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.

Answer 1

[Walter Roberson]

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.

For the non-uniform case, see http://www.mathworks.com/matlabcentral/answers/260083-frequency-to-time-domain-conversion#answer_203071

[Ken W]

Thank you.

If my data are equally spaced but with some missing data, can i apply the FFT?

Attached is my data.

[Walter Roberson]

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.