# 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

### 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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