How to perform a zoom FFT on certain portion of a signal

I have a signal with a sample rate of 200HZ, see attached. I am interested in a certain portion of the signal, so I zoomed in that section from the original signal (see figure below). I want to compute FFT (maybe zoom FFT) on that portion to see the frequencies within that region. Please I need help, thank you.

[Matt Gaidica]

@Star Strider has you covered on the FFT. I would offer that unless your hypothesis is that this phenomenon is rhythmic, an FFT may not be what you want. If your question is, rather, about how many times does my subject spawn over some amount of time, I think you would want to identify peaks in your data and come up with a mean rate (i.e. occurences / time). See also peakseek.

[Austin Ukpebor]

Matt, your contributions on this platform and privately are a great deal to the success of my project. Thank you.

Answer 1

[Star Strider]

First, see if the pspectrum funciton using the 'spectrogram' option will do what you want. It will segment the Fourier transform as a function of time as well as frequency. You can then easily see how the frequency content changes over time.

If that does not do what you want, please be a bit more specific.

[Austin Ukpebor]

Thank you, Star. The pspectrum function without option produced exactly what I want. But I still need your assistance here:

1. Please how can I change the power spectrum axis to amplitude in millimeter? My data is millimeter.
2. I can't relate the fres = 195.3602 mHz in the graph. Please help explain.

D = csvread('sensor4.csv');
Fs = 200;                       % Sampling Frequency
Fn = Fs/2; 
pspectrum(D, Fs);               %Compute the power spectrum of the signal.
pspectrum(D, Fs,'persistence'); %Compute the persistence spectrum of the signal.

When I computed the Pspectrum spectrum of the signal, I could see both signal components visibly. Again, I could not relate these two parameters - Fres = 782.0137 mHz, Tres = 25.93 s. Please shed more light on this.

Thank you.

[Austin Ukpebor]

Okay, I will take a look at those links.

I actually wanted the amplitude in mm. Is there anything I can do to achieve this?

Thank you.

[Star Strider]

My pleasure!

Power is amplitude², so taking the square root should produce the absolute value of the amplitude. I am not certain how to be certain of the units. The file you posted does not have units associated with it (that I can determine), so while power here is likely mm², (units are irrelevant to the analysis routines), taking the square root should return amplitude units as mm. Note that these will be the amplitudes of the deflections, ignoring whatever the constant offset or low-frequency baseline variations are. (Those would be accounted for in the 0 Hz and low-frequency signals in the spectrum.)

The ‘Fres’ and ‘Tres’ results are the frequency resolution and time resolution, respectively, of the analysis. The documentation explains it better than I can. See the documentation sections on FrequencyResolution and TimeResolution for the necessary details.

[Austin Ukpebor]

I have been trying to analyze my graph, I do not expect the frequencies to peak at the same level based on the system I am studying. See graph below. Maybe I am doing something wrong. Are there other techniques to analyze the frequency components of my data?

We used hall-effect sensors with a combination of magnet to measure the gaping (opening and closing) of oyster shells. The collected data is what I am analyzing. Those peak frequencies are the freq at which the oyster spawns.

Thank you.

[Star Strider]

I do not know enough about oyster physiology to comment on that plot. However, it would appear to be a fundamental frequency of 20 Hz (which seems to me to be quite fast for a physiologic process other than perhaps nerve or muscle membrane depolarization), and the rest are harmonics of the fundamental frequency, every 20 Hz.

I would be tempted to do something like this:

D1 = readmatrix('sensor4.csv');
Fs = 200;                                               % Sampling Frequency (Hz)

figure
[p,f,t] = pspectrum(D1, Fs, 'spectrogram');
freq = 20;
p20 = interp1(f,p,freq);
plot(t, p20)
grid
xlabel('Time (s)')
ylabel('Power')
title(sprintf('Signal at %6.1f Hz As Function Of Time', freq))

That plots the 20 Hz power ‘slice’ of the pspectrum 'spectrogram' as a function of time. (The choice of frequencies is arbitrary, and is controlled by the third argument (‘freq’) of the interp1 call.)

.

[Austin Ukpebor]

You are absolutely right, those are harmonics. I learned a lot from your contributions. Thank you.

[Star Strider]

As always, my pleasure!