How to find the instantanous SNR over Fading channel

I would like to prove that the instantnous SNR of flat fading channel $$y=hx+n$$ is $$\gamma=\frac{|h|^2}{N_0}P$$

where $$E\{z\}$$ is the expectation of $$z$$ and

$$E\{x\}=P$$ $$E\{n\}=0$$ $$E\{n^2\}=N_0$$

We know that $$h$$ and $$n$$ are random varibles so

The noise power is \begin{align} E\{n^2\}&=N_0 \end{align}

The problem for me is in the Received signal power

\begin{align} E\{|hx|^2\}&=E\{hxh^*x^*\}\\ &=E\{|h|^2\}P \end{align}

Why they take $$E\{|h|^2\}=|h|^2$$ and we know that $$|h|^2$$ is random varible and it has it own expectation.

Also what if a noise $$n_1$$ is given by

$$n_1=hn+n$$

what is the noise power of $$n_1$$ where h is the fading and $$n$$ is nose with zero mean and power $$N_0$$.

Thanks.

• What is your definition of instantaneous SNR of flat fading channel? – AlexTP Jan 26 '19 at 9:25

You are trying to find the average SNR. In instantaneous SNR, the power of the channel is $$|h|^2$$.

Your SNR equation isn't accurate. $$N_0$$ is usually denote the PSD of the noise. The noise power is $$\sigma_n^2 = N_0\,W$$, where $$W$$ is the signal bandwidth. Or you define SNR as $$E_b/N_0$$ where $$E_b$$ is the signal energy.

• Ok so the power of channel is $|h|^2$, so if we have noise in the form $n_1=hn+n$ where $n$ is noise with PSD $N_0$ how we find the noise power using expectition note conclude directly. I kno is $N_0(|h|^2+1)$ but how we get using the power formula. Thanks – Mokh Tar Bou Jan 26 '19 at 16:43
• I don't think your result is correct. It should be in the form $$\sigma_n^2(|h|^2+h^*+h+1)$$ You just take the expected value of $(hn+n)^*(hn+n)$. Again the noise power is $\sigma_n^2$ not $N_0$. – BlackMath Jan 26 '19 at 18:29
• Unless you mean $n_1 = hn+1$, but even in this case the equivalent noise power is $$\sigma_n^2|h|^2+1$$ which is not what you wrote. – BlackMath Jan 26 '19 at 18:34
• Ok this is my quetio, I would like to write for me please how you get this power with explication by using the expectation method. THanks . – Mokh Tar Bou Jan 28 '19 at 0:19
• Because $E\{h^2n^2\}=E\{h^2\}E\{n^2\}=E\{h^2\}\delta^2$, and why we put $E\{h^2\}=h^2$ and $h$ is random varible?! – Mokh Tar Bou Jan 28 '19 at 0:22

To complement BlackMath's answer: the noise does not suffer fading, because it only exists at the receiver.

The fading channel model $$y = hx + n$$ assumes that $$x$$ is transmitted and suffers a gain $$h$$. The noise $$n$$, however, is solely produced by the receiver; it is not multiplied by $$h$$.

The intuition behind this model is as follows. The transmitted signal $$x$$ has a very large power compared to the transmitter electronics' thermal noise, which is ignored. However, the received signal $$hx$$ typically has an extremely low power, measured in nanowatts, because RF waves are attenuated over even small distances (this is large scale fading). The receiver's thermal noise can also be measured in nanowatts, so it becomes quite relevant to the system's performance. This is the reason why we include it in the model as being added to $$hx$$.

• Why do we ignore the noise? Even in AWGN channels it's not ignored. We call AWGN channel noise-limited because the performance and capacity are limited by the effect of noise mainly if not solely. As for the small distance and large scale fading, what do you mean by it? The distance between the transmitter and receiver is small, or the transmitter moves a small distance relative to its previous position? – BlackMath Jan 27 '19 at 1:29
• @BlackMath The question is about a fading channel, so one can assume normal far-field RF propagation, which implies a large attenuation. Regarding noise: the transmitter's thermal noise is ignored because it is so small compared to the transmitted power. The receiver's noise, however, is not ignored at all; it is the $n$ in $y=hx+n$. If you didn't ignore the transmitter's noise, the model would be $y=(x+n_t)h + n_r$, where $n_t$ and $n_r$ are the transmitter's and receiver's power respectively. – MBaz Jan 27 '19 at 3:52
• OK, I misunderstood then. I thought you were talking about ignoring the receiver noise power. – BlackMath Jan 27 '19 at 4:15