Aufgaben:Exercise 4.8Z: AWGN Channel

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Channel model  "AWGN"

We consider here an analog message signal  $s(t)$  whose amplitude values are Gaussian distributed.  The standard deviation of this zero mean signal is  $\sigma_s=1 \hspace{0.05cm} \rm V$.

During transmission  $s(t)$  is additively overlaid by noise  $n(t)$  which like  $s(t)$  can be assumed to be Gaussian distributed and zero mean.

• Let the standard deviation of the noise be generally  $\sigma_n$.
• It is assumed that there are no statistical dependencies between the signals  $s(t)$  and  $n(t)$.
• Such a constellation is called  "Additive White Gaussian Noise"  $\rm (AWGN)$. 
• The quality criterion for the received signal  $r(t)= s(t)+n(t)$  the  "signal-to-noise power ratio":

$${\rm SNR} = {\sigma_s^2}/{\sigma_n^2}.$$

Hints:

• The exercise belongs to the chapter  Linear Combinations of Random Variables.
• Reference is also made to the chapter  Two-dimensional Gaussian random variables.

Questions

1

Give the PDF $f_r(r)$  of the received signal  $r(t)$  in general.  What is the standard deviation  $\sigma_r$  when  $\sigma_n =0.75 \hspace{0.05cm} \rm V$?

$\sigma_r \ = \ $ $ \ \rm V$

2

Calculate the correlation coefficient  $\rho_{sr}$  between the two signals  $s(t)$  and  $r(t)$.  What value results for  $\sigma_n =0.75 \hspace{0.05cm} \rm V$?

$\rho_{sr} \ = \ $

3

Calculate the correlation coefficient  $\rho_{sr}$  depending on the SNR of the AWGN channel.  Derive an approximation for large SNR.

$\rho_{sr} \ = \ $

Solution

Solution

(1)  It holds  $r(t) = s(t)+n(t)$.  Thus $f_r(r)$  can be calculated from the convolution of the two density functions $f_s(s)$  and $f_n(n)$  .

• Since both signals are Gaussian distributed, the convolution also yields a Gaussian function:

$$f_r(r)= \frac {1}{\sqrt{2 \pi} \cdot \sigma_r} \cdot {\rm e}^{-r^2/(2 \sigma_r^2)}.$$

• The variances of  $s(t)$  and  $n(t)$  add up.  Therefore, with  $\sigma_s =1 \hspace{0.05cm} \rm V$  and  $\sigma_n =0.75 \hspace{0.05cm} \rm V$:

$$\sigma_r = \sqrt{\sigma_s^2 + \sigma_n^2} =\sqrt{{(\rm 1\hspace{0.1cm}V)^2} + {(\rm 0.75\hspace{0.1cm}V)^2}}\hspace{0.15cm}\underline{ = {\rm 1.25\hspace{0.1cm}V}}.$$

(2)  For the correlation coefficient, with the joint moment  $m_{sr}$:

$$\rho_{sr } = \frac{m_{sr }}{\sigma_s \cdot \sigma_r}.$$

• This takes into account that  $s(t)$  and also  $r(t)$  are zero mean, so that  $\mu_{sr} =m_{sr}$  holds.
• Since  $s(t)$  and  $n(t)$  were assumed to be statistically independent of each other and thus uncorrelated, it further holds:

$$m_{sr} = {\rm E}\big[s(t) \cdot r(t)\big] = {\rm E}\big[s^2(t)\big] + {\rm E}\big[s(t) \cdot n(t)\big] ={\rm E}\big[s^2(t)\big] = \sigma_s^2.$$

$$\rightarrow \hspace{0.3cm} \rho_{sr } = \frac{\sigma_s}{ \sigma_r} = \sqrt{\frac{\sigma_s^2}{\sigma_s^2 + \sigma_n^2}} = \left (1+ {\sigma_n^2}/{\sigma_s^2}\right)^{-1/2}.$$

• With  $\sigma_s =1 \hspace{0.05cm} \rm V$,  $\sigma_n =0.75 \hspace{0.05cm} \rm V$  and  $\sigma_r =1.25 \hspace{0.05cm} \rm V$  one obtains  $\rho_{sr }\hspace{0.15cm}\underline{ = 0.8}$.

(3)  The expression calculated in the last subtask can be represented by the abbreviation  ${\rm SNR} =\sigma_s^2/\sigma_n^2$  as follows:

$$\rho_{sr } = \rm \frac{1}{ \sqrt{1 + \frac{1}{SNR}}} \approx \frac{1}{ {1 + \frac{1}{2 \cdot SNR}}} \approx 1 - \frac{1}{2 \cdot SNR}.$$

• The signal-to-noise ratio  $10 \cdot {\rm lg \ SNR = 30 \ dB}$  leads to the absolute value  $\rm SNR = 1000$.
• Inserted into the above equation, this gives an approximate correlation coefficient of  $\rho_{sr }\hspace{0.15cm}\underline{ = 0.9995}$.