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Signal Processing Problem
- Thread starter Balplus
- Start date
Correct. Mostly.
The frequencies are true, since the transform shows or lists all of the frequencies possible. The amplitudes are incorrect, but the best based on the assumptions about what happens outside of the time sample we used.
This happens even when there is no modulation, so modulation has nothing to do with it.
They are sometimes called sidelobes instead of sidebands. The Fast in Fast FT has nothing to do with this problem. It happens because we take a finite time data sample. To get a complete view in the frequency domain we would have to look at an infinite amount of time. So we make an approximation. In this approximation process we start at a different value than we end with. This head to tale matching problem is the cause of the energy making this splatter. You see this problem by trying it with privileged frequencies. A privileged frequency could be one that has an exact (integer) number of cycles in the time aperture. For instance, if you sampled a 1 kHz sine wave at 1us samples for 0.1 second, you would have 100 cycles captured. Because the voltage at the start exactly matches the voltage at the end, there is no transient energy to create those sidelobes. This happens in non sampled (i.e.) analog system analysis also. It is an artifact.
You can sometimes ameliorate this affect by attempting head to tail match. You can search in from the two ends to find a better match in the sample values. Or you could make a larger aperture, o you could use what is called a window technique. For example, if you multiplied the very first and the very last samples by 0.1, then they would have a better match, and your sidelobes would be reduced.
The frequencies are true, since the transform shows or lists all of the frequencies possible. The amplitudes are incorrect, but the best based on the assumptions about what happens outside of the time sample we used.
This happens even when there is no modulation, so modulation has nothing to do with it.
They are sometimes called sidelobes instead of sidebands. The Fast in Fast FT has nothing to do with this problem. It happens because we take a finite time data sample. To get a complete view in the frequency domain we would have to look at an infinite amount of time. So we make an approximation. In this approximation process we start at a different value than we end with. This head to tale matching problem is the cause of the energy making this splatter. You see this problem by trying it with privileged frequencies. A privileged frequency could be one that has an exact (integer) number of cycles in the time aperture. For instance, if you sampled a 1 kHz sine wave at 1us samples for 0.1 second, you would have 100 cycles captured. Because the voltage at the start exactly matches the voltage at the end, there is no transient energy to create those sidelobes. This happens in non sampled (i.e.) analog system analysis also. It is an artifact.
You can sometimes ameliorate this affect by attempting head to tail match. You can search in from the two ends to find a better match in the sample values. Or you could make a larger aperture, o you could use what is called a window technique. For example, if you multiplied the very first and the very last samples by 0.1, then they would have a better match, and your sidelobes would be reduced.
Skogsgurra
Electrical
Google "Hamming window" and "Hanning window" (names of the more common windows used to reduce the sidebands). There are also simpler windows like trapezoidal and others.
Gunnar Englund
Gunnar Englund
The is a hell of a difference between sidelobes in an FFT and sidebands. Genuine sidebands resulting from modulation will certainly be visible on an FFT if present.
Sidelobes are really not much of a problem on modern equipment because FFT size has been increasing due to the low cost of processing power. Oscilloscopes, for example, used to only be able to handle 256 point FFTs in real time. Nowadays they can handle 8K FFTs without slowing down the system. Since sidelobes are measured in terms of "bins" offset from the central tones, a larger FFT shows a relatively smaller proportion of sidelobes.
Jsolar’s description of spectral leakage due to non-coherent sampling is correct. You would never use a Hamming window nowadays. Use von Hann (Hanning), flat top, windowed sinc etc. These give excellent response so that you can minimise spectral leakage even on 16 bit acquisition systems.
Sidelobes are really not much of a problem on modern equipment because FFT size has been increasing due to the low cost of processing power. Oscilloscopes, for example, used to only be able to handle 256 point FFTs in real time. Nowadays they can handle 8K FFTs without slowing down the system. Since sidelobes are measured in terms of "bins" offset from the central tones, a larger FFT shows a relatively smaller proportion of sidelobes.
Jsolar’s description of spectral leakage due to non-coherent sampling is correct. You would never use a Hamming window nowadays. Use von Hann (Hanning), flat top, windowed sinc etc. These give excellent response so that you can minimise spectral leakage even on 16 bit acquisition systems.
- Thread starter
- #5
Thanks for the help so far. I am still not clear on the subject. My background is in vibration analysis and this questions relates to mechanical vibrations that produce an electrical signal from an accelerometer. Keep in mind that this signal represents the mechanical motion of a machine and not the results of the test from an electronic circuit. With this in mind, do the all the frequencies represented by sideband frequencies in an FFT from mechanical vibrations represent real frequencies are some or all of these sidebands false frequences produced by the FFT process? Thanks for any help you can give me.
Skogsgurra
Electrical
If there are distinct peaks, I would say that they also exist IRL.
It is the "curved sloping sides" that are artefacts and that can be removed/reduced by either taking many samples (which makes the abrubt start/stop of the signal less disturbing) or by using a suitable window.
My understanding is that vibration guys take short samples - probably because the frequencies involved are low and it simply would take too long time to collect 16k or 64k samples and it would also blur the spectra because speed is not always constant over extended periods. I think that 1k or even 512 samples are used in some cases.
With such short samples, a good window is needed if you want to discriminate between close peaks.
Gunnar Englund
It is the "curved sloping sides" that are artefacts and that can be removed/reduced by either taking many samples (which makes the abrubt start/stop of the signal less disturbing) or by using a suitable window.
My understanding is that vibration guys take short samples - probably because the frequencies involved are low and it simply would take too long time to collect 16k or 64k samples and it would also blur the spectra because speed is not always constant over extended periods. I think that 1k or even 512 samples are used in some cases.
With such short samples, a good window is needed if you want to discriminate between close peaks.
Gunnar Englund
- Thread starter
- #7
Logbook, thanks for the reply.
Yes, the data we collect for vibration analysis is very low frequency compared to the electronic frequencies you electornic wizards work with. I usually take 2048 samples (800 line FFT). The frequency span is 0 to 2 KHz. The sidebands show up even at 8192 samples (3200 line FFT). A common example of the sidebands that I see on an electric motor spectrum (FFT)is rotorbar pass frequency (number of rotorbars in the rotor times rotor rpm) with 7200 cpm (120 Hz) upper and lower sidebands around the center frequency (rotorbar pass). The 120 Hz frequency is a true frequency (motor torque pulse frequency or 2 times line frequency). My question is whether the rotorbar pass plus 120 Hz and rotorbar pass minus 120 Hz frequencies are true or are they a false creation of the FFT process. Thanks.
Yes, the data we collect for vibration analysis is very low frequency compared to the electronic frequencies you electornic wizards work with. I usually take 2048 samples (800 line FFT). The frequency span is 0 to 2 KHz. The sidebands show up even at 8192 samples (3200 line FFT). A common example of the sidebands that I see on an electric motor spectrum (FFT)is rotorbar pass frequency (number of rotorbars in the rotor times rotor rpm) with 7200 cpm (120 Hz) upper and lower sidebands around the center frequency (rotorbar pass). The 120 Hz frequency is a true frequency (motor torque pulse frequency or 2 times line frequency). My question is whether the rotorbar pass plus 120 Hz and rotorbar pass minus 120 Hz frequencies are true or are they a false creation of the FFT process. Thanks.
Skogsgurra
Electrical
Do not know if it is logbook or me you address with that question. But my answer is that the rotor bar frequency is very real. Mechanically, the phenomenon is known as "cogging". It can be reduced by slanting the rotor bars.
Gunnar Englund
Gunnar Englund
- Thread starter
- #9
Skogsgurra,
I was addressing you in my last post -- sorry for the mixup. I understand that rotorbar pass is a real frequency as well as the 120 Hz torque pulse frequecy. I would expect a peak in the spectrum at 120 Hz and a peak at rotorbar pass frequency. My question is ---- Are the 120 Hz sidebands around the rotorbar pass frequency real? Thanks
I was addressing you in my last post -- sorry for the mixup. I understand that rotorbar pass is a real frequency as well as the 120 Hz torque pulse frequecy. I would expect a peak in the spectrum at 120 Hz and a peak at rotorbar pass frequency. My question is ---- Are the 120 Hz sidebands around the rotorbar pass frequency real? Thanks
Skogsgurra
Electrical
They are not a result of the FFT per se. But they CAN be (I'm not saying that they are) a result of non-linearities in your measuring set-up so that the two components mix and produce sidebands. One typical non-linearitiy is amplifier overrange (clipping). If a 60 Hz component gets into the system and saturates an amplifier (non-symmetric clipping), you will get 120 Hz mixed with your "pay-signal".
But it is not at all unreal that the mixing happens in the motor and what you see is what you have. You just have to eliminate all other possibilities to be sure.
Gunnar Englund
But it is not at all unreal that the mixing happens in the motor and what you see is what you have. You just have to eliminate all other possibilities to be sure.
Gunnar Englund
- Thread starter
- #11
Vibration signals often looks "clipped" and/or distorted because, as far as I know, there is no such thing an a linear machine. Because of the different stiffnesses about the circumference of a machine, along with different types of preloading on the machine, such as misalignment and distorted mountings, signals show considerable distortion at times. This distortion is not from the measurement setup, but from the machine itself. This distortion can be from faults in the machine or just a design characteristic of the mahcine. Under these conditions, would I expect the first set of sidebands to be true frequencies and the multiple sidebands (sidebands beyond the first set) to be false frequencies produced by the FFT process? Could all the sidebands be false? Or are all the sideband real frequencies.
Skogsgurra
Electrical
If there is mixing - there are sidebands. And if there are sidebands - there will be interharmonics between these and the original signal.
I would say that all sidebands do exist, they are not FFT artefacts (if you are absolutely sure that your analog circuitry and your A/D are linear). Then, if you shall regard all sidebands as important and information carriers is mostly a question of experience and judgment.
You can do an easy test by adding a couple of known pure sinewaves with different frequencies and use your FFT equipment to analyse the composite signal. You should have nothing but the two original signals in your spectrum (addition is a linear operation that does not produce sidebands). Addition is best made by just feeding the signals in parallel via two equal resistors (1 or 10 kohms) into the analyzer. Normal metal film resistors have very low distortion and produce no harmonics. Remember that you will get a 6 dB attenuation as a result of the addition.
Gunnar Englund
I would say that all sidebands do exist, they are not FFT artefacts (if you are absolutely sure that your analog circuitry and your A/D are linear). Then, if you shall regard all sidebands as important and information carriers is mostly a question of experience and judgment.
You can do an easy test by adding a couple of known pure sinewaves with different frequencies and use your FFT equipment to analyse the composite signal. You should have nothing but the two original signals in your spectrum (addition is a linear operation that does not produce sidebands). Addition is best made by just feeding the signals in parallel via two equal resistors (1 or 10 kohms) into the analyzer. Normal metal film resistors have very low distortion and produce no harmonics. Remember that you will get a 6 dB attenuation as a result of the addition.
Gunnar Englund
- Thread starter
- #13
Thanks for all your help and patients. I do consider sidebands important. The sideband separation frequency identifies the problem component in the machine. Like I said in my original post, I was told in a class once that sidebands are false frequencies in machine vibration signals (a false creation due to the way the FFT handles modulation between two frequencies). I just wanted to clear things up. Thanks again for your help.
John J
John J
It is really simple--FFTs are great but you need to understand what they are doing. The are a way of calculating a Discrete Fourier Transform (DFT) using some calculation tricks to compute the transform with a Ncalc = cN*LOG(kN)type relationship between the size of the transform (N) and the number of multiplies typically (Ncalc).
Second the transform is exactly equivalent to taking the time waveform (real or complex) you capture and repeating it forever and then taking the Fourier Series. Obviously if the start and end points have discontinuities you will get artifacts due to the FFT alone--called leakage. The windows try to gradually shape the waveform so that the start and end samples of the FFT window are identical to eliminate the discontinuities (usually zero at start and end with some sort of lowpass window). Due to the convolution property this will distort the spectrum based on the window used to impart periodic continuity that reduces the leakage problem.
Third, The frequency to bin mapping of the FFT is important. The FFT samples the digital domain (Z-domain) unit circle. bin 0 = zero Hz, bin N/2 = both Fs/2 and -Fs/2 Hz since it is the point -1 + j0 on the unit circle then bin N/2 + 1 = -Fs/2 + delta Hz and finally bin N-1 = smallest negative frequency ( -delta Hz). Most people do not understand the "branch cut" at bin N/2 and that you actually cannot tell if the true signal is at Fs/2 or -Fs/2 from the nominal tuning frequency at the input to the FFT processor.
Other than these properties the FFT aside from scaling the output by the factor N there is no distortion. To allow parseval's theorem to be true (sum of squares of samples into FFT = sum of squares of frequency bins at output of FFT) you need to scale the outputs by 1/sqrt(N)
Hope this stuff helps,
John
Second the transform is exactly equivalent to taking the time waveform (real or complex) you capture and repeating it forever and then taking the Fourier Series. Obviously if the start and end points have discontinuities you will get artifacts due to the FFT alone--called leakage. The windows try to gradually shape the waveform so that the start and end samples of the FFT window are identical to eliminate the discontinuities (usually zero at start and end with some sort of lowpass window). Due to the convolution property this will distort the spectrum based on the window used to impart periodic continuity that reduces the leakage problem.
Third, The frequency to bin mapping of the FFT is important. The FFT samples the digital domain (Z-domain) unit circle. bin 0 = zero Hz, bin N/2 = both Fs/2 and -Fs/2 Hz since it is the point -1 + j0 on the unit circle then bin N/2 + 1 = -Fs/2 + delta Hz and finally bin N-1 = smallest negative frequency ( -delta Hz). Most people do not understand the "branch cut" at bin N/2 and that you actually cannot tell if the true signal is at Fs/2 or -Fs/2 from the nominal tuning frequency at the input to the FFT processor.
Other than these properties the FFT aside from scaling the output by the factor N there is no distortion. To allow parseval's theorem to be true (sum of squares of samples into FFT = sum of squares of frequency bins at output of FFT) you need to scale the outputs by 1/sqrt(N)
Hope this stuff helps,
John
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