Mathematical Models of Systems
Gs
121
Y s
K
U s sTs 1
3:12-35
where
K
!
BK2 A 1
AK1 K1
and
T
mK2
BK2 A2
If we assume that T 0, then Gs K =s. Therefore, we observe that the hydraulic
actuator of Figu
Time-Domain Analysis
Y s
157
!2
n
ss 2 !2
d
If we expand Y s in partial fractions, we have
1
s 2
1
s
Y s
2
2
2
2
s s ! d s s ! d
!d
!"
!d
s 2 !2
d
#
Taking the inverse Laplace transform, we have
yt
Mathematical Models of Systems
12.
13.
14.
15.
16.
17.
145
T Kailath. Linear Systems. Englewood Cliffs, New Jersey: Prentice Hall, 1980.
BC Kuo. Automatic Control Systems. London: Prentice Hall, 1995.
Mathematical Models of Systems
133
3:13-29
qy qv q
Since qv ) q , we have that qy qv K constant. Hence Eq. (3.13-28) becomes
dxy K
x
x q
dt
Vy V
3:13-30
Relation (3.13-30) describes the open-loop syst
Time-Domain Analysis
H
SK
K
H
!
169
!
!
dH
K ss a K ss a K K
s s a
2
dK
K
ss a K
s s a K
!
Similarly, the sensitivity of H s with respect to a is given by
!
!
h a i dH !
ass a K
sK
as
H
Sa
2
H da
Time-Domain Analysis
Figure 4.19
181
System with input rt and disturbance d t.
ps s3 s2 K1 K2 1s K1 K2
we form the Routh table
K1 K2 1
s3 1
K1 K2
s2 1
0
s1 1
0
K1 K2
0
s
Hence, in order for ps to be s
State-Space Analysis
229
Figure 5.9
2. For the RLC network shown in Figure 5.10a, determine (a) the state equations,
using the denitions of input, state, and output variables shown in Figure 5.10b,
(b
State-Space Analysis
217
t
rt Bu d
xt rt t0 xt0
5:6-4
t0
To simplify the proof, let xtf 0. Then, solving Eq. (5.6-4) for xt0 gives
t
tf
f
1
xt0 r tf t0 rtf Bu d rtf t0 tf Bu d
t0
tf
t0
rt0 Bu d
5:6
State-Space Analysis
205
C* CT
5:4-3c
D* D
5:4-3d
z0 T1 x0
5:4-3e
For brevity, systems (5.1-1) and (5.4-2) are presented as A; B; C; Dn and as
A* ; B* ; C* ; D* n , respectively. System A* ; B* ; C* ;
5
State-Space Analysis of Control
Systems
5.1
INTRODUCTION
The classical methods of studying control systems are mostly referred to singleinputsingle-output (SISO) systems, which are described in the