Time resolution of Spectral Entropy - How could I modify it?

Hello everyone,

I have a signal (a timetable of 307,200 data points) with sampling rate of 20480 Hz (0.00005s) and total length of 15.0 seconds.

When I apply the command pentropy I get a time vector te with length 500 points, equivalent to a time resolution of 0.03 seconds

I confirmed this value as shown below. The problem is that I need the spectral entropy of the signal every 0.02 seconds and every 0.05 seconds.

Is any way where I can adjust or define this "time resolution"?

[se,te] = pentropy(Datos01.Sensor1,Fs)
mean(diff(te))

ans =
     0.02997

Could someone help me? Thanks

P.S. I was planning to use the retime command and @pentropy as input but it sends me an error. See Below

Other alternative would be using a for and a moving window but not sure how to code it

DatosA = retime(Datos01,"regular",@pentropy,"TimeStep",seconds(windowLength));

% Error using timetable/retime (line 140)
% Applying the function 'pentropy' to the 1st group in the variable 'Sensor01' generated the following error:
 
% Expected input argument 2 to be time information in the form of a numeric scalar as sampling frequency, a duration
% scalar as sampling time or a numeric/duration/datetime array as time values.

Answer 1

[Yazan]

You need to provide the spectrogram to the Matlab function, and this spectrogram should have your desired time resolution.

See the example below.

clc, clear

fs = 20480;
f = 0.01*fs;
ts = 1/fs;
t = 0:ts:15-ts;

x = cos(2*pi*f*t);

% compute power spectrogram with time resolution equal to 0.02 sec
% note that the overlap between the spectrogram windows is set to zero
% if you introduce overlap between the windows, the time resolution should
% be changed to guarantee that 'tp' is sampled every 0.02 sec
[p, fp, tp] = pspectrum(x, fs, 'TimeResolution', 0.02, 'OverlapPercent', 0, 'spectrogram');

% compute spectral entropy
[se, te] = pentropy(p, fp, tp);

fprintf('Spectral entropy estimated every %g sec\n', mean(diff(te)));

Spectral entropy estimated every 0.0199707 sec

[John Navarro]

My new question about it at: Identification of fundamental frequency and not harmonics in a FFT - using peak locations - MATLAB Answers - MATLAB Central (mathworks.com)