solution of equation code of transcedental equation

1 次查看(过去 30 天)
function kps3
p0 = 0.5;
p1 = 1;
p2 = 1.5;
TOL = 10^-8;
N0 = 100; format long
h1 = p1 - p0;
h2 = p2 - p1;
DELTA1 = (f(p1) - f(p0))/h1;
DELTA2 = (f(p2) - f(p1))/h2;
d = (DELTA2 - DELTA1)/(h2 + h1);
i=3;
while i <= N0
b = DELTA2 + h2*d;
D = (b^2 - 4*f(p2)*d)^(1/2);
if abs(b-D) < abs(b+D)
E = b + D;
else
E = b - D;
end
h = -2*f(p2)/E;
p = p2 + h;
if abs(h) < TOL
disp(p)
break
end
p0 = p1;
p1 = p2;
p2 = p;
h1 = p1 - p0;
h2 = p2 - p1;
DELTA1 = (f(p1) - f(p0))/h1;
DELTA2 = (f(p2) - f(p1))/h2;
d = (DELTA2 - DELTA1)/(h2 + h1);
i=i+1
end
if i > N0
formatSpec = string('The method failed after N0 iterations,N0= %d \n');
// fprintf(formatSpec,N0);
end
function y=f(x)
t2=1e-9
k0=(2*pi/0.6328)*1e6;
n1=1.521;
n2=2.66;
ns=1.512;
nc=0.15-1i*3.2;
k1=k0*sqrt(n1.^2-x.^2);
k2=k0*sqrt(n2.^2-x.^2);
t1=1.5e-6;
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);
gs=(x.^2-ns.^2)*k0.^2;
gc= (x.^2-nc.^2)*k0.^2;
y= 1i*(gs*m11+gc*m22)-m21+gc*gs*m12 ;
end
end
  6 个评论
shiv gaur
shiv gaur 2021-12-12
plot is the problem and loop and store the value if any one have idea plot the graph

请先登录,再进行评论。

采纳的回答

Torsten
Torsten 2021-12-12
编辑:Torsten 2021-12-12
function kps3
T2 = 1e-9:1e-9:1e-6;
for j=1:numel(T2)
t2 = T2(j);
p0 = 0.5;
p1 = 1;
p2 = 1.5;
TOL = 10^-8;
N0 = 100; format long
h1 = p1 - p0;
h2 = p2 - p1;
DELTA1 = (f(p1,t2) - f(p0,t2))/h1;
DELTA2 = (f(p2,t2) - f(p1,t2))/h2;
d = (DELTA2 - DELTA1)/(h2 + h1);
i=3;
while i <= N0
b = DELTA2 + h2*d;
D = (b^2 - 4*f(p2,t2)*d)^(1/2);
if abs(b-D) < abs(b+D)
E = b + D;
else
E = b - D;
end
h = -2*f(p2,t2)/E;
p = p2 + h;
if abs(h) < TOL
%disp(p)
break
end
p0 = p1;
p1 = p2;
p2 = p;
h1 = p1 - p0;
h2 = p2 - p1;
DELTA1 = (f(p1,t2) - f(p0,t2))/h1;
DELTA2 = (f(p2,t2) - f(p1,t2))/h2;
d = (DELTA2 - DELTA1)/(h2 + h1);
i=i+1;
end
if i > N0
formatSpec = string('The method failed after N0 iterations,N0= %d \n');
fprintf(formatSpec,N0);
end
P(j)=real(p);
end
plot(T2,P)
end
function y=f(x,t2)
%t2=1e-9;
k0=(2*pi/0.6328)*1e6;
n1=1.521;
n2=2.66;
ns=1.512;
nc=0.15-1i*3.2;
k1=k0*sqrt(n1.^2-x.^2);
k2=k0*sqrt(n2.^2-x.^2);
t1=1.5e-6;
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);
gs=(x.^2-ns.^2)*k0.^2;
gc= (x.^2-nc.^2)*k0.^2;
y= 1i*(gs*m11+gc*m22)-m21+gc*gs*m12 ;
end
Maybe you can use p_old = P(j-1) as starting point for the solution of your equation with t2_new = T2(j).
  5 个评论

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Programming 的更多信息

标签

产品


版本

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by