chapter 2 stability testing routh hurwitz

8/22/2019 Chapter 2 Stability Testing Routh Hurwitz

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Chapter 2 (page#143)

2.5 Stability Testing

2.5.1 Coefficient Tests

2.5.4 Case 1

Introduction

2.5.2 Routh-Hurwitze Stability Testing

2.5.5 Case 2

2.5.6 Case 3

2.6 Parameter Shifting (Page 159)


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System should be

Stable: BIBO stable if, for every bounded input, the

output is bounded for all time

LTI system must have all poles in the left-half of the s-

plane (negative real parts)

Transfer function poles are same as roots of the

characteristic polynomial are the same as system

Eigen- values

All Eigen-values must have negative real parts for BIBOstability

Poles on imaginary axis are not stable by this definition


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Stability

Example: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 (poles of the

closed-loop transfer function) lie in the left

half of the s-plane.

x x

-2 -1

S-plane

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

-2t

22

2

2232

2

nn

n

sss

Over Damped System? Critically Damped System?


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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 ,k2e

3t,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

3


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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


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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 t

sint 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


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Necessary condition

If any coefficientai of Q(s) is zero or negative

then not all roots lie in the left half of the s-plane

Otherwise: set up Routharray and use Routh-

Hurwitz criterion:


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It requires that all poles of the closed-loop TF lie in the

LHP.

HenceAbsoluteStability analysis requires determiningif any poles are in the RHP or on the jwaxis.

Routh-Hurwitz Criterion (Doesnt actually calculate

roots)

Routh-Hurwitz Stability Testing (Criterion)

Gives the number of roots with positive real parts.

01

1

1)( asasasasQn

n

n

n


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First step make the Routh Array:

s

n

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(Qn

nn

nn

nn

nn

n

31

2

1

1 1

nn

nn

naaaa

ab

51

4

1

2

1

nn

nn

naa

aa

ab

21

31

1

1

1

bb

aa

b

cnn

31

51

1

2

1

bb

aa

bc

nn


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014

43

32

21

1 asasasasasasa)s(Qn

nn

nn

nn

nn

n


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Example: (Case: 1)

Number of sign changes in the first column = number of

unstable poles

Note: Case 1

(No zero element in first column)


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Example: (Case: 2) 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 6

s0

10Two sign changes whether was

assumed +ve orve.

0)2241(21

4221

21

1 xxb

0b

1s

3

21

31

1

10

1

bb

aa

bcthen

nn

12412

1

6

4211 )(cthen

)valuesmallveryvery(blet 1


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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 s4. Coefficients of the Polynomial replaces the zero row

Ex. Q(s) = s2 + 1

s2 1 1

s1 0

s0

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


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What is Auxiliary Polynomial

Q(s) contains an even

polynomial as factor. An evenpolynomial is one in which the

exponents of s are even

integers or zero

This even polynomial is called

AuxiliaryPolynomial

In Routh array, coefficients of

Aux Poly are those directly

above the zero row. (See

examples above)

s2 1 1

s1 0s0

Aux Pol =s2 + 1

Aux Pol =s2

+ 2

s2 1 2

s1 0s0

s3 1 2


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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 s4. 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)

2s2

1

RHP=2

LHP=2

IA = 0


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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)

2s2

2

RHP=0

LHP=2

IA = 2


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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 signchanges

+

-

+

Twounstable

poles

RHP=2

LHP=2

IA = 0


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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 orve.

0b1

s2


RHP=2

LHP=2

IA = 0

E l C 2 f B k (P 149)


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

s5 1 2 1

s4 1 3 4

s3 -1 -3

s2 4

s1 4/

s0 4

0


Two sign changes whether was

assumed +ve orve.

RHP=2

LHP=3

IA = 0

E l C 3 f B k (P 150)


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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


Aux Pol = 3s2 + 12

d/ds(3s2 + 12)

6s

6

No sign change in the first column

MARGINALLY STABLE.

Aux poly indicates roots on jwaxis

3s2=-12 s2=-4 s1,2=2

E l E l C 3 f B k (P 152)


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

s6 1 5 2 -8

s5 1 1 -2

s4 4 4 -8

s

3

0 0

s2 2 -8

s1 72

s0 -8

Aux Pol = 4s4 +4s2 + 8

16

Example Example_Case_3 from Book (Page 152)

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

16s3 +8s2

8

RHP=1

LHP=3

IA = 2

E l f B k (P 160)


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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

E l f B k (P 160)


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S3 2 K

S

2

4-K/2 6S1

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)


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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 -1

Stable

Stable

pcs

p GGlim0

K Ksss

K

sp

254

2lim

230K

50

1

1

1

K

ess

254

223

sss

KKGp

49K

For ess49system to be stable -1 < K


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_

K

2

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 0

Stable

Stable

s

ksKGwithreplacedisKtheexampleprevioustheIn

ipc

s

ksK ip

ksKK


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

s4 1 5 2Ki

s3 4 2+2Kp

s0 2Ki

Kp 0

Stable

Stable

s

ksK

s

KKGwithreplacedisKexampleprevioustheIn

ipppc

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

_

K

2

s3 + 4s2 + 5s + 2

_

K

2

s3 + 4s2 + 5s + 2

_

K

2

s3 + 4s2 + 5s + 2

chapter 2 stability testing routh hurwitz

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