- #1
iVenky
- 212
- 12
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?
Thanks in advance.
(As latex doesn't seem to work now I have even attached the image of the formula for Capacity)
[ 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?
Thanks in advance.
(As latex doesn't seem to work now I have even attached the image of the formula for Capacity)
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