abs(x.^2)
조회 수: 20 (최근 30일)
이전 댓글 표시
Hello everyone
This was my code and i was trying to find peak power of the given sin signal.I dont understand how does it really work mathametically?
clc;f=1;
t=1/f;
x=1*sin(2*pi*f*t);
y=max (abs(x).^2);
댓글 수: 2
Martti Ilvesmäki
2022년 5월 3일
.^2 is element-wise power function: https://se.mathworks.com/help/matlab/ref/power.html
abs() is absolute value and complex value function: https://se.mathworks.com/help/matlab/ref/abs.html?s_tid=doc_ta
x.^2 will raise all elements of vector or matrix x to the second power while abs() will take the absolute value of each element in vector or matrix. Using them both together in this case is unnecessary, because using x.^2 will remove all negative values already.
Walter Roberson
2022년 5월 3일
That is a formula for peak power, but not with that input. For it to work, the input has to be the fft of the signal (after subtracting the mean)
Your x only has a single sample.
That code with sin() looks like the bare outline of an attempt to calculate fft.
채택된 답변
Malar Vizhi
2022년 5월 3일
댓글 수: 3
Martti Ilvesmäki
2022년 5월 3일
% Example
x = [1 2 3];
x_squared = x.^2; % --> [1 4 9]
x_squared_mean = mean(x_squared); % --> 4.6667
추가 답변 (0개)
참고 항목
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!