# Capacity and Nyquist Theorem

1. Jul 2, 2011

### fisico30

Hello Forum,

the "maximum" data rate that can be used over a physical "noiseless" channel is given by

bits/s= 2*B* log2 M

where B is the channel bandwidth (set of frequencies that are passed with little attenuation), and M is the number of levels we can change the chosen parameter of the carrier signal that we modulate. We can increase the max bit rate by either increasing B or M.

If B is maintained fixed, we can increase the max data rate to infinity (in theory only of course). Let's say there is no noise and we have a baseband signal m(t) to be transmitted whose bandwidth is larger than the channel bandwidth: W>B...

Even if we can push the max data rate to whatever limit we want, what happens to the transmitted signal? Will it be smeared out regardless of the max data rate we have?

Of course we digitize the signal m(t) by sampling it at f>2W, and quantize the samples to get a stream of 1s and 0s. Example: W=30 Hz, B=5 Hz and the duration of m(t) is 20 seconds, and we quantize the sample so each one has 10 bits....
We get a total of 60*20*10=1200 bits...Now we need to transmit them through the channel whose bandwidth is simply 5 Hz and M=1000...What happens to the received bits?

thanks!
fisico30

2. Jul 2, 2011

### Floid

The receiving end has to be aware of the format of the data it is receiving if you want to digitize it and then decode it with as little loss of information as possible.

So the receiving end would know that it was receiving 10 bit "packets" of data, so that every 10 bits represents a quantized level.

So for the 1200 bits it received it divides that up into 120 packets of 10 bits each.

Then if you want to maintain the spectral content, it has to know the sample rate at which the packets were sampled, in this case 60Hz. So if you wanted to convert the signal back to analog the packets would be D/A at a 60Hz rate.