# Capacity of a discrete channel

1. Mar 26, 2013

### iVenky

Well I was reading this " A Mathematical Theory of Communication" paper by Claude Shannon. He says that the capacity of a discrete channel is given by

[ tex ] C= \ \lim_{T \to +\infty} \ \frac{\log ( N(T) )} {T} [ \tex ]

Here N(T) is the number of possible sequences in a duration of T seconds.

He gives one example before talking about this capacity formula. Here's the example- Consider 32 symbols. You have a system where you can transmit 2 symbols per second. Now he says it is clear that if each symbol has 5 bits then we can send 2 x 5 =10 bits per second (or 2 symbols per second). This is the capacity of the channel. But when I tried it out using the above expression I couldn't get the answer as 2 symbols/ second. I got it as 3 symbols per second. Can you help me with this?

(As latex doesn't seem to work now I have even attached the image of the formula for Capacity)

#### Attached Files:

• ###### capacity.jpg
File size:
2.4 KB
Views:
102
Last edited: Mar 26, 2013
2. Mar 26, 2013

### eq1

3. Mar 26, 2013

### marcusl

You must have made an error somewhere. In 2 seconds you transmit 2 symbols, so there are 32^2=1024 possible message combinations. log_2(1024)/2 = 10/2 = 5 bits/second which is one symbol per second, which is correct.

4. Mar 26, 2013

### iVenky

Ya ya I found it later. Thanks marcusl btw