清空筛选条件
清空筛选条件

Taking integral of a continuous pulse??? f: y=1, 0<x<1

1 次查看（过去 30 天）
Serhat
Serhat 2013-3-17
Hello,
I am generating a continuous pulse which is equal to 1 (y=1) while 0<x<1 such that:
t=0:0.001:1;
p1=ones(1,numel(t));
Now, I want to take its square integral (integral(f^2)). How can I do that?

请先登录，再回答此问题。

回答（1 个）

Walter Roberson
Walter Roberson 2013-3-17
The square of 1 is 1, the square of 0 is 0, so f^2 is the same as f.
The value of the integral is going to depend on the bounds of integration. Over A to B, A <= B, I figure it is
min(B,1) - min(max(A,0),1)

类别

在 Help Center 和 File Exchange 中查找有关 Electrical Block Libraries 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