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


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



At what bit rate do the two systems operate?

$R \ = \ $

$\ \rm Mbit/s$


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

$E_{\rm B} \ = \ $

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


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  $\rm B$. 


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

$f_{0} \ = \ $

$\ \rm MHz$


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} \ = \ $

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


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

System  $\rm A$,
System  $\rm B$.


(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 integrator has a rectangular impulse response and would be suitable for the realization of system  $\rm A$,  but not for system  $\rm B$.

(4)  For system  $\rm B$   ⇒   $G_{d}(f)$  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  $\rm A$  represents the optimal system even with peak limitation.
  • On the other hand,  system  $\rm B$  would be unsuitable due to the extremely unfavorable crest factor.