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|]]
+
[[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.
  
:* <i>H</i><sub>S</sub>(<i>f</i>), <i>H</i><sub>K</sub>(<i>f</i>) und <i>H</i><sub>E</sub>(<i>f</i>) beschreiben die Komponenten &bdquo;Sender&rdquo;, &bdquo;Kanal&rdquo; und &bdquo;Empfänger&rdquo; im Frequenzbereich.
+
*The transmitter, channel and receiver components are described in the frequency domain by&nbsp; $H_{\rm S}(f)$,&nbsp; $H_{\rm K}(f)$&nbsp; and&nbsp; $H_{\rm E}(f)$&nbsp;.
  
:* Der Gesamtfrequenzgang <i>H</i>(<i>f</i>) = <i>H</i><sub>S</sub>(<i>f</i>) &middot; <i>H</i><sub>K</sub>(<i>f</i>) &middot; <i>H</i><sub>E</sub>(<i>f</i>) soll einen cos<sup>2</sup>&ndash;förmigen Verlauf haben:
+
*The overall frequency response&nbsp; $H(f) = H_{\rm S}(f) \cdot H_{\rm K}(f) \cdot H_{\rm E}(f)$&nbsp; has a&nbsp; $\cos^2$&ndash;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}$$
 +
*The signal&nbsp; $y(t)$&nbsp; before the decision circuit thus exhibits equidistant zero crossings at intervals of&nbsp; $T$&nbsp;.
 +
*It is assumed here that the source transmits a&nbsp; [[Signal_Representation/Special_Cases_of_Impulse_Signals#Dirac_delta_or_impulse|Dirac-delta]]&nbsp; $x(t)$&nbsp; with weight&nbsp; $T$&nbsp; (see graph).
  
:* Das Signal <i>y</i>(<i>t</i>) vor dem (Schwellenwert&ndash;)Entscheider weist deshalb äquidistante Nulldurchgänge im Abstand <i>T</i> auf.
 
  
:* Vorausgesetzt wird dabei, dass die Quelle einen Diracimpuls <i>x</i>(<i>t</i>) mit Gewicht <i>T</i> abgibt (siehe Grafik).
+
It is pointed out that this is a so-called&nbsp; "Nyquist system".  
  
:Es wird darauf hingewiesen, dass es sich hierbei um ein so genanntes <i>Nyquistsystem</i> handelt. Wie im Buch &bdquo;Digitalsignalübertragung&rdquo; noch ausführlich diskutiert werden wird, stellen diese Nyquistsysteme eine wichtige Klasse digitaler Übertragungssysteme dar, da sich bei diesen die sequenziell übertragenen Symbole nicht gegenseitig beeinflussen.
+
As will be discussed in detail in the book&nbsp; [[Digital_Signal_Transmission/Linear_Nyquist_Equalization|Digital Signal Transmission]],&nbsp; these Nyquist systems represent an important class of digital transmission systems since the sequentially transmitted symbols do not influence each other in such systems.
  
:Für die Lösung dieser Aufgabe werden diese weiterreichenden Aspekte jedoch nicht benötigt. Es wird hier lediglich vorausgesetzt, dass
+
However, these far-reaching aspects are not needed for the solution of this task.  
  
:* der Sendeimpuls <i>s</i>(<i>t</i>) rechteckförmig sei mit Impulsdauer <i>T</i>:
+
Here, it is only assumed that
:$$H_{\rm S}(f) = {\rm si}(\pi f T),$$
 
  
:* der Kanal bis einschließlich Teilaufgabe 2) als ideal vorausgesetzt wird, während für die letzte Teilaufgabe gelten soll:
+
*the transmission pulse&nbsp; $s(t)$&nbsp; be rectangular with pulse duration&nbsp; $T$:
:$$H_{\rm K}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$
+
:$$H_{\rm S}(f) = {\rm si}(\pi f T)= {\rm sinc}(f T)\hspace{1.0cm} \Rightarrow \hspace{0.5cm}
 +
{\rm si}(x) = \sin(x)/x,\hspace{0.5cm}{\rm sinc}(x) = \sin(\pi x)/(\pi x),$$
 +
*the channel is assumed to be ideal up to and including subtask&nbsp; '''(2)'''&nbsp; while for the last subtask&nbsp; '''(3)'''&nbsp; the following shall hold:
 +
:$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$
  
:Gesucht ist für beide Kanäle der Empfänger- und gleichzeitig Entzerrerfrequenzgang <i>H</i><sub>E</sub>(<i>f</i>), damit der Gesamtfrequenzgang die gewünschte Nyquistform aufweist.
+
For both channels, the receiver and the equalizer frequency response&nbsp; $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.
  
:<b>Hinweis:</b> Diese Aufgabe bezieht sich auf die theoretischen Grundlagen von Kapitel 2.3. 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&nbsp; [[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 <i>t</i> = 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 <i>H</i><sub>K</sub>(<i>f</i>) = 1. Berechnen Sie für diesen Fall den Frequenzgang <i>H</i><sub>E</sub>(<i>f</i>). 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="{}"}
idealer Kanal: $|H_E(f \cdot T = 0)|$ = { 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% }
$|H_E(f \cdot T = 0.5)|$ = { 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% }
$|H_E(f \cdot T = 1)|$ = { 0 3% }
+
$|H_{\rm E}(f \cdot T = 1.00)|\ = \ $ { 0. }
  
  
{Berechnen Sie <i>H</i><sub>E</sub>(<i>f</i>) 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="{}"}
GAusßkanal: $|H_E(f \cdot T = 0)|$ = { 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% }
$|H_E(f \cdot T = 0.5)|$ = { 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% }
$|H_E(f \cdot T = 1)|$ = { 0 3% }
+
$|H_{\rm E}(f \cdot T = 1.00)|\ = \ $ { 0. }
  
  
Line 66: Line 77:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
[[File:P_ID922__LZI_Z_2_5_a.png|right|]]
+
[[File:P_ID922__LZI_Z_2_5_a.png|right|frame|Cosine-square spectrum]]
:<b>1.</b>&nbsp;&nbsp;Mit dem konstanten Spektrum <i>X</i>(<i>f</i>) = <i>T</i> erhält man für die Spektralfunktion des Empfängerausgangssignals <i>y</i>(<i>t</i>):
+
'''(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 <i>t</i> = 0 ist gleich der Fläche unter <i>Y</i>(<i>f</i>). Wie aus der nebenstehenden Skizze hervorgeht, ist diese gleich 1. Daraus folgt: <u><i>y</i>(<i>t</i> = 0) = 1</u>.
+
*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:
 +
:$$y(t = 0)\; \underline{= 1}.$$
 +
 
 +
 
  
:<b>2.</b>&nbsp;&nbsp;Aus der Bedingung <i>H</i><sub>S</sub>(<i>f</i>) &middot; <i>H</i><sub>E</sub>(<i>f</i>) = <i>H</i>(<i>f</i>) folgt im betrachteten Bereich:
+
[[File:EN_LZI_Z_2_5c.png|right|frame|Frequency response of the Nyquist equalizer]]
[[File:P_ID923__LZI_Z_2_5_c_neu.png|right|]]
+
'''(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, si(0) = 1 gilt auch <u><i>H</i><sub>E</sub>(<i>f</i> = 0) = 1</u>. Mit der gegebenen trigonometrischen Umformung gilt weiter:
+
*Due to&nbsp; $\cos(0) = 1$&nbsp; and&nbsp; ${\rm si}(0) = 1$,&nbsp; &nbsp; $H_{\rm E}(f = 0)\;\underline{=1}$ also holds.
 +
 
 +
*Considering the given trigonometric transformation it further holds:
 +
:$$H_{\rm  E}(f) = {\pi f T}/{2} \cdot {\rm cot}\left( {\pi f
 +
T}/{2}\right),$$
 +
 
 +
:$$H_{\rm  E}(f \cdot T = 0.25)  = {\pi }/{8} \cdot {\rm cot}\left( 22.5^{\circ}\right)  
 +
= {\pi }/{8} \cdot 2.414 =
 +
\hspace{0.15cm}\underline{0.948},$$
  
$$H_E(f) = \frac{\pi f T}{2} \cdot {\rm cot}\left( \frac{\pi f
+
:$$H_{\rm  E}(f \cdot T = 0.50) = {\pi }/{4} \cdot {\rm cot}\left( 45^{\circ}\right)  
T}{2}\right),\\
+
  = {\pi }/{4} \cdot  1 \hspace{0.15cm}\underline{=
H_{\rm  E}(f \cdot T = 0.25) = \frac{\pi }{8} \cdot {\rm cot}\left( 22.5^{\circ}\right)=\\
 
= \frac{\pi }{8} \cdot 2.414 = \underline{0.948},\\
 
H_{\rm  E}(f \cdot T = 0.5) \hspace{-0.15cm} = \hspace{-0.15cm} \frac{\pi }{4} \cdot {\rm cot}\left( 45^{\circ}\right) =\\
 
  = \hspace{-0.15cm}\frac{\pi }{4} \cdot  1 \hspace{0.15cm}\underline{=
 
 
  0.785},$$
 
  0.785},$$
  
$$ H_{\rm  E}(f \cdot T = 0.75) \hspace{-0.15cm} = \hspace{-0.15cm}  \frac{3 \pi }{8} \cdot {\rm cot}\left( 67.5^{\circ}\right) =\frac{3 \pi }{8} \cdot 0.414 \hspace{0.15cm}\underline{=
+
:$$ 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) \hspace{-0.15cm} = \hspace{-0.15cm}  \frac{ \pi }{2} \cdot {\rm cot}\left( 90^{\circ}\right) =\frac{ \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}.$$
  
:<b>3.</b>&nbsp;&nbsp;Unter Berücksichtigung des Gaußkanals gilt:
+
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
 
:$$H_{\rm  E}(f)=  \frac{H(f)}{H_{\rm  S}(f) \cdot H_{\rm  K}(f)} =  H_{\rm
  E}^{(b)}(f)\cdot {\rm e}^{\pi (f T)^2}.$$
+
  E}^{(2)}(f)\cdot {\rm e}^{\pi (f\hspace{0.05cm}\cdot \hspace{0.05cm} T)^2}.$$
 +
 
 +
Here,&nbsp; $H_{\rm E}^{(2)}(f)$&nbsp; denotes the equalizer frequency response computed in the subtask&nbsp; '''(2)'''&nbsp; assuming an ideal channel.
  
:<i>H</i><sub>E</sub><sup>(b)</sup>(<i>f</i>) bezeichnet den unter Punkt b) 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},$$
H_{\rm  E}(f \cdot T = 0.5) =  0.785 \cdot 2.193 \hspace{0.15cm}\underline{=  1.722},\\
+
:$$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},$$
H_{\rm  E}(f \cdot T = 1) =  0 \cdot 23.141 \hspace{0.15cm}\underline{=  0}.$$
+
:$$H_{\rm  E}(f \cdot T = 1.00) =  0 \cdot 23.141 \hspace{0.15cm}\underline{=  0}.$$
  
:Die obige 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.