Difference between revisions of "Aufgaben:Exercise 2.5Z: Nyquist Equalization"

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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_ID910__LZI_Z_2_5.png|right|Zur Nyquistentzerrung]]
+
[[File:EN_LZI_Z_2_5.png|right|frame|Block diagram for the considered Nyquist system]]
Ein digitales Basisbandübertragungssystem kann durch das dargestellte Blockschaltbild modelliert werden.
+
A digital baseband transmission system is modelled by the depicted block diagram.
  
* Die Komponenten „Sender”, „Kanal” und „Empfänger” werden im Frequenzbereich durch $H_{\rm S}(f)$, $H_{\rm K}(f)$ und $H_{\rm E}(f)$ beschrieben .
+
*The transmitter, channel and receiver components are described in the frequency domain by  $H_{\rm S}(f)$,  $H_{\rm K}(f)$  and  $H_{\rm E}(f)$ .
  
* Der Gesamtfrequenzgang $H(f) = H_{\rm S}(f) \cdot H_{\rm K}(f) \cdot H_{\rm E}(f)$ soll einen $\cos^2$–förmigen Verlauf haben:
+
*The overall frequency response  $H(f) = H_{\rm S}(f) \cdot H_{\rm K}(f) \cdot H_{\rm E}(f)$  has a  $\cos^2$–shaped curve:
:$$H(f) = \left\{ \begin{array}{c} \cos^2\left(\frac{\pi}{2} \cdot  f \cdot  T \right)  \\
+
:$$H(f) = \left\{ \begin{array}{c} \cos^2\left({\pi}/{2} \cdot  f \cdot  T \right)  \\
 
  0 \\  \end{array} \right.\quad \quad
 
  0 \\  \end{array} \right.\quad \quad
 
\begin{array}{*{10}c}  {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}}
 
\begin{array}{*{10}c}  {\rm{f\ddot{u}r}}  \\ {\rm{f\ddot{u}r}}
Line 16: Line 16:
 
{\left|\hspace{0.005cm} f \hspace{0.05cm} \right| \ge 1/T.}  \\
 
{\left|\hspace{0.005cm} f \hspace{0.05cm} \right| \ge 1/T.}  \\
 
\end{array}$$
 
\end{array}$$
* Das Signal $y(t)$ vor dem (Schwellenwert–)Entscheider weist deshalb äquidistante Nulldurchgänge im Abstand $T$ auf.
+
*The signal  $y(t)$  before the decision circuit thus exhibits equidistant zero crossings at intervals of  $T$ .
* Vorausgesetzt wird dabei, dass die Quelle einen [[Signaldarstellung/Einige_Sonderfälle_impulsartiger_Signale#Diracimpuls|Diracimpuls]] $x(t)$ mit Gewicht $T$ abgibt (siehe Grafik).
+
*It is assumed here that the source transmits a  [[Signal_Representation/Special_Cases_of_Impulse_Signals#Dirac_delta_or_impulse|Dirac-delta]]  $x(t)$  with weight  $T$  (see graph).
  
  
Es wird darauf hingewiesen, dass es sich hierbei um ein so genanntes „Nyquistsystem” handelt. Wie im Buch [[Digitalsignalübertragung]] noch ausführlich diskutiert werden wird, stellen diese Nyquistsysteme eine wichtige Klasse digitaler Übertragungssysteme dar, da sich bei ihnen die sequenziell übertragenen Symbole nicht gegenseitig beeinflussen.
+
It is pointed out that this is a so-called  "Nyquist system".  
  
Für die Lösung dieser Aufgabe werden diese weiterreichenden Aspekte jedoch nicht benötigt. Es wird hier lediglich vorausgesetzt, dass
+
As will be discussed in detail in the book  [[Digital_Signal_Transmission/Linear_Nyquist_Equalization|Digital Signal Transmission]],  these Nyquist systems represent an important class of digital transmission systems since the sequentially transmitted symbols do not influence each other in such systems.
  
* der Sendeimpuls $y(t)$ rechteckförmig sei mit Impulsdauer $T$:
+
However, these far-reaching aspects are not needed for the solution of this task.
:$$H_{\rm S}(f) = {\rm si}(\pi f T),$$
+
 
:* der Kanal bis einschließlich Teilaufgabe (2) als ideal vorausgesetzt wird, während für die letzte Teilaufgabe gelten soll:
+
Here, it is only assumed that
 +
 
 +
*the transmission pulse  $s(t)$  be rectangular with pulse duration  $T$:
 +
:$$H_{\rm S}(f) = {\rm si}(\pi f T)= {\rm sinc}(f T)\hspace{1.0cm} \Rightarrow \hspace{0.5cm}
 +
{\rm si}(x) = \sin(x)/x,\hspace{0.5cm}{\rm sinc}(x) = \sin(\pi x)/(\pi x),$$
 +
*the channel is assumed to be ideal up to and including subtask  '''(2)'''  while for the last subtask  '''(3)'''  the following shall hold:
 
:$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$
 
:$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$
  
Gesucht ist für beide Kanäle der Empfänger- und gleichzeitig Entzerrerfrequenzgang $H_{\rm E}(f)$, damit der Gesamtfrequenzgang die gewünschte Nyquistform aufweist.
+
For both channels, the receiver and the equalizer frequency response  $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.
 +
 
  
''Hinweise:''
 
*Die Aufgabe gehört zum Kapitel  [[Lineare_zeitinvariante_Systeme/Lineare_Verzerrungen|Lineare Verzerrungen]].
 
*Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
 
*Als bekannt vorausgesetzt wird die folgende trigonometrische Beziehung:
 
:$$\frac{\cos^2(\alpha /2)}{\sin(\alpha )} = \frac{1}{2} \cdot {\rm cot}(\alpha /2) .$$
 
  
  
===Fragebogen===
+
 
 +
 
 +
''Please note:''
 +
*The task belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]].
 +
 +
*The following trigonometric relation is assumed to be known:
 +
:$$\frac{\cos^2(\alpha /2)}{\sin(\alpha )} = {1}/{2} \cdot {\rm cot}(\alpha /2) .$$
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Berechnen Sie den Ausgangssignalwert zum Zeitpunkt $t = 0$.
+
{Compute the output signal value at time&nbsp; $t = 0$.
 
|type="{}"}
 
|type="{}"}
$y(t = 0) \ = $  { 1 3% }
+
$y(t = 0) \ = \ $  { 1 3% }
  
  
{Zunächst sei $H_{\rm K}(f) = 1$. Berechnen Sie für diesen Fall den Frequenzgang $H_{\rm E}(f)$. Welche Werte ergeben sich bei den nachfolgend genannten Frequenzen?
+
{First,&nbsp; let&nbsp; $H_{\rm K}(f) = 1$&nbsp; hold &nbsp; &rArr; &nbsp; <u>ideal channel</u>.&nbsp; Compute the frequency response&nbsp; $H_{\rm E}(f)$ for this case. <br>What values are obtained at the frequencies given below?
 
|type="{}"}
 
|type="{}"}
$\rm idealer \; Kanal:$ &nbsp;  $|H_E(f \cdot T = 0)| \ =$  { 1 3% }
+
$|H_{\rm E}(f \cdot T = 0)| \ = \ $  { 1 3% }
$|H_E(f \cdot T = 0.25)|\ =$ { 0.948 3% }
+
$|H_{\rm E}(f \cdot T = 0.25)|\ = \ $ { 0.948 3% }
$|H_E(f \cdot T = 0.50)|\ =$ { 0.785 3% }
+
$|H_{\rm E}(f \cdot T = 0.50)|\ = \ $ { 0.785 3% }
$|H_E(f \cdot T = 0.75)|\ =$ { 0.488 3% }
+
$|H_{\rm E}(f \cdot T = 0.75)|\ = \ $ { 0.488 3% }
$|H_E(f \cdot T = 1.00)|\ =$ { 0. }
+
$|H_{\rm E}(f \cdot T = 1.00)|\ = \ $ { 0. }
  
  
{Berechnen Sie $H_{\rm E}(f)$ für den gaußförmigen Kanal entsprechend der Angabe.
+
{Compute&nbsp; $H_{\rm E}(f)$&nbsp; for the Gaussian-shaped channel&nbsp; $H_{\rm K}(f) = H_{\rm G}(f)$&nbsp; according to the description.
 
|type="{}"}
 
|type="{}"}
$\rm Gausskanal:$ &nbsp; $|H_E(f \cdot T = 0)|\ =$ { 1 3% }
+
$|H_{\rm E}(f \cdot T = 0)|\ = \ $ { 1 3% }
$|H_E(f \cdot T = 0.25)| \ =$ { 1.154 3% }
+
$|H_{\rm E}(f \cdot T = 0.25)| \ = \ $ { 1.154 3% }
$|H_E(f \cdot T = 0.50)|\ =$ { 1.722 3% }
+
$|H_{\rm E}(f \cdot T = 0.50)|\ = \ $ { 1.722 3% }
$|H_E(f \cdot T = 0.75)|\ =$ { 2.857 3% }
+
$|H_{\rm E}(f \cdot T = 0.75)|\ = \ $ { 2.857 3% }
$|H_E(f \cdot T = 1.00)|\ =$ { 0. }
+
$|H_{\rm E}(f \cdot T = 1.00)|\ = \ $ { 0. }
  
  
Line 67: Line 77:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
[[File:P_ID922__LZI_Z_2_5_a.png|right|Cosinus–Quadrat–Spektrum]]
+
[[File:P_ID922__LZI_Z_2_5_a.png|right|frame|Cosine-square spectrum]]
'''(1)'''&nbsp; Mit dem konstanten Spektrum $X(f) = T$ erhält man für die Spektralfunktion des Empfängerausgangssignals $y(t)$:
+
'''(1)'''&nbsp; Using the constant spectrum&nbsp; $X(f) = T$&nbsp; the following is obtained for the spectral function of the output signal&nbsp; $y(t)$:
 
:$$Y(f)=  T \cdot {H(f)}.$$
 
:$$Y(f)=  T \cdot {H(f)}.$$
  
:Der Signalwert bei $t = 0$ ist gleich der Fläche unter $`Y(f)$. Wie aus der nebenstehenden Skizze hervorgeht, ist diese gleich $1$. Daraus folgt: $y(t = 0)\; \underline{= 1}$.
+
*The signal value at&nbsp; $t = 0$&nbsp; is equal to the area under $Y(f)$.  
 +
*As can be seen from the adjacent sketch, this is equal to&nbsp; $1$.  
 +
*From this it follows that:
 +
:$$y(t = 0)\; \underline{= 1}.$$
  
  
[[File:P_ID923__LZI_Z_2_5_c_neu.png|right|Frequenzgang des Nyquistentzerrers]]
+
 
'''(2)'''&nbsp; Aus der Bedingung $H_{\rm S}(f) \cdot H_{\rm E}(f) = H(f)$ folgt im betrachteten Bereich:
+
[[File:EN_LZI_Z_2_5c.png|right|frame|Frequency response of the Nyquist equalizer]]
 +
'''(2)'''&nbsp; From the condition&nbsp; $H_{\rm S}(f) \cdot H_{\rm E}(f) = H(f)$&nbsp; it follows in the considered range:
 
:$$H_{\rm  E}(f)=  \frac{H(f)}{H_{\rm  S}(f)} =  \frac{\cos^2(\pi f T/2)}{\sin(\pi f T)/(\pi f T)}.$$
 
:$$H_{\rm  E}(f)=  \frac{H(f)}{H_{\rm  S}(f)} =  \frac{\cos^2(\pi f T/2)}{\sin(\pi f T)/(\pi f T)}.$$
  
Wegen $\cos(0) = 1$ und ${\rm si}(0) = 1$ gilt auch $H_{\rm E}(f = 0)\;\underline{=1}$.  
+
*Due to&nbsp; $\cos(0) = 1$&nbsp; and&nbsp; ${\rm si}(0) = 1$,&nbsp; &nbsp; $H_{\rm E}(f = 0)\;\underline{=1}$ also holds.  
  
Mit der gegebenen trigonometrischen Umformung gilt weiter:
+
*Considering the given trigonometric transformation it further holds:  
 +
:$$H_{\rm  E}(f) = {\pi f T}/{2} \cdot {\rm cot}\left( {\pi f
 +
T}/{2}\right),$$
  
$$H_{\rm  E}(f) = \frac{\pi f T}{2} \cdot {\rm cot}\left( \frac{\pi f
+
:$$H_{\rm  E}(f \cdot T = 0.25)  = {\pi }/{8} \cdot {\rm cot}\left( 22.5^{\circ}\right)  
T}{2}\right),$$
 
 
 
$$H_{\rm  E}(f \cdot T = 0.25)  = {\pi }/{8} \cdot {\rm cot}\left( 22.5^{\circ}\right)  
 
 
  = {\pi }/{8} \cdot 2.414 =
 
  = {\pi }/{8} \cdot 2.414 =
 
  \hspace{0.15cm}\underline{0.948},$$
 
  \hspace{0.15cm}\underline{0.948},$$
  
$$H_{\rm  E}(f \cdot T = 0.50) = {\pi }/{4} \cdot {\rm cot}\left( 45^{\circ}\right)  
+
:$$H_{\rm  E}(f \cdot T = 0.50) = {\pi }/{4} \cdot {\rm cot}\left( 45^{\circ}\right)  
 
  = {\pi }/{4} \cdot  1 \hspace{0.15cm}\underline{=
 
  = {\pi }/{4} \cdot  1 \hspace{0.15cm}\underline{=
 
  0.785},$$
 
  0.785},$$
  
$$ H_{\rm  E}(f \cdot T = 0.75) = {3 \pi }/{8} \cdot {\rm cot}\left( 67.5^{\circ}\right) = {3 \pi }/{8} \cdot 0.414 \hspace{0.15cm}\underline{=
+
:$$ H_{\rm  E}(f \cdot T = 0.75) = {3 \pi }/{8} \cdot {\rm cot}\left( 67.5^{\circ}\right) = {3 \pi }/{8} \cdot 0.414 \hspace{0.15cm}\underline{=
 
  0.488},$$
 
  0.488},$$
  
$$ H_{\rm  E}(f \cdot T = 1.00)= { \pi }/{2} \cdot {\rm cot}\left( 90^{\circ}\right) ={ \pi }/{2} \cdot 0 \hspace{0.15cm}\underline{ = 0}.$$
+
:$$ H_{\rm  E}(f \cdot T = 1.00)= { \pi }/{2} \cdot {\rm cot}\left( 90^{\circ}\right) ={ \pi }/{2} \cdot 0 \hspace{0.15cm}\underline{ = 0}.$$
 +
 
 +
The red curve in the graph summarises the results of this subtask.
 +
 
 +
 
 +
 
 +
'''(3)'''&nbsp; Considering the Gaussian channel the following holds:
 +
:$$H_{\rm  E}(f)=  \frac{H(f)}{H_{\rm  S}(f) \cdot H_{\rm  K}(f)} =  H_{\rm
 +
E}^{(2)}(f)\cdot {\rm e}^{\pi (f\hspace{0.05cm}\cdot \hspace{0.05cm} T)^2}.$$
  
'''(3)'''&nbsp; Unter Berücksichtigung des Gaußkanals gilt:
+
Here,&nbsp; $H_{\rm E}^{(2)}(f)$&nbsp; denotes the equalizer frequency response computed in the subtask&nbsp; '''(2)'''&nbsp; assuming an ideal channel.  
$$H_{\rm  E}(f)=  \frac{H(f)}{H_{\rm  S}(f) \cdot H_{\rm  K}(f)} =  H_{\rm
 
E}^{(2)}(f)\cdot {\rm e}^{\pi (f T)^2}.$$
 
  
Hierbei bezeichnet $H_{\rm E}^{(2)}(f)$ den bei der Teilaufgabe (2) berechneten Entzerrerfrequenzgang unter der Voraussetzung eines idealen Kanals. Man erhält folgende numerische Ergebnisse:
+
The following numerical results are obtained:
$$H_{\rm  E}(f\cdot T = 0) =  1 \cdot {\rm e}^{0} \hspace{0.15cm}\underline{= 1},$$
+
:$$H_{\rm  E}(f\cdot T = 0) =  1 \cdot {\rm e}^{0} \hspace{0.15cm}\underline{= 1},$$
$$H_{\rm  E}(f \cdot T = 0.25) =  0.948 \cdot 1.217 \hspace{0.15cm}\underline{=  1.154},$$
+
:$$H_{\rm  E}(f \cdot T = 0.25) =  0.948 \cdot 1.217 \hspace{0.15cm}\underline{=  1.154},$$
$$H_{\rm  E}(f \cdot T = 0.50) =  0.785 \cdot 2.193 \hspace{0.15cm}\underline{=  1.722},$$
+
:$$H_{\rm  E}(f \cdot T = 0.50) =  0.785 \cdot 2.193 \hspace{0.15cm}\underline{=  1.722},$$
$$H_{\rm  E}(f \cdot T = 0.75) =  0.488 \cdot 5.854 \hspace{0.15cm}\underline{=  2.857},$$
+
:$$H_{\rm  E}(f \cdot T = 0.75) =  0.488 \cdot 5.854 \hspace{0.15cm}\underline{=  2.857},$$
$$H_{\rm  E}(f \cdot T = 1.00) =  0 \cdot 23.141 \hspace{0.15cm}\underline{=  0}.$$
+
:$$H_{\rm  E}(f \cdot T = 1.00) =  0 \cdot 23.141 \hspace{0.15cm}\underline{=  0}.$$
  
Die grüne Kurve in der zweiten Grafik fasst die Ergebnisse dieser Aufgabe zusammen.
+
The green curve in the graph above summarises the results of this subtask.
 
{{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 09:36, 6 October 2021

Block diagram for the considered Nyquist system

A digital baseband transmission system is modelled by the depicted block diagram.

  • The transmitter, channel and receiver components are described in the frequency domain by  $H_{\rm S}(f)$,  $H_{\rm K}(f)$  and  $H_{\rm E}(f)$ .
  • The overall frequency response  $H(f) = H_{\rm S}(f) \cdot H_{\rm K}(f) \cdot H_{\rm E}(f)$  has a  $\cos^2$–shaped curve:
$$H(f) = \left\{ \begin{array}{c} \cos^2\left({\pi}/{2} \cdot f \cdot T \right) \\ 0 \\ \end{array} \right.\quad \quad \begin{array}{*{10}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < 1/T,} \\ {\left|\hspace{0.005cm} f \hspace{0.05cm} \right| \ge 1/T.} \\ \end{array}$$
  • The signal  $y(t)$  before the decision circuit thus exhibits equidistant zero crossings at intervals of  $T$ .
  • It is assumed here that the source transmits a  Dirac-delta  $x(t)$  with weight  $T$  (see graph).


It is pointed out that this is a so-called  "Nyquist system".

As will be discussed in detail in the book  Digital Signal Transmission,  these Nyquist systems represent an important class of digital transmission systems since the sequentially transmitted symbols do not influence each other in such systems.

However, these far-reaching aspects are not needed for the solution of this task.

Here, it is only assumed that

  • the transmission pulse  $s(t)$  be rectangular with pulse duration  $T$:
$$H_{\rm S}(f) = {\rm si}(\pi f T)= {\rm sinc}(f T)\hspace{1.0cm} \Rightarrow \hspace{0.5cm} {\rm si}(x) = \sin(x)/x,\hspace{0.5cm}{\rm sinc}(x) = \sin(\pi x)/(\pi x),$$
  • the channel is assumed to be ideal up to and including subtask  (2)  while for the last subtask  (3)  the following shall hold:
$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$

For both channels, the receiver and the equalizer frequency response  $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.




Please note:

  • The following trigonometric relation is assumed to be known:
$$\frac{\cos^2(\alpha /2)}{\sin(\alpha )} = {1}/{2} \cdot {\rm cot}(\alpha /2) .$$


Questions

1

Compute the output signal value at time  $t = 0$.

$y(t = 0) \ = \ $

2

First,  let  $H_{\rm K}(f) = 1$  hold   ⇒   ideal channel.  Compute the frequency response  $H_{\rm E}(f)$ for this case.
What values are obtained at the frequencies given below?

$|H_{\rm E}(f \cdot T = 0)| \ = \ $

$|H_{\rm E}(f \cdot T = 0.25)|\ = \ $

$|H_{\rm E}(f \cdot T = 0.50)|\ = \ $

$|H_{\rm E}(f \cdot T = 0.75)|\ = \ $

$|H_{\rm E}(f \cdot T = 1.00)|\ = \ $

3

Compute  $H_{\rm E}(f)$  for the Gaussian-shaped channel  $H_{\rm K}(f) = H_{\rm G}(f)$  according to the description.

$|H_{\rm E}(f \cdot T = 0)|\ = \ $

$|H_{\rm E}(f \cdot T = 0.25)| \ = \ $

$|H_{\rm E}(f \cdot T = 0.50)|\ = \ $

$|H_{\rm E}(f \cdot T = 0.75)|\ = \ $

$|H_{\rm E}(f \cdot T = 1.00)|\ = \ $


Solution

Cosine-square spectrum

(1)  Using the constant spectrum  $X(f) = T$  the following is obtained for the spectral function of the output signal  $y(t)$:

$$Y(f)= T \cdot {H(f)}.$$
  • The signal value at  $t = 0$  is equal to the area under $Y(f)$.
  • As can be seen from the adjacent sketch, this is equal to  $1$.
  • From this it follows that:
$$y(t = 0)\; \underline{= 1}.$$


Frequency response of the Nyquist equalizer

(2)  From the condition  $H_{\rm S}(f) \cdot H_{\rm E}(f) = H(f)$  it follows in the considered range:

$$H_{\rm E}(f)= \frac{H(f)}{H_{\rm S}(f)} = \frac{\cos^2(\pi f T/2)}{\sin(\pi f T)/(\pi f T)}.$$
  • Due to  $\cos(0) = 1$  and  ${\rm si}(0) = 1$,    $H_{\rm E}(f = 0)\;\underline{=1}$ also holds.
  • Considering the given trigonometric transformation it further holds:
$$H_{\rm E}(f) = {\pi f T}/{2} \cdot {\rm cot}\left( {\pi f T}/{2}\right),$$
$$H_{\rm E}(f \cdot T = 0.25) = {\pi }/{8} \cdot {\rm cot}\left( 22.5^{\circ}\right) = {\pi }/{8} \cdot 2.414 = \hspace{0.15cm}\underline{0.948},$$
$$H_{\rm E}(f \cdot T = 0.50) = {\pi }/{4} \cdot {\rm cot}\left( 45^{\circ}\right) = {\pi }/{4} \cdot 1 \hspace{0.15cm}\underline{= 0.785},$$
$$ H_{\rm E}(f \cdot T = 0.75) = {3 \pi }/{8} \cdot {\rm cot}\left( 67.5^{\circ}\right) = {3 \pi }/{8} \cdot 0.414 \hspace{0.15cm}\underline{= 0.488},$$
$$ H_{\rm E}(f \cdot T = 1.00)= { \pi }/{2} \cdot {\rm cot}\left( 90^{\circ}\right) ={ \pi }/{2} \cdot 0 \hspace{0.15cm}\underline{ = 0}.$$

The red curve in the graph summarises the results of this subtask.


(3)  Considering the Gaussian channel the following holds:

$$H_{\rm E}(f)= \frac{H(f)}{H_{\rm S}(f) \cdot H_{\rm K}(f)} = H_{\rm E}^{(2)}(f)\cdot {\rm e}^{\pi (f\hspace{0.05cm}\cdot \hspace{0.05cm} T)^2}.$$

Here,  $H_{\rm E}^{(2)}(f)$  denotes the equalizer frequency response computed in the subtask  (2)  assuming an ideal channel.

The following numerical results are obtained:

$$H_{\rm E}(f\cdot T = 0) = 1 \cdot {\rm e}^{0} \hspace{0.15cm}\underline{= 1},$$
$$H_{\rm E}(f \cdot T = 0.25) = 0.948 \cdot 1.217 \hspace{0.15cm}\underline{= 1.154},$$
$$H_{\rm E}(f \cdot T = 0.50) = 0.785 \cdot 2.193 \hspace{0.15cm}\underline{= 1.722},$$
$$H_{\rm E}(f \cdot T = 0.75) = 0.488 \cdot 5.854 \hspace{0.15cm}\underline{= 2.857},$$
$$H_{\rm E}(f \cdot T = 1.00) = 0 \cdot 23.141 \hspace{0.15cm}\underline{= 0}.$$

The green curve in the graph above summarises the results of this subtask.