Do resonant frequencies produce harmonics 2

[GMarsh]

Hi,

I have a plate vibrating due to some forced excitation. When I take FFT of measured acceleration signal on the plate, I am getting a dominant mode and its harmonics. I see no link to this dominant frequency and exciting frequency (they are very wide and far). But the dominant frequency in FFT is very near to one of the resonant frequencies of the structure.

Before I conclude that the external excitation is exciting one of the resonant modes, I want to know if a resonant mode ever displays harmonics in FFT of acceleration signal.

Thank you.

Regards
Geoff

 

[GregLocock]

The +/-200 comes from the frequency resolution of your fft. You know absolutely nothing about that signal to a resolution better than +/-200 Hz, with confidence.

Incidentally, are you using a hanning window? and an AA filter?

Stick the accelerometer on the workpiece and hit the workpiece with your hammer. Do this with the tool touching the workpiece, without the tool touching the piece, and ideally, with the cutter running, ie 3 tests.

You are looking for frequencies, not amplitudes, so this last one can be rough and ready. You might look into a gated (synchronous) analysis if you want to get fancy.



Cheers

Greg Locock


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[GMarsh]

Greg, I think +/-200 sounds too wide. I sampled the data at 1e6 points/sec. Then how come the resolution can be so low? Or do you mean something else?

Yes I am using hanning window. No filter.

I understood first two cases - tool touching and not touching. But what about 3rd one? If tool is touching and rotating then within no time it removes material and then it is equivalent to not touching (except for some rubbing at times).

I did synchronous analysis earlier with tool cutting force and acceleration signal and saw the evolution of 4870Hz as the tool excites the workpiece, and then gradual diminishing of the peak. But it didn't answer my question as to why that mode only is getting excited and mainly why harmonics are present for a resonant mode.

Thank you.
Geoff

 

[electricpete]

On the zoomed-in time waveform, you can see the binwidth of the spectrum is 400hz. For example 5204-4804 = 400hz.
This is 1/T where T = 17.015-17.0125 = 0.0025 sec (1/0.0025sec = 400hz).

Apparently when you zoomed in you limited the duration of the data input to the FFT, which made your bandwidth bigger.

You can get much higher resolution if you keep the time record longer. In fact you can get much higher resolution (theoretically unlimited except by numerical error) simply be zero padding.
thread384-208992
Note resolution is not the same as accuracy. Accuracy in estimating frequency is complicated.

1 (unlikely imo) - the system may have two resonant frequencies which happen to differ by a factor of 2 (would be quite a coincidence for anything but very simple system).

2 - the high vibration at 4870 is interacting with some non-linearity of the system to create non-sinusoidal motion with fundamental frequency of 4870 and (since non-sinusoidal), harmonics of that fundamental frequency so 2*4870 = 9740.

Looking at the zoomed-in time waveform, it doesn't look like a non-sinusoidal waveform repeating at interval of 1/4870 (which would be #2). The phase of the 9740 and higher frequencies seems to be drifting with respect to the 4870, as if they are not exact multiples of 4870. That would seem to argue for #1, which doesn't make much sense sense to me. Still thinking.

The reason I asked about accelerometer mounting. There could be looseness introduced there. Ordinarily I'd think that would fall in the category of #1 and so doesn't really seem to be the case from your time wavform, but I don't to be too clever and rule it out by what I think it should look like. Sometimes looseness acts strangely.

Bottom line, I don't know what's going on with your data.


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[electricpete]

Correction
which made your bandwidth bigger
should have been
which made your bin-width bigger

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[GMarsh]

electricpete,

Many thanks for your detailed reply. It's very useful. Though I understood your bin width concept, I will look into the thread your forwarded.

I use to do zero padding to reaching to nearest 2^n when using FFT. But I never realised we can increase resolution. I am still wondering if it is ok to keep on adding zeros to increase resolution. Is there no limit to this? Anyway I always have a basic doubt about amplitude of FFT which I will post as a separate thread in this forum.

You commented
The phase of the 9740 and higher frequencies seems to be drifting with respect to the 4870, as if they are not exact multiples of 4870.

Can you briefly explain how you made this observation ? I see that there is some drift in double peaks and reduction in amplitude. But not clear and sure.

Accelerometer was mounted with superglue. I think the bond was very good and no looseness here.

Bit saddened to see your bottom line ! I was also so much puzzled with this that after all analyses failed I resorted here. Anyway learnt many good points here. Thanks a lot for that.

Kind regards
Geoff

 

[GregLocock]

You don't get something for nothing. Zero padding does not affect the confidence of the estimate, although the cynic in me suggests it increases the likelihood of it increasing my confidence that the answer is wrong.

dF=1/T is an energy or information based statement. Adding zeroes adds neither energy nor information to the data.

Cheers

Greg Locock


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[electricpete]

You're right, you don't get something for nothing. With the original short length FFT resuls (including phase), we can reconstruct a continuous function of magnitude and phase vs frequency. It is a sum of weighted, frequency-shifted continuos sync functions, one term for each FFT point). As a continuous function, the peaks are single values (not intervals). The individual FFT magnitudes represent samples of that continuos function (because all the other sync functions contribute zero at that particular point).

The resolution (not accuracy) is virtually unlimited because we can determine that single value. Accuracy is a different but related question. On average, the single value which represents the peak of the underlying continuous function is our most accurate estimate of the true frequency of the peak . If and engineer should just simplistically look for the highest peak and declares that bin center to be the peak frequency, he is using a less accurate estimate. It may be good enough, but it is not the most accurate estimate available from the data.

Do we gain something for nothing when we use the more exact estimate? No, we simply don't throw away information. The engineer who looks only at the single highest peak and ignores the rest of the spectrum is throwing information away.

A simple example.
Bin Center / Magnitude
100hz / 0.1
101hz / 0.2
102hz / 1.0
103hz / 1.1
104hz / 0.2
105hz / 0.1.

Where is the peak? The simplistic approach is pick the highest value. That would be 103hz.
I think a little intuition would tell us that 102.5 is a more accurate, and intuition would be correct.
We improved our estimate by not just looking at one peak but by looking at the neighboring peak. Although the peaks furhter away have less inflence on our estimate (the sync function disappears as we go away from the center), each additional point we consider from the FFT result improves our estimate. Best possible (most accurate) estimate comes from adding all the FFT contributions to determine the underlying continuos function and determine its peak.


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[electricpete]

Just to step back, starting with a given time record, we can improve our estimate of peak frequency two roughly equivalent ways
1 - Time domain zero padding prior to FFT.
OR
2 - Take the original FFT results (a function over a discrete domain of frequencies) and convert it into a DTFT (a function over a continuous domain of frequencies). This can be accomplished using equation 6.17 and 6.18 here:

http://ens.ewi.tudelft.nl/education/courses/et2405/notes/champagne04.pdf

Once we have the DTFT (a complex valued function over continuos frequencies), we can take it's magnitude (real valued) and apply the same peak-finding techniques we would use for any real-valued function on a continuos domain.

Further down on the page the expression P(w) is shown to be a sync function.
The DTFT will be sum of weighted frequency-shifted sync functions, one per FFT output point.
The FFT output magnitude and phase is the weight.
The sync function is centered on the bin center and decays away from there.
The first zero of the sync occurs one bin width away, then two bin-widths etc.
Maybe this supports my previous discussion better.

But also we should note the sync function so defined is relatively smooth. Therefore simpler interpolation mechanisms that fit a smooth curve through a few neighboring FFT points (for example quadratic interpolation) work pretty darned well (a lot more accurate than just looking for the highest FFT ouptut magnitude, not quite the best estimate, but but a lot less work then the full calculation described previously)





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[hacksaw]

all well and good, but you are no better off unless you have more data,

 

[electricpete]

I agree 100%, longer time record data will of course give better ability to determine frequency peak accurately. The point I was trying to make is that for a given time record (and associated associated FFT output), the approach to obtain the most accurate estimate of the frequency of a peak is NOT to simply look for the highest FFT magnitude and select the center of the corresponding frequency bin. It is a common approach, good enough for many purposes, but it is not the most accurate approach. When reviewing FFT's from already-collected condition monitoring data in our E-monitor database, attempting to determine the source of peaks such as bearing defect-related peaks, having a "peak label" feature which interpolates rather than giving the bin center is simply invaluable in getting the most out of the data you have available. I'm sure that anyone who has used such a tool in such a task would agree.

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(2B)+(2B)' ?

 

[electricpete]

The phase of the 9740 and higher frequencies seems to be drifting with respect to the 4870, as if they are not exact multiples of 4870.

Can you briefly explain how you made this observation ? I see that there is some drift in double peaks and reduction in amplitude. But not clear and sure.

Attached is an example of what I mean.
The tab labeled "calc" accepts inputs A1, A2, A3, F1, F2, F3 and plots the waveforms:
y1 =A1_*SIN(2*PI*F1_*t)
y2=A2_*SIN(2*PI*F2_*t)
y3==A3_*SIN(2*PI*F3_*t)
Ytotal=y1+y2+y3

In tab "HarmInPhase", the harmonics F2 and F3 are exact multiples of F1 and you can see they maintain a constant phase relationship to the fundamental and as a result the waveform repeats itself identically for each time period T = 1/F1.

In tab "HarmInPhase", the "harmonics" (not a great term) F2 and F3 are close but exact multiples of F1 and you can see they that their phase with respect to the fundamental drifts continuuously. As a result, the waveform does not repeat itself, but evolves slightly each period 1/T1.

Your waveform looks more like HarmDrift than HarmInPhase to me.


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(2B)+(2B)' ?

 

[electricpete]

all well and good, but you are no better off unless you have more data,

More data is better.
For a given amount of data, you are in fact better off (in terms of accuracy in estimating frequency) to use one of the approaches I have suggested (for example quadratic interpolation based on the highest magnitude bin and one neighbor on each side) than to simply select the frequency of the center of the bin with the highest magnitude.

Out of respect to OP (unless he has interest to discuss this further), may I suggest any further discussion on the question of frequency resolution be in a new thread (I would be very happy to particpate).


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[GMarsh]

electricpete,

I gave a try to zero padding concept with my data. Attached are the results in excel sheet.

As you can see, padding with zeros has effect on frequency and amplitude both. I used my data which we looked at earlier i.e. 17.0125 to 17.0150. Added zeros on left, right and equal on both sides. Looked at amplitude and frequency of first peak i.e 4804 in unpadded data.

While padding zeros on any side has no effect on frequency, amplitude is less effected (relatively) with equal padding on both sides. Still there is a significant reduction in amplitude.

With all these I feel zero padding may not be useful as neither the frequency nor the amplitude reduction converges.

I have another basic doubt on amplitude of FFT which I will post in a separate thread.

Thank you.

Kind regards
Geoff

 

[GMarsh]

Sorry electricpete, just now I saw that you advised posting in new thread on 'Frequency resolution and amplitude'. I will create one and please reply there.

Thank you.

Geoff