Solving Linear Equation of a Scattering Problem
1 view (last 30 days)
Show older comments
There is an equation I need to solve like this;
Ln+an*Mn=∑Nnm*cm On+an*Pn=∑Qnm*cm summation for m's -infinity to infinity
Ln,Mn,Nnm,On,Pn are known.an,cm are unknown. I need to get an for calculation. If anyone can help me , I'll appreciate.
Here is my code:
clear all
format long
tic
N_cut=20;
eps0=(10^-9)/(36*pi);
mu0=4*pi*10^-7;
epsr1=1.;
epsr2=2.7;
mur1=1.;
mur2=1.;
eps1=epsr1*eps0;
eps2=epsr2*eps0;
mu1=mur1*mu0;
mu2=mur2*mu0;
freq=900*10^6;
omeg=2*pi*freq;
sigma2=3e-7;
k0=omeg*sqrt(eps0*mu0);
k1=sqrt(omeg*omeg*eps2*mu2+1i*omeg*sigma2*mu2);
lambda=2*pi/k0;
a0=4.25e-2;
a1=3e-3;
d=1e-2;
phi_prime=pi/4;
nu0=sqrt(mu0/eps0);
nu1=sqrt((1i*omeg*mu0)/(sigma2+1i*omeg*eps2));
M_phi=180;
phibegin=0;
phiend=2*pi;
deltaphi=(phiend-phibegin)/M_phi;
phig=phibegin:deltaphi:phiend;
R_obs=4.2555e-2;
for n=1:N_cut L(n)=((-1j)^n)*besselj(n,k0*a0); M(n)=((-1j)^n)*besselh(n,2,k0*a0); bessel_der_1(n)=0.5*(besselj(n-1,k0*a1)-besselj(n+1,k0*a1)); hankel_der_1(n)=0.5*(besselh(n-1,2,k0*a1)-besselh(n+1,2,k0*a1)); bessel_der_2(n)=0.5*(besselj(n-1,k1*a1)-besselj(n+1,k1*a1)); hankel_der_2(n)=0.5*(besselh(n-1,2,k1*a1)-besselh(n+1,2,k1*a1)); O(n)=(1/nu0)*((-1j)^n)*bessel_der_1(n); P(n)=(1/nu0)*((-1j)^n)*hankel_der_1(n); for m=1:N_cut K(m)=(-besselh(m,2,k1*a1))/(besselj(m,k1*a1)); N(n,m)=((-1j)^m)*besselj(n-m,k1*d)*besselj(n,k1*a1)*exp(-1j*(n-m)*phi_prime)*K(m)+((-1j)^m)*besselj(n-m,k1*d)*besselh(n,2,k1*a1)*exp(-1j*(n-m)*phi_prime); Q(n,m)=((1/nu1)*((-1j)^m)*besselj(n-m,k1*d)*bessel_der_2(n)*exp(-1j*(n-m)*phi_prime)*K(m)+(1/nu1)*((-1j)^m)*besselj(n-m,k1*d)*hankel_der_2(n)*exp(-1j*(n-m)*phi_prime));
end
end
for mg=1:M_phi+1 for n=1:N_cut Es(mg,n)=((-1j)^n)*(a(n)*besselh(n,2,k0*R_obs))*exp(-1j*n*phig(mg)); end end F_Es=sum(Es,2); figure plot(rad2deg(phig),abs(F_Es),'r') grid on
0 Comments
Answers (0)
See Also
Categories
Find more on Linear Algebra in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!