INTEGARTION OF BESSELH function

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george veropoulos
george veropoulos am 19 Nov. 2024
Kommentiert: Walter Roberson am 22 Nov. 2024
Hi
i make a code where i integrate the hankel function besselh(0, 2)
function z = Escat(r,phi)
[f,N,ra,k0,Z0] =parameter();
[Is]=currentMoM()
Phim=zeros(N+1);
FINAL=0;
for jj=1:N
%Phi0(jj)=(jj-1).*(pi./N);
Phi0(jj)=(jj-1).*(2.*pi./N);
FINAL=FINAL-(k0.*Z0./4).*Is(jj).*ra.*integral(@(xx)besselh(0,2,k0.*sqrt(ra.^2+r.^2-2.*r.*ra.*(phi-xx))),Phi0(jj),2*pi/N+Phi0(jj));
end
z=FINAL;
end
when i plot the function in the range φ=0 : 2π with r=const the valus of Escat are extremely big value ...
clear all
clc
[f,N,ra,k0,Z0] = parameter();
%ph_i=pi/2;
rho=10.*ra ;
phi=0:pi/180:2*pi;
Es=zeros(length(phi));
%phi=0:pi/200:pi/3
for jj=1:length(phi)
Es(jj)=abs(Escat(rho,phi(jj)));
end
plot(phi*(180./pi),Es,'b--')
hold on
%plot(phi*(180./pi),abs(integ),'b--')
plot(phi,abs(Escattheory_new(rho,phi)),'r-')
xlabel('$\phi$','Interpreter','latex')
ylabel('$|E_{s}|$','Interpreter','latex' )
%legend('MoM', 'Theory')
hold off
can i check if the the integration is ok ?
the parameter fucntion is
function [f,N,ra,k0,Z0] = parameter()
%UNTITLED Summary of this function goes here
c0=3e8;
Z0=120.*pi;
ra=1;
N=80;
f=300e6;
lambda=c0./f;
k0=2*pi./lambda;
end
thank you
  6 Kommentare
Steven Lord
Steven Lord am 20 Nov. 2024
If you're adding information, please put it as a comment rather than a separate answer. Leave the answer for posts that may be a solution to the problem, to make them easier to distinguish from commentary.
Walter Roberson
Walter Roberson am 22 Nov. 2024
Note that function e_n leaves y undefined in the case where k is NaN.
If k cannot be NaN then there is no point in using elseif -- just use else

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Akzeptierte Antwort

David Goodmanson
David Goodmanson am 22 Nov. 2024
Bearbeitet: David Goodmanson am 22 Nov. 2024
Hi george,
I don't know what the values of your parameters are, but it appears that the problem with Escat is that
besselh(0,2,k0.*sqrt(ra.^2+r.^2-2.*r.*ra.*(phi-xx)))
is missing a cosine factor and should be
besselh(0,2,k0.*sqrt(ra.^2+r.^2-2.*r.*ra.*cos(phi-xx)))
Without the cosine, it's possible for the argument of the sqrt to become negative, which causes the argument of besselh to become imaginary, which leads to very large values in the result.
  1 Kommentar
george veropoulos
george veropoulos am 22 Nov. 2024
thank you very much!!! the cos was the problem

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