image plot after is calculating hilbert spectrum
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I have used th matlab function [hs,f,t,imfinsf,imfinse] = hht(imf,fs); to do hilbert haung transform.
The hilbert spectrum (hs) is a sparse matrix. How could this be plotted as an image plot? I tried to convert this to a double matrix using full(hs). But this doesn't give a hilbert spectrum. Its just blank. How to resolve this?
1 Comment
jiayun zhang
on 20 Apr 2020
I am also studying HHT transformation recently. I have met the same problem as you. I want to plot the instantaneous energy spectrum, but I don't know how to achieve it. Have you solved the problem
Answers (1)
Michael Madelaire
on 17 Jul 2019
Edited: Michael Madelaire
on 17 Jul 2019
There shouldn't be any problem. But you have attached no code, so it is hard to know.
Without for information many things could be the problem. Below is a test. There show be no problem in plotting a sparse matrix.
a = diag(1:20); % Create diagonal matrix
b = sparse(a); % Convert into sparse matrix
c = full(b); % Convert sparse matrix back into full matrix
figure();
imagesc(a); title('Original full matrix'); colorbar();
figure();
imagesc(b); title('Sparse matrix'); colorbar();
figure();
imagesc(c); title('Converted full matrix'); colorbar();
I do not know anything about the Hilbert-Huang Transformation. But by taking the initial example from: https://se.mathworks.com/help/signal/ref/hht.html
I converted the sparse matrix into full. No problem
Plotted the sparse matrix. No problem
Plotted the converted matrix (into full). No problem
load('sinusoidalSignalExampleData.mat','X','fs')
t = (0:length(X)-1)/fs;
figure();
plot(t,X)
xlabel('Time(s)')
[imf,residual,info] = emd(X,'Interpolation','pchip');
hs = hht(imf,fs);
q = full(hs);
figure();
imagesc(hs); colorbar()
figure();
imagesc(q); colorbar()
2 Comments
Michael Madelaire
on 17 Jul 2019
It is unclear to me what it is you want illustrated?
imagesc(imfinsf(:,1),f,q);
Does not make any sense.
The imagesc specification states:
"imagesc(x,y,C) specifies the image location. Use x and y to specify the locations of the corners corresponding to C(1,1) and C(m,n). To specify both corners, set x and y as two-element vectors. To specify the first corner and let imagesc determine the other, set x and y as scalar values. The image is stretched and oriented as applicable."
When you say "I directly plot hht(imf,fs); without getting the return values" what does that mean? There has to be a return value otherwise nothing is plotted. Do you mean hs?
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