How to understand spectrogram function

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Mark Golberg
Mark Golberg 2016 年 1 月 31 日
回答済み: Lazaros Moysis 2023 年 3 月 14 日
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

採用された回答

Vidya Viswanathan
Vidya Viswanathan 2016 年 2 月 4 日
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 件のコメント
Mark Golberg
Mark Golberg 2016 年 2 月 24 日
Thank you Vidya Viswanathan!!!
Mohannad suleiman
Mohannad suleiman 2021 年 2 月 27 日
dear vidya how can we show the (power / frequency ) in the figures

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その他の回答 (5 件)

Jonathan Grimbert
Jonathan Grimbert 2018 年 5 月 23 日
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 2018 年 12 月 18 日
Is it possible to get rid of "colorbar", which is automatically there with "spectrogram" command ?
  1 件のコメント
Pawan Sharma
Pawan Sharma 2019 年 2 月 12 日
%% 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 2021 年 2 月 27 日
dear vidya how can we show the (power / frequency ) in the figures
  1 件のコメント
Aditya Ramesh
Aditya Ramesh 2021 年 12 月 1 日
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 2021 年 8 月 25 日
You can also have a look here --> https://github.com/Christoph-Lauer/Sonogram

Lazaros Moysis
Lazaros Moysis 2023 年 3 月 14 日
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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