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FURTHER READING |
185 |
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[Evans, 1954] |
Evans, W.R. Control System Dynamics, McGraw-Hill, New York, 1954. |
|
[Houpis and Lamont, 1962] |
Houpis, C.H. and Lamont, G.B. Digital Control Systems: Theory, Hardware, Soft- |
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ware, 2nd edn., McGraw-Hill, New York, 1962. |
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[Hsu and Meyer, 1968] |
Hsu, J.C. and Meyer, A.U. Modern Control Principles and Applications. McGraw- |
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Hill, New York, 1968. |
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[Jury, 1958] |
Jury, E.I. Sampled-Data Control Systems. John Wiley & Sons, Inc., New York, 1958. |
|
[Katz, 1981] |
Katz, P. Digital Control Using Microprocessors. Prentice Hall, Englewood Cliffs, NJ, |
|
1981. |
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[Kuo, 1963] |
Kuo, B.C. Analysis and Synthesis of Sampled-Data Control Systems. Prentice Hall, |
|
Englewood Cliffs, NJ, 1963. |
||
[Lindorff, 1965] |
Lindorff, D.P. Theory of Sampled-Data Control Systems. John Wiley & Sons, Inc., |
|
New York, 1965. |
||
[Ogata, 1990] |
Ogata, K. Modern Control Engineering, 2nd edn., Prentice Hall, Englewood Cliffs, |
|
NJ, 1990. |
||
[Phillips and Harbor, 1988] |
Phillips, C.L. and Harbor R.D. Feedback Control Systems. Englewood Cliffs, NJ, |
|
Prentice Hall, 1988. |
||
[Raven, 1995] |
Raven, F.H. Automatic Control Engineering, 5th edn., McGraw-Hill, New York, 1995. |
|
[Strum and Kirk, 1988] |
Strum, R.D. and Kirk D.E. First Principles of Discrete Systems and Digital Signal |
|
Processing. Addison-Wesley, Reading, MA, 1988. |
||
8
System Stability
This chapter is concerned with the various techniques available for the analysis of the stability of discrete-time systems.
Suppose we have a closed-loop system transfer function
Y (z) |
= |
G(z) |
= |
N (z) |
|||||
, |
|||||||||
R(z) |
1 |
+ |
GH(z) |
D(z) |
|||||
where 1 + GH(z) = 0 is also known as the characteristic equation. The stability of the system depends on the location of the poles of the closed-loop transfer function, or the roots of the characteristic equation D(z) = 0. It was shown in Chapter 7 that the left-hand side of the s-plane, where a continuous system is stable, maps into the interior of the unit circle in the z- plane. Thus, we can say that a system in the z-plane will be stable if all the roots of the characteristic equation, D(z) = 0, lie inside the unit circle.
There are several methods available to check for the stability of a discrete-time system:
Factorize D(z) = 0 and find the positions of its roots, and hence the position of the closedloop poles.
Determine the system stability without finding the poles of the closed-loop system, such as Jury’s test.
Transform the problem into the s-plane and analyse the system stability using the wellestablished s-plane techniques, such as frequency response analysis or the Routh–Hurwitz criterion.
Use the root-locus graphical technique in the z-plane to determine the positions of the system poles.
The various techniques described in this section will be illustrated with examples.
8.1 FACTORIZING THE CHARACTERISTIC EQUATION
The stability of a system can be determined if the characteristic equation can be factorized. This method has the disadvantage that it is not usually easy to factorize the characteristic equation. Also, this type of test can only tell us whether or not a system is stable as it is. It does not tell us about the margin of stability or how the stability is affected if the gain or some other parameter is changed in the system.
Microcontroller Based Applied Digital Control D. Ibrahim
C 2006 John Wiley & Sons, Ltd. ISBN: 0-470-86335-8
188 System Stability
e(s) |
e*(s) |
1 − e−Ts |
y(s) |
||||||||||||||
r(s) |
4 |
||||||||||||||||
s |
|||||||||||||||||
− |
s + |
2 |
|||||||||||||||
+ |
|||||||||||||||||
Figure 8.1 Closed-loop system
Example 8.1
The block diagram of a closed-loop system is shown in Figure 8.1. Determine whether or not the system is stable. Assume that T = 1 s.
Solution
The closed-loop system transfer function is
Y (z) |
= |
G(z) |
(8.1) |
|||
, |
||||||
R(z) |
1 |
+ |
G(z) |
|||
where
1 − e−T s
G(z) = Z
s
= 2(1 − e−2T ) . z − e−2T
For T = 1 s,
4 |
(1 |
z−1)Z |
4 |
(1 |
z−1) |
2z(1 − e−2T ) |
|||||||||||||||||||||
s |
2 |
= |
− |
s(s |
2) |
= |
− |
||||||||||||||||||||
+ |
+ |
(z |
− |
1)(z |
− |
e |
− |
2T ) |
|||||||||||||||||||
(8.2) |
|||||||||||||||||||||||||||
G(z) = |
1.729 |
. |
|||||||||||||||||||||||||
z |
− |
0.135 |
|||||||||||||||||||||||||
The roots of the characteristic equation are 1 + G(z) = 0, or 1 + 1.729/(z − 0.135) = 0, the solution of which is z = −1.594 which is outside the unit circle, i.e. the system is not stable.
Example 8.2
For the system given in Example 8.1, find the value of T for which the system is stable.
Solution
From (8.2),
G(z) = 2(1 − e−2T ) . z − e−2T
The roots of the characteristic equation are 1 + G(z) = 0, or 1 + 2(1 − e−2T )/(z − e−2T ) = 0, giving
z − e−2T + 2(1 − e−2T ) = 0
JURY’S STABILITY TEST |
189 |
or
z = 3e−2T − 2.
The system will be stable if the absolute value of the root is inside the unit circle, i.e.
|3e−2T − 2|< 1,
from which we get
2T < ln |
31 |
or T < 0.549. |
Thus, the system will be stable as long as the sampling time T < 0.549.
8.2 JURY’S STABILITY TEST
Jury’s stability test is similar to the Routh–Hurwitz stability criterion used for continuoustime systems. Although Jury’s test can be applied to characteristic equations of any order, its complexity increases for high-order systems.
To describe Jury’s test, express the characteristic equation of a discrete-time system of order n as
F (z) = an zn + an−1zn−1 + . . . + a1z + a0 = 0, |
(8.3) |
where an > 0. We now form the array shown in Table 8.1. The elements of this array are defined as follows:
The elements of each of the even-numbered rows are the elements of the preceding row, in reverse order.
The elements of the odd-numbered rows are defined as:
bk |
an |
ak |
, |
ck |
nn 1 |
bk |
, |
dk |
cn 2 |
ck |
, |
. |
|||||||
= |
a0 |
an−k |
= |
b0 |
bn−k−1 |
= |
c0 |
cn−2−k |
· · · |
||||||||||
− |
− |
||||||||||||||||||
Table 8.1 Array for Jury’s stability tests
z0 |
z1 |
z2 |
. . . |
zn−k |
. . . |
zn−1 |
zn |
a0 |
a1 |
a2 |
. . . |
an−k |
. . . |
an−1 |
an |
an |
an−1 |
an−2 |
. . . |
ak |
. . . |
a1 |
a0 |
b0 |
b1 |
b2 |
. . . |
bn−k |
. . . |
bn−1 |
|
bn−1 |
bn−2 |
bn−3 |
. . . |
bk−1 |
. . . |
b0 |
|
c0 |
c1 |
c2 |
. . . |
cn−k |
. . . |
||
cn−2 |
cn−3 |
cn−4 |
. . . |
ck−2 |
. . . |
||
. . . |
. . . |
. . . |
. . . |
. . . |
|||
. . . |
. . . |
. . . |
. . . |
. . . |
|||
l0 |
l1 |
l2 |
l3 |
||||
l3 |
l2 |
l1 |
l0 |
||||
m0 |
m1 |
m2 |
190 System Stability
The necessary and sufficient conditions for the characteristic equation (8.3) to have roots inside the unit circle are given as
F (1) > 0, (−1)n F (−1) > 0, |a0|< an ,
|b0| > bn−1 |c0| > cn−2 |d0| > dn−3
. . .
. . .
|m0| > m2.
Jury’s test is then applied as follows:
(8.4)
(8.5)
Check the three conditions given in (8.4) and stop if any of these conditions is not satisfied.
Construct the array given in Table 8.1 and check the conditions given in (8.5). Stop if any condition is not satisfied.
Jury’s test can become complex as the order of the system increases. For systems of order 2 and 3 the test reduces to the following simple rules. Given the second-order system characteristic equation
F (z) = a2z2 + a1z + a0 = 0, where a2 > 0,
no roots of the system characteristic equation will be on or outside the unit circle provided that
F (1) > 0, F (−1) > 0, |a0|< a2..
Given the third-order system characteristic equation
F (z) = a3z3 + a2z2 + a1z + a0 = 0, where a3 > 0,
no roots of the system characteristic equation will be on or outside the unit circle provided that
F (1) > 0, |
F (−1) < 0, |
|a0|< a3, |
||||||
det |
a3 |
a0 |
> |
det |
a3 |
a2 |
. |
|
a0 |
a3 |
a0 |
a1 |
|||||
Examples are given below.
Example 8.3
The closed-loop transfer function of a system is given by
G(z) |
||||||
, |
||||||
where |
1 + G(z) |
|||||
G(z) |
0.2z + 0.5 |
. |
||||
= z2 |
||||||
− 1.2z + 0.2 |
||||||
Determine the stability of this system using Jury’s test.