Numerical integral where the bounds change with each evaluation

Question

L'O.G. 2023-5-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1964624-numerical-integral-where-the-bounds-change-with-each-evaluation

评论： Walter Roberson 2023-5-16

I'm trying to take the following integral:

I tried this symbolically, but I think that was throwing me off. I am attaching some numerical data for the function A(t) and the time vector. How would you do this numerically? I imagine it would be something like the following, but I can't quite get it to work.

omega = logspace(-1,1,1000);
fun = @(t,omega) A*sin(omega*t);
for i = 1:1000;
    w=omega(i);
    f(i) = integral(@(t) fun(t,w),0,2*pi/w);
end

3 个评论
显示 1更早的评论隐藏 1更早的评论

FannoFlow 2023-5-16

Does A(t) have a special meaning in this context?

Walter Roberson 2023-5-16

A(t) just means that A is a function of t.

It is the expression for some kind of tranform of A(t), but I am not sure what the name of this transform is.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2023-5-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1964624-numerical-integral-where-the-bounds-change-with-each-evaluation#answer_1236309

在 MATLAB Online 中打开

A = rand

A = 0.8252

omega = logspace(-1,1,1000);

fun = @(t,omega) A*sin(omega*t);

for i = 1:1000;

w=omega(i);

f(i) = integral(@(t) fun(t,w),0,2*pi/w);

end

plot(omega, f)

All of the values are within round-off error of 0.

However... there is a difference between what your mathematical expression shows here, compared to the integral you were calculating there and here.

In previous discussion, A was a constant. In the expression here, A is a function of t. That makes a big difference.

For example,

A = @(t) t.^2 - t + 1;

omega = logspace(-1,1,1000);

fun = @(t,omega) A(t).*sin(omega*t);

for i = 1:1000;

w=omega(i);

FUN = @(t) fun(t,w);

f(i) = integral(FUN,0,2*pi/w);

end

plot(omega, f)

syms t Omega

A = @(t) t.^2 - t + 1;

omega = logspace(-1,1,1000);

fun = @(t,omega) A(t).*sin(omega*t);

F = int(fun(t,Omega), t, 0, 2*pi/Omega)

F =

f = subs(F, Omega, omega);

plot(omega, f)

4 个评论
显示 2更早的评论隐藏 2更早的评论

Walter Roberson 2023-5-16

"how would you modify it to use a vector of type double which is also a function of t"

Vectors are not functions.

If you have the value of A(t) sampled at particular t, then calculate A(t).*sin(omega*t) at those t, and use trapz() or similar to do numeric integration, making sure to pass in the appropriate t values to trapz()

Walter Roberson 2023-5-16

Hmmm, first, could you confirm for me that you want omega to range from -1 to 1, and the bounds of integration is 0 to 2*pi/omega -- and since omega ranges from -1 to +1, that means that for negative omega you want negative upper bound, and you want to go right through to infinite upper bound as omega passes through 0 ?

请先登录，再进行评论。