For example if I perform a DFT on the function cos(6*pi*t) over the range t=0 to t=0.2 sec, with a sampling frequency of 1000Hz, the FFT in MATLAB results in a frequency of 5 Hz. This is due to the length of the sample range (1/0.2 sec = 5 Hz). However if I take the same function and sampling rate, but instead take a sample from 0 to 5 seconds I would get the actual frequency of the function (3Hz), but not one related to the sample length (1/5=0.2 Hz)? Can someone explain how I can determine whether I have an adequate sample rage and rate to correctly perform a FFT?
The Fourier-transformation (and FFT) supposes periodical infinite prolongation of observed function with period Tp equal to observed interval length. In the case of prolongation period is differ from real function period T0=1/f0 You get strange results.
The main rules to select true observation perid are:
- Tp=n*T0 (n-integer) n>1, if T0 is known;
- or
- Tp>>T0, if T0 is unknown. In addtion You needs to use Window function
Here is illustration of Your's situation
https://en.wikipedia.org/wiki/Window_function
James,
Basically, the result you get, is the result of 2 operations you do one after the other:
(1). A windowing operation. You multiply the function to be analyzed with a rectangular window, that is one in the range 0-0.2 sec or 0-5 sec, and zero outside. This windowing is done 'implicit'. You dont see it, it is part of the way you compute the FFT, but to analyse the result, it should be made explicit.
(2.) The actual FFT.
Now, what happens is that the windowing operation distorts your FFT result. Basically, signal processing wise, the frequency response you measure is the result of the convolution of the frequency response of the window with the frequency response of the signal.
Now, the frequency response of the window has sidebands. These sidebands are more pronounced if the window is smaller, and if its edges are sharper. It are these sidebands that cause the distortion you see.
In order to combat this effect, you need :
(1.) Make the analysis interval sufficiently long.
(2.) Select an appropriate window. A rectangular function gives usually quite poor results (even with the 5 seconds window). Follow the link in the answer above, and read there. The window is a compromise between main lobe width (determining resolution), and sidelobes suppression (determining spurious interaction.) I dont give any recommendation on how wide the FFT length should be, or which window is best, as this depends on the use you want to make of the data.
Best Regards,
Henri.
Following may help
Ref Roberto Cristi Modern Digital Processing P126 :This is due to loss of resolution based on samples N over 2pi (ie window of signal).The digital freq w0 =2pi*F/Fs (F is analog signal freq Fs =sampling freq ) lies in between +- 2pi/N . The reason is DFT is basically sampling the freq domain based on length of data . Uncertainty is given by delta F = +-1/T where T= data length in seconds which is independent of Fs.
Good Luck
Cheers
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