Using solution from vpasolve and syms for another function

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Justin
Justin el 2 de Nov. de 2022
Comentada: Torsten el 2 de Nov. de 2022
A =[ 0 0 0 1.0000 0 0
0 0 0 0 1.0000 0
0 0 0 0 0 1.0000
0 6.5000 -10.0000 -17.5000 0 0
-21.1185 18.1040 5.1540 -3.6920 -2.8128 0.3159
5.0000 -3.6000 0 0 0 -10.9545]
B = [ 0 0
0 0
0 0
26.5000 11.2000
5.5907 -1.6525
40.0000 57.3000]
C = [ 1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0]
syms T1 T2 T3 T4 T5 T6
T = [T1 T2 T3 T4 T5 T6];
d = -5;
G = [10 1 5];
vpasolve([T*A - d*T == G*C])
After solving for the values of T1 to T6, i want to use those values to solve for another equation, but when I input T1,T2... into the other equation, it returns the answer in terms of T1 - T6 instead of actual values
e = T*B
how do i get it to return the actual numbers instead of in terms of T1 etc.?

Respuesta aceptada

KSSV
KSSV el 2 de Nov. de 2022
A =[ 0 0 0 1.0000 0 0
0 0 0 0 1.0000 0
0 0 0 0 0 1.0000
0 6.5000 -10.0000 -17.5000 0 0
-21.1185 18.1040 5.1540 -3.6920 -2.8128 0.3159
5.0000 -3.6000 0 0 0 -10.9545] ;
B = [ 0 0
0 0
0 0
26.5000 11.2000
5.5907 -1.6525
40.0000 57.3000] ;
C = [1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0] ;
syms T1 T2 T3 T4 T5 T6
T = [T1 T2 T3 T4 T5 T6];
d = -5;
G = [10 1 5];
s = vpasolve([T*A - d*T == G*C]) ;
T = [double(s.T1) double(s.T2) double(s.T3) double(s.T4) double(s.T5) double(s.T6)];
e = T*B
e = 1×2
12.7215 13.2144
  2 comentarios
Torsten
Torsten el 2 de Nov. de 2022
You don't need symbolic computations.
A =[ 0 0 0 1.0000 0 0
0 0 0 0 1.0000 0
0 0 0 0 0 1.0000
0 6.5000 -10.0000 -17.5000 0 0
-21.1185 18.1040 5.1540 -3.6920 -2.8128 0.3159
5.0000 -3.6000 0 0 0 -10.9545] ;
B = [ 0 0
0 0
0 0
26.5000 11.2000
5.5907 -1.6525
40.0000 57.3000] ;
C = [1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0] ;
d = -5;
G = [10 1 5];
T = ((A.'-d*eye(6))\(G*C).').';
e = T*B
e = 1×2
12.7215 13.2144

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