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Math Help - Communication channel: baud rate, bit rate, etc.

  1. #1
    Member courteous's Avatar
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    Question Communication channel: baud rate, bit rate, etc.

    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?
    Please, help me correct my "answers". Thank you.

    (a) \text{baud rate} \leq 2*\text{width}_\text{frequency}? Reasoning: you only get 1 crest and 1 trough, hence this theoretical limit? So, if my reasoning is sound, the baud rate B\leq 2\omega = 2(3250\text{ Hz}-450\text{ Hz})=5600\text{ Bd}

    (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=\omega * \underbrace{\log_2(1+\frac{S}{N})}_\textrm{the definition of dB?}\stackrel{?}{=} (3250\text{ Hz}-450\text{ Hz})*42\text{ dB}

    (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].
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  2. #2
    Lord of certain Rings
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    Re: Communication channel: baud rate, bit rate, etc.

    The equation for channel capacity of a band limited channel with bandwidth W is C = W\log_2\left(1 + \frac{P}{N_0W}\right) where \frac{P}{N_0W} is SNR. But this SNR is not in dB. So substitute 10^4.2 for \frac{P}{N_0W} in the equation to get the maximum bit rate.
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  3. #3
    Member courteous's Avatar
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    Re: Communication channel: baud rate, bit rate, etc.

    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 39066\text{ }\frac{\text{bit}}{\text{s}}. So, given the theoretical limit of 5600\text{ bauds} from (a), what is the answer to (c)? Is it, that (if we ever wanted to achieve theoretical throughput) the signals should have 2^7 possible states and we would hence have 7\text{ }\frac{\text{bits}}{\text{baud}} ?
    Last edited by courteous; July 3rd 2011 at 11:22 AM.
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