# How can we find the channel capacity from a BSC? (information theory) [closed]

I'm struggling to find how to calculate the channel capacity from a binary symmetric channel, given

alpha(mean error) = 0.25
p(x1) = 0.25,
p(x2) = 0.75
r = 1.25MBits/sec

I found this in the course:

C = r . { 1 - omega(alpha) }

So I'd have

C = 1.25Mbits/sec * { 1 - omega(0.25) }


However, I'm stuck on how to calculate the omega(0.25) (binary entropy of the mean error).

We've never seen how to solve these exercises during the course, so any help would be very appreciated.

• despite my attempt at answering, this question is quite unclear. Try to edit it for better clarity. – Ran G. Jan 25 '16 at 23:12
• Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. We don't want to do your exercise for you, and questions that require us to do arithmetic aren't very interesting. It sounds like you have an exercise and there is some specific conceptual aspect you don't understand. Rather than copy-pasting your entire exercise, I suggest you just ask about that one concept. That's more likely to be useful to others in the future. – D.W. Jan 27 '16 at 0:21

In any case, for a BSC channel with error probability $p$
(i.e., the channel $$P_{Y\mid X} (y \mid x) = \begin{cases} p & y\ne x \\ 1-p & y=x \end{cases}$$),
it is known that the capacity of the channel is $1-H(p)$ where $H()$ is the binary entropy function, $H(x) = x \log_2 (\frac1x) + (1-x)\log_2 (\frac1{1-x})$.
That is, for a BSC with $p=0.25$, the capacity is $1-H(0.25)=1-0.811=0.188$. This is per channel instantiation, so multiply by $r$ to get what it seems you want.