On the Channel capacity

Question

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 \$.

Answer 1

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

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