Problem 8.1
Consider a linear dynamic system described by the following Differential Equation (DE)
with constant coefficients:
y (t ) 5 y(t ) 7 y(t ) 3 y(t ) u(t )
Find the state variable model associated with this system. Verify that the eigenvalues of
the state matrix A are coincident with the roots of the characteristic equation of the above
system.
Solution of Problem 8.1.
The system is a 3rd order system with 1 input and 1 output n 3, m 1, l 1 . A suitable
selection of state variables is given by:
x1 (t ) y (t )
d y (t )
x2 (t ) y (t )
dt
d 2 y (t )
x3 (t ) y (t ) x3 (t ) y (t )
dt 2
Using the system DE equation,
x3 (t ) y (t ) 5 y(t ) 7 y(t ) 3 y(t ) u(t )
Rewriting the above equations using the state variables will lead to the following:
x1 (t ) y (t )
x1 (t ) x2 (t )
x2 (t ) x3 (t )
x3 (t ) 5 x3 (t ) 7 x2 (t ) 3 x1 (t ) u (t )
Thus, the state variable model in a matrix format is given by:
x1 (t ) 0 1 0 x1 (t ) 0
x2 (t ) 0 0 1 x2 (t ) 0 u (t )
x (t ) 3 7 5 x (t ) 1
3 3
,The characteristic equation is given by:
s3 5s 2 7s 3 0
with roots :
1 1, 1 1, 1 3
The eigenvalues of the system matrix are,
0 1 0
eig 0 0 1 1, 1, 3
3 7 5
The SV output equation is given by,
x1 (t )
y 1 0 0 x2 (t ) 0u (t )
x (t )
3
,Problem 8.2
Consider a linear dynamic system described by the following set of Differential
Equations (DEs) with constant coefficients:
y1 (t ) 5 y2 (t ) u1 (t )
y2 (t ) 3 y1 (t ) 2 y2 (t ) u2 (t )
- Find the state variable model of the above system.
- The system is subjected to 2 inputs u1 (t ), u2 (t ) and has two measurable outputs
y1 (t ), y2 (t ) . Therefore, a transfer function-based modeling of the system will require
the evaluation of a (2 x 2) matrix of transfer functions. Find the matrix of transfer
functions under the following conditions:
- Derivation directly from the DEs
- Derivation from the state variable model.
Solution of Problem 8.2
State Variable Model
The system is a 2nd order system with 2 inputs and 2 outputs n 2, m 2, l 2 . A
suitable selection of state variables is given by:
x1 (t ) y1 (t ) x1 (t ) y1 (t )
x2 (t ) y2 (t ) x2 (t ) y2 (t )
Using the system DE equation,
x1 (t ) 5 x2 (t ) u1 (t )
x2 (t ) 3x1 (t ) 2 x2 (t ) u2 (t )
Rearranging:
x1 (t ) 5 x2 (t ) u1 (t )
x2 (t ) 3x1 (t ) 2 x2 (t ) u2 (t )
Thus, the SV state equation is given by:
, x1 (t ) 0 5 x1 (t ) 1 0 u1 (t )
x2 (t ) 3 2 x2 (t ) 0 1 u2 (t )
The SV output equation is given by,
y1 (t ) 1 0 x1 (t ) 0 0 u1 (t )
y2 (t ) 0 1 x2 (t ) 0 0 u2 (t )
Derivation of Transfer Function (TF) matrix DIRECTLY from the DEs
Following the application of the Laplace transformation to the above system we have:
sY1 ( s) 5Y2 ( s) U1 ( s)
sY2 ( s) 3Y1 ( s) 2Y2 ( s) U 2 ( s)
leading to the following matrix format:
s 5 Y1 ( s) U1 ( s)
3 ( s 2) Y ( s) U ( s)
2 2
By invoking the „superimposition of effect‟ property of linear systems, we can solve the
above system first considering u2 (t ) 0 U 2 (s) 0 and, next, considering
u1 (t ) 0 U1 ( s) 0 . The goal is to find all the Gij ( s) transfer functions in the
relationship:
Y1 ( s) G11 ( s) U1 ( s) G12 ( s) U 2 ( s)
Y2 ( s) G21 ( s) U1 ( s) G22 ( s) U 2 ( s)
Therefore, consider the system:
s 5 Y1 ( s) U1 ( s)
3 ( s 2) Y ( s) 0
2
Using Cramer‟s rule we have: