I am trying to understand the following thing:

Does the average signal to noise ratio (SNR) depends on fading channel coefficients ?

Any help in this regard will be highly appreciated.

  • $\begingroup$ What do you mean by "fading channel coefficients"? How is your SNR defined? Writing them down with mathematical expressions would help us. $\endgroup$
    – AlexTP
    Commented Oct 29, 2021 at 8:28
  • $\begingroup$ Fading channel coefficients: complex channel gains $\endgroup$ Commented Oct 29, 2021 at 12:52

1 Answer 1


In the literature, the SNR definition may alter due to its use. In fading channel, the performance of the receiver depends upon the received signal power to thermal noise power Let complex message signal $x(t)$ with power $P$ is transmitted over a fading channel with complex channel gain $h$. The received signal at the receiver can be written as $$y(t) = hx(t) + n_w(t),$$ where $n_w(t)$ is the complex additive white Gaussian noise (AWGN) with power spectral density $N_o$. Actually the power of thermal noise (AWGN) is infinite theoretically!!! After matched filtering and sampling at the receiver, the $k$-th sampling can be written as $$y_k = h x_k +n_k$$ the power of the complex noise per sample ($n_k$) becomes $N_o\times 1/T$ and $T$ is the symbol period. Similarly, the power of $x_k$ becomes $P$. Therefore the instantaneous received SNR can be written as $$ \gamma = |h|^2\frac{PT}{N_o} = |h|^2 \frac{E_s}{N_o}$$ where $E_s$ is the energy per symbol. Some of the studies defines the average SNR as $\bar{\gamma}= Es/N_o$ and some studies assumes that $$ \bar{\gamma} = \mathbb{E} [\gamma] = \Omega \frac{E_s}{N_o}$$ where $\Omega = \mathbb{E} [|h|^2]$.

Sometimes defining received SNR is not useful and trivial(for spatial modulation, space shift keying, and MIMO techniques) since there are multiple receive antennas consequently multiple SNR values. Therefore, directly using $\bar{\gamma} = E_s/N_o$ is simply a better option.


Your Answer

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

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