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fsolve inputs

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Alex
Alex on 28 Feb 2012
Hey all,
I'm trying to pass some equations to fsolve, but I'm getting nothing but errors. I have 3 complex equations, and to simplify input, I broke down the equations into component parts like so:
y_1 = 309;
n0_1 = 1;
n1_1 = x(1);
k1_1 = x(2);
n2_1 = 5.07;
k2_1 = 3.62;
d1_1 = x(3);
R_1 = .4335;
g1_1 = (n0_1.^2 - n1_1.^2 - k1_1.^2)./((n1_1 + n2_1).^2 + k1_1.^2);
g2_1 = (n1_1.^2 - n2_1.^2 + k1_1.^2 - k2_1.^2)./((n1_1 + n2_1).^2 + (k1_1 + k2_1).^2);
h1_1 = (2.*n0_1.*k1_1)./((n0_1 + n1_1).^2 + k1_1.^2);
h2_1 = (2.*(n1_1.*k2_1 - n2_1.*k1_1))./((n1_1 + n2_1).^2 + (k1_1 + k2_1).^2);
a_1 = (2.*pi().*k1_1.*d1_1)./y_1;
b_1 = (2.*pi().*n1_1.*d1_1)./y_1;
A_1 = 2.*(g1_1.*g2_1 + h1_1.*h2_1);
B_1 = 2.*(g1_1.*h2_1 - g2_1.*h1_1);
C_1 = 2.*(g1_1.*g2_1 - h1_1.*h2_1);
D_1 = 2.*(g1_1.*h2_1 + g2_1.*h1_1);
g1_1.^2 + h1_1.^2).*(exp(2.*a_1)) + (g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + A_1.*(cos(2.*b_1)) + B_1.*(sin(2.*b_1)))./((exp(2.*a_1)) + (g1_1.^2 + h1_1.^2).*(g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + C_1.*(cos(2.*b_1)) + D_1.*(sin(2.*b_1))) - R_1
This equations repeats itself two more times, with different values as constants. The problem I seem to be having is that Matlab doesn't like my definitions of the variables. I don't know how to express this information in a way that Matlab is able to understand, and despite all of the help I've had from the Mathworks community (thanks so much), I can't seem to get this to work. I've tried syms, I've troubleshot syms, I've tried setting up a seperate function file with no results, and I'm stuck at this point. How would you pass this through? I'm sure it would be easy enough to just create the entire system of equations without all of the substitution, and then pass that on to fsolve, but this is much user friendly, and I would like to learn how to make this work.
Thanks!
  2 Comments
Walter Roberson
Walter Roberson on 28 Feb 2012
Duplicate is at http://www.mathworks.com/matlabcentral/answers/30540-passing-data-to-fsolve

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Accepted Answer

Walter Roberson
Walter Roberson on 28 Feb 2012
I cannot test this as I do not have the toolbox with fsolve in it.
This is the structure I told you to adapt several threads ago.
function nkd = g3driver
Guess = [1.58, 0, 25];
nkd = fsolve(@g3, Guess);
end
function F = g3(X)
% This program aims to back-solve for reflectance (R) using given values
% input by hand.
x = X(1); y = X(2); z = X(3);
y_1 = 309;
n0_1 = 1;
%n1_1 = 1.580086;
n1_1 = x;
%k1_1 = 0;
k1_1 = y;
n2_1 = 5.07;
k2_1 = 3.62;
%d1_1 = 25;
d1_1 = z;
R_1 = .4335;
y_2 = 310;
n0_2 = 1;
%n1_2 = 1.579925;
n1_2 = x;
%k1_2 = 0;
k1_2 = y;
n2_2 = 5.07;
k2_2 = 3.56;
%d1_2 = 25;
d1_2 = z;
R_2 = .4294;
y_3 = 311;
n0_3 = 1;
%n1_3 = 1.579764;
n1_3 = x;
%k1_3 = 0;
k1_3 = y;
n2_3 = 5.08;
k2_3 = 3.53;
%d1_3 = 25;
d1_3 = z;
R_3 = .4277;
g1_1 = (n0_1.^2 - n1_1.^2 - k1_1.^2)./((n1_1 + n2_1).^2 + k1_1.^2);
g2_1 = (n1_1.^2 - n2_1.^2 + k1_1.^2 - k2_1.^2)./((n1_1 + n2_1).^2 + (k1_1 + k2_1).^2);
h1_1 = (2.*n0_1.*k1_1)./((n0_1 + n1_1).^2 + k1_1.^2);
h2_1 = (2.*(n1_1.*k2_1 - n2_1.*k1_1))./((n1_1 + n2_1).^2 + (k1_1 + k2_1).^2);
a_1 = (2.*pi().*k1_1.*d1_1)./y_1;
b_1 = (2.*pi().*n1_1.*d1_1)./y_1;
A_1 = 2.*(g1_1.*g2_1 + h1_1.*h2_1);
B_1 = 2.*(g1_1.*h2_1 - g2_1.*h1_1);
C_1 = 2.*(g1_1.*g2_1 - h1_1.*h2_1);
D_1 = 2.*(g1_1.*h2_1 + g2_1.*h1_1);
g1_2 = (n0_2.^2 - n1_2.^2 - k1_2.^2)./((n1_2 + n2_2).^2 + k1_2.^2);
g2_2 = (n1_2.^2 - n2_2.^2 + k1_2.^2 - k2_2.^2)./((n1_2 + n2_2).^2 + (k1_2 + k2_2).^2);
h1_2 = (2.*n0_2.*k1_2)./((n0_2 + n1_2).^2 + k1_2.^2);
h2_2 = (2.*(n1_2.*k2_2 - n2_2.*k1_2))./((n1_2 + n2_2).^2 + (k1_2 + k2_2).^2);
a_2 = (2.*pi().*k1_2.*d1_2)./y_2;
b_2 = (2.*pi().*n1_2.*d1_2)./y_2;
A_2 = 2.*(g1_2.*g2_2 + h1_2.*h2_2);
B_2 = 2.*(g1_2.*h2_2 - g2_2.*h1_2);
C_2 = 2.*(g1_2.*g2_2 - h1_2.*h2_2);
D_2 = 2.*(g1_2.*h2_2 + g2_2.*h1_2);
g1_3 = (n0_3.^2 - n1_3.^2 - k1_3.^2)./((n1_3 + n2_3).^2 + k1_3.^2);
g2_3 = (n1_3.^2 - n2_3.^2 + k1_3.^2 - k2_3.^2)./((n1_3 + n2_3).^2 + (k1_3 + k2_3).^2);
h1_3 = (2.*n0_3.*k1_3)./((n0_3 + n1_3).^2 + k1_3.^2);
h2_3 = (2.*(n1_3.*k2_3 - n2_3.*k1_3))./((n1_3 + n2_3).^2 + (k1_3 + k2_3).^2);
a_3 = (2.*pi().*k1_3.*d1_3)./y_3;
b_3 = (2.*pi().*n1_3.*d1_3)./y_3;
A_3 = 2.*(g1_3.*g2_3 + h1_3.*h2_3);
B_3 = 2.*(g1_3.*h2_3 - g2_3.*h1_3);
C_3 = 2.*(g1_3.*g2_3 - h1_3.*h2_3);
D_3 = 2.*(g1_3.*h2_3 + g2_3.*h1_3);
F = [((g1_1.^2 + h1_1.^2).*(exp(2.*a_1)) + (g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + A_1.*(cos(2.*b_1)) + B_1.*(sin(2.*b_1)))./((exp(2.*a_1)) + (g1_1.^2 + h1_1.^2).*(g2_1.^2 + h2_1.^2).*(exp(-2.*a_1)) + C_1.*(cos(2.*b_1)) + D_1.*(sin(2.*b_1))) - R_1; ...
((g1_2.^2 + h1_2.^2).*(exp(2.*a_2)) + (g2_2.^2 + h2_2.^2).*(exp(-2.*a_2)) + A_2.*(cos(2.*b_2)) + B_2.*(sin(2.*b_2)))./((exp(2.*a_2)) + (g1_2.^2 + h1_2.^2).*(g2_2.^2 + h2_2.^2).*(exp(-2.*a_2)) + C_2.*(cos(2.*b_2)) + D_2.*(sin(2.*b_2))) - R_2; ...
((g1_3.^2 + h1_3.^2).*(exp(2.*a_3)) + (g2_3.^2 + h2_3.^2).*(exp(-2.*a_3)) + A_3.*(cos(2.*b_3)) + B_3.*(sin(2.*b_3)))./((exp(2.*a_3)) + (g1_3.^2 + h1_3.^2).*(g2_3.^2 + h2_3.^2).*(exp(-2.*a_3)) + C_3.*(cos(2.*b_3)) + D_3.*(sin(2.*b_3))) - R_3];
end

More Answers (1)

Walter Roberson
Walter Roberson on 28 Feb 2012
Your line that starts g1_1.^2 is missing an opening '(', and has an extra ')' before the './'
  4 Comments
Walter Roberson
Walter Roberson on 28 Feb 2012
That still leaves the extra ')' before the './' .
When I put in the needed '(' and remove the unneeded ')', I have no problems with the code when I initialize x = rand(3,1)
Your code shown is not doing any substitution.

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