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\}&amp;=N_0
\end{align}

The problem for me is in the Received signal power 

\begin{align}
E\{|hx|^2\}&amp;=E\{hxh^*x^*\}\\
&amp;=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.
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.