In one of the research paper following equations are given :

$y(n) = h_0s(n)+w(n),d = 0$

$y(n) = h_1s(n)+w(n),d = 1$

where $n$ ranges from $1$ to $N$ and

$h_0,h_1$ are wireless channel assumed as $\sim\mathcal{C}{N}(0,\sigma^2)$. $s(n)\sim\mathcal{C}{N}(0,P_s)$ and it represents transmitted signal. $w(n)\sim\mathcal{C}{N}(0,N_w)$ and it represents transmitted AWGN.

In the paper it is written that distribution of received signal, $y(n)$ when $d = 0$ is $\textbf{y}(n)\sim\mathcal{C}{N}(\textbf{0},\sigma^2_0I_N)$.----(1)

I am not getting how the distribution of received signal, $y(n)$, in (1) is zero mean complex Gaussian.

Any help in this regard will be highly appreciated.


1 Answer 1

  • Since $w$ and $h_0s$ are independent (in the statistical sense), the variance of their sum must be the sum of their variances.

  • Sums of Gaussians are Gaussians.

Therefore, in the formula $\sigma_0^2 = \sigma^2+P_s$, and the diagonal shape of the covariance implies that $s(n_1)$ is uncorrelated to $s(n_2)$ for $n_1\ne n_2$. That's it.

  • $\begingroup$ Ok..I understood your answer to some extent. But I had a very basic doubt . In $y(n)$ expression, $h_0$, $s(n)$ and $w(n)$ all are zero mean complex Gaussian distribution. And $h_0$, $s(n)$ are in product. So how Gaussian x Gaussian is Gaussian? $\endgroup$
    – paru
    Aug 16, 2021 at 11:37
  • $\begingroup$ you're confusing what is random for how long. Assume $h_0$ and $h_1$ to be constants for the duration of your transmission (just that their value is randomly chosen before): that's why they are $h_0$ and not $h_0(n)$. $\endgroup$ Aug 16, 2021 at 11:48
  • $\begingroup$ Ok.. understood perfectly...Thanks for your brilliant response.. $\endgroup$
    – paru
    Aug 16, 2021 at 11:51
  • $\begingroup$ What will be the distribution of $y(n)$ if $s(n)$ is PSK signal instead of zero mean complex Gaussian... $\endgroup$
    – paru
    Aug 16, 2021 at 11:52
  • $\begingroup$ Don't ask new questions in the comments. Also, I think you should be able to answer that yourself! $\endgroup$ Aug 16, 2021 at 11:53

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.