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.