Question about converting integral formula to matlab code
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I'm trying to code an integral formula as a sum of discrete data.
That seems correct, but the result is different.
Please let me know if there are any changes to my code.
Please let me know if there is a new way.
The code I made is below.
~~~~~~~~~~~~~~~~~~~~
for r = 1:M
for p = 1:N
for f0 = 1:M0
for p0 = 1:N0-1
PSI(r,p) = PSI(r,p)+...
(1/(pi^2*Rho0^2))*...
K0(f0)*...
d_K0*d_Phi0_rad*...
U(f0,p0)*...
(Rho0*(Rho0-Rho(r)*cos(Phi_rad(p)-Phi0_rad(p0)))/(Rho0^2-2*Rho0*Rho(r)*cos(Phi_rad(p)-Phi0_rad(p0))+Rho(r)^2))*...
(norm([Rho(r)*cos(Phi_rad(p)) Rho(r)*sin(Phi_rad(p))]-[Rho0*cos(Phi0_rad(p0)) Rho0*sin(Phi0_rad(p0))])^2)*...
(exp(2*1i*K0(f0)*((norm([Rho(r)*cos(Phi_rad(p)) Rho(r)*sin(Phi_rad(p))]-[Rho0*cos(Phi0_rad(p0)) Rho0*sin(Phi0_rad(p0))]))-Rho0)));
end
end
end
end
6 Comments
Torsten
on 23 Sep 2022
Any reason why you don't use "integral2" ?
형준 이
on 23 Sep 2022
Torsten
on 23 Sep 2022
I think you should set up your problem for the use with "integral2".
형준 이
on 23 Sep 2022
형준 이
on 23 Sep 2022
Star Strider
on 23 Sep 2022
It may be necessary to use integral iteratively with the 'ArrayValued' flag set if you are integrating an array. Check the inner integral before calculating the outer integral to be certain that it is doing what you want.
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