hspace{0.05cm} (f/{1 \, \rm kHz})^2}  .$$
  
  
Weiterhin betrachten wir
+
Furthermore, we consider
*ein linear verzerrendes System&nbsp; $S_{\rm V}$&nbsp; mit&nbsp; $X(f)$&nbsp; am Eingang und&nbsp; $Y(f)$&nbsp; am Ausgang, sowie
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*a linearly distorting system&nbsp; $S_{\rm V}$&nbsp; with&nbsp; $X(f)$&nbsp; at the input and&nbsp; $Y(f)$&nbsp; at the output, and
*das Entzerrungssystem&nbsp; $S_{\rm E}$&nbsp; mit Eingangsspektrum&nbsp; $Y(f)$&nbsp; und Ausgangsspektrum&nbsp; $Z(f)$.
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*the equalization system&nbsp; $S_{\rm E}$&nbsp; with the input spectrum&nbsp; $Y(f)$&nbsp; and output spectrum&nbsp; $Z(f)$.
  
  
Die Frequenzgänge der beiden Systeme&nbsp; $S_{\rm V}$&nbsp; und&nbsp; $S_{\rm E}$&nbsp; lauten:
+
The frequency responses of the two systems&nbsp; $S_{\rm V}$&nbsp; and&nbsp; $S_{\rm E}$&nbsp; are:
 
:$$H_{\rm V}(f)  = \frac{Y(f)}{X(f)} , \hspace{0.3cm}$$
 
:$$H_{\rm V}(f)  = \frac{Y(f)}{X(f)} , \hspace{0.3cm}$$
 
:$$H_{\rm E}(f)  = \frac{Z(f)}{Y(f)} .$$
 
:$$H_{\rm E}(f)  = \frac{Z(f)}{Y(f)} .$$
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''Hinweise:''  
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''Please note:''  
*Die Aufgabe gehört zum Kapitel&nbsp;  [[Linear_and_Time_Invariant_Systems/Klassifizierung_der_Verzerrungen|Klassifizierung der Verzerrungen]].  
+
*The exercise belongs to the chapter&nbsp;  [[Linear_and_Time_Invariant_Systems/Classification_of_the_Distortions|Classification of the Distortions]].  
*Eine vollständige Entzerrung bedeutet, dass&nbsp; $Z(f) = X(f)$&nbsp; gilt.
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*A complete equalization means that&nbsp; $Z(f) = X(f)$&nbsp; holds.
  
  
  
===Fragebogen===
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Ist mit einem linearen System die Konstellation &nbsp;$X(f) = A(f)$&nbsp; und &nbsp;$Y(f) = B(f)$&nbsp; möglich? Begründen Sie Ihre Antwort.
+
{Is the constellation&nbsp;$X(f) = A(f)$&nbsp; and &nbsp;$Y(f) = B(f)$&nbsp; possible with a linear system? Justify your answer.
 
|type="()"}
 
|type="()"}
+ Ja.
+
+ Yes.
- Nein.
+
- No.
  
  
{Es gelte weiterhin &nbsp;$X(f) = A(f)$&nbsp; und &nbsp;$Y(f) = B(f)$. Ist mit einem linearen Filter &nbsp;$H_{\rm E}(f)$&nbsp; eine vollständige Entzerrung möglich? <br>Wenn JA, so geben Sie bitte &nbsp;$H_{\rm E}(f)$ an.
+
{$X(f) = A(f)$&nbsp; and &nbsp;$Y(f) = B(f)$ still hold true. Is complete equalization possible with a linear filter&nbsp;$H_{\rm E}(f)$&nbsp;? <br>If YES, please specify &nbsp;$H_{\rm E}(f)$.
 
|type="()"}
 
|type="()"}
+ Ja.
+
+ Yes.
- Nein.
+
- No.
  
  
{Ist mit einem linearen System die Konstellation  &nbsp;$X(f) = C(f)$&nbsp; und &nbsp;$Y(f) = B(f)$&nbsp; möglich? Begründen Sie Ihre Antwort.
+
{Is the constellation&nbsp;$X(f) = C(f)$&nbsp; and &nbsp;$Y(f) = B(f)$&nbsp; possible with a linear system? Justify your answer.
 
|type="()"}
 
|type="()"}
+ Ja.
+
+ Yes.
- Nein.
+
- No.
  
  
{Es gelte weiterhin &nbsp;$X(f) = C(f)$&nbsp; und &nbsp;$Y(f) = B(f)$. Ist mit einem linearen Filter &nbsp;$H_{\rm E}(f)$&nbsp; eine vollständige Entzerrung möglich? <br>Wenn JA, so geben Sie bitte &nbsp;$H_{\rm E}(f)$&nbsp; an.
+
{$X(f) = C(f)$&nbsp; und &nbsp;$Y(f) = B(f)$ still hold true. Is complete equalization possible with a linear filter&nbsp;$H_{\rm E}(f)$&nbsp;? <br>If YES, please specify &nbsp;$H_{\rm E}(f)$&nbsp;.
 
|type="()"}
 
|type="()"}
- Ja.
+
- Yes.
+ Nein.
+
+ No.
  
  
{Ist mit einem linearen System die Konstellation &nbsp;$X(f) = A(f)$&nbsp; und &nbsp;$Y(f) = C(f)$&nbsp; möglich? <br>Begründen Sie Ihre Antwort.
+
{Is the constellation&nbsp;$X(f) = A(f)$&nbsp; and &nbsp;$Y(f) = C(f)$&nbsp; possible with a linear system? <br>Justify your answer.
 
|type="()"}
 
|type="()"}
- Ja.
+
- Yes.
+ Nein.
+
+ No.
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig ist <u>Ja</u>:
+
'''(1)'''&nbsp; <u>Yes</u> is correct:
*Diese Konstellation ist möglich, da für alle&nbsp; $Y(f) \ne 0$&nbsp; auch&nbsp; $X(f)$&nbsp; stets von Null verschieden ist.  
+
*This constellation is possible because&nbsp; $X(f)$&nbsp; is also always different from zero for all&nbsp; $Y(f) \ne 0$&nbsp;.  
*Für alle Frequenzen kleiner als&nbsp; $0.5 \ \rm kHz$&nbsp; bewirkt&nbsp; $H_{\rm V} = B(f)/A(f) < 1$&nbsp; eine Dämpfung.
+
*For all frequencies less than&nbsp; $0.5 \ \rm kHz$&nbsp;, &nbsp; $H_{\rm V} = B(f)/A(f) < 1$&nbsp; causes an attenuation.
* Die Frequenzen zwischen&nbsp; $0.5 \ \rm kHz$&nbsp; und&nbsp; $1 \ \rm kHz$&nbsp; werden dagegen durch das System angehoben.
+
*In contrast, the frequencies between&nbsp; $0.5 \ \rm kHz$&nbsp; and&nbsp; $1 \ \rm kHz$&nbsp; are amplified by the system.
  
  
  
'''(2)'''&nbsp; Richtig ist <u>Ja</u>:
+
'''(2)'''&nbsp; <u>Yes</u> is correct:
*Bei dieser Konstellation ist auch eine vollständige lineare Entzerrung mit
+
*For this constellation, a complete linear equalization is also possible with
 
:$$H_{\rm E}(f)  = \frac{Z(f)}{Y(f)} = \frac{A(f)}{B(f)} = \frac{1}{H_{\rm V}(f)}$$
 
:$$H_{\rm E}(f)  = \frac{Z(f)}{Y(f)} = \frac{A(f)}{B(f)} = \frac{1}{H_{\rm V}(f)}$$
  
:möglich, da beide Spektren genau bis&nbsp; $1 \ \rm kHz$&nbsp; reichen.
+
:since both spectra extend exactly up to&nbsp; $1 \ \rm kHz$&nbsp;.
  
  
  
'''(3)'''&nbsp; Richtig ist <u>Ja</u>:
+
'''(3)'''&nbsp; <u>Yes</u> is correct:
*Auch diese Konstellation ist möglich. Das Filter&nbsp;  $H_{\rm V}(f)$&nbsp; muss für die Frequenzen&nbsp; $|f| <1 \ \rm kHz$&nbsp; aus dem Gaußspektrum ein Dreieckspektrum formen und alle Frequenzen&nbsp; $|f| > 1 \ \rm kHz$&nbsp; unterdrücken.
+
*This constellation is possible, too. The filter&nbsp;  $H_{\rm V}(f)$&nbsp; must form a triangular spectrum for the frequencies&nbsp; $|f| <1 \ \rm kHz$&nbsp; out of the Gaussian spectrum and suppress all frequencies&nbsp; $|f| > 1 \ \rm kHz$&nbsp;.
  
  
  
'''(4)'''&nbsp; Richtig ist <u>Nein</u>:
+
'''(4)'''&nbsp; <u>No</u> is correct:
*Eine vollständige Entzerrung ist hier nicht möglich:
+
*A complete equalization is not possible here:
*Die Anteile des Gaußspektrums, die durch&nbsp;  $H_{\rm V}(f)$&nbsp; vollständig eliminiert wurden, können durch das lineare System nicht wieder hergestellt werden.
+
*The parts of the Gaussian spectrum, which are completely eliminated by&nbsp;  $H_{\rm V}(f)$&nbsp;, cannot be recovered by the linear system.
  
  
  
'''(5)'''&nbsp; Richtig ist <u>Nein</u>:
+
'''(5)'''&nbsp; <u>No</u> is correct:
*Diese Konstellation ist mit einem linearen System nicht möglich, da im Spektrum &nbsp;$C(f) = A(f) \cdot H_{\rm V}(f)$&nbsp; keine Spektralanteile enthalten sein können, die es in &nbsp;$A(f)$&nbsp; nicht gibt.
+
*This constellation is not possible with a linear system since there cannot be any spectral components included in the spectrum &nbsp;$C(f) = A(f) \cdot H_{\rm V}(f)$&nbsp; that do not exist in&nbsp;$A(f)$&nbsp;.
*Die Frage, ob es ein nichtlineares System gibt, das aus dem&nbsp; $\cos^2$-Spektrum ein Gaußspektrum formt, ist nicht gestellt und muss deshalb auch nicht beantwortet werden: &nbsp; Die Autoren glauben eher &bdquo;Nein&rdquo;.
+
*The question whether there is a non-linear system which forms a Gaussian spectrum out of the&nbsp; $\cos^2$-spectrum is not asked and therefore does not need to be answered: &nbsp; The authors rather believe "no".
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Linear and Time-Invariant Systems: Exercises|^2.1 Klassifizierung der Verzerrungen^]]
+
[[Category:Linear and Time-Invariant Systems: Exercises|^2.1 Classification of the Distortions^]]

Latest revision as of 14:50, 21 September 2021

Three continuous spectral functions

Three continuous spectral functions are depicted in the graph:

  • a cos2–spectrum that has components only in the range  $|f| < 1 \ \rm kHz$  where the following holds:
$$A(f) = 10^{\rm -3} \ {\rm V}/{\rm Hz} \cdot \cos^2(\frac{|f|}{1 \, \rm kHz} \cdot \frac{\pi}{ 2} ) ,$$
  • a triangular spectrum which is also limited to the frequency range  $|f| < 1 \ \rm kHz$:
$$B(f) = 10^{\rm -3} \ {\rm V}/{\rm Hz} \cdot \left(1-\frac{|f|}{1 \, \rm kHz} \right),$$
  • a so-called Gaussian spectrum:
$$C(f) = 10^{\rm -3} \ {\rm V}/{\rm Hz} \cdot {\rm e}^{-\pi \hspace{0.05cm}\cdot \hspace{0.05cm} (f/{1 \, \rm kHz})^2} .$$


Furthermore, we consider

  • a linearly distorting system  $S_{\rm V}$  with  $X(f)$  at the input and  $Y(f)$  at the output, and
  • the equalization system  $S_{\rm E}$  with the input spectrum  $Y(f)$  and output spectrum  $Z(f)$.


The frequency responses of the two systems  $S_{\rm V}$  and  $S_{\rm E}$  are:

$$H_{\rm V}(f) = \frac{Y(f)}{X(f)} , \hspace{0.3cm}$$
$$H_{\rm E}(f) = \frac{Z(f)}{Y(f)} .$$




Please note:


Questions

1

Is the constellation $X(f) = A(f)$  and  $Y(f) = B(f)$  possible with a linear system? Justify your answer.

Yes.
No.

2

$X(f) = A(f)$  and  $Y(f) = B(f)$ still hold true. Is complete equalization possible with a linear filter $H_{\rm E}(f)$ ?
If YES, please specify  $H_{\rm E}(f)$.

Yes.
No.

3

Is the constellation $X(f) = C(f)$  and  $Y(f) = B(f)$  possible with a linear system? Justify your answer.

Yes.
No.

4

$X(f) = C(f)$  und  $Y(f) = B(f)$ still hold true. Is complete equalization possible with a linear filter $H_{\rm E}(f)$ ?
If YES, please specify  $H_{\rm E}(f)$ .

Yes.
No.

5

Is the constellation $X(f) = A(f)$  and  $Y(f) = C(f)$  possible with a linear system?
Justify your answer.

Yes.
No.


Solution

(1)  Yes is correct:

  • This constellation is possible because  $X(f)$  is also always different from zero for all  $Y(f) \ne 0$ .
  • For all frequencies less than  $0.5 \ \rm kHz$ ,   $H_{\rm V} = B(f)/A(f) < 1$  causes an attenuation.
  • In contrast, the frequencies between  $0.5 \ \rm kHz$  and  $1 \ \rm kHz$  are amplified by the system.


(2)  Yes is correct:

  • For this constellation, a complete linear equalization is also possible with
$$H_{\rm E}(f) = \frac{Z(f)}{Y(f)} = \frac{A(f)}{B(f)} = \frac{1}{H_{\rm V}(f)}$$
since both spectra extend exactly up to  $1 \ \rm kHz$ .


(3)  Yes is correct:

  • This constellation is possible, too. The filter  $H_{\rm V}(f)$  must form a triangular spectrum for the frequencies  $|f| <1 \ \rm kHz$  out of the Gaussian spectrum and suppress all frequencies  $|f| > 1 \ \rm kHz$ .


(4)  No is correct:

  • A complete equalization is not possible here:
  • The parts of the Gaussian spectrum, which are completely eliminated by  $H_{\rm V}(f)$ , cannot be recovered by the linear system.


(5)  No is correct:

  • This constellation is not possible with a linear system since there cannot be any spectral components included in the spectrum  $C(f) = A(f) \cdot H_{\rm V}(f)$  that do not exist in $A(f)$ .
  • The question whether there is a non-linear system which forms a Gaussian spectrum out of the  $\cos^2$-spectrum is not asked and therefore does not need to be answered:   The authors rather believe "no".