Communication channel: baud rate, bit rate, etc.

Quote:

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

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

**(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? (Wondering)

PS: I see that, instead of repairing the [tex] tag, you've just added the [TEX]. (Rofl)

Re: Communication channel: baud rate, bit rate, etc.

The equation for channel capacity of a band limited channel with bandwidth W is $\displaystyle C = W\log_2\left(1 + \frac{P}{N_0W}\right)$ where $\displaystyle \frac{P}{N_0W}$ is SNR. But this SNR is not in dB. So substitute 10^4.2 for $\displaystyle \frac{P}{N_0W}$ in the equation to get the maximum bit rate.

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 $\displaystyle 39066\text{ }\frac{\text{bit}}{\text{s}}$. So, given the theoretical limit of $\displaystyle 5600\text{ bauds}$ from **(a)**, what is the answer to **(c)**? Is it, that (if we ever wanted to achieve theoretical throughput) the signals *should *have $\displaystyle 2^7$ possible states and we would hence have $\displaystyle 7\text{ }\frac{\text{bits}}{\text{baud}}$ ?