how to calculate the value from function define

1 view (last 30 days)
la0 = 650; k0 = 2*%pi/la0; ef = 1; ec = -19.6224-0.443*%i; es=ec;
pc = ef/ec; ps = ef/es;
bcinf = sqrt(ec*ef/(ec+ef));
tol = 1e-12;
acut = -real(1/pc/sqrt(ef-ec))/k0;
wc = 2*acut;
nc = sqrt(ef);
w = 5:10:3500; //% thickness range for TM0
w1 = 1.001*wc:10:3500; //% thickness range for TM1
function be=f(la0, ef, ec, w, bcinf, m, tol);
for j=1:length(w)
be(j) = f(la0,ef,ec,w(j),bcinf,0,tol);
end
for j=1:length(w1)
be1(j) = f(la0,ef,ec,w1(j),bcinf,1,tol);
end
neff0 = real(be0); L0 = -1/2./imag(be0)/k0/1000; //% TM0 index & distance
neff1 = real(be1); L1 = -1/2./imag(be1)/k0/1000;// % TM1 index & distance
w2=1000; a = w2/2;// % specific solutions
b = f(la0,ef,ec,a,bcinf,0,tol);// % TM0
p = f(la0,ef,ec,a,bcinf,1,tol);// % TM1
neff02 = real(be02); L02 = -1/2/imag(be02)/k0/1000;
end
disp(be,b,p)
pl help to solve the problem to disp(be,b,p)
  1 Comment
Walter Roberson
Walter Roberson on 16 Feb 2022
Your function f calls itself multiple times, and does so unconditionally. That is an infinite loop.

Sign in to comment.

Answers (0)

Categories

Find more on Stress and Strain in Help Center and File Exchange

Tags

Products


Release

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!