DIGITAL CONTROL SYSTEM EEE354

Stability ConceptBilinear TransformationRouth-Hurwitz Criterion

Jury’s Stability test

DIGITAL CONTROL SYSTEMEEE354

Stability Analysis Techniques

MUHAMMAD NASIRUDDIN MAHYUDDIN

School of Electrical and Electronic Engineering,UNIVERSITI SAINS MALAYSIA

week 9Semester II 2007/2008

[email protected] EEE 354 : STABILITY ANALYSIS TECHNIQUES



Outline

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

1 Stability ConceptIntroduction: Stability Concept in Digital Control System

2 Bilinear TransformationDefinitionApplication of Bilinear Transformation

3 Routh-Hurwitz CriterionSteps taken before using RH criterion

4 Jury’s Stability testno need to do bilinear transformationJury’s Stability Criterion




Introduction: Stability Concept in Digital Control System

Introduction to Stability Concept

Generally, the stability analysis techniques applicable to LTIcontinuous-time systems may also be applied to the analysis ofLTI discrete time systems, if certain modifications are made.These techniques include:-

Routh-Hurwitz criterionroot-locus proceduresfrequency response method such as:-

bode plotnichols chart






Consider the following closed-loop pulse transfer functionsystem:

C(z)R(z) = G(z)

1+GH(z)

The stability of the system defined by the above equation maybe determined from the locations of the closed loop in thez-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 thecharacteristic equation must lie within the unit circle in thez-plane.

2 The system become critically stable if a simple pole lies at z=1or a single pair of conjugate complex poles lies on the unit circlein the z-plane.

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

4 Closed-loop zero do not affect the absolute stability andtherefore may be located anywhere in the z-plane




DefinitionApplication of Bilinear Transformation

Definition of bilinear transformation

The bilinear transformation is defined by:

z = w+1w−1

Solving for w, gives:

w = z+1z−1

The transformation maps the interior of the unit circle in the z-planeinto the left-half of the w plane.





mapping of z-plane

Figure 1 below shows the mapping of the unit circle in thez-plane onto the imaginary axis of the w-plane.

Figure 1: mapping of the unit circle in two planes





bilinear transformation

Once the characteristic equation P(z) = 0 of the digital systemis obtained, w = z+1

z−1 is substituted for z in the characteristicequation as follows:

P(z) = a0zn + a1zn−1 + a2zn−2 + · · · + an−1z + an = 0 (1)





bilinear transformation I

The characteristic equation in w -domain then becomes:

P(w) = a0

[w + 1w − 1

]n

+ a1

[w + 1w − 1

]n−1

+ a2

[w + 1w − 1

]n−2

+

. . . +an−1

[w + 1w − 1

]+ an = 0 (2)

P(w) = b0wn + b1wn−1 + · · · + bn−1w + bn = 0 (3)





bilinear transformation

The bilinear transformation transforms the z- plane into w plane. Thefrequency response of G(z) is G(jω), that is, by replacing z with jω.Equivalently, for w plane, the frequency response of replaced by afictitious frequency jv .The relationship between the fictitious frequency and the desiredfrequency v can be obtained as follows:

w∣∣∣w=jv

= z−1z+1

∣∣∣z=ejωT

jv = j tan ωT2

v = tan ωT2





Application of Bilinear transformation I

For stability analysis using frequency methods, such as the Bodediagram technique, another practical version of w- transformation wasproposed:

w′

=2T

w

w′

=2T

z − 1z + 1

(4)

where T is the sampling period. This implies that,

z =1 + T

2 w′

1 − T2 w′ =

2T + w

′

2T − w′

(5)





Application of bilinear transformation

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

factor of T2 .

The relationship between the actual/desired frequency ω and thefictitious frequency v

′is:

v′=

2T

tan(

ωT2

)(6)

(7)




Steps taken before using RH criterion

Routh- Hurwitz Criterion

The Routh-Hurwitz criterion cannot be applied directly to thez-domain since the stability boundary is now different. Themethod requires transformation from z-plane to another plane,the w plane.Once the characteristic equation P(z) = 0 is transformed into apolynomial of the same order in w : P(w) = 0, the R-H criterioncan then be applied directly as in the case of continuous datasystem.




Steps taken before using RH criterion

Routh-Hurwitz Criterion

Worked Examples onRouth-Hurwitz Criterion




no need to do bilinear transformationJury’s Stability Criterion

Jury’s Stability Test

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

P(z) = anzn + an−1zn−1 + an−2zn−2 + · · · + a2z2 + a1z + a0 = 0 (8)

where an > 0 or it can be made positive by changing the sign of all thecoeffecients, and ai, are real coeffecients.





Jury’s Stability Test

In applying the Jury’s stability test, a Jury Stability table mustfirst be constructed as shown in Table 1

Table 1: Jury’s Table

row z0 z1 z2 z3 . . . zn−1 zn−1 zn

1 a0 a1 a2 a3 . . . an−2 an−1 an

2 an an−1 an−2 a2 . . . a1 a0

3 b0 b1 b2 b3 . . . bn−2 bn−1 bn

4 bn bn−1 bn−2 bn−3 . . . b2 b1 b0

2n-5 p0 p1 p2 p3 . . .2n-4 p3 p2 p1 p0 . . .2n-3 q0 q1 q2





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 thefollowing determinants:

bk =∣∣∣∣ a0 an−k

an ak

∣∣∣∣ , k = 0, 1, 2, . . . , n − 1. (9)

ck =∣∣∣∣ b0 bn−1−k

bn−1 bk

∣∣∣∣ , k = 0, 1, 2, . . . , n − 2. (10)

dk =∣∣∣∣ c0 cn−2−k

cn−2 ck

∣∣∣∣ , k = 0, 1, 2, . . . , n − 3. (11)

q0 =∣∣∣∣ p0 p3

p3 p0

∣∣∣∣ , q2 =∣∣∣∣ p0 p1

p3 p2

∣∣∣∣ (12)





Jury’s stability test

1 The elements in any even-numbered row ( row 2,4, . . . )are simply the reverse of the immediately precedingodd-numbered row ( row 1,3, . . . )

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





Jury’s Stability Criterion

The stability criterion by the Jury’s Stability Test: For thepolynomial P(z) to have no roots on and outside the unit circlein the z-plane (i.e. for the digital or discrete data system to bestable), the following conditions must be satisfied:

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

b) P(z)|z=−1 = P(−1){ > 0 for n even

< 0 for n odd(14)






c)

|a0| < |an||b0| > |bn−1||c0| > |cn−2||d0| > |dn−3|

...|q0| > |q2|

}(n − 1) constraints (15)






For example, a second-order system would have n = 2.Therefore, Jury’s stability table contains only 1 row. Thus, forstability:

P(1) > 0

P(−1) > 0 (16)|a0| < |a2|

(17)


DIGITAL CONTROL SYSTEM EEE354(Sistem Kawalan Digit)

Worked Examples

prepared by Nasiruddin ∗

semester II 2007/2008

1 Example on Routh HurwitzCriterion

By using Routh-Hurwitz stability criterion, deter-mine the stability of the following digital systemswhose characteristic are given as:

a) z2 − 0.25 = 0

b) z3 − 1.2z2 − 1.375z − 0.25

1.1 Solution to Example 1(a)

Transforming the characteristic equation z2− 0.25 =0 into w-domain by using the bilinear transformationz = w+1

w−1 , gives:

0.75w2 + 2.5w + 0.75 = 0 (1)

It can be observed that all the coeffecients are of thesame sign, and none of the coeffecients is zero. Thus,all the roots of the tranformed equation 1 are in theleft-half of the w-plane. Hence, all the roots of the

∗Muhammad Nasiruddin Mahyuddin (Mechatronic Pro-gramme, School Of Electrical and Electronic Engineering, USM)

characteristic equation in the z-domain are inside theunit-circle in the z-plane.

Thus, the system is stable.

1.2 Solution to Example 1(b)

Transforming the characteristic equation z3−1.2z2−1.375z−0.25 = 0 into w-domain by using the bilineartransformation z = w+1

w−1 , gives:

− 1.875w3 + 3.875w2 + 4.875w + 1.125 = 0 (2)

Based on the transformed CE, the Routh tabula-tion is constructed as follows:

w3 -1.875 4.875

w2 3.875 1.125

w1 5.419 0

w0 1.125

From the table above, since there is one sign changein the first column, equation 2 has one root in theright-half of the w-plane. This, in turn, implies thatthere will be one root of the characteristic equationoutside 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 adigital system whose CE is given as P (z) = z3 +3.3z2 + 3z + 0.8 = 0.

1.3.1 Solution to Example 2

Transforming the CE P (z) into w-domain by usingbilinear transformation z = w+1

w−1 , gives:

8.1w3 + 0.9w2 − 0.9w − 0.1 = 0 (3)

The RH tabulation for the transformed CE 3 is asfollows:

w3 8.1 -0.9w2 0.9 -0.1w1 0 0w0

The routh test could not be continued in the usualmanner since row w1 contains all zeros. The tabula-tion is proceeded by the auxillary equation from thecoeffecients in the w2 row:

A(w) = 0.9w2 − 0.1 = 0 (4)

Taking the derivative of A(w) in equation 4 withrespect to w , gives:

A(w)dw

= 1.8w (5)

The coeffecient in the w1 row are then filled withthe coeffecient of A(w)

dw , and the Routh tabulationthen becomes:

w3 8.1 -0.9w2 0.9 -0.1w1 1.8 0w0 -0.1

From the Routh table, it can be see that there isone sign change, which implies that the CE has oneroot in the right half of the w-plane, or P (z) has oneroot outside the unit circle in the z-plane.

1.4 Example 3

By using Routh-Hurwitz stability criterion, deter-mine the range of the gain K and the sampling periodT, so that the digital system whose CE is given asP (z) = z3 + a2z

2 + a1z + a0 is asymptotically stable.The coeffecients of the CE are as follows:

a2 = 111.6T 2 + 16.74T − 3 (6)

a1 = 1.395 ∗ 10−4KT 3 − 33.48T + 3 (7)

a0 = 1.395 ∗ 10−4KT 3

− 16.74T − 111.6T 3 − 1 (8)

1.4.1 Solution to example 3

Transforming the CE P(z) into w-domain by usingthe bilinear transformation z = w+1

w−1 , gives

A3w3 + A2w

2 + A1w + A0 = 0, (9)

where,

A3 = a2 + a1 + a0 + 1 = 2.79 ∗ 10−4KT 3(10)

A2 = a2 − a1 + 3a0 + 3 = 446.4T 2

− 5.58 ∗ 10−4KT 3 (11)

A1 = −446.4T 2 + 66.96T

+ 2.79 ∗ 10−4KT 3 (12)

A0 = −a2 + a1 − a0 + 1 = 8− 66.96T (13)

2

The routh tabulation for the transformed CE willbe:

w3 A3 A1

w2 A2 A0

w1 A1A2−A0A3A2

w0 A0

For stability, all the coffecients in the first columnmust be of the same sign, Thus from,

w0 : A0 = 8− 66.96T > 0. ⇒ T < 0.1195(14)

w1 : A1A2 −A0A3 > 0.

⇒ −15.568 ∗ 10−8T 3K2

+ (0.3736T 2 − 0.01868T − 0.002232)K

+ (29890.94− 199272.96T ) > 0 (15)

w2 : A2 = 446.4T 2 − 5.58 ∗ 10−4KT 3 > 0.

⇒ K < 800000/T. (16)

w3 : A3 = 2.79 ∗ 10−4KT 3 > 0. (17)

Thus, the stability of the system is controlled bythe four inequality conditions above. Hence, theranges of T and K for stability can be written asfollows:

0 < T < 0.11950 < K < 800000/T

−15.568 ∗ 10−8T 3K2

+(0.3736T 2 − 0.01868T − 0.002232)K+(29890.94− 199272.96T ) > 0

2 Example on Jury’s StabilityTest

2.1 solution using Jury’s StabilityTest

By using question from example 1(a), we can solveusing Jury’s stability test,

(a) The equation P (z) = z2−0.25 = 0 is of secondorder. Under the Jury’s Stability test, for the systemto be stable, the necessary condition must be satisfiedaccording to Table 2.4:

Table 1: Jury’s TableConditions System satisfied

or not?1 P (1) > 0 P (1) = 1− 0.25 = 0.75 > 0 satisfied2 P (−1) > 0 P (−1) = 1− 0.25 = 0.75 > 0 satisfied3 |a0| < a2 |a0| = 0.25 < a2 = 1 satisfied

Since all the conditions are satisfied, the systemis stable.

2.2 Example 2

Determine the stability of a discrete data system de-scribed by the following CE by using Jury’s Stabilitycriterion.

P (z) = z3 − 1.2z2 − 1.375z − 0.25 = 0

3

2.3 Solution to Example 2

Under the Jury’s stability test for the system to bestable, the following three necessary conditions mustbe satisfied first:

Table 2: Jury’s Table

Conditions System satisfied

or not?

1 P (1) > 0 P (1) = 1− 1.2− 1.375

−0.25 = −1.875 > 0 not

satisfied

2 P (−1) < 0 P (−1) = −1− 1.2

+1.375− 0.25 = −1.125 > 0 satisfied

3 |a0| < a3 |a0| = 0.25 < a3 = 1 satisfied

Since the first condition is not satisfied, the sys-tem is not stable.

2.4 Example 3

Determine the stability of a discrete data system de-scribed by the following CE by using Jury’s stabilitycriterion.

P (z) = z3 + 3.3z2 + 4z + 0.8 = 0

Solution to Example 3Under the Jury’s stability test, for the system to

be stable, the following three necessary conditionsmust be satisfied first:

Table 3: Jury’s TableConditions System satisfied

or not?1 P (1) > 0 P (1) = 1 + 3.3

+4 + 0.8 = 9.1 satisfied2 P (−1) < 0 P (−1) = −1 + 3.3

−4 + 0.8 = −0.9 satisfied3 |a0| < a3 |a0| = 0.8 < a3 = 1 satisfied

All the necessary conditions are satisfied, thus wehave to carry out the Jury tabulation to determinethe stability of the system as follows:

Table 4: Jury’s Tablerow z0 z1 z2 z3

1 0.8 4.0 3.3 1.0

2 1.0 3.3 4.0 0.8

3 b0 b1 b2

where

b0 =∣∣∣∣ a0 a3

a3 a0

∣∣∣∣ = a20 − a2

3 = −0.36 (18)

b1 =∣∣∣∣ a0 a2

a3 a1

∣∣∣∣ = a0a1 − a2a3 = −0.1 (19)

b2 =∣∣∣∣ a0 a1

a3 a2

∣∣∣∣ = a0a2 − a1a3 = −1.36

(20)

The sufficient condition for stability is |b0| > |b2|.Thus, it can be seen that from the values of b0 andb2 obtained, this condition is not satisfied.Hence, thesystem is not stable. In fact the roots are at z =−0.2463.− 1.5268± j0.9574.

4

DIGITAL CONTROL SYSTEM EEE354

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