Greetings dear friends, I am trying to graph this integral so that I can obtain a graph x =f(t), thanks for your help!
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![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/906965/image.png)
I need to get this curve in my 3D graph:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/906970/image.png)
My code which I was working is this:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/906975/image.png)
Thank you dear friends!
댓글 수: 4
Paul
2022년 2월 26일
How does exp(i*p*x) in the equation become just cos(p*x) in the code? Unless of course only the real part of the integral is goal.
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Voss
2022년 2월 25일
편집: Voss
2022년 2월 25일
Note that t > 0 and the grids on the surface in the desired image are more widely spaced than the actual points where the surface has been calculated (i.e., there is curvature in between grid lines).
t = linspace(0.0001,2,50);
x = linspace(-2,2,50);
[T,X] = meshgrid(t,x);
for i = 1:numel(x)
for j = 1:numel(t)
% f = @(p)1/pi*sqrt(2*pi)*sin(p)./p.*exp(-T(i,j).*p.^2).*cos(p.*X(i,j));
f = @(p)1/(pi*sqrt(2*pi))*sin(p)./p.*exp(-T(i,j).*p.^2).*cos(p.*X(i,j));
F(i,j) = integral(f,-Inf,Inf);
end
end
surf(T,X,F);
colormap(flip(autumn()));
xlabel('t');
ylabel('x');
zlabel('u(x,t)');
댓글 수: 2
Torsten
2022년 2월 25일
f = @(p)1/(pi*sqrt(2*pi))*sin(p)./p.*exp(-T(i,j).*p.^2).*cos(p.*X(i,j));
instead of
f = @(p)1/pi*sqrt(2*pi)*sin(p)./p.*exp(-T(i,j).*p.^2).*cos(p.*X(i,j));
Voss
2022년 2월 25일
Oh yeah! Thanks!
I saw your comment before, but then I just typed in the code from the screenshot in the question anyway. D'oh!
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