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?
댓글 수: 12
Matt J
2018년 11월 2일
This looks like only part of the message. The full error message should give a reason, as well as the error location.
madhan ravi
2018년 11월 2일
편집: madhan ravi
2018년 11월 2일
when I ran your code it didn't have any errors instead it returned all zeros
Skill_s
2018년 11월 2일
Skill_s
2018년 11월 2일
madhan ravi
2018년 11월 2일
do you have symbolic toolbox? type ver in command window to check whether you have that toolbox
Skill_s
2018년 11월 2일
Star Strider
2018년 11월 2일
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
2018년 11월 2일
채택된 답변
Bruno Luong
2018년 11월 3일
편집: Bruno Luong
2018년 11월 3일
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.
댓글 수: 4
Skill_s
2018년 11월 3일
Walter Roberson
2018년 11월 3일
편집: Walter Roberson
2018년 11월 4일
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
2018년 11월 4일
Bruno Luong
2018년 11월 4일
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.
추가 답변 (1개)
Florian Augustin
2018년 11월 2일
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
댓글 수: 7
Skill_s
2018년 11월 2일
Matt J
2018년 11월 2일
@Skill93,
If Florian's answer was what you were looking for, then you should Accept-click it.
Walter Roberson
2018년 11월 2일
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
2018년 11월 2일
Stephan
2018년 11월 2일
Executing your code in 2018b gives the same result. So you need a better approach.
Walter Roberson
2018년 11월 2일
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
2018년 11월 3일
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