How to solve complex equations?

11 次查看(过去 30 天)
I am trying to solve the following equation.
I have the following script to solve for theta, but getting erro when I am running the program.
Any suggestions is much appreciated.
Thank you.
clc;
clear;
close all;
syms theta
assume(0 <= theta <= 2*pi)
t = 0.4e-3;
n = 1.523;
delta_y = 3;
S1 = solve(delta_y == t*sind(theta)*(1 - sqrt((1-sind(theta)^2)/n^2 - sind(theta)^2)),theta);
  3 个评论
Torsten
Torsten 2024-2-13
You forgot a bracket around the denominator under the square root.
Navaneeth
Navaneeth 2024-2-14
Sorry that I left out some details. N here is refractive index of the material and t is the thickness of the slab...I want to find the theta in angles for a fixed shift of delta_y.
Thank you.

请先登录,再进行评论。

采纳的回答

Torsten
Torsten 2024-2-13
编辑:Torsten 2024-2-13
Your equation has only complex solutions.
Note that you used sind instead of sin in the function definition. Thus assuming theta in radians is wrong.
clc;
clear;
close all;
syms theta
%assume(0 <= theta <= 2*pi)
t = 0.4e-3;
n = 1.523;
delta_y = 3;
S1 = solve(delta_y == t*sind(theta)*(1 - sqrt((1-sind(theta)^2)/(n^2 - sind(theta)^2))),theta,'ReturnConditions',1,'MaxDegree',3)
S1 = struct with fields:
theta: [2×1 sym] parameters: k conditions: [2×1 sym]
vpa(S1.theta)
ans = 
S1.parameters
ans = 
k
S1.conditions
ans = 
  5 个评论
Torsten
Torsten 2024-2-14
编辑:Torsten 2024-2-14
If you use the new parameters, you get back some real solutions:
clc;
clear;
close all;
syms theta
t = 0.7e-3;
n = 1.533;
delta_y = 21.2991*1e-6;
S1 = vpa(solve(delta_y == t*sind(theta)*(1 - sqrt((1-sind(theta)^2)/(n^2 - sind(theta)^2))),theta,'MaxDegree',3))
S1 = 
format long
S1 = double(S1(abs(imag(S1))<1e-6))
S1 =
1.0e+02 * 1.749999973612585 - 0.000000000000000i 0.050000026387415 + 0.000000000000000i
Navaneeth
Navaneeth 2024-2-14
Thank you, this is exactly what I wanted.

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

产品


版本

R2023a

Community Treasure Hunt

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

Start Hunting!

Translated by