필터 지우기
필터 지우기

Error 'To Many Output Arguments

조회 수: 3 (최근 30일)
Ivan Dwi Putra
Ivan Dwi Putra 2020년 6월 11일
댓글: Ivan Dwi Putra 2020년 6월 11일
This is my code
clear; close; clc;
syms a1_head a2_head b hstar
%Parameter Massa
m1 = 8095; % massa train set 1 dalam kg
m2 = 8500; % massa train set 2 dalam kg
g = 10;
c_0_1 = 0.01176;
c_1_1 = 0.00077616;
c_2_1 = 4.48 ;
c_0_2 = 0.01176 ;
c_1_2 = 0.00077616;
c_2_2 = 4.48;
v_0 = 300;
hstar = 120;
a_1 = -1./m1.*(c_1_1 + 2.*c_2_1.*v_0);
a_2 = -1./m2.*(c_1_2 + 2.*c_2_2.*v_0);
a_1_head = 1-(a_1.*hstar);
a_2_head = 1-(a_2.*hstar);
b = 1;
% Model data
A = sym(zeros(4,4));
A(1,2) = a_1_head;
A(3,2) = (a_2_head) - 1; A(3,4) = a_2_head;
display(A);
B = sym(zeros(4,2));
B(1,1) = -b*hstar;
B(2,1) = b;
B(3,2) = -b*hstar ;
B(4,1) = -b; B(4,2) = b;
display(B);
% Q and R matrices for ARE
Q = sym(eye(4)); display(Q);
R = sym(zeros(2,2)); R(1,:) = [1 2]; R(2,:) = [2 3]; display(R);
% Matrix S to find
svar = sym('s',[1 16]);
S = [svar(1:4); svar(5:8); svar(9:12); svar(13:16)];
S(2,1) = svar(2);
S(2,2) = svar(1);
S(2,4) = svar(2);
S(3,1) = svar(3);
S(3,2) = svar(7);
S(3,3) = svar(1);
S(4,1) = svar(4);
S(4,2) = svar(2);
S(4,3) = svar(12);
S(4,4) = svar(1);
display(S);
% LHS of ARE: A'*S + S*A' - S*B*Rinv*B'*S
left_ARE = transpose(A)*S + S*A - S*B*inv(R)*transpose(B)*S;
display(left_ARE);
% RHS of ARE: -Q
right_ARE = -Q;
display(right_ARE);
% subs(left_ARE,{s1, s2, s3}, {1, 5, 10})
% Find S in ARE
syms s1 s2 s3 s4 s7 s12
[Sol_S] = solve(left_ARE == right_ARE,s1,s2,s3,s4,s7,s12,'Real',true,'returnconditions',true)
The result like this
Sol_S =
struct with fields:
s1: [0×1 sym]
s2: [0×1 sym]
s3: [0×1 sym]
s4: [0×1 sym]
s7: [0×1 sym]
s12: [0×1 sym]
I want to find my value s using this code
D = [s1 s2 s3 s4 s7 s12];
D = [S.s1 S.s2 S.s3 S.s4 S.s7 S.s12];
But it said
Error using sym/subsref
Too many output arguments.
Error in ARE (line 78)
D = [S.s1 S.s2 S.s3 S.s4 S.s7 S.s12];

채택된 답변

Ameer Hamza
Ameer Hamza 2020년 6월 11일
The values of S are stored in the struct returned by solve(). The actual value of the symbolic variable does not change. So the correct syntax is
D = [Sol_S.s1 Sol_S.s2 Sol_S.s3 Sol_S.s4 Sol_S.s7 Sol_S.s12];
However, the result shows
Sol_S =
struct with fields:
s1: [0×1 sym]
s2: [0×1 sym]
s3: [0×1 sym]
s4: [0×1 sym]
s7: [0×1 sym]
s12: [0×1 sym]
that MATLAB is not able to find a solution. It is probably the case that a solution does not exist. Also, note that there are 16 equations and only 6 variables, so I guess the system is over-determined.
  댓글 수: 15
Ameer Hamza
Ameer Hamza 2020년 6월 11일
In this case, I think the equation is unsolvable for these particular matrices A, B, Q, and R.
Ivan Dwi Putra
Ivan Dwi Putra 2020년 6월 11일
ok thank you for helping me

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Linear Algebra에 대해 자세히 알아보기

태그

제품


릴리스

R2020a

Community Treasure Hunt

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

Start Hunting!

Translated by