0
\$\begingroup\$

If we are told that the instantaneous SNR at a specific frequency \$f\$ is \$\gamma\$ (We do NOT assume that the channel is AWGN but we are only interested in the capacity in bit/s/Hz at this specific frequency), can we safely say that the instantaneous capacity is:

\$C = log_2(1+\gamma)\$.

If we are further told that the the instantaneous SNR at this specific frequency is \$f_\gamma(\gamma)\$, can we say that the ergodic capacity can be calculated from

\$C_{erg} = \int log_2(1+\gamma) \cdot f_\gamma(\gamma)\,d\gamma \$.

\$\endgroup\$
1
\$\begingroup\$

Channel capacity gives us a tight upper bound on the rate at which error free communication can take place. The instantaneous channel capacity over AWGN (channel capacity per sample) is:

$$ C_s = \frac{1}{2} \log_{2} (1 + SNR) $$

and the channel capacity over AWGN (channel capacity per second) is:

$$ C = (\text{Number of Samples Per Second)} \times \text{(Capacity Per Sample)} $$

$$ \implies C = f_sC_s = \frac{f_s}{2} \log_{2} (1 + SNR) = B \log_{2} (1 + SNR) $$

If the noise is not AWGN and is additive but some other colour of noise then your SNR will be a function of the carrier frequency (i.e \$SNR(f_c)\$) but in wireless comms the additive noise that we are most concerned about in white noise.

The ergodic capacity is the average capacity, so it is the expected value of \$C\$, where \$C\$ is a function of the SNR (\$\gamma\$):

$$ C_{erg} = E[C(\gamma)] $$

and from probability we know that the expected value of a random variable \$X\$ with a probability density function \$p(x)\$ is

$$ E[X] = \int_{-\infty}^{\infty} x\hspace{1mm}p(x)\hspace{1mm} dx $$

so

$$ C_{erg} = \int_{-\infty}^{\infty} C(\gamma)\hspace{1mm}p(\gamma)\hspace{1mm} d\gamma $$

if the SNR is always greater than zero then we have

$$ C_{erg} = \int_{0}^{\infty} C(\gamma)\hspace{1mm}p(\gamma)\hspace{1mm} d\gamma $$

If you are instead given the SNR as a function of the sampling frequency (i.e \$\gamma (f_s)\$) then:

$$ C_{erg} = \int_{0}^{\infty} C(f_s)\hspace{1mm}p(f_s)\hspace{1mm} df_s $$

where

$$ C(f_s) = \frac{f_s}{2} \log_{2} (1 + \gamma(f_s)) $$

\$\endgroup\$
  • \$\begingroup\$ Hi, KillaKem. Thanks a lot for the answer. One question is if the noise is not AWGN, let's say the SNR is SNR(f), is it still OK to say that the capacity at the frequency f is $\log(1+SNR(f))$? Is this technically sound according to the Shannon's theorem? \$\endgroup\$ – Vic Jun 20 '16 at 3:12
  • \$\begingroup\$ No, the capacity will be \$ B \log (1 + SNR(f) )\$. \$\endgroup\$ – KillaKem Jun 20 '16 at 18:03
  • \$\begingroup\$ Hi, Killakem, thanks again for the answer. Just to be more clear for me, as you have pointed out, the capacity in the form of log(1+SNR) is derived by Shannon based on the assumption that the noise is Gaussian. So if the noise is not Gaussian, the capacity is still in the form of log(1+SNR)? Thanks in advance. \$\endgroup\$ – Vic Jun 21 '16 at 1:06
  • \$\begingroup\$ If the noise is additive, then the formula will still remain the same the only thing that will change with other types of additive noises is that the SNR will be a function of frequency band you are operating in. With AWGN noise power is constant across all frequencies but with other types of additive noises the noise power is a function of frequency. \$\endgroup\$ – KillaKem Jun 23 '16 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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