solve 4 equations with some unknowns parameters.
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I'm attempting to solve this problem which is attached. But I faced ERROR. Any suggestions?
12 件のコメント
Matt J
2018 年 11 月 2 日
This looks like only part of the message. The full error message should give a reason, as well as the error location.
madhan ravi
2018 年 11 月 2 日
編集済み: madhan ravi
2018 年 11 月 2 日
when I ran your code it didn't have any errors instead it returned all zeros
Skill_s
2018 年 11 月 2 日
Skill_s
2018 年 11 月 2 日
madhan ravi
2018 年 11 月 2 日
do you have symbolic toolbox? type ver in command window to check whether you have that toolbox
Skill_s
2018 年 11 月 2 日
Star Strider
2018 年 11 月 2 日
The ver output disagrees with you:
MATLAB Version 7.11.0.584 (R2010b)
Operating System: Microsoft Windows 7 Version 6.2 (Build 9200)
...
Symbolic Math Toolbox Version 5.5 (R2010b)
...
Skill_s
2018 年 11 月 2 日
採用された回答
Bruno Luong
2018 年 11 月 3 日
編集済み: Bruno Luong
2018 年 11 月 3 日
Your problem looks like you find the EM field with 3 layers and 2 interfaces. The coefficients A, B, C, D gives the strength of different field components, they are linearly related because EM is linear. In order to solve with without getting trivial solution
A=B=C=D=0
You must for example fix one of them arbitrary, e.g.
A = 1
That left you with 4 linear equations and 3 unknowns. Then in order such system to have solution, you must have the linear system to have dependent equation, meaning determinant of the matrix is 0.
Write it down this probably gives the equation you looks for.
I don't have symbolic tbx to do this kind of calculation to confirm, it would be possible to carry out numerically me think.
4 件のコメント
Skill_s
2018 年 11 月 3 日
Walter Roberson
2018 年 11 月 3 日
編集済み: Walter Roberson
2018 年 11 月 4 日
If you solve the first three equations for B, C, D, then you are left with
-A*(-(eps1*k3+eps3*k1)*(eps1*k2+eps2*k1)*exp(a*(2*k1-k3))+exp(-a*(2*k1+k3))*(-eps1*k3+eps3*k1)*(-eps1*k2+eps2*k1))/(2*eps3*k1*eps2*eps1) == 0
The obvious solution is A == 0, which then causes B, C, D to all be 0. If, on the other hand, (-(eps1*k3+eps3*k1)*(eps1*k2+eps2*k1)*exp(a*(2*k1-k3))+exp(-a*(2*k1+k3))*(-eps1*k3+eps3*k1)*(-eps1*k2+eps2*k1))/(2*eps3*k1*eps2*eps1) can be 0, then the equations would hold for all finite A, in which case there would be a family of solutions
B = A*((eps1*k3+eps3*k1)*exp(4*a*k1)+eps3*k1-eps1*k3)*exp(-a*(2*k1-k2+k3))/(2*eps3*k1)
C = -A*(eps1*k3-eps3*k1)*exp(-a*(k1+k3))/(2*eps3*k1)
D = A*(eps1*k3+eps3*k1)*exp(a*(k1-k3))/(2*eps3*k1)
which are linear in A.
Skill_s
2018 年 11 月 4 日
Bruno Luong
2018 年 11 月 4 日
Actually solving linear system is not what Skill interested, what he is interested is the the condition under which the 4 linear equation is dependent thus solvable.
This boils down to single equation of determinant. I shows here the solution he gives makes DET(M) = 0.
% Fake k, epsilon 1,2,3
k = rand(1,3);
eps = rand(1,3);
p = k./eps;
% compute solution a
r = ((p(1)+p(2))*(p(1)+p(3)))/((p(1)-p(2))*(p(1)-p(3)));
a = log(r)/(-4*k(1))
% Forming linear matrix
ep = exp(k*a);
em = exp(-k*a);
M = [em(3) 0 -ep(1) -em(1);
p(3)*em(3) 0 p(1)*ep(1) -p(1)*em(1);
0 em(2) -em(1) -ep(1);
0 -p(2)*em(2) p(1)*em(1) -p(1)*ep(1)];
% verifies the 4 x 4 system is rank 3
rank(M)
det(M)
Now if Walter who has access to Symb Tbx can show the opposite, started from
det(M) = 0
shows the solution of it is
r = ((p(1)+p(2))*(p(1)+p(3)))/((p(1)-p(2))*(p(1)-p(3)));
a = log(r)/(-4*k(1))
Then the problem is completely solved.
その他の回答 (1 件)
Florian Augustin
2018 年 11 月 2 日
Hi,
I think you are using a modern syntax to call 'solve' that was not supported in R2010b. The equivalent call in R2010b would be
syms a A B C D eps1 eps2 eps3 k1 k2 k3;
eqn1 = (C*exp(k1*a))+(D*exp(-k1*a))-(A*exp(-k3*a));
eqn2 = ((-(C*k1)/eps1)*exp(k1*a))+(((D*k1)/eps1)*exp(-k1*a))-(((A*k3)/eps3)*exp(-k3*a));
eqn3 = (C*exp(-k1*a))+(D*exp(k1*a))-(B*exp(-k2*a));
eqn4 = ((-(C*k1)/eps1)*exp(-k1*a))+(((D*k1)/eps1)*exp(k1*a))+(((B*k2)/eps2)*exp(-k2*a));
sol = solve(eqn1, eqn2, eqn3, eqn4, A, B, C, D);
ASol = sol.A
BSol = sol.B
CSol = sol.C
DSol = sol.D
You can access the documentation for your release of MATLAB by typing 'doc solve' in the MATLAB interface.
Hope this helps,
-Florian
7 件のコメント
Skill_s
2018 年 11 月 2 日
Matt J
2018 年 11 月 2 日
@Skill93,
If Florian's answer was what you were looking for, then you should Accept-click it.
Walter Roberson
2018 年 11 月 2 日
If I recall correctly, in R2010b, the syntax used for solve() was available but not clearly documented.
The problem was elsewhere: before R2011b, using relational expressions with symbolic expressions caused the expressions to be evaluated as a logical immediately. So the line
eqn1 = (C*exp(k1*a))+(D*exp(-k1*a))-(A*exp(-k3*a))==0;
built the expression on the left side, and immediately compared it to 0, found it was not identical, and immediately returned false into eqn1.
The workaround in those days was to not do the ==0 part on the equations: convert
eqn = A == B
to
eqn = (A) - (B)
And in those days, if you had an inequality such as
eqn1 = (C*exp(k1*a))+(D*exp(-k1*a))-(A*exp(-k3*a)) > 3
then you would have to build it using character vectors,
eqn1 = '(C*exp(k1*a))+(D*exp(-k1*a))-(A*exp(-k3*a)) > 3'
This would pretty much force you to use separate entries in the solve() command, like the
sol = solve(eqn1, eqn2, eqn3, eqn4, A, B, C, D);
that Florian shows, since [eqn1, eqn2, eqn3, eqn4] with character vectors would result in a single long character vector instead of multiple equations.
So, Florian's suggested code should work for the purpose, just not for exactly the reason that Florian indicated.
Skill_s
2018 年 11 月 2 日
Stephan
2018 年 11 月 2 日
Executing your code in 2018b gives the same result. So you need a better approach.
Walter Roberson
2018 年 11 月 2 日
MATLAB gives all 0.
If you work through the equations one by one doing stepwise elimination, then all 0 is the only solution.
Skill_s
2018 年 11 月 3 日
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