CODEWITHSUNDEEP
matlabnumerical-integration
gallamine
gallamine

Reputation: 875

Double integral in MATLAB

I'm trying to find the power in an aperture from a Gaussian beam, where the aperture is offset from the beam center. The solution is the following equation (reference) (sorry, no LaTeX here):

enter image description here

Wz is a constant, along with a and r. I'm not sure how I can do something like this with MATLAB. Does anyone have a suggestion? I know there's a dblquad() function, but it assumes that the limits of integration are fixed, and not dependent on each other.

Upvotes: 2

Views: 7540

Answers (4)

Marc
Marc

Reputation: 5625

Generally speaking, for numeric integration, you can transform an integral with dependent boundary conditions to one with independent boundaries by multiplying by a function that is 1 if you are inside the original boundary and 0 if you are outside. Then take your limits to be a square that contains your original conditions. In other words here you would multiply by

g(x,y) = ((x^2 + y^2) < a^2)

and your limits would be -a

You have to be a little careful about the continuity assumptions in your integration method, but you should be OK unless something is very weird. You can always check by changing your cell size and making sure the computed integral value doesn't change.

In this particular case, you could also make the transformation from cartesian to polar coordinates;

x = rcos(t)

y = rsin(t)

dxdy = rdrdt

Then your limits of integration would be r from 0 to a and t from 0 to 2*pi

Upvotes: 0

gallamine
gallamine

Reputation: 875

It turns out that more recent versions of MATLAB now have a quad2d() function, which does a 2d integral over a surface. Example 2 on the reference page details an example of doing this type of integration.

My code ended up looking something like this:

powerIntegral = @(x,y) 2/(pi*W^2)*exp(-2*((x - offsetDist).^2 + y.^2)/(W^2)); 
ymin = @(x) -sqrt(radius.^2 - x.^2);
ymax = @(x) sqrt(radius.^2 - x.^2);
powerRatioGaussian = quad2d(powerIntegral,-radius,radius,ymin,ymax);

Pretty nifty. Thanks for the help.

Upvotes: 3

You
You

Reputation: 23824

Using a bit of mathematical footwork, you could reduce the double integral to a single one (albeit containing the error function) which should be easier to calculate numerically in MATLAB:

enter image description here

(With reservation for errors; check the calculations yourself if possible.)

Upvotes: 3

Andrey Rubshtein
Andrey Rubshtein

Reputation: 20915

I am not sure, but I think that the symbolic toolbox can help you here. It is suited for this kind of problems. You can define your variables as symbolic vars using the syms command, and compute the integral symbolically. Then, you can assign the variables values and find the actual value.

Disclaimer : I have never actually used it myself.

Upvotes: 1

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