A problem about comm.RicianChannel

Hello,everyone.

Recently I am working on a project of OFDM simulation. I want to use comm.RicianChannel object for the fading channel. After I read the docuument, I am confused that why the output signal of the object have the same length with the input signal? I mean that if we have a fading channel response, the conv result will make the signal longer(for delay reason), is that right?

Answer 1

[Divyam]

Hi, the size of the input argument is equal to the size of number of samples to be simulated by the "comm.RicianChannel" object. The output of the "comm.RicianChannel" object contains two arguments, the "output signals" and the "pathgains". The size of the "output signals" is the same as that of the input argument since the number of samples() remain same throughout the fading.

However, the size of the "pathgains" is a matrix where is the number of discrete path delays. You are correct in stating that the conv result will result in path delays. The path delays are captured in the "pathgains" output argument whose size is not equal to the size of input argument ().

To read more about the output arguments, you can refer to this documentation: https://www.mathworks.com/help/comm/ref/comm.ricianchannel-system-object.html#:~:text=%7C%20nonnegative%20scalar-,Output%20Arguments,-expand%20all

[lc]

Thanks for your help.

I read the document and I see an example of comupting the output result by using ''pathgains''. So if I want to get a result wich has a longer length than the input signal (in order to show the delay influence and the use of CP ), I can use the ''pathgains'' and add zeros before the data matrix manually (what i mean is that add zeros in front of the column representing the delay path) and sum the matrix by column?

[Divyam]

@lc Yes that is correct.

[lc]

Thanks very much!