INTEGARTION OF BESSELH function

Question

george veropoulos 2024-11-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2167338-integartion-of-besselh-function

评论： Walter Roberson 2024-11-22

采纳的回答： David Goodmanson

在 MATLAB Online 中打开

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

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george veropoulos 2024-11-20

编辑：Walter Roberson 2024-11-22

在 MATLAB Online 中打开

and the cuurentMoM

function   [Is]=currentMoM()
%UNTITLED2 Summary of this function goes here
%   Detailed explanation goes here
[f,N,ra,k0,Z0] = parameter();
gamma_const=1.781;
Phi0=zeros(N);
e=exp(1);
dfpm=2.*pi./N;
for jj = 1:N
    Phi0(jj)=(jj-1).*(2.*pi./N);
    
end
% delta_c(i) = sqrt((pos(i,1) - pos(i+1,1))^2 + (pos(i,2) - pos(i+1,2))^2);
lmn = zeros(N);
%zmn = zeros(N);
gm = zeros(1,N);
zmn = zeros(N);
%vim = zeros(1,N);
%vsn = zeros(1,N);
coeif=(Z0.*k0./4).*ra.*dfpm;
coeifn=(Z0./2).*sin(k0.*ra.*dfpm./2);
for index_i = 1:N
    for index_j = 1:N
        if index_i == index_j
            lmn(index_i,index_j) =coeif.*(1 - 1i.*(2./pi).*(log((gamma_const.*k0.*ra.*dfpm)./4)-1));
        else
            R(index_i,index_j)=ra.*sqrt(2.-2.*cos(Phi0(index_j)-Phi0(index_i)));
            lmn(index_i,index_j) =coeifn.*besselh(0,2,k0.*R(index_i,index_j));
        end
        zmn(index_i,index_j) = lmn(index_i,index_j);
        gm(index_i) =Efieldin(Phi0(index_i));
    end
    
    %vim(index_i) = delta_c(index_i) * exp(j*k0*(xm(index_i)*cos(phi_i)+ym(index_i)*sin(phi_i)));
    %vsn(index_i) = delta_c(index_i) * exp(j*k0*(xm(index_i)*cos(phi_s)+ym(index_i)*sin(phi_s)));
end
W = linsolve(zmn,gm');
for ii=1:N 
    Is(ii)=W(ii); 
end 
y= Is
end

Steven Lord 2024-11-20

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 2024-11-22

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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Answer 1

David Goodmanson 2024-11-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2167338-integartion-of-besselh-function#answer_1548303

编辑：David Goodmanson 2024-11-22

在 MATLAB Online 中打开

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.

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george veropoulos 2024-11-22

thank you very much!!! the cos was the problem

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INTEGARTION OF BESSELH function

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