Exercise 5.3: AWGN and BSC Model

• Return to book

AWGN channel and BSC model

The graphic above shows the analog channel model of a digital transmission system, where the additive noise signal  $n(t)$  with the (two-sided) noise power density  $N_0/2$  is effective. This is AWGN noise. The variance of the noise component before the decision (after the matched filter) is then

$$\sigma^2 = \frac{N_0}{2T} \hspace{0.05cm}.$$

Further, let hold:

• No intersymbol interference occurs. If the symbol  $q_{\nu} = \mathbf{H}$  was sent, the useful component of the detection signal is equal to  $+s_0$, while for  $q_{\nu} = \mathbf{L}$,  it is equal to  $-s_0$.
• The threshold decision takes into account a threshold drift, that is, the threshold  $E$  may well deviate from the optimal value  $E = 0$.  The decision rule is:

$$\upsilon_\nu = \left\{ \begin{array}{c} \mathbf{H} \\ \mathbf{L} \end{array} \right.\quad \begin{array}{*{1}c} {\rm if}\hspace{0.15cm}d (\nu \cdot T) > E \hspace{0.05cm}, \\ {\rm if} \hspace{0.15cm} d (\nu \cdot T) \le E\hspace{0.05cm}.\\ \end{array}$$

• With the threshold value  $E = 0$,  the mean error probability is given by

$$p_{\rm M} = {\rm Q} \left ( {s_0}/{\sigma} \right ) = 0.01\hspace{0.05cm}.$$

The bottom graph shows a digital channel model characterized by the four transition probabilities  $p_1,  p_2,  p_3$  and  $p_4$.  This is to be fitted to the analog channel model.

Notes:

• The exercise belongs to the chapter  "Binary Symmetric Channel (BSC)".
• Numerical values of the Q–function can be determined with the interactive applet  "Complementary Gaussian Error Functions". 

Questions

1

Which quotient  $s_0/\sigma$  is the basis of this exercise?

$s_0/\sigma\ = \ $

2

For the threshold, let  $E = 0$. Is the digital transmission system at hand describable by the BSC model, assuming that

	the source symbols  $\mathbf{L}$  and  $\mathbf{H}$  are equally probable,
	the source symbol  $\mathbf{L}$  occurs significantly more frequently than $\mathbf{H}$?

3

Calculate the transition probabilities for  $E = +s_0/4$.

$p_1 \ = \ $

$p_2 \ = \ $

$p_3 \ = \ $

$p_4 \ = \ $

4

Now let  $E = +s_0/4$. Is the present digital transmission system describable by the BSC model under the condition that

	the source symbols  $\mathbf{L}$  and  $\mathbf{H}$  are equally probable,
	the source symbol  $\mathbf{L}$  occurs significantly more frequently than  $\mathbf{H}$?

5

Let  $p_{\rm L} = {\rm Pr}(q_{\nu} = \mathbf{L})$  and  $p_{\rm H} = {\rm Pr}(q_{\nu} = \mathbf{H})$. Which of the following statements is then true for the mean error probability  $p_{\rm M}$? 

	$p_{\rm M}$  in the BSC model  $($valid for  $E = 0)$  is independent of  $p_{\rm L}$  and  $p_{\rm H}$.
	$p_{\rm M}$  in the BSC model  $($valid for  $E = 0)$  is smallest for  $p_{\rm L} = p_{\rm H}$.
	For  $p_{\rm L} = 0.9$,  $p_{\rm H} = 0.1$  and  $E = +s_0/4$  is  $p_{\rm M} < 1\%$.

Solution

Solution

(1)  The average error probability is $p_{\rm M} = {\rm Q}(s_0/\sigma) = 0.01$.

• From this it follows for the quotient of the detection useful sample value and the detection noise rms value:

$${s_0}/{\sigma}= {\rm Q}^{-1} \left ( 0.01 \right ) \hspace{0.15cm}\underline {\approx 2.32}\hspace{0.05cm}.$$

(2)  With $E = 0$, the probabilities of the given digital channel model are given by:

$$p_2 = p_3 = p = 0.01 \hspace{0.05cm}, \hspace{0.2cm}p_1 = p_4 = 1-p = 0.99\hspace{0.05cm}.$$

• Ein Vergleich mit dem Theorieteil zeigt, dass dieses Kanalmodell dem BSC–Modell entspricht, und zwar unabhängig von der Statistik der Quellensymbole.
• Richtig sind also beide Lösungsvorschläge.

(3)  Die Übergangswahrscheinlichkeit $p_2$ beschreibt nun den Fall, dass die Enscheiderschwelle $E = 0.25 \cdot s_0$ fälschlicherweise unterschritten wurde.

• Dann ist $v_{\nu} = \mathbf{L}$, obwohl $q_{\nu} = \mathbf{H}$ gesendet wurde. Der Abstand von der Schwelle beträgt somit nur $0.75 \cdot s_0$ und es gilt:

$$p_{\rm 2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Q} \left ( \frac{0.75 \cdot s_0}{\sigma} \right ) = {\rm Q} \left ( 0.75 \cdot 2.32 \right ) = {\rm Q} \left ( 1.74 \right )\hspace{0.15cm}\underline {\approx 0.041}\hspace{0.05cm}, \hspace{0.5cm} p_{\rm 1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}1 - p_{\rm 2} \hspace{0.15cm}\underline {= 0.959}\hspace{0.05cm}.$$

• In ähnlicher Weise können die Übergangswahrscheinlichkeiten $p_3$ und $p_4$ berechnet werden, wobei nun vom Schwellenabstand $1.25 \cdot s_0$ auszugehen ist:

$$p_{\rm 3} = {\rm Q} \left ( 1.25 \cdot 2.32 \right ) = {\rm Q} \left ( 2.90 \right )\hspace{0.15cm}\underline {\approx 0.002}\hspace{0.05cm}, \hspace{0.2cm} p_{\rm 4} = 1 - p_{\rm 3}\hspace{0.15cm}\underline { = 0.998}\hspace{0.05cm}.$$

(4)  Keiner der beiden Lösungsvorschläge trifft zu:

• Mit der Entscheiderschwelle $E ≠ 0$ ist das BSC–Modell unabhängig von der Symbolstatistik nicht anwendbar,
• da die Symmetrieeigenschaft des Kanals (das Kennzeichen "S" in "BSC") nicht gegeben ist.

(5)  Die Aussagen 1 und 3 treffen zu, nicht aber die Aussage 2:

• Beim BSC–Modell ist $p_{\rm M} = 1\%$ unabhängig von den Symbolwahrscheinlichkeiten $p_{\rm L}$ und $p_{\rm H}$.
• Dagegen gilt für $p_{\rm L} = 0.9$, $p_{\rm H} = 0.1$ sowie $E = +s_0/4$:

$$p_{\rm M} = 0.9 \cdot p_{\rm 3} + 0.1 \cdot p_{\rm 2}= 0.9 \cdot 0.2\% + 0.1 \cdot 4.1\% \approx 0.59\% \hspace{0.05cm}.$$

• Das Minimum ergibt sich für $p_{\rm L} = 0.93$ und $p_{\rm H} = 0.07$ zu $p_{\rm M} \approx 0.45\%$.