Matrix is Singular to working precision
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Hello, I am working on state space design, and have read through all the trouble shooting advice on this website for the above error. I have attached my code. The error is in line 43, the last one. I am not sure what I am doing that is causing this error. Could somebody guide me on what I am doing wrong? Please forgive the horrendous formating. I will also attach the file if that is easier to see.
% Defining parameters and state space model
R1 = 3.5e-3;
R2 = 0.45;
L1 = 1.8e-3;
L2 = 12e-3;
C = 50e-3;
A = [-R1/L1 -1/L1 0 0;
1/C 0 -1/C 0;
0 1/L2 -R2/L2 0;
0 1 0 0];
E = [-3-(R1/L1) -1/L1 0 0 0;
1/C -3.5 -1/C 0 0;
0 1/L2 -33-(R2/L2) 0 1/L1;
0 1 0 -0.5 0];
Ao = [-3 0 0 0 0;
0 -3.5 0 0 0;
0 0 -33 0 0;
0 0 0 -0.5 0;
0 0 0 0 -3
];
B = [1/L1; 0; 0; 0];
A_bar = [A E;zeros(5,4) Ao];
B_bar = [B;zeros(5,1)];
z = 0.9;
wn = 10;
p1 = roots([1 2*z*wn wn^2]);
poles = [-20 -40 p1'];
K = place(A,B,poles);
check = eig(A-B*K);
C = [0 1 0 0]; % C is Vc as given
Go =inv(C*inv(A - B*K)*B)*C*inv(A - B*K)*E
1 件のコメント
Stephen23
2017 年 5 月 1 日
編集済み: Stephen23
2017 年 5 月 1 日
The inv documentation explains that mldivide and mrdivide are faster and numerically more accurate compared to using inv and matrix multiplication.
The documentation clearly states "It is seldom necessary to form the explicit inverse of a matrix. A frequent misuse of inv arises when solving the system of linear equations Ax = b. One way to solve the equation is with x = inv(A)*b. A better way, from the standpoint of both execution time and numerical accuracy, is to use the matrix backslash operator x = A\b. This produces the solution using Gaussian elimination, without explicitly forming the inverse. See mldivide for further information."
回答 (3 件)
Jayaram Theegala
2017 年 5 月 1 日
The warning "Matrix is singular to working precision" occurs if the matrix for which you are trying to calculate the inverse is singular and hence the inverse does not exist.
From your MATLAB script, line 30, I can see that the value of "C*inv(A - B*K)*B" is zero, and hence if you try to calculate inverse for it, you will get that warning message.
0 件のコメント
Hafiz Qasim Ali
2021 年 4 月 30 日
Can anyone guide me how to resolve the issue? In case below code is not readable, please see the attachments.
function [f] = Objective_Function(x)
% First are defined the influent flow and depth of the pond.
Qi=209;
z=1.0;
% f = Total Cost and the retention time and number of screens are represented by x(1)and x(2), respectively;
f=5850*(x(1)*Qi/(3*z))+9600*sqrt(x(1)*Qi/(3*z))+2520*x(2)*sqrt(x(1)*Qi/(3*z));
function [c, ceq]= Constraints_Function(x)
Ni=58383.9;
T=11.8;
e=5;
z=1.0;
Qi=209;
DBOi=220;
kb=0.841*(1.075)^(T-20);
%c(1) represents the inequality constraint for the fecal coliform:c(1)= Ne-1000 <= 0
c(1)=(4*Ni*sqrt(1+4*kb*x(1)*3*0.7*(x(2)+1)^(2)/(-0.26118+0.25392*...
(3*0.7*(x(2)+1)^(2))+1.0136*(3*0.7*(x(2)+1)^(2))^(2)))...
*exp((1-sqrt(1+4*kb*x(1)*3*0.7*(x(2)+1)^(2)/(-0.26118+0.25392*...
(3*0.7*(x(2)+1)^(2))+1.0136*(3*0.7*(x(2)+1)^(2))^(2))))/(2*3*0.7*...
(x(2)+1)^(2)/(-0.26118+0.25392*(3*0.7*(x(2)+1)^(2))+1.0136*(3*0.7*...
(x(2)+1)^(2))^(2)))))/((1+sqrt(1+4*kb*x(1)*3*0.7*(x(2)+1)^(2)/...
(-0.26118+0.25392*(3*0.7*(x(2)+1)^(2))+1.0136*(3*0.7*(x(2)+1)^(2))...
^(2))))^(2))*(Qi/(Qi-0.001*3*x(1)*Qi/(3*z)*e))-1000;
%c(2) represents the inequality constraint for the BOD: %c(2)=BODe-75 <=0
c(2)=(DBOi/((1.2* x(1)/(1.085)^(35-T))+1))*(Qi/(Qi-0.001*3*x(1)*Qi/(3*z)*e))-75;
%ceq represent the equality constraints, but as there is none. Ceq is equal to cero.
ceq=0;
end
Warning: Matrix is singular to working precision.
> In backsolveSys
In solveKKTsystem
In computeTrialStep
In barrier
In fmincon (line 834)
In callSolver (line 32)
In optimguirun (line 40)
In optimguiswitchyard (line 14)
0 件のコメント
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