# Normalizing scilab's FFT

I want to normalize the FFT used in scilab in a way so that the absolute values of the coefficients equal to the amplitudes of the time domain signal with that frequency.

Example: I want an input sine [0, 5.5, 0, -5.5] to transform to [0, 5.5, 0, 5.5] (absolute values).

For that I used some very simple sines and cosines with an amplitude of 1, and transformed them to see what the fft yields. N is the sample size.

// cos, N = 4
-->fft([1  0  -1  0])
ans  =
0.    2.    0.    2.    // looks like I have to divide by N/2

// sin, N = 4
-->fft([0  1  0  -1])
ans  =
0  - 2.i    0    2.i    // same conclusion

// sin, N = 8 (2 periods)
-->abs(fft([0  1  0  -1  0  1  0  -1]))
ans  =
0.    0.    4.    0.    0.    0.    4.    0.      // same conclusion

// cos, N = 8
-->abs(fft([1  0.7  0  -0.7  -1  -0.7  0  0.7]))
ans  =
0.    3.98    0.    0.02    0.     0.02    0.    3.98  // same conclusion


But:

// cos, N = 2
-->abs(fft([1  -1]))
ans  =
0.    2.      // looks like I have to divide by N

// cos, N = 1
-->abs(fft([1]))
ans  =
1.      // looks like I have to divide by N


The N/2 rule also applies to much larger values for N, but not for N = 2 and N = 1. Is this mathematically explainable, or is scilab's FFT scaling just arbitrary? Or am I missing something basic?

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## 1 Answer

It's not a matter of $N$ being $1$ or $2$. What's going on is that for most of your examples, the FFT has two nonzero entries at $k$ and $N-k$, where $k$ corresponds to the frequency of your sinusoid. But when $k=0$ (your last example) or $k=N/2$ (your second-last example), these two entries coincide, so the value appears to be twice what you expect.

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I had to read your answer 5 times, but when I looked at the FFT results, the scales fell from my eyes. Thank you! –  dialer Dec 26 '12 at 16:36