how to find magnitude and phase of fft output signal?

Hi How can I get the amplitude and phase of fft output signal for specific frequency? let say we have this signal: y=A*sin(w1t+phi1)+B*sin(w2t+phi2) using fast Fourier transform we will get the two frequencies (w1 and w2). How can I get those frequencies and phases?

Answer 1

[David Goodmanson]

Hello Khaled, If you have t=0 as the first element of a uniform time array, then calculating y from that and taking the fft will give two peaks per frequency. For a given frequency w, assuming your normalization is correct, the left complex amplitude [C exp(i phi)] is the positive freq. contribution, to be multiplied by exp(iwt), and the right complex amplitude [C exp(-i phi)] is the negative freq. contribution, to be multiplied by exp(-iwt). Adding these you get the result 2C cos(wt + phi). It's most often stated in terms of cos but of course it is also 2C sin(wt + phi + pi/2).

This all assumes you have fft'd a real function as in your example.

[Khaled Abojlala]

Thanks David For two frequencies it ok. But how can I make it for number of frequencies, let say 10, 20 frequencies?

[David Goodmanson]

First the frequency grid has to be fine enough so the peaks are all resolved. If you do an fftshift on the fft, it puts zero frequency at the center of a new array (at array index (N+1)/2 for N odd, (N/2)+1 for N even). Each pair of peaks corresponding to +-w0 for some frequency w0 is symmetric about the center so it's easy to pair them up. If your peaks start to overlap each other the problem gets harder.

Actually this makes it sound like you need both peaks, positive and negative,to solve the problem. The amplitudes of these are complex conjugates of each other so all you need is one peak with center at w0 and complex amplitude C exp(i phi). (C being real).

[Walter Roberson]

Small correction: the amplitudes of the pairs are complex conjugates iff and only if the inputs were real-valued. If the input signal was complex valued then the two halves are not complex conjugates.

[David Goodmanson]

I had mentioned this in the answer at the top but in an rather offhand way, so thanks for emphasizing this important point.