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About gain of Rayleigh fading channels

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wave1

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Hi,


can anyone please explain the next point -
Assume a flat fading Rayleigh channel, given by the relation

y = h*x + w %where h is a complex gaussian RV and w is noise

why does every textbook says E{|h|^2}=1 ??
where does this assumption comes from ? i mean, i understand the physical justification for the "Rayliness", because of the summation of many scatterers, none with dominant amplitude and so on, that gives according to law of large numbers to complex normal.

but why is the assumption that the mean power of the channel gain is usually 1 ?
 

paths add constructively and destructively with equal probabilities, thus the channel gain varies about the mean channel gain -> 0 dB (gain = 1).
 

Hi,
well the point that is misunderstood is - why does the mean channel gain has to be 1 ?
why not 0.5 for example ?
the fact that the paths adds constuctively/destructively doesn't explain this point.
 

The reason is that 1 simplifies analysis. In any case, the variance of h will be a scaling factor. For example if E[|h|^2]=a, then the average BER over Rayleigh fading channel and using BPSK is given by:

\[0.5\left[1-sqrt{\frac{aE_s/N_0}{aE_s/N_0+1}}\right]\]

which shows that a does not affect the diversity order of the system, which is the major concern in wireless communication systems.
 

Hi,
So if I understand correctly, you mean that it doesn't matter what is the variance of the channel gain ?
I mean, two RV with the same probability and different average gain will effect the system the same way ?
 

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