Page 367 - Modern Control Systems

Section 5.8 The Simplification of Linear Systems 341

EXAMPLE 5.9 A simplified model
Consider the third-order system

G 5 = - 5 57
"<> = w ^ n XA — r r ^ r - <- >
r + 6.r + 11,+6 U + ^ + 1 ,
Using the second-order model

1 + d\S + d 2S
we determine that

11 1
2
M(s) = 1 + d hs + d 2s , and A(.y) = 1 + — .v + s 2 + -s\
6 6
Then we know that
2
(0)
M (s) = 1 + rfj* + ^- , (5.59)
(0)
and M (0) = 1. Similarly, we have
! 2
M<> = — (1 + rf,.y + d 2s ) = </, + 2d 2s. (5.60)
as
(l
Therefore, M ^(0) = d h Continuing this process, we find that
(0)
(0)
M (0) = 1 A (0) = 1,
(l)
(1)
M (0) = d A A (0) = -^-,
6
(2)
(2)
M (0) = 2d 2 A (0) = 2, (5.61)
and
(3)
(3)
M (0) = 0 A (0) = 1.
We now equate M 2q = A 2q for q = 1 and 2. We find that, for q = 1,
(1)
(1)
(0)
(2)
(0)
(2)
M (0)M (0) M (0)M (0) M (0)M (0)
(
(
M 2 = -1) ^ ^ + ^ ^ + -1) ^ ^
2 2
= -rf 2 + a", - d 2 = -2d 2 + d x . (5.62)
Since the equation for A 2 is similar, we have
(2) (())
A(Q)( 0) A(2)( 0) A (i)( 0) d) ( 0 ) ^ A (0) A (0)
A