multiply by two matrix

2021 9 26
1 个回答
更新时间：2021 9 27
2 次查看（30 天）

0 个投票

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 个评论
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Image Analyst
Image Analyst 2021-9-26
And the question is. . . . 🤔
Steven Lord
Steven Lord 2021-9-26
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 2021-9-26
编辑：Walter Roberson 2021-9-27
%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 2021-9-26
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回答（1 个）

Walter Roberson
Walter Roberson 2021-9-27
Ran in:

0 个投票

Ran in:
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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