The equation for channel capacity of a band limited channel with bandwidth W is where is SNR. But this SNR is not in dB. So substitute 10^4.2 for in the equation to get the maximum bit rate.
Please, help me correct my "answers". Thank you.The communication channel (ideally) "lets through" frequencies from 450 Hz to 3250 Hz and has a signal / noise (S/N) ratio of 42 dB.
(a) How many signal elements per second (bauds) is it possible to transfer over this channel?
(b) What is the max. bit rate over this channel?
(c) How do results from (a) and (b) influence modem's properties?
(a) ? Reasoning: you only get 1 crest and 1 trough, hence this theoretical limit? So, if my reasoning is sound, the baud rate
(b) Depends on number of states each baud (signal) can have. E.g. if a signal can be in 4 states, then each baud "carries" 2 bits, and the max. bit rate = 2 * baud rate.
As we're given the S/N ratio, we can use the equation for the channel capacity:
(c) We divide the max. bit rate from (b) by the baud rate from (a) and get the number of bits/baud? That's all?
PS: I see that, instead of repairing the [tex] tag, you've just added the [TEX].
Is answer to (a) correct?
And what would you add to my (incomplete) answer to (c):
For (b), I got the max. bit rate of . So, given the theoretical limit of from (a), what is the answer to (c)? Is it, that (if we ever wanted to achieve theoretical throughput) the signals should have possible states and we would hence have ?