multiply by two matrix

Mj = [cos (kj hj) i*sin(kj hj) /kj ; i*kj*sin(kj hj) cos (kj hj) ];
kj=sqrt(nj^2*k02-x^2);
Take
n1=1.521;n2=2.66;k0=1;h1=1.5e-6;h2=1e-6
multiply M1*M2

4 comentarios

Image Analyst
Image Analyst el 26 de Sept. de 2021
And the question is. . . . 🤔
Steven Lord
Steven Lord el 26 de Sept. de 2021
This sounds like a homework assignment. If it is, show us the code you've written to try to solve the problem and ask a specific question about where you're having difficulty and we may be able to provide some guidance.
If you aren't sure where to start because you're not familiar with how to write MATLAB code, I suggest you start with the MATLAB Onramp tutorial (https://www.mathworks.com/support/learn-with-matlab-tutorials.html) to quickly learn the essentials of MATLAB.
If you aren't sure where to start because you're not familiar with the mathematics you'll need to solve the problem, I recommend asking your professor and/or teaching assistant for help.
shiv gaur
shiv gaur el 26 de Sept. de 2021
Editada: Walter Roberson el 27 de Sept. de 2021
%this is program
function kp1
close all;
clear all;
function y=F(x)
for t2=1
k0=1;
n1=1.521;
%n2=2.66;
n2=4.1-0.211*1i;
ns=1.512;
%nc=0.15-1i*3.2;
nc=1;
k1=sqrt(n1.^2*k0.^2-x.^2);
k2=sqrt(n2.^2*k0.^2-x.^2);
t1=1.5;
m11= cos(t1*k1)*cos(t2*k2)-(k2/k1)*sin(t1*k1)*sin(t2*k2);
m12=(1/k2)*(cos(t1*k1)*sin(t2*k2)*1i) +(1/k1)*(cos(t2*k2)*sin(t1*k1)*1i);
m21= (k1)*cos(t2*k2)*sin(t1*k1)*1i +(k2)*cos(t1*k1)*sin(t2*k2)*1i;
m22=cos(t1*k1)*cos(t2*k2)-(k1/k2)*sin(t1*k1)*sin(t2*k2);
A=[m11 m12 ; m21 m22];
disp(eig(A))
gs=x.^2-ns.^2*k0.^2;
gc= x.^2-nc.^2*k0.^2;
y= 1i*(gs*m11+gc*m22)-m21+gc*gs*m12 ;
%y=x^2+x+1;
end
end
p0=1;
p1=1.5;
p2=2;
tol = 10^-5;
n = 50;
h1 = p1 - p0;
h2 = p2 - p1;
del1 = (F(p1)-F(p0))./h1;
del2 = (F(p2)-F(p1))./h2;
d = (del2-del1)./(h2+h1);
I = 3;
%Step 2
while I <= n
%Step 3
b = del2+h2.*d;
D = sqrt(b.^2-4.*F(p2).*d); % could be complex
%Step 4
if abs(b - D) < abs(b + D)
E = b + D;
else
E = b - D;
end
%Step 5
h = -2.*F(p2)./E;
p = p2 + h;
if I == 3
table{1} = 'Muller''s Method Iterations';
table{2}=' I P f(P) ';
table{3}='-----------------------------------------------------';
end
str = sprintf('%3u: % 6.6f + %6.6fi % 6.6f + %6.6fi',I,real(p),imag(p),real(F(p)),imag(F(p)));
table{I + 1} = str; %#ok<*AGROW>
%Step 6
if abs(h) < tol
val = p;
table = char(table);
break
end
p0 = p1;
p1 = p2;
p2 = p;
h1 = p1 - p0;
h2 = p2 - p1;
del1 = (F(p1)-F(p0))./h1;
del2 = (F(p2)-F(p1))./h2;
d = (del2-del1)./(h2+h1);
I = I + 1;
end
disp(p)
end
Jan
Jan el 26 de Sept. de 2021
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Respuestas (1)

Walter Roberson
Walter Roberson el 27 de Sept. de 2021
syms x
n = [1.521, 2.66];
k0 = 1;
h = [1.5e-6, 1e-6];
k = @(j) sqrt(n(j).^2*k0.^2-x.^2);
M = @(j) [
cos(k(j) .* h(j)), i*sin(k(j) .* h(j))/k(j)
i .* k(j) .* sin(k(j).*h(j)), cos(k(j) .* h(j))
]
M = function_handle with value:
@(j)[cos(k(j).*h(j)),i*sin(k(j).*h(j))/k(j);i.*k(j).*sin(k(j).*h(j)),cos(k(j).*h(j))]
M12 = M(1) * M(2)
M12 = 

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Respondida:

el 27 de Sept. de 2021

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