# metric for signal to noise ratio in communication systems

I'm not quite sure about how to define a good measure of the quality of a communication channel with fading and interference. Let us assume the simplest case, where a node in a network receives the following quantity:
$$y = s + w$$
where $s$ is the amplitude (positive real) of the signal and $w$ the noise (gaussian with zero mean and variance $\sigma^2$). A simple measure of the channel quality is simply given by ${\rm SNR} = s^2 / \sigma^2$, i.e., the ratio of the power of the signal vs the power of the noise.
Now, suppose the received signal is $$y = h_s s + w$$
where $h_s$ and $w$ are complex gaussian with zero mean and unitary variance. A measure of the SNR that makes sense could be derived by looking at the power of the received signal $$|y|^2 = y y' = |h_s|^2 s^2 + 2{\rm Re}(h_ss~w') + |w|^2$$
now, since for practical applications the noise process is much faster than the fading (represented by $h_s$) what people normally do is to average over $w$, obtaining
$$E[|y|^2]_w = |h_s|^2 s^2 + \sigma^2$$
hence, a measure of the signal strength that makes sense would be ${\rm SNR} = |h_s|^2 s^2 / \sigma^2$.
What I am interested about is the case with interference. Suppose there is an extra term that takes into account for interfering signals
$$y = h_s s + h_{\rm I} s_{\rm I} + w$$
where $h_I$ is complex with zero mean and unitary variance. How could I possibly define the SNR in this case? By taking $|y|^2$ there are cross terms, i.e., products of $h_s$ and $h_{\rm I}$ that do not disappear even if I average over the noise $w$. So it seems not obvious to me how to decouple $|y|^2$ into two different pieces, one for the signal and the other for the noise. The final goal would be determining a good measure of the SNR in order to be able to compute probabilities like

$$P({\rm SNR} \geq \gamma_0).$$

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if one can assume that $h_I$ is uncorrelated with $w$ and $h_s$, then you could just average $∣y∣^2$ over both $h_I$ and $w$; cross terms would still vanish; this is not what you want? – Carlo Beenakker Aug 16 '11 at 13:47
yes I know, but typically, the $h$ terms vary more slowly than the noise $w$. So averaging over $h$ is not that practical. – Bob Aug 16 '11 at 22:37