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Topic: Capacity of a discrete channel
Replies: 0

 Venkatraman Posts: 1 From: Chennai,India Registered: 3/26/13
Capacity of a discrete channel
Posted: Mar 26, 2013 9:09 AM

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

C = log(N(T))/T where limit T->infinity [ is there a way to write equation in this forum eg: by using latex?]

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?