The book Computer Networks by Andrew S Tanenbaum mentions the following (paraphrased):
For a noiseless channel, Nyquist theorem states:
Maximum data rate = \$2H \space log_{2} V \$ bits/sec\$H\$ : channel bandwidth, \$V\$: no. of discrete levels in the signal
In the end of chapter exercises, there's a question:
A noiseless 4-kHz channel is sampled every 1 msec. What is the maximum data rate?
From what I understood, the maximum data rate is twice of the channel bandwidth for a two-level (binary) signal, which in this case is 8 kHz. However, I am unable to understand how the sampling rate comes into picture.
I think the sampling rate somehow influences the \$V\$ in the formula. Since we have 1000 samples/sec correspond to 8000 bits/sec (as per formula), this gives \$V\$ = 2, but I am not sure if this is correct, or even if it is required.
Could someone please explain this to me?
Best Answer
You can google that exact question to find several variations of this answer:
A noiseless channel can carry an arbitrary large amount of information, no matter how often it is sampled.
Just send a lot of data per sample.
For 4KHz channel, make 1000 samples/sec. If each sample is 16 bits, the channel can send 16 Kbps.
If each sample is 1024 bits, the channel can send 1000 samples/sec * 1024 bits = 1024 Mbps.
The key word here is “noiseless”. With a normal 4 KHz channel, Shannon limit would not allow this.
For the 4 KHz channel we can make 8000 samples/sec. In this case if each sample is 1024 bits this channel can send 8.2 Mbps.