# Symbol rate and bit rate

Nyquist's formula for maximum channel capacity (noiseless channel): $C=2B \log_2(M)$

Shannon's formula (noisy channel): $C=B \log(1+S/N)$.

• I need to distinguish between the symbol rate and the bit rate.
• Ragarding the symbol rate, do the two formulas says that one cycle of a signal (i.e 1 Hz) cannot carry no more than 2 symbol/cycle?
• Regarding the bit rate, Nyquist formula doesn't put a limit on the number of bits that can be put in each symbol, while Shannon formula takes that consideration.

Is my understanding correct?

• The sampling rate has nothing to do with capacity. Do you mean the symbol rate (or baud rate)? – MBaz Feb 21 '18 at 17:53
• Your "formula for maximum channel capacity (noiseless channel)" is wrong. A noiseless channel has infinite capacity. You may be confusing different concepts here – Hilmar Feb 21 '18 at 18:18
• I mean symbol rate @MBaz – user24907 Feb 21 '18 at 18:28
• @user24907 Could you edit your question accordingly? – MBaz Feb 21 '18 at 18:38
• @MBaz The question is edited – user24907 Feb 21 '18 at 19:13

What Nyquist says is that $$R_p = 2B\,\text{ Bd},$$ where $R_p$ is the number of pulses per second transmitted, $B$ is the available bandwidth, and the units are Bauds or symbols/sec ($\text{Bd}$). Here it is assumed that the pulses being used are sinc pulses; otherwise, the pulse rate will decrease. In this sense, $2B$ is an upper bound on the pulses per second that can be transmitted. It is correct to say that you can transmit two pulses for every Hz of bandwidth available.
Now, each pulse can carry several bits, by means of its amplitude. If you allow, say, two amplitudes (for example, -1 and 1), then each pulse carries one bit and $R_b=R_p$, where $R_b$ is the bit rate. In general, if you allow $M=2^k$ amplitudes, you'll be able to transmit at a bit rate $R_b=kR_p$.
• @user24907 Because sinc pulses have the shortest possible bandwidth for a given pulse rate. If you use a different pulse, its bandwidth will be larger for the same pulse rate. Let's say that you use raised cosine pulses with rolloff factor $\beta$; then $R_p = (2B)/(1+\beta)$. Or you use rectangular pulses; then $R_p \approx B/5$ (depending on how you account for the rectangular pulses theoretically infinite bandwidth). – MBaz Feb 21 '18 at 22:09