solution of transcedental equation

9 views (last 30 days)
shiv gaur on 14 Oct 2021
Commented: Matt J on 14 Oct 2021
function beta1 = trial0(beta, eps1, eps2, epsm,k0,h)
S1 = sqrt(power(beta,2)-(eps1*power(k0,2)));
S2 = sqrt(power(beta,2)-(epsm*power(k0,2)));
S3 = sqrt(power(beta,2)-(eps2*power(k0,2)));
beta1 = tanh(S2*h)*((eps1*eps2*power(S2,2))+(power(epsm,2)*S1*S3)) + (S2*((eps1*S3)+(eps2*S1))*epsm);
end
eps1 = 1.5471;
eps2 = 1.5431;
eps1m = -9.894 ;
epsm2 = 1.0458;
epsm = eps1m + i*eps2m;
lambda = 633;
k0 = (2*pi)/lambda;
h = 50;
find the value of beta
Catalytic on 14 Oct 2021
You have not accepted any answers previously provided to you. That is a bit of a disincentive for people to help you.

Matt J on 14 Oct 2021
Edited: Matt J on 14 Oct 2021
eps1 = 1.5471;
eps2 = 1.5431;
eps1m = -9.894 ;
eps2m = 1.0458;
epsm = eps1m + i*eps2m;
lambda = 633;
k0 = (2*pi)/lambda;
h = 50;
[b,fval]=fminsearch(@(b) abs(trial0(complex(b(1),b(2)), eps1, eps2, epsm,k0,h)) ,[1,1]);
beta=complex(b(1),b(2)), fval
beta = -0.0145 - 0.0004i
fval = 2.7735e-05
function beta1 = trial0(beta, eps1, eps2, epsm,k0,h)
S1 = sqrt(power(beta,2)-(eps1*power(k0,2)));
S2 = sqrt(power(beta,2)-(epsm*power(k0,2)));
S3 = sqrt(power(beta,2)-(eps2*power(k0,2)));
beta1 = tanh(S2*h)*((eps1*eps2*power(S2,2))+(power(epsm,2)*S1*S3)) + (S2*((eps1*S3)+(eps2*S1))*epsm);
end
Matt J on 14 Oct 2021

R2021a

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