Total energy in Time and Frequency Dimain

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I truly appreciated if somebody can direct me how to measure total energy in a certain frequency band?
I have a signal and use this code to measure total enery and the result is reasonable.
E1_timedomain=sum(abs(x.^2))
but it measuers total energy of whole signal. I just want to measure it between [0-100] Hz. Appreciared!

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Star Strider
Star Strider il 12 Dic 2020
If you have R2018a or later, and the Signal Processing Toolbox, use the bandpass function to selectively filter the frequencies-of-interest. Then do whatever calculations you want to do on the output. If you have an earlier version, it is easy to design filters in MATLAB. I will help you with that, if necessary.
  2 Commenti
Maria Amr
Maria Amr il 12 Dic 2020
Star Strider Thank you so much.
As usual, the best and the fast way.BIG like!
Star Strider
Star Strider il 12 Dic 2020
As always, my pleasure!

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Più risposte (3)

Maria Amr
Maria Amr il 14 Dic 2020
Star Strider I have applied a bandpass filter and get a reasonable result. Would you please direct me how may I retrive the filtered signal? Appreciated!
that is my code:
bandpass(amp1,[1 78],fs);

Maria Amr
Maria Amr il 14 Dic 2020
Star Strider Appreciated!
  1 Commento
Star Strider
Star Strider il 14 Dic 2020
As always, my pleaure!
(Also, in the future please post comments as Comments, not Answers.)

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Paul
Paul il 10 Nov 2024
Modificato: Paul il 10 Nov 2024
Define a real-valued, finite duration signal of length N, assumed to be uniformly spaced samples collected as Fs = 2000 Hz
rng(100);
N = 1000;
x = randn(1,N); Fs = 2000; Ts = 1/Fs;
The total energy in this signal is computed in the time domain as
E = sum(abs(x).^2)
E = 1.0063e+03
The total energy computed in the frequency domain is (taking advantage of the symmetry in the frequency domain because x is real-valued)
X = @(f) freqz(x,1,f,Fs);
E = 2*integral(@(f) abs(X(f)).^2,0,Fs/2)*Ts
E = 1.0063e+03
Hence, the energy in the signal between 0-100 Hz is
E100 = 2*integral(@(f) abs(X(f)).^2,0,100)*Ts
E100 = 95.8129
Can also approximate E100 from the DFT samples.
Xdft = fft(x);
fdft = (0:N-1)/N*Fs;
E100 = 2*sum(abs(Xdft(fdft<=100)).^2)/N
E100 = 99.5646
Or better yet
E100 = 2*trapz(fdft(fdft<=100),abs(Xdft(fdft<=100)).^2)/Fs
E100 = 94.7997
Can also apply a very sharp, low pass filter to x and then compute the energy in the filter output as a rough approximation
y = lowpass(x,100,Fs,'Steepness',0.95);
sum(abs(y).^2)
ans = 112.4906

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