To scale values from Matlab's spectrogram to real range

Question

I run

spectrogram(data, 64, 60, 256, 2,  'yaxis');

I get

enter image description here

I want to get range from 0 to 180 Hz, instead of normalised range from 0 to 1. I can change axis by

axis([0 450 0 180])

but it does not change the values.

I run unsuccessfully

spectrogram(180 .* data, 64, 60, 256, 2,  'yaxis');

How can you scale the range from normalised to real range?

[`spectrogram`](http://www.mathworks.es/es/help/signal/ref/spectrogram.html) actually has the time as y-axis and the frequency as x-axis. So are you sure you haven't just labeled your axes wrong? — Robert Seifert, Dec 23 '13 at 17:28
@thewaywewalk You are right. I am plotting the fourier transform of the data, instead of time and frequency. — Léo Léopold Hertz 준영, Dec 23 '13 at 17:38

lennon310 · Accepted Answer · 2013-12-23T17:47:54.853

1

The x-axis should be frequency. You need to know the sampling rate fs (is it 2 ?) of data. Then the frequency range will be (0:63)/64*fs

Actually you can get the frequency output in the spectrogram function:

[B,F,T]=spectrogram(data,64,60,256,2);  % if fs = 2 in your case

F is the frequency range.

edited Dec 23 '13 at 17:47

answered Dec 23 '13 at 17:29

lennon310

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How can you plot time-frequency plot from those points in F and T? I get 129x1 double for F and 1x218 double for T. – Léo Léopold Hertz 준영 Dec 23 '13 at 17:39
use imagesc(abs(B)), and the F is for the column, while T for the row – lennon310 Dec 23 '13 at 17:41
**imagesc(T, F, abs(B))** seems to be working. Still the problem with range. Values are from [0,1], although should be [0,180] Hz. – Léo Léopold Hertz 준영 Dec 23 '13 at 17:44
F*fs is your actual frequency, (2*F) in your case. see my answer on (0:63)/64*fs. They are equivalent. But the problem is, your sampling rate is only 2 Hz, so you won't get 180 Hz signal. – lennon310 Dec 23 '13 at 17:47
Thank you very much for explanation! I changed **fs = 360** so I get **180 Hz** max, which I know should be the maximal frequency. – Léo Léopold Hertz 준영 Dec 23 '13 at 17:50
I need to now rotate the picture to get peaks from up to down. I see now this: https://dl.dropboxusercontent.com/u/62073194/Screen%20Shot%202013-12-23%20at%2018.52.35.png Do you have any idea? Multiplying by **-1** put also minus -mark on the y-axis which I do not want. – Léo Léopold Hertz 준영 Dec 23 '13 at 17:53
what do you mean by that because the peaks are already on the top? You can rotate the matrix by imagesc(abs(B')), I'm not sure whether that is what you want. – lennon310 Dec 23 '13 at 17:57
It puts those peaks to the left-hand-side: https://dl.dropboxusercontent.com/u/62073194/Screen%20Shot%202013-12-23%20at%2020.21.45.png I want those peaks down. – Léo Léopold Hertz 준영 Dec 23 '13 at 19:22
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imagesc(abs(B(end:-1:1,:))) – lennon310 Dec 23 '13 at 19:24
Thank you very much for your support and answers in this issue! – Léo Léopold Hertz 준영 Dec 23 '13 at 19:31
I am thinking now an extension of the problem but for Wigner-Ville distribution here: http://stackoverflow.com/questions/20750660/to-scale-values-from-wigner-ville-distribution-to-real-range-in-matlab I have apparently a problem in initialising windows. – Léo Léopold Hertz 준영 Dec 24 '13 at 15:04

To scale values from Matlab's spectrogram to real range

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