# Why is the Nyquist data rate lower than the Shannon data rate?

In the book Computer Networks, the author talks about the maximum data rate of a channel. He presents the Nyquist formula :

C = 2H log$_2$ V (bits/sec)

And gives an example for a telephone line :

a noiseless 3-kHz channel cannot transmit binary (i.e., two-level) signals at a rate exceeding 6000 bps.

He then explain the Shannon equation :

C = H log$_2$ (1 + S/N) (bits/sec)

And gives (again) an example for a telephone line :

a channel of 3000-Hz bandwidth with a signal to thermal noise ratio of 30 dB (typical parameters of the analog part of the telephone system) can never transmit much more than 30,000 bps

I don't understand why the Nyquist rate is much lower than the Shannon rate, since the Shannon rate takes noise into account. I'm guessing they don't represent the same data rate but the book doesn't explain it.

To understand this you first have to understand that bits transmitted don't have to be purely binary, as given in the example for the Nyquist capacity. Lets say you have a signal that ranges between 0 and 1V. You could map 0v to [00] .33v to [01] .66v to [10] and 1v to [11]. So to account for this in Nyquist's formula you would change 'V' from 2 discrete levels to 4 discrete levels thus changing your capacity from 6000 to 12000. This could then be done for any number of discrete values.

There is a problem with Nyquist's formula though. Since it doesn't account for noise, there is no way of to know how many discrete values are possible. So Shannon came along and came up with a method to essentially place a theoretical maximum on the number of discrete levels that you can read error free.

So in their example of being able to get 30,000 bps, you would have to have 32 discrete values that can be read to mean different symbols.

The Nyquist data rate (not the Nyquist frequency) is the maximum rate for a binary (2 discrete levels) signal.

The Shannon rate takes into account signal levels, as the maximum data rate is not just a function of the bandwidth - if an infinite number of signal levels can be used then the data rate can be infinite regardless of bandwidth.
Since the smallest level increment possible would depend on the signal to noise ratio, this is why it is included in the Shannon rate. So for the above example, it is shown for a 3000kHz bandwidth and a 30dB SNR, you can transmit levels that represent 5 bits of information each.

The power ratio of 30dB = 1000 to 1 can be converted back to voltage by sqrt(1000) = ~32 distinguishable levels (5 bits). If we apply this to Hartley's simpler theorem, we get 2B * log2(32) = 30kHz for B = 3Khz. So 5 bits of information times the Nyquist data rate of 2B (=6000 in this example) equals 30,000 bits/sec.

One describes how fast you sample, the other how much data you can transfer. The minimum required sample rate is only a function of the highest frequency you want to represent correctly. That is independent of the amount of noise on the channel. However, with less noise you can transfer more information per sample. Put another way, Nyquist says what the sample rate needs to be and Shannon says how many bits you then get per sample.

• Hmm, I'm not sure. The book clearly state its a data rate (bits/sec), not a sample rate (Hz). The minimum sampling rate is 2H, not 2H log$_2$ V, no? – subb Nov 5 '11 at 13:58