CLASSICAL CONTROL 27
2.3 Closed-lo op stability
One of the main issues in designing feedback controllers is stability. If the
feedback gain is to o large, then the controller may\overreact" and the closed-
loop system b ecomes unstable. This is illustrated next by a simple example.
0 5 10 15 20 25 30 35 40 45 5
0.5
0
0.5
1
1.5
2
K
c
=2
:
5
K
c
=1
:
5
K
c
=0
:
5
K
c
= 3 (Unstable)
P
u
Time sec]
Figure 2.5
: Eect of prop ortional gain
K
c
on the closed-loop response
y
(
t
)ofthe
inverse resp onse pro cess.
Example 2.1
Inverse response pro cess.
Consider the plant (time in seconds)
G
(
s
)=
3(
;
2
s
+1)
(5
s
+1)(10
s
+1)
(2.27)
This is one of two main example processes used in this chapter to il lustrate the
techniques of classical control. The model has a right-half plane (RHP) zero at
s
= 0
:
5
rad/s]. This imposes a fundamental limitation on control, and high
controller gains wil l induce closed-loop instability.
This is il lustrated for a proportional (P) controller
K
(
s
) =
K
c
in Figure 2.5,
where the response
y
=
Tr
=
GK
c
(1 +
GK
c
)
;
1
r
to a step change in the reference
(
r
(
t
)=1
for
t>
0
) is shown for four dierent values of
K
c
. The system is seen
to be stable for
K
c
<
2
:
5
, and unstable for
K
c
>
2
:
5
. The control ler gain at the
limit of instability,
K
u
= 2
:
5
, is sometimes called the ultimate gain and for this
value (
K
c
=
K
u
) the system is seen to cycle continuously with a period
P
u
=15
:
2
s,
corresponding to the frequency
!
u
=2
=P
u
=0
:
42
rad/s].
Two metho ds are commonly used to determine closed-lo op stability:
1. The poles of the closed-loop system are evaluated. That is, the zeros of
1+
L
(
s
) = 0 are found where
L
is the transfer function around the lo op.
The system is stable
if and only if
all the closed-lo op p oles are in the op en
left-half plane (LHP) (that is, p oles on the imaginary axis are considered
\unstable"). The poles are also equal to the eigenvalues of the state-space
A
-matrix, and this is usually how the p oles are computed numerically.
28 MULTIVARIABLE FEEDBACK CONTROL
2. The frequency resp onse (including region in frequencies) of
L
(
j!
) is
plotted in the complex plane and the number of encirclements it makes
of the critical p oint
;
1iscounted. By Nyquist's stability criterion (as is
illustrated in Figure 2.12 and for which a detailed statement is given in
Theorem 4.14) closed-loop stability is inferred by equating the number
of encirclements to the number of open-lo op unstable p oles (RHP-p oles).
For op en-loop stable systems where
6
L
(
j!
) falls with frequency such that
6
L
(
j!
) crosses
;
180
only once (from ab ove at frequency
!
180
), one may
equivalently use
Bode's stability condition
whichsays that the closed-loop
system is stable if and only if the lo op gain
j
L
j
is less than 1 at this
frequency, that is
Stability
, j
L
(
j!
180
)
j
<
1 (2.28)
where
!
180
is the phase crossover frequency dened by
6
L
(
j!
180
)=
;
180
.
Method 1, whichinvolves computing the p oles, is best suited for numerical
calculations. However, time delays must rst be approximated as rational
transfer functions, e.g., Padeapproximations. Metho d 2, which is based on
the frequency resp onse, has a nice graphical interpretation, and may also b e
used for systems with time delays. Furthermore, it provides useful measures
of relative stability and forms the basis for several of the robustness tests used
later in this b o ok.
Example 2.2
Stability of inverse response pro cess with prop ortional
control.
Let us determine the condition for closed-loop stability of the plant
G
in
(2.27) with proportional control, that is, with
K
(
s
)=
K
c
and
L
(
s
)=
K
c
G
(
s
)
.
1. The system is stable if and only if al l the closed-loop poles are in the LHP. The
poles are solutions to
1+
L
(
s
)=0
or equivalently the roots of
(5
s
+ 1)(10
s
+1)+
K
c
3(
;
2
s
+1) = 0
,
50
s
2
+ (15
;
6
K
c
)
s
+(1+3
K
c
)=0 (2.29)
But sinceweare only interested in the half plane
location
of the poles, it is not
necessary to solve (2.29). Rather, one may consider the coecients
a
i
of the
characteristic equation
a
n
s
n
+
a
1
s
+
a
0
= 0
in (2.29), and use the Routh-
Hurwitz test to check for stability. For second order systems, this test says that
we have stability if and only if al l the coecients have the same sign. This yields
the fol lowing stability conditions
(15
;
6
K
c
)
>
0 (1 + 3
K
c
)
>
0
or equivalently
;
1
=
3
< K
c
<
2
:
5
. With negative feedback (
K
c
0
) only the
upper bound is of practical interest, and we nd that the maximum al lowedgain
(\ultimate gain") is
K
u
= 2
:
5
which agrees with the simulation in Figure 2.5.
The poles at the onset of instability may be found by substituting
K
c
=
K
u
=2
:
5