1
$\begingroup$

Given an hidden law $$I(t)=\frac{1}{2}[A+B.cos(2\pi\nu_0t+\phi_0)]$$ I observe its signal with 256 samples and compute its FFT over this interval, and represent the FFT result from 0 to 1 (normalised frequency).

I get a figure whose pic is at 0.83.

How can I retreive A,B and $\phi_0$ from this graphics ?

A second question is : is it normal that this figure is not symmetric ?

A final question is : doing an analytical computation on classical Fourier transform gives a result. How much does what I observe from FFT and the analytical result differ ?

| improve this question | |
$\endgroup$
  • $\begingroup$ Is this homework? If so, you should specify it using the homework tag. It seems like that to me, so here are some hints: (1) A is like a DC component. Its frequency is 0. (2) A real signal's FT is symmetric. A complex signals doesn't have to. (3) A FT has both magnitude (which I believe you are plotting) and phase. $\endgroup$ – Serge Mar 8 '13 at 16:58
  • $\begingroup$ it is not a homework per se. $\endgroup$ – JCLL Mar 8 '13 at 17:02
  • 1
    $\begingroup$ How closely spaced are the $256$ samples? For best results, take the sample spacing to be $(256\nu)^{-1}$ seconds so that the $256$ samples fill up the time interval of one period, e.g. $\[0,\nu^{-1})$. Note that the $257$-th sample will occur at $t = \nu^{-1}$ and is not part of the data set. See, for example, this answer to understand why not. $\endgroup$ – Dilip Sarwate Mar 8 '13 at 17:19
2
$\begingroup$

If you feed an FFT a finite set of samples from a function of theoretically infinite range, you have implicitly windowed your function with a rectangular window before the FFT. Thus, the FFT result will be the convolution of the spectrum of your function and the transform of a rectangular window (a periodic Sinc). This convolution will be an identity operation only if the samples happen to span exactly an integer multiple of periods of a purely periodic function. So how much your FFT result will change or differ will depend on the offset and length of your FFT aperture/window with respect to some reference point on and the scale of your function.

In the case of a single sinusoid in zero noise, I reference 2 papers by Clay Turner on solving for the 3 unknowns using only 3 or 4 non-aliased samples, somewhere on my web page: http://www.nicholson.com/rhn/dsp.html No FFT needed.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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