Stability TestingStability Testing

Coefficient Tests

Case 1No Zero in Ist Column

Introduction

Routh-Hurwitze Stability Testing

Case 2Zero in Ist Column

Case 3All Element of a row are Zero

StabilityExample:1

)s)(s()s(R

)s(C)s(G)s(T

21

2

Stable or Unstable? Why Characteristic equation (s+1)(s+2)=0

Hence s+1=0 or s=-1 and s+2=0 or s=-2

system to be stable is that all roots (-1 & -2) of the characteristic equation (Over-damped)

x x-2 -1

S-plane

The natural-response terms for the system are k1e-t and k2e-2t

22

2

2232

2nn

n

sss

Example 2:12112

241023

sss

s)s(T

)s)s)(s(

s

431

2410

Hence roots are s= -1, s= 3, & s= -4

The natural-response terms for the system are k1e-t ,k2e3t,and k3e-4t

Stable or Unstable? Unstable due to s=3

Unstable due to term k2e3t

MATLAB Program:

>> p = [1 2 -11 -12];

>> r = roots(p)

Note: Numerator 10s+24 has NO role in the stability of the system

x x-4 -1

x 32.4

Example:312

s

s)s(T

)js)(js(

s

Hence roots are s= -j, s=j

The natural-response terms for the system are ksin(t+)

Stable or Unstable? Marginally stable

Marginally stable because no exponential term

MATLAB Program:

>> p = [1 0 1];

>> r = roots(p)

x

x

j

-j

Example:4 (Example 3)12

s

s)s(T

)s()s(

s)s(R)s(T)s(C

1

1

1 22

The response of the system c(t) = tsint

Marginally stable

Let the input r(t)=sint1

12

s

)s(R

)js)(js(

s

Hence roots are s= -j, s=j

x

x

j

-j

x

x

j

-j

Stable or Unstable? Unstable due to tsint is marginally stable but the multiplication of time (t) it makes it unstable

NOTE: The natural-response of the system are ksin(t+) bounded (marginally stable) but for unbounded output (unstable) for certain bounded input

2.3 Routh-Hurwitz Stability Criterion (Doesn’t actually calculate roots)

The Routh-Hurwitz criterion is an analytical procedure for determining if all roots of a polynomial have negative real parts and is used in the stability analysis of linear time-invariant systems for a closed loop system.

The criterion gives the number of roots with positive real parts, and does not give the exact value of the roots.

Therefore this stability criterion gives us the absolute stability

analysis requires determining if any poles are in the RHP or on the jw axis.. The Routh-Hurwitz criterion applies to a polynomial of the form

The first step in the application of the Routh-Hurwitz criterion is to form the array below, called the Routh array, where the first two rows are the coefficients of the above polynomial.

011

1)( asasasasQ nn

nn

Relatively More Stable

Absolutely Stable

X

Relatively Less Stable

Absolutely Stable

X

Note: No change in Absolute stability – no pole(s) location

Note: From Stable to unstable

Absolutely Stable

Unstable

First step make the Routh Array:

sn an an-2 an-4 an-6 ……

sn-1 an-1 an-3 an-5 an-7 ……

sn-2 b1 b2 b3 b4 ……

sn-3 c1 c2 c3 c4 ……

. .

. .

. .s2 k1 k2

s1 l1

s0 m1

014

43

32

21

1 asasasasasasa)s(Q nn

nn

nn

nn

nn

31

2

11

1

nn

nn

n aa

aa

ab

51

4

12

1

nn

nn

n aa

aa

ab

21

31

11

1

bb

aa

bc nn

31

51

12

1bb

aa

bc nn

014

43

32

21

1 asasasasasasa)s(Q nn

nn

nn

nn

nn

Next

Case 1

(No Zero element in first column)

Example: (Case: 1) (No zero element in first column)

–Number of sign changes in the first column = number of unstable poles

Note: Case 1(No zero element in first column)

Example_Case_1 from Book (Page 146)

P(s) = 2s4 + 3s3 + 5s2 + 2s + 6 (2.14)

s4 2 5 6

s3 3 2

s2 +11/3 6

s1 -32/11

s0 + 6

–Number of sign changes in the first column = number of unstable poles

Two sign changes

+

-

+

Two unstable

poles

RHP=2

LHP=2

IA = 0

Next

Case 2(Zero element in first column)

Un-stable system

Example: (Case: 2) (Zero element in first column) Un-stable system

Case 2 (Steps)1. First Element of a row is Zero (0)2. Replace “0” by small number . (=0.00000000..01)

3. Continue the array

Q(s) = s5 + 2s4 + 2s3 + 4s2 + 11s + 10

s5 1 2 11s4 2 4 10

6

s2 -12/ 10s1 6s0 10 • Two sign changes whether ‘’ was

assumed +ve or –ve. 2-roots in RHS

0)2241(2

1

42

21

2

11 xxb

0b1s3

21

31

11 0

1

bb

aa

bcthen nn

12412

1

6

4211 )(cthen

)valuesmallveryvery(blet 1

Case 2 (Steps)1. First Element of a row is Zero (0)2. Replace “0” by small number . (=0.00000000..01)

3. Continue the array

Q(s) = 3s4 + 6s3 + 2s2 + 4s +5 (2.16)

s4 3 2 5s3 6 4

2

s1 -12/s0 2

• Two sign changes whether ‘’ was assumed +ve or –ve.

0b1s2

Example_Case_2 from Book (Page 149)

RHP=2

LHP=2

IA = 0

Q(s) = s5 + s4 + 2s3 + 3s2 + s + 4 (2.20)

s5 1 2 1s4 1 3 4s3 -1 -3

s2 4

s1 4/

s0 4

0

Example_Case_2 from Book (Page 149)

• Two sign changes whether ‘’ was assumed +ve or –ve.

RHP=2

LHP=3

IA = 0

Next

Case 3(All Element of a row are Zero

(Premature Termination )

Example: (Case: 3)Case 3: (Steps)

1. All Element of a row are Zero. (Premature Termination )2. Auxiliary Polynomial 3. Aux Poly is differentiated with respect to ‘s’ 4. Coefficients of the Polynomial replaces the zero row

Ex. Q(s) = s2 + 1

s2 1 1s1 0s0

No sign change in the first column implies that system is STABLE. Aux poly indicates roots on jw axis, which implies MARGINALLY STABLE

Aux Pol = s2 + 1

d/ds(s2 + 2)

2s

2 1

S2=-1; S=j

What is Auxiliary Polynomial

Q(s) contains an even polynomial as factor. An even polynomial is one in which the exponents of s are even integers or zero

This even polynomial is called “Auxiliary Polynomial”

In Routh array, coefficients of “Aux Poly” are those directly above the zero row. (See examples above)

s2 1 1s1 0s0

Aux Pol = s2 + 1

Aux Pol = s2 + 2

s2 1 2s1 0s0

s3 1 2

Example: (Case: 3)Case 3: (Steps)

1. All Element of a row are Zero. (Premature Termination )2. Auxiliary Polynomial 3. Aux Poly is differentiated with respect to ‘s’ 4. Coefficients of the Polynomial replaces the zero row

Ex. Q(s) = (s+1) (s2 + 2) = s3 + s2 + 2s + 2

s3 1 2

s2 1 2

s1 0

s0

Aux Pol = s2 + 2

d/ds(s2 + 2)

2s 2

2

RHP=0

LHP=1

IA = 2

Example: (Case: 3)

Ex. Q(s) = s4 + s3 + 3s2 + 2s + 2

s4 1 3 2

s3 1 2

s2 1 2

s1 0

s0

Aux Pol = s2 + 2

d/ds(s2 + 2)

2s 2

2

RHP=0

LHP=2

IA = 2

Q(s) = s5 +2s4 + 8s3 + 11s2 + 16s + 12 (2.24)

s5 1 8 16

s4 2 11 12

s3 2.5 10

s2 3 12

s1 0

s0 12

Example_Case_3 from Book (Page 150)

Aux Pol = 3s2 + 12

d/ds(3s2 + 12)

6s

6

No sign change in the first column MARGINALLY STABLE.

Aux poly indicates roots on jw axis 3s2 = -12 s2 = -4 s1,2=j2

s6 + s5 + 5s4 + s3 + 2s2 - 2s – 8 (2.26)

s6 1 5 2 -8

s5 1 1 -2

s4 4 4 -8

s3 0 0

s2 2 -8s1 72s0 -8

Aux Pol = 4s4 +4s2 + 8

16

Example Example_Case_3 from Book (Page 152)

d/ds(4s4 +4s2 + 8)

16s3 +8s8

RHP=1

LHP=3 or 1

IA =2 or 4

S3 2 2

S2 3 K

S1 (6-2K)/3

S0 K

→ (6-2K)/3 > 0

→ K> 0

Stable

Stable

For what value of K the system will be marginally stable?

S4 1 4 K

Stability range 0 < K < 3

Example_ from Book (Page 160)

P(s) = S4 + 2s3 + 4s2 + 2s + K

S3 2 K

S2 4-K/2 6

S1

S0 6

→ > 0

→ K < 8

Stable

Stable

S4 1 4 6

Stability range can not be satisfied for any value of K (complex roots). Polynomial has RHP roots for all K

Example_ from Book (Page 160)

complexroots).(

K

502

241640124

2

2

KK

P(s) = S4 + 2s3 + 4s2 + Ks + 6

24

12

/KK

Example (Important – used in designing)

Design specification that ess must be less than 2% of the constant unit step input. Find value of K that will produce error > 2%. Using Routh Hurwitze criteria, verify the values of K for stability.

S3 1 5

S2 4 2+2K

S1 1/4(18-2K)

S0 2+2K

→ K<9

→ K> -1

Stable

Stable

pcs

p GGlim0

K Ksss

Ks

p

254

2lim

230K

50

1

1

1

Kess

254

223

sss

KKGp

49K

For ess<2% K>49 system to be stable -1 < K<9 Hence Not possible

_

K2

s3 + 4s2 + 5s + 2

492 Kofvalue%forHenc

Q(s) = s3 + 4s2 + 5s +2+2K

_

K2

s3 + 4s2 + 5s + 2

Q(s) = s4 + 4s3 + 5s2 + (2 + 2Kp) s + 2 Ki

s4 1 5 2Ki

s3 4 2+2Kp

s0 2Ki

→ Kp<9

→ Ki> 0

Stable

Stable

s

ksKGwithreplacedisKtheexampleprevioustheIn ip

c

s

ksK ip

Q(s) = s4 + 4s3 + 5s2 + (2 + 2Kp) s + 2 Ki

s4 1 5 2Ki

s3 4 2+2Kp

s0 2Ki

→ Kp<9

→ Ki> 0

Stable

Stable

s

ksK

s

KKGwithreplacedisKexampleprevioustheIn ipp

pc

Choose Kp &Ki = 3; using simulink simulate the problem

_

K2

s3 + 4s2 + 5s + 2

_

K2

s3 + 4s2 + 5s + 2

_

K2

s3 + 4s2 + 5s + 2