Fourier transform of oscillating signals

Ankur,

First, it would be good to separate the structural code from the analysis code - just to make it easier to read and to help you in debugging. It would also be helpful to comment the code so that someone could tell what it is you're trying to do at each step.

Finding the peak of the fft energy is probably an OK way to find the fundamental frequency. Once you have that, it should be easy to calculate the period as long as you know the frequency scale to begin with.

The frequency index I at which you find the peak represents:

fp=(I-1)*fs/N Hz

so, as a check, the maximum frequency that you can find is (N-1)*fs/N or fs-fs/N which is the expected result. Of course, since the data is sampled, the actual peak will be below fs/2 and will repeat above fs/2.

The period of the fundamental estimated this way is:

So much for the period in #1.

I don't understand your objective #2. It makes no sense to me. Perhaps you can clarify. Why is it different than #1?

You didn't address #3. You might A) remove the mean so the result is zero mean. then B) compute a suitably short-term average of the signal using an odd number of elements and find where the zero-crossings are relative to the center of the averaging filter.

Fred

I am appending comments about the codes used below.

Nc = length(Yftc); % Calculates the number of samples in the controlled signal N = length(Yft); % Calculates the number of samples in the chaotic signal

Yftc(1) = []; % Removes the first element (mean value) of the FFT transformed signal Yft(1) = []; % Removes the first element (mean value) of the FFT transformed signal

powerc = abs(Yftc(1:Nc/2)).^2; % Calculates the power of the controlled signal power = abs(Yft(1:N/2)).^2; % Calculates the power of the controlled signal

nyquistc = 1/2; % Dont know why this stamement is used?? Copied from the documentation file nyquist = 1/2; % Dont know why this stamement is used?? Copied from the documentation file

freqc = (1:Nc/2)/(Nc/2)*nyquistc; % Calculates the frequencies of the controlled signal freq = (1:N/2)/(N/2)*nyquist; % Calculates the frequencies of the chaotic signal

periodc = 1./freqc; % Calculates the period of oscillations for the controlled signal period = 1./freq; % Calculates the period of oscillations for the chaotic signal

************************************************

As for the statement #2: I wanted to find the frequency component of the signal if its a period 2 oscillation, or higher. So I guess the power spectrum would show 2 peaks if its a period 2 oscillation. In that case, would the fundamental frequency still give me the appropriate time period.

For statement #3: I didn't understand your statement: "compute a suitably short-term average of the signal using an odd number of elements and find where the zero-crossings are relative to the center of the averaging filter". Could you please elaborate it further.

Thanks for your help and suggestions.

I don't know what you mean by a "period 2 oscillation"

nyquist = 1/2 is the normalized case where we assume that fs=1 and fs/2=1/2 and fs/2 is always the nyquist frequency.

So, it appears if you were to use nyquist=1000 for fs=2000 that would fit into the scheme of the code. Or, whatever numbers suit your implementation.

I don't think you *need* to remove the mean after the fft but you may want to remove the mean *before* the fft depending on what you intend to do next.

My comments about finding the location of the zero crossings apply to processing in the time domain.

Fred

Thanks Fred for your reply.

Period 2 oscillations are those type of oscillations which has 2 distinct loops in a phase phase, unlike a period 1 oscillation (more commonly known as the limit cycle) which has a single loop in the phase plane. Higher period oscillations would have similar higher number of loops in the phase plane.

Regards Ankur

In other words it's a chaotic systems thing. One of the 'routs to chaos' as a system's gain goes up is the bifurcation of the period, until finally things just go to hell.