Exercise 1.6Z: Two Optimal Systems

Optimal systems in time and frequency domain

Consider two binary transmission systems $\rm A$ and $\rm B$, which have the same error behavior for an AWGN channel with noise power density $N_{0}$. In both cases, the bit error probability is:

$$p_{\rm B} = {\rm Q} \left( \sqrt{{2 \cdot E_{\rm B}}/{N_0}}\right)\hspace{0.05cm}.$$

System $\rm A$ uses the NRZ basic transmission pulse $g_{s}(t)$ according to the upper sketch with amplitude $s_{0} = 1 \ \rm V$ and duration $T = 0.5\ \rm µ s$.
In contrast, system $\rm B$, which is to operate at the same bit rate as system $\rm A$, has a rectangular basic transmission pulse spectrum:

$$G_s(f) = \left\{ \begin{array}{c} G_0 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} |f| < f_0 \hspace{0.05cm}, \\ |f| > f_0 \hspace{0.05cm}.\\ \end{array}$$

Notes:

The exercise belongs to the chapter "Optimization of Baseband Transmission Systems".

Here, the pulse amplitude is given in "volts", so that the average energy per bit $(E_{\rm B})$ has the unit $\rm V^{2}/Hz$.

Questions

$R \ = \ $

$\ \rm Mbit/s$

$E_{\rm B} \ = \ $

$\ \cdot 10^{-6} \ \rm V^{2}/Hz$

	For system $\rm A$, $H_{\rm E}(f)$ has a sinc-shaped curve.
	For system $\rm B$, $H_{\rm E}(f)$ is an ideal rectangular low-pass filter.
	$H_{\rm E}(f)$ can be realized by an integrator in system $\rm B$.

$f_{0} \ = \ $

$\ \rm MHz$

$G_{0} \ = \ $

$\ \cdot 10^{-6} \ \rm V/Hz$

	System $\rm A$,
	System $\rm B$.

Solution

(1) Both systems operate according to the specification with the same bit rate.

The NRZ basic transmission pulse of system $\rm A$ has the symbol duration $T = 0.5\ \rm µ s$.
This results in the bit rate $R = 1/T$ $ \underline{= 2\ \rm Mbit/s}$.

(2) The energy of the NRZ basic transmission pulse of system $\rm A$ is given by

$$E_{\rm B} = \int_{-\infty}^{+\infty}g_s^2 (t)\,{\rm d} t = s_0^2 \cdot T = {1\,{\rm V^2}}\cdot {0.5 \cdot 10^{-6}\,{\rm s}}\hspace{0.1cm}\underline { = 0.5 \cdot 10^{-6}\,{\rm V^2/Hz}}\hspace{0.05cm}.$$

(3) The first two statements are true:

In both cases $h_{\rm E}(t)$ must be equal in form to $g_{s}(t)$ and $H_{\rm E}(f)$ must be equal in form to $G_{s}(f)$.
Thus, for system $\rm A$, the impulse response $h_{\rm E}(t)$ is rectangular and the frequency response $H_{\rm E}(f)$ is sinc-shaped.
For system $\rm B$, $H_{\rm E}(f)$ is rectangular like $G_{s}(f)$ and thus the impulse response $h_{\rm E}(t)$ is an sinc-function.
Statement 3 is false: An in $\rm A$, B.

(4) For system B $G_{d}(f)$ nearly coincides with $G_{s}(f)$.

There is only a difference in the Nyquist frequency, but this does not affect the considerations here:
While $G_{s}(f_{\rm Nyq}) = 1/2$, $G_{d}(f_{\rm Nyq}) = 1/4$.

This results in a Nyquist system with rolloff factor $r = 0$.
From this follows for the Nyquist frequency from the condition that the symbol duration should also be $T = 0.5\ \rm µ s$:

$$f_{\rm 0} = f_{\rm Nyq} = \frac{1 } {2 \cdot T} = \frac{1 } {2 \cdot 0.5 \cdot 10^{-6}\,{\rm s}}\hspace{0.1cm}\underline {= 1\,{\rm MHz}}\hspace{0.05cm}.$$

(5) For the energy of the basic transmission pulse can also be written:

$$E_{\rm B} = \int_{-\infty}^{+\infty}|G_s(f)|^2 \,{\rm d} f = G_0^2 \cdot 2 f_0\hspace{0.05cm}.$$

Using the results from (2) and (4), it follows:

$$G_0^2 = \frac{E_{\rm B}}{2 f_0} = \frac{5 \cdot 10^{-7}\,{\rm V^2/Hz}}{2 \cdot 10^{6}\,{\rm Hz}}= 2.5 \cdot 10^{-13}\,{\rm V^2/Hz^2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_0 \hspace{0.1cm}\underline {= 0.5 \cdot 10^{-6}\,{\rm V/Hz}} \hspace{0.05cm}.$$

(6) Solution 1 is correct:

System A represents the optimal system even with peak limitation.
On the other hand, system B would be unsuitable due to the extremely unfavorable crest factor.