How to understand spectrogram function

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Hello, can someone explain please (in plain english as much as possible) what's the difference between:
1) [s,w,t] = spectrogram(x,window,noverlap,w) returns the spectrogram at the normalized frequencies specified in w.
and
2) [s,f,t] = spectrogram(x,window,noverlap,f,fs) returns the spectrogram at the cyclical frequencies specified in f.
maybe someone can provide a simple example.
THANK YOU

Risposta accettata

Vidya Viswanathan
Vidya Viswanathan il 4 Feb 2016
The primary difference between the two is in the way the input signal is specified. In the first statement, the signal 'x' is expected to be specified in terms of normalized frequency while the latter uses the actual frequency of the signal in Hz and the sampling frequency. To understand the difference better, consider the following code snippet:
n = 0:999;
x1 = cos(pi/4*n)+0.5*sin(pi/2*n);
fs = 1000;
t = 0:0.001:1-0.001;
x2 = cos(2*pi*125*t)+0.5*sin(2*pi*250*t);
freq = [125 250];
In the above case, both the signals x1 and x2 are basically the same signal but expressed using the two different representations (x1 uses normalized frequency and x2 uses the cyclical frequency).
The spectrogram for the two signals can be visualized using:
[s1,w,t1]=spectrogram(x1,[],8,[pi/4 pi/2]);
figure,
spectrogram(x1,'yaxis')
[s2,f,t2]=spectrogram(x2,[],8,freq,fs);
figure,
spectrogram(x2,[],[],[],fs,'yaxis')
You can notice that the two spectrograms are similar except for a difference in the magnitudes of the power because they are represented in different units. In the first case, the units is dB/rad/sample while the second representation uses dB/Hz.
Computing the power spectral density of the two signals will give you a better insight about the difference. Consider the following lines of code:
[pxx1,w] = periodogram(x1,[],[pi/4 pi/2]);
pxx2 = periodogram(x2,[],freq,fs)
The magnitudes of pxx1 and pxx2 are different only due to the difference in the units. The conversion from one to another follows the relation:
PSD (in normalized frequency)= (PSD (in cyclical frequency) * Sampling frequency) /(2*pi).
  2 Commenti
Mark Golberg
Mark Golberg il 24 Feb 2016
Thank you Vidya Viswanathan!!!
Mohannad suleiman
Mohannad suleiman il 27 Feb 2021
dear vidya how can we show the (power / frequency ) in the figures

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Più risposte (5)

Jonathan Grimbert
Jonathan Grimbert il 23 Mag 2018
thanks Vidya Viswanathan for your answer. However I got a question on the X axis... How do you know it is in ms for the second spectrogram ?
Thanks for your answer

Zulfidin Khodzhaev
Zulfidin Khodzhaev il 18 Dic 2018
Is it possible to get rid of "colorbar", which is automatically there with "spectrogram" command ?
  1 Commento
Pawan Sharma
Pawan Sharma il 12 Feb 2019
%% use this to get rid of the colorbar at the end of your plot syntax
colorbar off

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Mohannad suleiman
Mohannad suleiman il 27 Feb 2021
dear vidya how can we show the (power / frequency ) in the figures
  1 Commento
Aditya Ramesh
Aditya Ramesh il 1 Dic 2021
colorbar off
This disables the colorbar
colorbar on
This shows the colorbar without the bar label (power/frequency)
Colorbar with the legend label is shown by default even if no colorbar properties are defined. SO basically write nothign after spectrogram(.......) to get what you want.

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Christoph
Christoph il 25 Ago 2021
You can also have a look here --> https://github.com/Christoph-Lauer/Sonogram

Lazaros Moysis
Lazaros Moysis il 14 Mar 2023
Based on your comments and feedback, I want to understand, how could I compute the Spectrogram according to the Bark scale, which is
BandBarks = [20, 100, 200, 300, 400, 510, 630, 770, 920, 1080, 1270, 1480, 1720, 2000, 2320, 2700, 3150, 3700, 4400, 5300, 6400, 7700];

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