Difference between revisions of "Aufgaben:Exercise 2.5: Distortion and Equalization"

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[[File:P_ID907__LZI_A_2_5.png|right|frame|Trapezoidal spectrum (top) and <br>associated impulse response (bottom)]]
+
[[File:P_ID907__LZI_A_2_5.png|right|frame|Trapezoidal spectrum (top), <br>associated impulse response)]]
A communication system with input&nbsp; $x(t)$&nbsp; and output&nbsp; $y(t)$, which is fully described by the trapezoidal frequency response&nbsp; $H(f)$&nbsp; according to the top graph, is considered. Using the roll-off factor&nbsp; $r = 0.5$&nbsp; and the equivalent bandwidth&nbsp; $\Delta f = 16 \ \rm kHz$&nbsp; the corresponding impulse response, which is computable by applying the inverse Fourier transform, is:
+
A communication system with input&nbsp; $x(t)$&nbsp; and output&nbsp; $y(t)$,&nbsp; which is fully described by the trapezoidal frequency response&nbsp; $H(f)$&nbsp; according to the top graph, is considered.&nbsp; Using the roll-off factor&nbsp; $r = 0.5$&nbsp; and the equivalent bandwidth&nbsp; $\Delta f = 16 \ \rm kHz$&nbsp; the corresponding impulse response,&nbsp; which is computable by applying the inverse Fourier transform, is:
 
:$$h(t) = \Delta f \cdot {\rm si}(\pi \cdot \Delta f \cdot t )\cdot
 
:$$h(t) = \Delta f \cdot {\rm si}(\pi \cdot \Delta f \cdot t )\cdot
 
{\rm si}(\pi \cdot r \cdot \Delta f \cdot t
 
{\rm si}(\pi \cdot r \cdot \Delta f \cdot t
) .$$
+
) = \Delta f \cdot {\rm sinc}(\Delta f \cdot t )\cdot
 
+
{\rm sinc}(r \cdot \Delta f \cdot t
 +
).$$
 +
Here the following functions which can be converted into each other are used:
 +
:$${\rm si}(x) = \sin(x)/x,\hspace{0.5cm}{\rm sinc}(x) = \sin(\pi x)/(\pi x).$$
 
The available input signals are:
 
The available input signals are:
 
*The sum of two harmonic oscillations:
 
*The sum of two harmonic oscillations:
 
:$$x_1(t) =  {1\, \rm V} \cdot \cos(\omega_1 \cdot  t) + {1\, \rm V} \cdot \sin(\omega_2 \cdot
 
:$$x_1(t) =  {1\, \rm V} \cdot \cos(\omega_1 \cdot  t) + {1\, \rm V} \cdot \sin(\omega_2 \cdot
 
  t).$$
 
  t).$$
:Here, the following holds:&nbsp; $\omega_1 = 2\pi \cdot 2000 \ {\rm 1/s}$&nbsp; and&nbsp; $\omega_2 \gt \omega_1$.
+
:Here, the following holds: &nbsp; $\omega_1 = 2\pi \cdot 2000 \ {\rm 1/s}$&nbsp; and&nbsp; $\omega_2 \gt \omega_1$.
 
*A periodic triangular signal:
 
*A periodic triangular signal:
 
:$$x_2(t) =  \frac{8\, \rm V}{\pi^2} \cdot \big[\cos(\omega_0  t) + {1}/{9} \cdot \cos(3\omega_0  t)
 
:$$x_2(t) =  \frac{8\, \rm V}{\pi^2} \cdot \big[\cos(\omega_0  t) + {1}/{9} \cdot \cos(3\omega_0  t)
 
  + {1}/{25} \cdot \cos(5\omega_0  t) + \hspace{0.05cm}\text{...}\big].$$
 
  + {1}/{25} \cdot \cos(5\omega_0  t) + \hspace{0.05cm}\text{...}\big].$$
:It should be noted that the fundamental frequency is&nbsp; $f_0 = 2 \ \rm kHz$&nbsp; or&nbsp; $3\ \rm kHz$&nbsp; beträgt. At time&nbsp; $t = 0$&nbsp; the signal value in both cases is&nbsp; $1 \ \rm V$.
+
:It should be noted that the basic frequency is&nbsp; $f_0 = 2 \ \rm kHz$&nbsp; or&nbsp; $3\ \rm kHz$.&nbsp; At time&nbsp; $t = 0$&nbsp; the signal value in both cases is&nbsp; $1 \ \rm V$.
*A rectangular pulse&nbsp; $x_3(t)$&nbsp; with amplitude&nbsp; $A = 1 \ \rm V$&nbsp; and duration&nbsp; $T = 1 \ \rm ms$. Since its spectrum&nbsp; $X_3(f)$&nbsp; extends to infinity, &nbsp; $H(f)$&nbsp; always results in linear distortions here.
+
*A rectangular pulse&nbsp; $x_3(t)$&nbsp; with amplitude&nbsp; $A = 1 \ \rm V$&nbsp; and duration&nbsp; $T = 1 \ \rm ms$. <br>Since its spectrum&nbsp; $X_3(f)$&nbsp; extends to infinity, &nbsp; $H(f)$&nbsp; always results in linear distortions here.
  
  
From subtask&nbsp; '''(6)'''&nbsp; onwards, it shall be attempted to eliminate the distortions possibly generated by&nbsp; $H(f)$&nbsp; by means of a downstream equaliser with
+
From subtask&nbsp; '''(6)'''&nbsp; onwards, it shall be attempted to eliminate the distortions possibly generated by&nbsp; $H(f)$&nbsp; by means of a downstream equalizer with
 
* frequency response&nbsp; $H_{\rm E}(f)$,
 
* frequency response&nbsp; $H_{\rm E}(f)$,
 
* input signal&nbsp; $y(t)$,&nbsp; and
 
* input signal&nbsp; $y(t)$,&nbsp; and
Line 32: Line 35:
 
''Please note:''  
 
''Please note:''  
 
*The task belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]].
 
*The task belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]].
*In particular, reference is made to the page&nbsp;  [[Linear_and_Time_Invariant_Systems/Linear_Distortions#Entzerrungsverfahren|Entzerrungsverfahren]].
+
*In particular, reference is made to the page&nbsp;  [[Linear_and_Time_Invariant_Systems/Linear_Distortions#Equalization_methods|Equalization methods]].
*The term "overall distortion" used in the formulation of the questions refers to the input signal&nbsp; $x(t)$&nbsp; and the output signal&nbsp; $z(t)$.
+
*The term&nbsp; "overall distortion"&nbsp; used in the formulation of the questions refers to the input signal&nbsp; $x(t)$&nbsp; and the output signal&nbsp; $z(t)$.
 
   
 
   
  
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{For the test signal&nbsp; $x_2(t)$&nbsp;  with&nbsp; $\underline{f_0 = 3 \ \rm kHz}$,&nbsp; what is the maximum deviation&nbsp;$\varepsilon_{\rm max} = |y_2(t_0) - x_2(t_0)|$. <br>At what point&nbsp;$t_0$&nbsp; does&nbsp; $\varepsilon_{\rm max}$&nbsp; occur for the first time?
+
{For the test signal&nbsp; $x_2(t)$&nbsp;  with&nbsp; $\underline{f_0 = 3 \ \rm kHz}$,&nbsp; what is the maximum deviation&nbsp; $\varepsilon_{\rm max} = |y_2(t_0) - x_2(t_0)|$. <br>At what time&nbsp; $t_0$&nbsp; does&nbsp; $\varepsilon_{\rm max}$&nbsp; occur for the first time?
 
|type="{}"}
 
|type="{}"}
 
$\varepsilon_\text{max} \ = \ $ { 0.156 3% } $\ \rm V$
 
$\varepsilon_\text{max} \ = \ $ { 0.156 3% } $\ \rm V$
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{What is the maximum deviation&nbsp;$\varepsilon_{\rm max}$&nbsp; with&nbsp;$\underline{f_0 = 2 \ \rm kHz}$?
+
{What is the maximum deviation&nbsp; $\varepsilon_{\rm max}$&nbsp; with&nbsp; $\underline{f_0 = 2 \ \rm kHz}$?
 
|type="{}"}
 
|type="{}"}
 
$\varepsilon_\text{max} \ = \ $ { 0.114 3% } $\ \rm V$
 
$\varepsilon_\text{max} \ = \ $ { 0.114 3% } $\ \rm V$
  
  
{What curve shape should the equaliser&nbsp; $H_{\rm E}(f)$&nbsp; have to compensate all distortions of&nbsp; $H(f)$&nbsp; in the best possible way. <br>What magnitude value arises as a result for&nbsp; $\underline{f = 10 \ \rm kHz}$?
+
{What curve shape should the equalizer&nbsp; $H_{\rm E}(f)$&nbsp; have to compensate all distortions of&nbsp; $H(f)$&nbsp; in the best possible way? <br>What magnitude value arises as a result for&nbsp; $\underline{f = 10 \ \rm kHz}$?
 
|type="{}"}
 
|type="{}"}
 
$|H_E(f = 10 \ \rm kHz)| \ = \ $ { 4 3% }
 
$|H_E(f = 10 \ \rm kHz)| \ = \ $ { 4 3% }
  
  
{For which of the signals listed is complete equalisation possible? <br>$z(t) = x(t)$&nbsp; should be understood by "complete equalisation".
+
{For which of the listed signals is complete equalization possible? <br>$z(t) = x(t)$&nbsp; should be understood by "complete equalization".
 
|type="[]"}
 
|type="[]"}
 
+ For signal&nbsp; $x_1(t)$&nbsp; with&nbsp; $f_2 = 10 \ \rm kHz$,
 
+ For signal&nbsp; $x_1(t)$&nbsp; with&nbsp; $f_2 = 10 \ \rm kHz$,
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===Solution===
 
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; <u>Proposed solutions 1 and 3</u> are correct:
+
'''(1)'''&nbsp; <u>Proposed solutions 1 and 3</u>&nbsp; are correct:
 
*A linear system is already implicitly assumed by specifying a frequency response so that nonlinear distortions cannot occur.  
 
*A linear system is already implicitly assumed by specifying a frequency response so that nonlinear distortions cannot occur.  
 
*Since&nbsp; $H(f)$&nbsp; is purely real, phase distortions can also be ruled out.
 
*Since&nbsp; $H(f)$&nbsp; is purely real, phase distortions can also be ruled out.
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'''(3)'''&nbsp; <u>Proposed solution 3</u> is correct:
+
'''(3)'''&nbsp; <u>Proposed solution 3</u>&nbsp; is correct:
 
*In this case, the following is obtained for the output signal:
 
*In this case, the following is obtained for the output signal:
 
:$$y_1(t)=  1\,{\rm  V}\cdot \cos(2 \pi \cdot f_1 \cdot t) + {1}/{4}\cdot 1\,{\rm  V}\cdot \sin(2 \pi \cdot f_2 \cdot t).$$
 
:$$y_1(t)=  1\,{\rm  V}\cdot \cos(2 \pi \cdot f_1 \cdot t) + {1}/{4}\cdot 1\,{\rm  V}\cdot \sin(2 \pi \cdot f_2 \cdot t).$$
*While the component at&nbsp; $f_1$&nbsp; is transmitted unchanged, the sinusoidal component with&nbsp; $f_2$&nbsp; is attenuated to one-quarter.  
+
*While the component at&nbsp; $f_1$&nbsp; is transmitted unchanged, the sinusoidal component at&nbsp; $f_2$&nbsp; is attenuated and one-quarter of the original sinusoidal component.  
 
*So, there are attenuation distortions.
 
*So, there are attenuation distortions.
  
  
  
'''(4)'''&nbsp; Das Ausgangssignal&nbsp; $y_2(t)$&nbsp; hat die folgende Form, wenn man die Grundfrequenz&nbsp; $f_0 = 3 \ \rm kHz$&nbsp; berücksichtigt:
+
'''(4)'''&nbsp; The output signal&nbsp; $y_2(t)$&nbsp; has the following form taking into account the basic frequency&nbsp; $f_0 = 3 \ \rm kHz$:
 
:$$y_2(t)=  \frac{8\,{\rm  V}}{\pi^2} \left( \cos(\omega_0  t) +
 
:$$y_2(t)=  \frac{8\,{\rm  V}}{\pi^2} \left( \cos(\omega_0  t) +
 
  \frac{3}{8}\cdot \frac{1}{9} \cdot \cos(3\omega_0  t)\right)
 
  \frac{3}{8}\cdot \frac{1}{9} \cdot \cos(3\omega_0  t)\right)
 
.$$
 
.$$
  
*Der Faktor&nbsp; $3/8$&nbsp; beschreibt&nbsp; $H(f = 9 \ \rm kHz)$. Alle weiteren Spektralanteile bei&nbsp; $15 \ \rm kHz$,&nbsp; $21 \ \rm kHz$,&nbsp; usw. werden vom System unterdrückt.
+
*The factor&nbsp; $3/8$&nbsp; describes&nbsp; $H(f = 9 \ \rm kHz)$.&nbsp; All other spectral components at&nbsp; $15 \ \rm kHz$,&nbsp; $21 \ \rm kHz$,&nbsp; etc. are suppressed by the system.
  
*Die stärksten Abweichungen zwischen&nbsp; $x_2(t)$&nbsp; und&nbsp; $y_2(t)$&nbsp; wird es bei den Dreieckspitzen geben, da sich hier die fehlenden hohen Frequenzen am stärksten auswirken. Zum Beispiel erhält man für den Zeitpunkt&nbsp; $\underline{t= 0}$:
+
*The strongest deviations between&nbsp; $x_2(t)$&nbsp; and&nbsp; $y_2(t)$&nbsp; will occur at the triangle peaks since the missing high frequencies have the strongest effect here.
 +
*For example, for the time&nbsp; $\underline{t= 0}$ one obtains:
 
:$$y_2(t=0)=  \frac{8\,{\rm  V}}{\pi^2} \left( 1 +
 
:$$y_2(t=0)=  \frac{8\,{\rm  V}}{\pi^2} \left( 1 +
 
  {3}/{72}\right)= 0.844\,{\rm
 
  {3}/{72}\right)= 0.844\,{\rm
Line 125: Line 129:
  
  
'''(5)'''&nbsp; Mit der Grundfrequenz&nbsp; $f_0 = 2 \ \rm kHz$&nbsp; sowie den Übertragungswerten&nbsp;  $H(3f_0) = 0.75$,&nbsp; $H(5f_0) = 0.25$,&nbsp; $H(7f_0) = 0$&nbsp; ergibt sich:
+
'''(5)'''&nbsp; With the basic frequency&nbsp; $f_0 = 2 \ \rm kHz$&nbsp; and the values&nbsp;  $H(3f_0) = 0.75$,&nbsp; $H(5f_0) = 0.25$,&nbsp; $H(7f_0) = 0$&nbsp; the following is obtained:
 
:$$y_2(t=0)=  \frac{8\,{\rm  V}}{\pi^2} \left( 1 +
 
:$$y_2(t=0)=  \frac{8\,{\rm  V}}{\pi^2} \left( 1 +
 
  \frac{3}{4}\cdot \frac{1}{9} + \frac{1}{4} \cdot\frac{1}{25}\right)= 0.886\,{\rm
 
  \frac{3}{4}\cdot \frac{1}{9} + \frac{1}{4} \cdot\frac{1}{25}\right)= 0.886\,{\rm
Line 134: Line 138:
  
  
'''(6)'''&nbsp; Im Bereich bis&nbsp; $4  \  \rm kHz$&nbsp; ist&nbsp; $H_{\rm E}(f) = H(f) = 1$&nbsp; zu setzen. Dagegen gilt im Bereich von&nbsp; $4  \  \rm kHz$&nbsp; bis&nbsp; $12  \  \rm kHz$:
+
'''(6)'''&nbsp; In the range up to&nbsp; $4  \  \rm kHz$,&nbsp; $H_{\rm E}(f) = H(f) = 1$&nbsp; is to be set.&nbsp; In contrast, in the range from&nbsp; $4  \  \rm kHz$&nbsp; to&nbsp; $12  \  \rm kHz$ the following holds:
 
:$$H_{\rm  E}(f)=  \frac{1}{H(f)} =
 
:$$H_{\rm  E}(f)=  \frac{1}{H(f)} =
 
  \frac{1}{1.5 \cdot \big[1 - f/(12\,{\rm  kHz})\big]}
 
  \frac{1}{1.5 \cdot \big[1 - f/(12\,{\rm  kHz})\big]}
Line 141: Line 145:
 
.$$
 
.$$
  
Der Nennerausdruck beschreibt hierbei die Geradengleichung des Flankenabfalls.
+
Here, the denominator expression describes the equation of the straight line of the frequency roll-off.
  
  
  
'''(7)'''&nbsp; Richtig ist der <u>Lösungsvorschlag 1</u>:
+
'''(7)'''&nbsp; <u>Proposed solution 1</u> is correct:
*Sowohl&nbsp; $x_2(t)$&nbsp; als auch&nbsp; $x_3(t)$&nbsp; beinhalten auch Spektralanteile bei Frequenzen größer als&nbsp; $12  \  \rm kHz$.  
+
*Both&nbsp; $x_2(t)$&nbsp; and&nbsp; $x_3(t)$&nbsp; also contain spectral components at frequencies greater than&nbsp; $12  \  \rm kHz$.  
*Wurden diese von&nbsp; $H(f)$&nbsp; abgeschnitten &nbsp; &rArr; &nbsp; Bandbegrenzung, so können sie durch den Entzerrer nicht mehr rekonstruiert werden.  
+
*If these have been truncated by&nbsp; $H(f)$&nbsp; &rArr; &nbsp; band limitation, they can no longer be reconstructed by the equalizer.  
*Das heißt, dass nur das Signal&nbsp; $x_1(t)$&nbsp; durch&nbsp; $H_{\rm E}(f)$&nbsp; wieder hergestellt werden kann, allerdings nur dann, wenn&nbsp; $f_2 < 12  \  \rm kHz$ gilt:
+
*This means that only the signal&nbsp; $x_1(t)$&nbsp; can be recovered by&nbsp; $H_{\rm E}(f)$&nbsp; but only if&nbsp; $f_2 < 12  \  \rm kHz$ holds:
 
:$$z_1(t)=  \underline{1} \cdot 1\,{\rm  V}\cdot \cos(2 \pi \cdot f_1 \cdot t) +  \underline{4} \cdot \frac{1}{4}\cdot 1\,{\rm  V}\cdot \sin(2 \pi \cdot f_2 \cdot t).$$
 
:$$z_1(t)=  \underline{1} \cdot 1\,{\rm  V}\cdot \cos(2 \pi \cdot f_1 \cdot t) +  \underline{4} \cdot \frac{1}{4}\cdot 1\,{\rm  V}\cdot \sin(2 \pi \cdot f_2 \cdot t).$$
*Die jeweils ersten (unterstrichenen) Faktoren geben jeweils die Verstärkungswerte von&nbsp; $H_{\rm E}(f)$&nbsp; an.
+
*The first (underlined) factors indicate the gain values of&nbsp; $H_{\rm E}(f)$&nbsp; respectively.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Linear and Time-Invariant Systems: Exercises|^2.3 Lineare Verzerrungen^]]
+
[[Category:Linear and Time-Invariant Systems: Exercises|^2.3 Linear Distortions^]]

Latest revision as of 15:56, 5 October 2021

Trapezoidal spectrum (top),
associated impulse response)

A communication system with input  $x(t)$  and output  $y(t)$,  which is fully described by the trapezoidal frequency response  $H(f)$  according to the top graph, is considered.  Using the roll-off factor  $r = 0.5$  and the equivalent bandwidth  $\Delta f = 16 \ \rm kHz$  the corresponding impulse response,  which is computable by applying the inverse Fourier transform, is:

$$h(t) = \Delta f \cdot {\rm si}(\pi \cdot \Delta f \cdot t )\cdot {\rm si}(\pi \cdot r \cdot \Delta f \cdot t ) = \Delta f \cdot {\rm sinc}(\Delta f \cdot t )\cdot {\rm sinc}(r \cdot \Delta f \cdot t ).$$

Here the following functions which can be converted into each other are used:

$${\rm si}(x) = \sin(x)/x,\hspace{0.5cm}{\rm sinc}(x) = \sin(\pi x)/(\pi x).$$

The available input signals are:

  • The sum of two harmonic oscillations:
$$x_1(t) = {1\, \rm V} \cdot \cos(\omega_1 \cdot t) + {1\, \rm V} \cdot \sin(\omega_2 \cdot t).$$
Here, the following holds:   $\omega_1 = 2\pi \cdot 2000 \ {\rm 1/s}$  and  $\omega_2 \gt \omega_1$.
  • A periodic triangular signal:
$$x_2(t) = \frac{8\, \rm V}{\pi^2} \cdot \big[\cos(\omega_0 t) + {1}/{9} \cdot \cos(3\omega_0 t) + {1}/{25} \cdot \cos(5\omega_0 t) + \hspace{0.05cm}\text{...}\big].$$
It should be noted that the basic frequency is  $f_0 = 2 \ \rm kHz$  or  $3\ \rm kHz$.  At time  $t = 0$  the signal value in both cases is  $1 \ \rm V$.
  • A rectangular pulse  $x_3(t)$  with amplitude  $A = 1 \ \rm V$  and duration  $T = 1 \ \rm ms$.
    Since its spectrum  $X_3(f)$  extends to infinity,   $H(f)$  always results in linear distortions here.


From subtask  (6)  onwards, it shall be attempted to eliminate the distortions possibly generated by  $H(f)$  by means of a downstream equalizer with

  • frequency response  $H_{\rm E}(f)$,
  • input signal  $y(t)$,  and
  • output signal  $z(t)$.



Please note:

  • The task belongs to the chapter  Linear Distortions.
  • In particular, reference is made to the page  Equalization methods.
  • The term  "overall distortion"  used in the formulation of the questions refers to the input signal  $x(t)$  and the output signal  $z(t)$.



Questions

1

What types of distortion can be ruled out for this system??

Nonlinear distortions.
Attenuation distortions.
Phase distortions.

2

What characteristics does the system exhibit for the test signal  $x_1(t)$  with  $\underline{f_2 = 4 \ \rm kHz}$?

It acts like an ideal system.
It acts like a distortion-free system.
It can be seen that the system at hand is a distorting system.

3

What characteristics does the system exhibit for the test signal  $x_1(t)$  with  $\underline{f_2 = 10 \ \rm kHz}$?

It acts like an ideal system.
It acts like a distortion-free system.
It can be seen that the system at hand is a distorting system.

4

For the test signal  $x_2(t)$  with  $\underline{f_0 = 3 \ \rm kHz}$,  what is the maximum deviation  $\varepsilon_{\rm max} = |y_2(t_0) - x_2(t_0)|$.
At what time  $t_0$  does  $\varepsilon_{\rm max}$  occur for the first time?

$\varepsilon_\text{max} \ = \ $

$\ \rm V$
$t_0 \ = \ $

$\ \rm ms$

5

What is the maximum deviation  $\varepsilon_{\rm max}$  with  $\underline{f_0 = 2 \ \rm kHz}$?

$\varepsilon_\text{max} \ = \ $

$\ \rm V$

6

What curve shape should the equalizer  $H_{\rm E}(f)$  have to compensate all distortions of  $H(f)$  in the best possible way?
What magnitude value arises as a result for  $\underline{f = 10 \ \rm kHz}$?

$|H_E(f = 10 \ \rm kHz)| \ = \ $

7

For which of the listed signals is complete equalization possible?
$z(t) = x(t)$  should be understood by "complete equalization".

For signal  $x_1(t)$  with  $f_2 = 10 \ \rm kHz$,
for signal  $x_2(t)$,
for signal  $x_3(t)$.


Solution

(1)  Proposed solutions 1 and 3  are correct:

  • A linear system is already implicitly assumed by specifying a frequency response so that nonlinear distortions cannot occur.
  • Since  $H(f)$  is purely real, phase distortions can also be ruled out.


(2)  Proposed solutions 1 and 2 are correct:

  • The output signal is  $y_1(t) = x_1(t)$.
  • Thus, the system is not only distortion-free but can also be termed ideal for this application.


(3)  Proposed solution 3  is correct:

  • In this case, the following is obtained for the output signal:
$$y_1(t)= 1\,{\rm V}\cdot \cos(2 \pi \cdot f_1 \cdot t) + {1}/{4}\cdot 1\,{\rm V}\cdot \sin(2 \pi \cdot f_2 \cdot t).$$
  • While the component at  $f_1$  is transmitted unchanged, the sinusoidal component at  $f_2$  is attenuated and one-quarter of the original sinusoidal component.
  • So, there are attenuation distortions.


(4)  The output signal  $y_2(t)$  has the following form taking into account the basic frequency  $f_0 = 3 \ \rm kHz$:

$$y_2(t)= \frac{8\,{\rm V}}{\pi^2} \left( \cos(\omega_0 t) + \frac{3}{8}\cdot \frac{1}{9} \cdot \cos(3\omega_0 t)\right) .$$
  • The factor  $3/8$  describes  $H(f = 9 \ \rm kHz)$.  All other spectral components at  $15 \ \rm kHz$,  $21 \ \rm kHz$,  etc. are suppressed by the system.
  • The strongest deviations between  $x_2(t)$  and  $y_2(t)$  will occur at the triangle peaks since the missing high frequencies have the strongest effect here.
  • For example, for the time  $\underline{t= 0}$ one obtains:
$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 + {3}/{72}\right)= 0.844\,{\rm V} \hspace{0.3cm}\Rightarrow\hspace{0.3cm} \varepsilon_{\rm max} = |y_2(t=0)- x_2(t=0)| \hspace{0.15cm}\underline{= 0.156\,{\rm V}}.$$


(5)  With the basic frequency  $f_0 = 2 \ \rm kHz$  and the values  $H(3f_0) = 0.75$,  $H(5f_0) = 0.25$,  $H(7f_0) = 0$  the following is obtained:

$$y_2(t=0)= \frac{8\,{\rm V}}{\pi^2} \left( 1 + \frac{3}{4}\cdot \frac{1}{9} + \frac{1}{4} \cdot\frac{1}{25}\right)= 0.886\,{\rm V}\hspace{0.5cm} \Rightarrow \hspace{0.5cm}\varepsilon_{\rm max} \hspace{0.15cm}\underline{= 0.114\,{\rm V}}.$$


(6)  In the range up to  $4 \ \rm kHz$,  $H_{\rm E}(f) = H(f) = 1$  is to be set.  In contrast, in the range from  $4 \ \rm kHz$  to  $12 \ \rm kHz$ the following holds:

$$H_{\rm E}(f)= \frac{1}{H(f)} = \frac{1}{1.5 \cdot \big[1 - f/(12\,{\rm kHz})\big]} \hspace{0.5cm} \Rightarrow \hspace{0.5cm} H_{\rm E}(f = 10\,{\rm kHz})\hspace{0.15cm}\underline{= 4} .$$

Here, the denominator expression describes the equation of the straight line of the frequency roll-off.


(7)  Proposed solution 1 is correct:

  • Both  $x_2(t)$  and  $x_3(t)$  also contain spectral components at frequencies greater than  $12 \ \rm kHz$.
  • If these have been truncated by  $H(f)$  ⇒   band limitation, they can no longer be reconstructed by the equalizer.
  • This means that only the signal  $x_1(t)$  can be recovered by  $H_{\rm E}(f)$  but only if  $f_2 < 12 \ \rm kHz$ holds:
$$z_1(t)= \underline{1} \cdot 1\,{\rm V}\cdot \cos(2 \pi \cdot f_1 \cdot t) + \underline{4} \cdot \frac{1}{4}\cdot 1\,{\rm V}\cdot \sin(2 \pi \cdot f_2 \cdot t).$$
  • The first (underlined) factors indicate the gain values of  $H_{\rm E}(f)$  respectively.