Difference 'bandpower' with time-dependent signal or 'bandpower' with PSD

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Inti Vanmechelen 2020-7-13

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/564050-difference-bandpower-with-time-dependent-signal-or-bandpower-with-psd

评论： Inti Vanmechelen 2020-7-13

采纳的回答： Deepak Gupta

在 MATLAB Online 中打开

Hi all,

I have a question with respect to matlab's built-in function "bandpower".

The power of a certain signal can be calculated starting from a time-dependent signal, or from a PSD outcome.

In both cases, the result is the same:

t = 0:0.001:1-0.001;
fs = 1000;
x = cos(2*pi*100*t)+randn(size(t));
p = bandpower(x)
p = 1.4368
[pxx,f] = periodogram(x,rectwin(length(x)),length(x),fs);
p2 = bandpower(pxx,f,'psd')
p2 = 1.4368

However, when adding a frequency range (since I am interested in the average power in a specific frequency range), the result is not the same with the two methods:

freqrange = [0,5];
p = bandpower(x,fs,freqrange)
p = 0.0122
[pxx,f] = periodogram(x,rectwin(length(x)),length(x),fs);
p2 = bandpower(pxx,f,freqrange,'psd')
p2 = 0.0136

Any suggestion as to why these results are not equal would be greatly appreciated!

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Answer 1

Deepak Gupta 2020-7-13

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/564050-difference-bandpower-with-time-dependent-signal-or-bandpower-with-psd#answer_464984

在 MATLAB Online 中打开

Hi Inti,

Bandpower function by default uses hamming window and in the psd calculation you have used rectangular window which is creating the difference in two methods so if you use same windows in both methods, results should be same.

t = 0:0.001:1-0.001;
fs = 1000;
x = cos(2*pi*100*t)+randn(size(t));
freqrange = [0,5];
p1 = bandpower(x,fs,freqrange)
[pxx,f] = periodogram(x,hamming(length(x)),length(x),fs);
p2 = bandpower(pxx,f,freqrange,'psd')