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

1. ## 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?

(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].

2. ## 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.

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

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 $7\text{ }\frac{\text{bits}}{\text{baud}}$ ?