Fourier Transform: getting mag + phase then using those to plot original signal

Question

Hi guy\'s I\'m working on simple signals and I want to calculate the Fourier transform of a signal, get the magnitude and phases, then reconstruct the original signal from that.

Answer 1

There is no such numerical instability in the FFT. The problem is that roundoff errors give a very small imaginary part in the recovered signal x_i. That's normal. The real part of x_i correctly reproduces x, whereas the imaginary part of x_i is very small. You can check with stem(real(x_i)) and stem(imag(x_i))

Now stem(x_i) with complex x_i plots the imaginary part versus the real part. On the othe hand, stem(1:length(x_i),x_i) (or of course stem(real(x_i))) plots the real part of x_i. Note that this is consistent with plot's behaviour (see also the answer to this question)

So just plot the real imaginary parts of x_i separately with stem. You'll see that the real part reproduces x as expected, and the imaginary part is negligible (of the order of eps).

Answer 2

The problem is caused by round-off errors in phase, so don't use them when calulating the sine and consine of the phase angles. Instead, use trig identities cos(atan(A))=(1+A^2)^(-1/2), and sin(atan(A))=A*(1+A^2)^(-1/2), and so

re = mag .* real(F)./sqrt(real(F).^2+imag(F).^2);
im = mag .* imag(F)./sqrt(real(F).^2+imag(F).^2);

EDIT: I think that if you want to change the phase angle by S, this will do the trick:

re = mag .* (real(F)*cos(S)-imag(F)*sin(S))./sqrt(real(F).^2+imag(F).^2);
im = mag .* (real(F)*sin(S)+imag(F)*cos(S))./sqrt(real(F).^2+imag(F).^2);

EDIT2: You will still sometimes get bad results with non-zero imaginary part (e.g. if S=pi), and you will need to plot either stem(real(x_i)) or stem(1:length(x_i),x_i) as Luis suggested.