Lecture 14: Routh's Stability Criterion

Lecture 14: Routh’s Stability Criterion

Concept of Stability:

Absolute vs. Relative Stability

 A system is either stable or unstable. This type of stable/unstable is referred as absolute stability.

 If the system is stable, we can further investigate the degree of stability. This is referred as relative stability.

Definition of System Stability

* BIBO Stability : A stable system is a dynamic system with a bounded response to a bounded input. This is called a BIBO stability.

 Ex. Step response of a 2

^nd

-order system:

Stability depends on the pole locations in the s-plane

General Stability Condition:

A necessary and sufficient condition for a feedback system to be stable is that all the poles of the system transfer function have negative real parts.

Effect of Pole Locations of 2

^nd

-Order Systems:

 Standard transfer function:

^{2 2}

2

) 2 (

n n

n

s s

G  





 

 Poles of G(s) are: s

1,2

  

_n

 j 1  

²

 Unit Step Response:

Y(s)=G(s)*R(s), R(s)=1/s

=[ω

n2

/s(s

²

+2ζ ω

n

s+ ω

n2

)]

=>

y(t) = 1 – (1/β)e

^{- ζ ωnt}

sin(ω

n

β t + θ)

=> System is stable if all poles have negative real parts.

Effect of Pole Locations on Impulse Responses:

Example:

 Let G(s)=2/(s+1)(s+2) =p(s)/q(s)

 Then, poles are -1,-2

=>

System impulse (or natural) response is y(t)=k

1

e

^-t

+ k

2

e

^-2t

=>

System is stable.

 Show that

 1. G(s)=s/(s

²

+1) is stable

 2. G(s)=10(s+2)/(s+1)(s-3)(s+4) is unstable.

System Stability Criterion:

Recall: Let the system transfer function be G(s)=p(s)/q(s).

Then,

The (System) Characteristic Eqn. is q(s)=0, and Poles are the roots of q(s)=o.

 Necessary condition: It is necessary for a stable system to have all the coefficients of the characteristic equation have the same sign.

 Proof: Yourself , check book

 Necessary condition: It is necessary for a stable system to have all the coefficients of the characteristic equation have the same sign.

Ex.

1. q(s)=s

³

+s

²

+2s+8 2. q(s)=s

⁴

+4

 Necessary condition is satisfied for both.

 Are they stable?

 1. q(s)=(s+2)(s

²

-s+4) => poles: -2, ½±j /2 => stable

 2. q(s)=(s

²

+2s+2)(s

²

-2s+2) => poles: -1±j1, 1±j1 => unstable

Routh-Hurwitz Stability Test:

 Necessary and Sufficient condition:

 R-H test

The Routh*-Hurwitz** stability criterion states that the number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array.

Proof: See Book.

*John Routh, English mathematician **Adolf Hurwitz, German mathmatician

The Routh-Hurwitz Stability Test for SISO-LTI Systems:

 A necessary condition for a feedback system to be stable is that all the poles of the system transfer function have negative real parts. How about a sufficient condition?

=> Use the Routh-Hurwitz test.

 Step 1. In order to test system stability, first construct the Routh array from the system characteristic equation q(s)=0.

 Step2. The number of changes in sign of the first column of the Routh array is equal to the number of poles in the right half-plane.

Routh Array:

The response transform X₂(s) has the following form. X₁(s) is the driving transform

0 1 1

1

1 1 2

) ( ) (

) ) (

( ) ) (

(

b s b s

b s b

s X s P

s s X

Q s s P

X

n n n

n

   





 



(1)

The characteristic equation is

0 )

( s  b s  b

_1

s

^1

  b

₁

s  b

₀



Q

_n ⁿ _n ⁿ



We can get the Routhian array by

The constants c

1

, c

2

, c

3

etc., in the third row are evaluated as follows:

This pattern is continued until the rest of the c’s are all equal to zero. Then the d row is formed by using the s

^n_1

and s

^n_2

rows. The constants are

This process is continued until no more d terms are present. The rest of the rows are formed in this way

down to the s

⁰

row. The complete array is triangular, ending with the s

⁰

row. Notice that the s

¹

and s

⁰

rows contain only one term each. Once the array has been found, Routh’s criterion states that the

number of roots of the characteristic equation with positive real parts is equal to the

number of changes of sign of the coefficients in the first column . Therefore, the system is

stable if all terms in the first column have the same sign.*

Example 1:

Example 2:

Routhian Array

Example 3:

Theorem 2: Zero Coefficient in the First Column. When the first term in a row is zero but not all the other terms are zero, the following methods can be used:

1. Substitute s = 1/x in the original equation; then solve for the roots of x with positive real parts. The number of roots x with positive real parts will be the same as the number of s roots with positive real parts.

2. Multiply the original polynomial by the factor (s+1), which introduces an additional negative root. Then form the Routhian array for the new polynomial.

Two sign changes → Two roots in RHP in the first column

1→ – 62 → 70.6

There are two changes of sign in the first column. Therefore, there are two roots of x in the RHP. The number of roots s with positive real parts is also two. This method does not work when the coefficients of Q(s) and of Q(x) are identical.

Method 2:

The same result is obtained by both methods. There are two changes of sign in the first column, so there are two zeros of Q(s) with positive real parts.

Method3:

Example 4:

Theorem 3: A Zero Row. When all the coefficients of one row are zero, the procedure is as follows:

1. The auxiliary equation can be formed from the preceding row, as shown below.

2. The Routhian array can be completed by replacing the all-zero row with the coefficients obtained by differentiating the auxiliary equation.

3. The roots of the auxiliary equation are also roots of the original equation. These roots occur in pairs and are the negative of each other. Therefore, these roots may be imaginary (complex conjugates) or real (one positive and one negative), may lie in quadruplets (two pairs of complex-conjugate roots), etc.

Consider the system that has the characteristic equation

The presence of a zero row (the s

¹

row) indicates that there are roots that are the negatives of each other. The next step is to form the auxiliary equation from the preceding row, which is the s

²

row. The highest power of s is s

²

, and only even powers of s appear. Therefore, the auxiliary equation is

0

2

 9 

 s AE

The roots of this equation are

s

_1,2

  j 3

These are also roots of the original equation. The presence of imaginary roots indicates that the output includes a sinusoidally oscillating component.

To complete the Routhian array, the auxiliary equation is differentiated

0 ) 2

(  s 

ds AE d

The coefficients of this equation are inserted in the s¹ row, and the array is then completed:

Since there are no changes of sign in the first column, there are no roots with positive real parts.

Example:

Example: Find the range of K for which the system is stable.

Answer:

The characteristic equation is

Routhian array

Here

(i)

From s

²

row, 80 – K >0, so that K < 80, and

(ii)

From s

⁰

row 14K > 0, so that K > 0.

(iii)

The numerator of the first term in the s

¹

row is equal to – K

²

– 43K + 2000 =0 , and this function must be positive for a stable system. From above equation,

(K – 28.1)(K + 71.1) =0, gives K = 28.1 or K= – 71.1

Therefore, the combined restriction gives 0 < K < 28.1

Example:

Simple & important criteria for stability:

(i) ¹

^st

Order polynomial

0 )

( s  a

₁

s  a

_o



Q , All roots are in LHP ↔ a

₁

and a

₀

have the same sign

(ii) 2nd Order polynomial

0 )

( s  a

₂

s

²

 a

₁

s  a

_o



Q , All roots are in LHP ↔ a

2

, a

1

and a

0

have the same sign

(iii) Higher Order polynomial

0 )

( s  a

_n

s

ⁿ

 a

_n1

s

^n1

  a

₁

s  a

_o



Q  , All roots are in LHP ↔ All a

_k

have

the same sign

Routh-Hurwitz criterion is applicable to only polynomials (so, it is not possible to deal with exponential, sin, cos etc.).

Example: Whether system is stable or unstable?

Solution:

Example: Antenna position control system

Solution:

 q(s)=s

³

+101.7s

²

+171s+6.63K

=>

Routh array:

s

³

1 171 s

²

101.7 6.63K s

¹

17392.4 – 6.63K 0 s

⁰

6.63K

=>

17392.4-6.63K=0 when K=2623

=> Stable for 0<K<2623.

Stability of a state variable system Steps:

(i) From the state equation find the transfer equation of the system (ii) from transfer equation, find the characteristic equation

(iii) From characteristic equation, Routhian array can be obtained The linear time invariant (LTI) state-space form is

Du Cx

y

Bu Ax

x











The transfer function is given by

) (

) (

|

|

|

| )]

( [

|

|

)]

( ) [

( )

(

¹

s D

s N A

sI

A sI D B A sI adj C

A D sI

B A sI adj D C

B A sI C s

H

c



c







 

 

 







^

Here subscript ‘c’ denotes variables after pole/zero cancellation.

The denominator of this is the characteristic polynomial

.

|

| )

( s sI A

D  

The system poles are the roots of the characteristic equation Δ(s) =| sI – A| = 0.

The transfer function poles are the roots of D

c

(s)=0.

Example1:

so that both poles are at s = –1. Therefore the system is stable.

Example2: