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How to find the energy of a discrete signal?

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RoBoTBoY
RoBoTBoY on 24 Jan 2021
Answered: Shubham Khatri on 5 Feb 2021
I want to find the energy of this signal in rolling windows.
Note that the energy of a signal x[n] overlapping by a window w[n] is given by the following formula:
where as signal w[n] I will use the Hamming window given by the equation:
I will use window length N = 1000 samples.
I run the below code but without result.
[y,Fs] = audioread('viola_series.wav');
figure(26);
plot(y);
title('Audio viola series.wav');
sound(y,Fs);
N = 1000;
% Normalization
min(y);
max(y);
y_norm = (y-min(y))/range(y)*2-1;
figure(27);
plot(y_norm);
title('Normalized audio viola series.wav');
syms m
w = hamming(1001);
energy_signal = symsum(conv(y_norm.^2,w),m,0,10);
plot(energy_signal);
How to do that?
Thanks in advance
  3 Comments
Walter Roberson
Walter Roberson on 25 Jan 2021
w = 0.54 + 0.46 * cos(2*pi*(0:n)/N);
E = conv(x.^2, w, 'valid')
or perhaps it should be 'same' instead of 'valid'.

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Answers (1)

Shubham Khatri
Shubham Khatri on 5 Feb 2021
Hello,
You can remove the syms and symsum from the code and directly use convolution. Please take a look at the following code. For more information please refer to the documentation for conv here.
[y,Fs] = audioread('viola_series.wav');
figure(26);
plot(y);
title('Audio viola series.wav');
sound(y,Fs);
N = 1000;
% Normalization
min(y);
max(y);
y_norm = (y-min(y))/range(y)*2-1;
figure(27);
plot(y_norm);
title('Normalized audio viola series.wav');
w = 0.54 + 0.46 * cos(2*pi*(0:n)/N);
energy_signal = conv(x.^2, w, 'valid');
plot(energy_signal);
Hope it helps

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