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Continuous Power Spectral Density (PSD) from a Discrete PSD

ClaudioPedrazzi
11-Garnet

Continuous Power Spectral Density (PSD) from a Discrete PSD

Hi everyone,

it is possible that I already posted a similar question in the past , but this one is a slightly different problem.  The physical background is the fatigue calculation derived from random vibration with a known PSD. There are many literature sources for that, I will just mention a good practical summary by Tom Irvine (hope the link will work), or just google "Estimating Fatigue Damage from Stress Power Spectral Density Function".

Now my problem is: the input for these approaches is always a continuous integrable function like for example this example from that paper:

Unbenannt.JPG

Obviously the units of these function are the square of the input quantity divided by Hz (for stress for example ksi**2/Hz or MPa**2/Hz). In this way the square root of the integral of the function will provide the RMS of the input quantity (in this example 6.64 ksi).

Now let's say I have a measured time signal (it is actually a vibration velocity, but it would be no problem to convert it in stress, so that is not the point).  I use pspectrum (from the Signal Analysis Extension Pack) to compute its "discrete PSD".  Using /cfft**2/ gives similar results, so the computation should be ok.

I say discrete, because while the sum of the "spectral lines" divided by their number is constant (and equal to the sum of the square of the input values, also divided by their number, see illustration from the "spectral analysis chapter of the e-book "Signal Analysis Extension Pack"),

Unbenannt4.JPG

well, now depending on the number of intervals that I chose, I get different "heights".  Here an example:

Unbenannt1.JPG

Unbenannt2.JPG

This is obvious, because the sum must remain constant, and so the function, though shown with a track, like it was continuous, is actually made from many vertical "lines" separated one from the other.  In other words, the units of this psd output are (I believe) the square of the input quantity, and not, as in the analytical/numerical PSD example above, the quantity divided by Hz.

Now I would like to be able to fit some kind of envelope to the measured PSD, in order to do some computations, something like this:

Unbenannt3.JPG

Possibly the integral of the fitted function should be only slightly above the RMS of the input signal, to avoid overconservativity, but that is another subject.

Now finally my question: how do I get the "right" units [MPa**2/Hz] or [ksi**2/Hz] on the vertical axis? It may be so a dummy question that the answer is: just divide the peaks by their frequency and that's it, but somehow I am unsure if that's all what I need to do.

Thanks a lot for any hints

Best regards

Claudio

1 ACCEPTED SOLUTION

Accepted Solutions
LucMeekes
23-Emerald III
(To:LucMeekes)

Ah,... but that would mean that your spectrum gets 'attenuated' at high frequencies much more than at low frequencies.

So I could be wrong, and you just have to divide all sample values with the sampling frequency.

Success!
Luc.

View solution in original post

10 REPLIES 10

Hello Claudio Pedrazzi,

you could use the genfit function, as in Help of the Prime and that I present briefly, here below:

Continuos PSD from a discret PSD.jpg

Greetings

F. M.

Hi F.M., thanks a lot.  I probably wrote too much and so the real question "disappeared" in the flood .

The "real question" is: how do I get from a pspectrum generated spectrum that has values in (for example) [ksi2] to a "normal" function having units of (for example) [ksi2/Hz].

But thanks for suggesting genfit, I will surely try it.  But I have to get to the right numerical values before I can even start to fit.

Best regards

Claudio

Hi Claudio,

The CFFT function produces the magnitude of the spectrum for each positive (>=0) frequency fn, where fn=Fs*n/N, where N is the total number of frequency points that you have.

And your Fs should be equal to the number of samples you took divided by the total observation time.

So I'd guess you divide each sample n by its Fn.

Hope this helps,

Luc

LucMeekes
23-Emerald III
(To:LucMeekes)

Ah,... but that would mean that your spectrum gets 'attenuated' at high frequencies much more than at low frequencies.

So I could be wrong, and you just have to divide all sample values with the sampling frequency.

Success!
Luc.

Exactly Luc, I do not know what is correct.  I hope some expert will come out with the right method.

I will certainly try both anyway.

Regards

Claudio

Claudio,

you may want to have a look at this Spectrum analyser‌, which I built a couple of years ago to have a Spectrum Analyser in Mathcad, to mimic a real-world spectrum analyser (eletronic equipment that you can use to investigate spectral properties of real world electrical signals).

I remember trying to deal with the units there, abut don't know if I was able to sort it out.

Unfortunately I'm limited in time now, so have no time to dig deeply into it.

Success!
Luc

Claudio,

This thread is loosening up some rust in my brain. I had to cover this issue for an informal class some years ago, and I found the note I had written about this at that time, now attached. It may help with the issues that you are addressing.

Mathcad has a few different discrete Fourier transform functions. In my work with Mathcad (mostly Mathcad 11) as an engineer, I almost always used CFFT/ICFFT as these had the scaling I considered the most useful (CFFT/ICFFT use C1=1 in the language of the note).

Lou

Hi Lou and Luc

this is just to say a big great thank you.  At the moment I had to suspend working on this question, but as soon as I can come back, I will (after using your suggestions) also post here what I hope will be the answer working for me.  Please excuse me if a little time goes by ...

Regards

Claudio

So I am posting here, as usual, what is apparently working for me.  I wished to use the "pspectrum" function because it is very general and has a few interesting input options, missing on cfft. For example the number of intervals, the optional windowing (you know, Hanning, Hamming, Rectangular, Tapered rectangular) and the fractional overlap.

Unbenannt.JPG

Having said this, pspectrum provides the "power spectrum", per definition [Units: Signal**2].  So "obviously" to obtain the "power spectral density" [Units: Signal**2/Hz] it has to be divided by the sampling frequency fs (and multiplied by two if one wants the "one sided" spectrum that makes sense in engineering applications.

psd = 2 * pspectrum / fs

(and use only the first half of the pspectrum output)

With this transformation I obtain a vector of psd points, that if integrated numerically for example with the trapeze integration rule, no matter what the frequency resolution, always correctly reproduces the variance of the original signal. I tested this with different values of n and r in the pspectrum input.  With lowering frequency resolution the peaks become flatter and wider, but the integral remains constant, as wished.  Here a few examples based on the same signal.  Note the ordinate scaling is changing. The integral of the curve is approximately constant and numerically equal to the variance of the signal (sigma**2).

Unbenannt.JPG

So I believe this should be the right approach.  I would be grateful for further hints and of course if someone can point to an error in my reasoning.

Also, thank you again to all answerers, they are highly appreciated.

Best regards

Claudio

PS: LucMeekes‌ I believe that the division by sampling frequency fs is correct, because if it was wrong I would not obtain the correct integral. Dividing by the frequency of each sample would "attenuate" the higher frequencies.  Also cfft, after all, has to be squared and divided by fs, in order to produce the psd.

Grazie mille, Claudio,

for sorting it out.

I guess I may have missed the factor of two, due to taking only the positive half of the spectrum.

Once I have more time , I'll check my spectrum analyser.

Chiao,

Luc

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