# Exercise 3.10Z: BSC Channel Capacity

The channel capacity  $C$  was defined by  Claude E. Shannon  as the maximum mutual information, whereby the maximization refers solely to the source statistics:

$$C = \max_{P_X(X)} \hspace{0.15cm} I(X;Y) \hspace{0.05cm}$$

In the binary channel with the probability function  $P_X(X) = \big [p_0, \ p_1 \big]$  only one parameter is optimizable, for example   $p_0$.  The probability for a  $1$  is thus also fixed:   $p_1 = 1 - p_0.$

The upper graph (reddish background) summarises the results for the  asymmetric binary channel  with  $ε_0 = 0.01$  and  $ε_1 = 0.2$ , which was also considered in the theory section.

The maximization leads to the result  $p_0 = 0.55$   ⇒   $p_1 = 0.45$,  and one obtains for the channel capacity:

$$C_{\rm BC} = \hspace{-0.05cm} \max_{P_X(X)} \hspace{0.1cm} I(X;Y) \big |_{p_0 \hspace{0.05cm} = \hspace{0.05cm}0.55} \hspace{0.05cm}=\hspace{0.05cm} 0.5779\,{\rm bit} \hspace{0.05cm}.$$

In the lower graph (blue background), the same information-theoretical quantities are given for the  Binary Symmetric Channel  $\rm (BSC)$  with the falsification probabilities  $ε_0 = ε_1 = ε = 0.1$,  which was also assumed for  Exercise 3.10.

In the present exercise you are

• to analyze the entropies  $H(X)$,  $H(Y)$,  $H(X|Y)$  and  $H(Y|X)$,
• to optimize the source parameter  $p_0$  with respect to maximum mutual information  $I(X; Y)$ ,
• to determine the channel capacity  $C(ε)$ ,
• to give a closed equation for  $C(ε)$  by generalization for the BSC channel model  $($initially for  $ε = 0.1)$.

Hints:

### Questions

1

Which statements are true for the conditional entropies in the BSC model?

 The equivocation results in  $H(X|Y) = H_{\rm bin}(ε)$. The irrelevance results in  $H(Y|X) = H_{\rm bin}(ε)$. The irrelevance results in  $H(Y|X) = H_{\rm bin}(p_0)$.

2

Which statements are true for the channel capacity  $C_{\rm BSC}$  of the BSC model?

 The channel capacity is equal to the maximum mutual information. For the BSC, maximization leads to the result  $p_0 = p_1 = 0.5$. For  $p_0 = p_1 = 0.5$   ,  $H(X) = H(Y) = 1 \ \rm bit$.

3

Which channel capacity  $C_{\rm BSC}$  results depending on the BSC parameter  $ε$?

 $ε = 0.0\text{:} \hspace{0.5cm} C_{\rm BSC} \ = \$ $\ \rm bit$ $ε = 0.1\text{:} \hspace{0.5cm} C_{\rm BSC} \ = \$ $\ \rm bit$ $ε = 0.5\text{:} \hspace{0.5cm} C_{\rm BSC} \ = \$ $\ \rm bit$

4

Which channel capacity  $C_{\rm BSC}$  results depending on the BSC parameter   $ε$?

 $ε = 1.0\text{:} \hspace{0.5cm} C_{\rm BSC} \ = \$ $\ \rm bit$ $ε = 0.9\text{:} \hspace{0.5cm} C_{\rm BSC} \ = \$ $\ \rm bit$

### Solution

#### Solution

(1)  The  proposed solution 2 is correct, as the following calculation shows:

$$H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) = p_0 \cdot (1 - \varepsilon) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{1 - \varepsilon} + p_0 \cdot \varepsilon \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{\varepsilon} +p_1 \cdot \varepsilon \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{\varepsilon} + p_1 \cdot (1 - \varepsilon) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{1 - \varepsilon}$$
$$\Rightarrow \hspace{0.3cm} H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) = (p_0 + p_1) \cdot \left [ \varepsilon \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{\varepsilon} + (1 - \varepsilon) \cdot {\rm log}_2 \hspace{0.1cm} \frac{1}{1 - \varepsilon} \right ] \hspace{0.05cm}.$$
• With  $p_0 + p_1 = 1$  and the binary entropy function  $H_{\rm bin}$,  one obtains the proposed result:   $H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X) = H_{\rm bin}(\varepsilon)\hspace{0.05cm}.$
• For  $ε = 0.1$  we get  $H(Y|X) = 0.4690 \ \rm bit$.  The same value stands for  $p_0=0.50$  in the given table.
• From the table one can also see that for the BSC model (blue background) as well as for the more general (asymmetric) BC model (red background)
the equivocation  $H(X|Y)$  depends on the source symbol probabilities  $p_0$  and  $p_1$.  It follows that proposed solution 1 cannot be correct.
• The irrelevance  $H(Y|X)$  is independent   of the source statistics, so that solution proposal 3 can also be excluded.

(2)  All given alternative solutions are correct:

• The channel capacity is defined as the maximum mutual information, where the maximization has to be done with respect to  $P_X = (p_0, p_1)$ :
$$C = \max_{P_X(X)} \hspace{0.15cm} I(X;Y) \hspace{0.05cm}.$$
• The equation is generally valid, i.e. also for the unbalanced binary channel highlighted in red.
• The mutual information can be calculated, for example, as  $I(X;Y) = H(Y) - H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X)\hspace{0.05cm}$,
where according to subtask  (1)  the term  $H(Y \hspace{-0.1cm}\mid \hspace{-0.1cm} X)\hspace{0.05cm}$  is independent of  $p_0$  or  $p_1 = 1- p_0$ .
• The maximum mutual information thus results exactly when the sink entropy  $H(Y)$  is maximum.  This is the case for  $p_0 = p_1 = 0.5$:
$$H(X) = H(Y) = 1 \ \rm bit.$$

(3)  According to the subtasks  (1)  and  (2)  one obtains for the BSC channel capacity:

$$C = \max_{P_X(X)} \hspace{0.15cm} I(X;Y) \hspace{0.05cm}.$$

The graph shows the binary entropy function on the left and the channel capacity on the right.  One obtains:

• for  $ε = 0.0$  (error-free channel):
$C = 1\ \rm (bit)$   ⇒   dot with yellow filling,
• for  $ε = 0.1$  (considered so far):
$C = 0.531\ \rm (bit)$   ⇒   dot with green filling,
• for  $ε = 0.5$  (completely disturbed):
$C = 0\ \rm (bit)$   ⇒   dot with grey filling.

(4)  From the graph one can see that from an information-theoretical point of view  $ε = 1$  is identical to  $ε = 0$ :

$$C_{\rm BSC} \big |_{\hspace{0.05cm}\varepsilon \hspace{0.05cm} = \hspace{0.05cm}1} \hspace{0.05cm}= C_{\rm BSC} \big |_{\hspace{0.05cm}\varepsilon \hspace{0.05cm} = \hspace{0.05cm}0} \hspace{0.15cm} \underline {=1\,{\rm (bit)}} \hspace{0.05cm}.$$
• The channel only carries out a renaming here.  This is called  "mapping".
• Every  $0$  becomes a  $1$  and every  $1$  becomes a  $0$.  Accordingly:
$$C_{\rm BSC} \big |_{\hspace{0.05cm}\varepsilon \hspace{0.05cm} = \hspace{0.05cm}0.9} \hspace{0.05cm}= C_{\rm BSC} \big |_{\hspace{0.05cm}\varepsilon \hspace{0.05cm} = \hspace{0.05cm}0.1} \hspace{0.15cm} \underline {=0.531\,{\rm (bit)}} \hspace{0.05cm}$$