lecture 6: stability of discrete-time systems 1. suppose that we have the following transfer...
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Lecture 6:
Stability of discrete-time systems
1
• Suppose that we have the following transfer function of a closed-loop discrete-time system:
• The system is stable if all the poles (roots of the characteristic equation D(z) = 0) lie inside the unit circle in the z-plane.
2
.)()(
)(1)(
)()(
zDzN
zGHzG
zRzY
Stability of discrete-time systems
There are several methods to check the stability of a discrete-time system such as:
• Factorizing D(z) = 0 and finding its roots.
• Jury’s test.
• Routh–Hurwitz criterion .
3
Stability of discrete-time systems
Factorizing the characteristic equation
• The direct method to check system stability is to factorize the characteristic equation, determine its roots, and check if their magnitudes are all less than 1.
• Although it is not usually easy to factorize the characteristic equation by hand, this is easy using MATLAB command “roots”.
4
Example 1
Check the stability of the following closed-loop discrete-time system. Assume that T = 1 s.
Solution:• The transfer function of the closed-loop system is
5,)(1
)()()(
zGzG
zRzY
• Where
• The characteristic equation is thus
6
.135.0
729.1)()1(2
))(1()1(2)1(
241)(
sec12
2
2
21
zeze
ezzezz
sseZzG
TT
T
T
T
Ts
unstable is system1||5941
0594101
z.z.z
G(z)
Example 2
In the previous example, find the value of T for which the system is stable.
Solution:• From the previous example, we found
7.)()1(2)( 2
2
T
T
ezezG
• The characteristic equation is
• For stability, the condition |z|<1 must be satisfied;
• Thus the system is stable as long as T < 0.549. 8
23
023
0)(1
2
2
T
T
ez
ez
zG
0.549 00)3/1ln(5.0
02)3/1ln(1231
1|23|||2
2
TT
Te
ezT
T
Jury’s stability test• Jury’s stability test is similar to the Routh–Hurwitz stability
criterion used for continuous systems. • To describe Jury’s test, express the characteristic equation of a
discrete-time system of order n as
• Then, we form the following:
9
,0,0)( 012
21
1 n
nn
nn awhereazazazazazF
• The elements of this array are defined as follows:
The elements of each even-numbered row are the elements of the preceding row, in reverse order.
The elements of the odd-numbered rows are defined as:
10
.,,1
100 kn
knk
kn
knk bb
bbc
aaaa
b
Jury’s stability test
The necessary and sufficient conditions for the characteristic equation to have all roots inside the unit circle are given as
Jury’s test is applied as follows:• Check the three conditions (I) and stop if any of them is not
satisfied.• Construct Jury’s array and check the conditions (II). Stop if any
condition is not satisfied.11
.
)(
20
30
20
10
mm
IIdd
cc
bb
n
n
n
,||)(,0)1()1(
,0)1(
0 n
n
aaIF
F
Jury’s stability test
Jury test for 2nd order polynomial
For 2nd characteristic equation:
Jury’s test reduces to the following simple rules: no roots of the system characteristic equation will be on or outside the unit circle provided that
12
.||,0)1(,0)1( 20 aaFF
,0,0)( 2012
2 awhereazazazF
Jury test for 3rd order polynomial
For 3rd order characteristic equation:
Jury’s test reduces to the following simple rules: no roots of the system characteristic equation will be on or outside the unit circle provided that
13.detdet
,||,0)1(,0)1(
23
10
03
30
30
aaaa
aaaa
aaFF
,0,0)( 3012
23
3 awhereazazazazF
Example 3
The closed-loop transfer function of a system is given by
Where
Determine the stability of this system using Jury’s test.
14
,)(1
)(zGzG
.2.02.1
5.02.0)( 2
zz
zzG
Solution
• The characteristic equation is
• Or
• Applying Jury’s test
• All conditions are satisfied and the system is stable. 15
,02.02.1
5.02.01)(1 2
zzzzG
.07.02 zz
.17.0||,07.2)1(,07.0)1( 20 aaFF
Determine the stability of the system having the following characteristic equation:
16
Example 4
.01.04.12)( 23 zzzzF
Example 5
The block diagram of a sampled data system is shown below. Use Jury’s test to determine the value of K for which the system is stable. Assume that K > 0 and T = 1 s.
17
18
• Apply Jury’s test:
• The third condition is
• Combining all inequalities together, the system is stable for K < 2.4 19
3.260104.0736.2)1(00632.0)1(
KKFKKF
4.218.51264.0368.01
1264.0368.0
KK
K
Determine the stability of the system having the following characteristic equation:
Example 6
.05.022)( 234 zzzzzF
40
4
15.0,05.02211)1()1(
,05.6)1(
aaF
F
75.05.11.6875-75.0015.15.11075.0
5.0221111225.0
43210
zzzzz
|b0|=0.75< |b3|=1.5
|c0|=1.6875> |c2|=0.75
System is unstable!20
Routh–Hurwitz Criterion• The stability of a sampled data system can be analyzed by
transforming the system characteristic equation into the s-plane and then applying the well-known Routh–Hurwitz criterion.
• A bilinear transformation is usually used to transform the interior of the unit circle in the z-plane into the left-hand s-plane (or w-plane). For this transformation, z is replaced by
giving the characteristic equation in w ,21
,11
z
.0)( 011
1 bbbbF nn
nn
Routh–Hurwitz Criterion• The Routh-Hurwitz array is formed as
shown.• The first two rows are obtained from
the equation directly and the other rows are calculated as shown.
• The Routh–Hurwitz criterion states that the number of roots of the characteristic equation in the right hand s-plane is equal to the number of sign changes of the coefficients in the first column of the array.
• Thus, for a stable system all coefficients in the first column must have the same sign. 22
1
5412
1
3211
n
nnnn
n
nnnn
bbbbbc
bbbbbc
Example 7• The characteristic equation of a sampled data system is given
by
• Determine the stability of the system using the Routh–Hurwitz criterion.
23
.012 23 zzz
.0537
.01)1(11)1(12
.0111
11
112
23
3223
23
• Now, we form Routh array:
• To check the answer, the roots of the characteristic equation,
are found using Matlab command roots([ 2 1 1 1]) to be0.1195 + 0.8138i0.1195 - 0.8138i
-0.7390 • As we are interested their magnitudes, we can write directly
abs(roots([ 2 1 1 1])). This gives 0.8226, 0.8226, 0.7390 which are all less than one, i.e. the roots lie inside the unit circle. Hence, we can conclude that the system is stable.
24
57/16
5731
0
1
2
3
,012 23 zzz
No sign change in the first column, so the system is stable.
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