1
$\begingroup$

Nyquist channel capacity(Cn) used for theoretical noiseless channel $C_n=2B\log_2(M)$

Shannon channel capacity(Cs) for noisy channel $C_s=B\log_2(1+\text{SNR})$

So is there a relationship between Nyquist and Shannon channel capacity? is Nyquist channel capacity($C_n$) the upper limit? so $C_s \le C_n$?

Improve this question
$\endgroup$
2
  • $\begingroup$ What is M in the above expression $\endgroup$
    – Dsp guy sam
    May 1, 2020 at 6:26
  • $\begingroup$ @Dspguysam it is the number of symbols, so $\text{log}_2(M)$ is the number of bits per symbol $\endgroup$
    – Engineer
    May 1, 2020 at 11:56

2 Answers 2

Reset to default
2
$\begingroup$

They become the same if $M = \sqrt{1+SNR}$

Nyquist simply says: you can send 2B symbols per second. Shannon extends that to: AND the number of bits per symbol is limited by the SNR. Shannon builds on Nyquist

Nyquist doesn't really tell you the actual channel capacity since it only makes an implicit assumption about the quality of the channel. Shannon makes that explicit by relating bit-per-sample to SNR.

Improve this answer
$\endgroup$
2
  • $\begingroup$ so is Shannon capacity tells how many bits per second(related to bandwidth&SNR), and Nyquist could be used to tell how many levels per symbol required to achieve the same capacity(Shannon capacity) with the same bandwidth and SNR? $\endgroup$
    – Computer_guy11
    May 1, 2020 at 13:09
  • $\begingroup$ Shannon is just an extension of Nyquist for a real-world channel. $\endgroup$
    – Hilmar
    May 2, 2020 at 12:07
2
$\begingroup$

Well, I think both are unrelated unless you have some constraints.

The Nyquist channel capacity says you can transmit $2B\log_2(M)$ bits per second for given channel bandwidth $B$. It does not say if these bits would be received reliably at the receiver in the presence of noise or channel distortion. I am assuming here that bandwidth limit is imposed by channel. Your transmitter can afford to have higher bandwidth but channel does not.

The Shannon Channel capacity theorem says, in order to receive the transmitted bits reliably (that is Probability of Error $\text{P}_e \rightarrow 0$) in the presence of additive noise, the maximum rate at which you can transmit is $B\log_2(1+\text{SNR})$ bits per second.

So if your requirement is reliable communication and the channel has a bandwidth limit of $B$, then $C_n \le C_s$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.