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

From LNTwww
 
(7 intermediate revisions by 3 users not shown)
Line 6: Line 6:
 
A digital baseband transmission system is modelled by the depicted block diagram.
 
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 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:
 
*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:
Line 17: Line 17:
 
\end{array}$$
 
\end{array}$$
 
*The signal  $y(t)$  before the decision circuit thus exhibits equidistant zero crossings at intervals of  $T$ .
 
*The signal  $y(t)$  before the decision circuit thus exhibits equidistant zero crossings at intervals of  $T$ .
*It is assumed here that the source emits a  [[Signal_Representation/Special_Cases_of_Impulse_Signals#Dirac_delta_or_impulse|Dirac-delta]]  $x(t)$  with weight  $T$  (see graph).
+
*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).
  
  
It is pointed out that this is a so-called "Nyquist system".  
+
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.
+
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.
  
 
However, these far-reaching aspects are not needed for the solution of this task.  
 
However, these far-reaching aspects are not needed for the solution of this task.  
Line 29: Line 29:
  
 
*the transmission pulse  $s(t)$  be rectangular with pulse duration  $T$:
 
*the transmission pulse  $s(t)$  be rectangular with pulse duration  $T$:
:$$H_{\rm S}(f) = {\rm si}(\pi f T),$$
+
:$$H_{\rm S}(f) = {\rm si}(\pi f T)= {\rm sinc}(f T)\hspace{1.0cm} \Rightarrow \hspace{0.5cm}
 +
{\rm si}(x) = \sin(x)/x,\hspace{0.5cm}{\rm sinc}(x) = \sin(\pi x)/(\pi x),$$
 
*the channel is assumed to be ideal up to and including subtask  '''(2)'''  while for the last subtask  '''(3)'''  the following shall hold:
 
*the channel is assumed to be ideal up to and including subtask  '''(2)'''  while for the last subtask  '''(3)'''  the following shall hold:
 
:$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$
 
:$$H_{\rm K}(f) = H_{\rm G}(f) = {\rm e}^{-\pi(f \cdot T)^2} .$$
  
For both channels, the receiver– and simultaneously the equaliser frequency response  $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.
+
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.
  
  
Line 55: Line 56:
  
  
{First, let&nbsp; $H_{\rm K}(f) = 1$ hold &nbsp; &rArr; &nbsp; <u>ideal channel</u>. Compute the frequency response&nbsp; $H_{\rm E}(f)$ for this case. <br>What values are obtained at the frequencies given below?
+
{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="{}"}
 
$|H_{\rm E}(f \cdot T = 0)| \ = \ $  { 1 3% }
 
$|H_{\rm E}(f \cdot T = 0)| \ = \ $  { 1 3% }
Line 79: Line 80:
 
{{ML-Kopf}}
 
{{ML-Kopf}}
 
[[File:P_ID922__LZI_Z_2_5_a.png|right|frame|Cosine-square spectrum]]
 
[[File:P_ID922__LZI_Z_2_5_a.png|right|frame|Cosine-square spectrum]]
'''(1)'''&nbsp; Mit dem konstanten Spektrum&nbsp; $X(f) = T$&nbsp; erhält man für die Spektralfunktion des Empfängerausgangssignals&nbsp; $y(t)$:
+
'''(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&nbsp; $t = 0$&nbsp; ist gleich der Fläche unter $Y(f)$.  
+
*The signal value at&nbsp; $t = 0$&nbsp; is equal to the area under $Y(f)$.  
*Wie aus der nebenstehenden Skizze hervorgeht, ist diese gleich&nbsp; $1$. Daraus folgt:  
+
*As can be seen from the adjacent sketch, this is equal to&nbsp; $1$.  
 +
*From this it follows that:  
 
:$$y(t = 0)\; \underline{= 1}.$$
 
:$$y(t = 0)\; \underline{= 1}.$$
  
  
  
[[File:EN_LZI_Z_2_5c.png|right|frame|Frequency Response of the Nyquist equaliser]]
+
[[File:EN_LZI_Z_2_5c.png|right|frame|Frequency response of the Nyquist equalizer]]
'''(2)'''&nbsp; Aus der Bedingung&nbsp; $H_{\rm S}(f) \cdot H_{\rm E}(f) = H(f)$&nbsp; folgt im betrachteten Bereich:
+
'''(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&nbsp; $\cos(0) = 1$&nbsp; und&nbsp; ${\rm si}(0) = 1$&nbsp; gilt auch&nbsp; $H_{\rm E}(f = 0)\;\underline{=1}$.  
+
*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
 
:$$H_{\rm  E}(f) = {\pi f T}/{2} \cdot {\rm cot}\left( {\pi f
 
  T}/{2}\right),$$
 
  T}/{2}\right),$$
Line 111: Line 113:
 
:$$ H_{\rm  E}(f \cdot T = 1.00)= { \pi }/{2} \cdot {\rm cot}\left( 90^{\circ}\right) ={ \pi }/{2} \cdot 0 \hspace{0.15cm}\underline{ = 0}.$$
 
:$$ H_{\rm  E}(f \cdot T = 1.00)= { \pi }/{2} \cdot {\rm cot}\left( 90^{\circ}\right) ={ \pi }/{2} \cdot 0 \hspace{0.15cm}\underline{ = 0}.$$
  
 +
The red curve in the graph summarises the results of this subtask.
  
'''(3)'''&nbsp; Unter Berücksichtigung des Gaußkanals gilt:
+
 
$$H_{\rm  E}(f)=  \frac{H(f)}{H_{\rm  S}(f) \cdot H_{\rm  K}(f)} =  H_{\rm
+
 
 +
'''(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}.$$
 
  E}^{(2)}(f)\cdot {\rm e}^{\pi (f\hspace{0.05cm}\cdot \hspace{0.05cm} T)^2}.$$
  
Hierbei bezeichnet&nbsp; $H_{\rm E}^{(2)}(f)$&nbsp; den bei der Teilaufgabe&nbsp; '''(2)'''&nbsp; berechneten Entzerrerfrequenzgang unter der Voraussetzung eines idealen Kanals. Man erhält folgende numerische Ergebnisse:
+
Here,&nbsp; $H_{\rm E}^{(2)}(f)$&nbsp; denotes the equalizer frequency response computed in the subtask&nbsp; '''(2)'''&nbsp; assuming an ideal channel.  
 +
 
 +
The following numerical results are obtained:
 
:$$H_{\rm  E}(f\cdot T = 0) =  1 \cdot {\rm e}^{0} \hspace{0.15cm}\underline{= 1},$$
 
:$$H_{\rm  E}(f\cdot T = 0) =  1 \cdot {\rm e}^{0} \hspace{0.15cm}\underline{= 1},$$
 
:$$H_{\rm  E}(f \cdot T = 0.25) =  0.948 \cdot 1.217 \hspace{0.15cm}\underline{=  1.154},$$
 
:$$H_{\rm  E}(f \cdot T = 0.25) =  0.948 \cdot 1.217 \hspace{0.15cm}\underline{=  1.154},$$
Line 123: Line 130:
 
:$$H_{\rm  E}(f \cdot T = 1.00) =  0 \cdot 23.141 \hspace{0.15cm}\underline{=  0}.$$
 
:$$H_{\rm  E}(f \cdot T = 1.00) =  0 \cdot 23.141 \hspace{0.15cm}\underline{=  0}.$$
  
Die grüne Kurve in obiger Grafik fasst die Ergebnisse dieser Teilaufgabe zusammen.
+
The green curve in the graph above summarises the results of this subtask.
 
{{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 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.