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As part of my uni assignment I have to figure out what is the impact of sampling frequency on actual channel impulse response. One obvious thing is that the higher sampling frequency is, the more taps the response has.

But there must have been something more here. When I increase the sampling frequency from 20MHz to 100MHz (and adjust my CP accordingly), I get much better performance. The actual CIR is generated using the following MATLAB code

function cir = getchannel(Trms, sampfreq)
if Trms == 0
    Kmax = 0;
    vark = 1;
else
    % Calculate the exponential decay envelope  
    Kmax = ceil( 10 * Trms * sampfreq * 1e-9);
    var0 = (1 - exp( - 1/(sampfreq*(Trms*(1e-9))))); 
    k = (0:Kmax);     
    env = var0 * exp( - k/(sampfreq*(Trms*(1e-9))));
end
stdDevReOrIm = sqrt(env/2);
cir = stdDevReOrIm .* (randn(1, Kmax+1) + j*randn(1, Kmax+1));

I see that in case of 100MHz has a slightly different shape and var0 is much smaller. I wonder why. The following code was implemented based on "IEEE 802.11 Handbook: A Designer's Companion", but I have no access to this book. Also, I was not able to google anything useful, so I decided to ask here.

Pawel
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    the channel impulse response doesn't care how you sample. It's a physical thing you're observing, not something you're effecting. Your estimate of the CIR is of course affected. I know that sounds nitpicky (it is!), but it helps get a few concepts straight when you don't confuse the thing you're observing and the observation. You say "you get much better performance", but you need to say "for what, doing what, measuring what as performance metric". – Marcus Müller Mar 10 '22 at 12:56
  • Can you explain then why the MATLAB code looks uses sampling frequency to generate CIR? – Pawel Mar 10 '22 at 13:00
  • you generate a discrete model of a CIR. – Marcus Müller Mar 10 '22 at 13:01
  • Perfect, and why sampling frequency is used when I generate a discrete model of CIR? – Pawel Mar 10 '22 at 13:10
  • that's in the word "discrete"! I think you want to tell me something, and plan to put it in a series of questions: please don't. Tell me what you think, directly, bluntly. I can take it, I promise! :) – Marcus Müller Mar 10 '22 at 13:16
  • @Pawel There are two different things here. One is the CIR of a particular channel. That is fixed and given by nature, as Marcus points out. Another is a Matlab program that generates random CIRs, and which takes the sampling time as a parameter. Don't confuse these two things! – MBaz Mar 10 '22 at 13:29
  • Yes, I fully understand what you are saying here. But my question here is why sampling frequency impacts the shape of CIR. In other words, why var0 depends of it? – Pawel Mar 10 '22 at 13:43
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    @Pawel because the matlab code tries to model the same physical thing observed at different sampling rates. That's like asking why a computer-generated image of a bee looks different when you tell the computer program that you want 10000 pixels per mm vs 10 pixels per millimeter. – Marcus Müller Mar 10 '22 at 15:14
  • Ii think this function also tries to normalize the CIR so that the received power (after convolution with the CIR) is independent of the sampling rate. – MBaz Mar 10 '22 at 16:20

1 Answers1

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I see that in case of 100MHz has a slightly different shape and var0 is much smaller. I wonder why.

Because your calculation depends on the sampling frequency.

If your question becomes "is this calculation/model reasonable?", read section 2.4 of this chapter. In short, the CIR is continuous, and you are generating a discrete model of the CIR that modelize each channel tap, being a sum of large number of independent unresolvable paths, as a circularly-symmetric complex normal random variable.

Because the higher the sampling frequency, the narrower the channel tap, the number of unresolvable paths in a channel tap is reduced in increasing sampling frequency and, therefore, its power is reduced too. The model you are using tries to modelize that decreasing in power by an exponential decay function in the way that normalizes the generated discrete CIR.

Note that because increasing sampling frequency results in reducing the number of unresolvable paths in a tap, the higher the sampling frequency, the less accurate modeling the taps by normal random variables.

AlexTP
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