2
$\begingroup$

I have to study the effect of low-pass filters (if any) on the outcome of the experiment. And the experiment here is to measure the vibration of a (metal) specimen by hitting the said specimen with impulse (or "knocking").

Parameters for measurement include:

  • Average complex (convert from time to frequency domain with FFT)
  • Hann (or Hanning) window function (this is the next parameter study, for now, bear with it)
  • 20kHz total bandwidth with 25600 FFT lines (or 25.6kHz sampling rate)

Theoretically speaking, applying the low-pass filter should lead to some differences in the frequency spectrum of the transfer function. Yet, in the image below, there is practically none. Sure, there is a drop of 1~2 dB from 350 to 500 Hz, but my instructor says that this frequency range is mostly noise, and they are not as important as the peaks (where there is no significant change).

Can anyone explain this to me, why there is no difference in the spectrum?

Further check on the components also say that there is only negligible differences in both the input (knocking force) and the output (vibration). My hunch is that for the force, because I only "knock", which means there is no frequency range, and I generate the knocking signal in time domain, so there is no frequency range to block. In other words, because of my setting for the input, the filter has no effect. However, this would not explain the situation with the output and the transfer function.

enter image description here

Edit: Additional graph, with frequency spectrum runs from 0 to 20kHz, comparing with and without 100 Hz filter (with 3 options: No filter (black), Filter on Output only (Yellow), Filter on both input and Output (Blue))

Linear scale for frequency enter image description here

Logarithmic scale for frequency enter image description here

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ If I am not mistaken you haven't added any information on the filter type (e.g. Butterworth, Chebychev). Also in the image if seems that you have several low pass filters and you are comparing them (100, and 200 Hz). Knowing the filter type, and parameter will help you find out its response. $\endgroup$ – NMech Dec 14 '20 at 10:23
  • $\begingroup$ @NMech There is no specific type of filter mentioned in the theory manual of the scanner. It only says "digital filters". And in the graph above, I want to see the effect of low-pass filters (if any). The original graphs have a few more lines, all show the same thing: low-pass filters have no visible effect in the spectrum $\endgroup$ – ComradeH Dec 14 '20 at 10:28
  • 1
    $\begingroup$ Seems like it's a very "soft" filter. Export the response plots as data, import into Octave, and you can multiply them with a response of any filter you want (just choose say a 4th order butterworth set for a frequency you want), and you will see the effect. The differences you're seeing means that the filters have a less-than-standard response, and perhaps assigning those frequencies to them may be slightly misleading. Those are quite soft filters, usually you'll see those when the processing is limited by CPU performance (e.g. use of low precision coefficients). $\endgroup$ – Kuba hasn't forgotten Monica Dec 14 '20 at 19:56
  • 1
    $\begingroup$ I implemented filters that look similar to the ones you see - not as bad, but that did ring a bell. In my case filters had to run in real time on an MCU that didn't have much in the way of DSP capability, i.e. lots of filter coefficients had to be truncated, and so on. $\endgroup$ – Kuba hasn't forgotten Monica Dec 14 '20 at 19:59
  • 1
    $\begingroup$ The action of your LPF is perfectly "observable" in your first graph. It has attenuated the signal by around 3dB. If you were hoping that it would somehow remove everything about 200Hz like an ideal "brick wall" LPF, your expectations were wrong (ideal filters implemented in the time domain don't exist in real life). $\endgroup$ – alephzero Dec 14 '20 at 20:07
2
$\begingroup$

This is more of a continuation of the comments. Maybe later with more information it will be converted to a proper answer.

Low pass filter frequency response

Low pass filters have a specific response with respect to frequency. A typical example for a low pass filter is presented below.

enter image description here

In the image above, the cutoff frequency is 1.

Notice that the x-axis is logarithmic. One of the most important features is that the filter attenuation is linear in this log scale.

As you might notice, that at a frequency which is double the cutoff frequency you have only a few db drop.

Your example

In your example, the x-axis is linear. The plot is not entirely clear and I can't be sure what the cutoff frequency is used. However, I will assume that you use 200Hz as a cutoff frequency. If you notice up to 200Hz hardly anything happens. Only above 200Hz you start seeing a difference. That difference becomes greater for higher frequencies. Around 400 Hz its in the order of a couple db.

This is consistent with the typical response graph, presented above. However lacking more specific data on the filter parameters its difficult -for me- to hazard another guess.

Given the sampling frequency, you could present data up to 12.8kHz. If you present that you should see greater attenuation of the higher frequencies.

Other filters

There are other filters with more pronounced cutoff regions. If you used one of these you could have higher attenuation at frequencies closer to the cutoff frequency. The main problem, is that some of them could mess the signal, and the phase.

Below is comparison between

  • different orders of a low pass Butterworth signal

enter image description here

  • different types of filters

enter image description here

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks for your comment. It seems to shed light on the "no difference with the 200 Hz filter". But how about the 100 Hz filter then. As you say "a frequency which is double the cutoff frequency you have only a few db drop.", which means that after 200 Hz (and the peaks at 300 and 330 Hz), the 100 Hz filter should still do something (decrease the spectrum by a few dB). Yet, nothing happens. And for reference, the filter I use here did not specify which type (no room to do that in the software). Only "low pass filter" $\endgroup$ – ComradeH Dec 14 '20 at 14:37
  • 1
    $\begingroup$ If you can plot the data in log plot with just the original data and the 100 Hz, so that they are more clear. If possible plot it from 0 to 12.8kHz. $\endgroup$ – NMech Dec 14 '20 at 14:48
  • $\begingroup$ I have updated the figures in the original question post with 2 graphs - one in linear and one in logarithmic scale for the frequency. You are right, there are differences at much higher frequency - fully visible from around 1kHz. Is there any plausible explaination on why there is little difference for <1kHz range? Thanks $\endgroup$ – ComradeH Dec 14 '20 at 15:05
  • 1
    $\begingroup$ Probably you should have something like 20dB per decade. If you look more closely around 1kHZ (which is a decade - 10 times) the cutoff frequency you get something in the order of the 20dB. However, there is not much more that I can say unless you know a bit more about the specifics of the filter used. Maybe someone else more experienced on the subject can be of more help. $\endgroup$ – NMech Dec 14 '20 at 15:11
  • 1
    $\begingroup$ Since you have the raw data, another thing you could do is try to do the filtering in Matlab/Octave or Python. That will give you better insight. $\endgroup$ – NMech Dec 14 '20 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.