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## FFT Frequency Conversion

I am new at the use of FFT's and am trying to analyze a set of periodic data on a machine we now use that has some feed rate variation coming from the drive system. I wish to convert the peak j values to frequencies of the coefficients of interest. The help screan says:

Frequencies corresponding to coefficients
Both fft and cfft return vectors whose elements are the complex amplitudes of the various frequencies in the original signal.

To recover the actual frequencies, you must know:

The sampling frequency of the original signal.

The number of samples in the original signal.

Given these parameters, the frequency associated with the jth element is given by:

C(k)=(j/N)*f(s)

where N is the number of data points in the analysis and f(s)
is the frequency of sampling.

My question is what the units are for the corresponding C(k) values calculated in this manner. I am trying to understand the results I have obtained in terms of cycles of rotation of the drive system. Can anyone help?

Thanks!

Gerry J. Dail
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## FFT Frequency Conversion

First some factoids.

The C(k) are the coefficients of the kth term of the sum of C(k)exp(ikx). Here I am in some trouble since I don't at this minute know whether k includes 2 Pi or not.

However, k = 0 is the average value of the function, F(x), corresponding to infinite wavelength.

k = 1 corresponds to 2Pi radians for the complex wave extending over the range of N times the delta x between samples. k=2 corresponds to 2 x 2Pi radians over the range of the data.

If your data shows five full cycles of real sinusoidal function over the range of N samples, then the frequency of your signal is 5 2Pi = freq T.

Since T = N delta x, freq = (5/N) (2pi/delta x)

Here, 2pi/delta x is the sampling frequency in radians per x. Sometimes the 2pi is included and sometimes not, depending upon whether you are talking with EE's or physics guys.

This equation states that freq = (k/N) sampling frequency.

If your |C(k)|maximum for N=100 with a sampling rate of 1000 samples per second is found at k=12 gives (12/100)1000 cycles per second, not accounting for the way 2 pi was handled.

Generate your own data set using sin(2 pi f t) You will then know exactly how the 2pi is handled.

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Physics: Common Sense made Obscure by Mathematics