How to find the energy of a discrete signal?

35 次查看（过去 30 天）
RoBoTBoY
RoBoTBoY 2021-1-24
回答： Shubham Khatri 2021-2-5
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 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2021-1-25
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'.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Shubham Khatri
Shubham Khatri 2021-2-5
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

类别

在 Help Center 和 File Exchange 中查找有关 Discrete Fourier and Cosine Transforms 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by