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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?

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

Shannon capacity does constrain the number of bits that you can transmit per second; furthermore, the usual formulation assumes sinc pulses.

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  • $\begingroup$ Why the assumption of sinc pulses? and why the symbol rate will decrease if this assumption is not valid? $\endgroup$ – user24907 Feb 21 '18 at 20:06
  • $\begingroup$ @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). $\endgroup$ – MBaz Feb 21 '18 at 22:09
  • $\begingroup$ The reason for choosing sinc pulses is to avoid intersymbol interference (ISI), right? Do you mean that for any type of transmission (baseband or passband using M ary modulation ) pulse shaping must be done? $\endgroup$ – user24907 Feb 22 '18 at 15:43
  • $\begingroup$ @user24907 You must always do pulse shaping, and you always avoid ISI (unless you're doing more advanced stuff like partial response signaling). Among the pulses that avoid ISI, sinc pulses use the least amount of bandwidth. See this other answer: dsp.stackexchange.com/a/22975/11256 $\endgroup$ – MBaz Feb 22 '18 at 15:54
  • $\begingroup$ A final question please, Why does the sampling rate has nothing to do with channel capacity? The channel capacity determines the maximum bit rate that can be transmitted over the channel, and the bit rate equals the sampling rate times the number of bits per sample. $\endgroup$ – user24907 Feb 22 '18 at 20:37

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