How to convert time domain to frequency domain to understand the vibration
Hi expert,
I want to know how to convert the time domain to the frequency domain to understand the vibration.
I tried to convert but my results are not the same as your report (I am using Fourier transform> Hanning)
I referred gearbox model:
Best regards,
Tuan.
Answers
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Hi Brian,
From what i can see, you are missing 2 more steps to get your FFT to match with the report.
To account for the Aliasing phenomena , the "fold" function is additionaly used
Also at the end the FFT is divided by the length of the vector (N) to appropriately scale/normalize it.
So your "y" math function in Hypergraph would look like this: fold(FFTmag(Hamming("function")))/"length of function"
More info regarding this can be found here Signal Processing (altair.com)
Hope this is helpfull, let us know if you have more questions !
Regards,
Spiros Mallios
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Hi Brian,
From what i can see, you are missing 2 more steps to get your FFT to match with the report.
To account for the Aliasing phenomena , the "fold" function is additionaly used
Also at the end the FFT is divided by the length of the vector (N) to appropriately scale/normalize it.
So your "y" math function in Hypergraph would look like this: fold(FFTmag(Hamming("function")))/"length of function"
More info regarding this can be found here Signal Processing (altair.com)
Hope this is helpfull, let us know if you have more questions !
Regards,
Spiros Mallios
How to know the length of the function?
And it is still not the same.
Thanks
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Brian_20847 said:
How to know the length of the function?
And it is still not the same.
Thanks
To follow the report, we first sample the data with a fixed interval of 0.00025s (2nd plot)
The length of the vector will be N = End_Time / Sample interval = 2/0.00025 = 8000
On the 3rd plot is the final result.
Hope this clears it up !
Kind regards,
Spiros.
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To follow the report, we first sample the data with a fixed interval of 0.00025s (2nd plot)
The length of the vector will be N = End_Time / Sample interval = 2/0.00025 = 8000
On the 3rd plot is the final result.
Hope this clears it up !
Kind regards,
Spiros.
Thanks for your help. But I have another question.
How to convert for the not uniform sample interval?
We cannot calculate the length of the function.
Thanks
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Brian_20847 said:
Thanks for your help. But I have another question.
How to convert for the not uniform sample interval?
We cannot calculate the length of the function.
Thanks
Hi Brian,
As stated in the Signal Processing (altair.com) guide:
"In order to perform a Fourier Transform on a computer, the signal to be transformed must be digitized. The basic requirement for all discrete Fourier Transforms is that the discrete input data be sampled at a constant frequency, such that all time intervals are identical. If the input data is not evenly sampled, the discrete Fourier transform will be incorrect."
So, if the signal is not sampled uniformly the result of the FFT can be misleading.
Hope this is helpful !
Thanks,
Spiros.
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Hi Brian,
As stated in the Signal Processing (altair.com) guide:
"In order to perform a Fourier Transform on a computer, the signal to be transformed must be digitized. The basic requirement for all discrete Fourier Transforms is that the discrete input data be sampled at a constant frequency, such that all time intervals are identical. If the input data is not evenly sampled, the discrete Fourier transform will be incorrect."
So, if the signal is not sampled uniformly the result of the FFT can be misleading.
Hope this is helpful !
Thanks,
Spiros.
Thanks for your help.
How to convert the signal that is not sampled uniformly?
We don't have any way to convert them, right?
Thanks
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Brian_20847 said:
Thanks for your help.
How to convert the signal that is not sampled uniformly?
We don't have any way to convert them, right?
Thanks
Hi Brian,
We used the "lininterp" function as shown in plot 2, to make sure the signal is sampled uniformly.
Thanks,
Spiros
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