Module-5: Stability Analysis of Linear Systems
Routh-Hurwitz Stability Criteria
Akshya Swain
Department of Electrical, Computer & Software Engineering, The University of Auckland,
Auckland, New Zealand.
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Learning Outcome of This Module
I After completion of this module, the student must be able to do the following:
1. Determine the absolute stability of a linear system. 2. Find the stability region of the linear systems
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Stability of Linear Systems: Concept of Stability
Consider the ball which is free to roll on the surface.
I The ball could be made to rest at points A,E,F,G and any where between the points B and D, such as at C.
I An infinitesimal perturbation away from A or F will cause the ball to diverge from these points. Thus A and F are unstable equilibrium points.
I After small perturbations away from E or G, the ball will eventually return to these points. Thus E and G are stable equilibrium points.
I If the ball is perturbed slightly away from C, it will remain at the new position. Points like C are sometimes said be neutrally stable.
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Stability of Linear Systems: Concept of Stability
I Roughly speaking, stability in a system implies that small changes in
i. System inputs, or
ii. Initial conditions, or
iii. System parameters
do not result in large changes in system behaviour. I A linear time invariant system is stable if
a. When the system is excited by a bounded input, the output is bounded. (BIBO stability)
b. With zero input and arbitrary initial conditions, the output tends towards zero – the equilibrium state of the system (Asymptotic stability).
I Consider a system having equilibrium point at x = xe. In stability studies we are concerned with the following questions:
1. If the system with zero input is perturbed from the equilibrium point xe at t = t0, will the state x(t)
i. Return to xe ?
ii. Remain close to xe ? or
iii. Diverge from xe ?
2. If the system is relaxed, will a bounded input produce a bounded output ?
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Concept of Stability
I The following figure shows different types of possible stability surfaces for globally stable, stable, unstable and locally stable systems.
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Concept of Stability
Question ?
I Let us assume that the system is modelled or represented by a transfer function.
I Then, it will be proved ( in following slides) that
How would we know whether a system is stable or unstable ?
For a linear time invariant system to be stable ( BIBO stable), the poles of the closed loop system i.e. the roots of the closed loop characteristic equation must lie in the left half of the s− plane.
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Transient Response of Higher-Order Systems-1
I Consider the system shown in the Figure.
I The closed loop transfer function is given by
C(s) = G(s) , where G(s) = p(s) and H(s) = n(s)
R(s) 1 + G(s)H(s) q(s) d(s) where p(s) and q(s), n(s) and d(s) are polynomials in s.
I The closed loop transfer therefore can be expressed as: C(s) = p(s)d(s)
R(s)
q(s)d(s) + p(s)n(s)
= b0sm +b1sm−1 +…+bm−1s+bm; (m≤n) a0sn +a1sn−1 +…+an−1s+an
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Transient Response of Higher-Order Systems-2
I If pi,i = 1,…,n, and zj,j = 1,…,m denote respectively the poles and zeros of the closed loop transfer function, then
C(s) = K(s + z1)(s + z2),…(s + zm) R(s) (s + p1)(s + p2),…(s + pn)
I Let us us examine the response behaviour of this system to step input. Case-1: Assume that all the closed loop poles are real and distinct . Thus
n
C(s)=a+ ai
s s+pi
i=1
where ai is the residue of the pole at s = −pi
I The unit step response c(t), the inverse Laplace transform of C(s), is then
n
c(t) = a + ai e−pi t
i=1
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Transient Response of Higher-Order Systems-3
Case-2: Assume that the closed loop poles consist of real poles and pairs of complex conjugate poles. Thus
m
K (s+zi)
j=1 k=1
where q + 2r = n.
I If the closed loop poles are distinct, the above can be expanded into partial
fractions as follows:
i=1 qr
C(s) =
s (s+pj) (s2 +2ζkωks+ωk2)
a q r bk(s+ζkωk)+ckωk1−ζ2 C(s)= +(s+pj)+ k
s j=1 k=1 s2 +2ζkωks+ωk2 I The unit step response c(t) is then given by
qr
c(t) = aje−pjt + bke−ζkωktcosωk1 − ζk2t + cke−ζkωktsinωk1 − ζk2t
j=1 k=1
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Transient Response of Higher Order Systems-4
I The unit step response c(t) is then given by
qr
c(t) = aje−pjt + bke−ζkωktcosωk1 − ζk2t + cke−ζkωktsinωk1 − ζk2t
j=1 k=1
I Thus the response curve of a stable, higher order system is the sum of number of exponential curves and damped sinusoidal curves.
1. If all the closed loop poles lie in the left half s-plane, then the exponential terms and damped sinusoidal exponential terms will approach to zero as time t increases ( approaches infinity).
2. The steady state output is then c(∞) = a.
3. However, if any of these poles lie in the right-half s-plane, then the transient
response increases monotonically or oscillates with increasing amplitude.
4. This represents an unstable system.
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Routh’s Stability Criterion
I System stability is one of the most important performance specification of a control system. A system is considered unstable if it does not return to its initial position; but continues to oscillate after it is subjected to any change in input or is subjected to undesirable disturbance.
I For any time invariant control system to be stable the following two conditions need to be satisfied.These are:
a. The system will produce a bounded output for every bounded input(BIBO Stability);
b. If there is no input, the output should tend to be zero, irrespective of any initial conditions.
I We know that a feedback control system is stable if and only if all the closed loop poles lie in the left-half of s-plane.
I The closed loop transfer function of most linear feedback systems are of the form C(s) = b0sm +b1sm−1 +…+bm−1s+bm; (m≤n)= B(s)
R(s) a0sn + a1sn−1 + … + an−1s + an A(s)
I One method is to factor the polynomial A(s) and find all its roots and determine
if there exist any roots on the right-half s-plane.
I Note that we are only interested to know whether there are some closed loop poles which lie in the right-half of s-plane.
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Routh’s Stability Criterion
Consider that the closed loop characteristic equation of a linear time invariant system is of the form
a0sn +a1sn−1 +…+an−1s+an = 0
where all the coefficients are real numbers.
I It is assumed that an ̸= 0, that is any zero root has been removed.
I In order that there be no roots of the characteristic equation with positive real parts it is necessary but not sufficient that
1. All the coefficients of the polynomial have the same sign. 2. None of the coefficients vanishes.
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Formation of Routh’s Table-1
1. If all the coefficients are positive, arrange the coefficients of polynomials into two rows.
2. The first row consists of the first,third, fifth,…coefficients, and the second row consists of the second, fourth,sixth,…coefficients as shown below
Table: Completed Routh Table
a0 a2 a4 a6 .
I
I
The b′s coefficients are evaluated as follows:
b1 = a1a2 −a0a3, b2 = a1a4 −a0a5, a1 a1
sn n−1
….. s2 e1 e2
s1 f s0 g1
b3 = a1a6 −a0a7,….. a1
Following similar procedure, the c′s coefficients are evaluated as follows:
s a1 a3 a5 a7 . sn−2 b1 b2 b3 b4 . sn−3 c1 c2 c3 c4 .
b1a3 − a1b2 b1
b1a5 − a1b3 b1
,
c1 = c3=b1a7−a1b4,…..
, c2 =
1 b1
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Formation of Routh’s Table-2
I The evaluation the b′s coefficients are evaluated as follows:
b1 = a1a2 −a0a3, b2 = a1a4 −a0a5, b3 = a1a6 −a0a7,…..
a1 a1 a1
I Following similar procedure, the evaluation the c′s coefficients are evaluated as
follows:
c1 = b1a3 −a1b2, c2 = b1a5 −a1b3, c3 = b1a7 −a1b4,….. b1 b1 b1
Routh’s Criterion:
1. The number of roots of the polynomial with positive real parts is equal to the number of changes in sign of the elements of the first column of the array.
2. If all the elements of the first column of the Routh tabulation are of the same sign, then the roots of the polynomial are all in the left half of the s-plane.
3. The number of sign changes in the elements of first column indicates the number of roots with positive real part.
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Routh-Hurwitz Stability Criteria for Linear Feedback Control System
I It is assumed that the polynomial used to construct the Routh’s table is the closed loop characteristic equation of a control system.
I Therefore, the roots of this polynomial correspond to the poles of the closed loop system.
Stability Criterion:
1. If there are no sign changes in the first column of the Routh table, then the poles of the closed loop system lie in the negative half of the s− plane and the closed loop system is stable.
2. If there are sign changes in the first column of Routh’s table, then some of the poles lie in the right half of s-plane and therefore the closed loop system is unstable
3. The number of sign changes in the first column of Routh’s table equals to number of roots which lie in the right half of s− plane.
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Example-1:Stability Analysis using Routh-Hurwitz Criteria Consider a system with closed loop characteristics equation:
s3 +4s2 +6s+4
I Construct the Routh array and determine the stability of the system
Table: Completed Routh Table
s3 1 6 s2 4 4 s1 −(1×4−6×4)/4 = 20/4 = 5
s0 −(4×0−4×5)/4 = 4
I Since there are no sign changes in the first column of Routh array, all the closed poles lie in the left half of s-plane and the system is stable.
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Example-2: Stability Analysis using Routh-Hurwitz Criteria Consider the following characteristics equation:
s4 +2s3 +3s2 +4s+5
I Construct the Routh array and determine the stability of the system
s4 s3
s2
s1
s0 Akshya Swain
Table: Completed Routh Table
1 3 5
2 4 0
1 3 1 5
− − 2 4 2 0
=1 =50
22
2 4
− 1 5
=−6 0 0
1 5 −
−6 0
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= 5
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Example-2: Solution (Contd)
Thus the Routh array is given by
Table: Completed Routh Table
s4 s3 s2 s1 s0
1 3 5 2 4 0 1 5
-6 5
I Since there are sign changes in the first column of Routh array, the system has closed poles in the right half of s-plane. Thus it is unstable.
I Further, there are 2-number of sign changes in the first column ( 1 to -6 and -6 to 5). This implies out of 4 roots, 2 number of roots have positive real part.
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Example-3: Stability via Routh-Hurwitz Criteria
I Determine whether the unity feedback system shown in the Figure is stable if
G(s) = 240 (s+1)(s+2)(s+3)(s+4)
1. Since there are sign changes in the first column of Routh array, the system has closed poles in the right half of s-plane. Thus it is unstable.
2. Further, there are 2-number of sign changes in the first column ( 30 to −114/3 and from −114/3 to 264). This implies out of 4 roots, 2 number of roots have positive real part.
I The closed loop characteristic equation is given by
s4 +10s3 +35s2 +50s+264 Table: Completed Routh Table
s4 s3
1 35 264 10 50
s2 30 264 s1 −114
3
s0 264
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Example-4: Stability via Routh-Hurwitz Criteria
I
The closed loop characteristic equation of a system is given by
s3 +3s2 +3s+1+K =0
Determine the range of values of the system parameter K which ensures system
stability
Solution:
Table: Completed Routh Table
s3 1 3
For system stability, it is necessary to fulfill the following conditions.
1. 8 − K > 0 i.e. 8 > K and
2. K+1>0i.eK>−1
1. Thus for the system to be stable, the parameter K must lie between -1 and 8 ( i.e. −1
Thus for the system to be stable,
the parameter K must lie s0 K 0 between 0 and 8 ( i.e.
0
I From the s1 row, the condition of stability is
3K2+6K−4>0, =⇒ K<−2.528orK>0.528.
I Comparing the conditions K < −2.528, K > 0 and K > 0.528, it is obvious that the latter is more stringent. Thus for the system to be stable, K > 0.528
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Example-2: Control System Analysis using Routh’s Stability Criterion
Consider the linear feedback system with integral controller shown
The closed loop characteristic equation of a system for ωn = 86.6025rad/sec and ζ = 0.1192 is described by
s3 + 34.5s2 + 7500s + 7500KI = 0
Determine the range of KI which ensures the stability of the system.
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Example-2:Solution(Contd)
The closed loop characteristic equation of a system is described by s3 + 34.5s2 + 7500s + 7500KI = 0
Solution: The Routh Table is shown below.
s3 1 7500
s2 34.5
s1 s0
7500KI 258,750−7500KI 0
34.5 7500KI
I For the system to be stable, all the coefficients in the first column of Routh array should be of same sign. This gives the following 2-conditions:
258, 750 − 7500KI > 0 (1) 34.5
7500KI >0 (2)
I Condition-1 gives KI < 34.5 and condition-2 gives KI > 0 I Thus for stability, KI must satisfy 0 < KI < 34.5
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Example-3: Control System Analysis using Routh’s Stability Criterion
The open loop transfer function of a unity feedback control system is
G(s) = K(s + 4) s(s+1)(s+2)
Determine
a. The range of K that keeps the system stable. b. The value of K that make the system oscillate.
c. The frequency of oscillation when K is set to the value that make the system oscillate.
Solution:
The closed loop transfer function of the system is given by
T(s)C(s) = G(s) = K(s+4)
R(s) 1+G(s) s3 +3s2 +(2+K)s+4K
The closed loop characteristic equation of this system is described by s3 +3s2 +(2+K)s+4K =0
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Example-3 : Solution (contd)
The closed loop characteristic equation of the system is described by
s3 +3s2 +(2+K)s+4K =0
The Routh Table is shown below.
s3 1 2+K s2 3 4K s1 6−K 0 s0 4K
a. For the system to be stable, all the coefficients in the first column of Routh array should be of same sign. This gives the following 2-conditions:
6−K>0, K>0, =⇒ 0