Aufgaben:Exercise 1.6Z: Two Optimal Systems

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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

1

At what bit rate do the two systems operate?

$R \ = \ $ $\ \rm Mbit/s$

2

Calculate the energy per bit for system  $\rm A$.

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

3

Which statements are true for the receiver filters of systems  $\rm A$  and  $\rm B$?

	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  $\rm B$.

4

For which cutoff frequency  $f_{0}$  does system  $\rm B$  have the symbol duration  $T$? 

$f_{0} \ = \ $ $\ \rm MHz$

5

How large should the constant height  $G_{0}$  of the spectrum  $\rm B$  be chosen so that the same energy per bit results as for system   $\rm A$?

$G_{0} \ = \ $ $\ \cdot 10^{-6} \ \rm V/Hz$

6

Would one of the two systems be suitable even with peak limitation?

	System  $\rm A$,
	System  $\rm B$.

Solution

Solution

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

• The NRZ basic transmission pulse of system 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 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 A, the impulse response $h_{\rm E}(t)$ is rectangular and the frequency response $H_{\rm E}(f)$ is si-shaped.
• For system B, $H_{\rm E}(f)$ is rectangular like $G_{s}(f)$ and thus the impulse response $h_{\rm E}(t)$ is an si-function.
• Statement 3 is false:   An integrator has a rectangular impulse response and would be suitable for the realization of system A, but not for system 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.