HIGH AMPLITUDE HARMONICS 3

[jcc123]

WE MONITOR VIBRATION ON THE DIE BLOCK OF A ROTARY PUNCH MACHINE PUNCHING .015IN STEEL AT 800/SEC.ACCL. HAS 35KHZ RESONANCE AND MONITOR IS FFT WITH UPTO 20KHZ BW,UPTO 32OO BINS AND TO 51.2KHZ SAMPLING RATE.PUNCH VELOCITY IS SUCH THAT SHEARING EDGE THEORITICALLY PASSES THRU STEEL IN .0000102 SECONDS.THE TIME TRACE IS CLASSIC IMPACT TRAIN. THE SPECTRUM HOWEVER IS TOTALLY DOMINATED BY HIGH AMPLITUDE HARMONICS USUALLY ABOVE 9KHZ.AMPLITUDES ARE 3 TO 10 TIMES FUNDAMENTAL IF IT CAN BE FOUND. VARIATIONS IN PUNCH FREQUENCY HAVE NOT BEEN FOUND.IT IS UNDERSTOOD THAT HARMONICS SHOULD EXIST BUT NOT WHY THE AMPLITUDES ARE SO HIGH RELATIVE TO FUNDAMENTAL.

 

[Hatch]

There are several issues here. First how is the accelerometer mounted. Stud mounting is considered the best with a mounted resonance >28kHz. Magnet mounting under ideal conditions is around 7 kHz. If you have magnet mounted the accelerometer you may be seeing the mounting resonance.

Next, depending on the size of your accelerometer (small has a higher frequency range), you may be at the top end of the accelerometer range which shows up as a high frequency band of peaks.

High frequency noise is also common from motors and other electrical systems. Measurements taken above 10kHz may show those peaks.

The Nyquist theorem also says that your sampling frequency is at least twice the frequency of interest. Does your FFT analyzer have a sufficient sample rate to go all the way up to 51.2 kHz.

C. Hugh (

http://www.Hatch.ca)

 

[anands78]

Accelerometers normally have an internal resonant frequency(usually of the order of 16 Khz). Check the specs of your accmeter fr the same.

 

[MikeyP]

What you describe is precisely what I would expect. The spectrum of an ideal implulse train spaced at 1/800 sec intervals is a is an "impulse train" in the frequency domain spaced at intervals of 800 Hz. All peaks will have the same amplitude.

In your case, we can assume that the force being applied to the machine by the punch is a nearly ideal implulse train. If it were possible to measure the applied force, you would see a series of peaks in the spectrum of the force each having the same amplitude.

However, you are measuring the acceleration RESPONSE of the machine at a particular point on the machine body. This response will be 'coloured' by the modal dynamic behaviour of the structure of the whole machine. This means that the peaks in the response will not be all the same amplitude.

I'm sure you are familiar with what a frequency response function (FRF) is: a graph with peaks corresponding to the natural resonant frequencies of the machine. The acceleration response spectrum that you measure on your machine will be the spectrum of the applied force (a train of "impulses" in the frequency domain with equal amplitude) multiplied by the FRF for the machine at the measurement point. So you can see that if one of your "impulses" in the force spectrum happens to coincide with a peak in the FRF then the peak in the response spectrum will be big. If it coincides with a dip in the FRF, then the peak in the response will be small.

I think you are confusing yourself by calling them "harmonics" and thinking "Harmonics usually decrease in amplitude at higher frequencies". If you think purely in the frequency domain, the punch is only putting energy into the structure of the machine at 800 Hz, 1600 Hz, 2400 Hz etc. Therefore, when you measure the response, you are only getting response information out at 800 Hz, 1600 Hz, 2400 Hz etc. The "harmonics" have nothing to do with the dynamic behaviour of the machine, they are present in the input force spectrum and so they are also present in the output response spectrum.

Hmm. Not sure that was very clear! let me know if you want clarification on anything.

M

 

[jcc123]

Thanks to all respondents. In reply the accl. is stud mounted,is usable above 20khz,no internal resonance below 20khz.Accls.usable to 75khz and other instrumentation confirm findings.Modal study has been done and now being restudied. MikeyP-your perception that I was confused by thinking harmonics should decay with frequency was correct.Thanks. I know that the F(f) transform of an impact will show harmonics with the fundamental=1/T where T=impulse width which in our case=1/.0000102=98khz which we cannot see. But I did not realize that harmonics of the impact frequency exist when no variation in fundamental exist.Can you explain how? I have read electricpete's explanation to the thread"Why do I see harmonics of Impact Frequency---"This was also very helpful.Thanks.

 

[MikeyP]

The "harmonics" at multiples of the impact frequency occur because you are finding the spectrum of a train of impulses rather than just a single impulse. The spectrum of a single ideal impulse has the same value at every frequency (ie it is white noise). But the spectrum of a TRAIN of impulses is also a train of impulses in the frequency domain at multiples of the impact frequency (A signal which is periodic in the time domain is periodic in the frequency domain as well). This idea forms the basis of Shannon's sampling theorem and many other fundamental signal processing ideas.

M

 

[electricpete]

I'm not sure if this will add anything to the discussion, but let me try.

As Mikey mentioned, the Fourier transform of an impulse train in time spaced at T is an impulse train in frequency spaced at 1/T.

You also may know that convolution (**) in time results in convolution in frequency.

From these two facts we can deduce the relationship between the Fourier transform of a single impact (call it Fsingle= Fourier[fsingle]) and the Fourier transform of repeeating impacts call it Fperiodic=Fourier[fperiodic]), assuming for simplicity that the time waveform f decays away between each impact.

The shifting propery of the impulse d(t):
fsingle(t)**d(t-k*t0) = fsingle(t-k*t0)

impulse train(t)**fsingle(t) = Sum{fsingle(t)**d(t-k*t0) from k=-Infinity..Infinity} = Sum {fsingle(t-k*t0) from k=-Infinity..Infinity} = fperiodic(t)

Fourier transform of the leftmost quantity is Fsingle(t)*IMPULSETRAIN(t).

Fourier transform of the rightmost quantity is Fperiodic(t).

That proves that the Fourier transform of the periodic impacts is the same as the Fourier transform of the signle impact, except that we multiply the spectrum by the frequency impulse train, which leaves only harmonics.

Fou