Exercise 2.5Z: Nyquist Equalization

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 emits 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 $T$:

$$H_{\rm S}(f) = {\rm si}(\pi f T),$$

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 simultaneously the equaliser frequency response $H_{\rm E}(f)$ are searched-for so that the overall frequency response has the desired Nyquist shape.

Please note:

The task belongs to the chapter 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

$y(t = 0) \ = \ $

$|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)|\ = \ $

$|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

(1) Mit dem konstanten Spektrum $X(f) = T$ erhält man für die Spektralfunktion des Empfängerausgangssignals $y(t)$:

$$Y(f)= T \cdot {H(f)}.$$

Der Signalwert bei $t = 0$ ist gleich der Fläche unter $Y(f)$.
Wie aus der nebenstehenden Skizze hervorgeht, ist diese gleich $1$. Daraus folgt:

$$y(t = 0)\; \underline{= 1}.$$

Frequency Response of the Nyquist equaliser

(2) Aus der Bedingung $H_{\rm S}(f) \cdot H_{\rm E}(f) = H(f)$ folgt im betrachteten Bereich:

$$H_{\rm E}(f)= \frac{H(f)}{H_{\rm S}(f)} = \frac{\cos^2(\pi f T/2)}{\sin(\pi f T)/(\pi f T)}.$$

Wegen $\cos(0) = 1$ und ${\rm si}(0) = 1$ gilt auch $H_{\rm E}(f = 0)\;\underline{=1}$.

Mit der gegebenen trigonometrischen Umformung gilt weiter:

$$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},$$

$$ H_{\rm E}(f \cdot T = 1.00)= { \pi }/{2} \cdot {\rm cot}\left( 90^{\circ}\right) ={ \pi }/{2} \cdot 0 \hspace{0.15cm}\underline{ = 0}.$$

(3) Unter Berücksichtigung des Gaußkanals gilt: $$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}.$$

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:

$$H_{\rm E}(f\cdot T = 0) = 1 \cdot {\rm e}^{0} \hspace{0.15cm}\underline{= 1},$$

$$H_{\rm E}(f \cdot T = 0.25) = 0.948 \cdot 1.217 \hspace{0.15cm}\underline{= 1.154},$$

$$H_{\rm E}(f \cdot T = 0.50) = 0.785 \cdot 2.193 \hspace{0.15cm}\underline{= 1.722},$$

$$H_{\rm E}(f \cdot T = 0.75) = 0.488 \cdot 5.854 \hspace{0.15cm}\underline{= 2.857},$$

$$H_{\rm E}(f \cdot T = 1.00) = 0 \cdot 23.141 \hspace{0.15cm}\underline{= 0}.$$

Die grüne Kurve in obiger Grafik fasst die Ergebnisse dieser Teilaufgabe zusammen.