Fast fourier transform (FFT) data points

Dear All,
I have a set of data with length of 17520 data points. I wish to know is that necessary to follow the rule of 2^n data points (2^14 = 16384). I found that FFT can also transform all the 17520 data points. What are the effects if i transform all the data points? The transform of all data points is equivalent to 2^14 (16384) data points?
I have another data set which not equally spaced. Can FFT applied to this data? Because FFT is applied to equally spaced data.
Please advise me.
Thank you.

 采纳的回答

There is no rule of 2^n data points. Certain computations can be faster if you have 2^n data points, but any number of data points can be used in fft(). The effect of using fft(YourArray,16384) when length(YourArray) > 16384 is the same as using fft(YourArray(1:16384)) -- that is, the remaining data is ignored, so you can get a higher resolution by using fft() of all of the data.

2 个评论

Ken W
Ken W 2015-12-16
编辑:Ken W 2015-12-16
Thank you.
If my data are equally spaced but with some missing data, can i apply the FFT?
Attached is my data.
If there is just the occasional missing value you might want to interpolate it; otherwise you need one of the non-uniform techniques I link to above.

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