Lecture notes - School Of Electrical & Electronic Engineering - USM

Stability Concept 

Bilinear Transformation 

Routh-Hurwitz Criterion 

Jury’s Stability test 

DIGITAL CONTROL SYSTEM 

EEE354 

Stability Analysis Techniques 

MUHAMMAD NASIRUDDIN MAHYUDDIN 

School of Electrical and Electronic Engineering, 

UNIVERSITI SAINS MALAYSIA 

week 9 

Semester II 2007/2008 

nasiruddin@eng.usm.my EEE 354 : STABILITY ANALYSIS TECHNIQUES

Outline 





week 9: 11/2/2008 - 15/2/2008 

1 Stability Concept 

Introduction: Stability Concept in Digital Control System 

2 Bilinear Transformation 

Definition 

Application of Bilinear Transformation 

3 Routh-Hurwitz Criterion 

Steps taken before using RH criterion 

4 Jury’s Stability test 

no need to do bilinear transformation 

Jury’s Stability Criterion 






Introduction to Stability Concept 


Generally, the stability analysis techniques applicable to LTI 

continuous-time systems may also be applied to the analysis of 

LTI discrete time systems, if certain modifications are made. 

These techniques include:- 

Routh-Hurwitz criterion 

root-locus procedures 

frequency response method such as:- 

bode plot 

nichols chart 








Consider the following closed-loop pulse transfer function 

system: 

C(z) 

R(z) = 

G(z) 

1+GH(z) 

The stability of the system defined by the above equation may 

be determined from the locations of the closed loop in the 

z-plane, or the roots of the characteristic equation: 

P(z) = 1 + GH(z) 








1 For the system to be stable, the closed-poles or the roots of the 

characteristic equation must lie within the unit circle in the 

z-plane. 

2 The system become critically stable if a simple pole lies at z=1 

or a single pair of conjugate complex poles lies on the unit circle 

in the z-plane. 

3 Any multiple closed-pole on the unit circle makes the system 

unstable 

4 Closed-loop zero do not affect the absolute stability and 

therefore may be located anywhere in the z-plane 






Definition of bilinear transformation 

Definition 


The bilinear transformation is defined by: 

Solving for w, gives: 

z = w+1 

w−1 

w = z+1 

z−1 

The transformation maps the interior of the unit circle in the z-plane 

into the left-half of the w plane. 


mapping of z-plane 





Definition 


Figure 1 below shows the mapping of the unit circle in the 

z-plane onto the imaginary axis of the w-plane. 

Figure 1: mapping of the unit circle in two planes 






bilinear transformation 

Definition 


Once the characteristic equation P(z) = 0 of the digital system 

is obtained, w = z+1 

z−1 

is substituted for z in the characteristic 

equation as follows: 

P(z) = a 0 z n + a 1 z n−1 + a 2 z n−2 + · · · + a n−1 z + a n = 0 (1) 






bilinear transformation I 

Definition 


The characteristic equation in w -domain then becomes: 

P(w) = a 0 

[ w + 1 

w − 1 

. . . +a n−1 

[ w + 1 

w − 1 

] n [ w + 1 

+ a 1 

w − 1 

] n−1 [ ] w + 1 n−2 

+ a 2 + 

w − 1 

] 

+ a n = 0 (2) 

P(w) = b 0 w n + b 1 w n−1 + · · · + b n−1 w + b n = 0 (3) 






bilinear transformation 

Definition 


The bilinear transformation transforms the z- plane into w plane. The 

frequency response of G(z) is G(jω), that is, by replacing z with jω. 

Equivalently, for w plane, the frequency response of replaced by a 

fictitious frequency jv . 

The relationship between the fictitious frequency and the desired 

frequency v can be obtained as follows: 

w∣ = z−1 ∣ 

w=jv 

z+1 

∣ 

z=e jωT 

jv = j tan ωT 

2 

v = tan ωT 

2 






Definition 


Application of Bilinear transformation I 

For stability analysis using frequency methods, such as the Bode 

diagram technique, another practical version of w- transformation was 

proposed: 

w ′ = 2 T w 

w ′ = 2 T 

z − 1 

z + 1 

where T is the sampling period. This implies that, 

(4) 

z = 1 + T 2 w′ 

1 − T 2 w′ = 

2 

T + w′ 

2 

T − w′ (5) 






Definition 


Application of bilinear transformation 

The w ′ -transformation is a bilinear transformation scaled with the 

factor of T 2 . 

The relationship between the actual/desired frequency ω and the 

fictitious frequency v ′ 

is: 

v ′ 

= 2 ( ) ωT 

T tan 2 

(6) 

(7) 






Routh- Hurwitz Criterion 


The Routh-Hurwitz criterion cannot be applied directly to the 

z-domain since the stability boundary is now different. The 

method requires transformation from z-plane to another plane, 

the w plane. 

Once the characteristic equation P(z) = 0 is transformed into a 

polynomial of the same order in w : P(w) = 0, the R-H criterion 

can then be applied directly as in the case of continuous data 

system. 








Worked Examples on 



Jury’s Stability Test 







Let the CE P(z) = 0 of a digital system given in polynomial of z: 

P(z) = a n z n + a n−1 z n−1 + a n−2 z n−2 + · · · + a 2 z 2 + a 1 z + a 0 = 0 (8) 

where a n > 0 or it can be made positive by changing the sign of all the 

coeffecients, and a i , are real coeffecients. 


Jury’s Stability Test 







In applying the Jury’s stability test, a Jury Stability table must 

first be constructed as shown in Table 1 

Table 1: Jury’s Table 

row z 0 z 1 z 2 z 3 . . . z n−1 z n−1 z n 

1 a 0 a 1 a 2 a 3 . . . a n−2 a n−1 a n 

2 a n a n−1 a n−2 a 2 . . . a 1 a 0 

3 b 0 b 1 b 2 b 3 . . . b n−2 b n−1 b n 

4 b n b n−1 b n−2 b n−3 . . . b 2 b 1 b 0 

2n-5 p 0 p 1 p 2 p 3 . . . 

2n-4 p 3 p 2 p 1 p 0 . . . 

2n-3 q 0 q 1 q 2 






Method to construct Jury’s Table 



The Jury’s table is constructed as follows: 

1 The table consists of (2n − 3) rows. For second-order system, 

the table has only 1 row. 

2 The first row: the elements consist of the coeffecients in P(z) 

arranged in the ascending order of powers in z 

3 The second row: the elements consist of the coeffecients of P(z) 

arranged in the descending order of powers in z 








The elements of rows 3 through (2n − 3) are given by the 

following determinants: 

b k = 

c k = 

d k = 

q 0 = 

∣ a ∣ 

0 a n−k ∣∣∣ 

, k = 0, 1, 2, . . . , n − 1. (9) 

a n a k ∣ b 0 b n−1−k ∣∣∣ 

∣ 

, k = 0, 1, 2, . . . , n − 2. (10) 

b n−1 b k ∣ c 0 c n−2−k ∣∣∣ 

∣ 

, k = 0, 1, 2, . . . , n − 3. (11) 

c n−2 c k ∣ p ∣ 0 p 3 ∣∣∣ , q 

p 3 p 2 = 

0 

∣ p ∣ 

0 p 1 ∣∣∣ 

(12) 

p 3 p 2 


Jury’s stability test 







1 The elements in any even-numbered row ( row 2,4, . . . ) 

are simply the reverse of the immediately preceding 

odd-numbered row ( row 1,3, . . . ) 

2 The last row (row 2n − 3) has 3 elements. 









The stability criterion by the Jury’s Stability Test: For the 

polynomial P(z) to have no roots on and outside the unit circle 

in the z-plane (i.e. for the digital or discrete data system to be 

stable), the following conditions must be satisfied: 

a) P(z)| z=1 = P(1) > 0 (13) 

{ > 0 for n even 

b) P(z)| z=−1 = P(−1) 

< 0 for n odd 

(14) 









c) 

|a 0 | < |a n | 

|b 0 | > |b n−1 | 

|c 0 | > |c n−2 | 

|d 0 | > |d n−3 | 

. 

|q 0 | > |q 2 | 

} 

(n − 1) constraints (15) 









For example, a second-order system would have n = 2. 

Therefore, Jury’s stability table contains only 1 row. Thus, for 

stability: 

P(1) > 0 

P(−1) > 0 (16) 

|a 0 | < |a 2 | 

(17) 


DIGITAL CONTROL SYSTEM EEE354 

(Sistem Kawalan Digit) 

Worked Examples 

prepared by Nasiruddin ∗ 

semester II 2007/2008 

1 Example on Routh Hurwitz 

Criterion 

By using Routh-Hurwitz stability criterion, determine 

the stability of the following digital systems 

whose characteristic are given as: 

a) z 2 − 0.25 = 0 

b) z 3 − 1.2z 2 − 1.375z − 0.25 

1.1 Solution to Example 1(a) 

Transforming the characteristic equation z 2 − 0.25 = 

0 into w-domain by using the bilinear transformation 

z = w+1 

w−1 , gives: 

0.75w 2 + 2.5w + 0.75 = 0 (1) 

It can be observed that all the coeffecients are of the 

same sign, and none of the coeffecients is zero. Thus, 

all the roots of the tranformed equation 1 are in the 

left-half of the w-plane. Hence, all the roots of the 

∗ Muhammad Nasiruddin Mahyuddin (Mechatronic Programme, 

School Of Electrical and Electronic Engineering, USM) 

characteristic equation in the z-domain are inside the 

unit-circle in the z-plane. 

Thus, the system is stable. 

1.2 Solution to Example 1(b) 

Transforming the characteristic equation z 3 −1.2z 2 − 

1.375z −0.25 = 0 into w-domain by using the bilinear 

transformation z = w+1 


− 1.875w 3 + 3.875w 2 + 4.875w + 1.125 = 0 (2) 

Based on the transformed CE, the Routh tabulation 

is constructed as follows: 

w 3 -1.875 4.875 

w 2 3.875 1.125 

w 1 5.419 0 

w 0 1.125 

From the table above, since there is one sign change 

in the first column, equation 2 has one root in the 

right-half of the w-plane. This, in turn, implies that 

there will be one root of the characteristic equation 

outside of the unit circle in the z-plane. 

1

Thus, the system is not stable. 

1.3 Example 2 

By using RH criterion, determine the stability of a 

digital system whose CE is given as P (z) = z 3 + 

3.3z 2 + 3z + 0.8 = 0. 

1.3.1 Solution to Example 2 

Transforming the CE P (z) into w-domain by using 

bilinear transformation z = w+1 


8.1w 3 + 0.9w 2 − 0.9w − 0.1 = 0 (3) 

The RH tabulation for the transformed CE 3 is as 

follows: 

w 3 8.1 -0.9 

w 2 0.9 -0.1 

w 1 0 0 

w 0 

The routh test could not be continued in the usual 

manner since row w 1 contains all zeros. The tabulation 

is proceeded by the auxillary equation from the 

coeffecients in the w 2 row: 

A(w) = 0.9w 2 − 0.1 = 0 (4) 

Taking the derivative of A(w) in equation 4 with 

respect to w , gives: 

A(w) 

dw 

= 1.8w (5) 

The coeffecient in the w 1 row are then filled with 

A(w) 

the coeffecient of 

dw 

, and the Routh tabulation 

then becomes: 

w 3 8.1 -0.9 

w 2 0.9 -0.1 

w 1 1.8 0 

w 0 -0.1 

From the Routh table, it can be see that there is 

one sign change, which implies that the CE has one 

root in the right half of the w-plane, or P (z) has one 

root outside the unit circle in the z-plane. 

1.4 Example 3 

By using Routh-Hurwitz stability criterion, determine 

the range of the gain K and the sampling period 

T, so that the digital system whose CE is given as 

P (z) = z 3 + a 2 z 2 + a 1 z + a 0 is asymptotically stable. 

The coeffecients of the CE are as follows: 

a 2 = 111.6T 2 + 16.74T − 3 (6) 

a 1 = 1.395 ∗ 10 −4 KT 3 − 33.48T + 3 (7) 

a 0 = 1.395 ∗ 10 −4 KT 3 

− 16.74T − 111.6T 3 − 1 (8) 

1.4.1 Solution to example 3 

Transforming the CE P(z) into w-domain by using 

the bilinear transformation z = w+1 

w−1 , gives 

where, 

A 3 w 3 + A 2 w 2 + A 1 w + A 0 = 0, (9) 

A 3 = a 2 + a 1 + a 0 + 1 = 2.79 ∗ 10 −4 KT 3 (10) 

A 2 = a 2 − a 1 + 3a 0 + 3 = 446.4T 2 

− 5.58 ∗ 10 −4 KT 3 (11) 

A 1 = −446.4T 2 + 66.96T 

+ 2.79 ∗ 10 −4 KT 3 (12) 

A 0 = −a 2 + a 1 − a 0 + 1 = 8 − 66.96T (13) 

2

The routh tabulation for the transformed CE will 

be: 

w 3 A 3 A 1 

w 2 A 2 A 0 

w 1 

A1A2−A0A3 

A 2 

w 0 A 0 

For stability, all the coffecients in the first column 

must be of the same sign, Thus from, 

w 0 : A 0 = 8 − 66.96T > 0. ⇒ T < 0.1195(14) 

w 1 : A 1 A 2 − A 0 A 3 > 0. 

⇒ −15.568 ∗ 10 −8 T 3 K 2 

+ (0.3736T 2 − 0.01868T − 0.002232)K 

+ (29890.94 − 199272.96T ) > 0 (15) 

w 2 : A 2 = 446.4T 2 − 5.58 ∗ 10 −4 KT 3 > 0. 

⇒ K < 800000/T. (16) 

w 3 : A 3 = 2.79 ∗ 10−4KT 3 > 0. (17) 

Thus, the stability of the system is controlled by 

the four inequality conditions above. Hence, the 

ranges of T and K for stability can be written as 

follows: 

2 Example on Jury’s Stability 

Test 

2.1 solution using Jury’s Stability 

Test 

By using question from example 1(a), we can solve 

using Jury’s stability test, 

(a) The equation P (z) = z 2 − 0.25 = 0 is of second 

order. Under the Jury’s Stability test, for the system 

to be stable, the necessary condition must be satisfied 

according to Table 2.4: 


Conditions System satisfied 

or not 

1 P (1) > 0 P (1) = 1 − 0.25 = 0.75 > 0 satisfied 

2 P (−1) > 0 P (−1) = 1 − 0.25 = 0.75 > 0 satisfied 

3 |a 0 | < a 2 |a 0 | = 0.25 < a 2 = 1 satisfied 

Since all the conditions are satisfied, the system 

is stable. 

2.2 Example 2 

Determine the stability of a discrete data system described 

by the following CE by using Jury’s Stability 

criterion. 

P (z) = z 3 − 1.2z 2 − 1.375z − 0.25 = 0 

0 < T < 0.1195 

0 < K < 800000/T 

−15.568 ∗ 10 −8 T 3 K 2 

+(0.3736T 2 − 0.01868T − 0.002232)K 

+(29890.94 − 199272.96T ) > 0 

3

2.3 Solution to Example 2 

Under the Jury’s stability test for the system to be 

stable, the following three necessary conditions must 

be satisfied first: 



1 P (1) > 0 P (1) = 1 − 1.2 − 1.375 

−0.25 = −1.875 > 0 not 

2 P (−1) < 0 P (−1) = −1 − 1.2 

or not 

satisfied 

+1.375 − 0.25 = −1.125 > 0 satisfied 

3 |a 0 | < a 3 |a 0 | = 0.25 < a 3 = 1 satisfied 

Since the first condition is not satisfied, the system 

is not stable. 

2.4 Example 3 

Determine the stability of a discrete data system described 

by the following CE by using Jury’s stability 

criterion. 

P (z) = z 3 + 3.3z 2 + 4z + 0.8 = 0 

Solution to Example 3 

Under the Jury’s stability test, for the system to 

be stable, the following three necessary conditions 

must be satisfied first: 

All the necessary conditions are satisfied, thus we 

have to carry out the Jury tabulation to determine 

the stability of the system as follows: 

where 

b 0 = 

b 1 = 

b 2 = 


row z 0 z 1 z 2 z 3 

1 0.8 4.0 3.3 1.0 

2 1.0 3.3 4.0 0.8 

3 b 0 b 1 b 2 

∣ a ∣ 

0 a 3 ∣∣∣ 

= a 2 

a 3 a 

0 − a 2 3 = −0.36 (18) 

0 ∣ a ∣ 

0 a 2 ∣∣∣ 

= a 

a 3 a 0 a 1 − a 2 a 3 = −0.1 (19) 

1 ∣ a ∣ 

0 a 1 ∣∣∣ 

= a 

a 3 a 0 a 2 − a 1 a 3 = −1.36 

2 

(20) 

The sufficient condition for stability is |b 0 | > |b 2 |. 

Thus, it can be seen that from the values of b 0 and 

b 2 obtained, this condition is not satisfied.Hence, the 

system is not stable. In fact the roots are at z = 

−0.2463. − 1.5268 ± j0.9574. 



or not 

1 P (1) > 0 P (1) = 1 + 3.3 

+4 + 0.8 = 9.1 satisfied 

2 P (−1) < 0 P (−1) = −1 + 3.3 

−4 + 0.8 = −0.9 satisfied 

3 |a 0| < a 3 |a 0| = 0.8 < a 3 = 1 satisfied 

4

Lecture notes - School Of Electrical & Electronic Engineering - USM

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