[SOLVED] Turbo code in Rayleigh fading channel

[emma2]

Hi there,
I'm considering flat fading Rayleigh channel. My problem is that , after multiplying the signal by h and adding the AWGN it has this form r=h*tx+n and its obvious that this is a complex values. and to retrieve my data I estimated h at the receiver and divided the received data by it like this y=r/h and this also complex and I need a real values as an input to the logmap decoder..is it enough for me to just take the real of y or there some fancy steps I should consider.

Any suggestions are highly appreciated

 

[kalyanasv]

In the case of a fading channel you need, it is the received signal attenuation that ends up having a Rayleigh distribution. Now if we consider the received signal as 'r', transmitted signal as 't' and AWGN as 'n', then consider 'x' as the Rayleigh dist. Assuming you have a BPSK signal then, you need a matched filter at the output (not division), since it is rayleigh channel you have multiplied "t" with the attenuation factor and phase factor. A detector-combiner compensates for the phase shift and weights the signal for rayleigh channel, this is done segregating 1's and 0's . Then a simple method is to use a "Brennan combiner" to combine the transmitted 1's and 0's into a single channel. Here we consider only the real parts. of the output i.e. r=t.atten_factor.phase factor+n -> matched filter ->(detector)maximum ratio combiner.

 

[emma2]

Thanks alot in the first place, then isn't the division equal to multiplying by the conjugate of h. for my case i will consider h is known at the receiver, and to explain my case:
Consider Tx: transmitted signal (coded and BPSK modulated)
n:awgn
h: channel gain or channel coefficient "I'm considering slow fading ".
now at the transmitter output will be : out_put_rec=h.*Tx+n
at the receiver: out_put_hat=out_put_rec./ h
now why do I need the maximum ratio combiner?

 

[kalyanasv]

i) at the receiver you need a matched filtering, which is not the same as division. ii) a combiner is required for the multipath assuming "M" channels over which the same data come through, then you combine the multipath data optimally to get a the correct output. (This you could probably ignore if you want a sub optimal solution, which would still be ok). (You can look more about this in typical communication books by Proakis, Haykin, Sklar,....

 

[emma2]

I got it! thank you