how to solve 4 state equation ?

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tomer polsky
tomer polsky 2017-10-23
编辑： Birdman 2017-10-24
A_a=[0 0 0 1/(C1);0 -1/(R*C) 0 0;0 0 -R1/(L1) 0;-1/(L2) 0 0 0]
B_a=[0;0;(1/L1);0]
u=12
dx/dt=(A_a)*x+B*u
% dx/dt is differential of x
% of course R1,C1,C,R,L1,L2 are constants
how do i solve this equation?

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回答（2 个）

Birdman
Birdman 2017-10-23
This system has a transfer function of
X(s)/U(s)=B/sI-A;
This means that this differential equation's solutions will be the root of the following equation.
sI-A=0;
In this situation, you can easily find the solution for this differential equation by typing
eig(A)
The code will be as follows:
syms C1 R C R1 L1 L2
A=[0 0 0 1/(C1);0 -1/(R*C) 0 0;0 0 -R1/(L1) 0;-1/(L2) 0 0 0];
B=[0;0;(1/L1);0];
u=12;
eig(A)
You will have four solutions depending on the values of C1, R, C, R1, L1 and L2. Hope this helps.
  2 个评论
tomer polsky
tomer polsky 2017-10-24
this is not what i am asking for , u are telling me how to find the roots . i dont need the roots i need to find vector x
Birdman
Birdman 2017-10-24
编辑：Birdman 2017-10-24
syms C1 R C R1 L1 L2 t
x1=sym('x1(t)');x2=sym('x2(t)');x3=sym('x3(t)');x4=sym('x4(t)');
x=[x1;x2;x3;x4];clc;
A=[0 0 0 1/C1;0 -1/(R*C) 0 0;0 0 -R1/L1 0;-1/L2 0 0 0];
B=[0;0;1/L1;0];
u=12;
eqn=diff(x,t)==A*x+B*u;
sol=dsolve(eqn);
disp(sol.x1)
disp(sol.x2)
disp(sol.x3)
disp(sol.x4)
x=[sol.x1;sol.x2;sol.x3;sol.x4]

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Torsten
Torsten 2017-10-24
编辑：Torsten 2017-10-24
syms a(t) b(t) c(t) d(t) C1 R C R1 L1 L2 u
eqns = [diff(a,t)==d/C1 , diff(b,t)=-b/(R*C) , diff(c,t)=-R1*c/L1+u/L1 , diff(d,t)=-a/L2];
[asol(t) bsol(t) csol(t) dsol(t)] = dsolve(eqns)
Best wishes
Torsten.

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