Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a wave that is a sum of sines and cosines:

$$x = A\sin(\omega t + \phi_1) + B \cos(2\omega t + \phi_2) + C\sin(2\omega t + \phi_3) + D\cos(2\omega t + \phi_4).$$

Now I use fft on $x$ and get the magnitude with abs(fft(x)). How do I get $A$, $B$, $C$, and $D$ back? The reason behind this is that I am new to fft and I am trying to understand the output that Matlab fft gives back in depth. Thank you PS: I can not use ifft in any of this. Also as you can tell I am new to FFT, any of you can recommend a good source to start with ? Thank you so much for your time...

share|cite|improve this question
Are you sure you want the $B$, $C$ and $D$ terms to all have the same frequency? Presumably the frequency is meant to increase linearly? In the present formulation the six constants $B,C,D,\phi_2,\phi_3,\phi_4$ can be collapsed into two (and hence in particular can't be reconstructed from $x$ or its Fourier transform). Also it's an unnecessary complication to write some of the sinusoidals as sines and others as cosines; you can write them all as cosines and shift the $\phi_i$ accordingly. – joriki Feb 18 '13 at 18:58
Thank you for responding, I guess the sin and cos can be changed to be all either sin or cos no problem, I guess my main thing is to get the correct way of extracting info from fft output, any suggestions on how to go about this? thank you again :) – Amani Lama Feb 18 '13 at 19:02
@Amani: That question can't be separated from the issues I addressed. No comment on those? – joriki Feb 18 '13 at 19:07
up vote 1 down vote accepted

Use fft(signal,M) with M a large power of 2 (e.g. 2^13). Multiply it with 2/N, where N is the number of samples in your signal: N=numel(signal). If you now plot the abs(), then the peak heights are an estimate of the amplitude of the corresponding periodic signal with that particular frequency.

An example for three signals with different amplitudes and frequencies:

close all


x = cos(2*pi*t) + .8*sin(4*pi*t) + .5*cos(6*pi*t);

M=2^13; % take a large power of 2

xh_amplitudes = 2*xh/N; % amplitude estimate in fft of x


xlabel f
ylabel amplitude

If you zoom in, however, you will see the peaks are not really on the actual amplitudes, but this is probably due to the leakage from the other peaks. Are these estimates good enough for you?

share|cite|improve this answer
Thank you very much, I will look into this and see if I understand it completely... – Amani Lama Apr 30 '14 at 15:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.