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Let's say I have a noisy raw signal with 1GS/s sampling rate (That means the time resolution is 1 ns). Now I apply a FIR lowpass filter (window filter) of 25 MHz on it. The length of Window filter will be 40 ns in time domain. Since it is doing convolution in time domain, does it mean that I get one "non-correlated" point every 40 points? Should I discard those points and get the filtered signal with actual time resolution 40 ns?

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    $\begingroup$ No, doesn't mean that, strictly, unless you can build an infinitely good filter with finite length. Why do you choose to restrict yourself to filters that are of this particular length? Do you have a specific type of filter in mind $\endgroup$ Commented Jan 24, 2023 at 8:43

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If the noise itself is white noise this by definition means every sample is uncorrelated. When each sample is independent of every other sample, all samples will be uncorrelated and the frequency spectrum will have a constant power spectral density. If we then filter the signal with such noise with a low pass filter, then yes adjacent samples will be correlated and if the filtering is sufficient as detailed below, can be discarded. Considering a simple example of an FIR filter makes it clear how the correlation occurs: In an N-tap lowpass FIR filter, the output depends on the current sample plus a weighted average of the $N-1$ samples, and thus there must be a correlation in each sample at the filter output with the prior $N-1$ samples. If we shift by 1 sample, the next output will still have $N-1$ samples that we had in the previous output in the weighted sum. The adjacent samples in this case are no longer necessary and can optionally be discarded. This is the fundamental property used in multirate signal processing and the decimation process: we low pass filter and then after filtering we can discard all correlated samples with no consequence to the signal quality, since now at the output each sample contains all the info we need within the memory depth of the filter. The "info we need" is implied to be in a much narrower bandwidth.

The ideal filter and the amount we can decimate is best viewed in the frequency domain with a clear understanding of sampling and aliasing as described further by Hilmar in his post, and depicted in the graphic below showing the ideal decimation filter for a decimate by 4.

decimate by 4

Below shows the low pass filter that would allow for perfect decimation. Because the alias regions are restricted to specific zones, filters with selective stopbands (multi-band) can also be used resulting in more efficient structures. The graphic says that "perfect decimation" is not achievable. But on practical terms, and knowing the regions that would alias into our main passband after down-sampling, we can design our filters so that the rejection of this aliasing is well enough below a noise floor of interest to be considered sufficient for our application.

ideal low pass

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  • $\begingroup$ Not sure that I agree with "yes adjacent samples will be correlated and can be discarded." If I make brown noise by filtering white noise with a 1st order lowpass filter, I can't discard samples without creating significant aliasing. So it really depends on the type and purpose of the low pass filter. $\endgroup$
    – Hilmar
    Commented Jan 24, 2023 at 14:09
  • $\begingroup$ @Hilmar Yes agreed, to be strict the filter would need to completely attenuate the upper portions of the spectrum that alias and in that case adjacent samples can be discarded. If we mean by "Low Pass" that we wish to attenuate the higher frequencies to our requirements, we will have adjacent sample correlated to the degree that matters to us. Even with the first order low pass, if our sampling rate was high enough and our lower frequencies were low enough, we can discard adjacent samples with no consequence depending on req'ts. Best clarified by understanding aliasing and the freq domain. $\endgroup$ Commented Jan 24, 2023 at 14:19
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The sampling theorem states that in order to avoid aliasing your sampling rate needs to be higher than twice the highest frequency in your signal.

In practice this is a little more complicated: in the physical world there is no such thing as a "bandlimited signal" and sampling always introduces some amount of aliasing. Hence the sampling parameters are typically chosen based on the properties of the signal and the requirements of the application to keep this error "sufficiently small".

Applying a low pass filter will reduce energy at higher frequencies. Whether that reduces the bandwidth of your raw signal sufficiently to allow reducing the sample rate as well depends again the details: The properties of the filter, signal and application requirements. Specifically you want to look at "stop band frequency" and "stop band attenuation". The filter length is not relevant in itself (but related to the other quantities).

Assuming that stop band attenuation is good enough for an acceptable aliasing error you can indeed reduce the sample rate to something that's slightly higher than the stop band frequency.

This whole process is called "down sampling": use a low pass filter to reduce bandwidth and then throw a away a certain number of samples (that are independent anymore). Obviously this only makes sense if there is extra bandwidth in your raw signal that you don't need for your application.

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  • $\begingroup$ Reply to Hilmar: I understand that. But what is confusing me is that when I apply FIR filter, in time domain it is basically convolution between the signal and the filter. Right? With the example I posted in y question, the first point of the filtered signal is convolution of filter and first 40 points of the raw signal; the second point of the filtered signal comes from convolution between the filter and 2nd till 41st points of the raw signal and so on. It seems that two consecutive points of the filtered signal is correlated to each other. I am having trouble with understanding this part. Ar $\endgroup$
    – user66275
    Commented Jan 24, 2023 at 11:52

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