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Is this MATLAB example wrong? - Dividing the FFT by the signal length

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azzendix
azzendix 2017 年 5 月 28 日
回答済み: Star Strider 2017 年 5 月 28 日
Someone in this link claim that
the MATLAB example is actually wrong in dividing the fft
by the signal length in the time domain (which is L)
The right scaling needed to adhere to Parseval's theorem would be
dividing the Fourier transform by the sampling frequency.
Is it correct? Should I divide the result from FFT by sampling frequency instead of signal length?
Here is the Matlab example of fft function: https://www.mathworks.com/help/matlab/ref/fft.html

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Star Strider
Star Strider 2017 年 5 月 28 日
The example is correct. The Fourier transform is a sum-of-sines (and cosines), so to normalise the coefficients, divide by the integration time or length of the summation. The division normalises for the total energy in the signal, so that the coefficients of a long (assumed periodic) signal have the same values of a shorter version of the same signal.
See the Wikipedia article on Fourier analysis (link) in the section on the Discrete Fourier transform (DFT) (link) for a full discussion.

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