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I am performing the following simulation:

  1. Complex chirp signal with bandwidth [-B, B] is transmitted through a multipath channel which has $N$ taps all being complex Gaussian variables (i.e., Rayleigh channel) with zero-mean AWGN: $$y=xh+w,$$ where $x$ and $y$ are the transmitted and received signals, $h$ denotes the channel and $w\sim \mathcal{CN}(0,\sigma^2)$ is the noise variable.

  2. I use a low-pass filter (assume ideal filter) with cut-off frequency $L \leq B$, and I filter the two variables $y$ and $xh$, separately. Let's call the outputs of the filter $y_{filt}$ and $xh_{filt}$.

  3. I evaluate the power of these signals, i.e., $P_{y}=||y_{filt}||^2$ and $P_{xh}=||xh_{filt}||^2$.

What I observed is that the error (which is a combination of noise and estimation error), i.e., $P_{y}-P_{xh}$, decreases when increasing $L$. Any idea on the reason behind that?

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  • $\begingroup$ Hi @Dan. Yes, I believe that this is the reason, i.e., by adding more samples and then averaging in order to measure the power we effectively reduce the noise power. This is because the noise is zero-mean and when we add more and more samples their average goes slowly to the mean (zero) and as a result $P_y$ gets closer and closer to $P_{xh}$. However, I am not sure what is how this averaging affects the $P_{xh}$, or more simply just $P_{h}$? Isn't the effect similar? $\endgroup$
    – Miroslav
    Nov 23, 2021 at 16:38

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The chirp signal bandwidth is [-B, B] and your cut-off frequency is $ L≤B$, so your filter is removing part of the desired signal. By increasing $L$ you will increase the power due to the signal component.

If the signal power in the increased bandwidth is more than the noise in that extra bandwidth then you would see an increase in the SNR. Once you set the cut-off $L>B$ you will see a decrease in the SNR.

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  • $\begingroup$ Hi @David. First, I agree that if $L>B$ we expect a decrease in the SNR. However, I do not fully agree with that: "If the signal power in the increased bandwidth is more than the noise in that extra bandwidth then you would see an increase in the SNR". What I mean is that even if the signal power is less than the noise, we can still see increase in the SNR by increasing the bandwidth (only if we are still within the region [-B, B]). Please see my comment above that summarizes the reason. $\endgroup$
    – Miroslav
    Nov 23, 2021 at 16:46

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