# How is the maximum theoretical data rate of a channel equal to $2B\log_2(V)$ bits/sec.?

According to Andrew S. Tanenbaum (in "Computer Networks", Chap. 2, Section 4 "The Maximum Data Rate of a Channel"), the Nyquist/sampling theorem states that "if an arbitrary signal has been run through a low-pass filter of bandwidth $$B$$, the filtered signal can be completely reconstructed by making only $$2B$$ (exact) samples per second. Sampling the line faster than $$2B$$ times per second is pointless because the higher-frequency components that such sampling could recover have already been filtered out. If the signal consists of $$V$$ discrete levels, Nyquist’s theorem states:"

$$\text{Maximum data rate} = 2B\log_2(V) \mbox{ bits/sec}$$

I'm not quite sure how that derives from the Nyquist theorem.

I (believe I) understood that the sampling rate of a signal with a maximum frequency of $$f_{\rm max}$$ has to be at least (strictly greater than) $$\gt 2 f_{\rm max}$$ for the signal to be fully reconstructed (a sampling rate of exactly $$2 f_{\rm max}$$ would not be sufficient if we measure a sin wave each point at an amplitude of zero) (assuming a noiseless channel).

Now, this makes me wonder how this leads to the maximum data rate. Say I'm sending a signal at a baud rate/with a bandwidth of $$B=4$$kHz and two levels ($$S=2$$). According to Tanenbaum, this means that I can at a theoretical maximum data rate twice as great as the baud rate. Why is that?

Is that because when reconstructing the signal, each peak and each trough is interpreted as a bit, so generally, each wave period consists of two bits?

• My advice: if you want to learn this properly, read a communications book, not a networking book. Networking people seem to be unable to explain this correctly for some reason.
– MBaz
Jul 10 at 1:03

Each sample is at one of $$V$$ total levels. The number of bits needed to encode $$V$$ levels is $$\log_2(V)$$.

If you send $$2B$$ of these samples per second, then the total number of bits sent per second (assuming no noise) will be

$$2B \log_2(V)\tag{1}$$

which is where your "maximum data rate" came from.

Be careful with using baud rate and sampling rate interchangeably. They cannot be, in general.

Baud rate is the number of $$\color{red}{\rm symbols}$$ per second sent.

Bit rate is the number of $$\color{magenta}{\rm bits}$$ per second sent.

Sample rate is the number of $$\color{orange}{\rm samples}$$ taken per second.

So if each symbol encodes one bit, then the baud rate and the bit rate are equal. However, this is not necessarily the case.

The sample rate will be (mostly / partly) disconnected from the two other rates.

Is that because when reconstructing the signal, each peak and each trough is interpreted as a bit, so generally, each wave period consists of two bits?

really depends on the digital modulation scheme used to transform the bits being sent into the signal being received.

So Tannenbaum is appealing to the Nyquist sampling criterion: that a signal of bandwidth $$B$$ can be sampled at $$2B$$ samples per second and be reconstructed perfectly. You can sample faster, but the minimum is (slightly higher than) $$2B$$ samples per second.
If each of those $$2B$$ samples per second contains $$V$$ levels, then that is where equation (1) comes from.