Question

A noiseless 8-kHz channel is sampled every 1 msec. What is the maximum data rate?​

Answer 1

Answer:

explanation

Explanation:

The book Computer Networks by Andrew S Tanenbaum mentions the following (paraphrased):

For a noiseless channel, Nyquist theorem states:

Maximum data rate = 2H log2V 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?