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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?

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  • \$\begingroup\$ What is H in the above formula? \$\endgroup\$
    – Andy aka
    Dec 28, 2013 at 12:09
  • \$\begingroup\$ @Andyaka Thanks for pointing that out. I have edited the question with the details. \$\endgroup\$
    – Masked Man
    Dec 28, 2013 at 12:27
  • \$\begingroup\$ dspcsp.com/pdf/shannon.pdf This demonstrates that the information carrying capacity of a noiseless channel is infinite. \$\endgroup\$
    – mstewart
    Feb 26, 2017 at 2:35

2 Answers 2

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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.

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  • \$\begingroup\$ I'm not sure I like this answer... I mean, it is the answer found in many answer sheets for that particular problem from chapter two of the book, but it doesn't discuss the reasons for the official answer. I myself spent quite a bit of time trying to write an answer to this question, but in the end, I couldn't understand myself how it would be possible to fit arbitrary amount of data into a single sample. \$\endgroup\$
    – AndrejaKo
    Dec 28, 2013 at 15:57
  • \$\begingroup\$ Oh, yes you can fit way more than just 2 levels into one sample. That's called modulation. Think of OFDM or QAM. \$\endgroup\$
    – Rolf Ostergaard
    Dec 28, 2013 at 16:19
  • \$\begingroup\$ Most of my confusion stems from the use of 'bits/sec' in that formula (in the book). If they had used 'samples/sec', things would have been less confusing. \$\endgroup\$
    – Masked Man
    Dec 29, 2013 at 3:13
  • \$\begingroup\$ bit/s is pretty clear and concrete to me. The "confusing" thing may be the notion of having multiple bits per sample (or symbol). Maybe read this on spectral efficiency in real world systems: en.wikipedia.org/wiki/Spectral_efficiency (note Nyquist is only about an ideal noiseless channel) \$\endgroup\$
    – Rolf Ostergaard
    Dec 29, 2013 at 8:48
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    \$\begingroup\$ You see, technically, each sample need not be represented by bits (they could be digits, for instance), so samples/sec is better. \$\endgroup\$
    – Masked Man
    Dec 30, 2013 at 16:41
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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.

Yes, that's exactly right. For such a binary signal, V = 2, so log2(V) = 1 bit, so the maximum data rate at a bandwidth of H = 4 kHz is Maximum data rate = 2H log2V bits/sec = 2*4 kHz * 1 bit = 8 kHz*bit = 8 kbit/s.

However, many kinds of signals have much more than 2 levels.

One popular system has a 16-level signal, which provides log2(16) = 4 bits every time a new sample is gathered.

A receiver that decodes 4 bits every time it grabs a sample, if it grabs a sample every 1 msec, will end up decoding at a data rate of:

4 bits every 1 msec = 4 bits / (1 msec) = 4 kbit/s.

A few systems have 256 points in their constellation diagram, providing log2(256) = 8 bits for every sample. That would give

8 bits every 1 msec = 8 bits / (1 msec) = 8 kbit/s.

A few systems give even more bits per sample.

In theory, a noiseless system could support any number of possible levels, resulting in any number of bits per sample. (In practice, we always have some noise).

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.

In this question, the"1 msec" sampling rate is a "red herring" -- it has no effect on V, the number of discrete levels in each sample, or on the maximum data rate.

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    \$\begingroup\$ Actually the 4kHz is the red herring. Well, the only important thing about it is that it is "large enough" to avoid inter symbol interference between the samples. \$\endgroup\$
    – Ben Voigt
    Jan 1, 2014 at 19:43

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