Topic 9. Stability Analysis of Linear
Time-Invariant Systems
Instructor: Prof. Kwang-Hyun Cho
Department of Bio and Brain Engineering
Korea Advanced Institute of Science and Technology
Web: http://sbie.kaist.ac.kr
BiS352 System Modeling in Bioengineering
Fall 2013
Stability analysis
Stability is the most important system specification. If a system is
unstable, transient response and steady-state errors are moot points.
An unstable system cannot be designed for a specific transient response
or steady-state error requirement.
What is stability?

There are many definitions for stability, depending upon the kind of system
or the point of view. In this section we limit ourselves to linear timeinvariant systems.
The total response of a system is the sum of the forced and natural
responses, or
c(t )  c forced (t )  cnatural (t )
Using these concepts, we present the following definitions of stability,
instability and marginal stability.


A linear, time-invariant system is stable if the natural response approaches
zero as time approaches infinity.
A linear, time-invariant system is unstable if the natural response grows
without bound as time approaches infinity.
Kwang-Hyun Cho
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Stability analysis

A linear, time-invariant system is marginally stable if the natural response
neither decays nor grows but remains constant or oscillates as time
approaches infinity.
Thus, the definition of stability implies that only the forced response
remains as the natural response approaches zero.
These definitions rely on a description of the natural response. When
one is looking at the total response, it may be difficult to separate the
natural response from the forced response.
However, if the input is unbounded, we see an unbounded total
response, and we cannot arrive at any conclusion about the stability of
the system; we cannot tell whether the total response is unbounded
because the forced response is unbounded or because the natural
response is unbounded.
Kwang-Hyun Cho
Stability analysis
Thus, an alternate definition of stability, one that regards the total
response and implies the first definition based upon the natural
response, is this:
A system is stable if every bounded input yields a bounded output.
We call this statement the bounded-input, bounded-output (BIBO)
definition of stability.
This definition helps clarify our previous definition of marginal stability,
which really means that the system is stable for some bounded inputs
and unstable for others.
Thus, marginally stable systems by the natural response definitions are
included as stable systems under the BIBO definitions.
Kwang-Hyun Cho
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Stability analysis
=
y(t)
=
Definition 1 of stability: The natural response approaches zero.
Definition 2 of stability (BIBO stability): If the input u(t) is bounded
(|u(t)|≤Bu), then the output y(t) is also bounded (|y(t)|≤By).
y (t )  u (t )  g (t )   u ( )  g (t   )d  B

0
y
u (t )  B
 g (t )dt  

0
u
Kwang-Hyun Cho
Stability analysis
Physically, an unstable system whose natural response grows without
bound can cause damage to the system, to adjacent property, or to
human life.
Systems are often designed with limit stops to prevent total runaway.
From the perspective of the time response plot of a physical system,
instability is displayed by transients that grow without bound and,
consequently, a total response that does not approach a steady-state
value or other forced response.

Care must be taken here to distinguish between a natural response growing
without bound and a forced response, such as a ramp or exponential
increase, that also grows without bound. A system whose forced response
approaches infinity is stable as long as the natural response approaches
zero (See the example in Topic 1).
How do we determine if a system is stable?

Let us focus on the natural response definitions of stability.
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Stability analysis


Recall from our study of system poles that poles in the left half-plane yield
either pure exponential decay or damped sinusoidal natural responses.
Thus, if the closed-loop system poles are in the left half-plane of the s-plane
and hence have a negative real part, the system is stable.
That is, stable systems have closed-loop transfer functions with poles only
in the left-half plane.
Definition 3 of stability : All poles of closed-loop transfer functions are
located in the left-half complex plane.
Kwang-Hyun Cho
Stability analysis



Poles in the right half-plane yield either pure exponentially increasing or
exponentially increasing sinusoidal natural responses. Thus, if the closedloop system poles are in the right half of the s-plane and hence have a
positive real part, the system is unstable.
Also, poles of multiplicity greater than one on the imaginary axis lead to the
n
sum of responses of the form At cos(t   ) , where n = 1, 2, …, which also
approaches infinity as time approaches infinity.
Thus, unstable systems have closed-loop transfer functions with at least
one pole in the right half-plane and/or poles of multiplicity greater than one
on the imaginary axis.
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Stability analysis


Finally, a system that has imaginary axis poles of multiplicity 1 yields pure
sinusoidal oscillations as a natural response. These responses neither
increase nor decrease in amplitude.
Thus, marginally stable systems have closed-loop transfer functions with
only imaginary axis poles of multiplicity 1 and poles in the left half-plane.
As an example, the unit step response of the stable system of Figure
6.1(a) is compared to that of the unstable system of Figure 6.1(b).
Figure 6.1 (from Control Systems Engineering book)
Kwang-Hyun Cho
Stability analysis
It is not always a simple matter to determine if a feedback control
system is stable. Unfortunately, a typical problem that arises is shown
in Figure 6.2.


Although we know the poles of the forward transfer function in Figure
6.2(a), we do not know the location of the poles of the equivalent closedloop system of Figure 6.2(b) without factoring or otherwise solving for the
roots.
There is a method to test for stability without having to solve for the roots
of the denominator.
Figure 6.2 (from Control Systems Engineering book)
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Stability analysis - Routh’s stability criterion
Routh’s stability criterion (Routh-Hurwits criterion )

Most linear closed-loop control systems have closed-loop transfer functions
of the form
C ( s ) b0 s m  b1s m 1    bm 1s  bm B( s )


Eq.10-31
R ( s ) a0 s n  a1s n 1    an 1s  an A( s )

where a’s and b’s are constants and m ≤ n.
The locations of the roots of the characteristic equation (the denominator of
the preceding equation) determine the stability of the closed-loop system.
A simple criterion, known as Routh’s stability criterion, enables us to
determine the number of closed-loop poles that lie in the right half s-plane
without having to factor the polynomial.
This criterion applies only to polynomials with a finite number of terms.

The procedure used in Routh’s stability criterion is as the following steps :


Kwang-Hyun Cho
Stability analysis - Routh’s stability criterion
1.
2.
Write the polynomial in s in the following form:
a0 s n  a1s n 1    an 1s  an  0
where the coefficients are real quantities. Assume that an ≠ 0; that is, any
zero root has been removed.
If any of the coefficients are zero or negative in the presence of at least
one positive coefficient, there is a root or roots that are imaginary or that
have positive real parts. In such a case, the system is not stable. Note
that all the coefficients must be positive. This is a necessary conditions,
as may be seen from the following argument:



A polynomial in s having real coefficients can always be factored into linear
and quadratic factors, such as (s+a) and (s2 + bs + c), where a, b, and c are
real.
The factor (s2 + bs + c) yields roots having negative real parts only if b and c
are both positive. For all roots to have negative real parts, the constants a, b,
c, and so on, in all factors must be positive.
It is important to note that the condition that all the coefficients be positive is
not sufficient to assure stability. (If all a’s are negative, they can be made
positive by multiplying both sides of the equation by -1. This operation does
not change the root location)
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Stability analysis - Routh’s stability criterion
3.
If all the coefficients of the polynomial are positive, arrange the
coefficients of the polynomial in rows and columns according to the
following pattern:
sn
s n 1
s n 2
s n 3
s n 4

a0
a1
b1
c1
d1

a2
a3
b2
c2
d2

s2
s1
s0
e1
f1
g1
e2
a4
a5
b3
c3
d3
a6
a7
b4
c4
d4





The coefficients b1, b2, b3, and so on, are evaluated as follows:
b1  
a0
a2
a1
a3
a1
a a a a
 1 2 0 3
a1
b2  
a0
a4
a1
a5
a1
a a a a
 1 4 0 5
a1
b3  
a0
a1
a6
a7
a1

a1a6  a0 a7
a1

Kwang-Hyun Cho
Stability analysis - Routh’s stability criterion
The evaluation of the b’s is continued until the remaining ones are all zero.
The same pattern of cross multiplying the coefficients of the two previous
rows is followed in evaluating the c’s, d’s, e’s and so on. That is:
c1  
a1
b1
a3
b2
b1
ba ab
 1 3 1 2
b1
c2  
a1
a5
b1
b3
b1
a1

b1a5  a1b3
b1

c1b3  b1c3
c1
c3  
a7
b1
b4
b1

b1a7  a1b4
b1

and
b1 b2
c c
c b bc
d1   1 2  1 2 1 2
c1
c1


d2  
b1 b3
c1 c3
c1

The process continues until the nth row has been completed. The finished
array of coefficients is triangular.
Note that, in developing the array, an entire row may be divided or
multiplied by a positive number in order to simplify the subsequent
numerical calculation without altering the stability conclusion.
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Stability analysis - Routh’s stability criterion



Routh’s stability criterion states that the number of roots of Equation (1031) with positive real parts is equal to the number of changes in sign of the
coefficients of the first column of the array. (The first column of the array
means the first numerical column.)
Note that the exact value of the terms in the first column need not be
known; only the signs are needed.
The necessary and sufficient condition that all roots of Equation (10-31) lie
in the left-half s-plane is that all the coefficients of Equation (10-31) be
positive and all terms in the first column of the array have positive signs.
Example 10-5

Let us apply Routh’s stability criterion to the third-order polynomial:
a0 s 3  a1s 2  a2 s  a3  0
s3
s2
where all the coefficients are positive
numbers. The array of coefficients
becomes:
s1
s0
a0
a1
a1a2  a0 a3
a1
a2
a3
a3
Kwang-Hyun Cho
Stability analysis - Routh’s stability criterion

The condition that all roots have negative real parts is given by:
a1a2  a0 a3
Example 10-6

Consider the polynomial:
s 4  2 s 3  3s 2  4 s  5  0

Let us follow the procedure just presented and construct the array of
coefficients. (If any coefficients are missing, they may be replaced by zeros
in the array.) The completed array is as follows:
s4
1
3 5
3
2
1
4 0
5
s
s2
1
s
s0

6
5
s4
s3
s
2
1
2
3 5
4 0
1
1
2 0
5
The second row is
divided by 2
s1 3
s0 5
In this example, the number of changes in sign of the coefficients in the
first column of the array is 2.
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Stability analysis - Routh’s stability criterion


This means that there are two roots with positive real parts.
Note that the result is unchanged when the coefficients of any row are
multiplied or divided by a positive number in order to simplify the
computation.
Special cases (1) - Zero only in the first column



In a term in the first column of the array in any row is zero, but the
remaining terms are not zero or there is no remaining term, then the zero
term is replaced by a very small positive number ε and the rest of the array
is evaluated.
For example, consider the following equation:
Eq. 10-32
s3  2s 2  s  2  0
The array of coefficients is:
s3
s2
1
2
s1
s0
0
2
1
2
Kwang-Hyun Cho
Stability analysis - Routh’s stability criterion


If the sign of the coefficient above the zero (ε) is the same as that below it,
it indicates that there is a pair of imaginary roots. Actually, Eq. 10-32 has
two roots at s = ±j.
If, however, the sign of the coefficient above the zero (ε) is opposite that
below it, then there is one sign change. For example, for the equation
s 3  3s  2  ( s  1) 2 ( s  2)  0

The array of coefficients is as follows:
1
3
2
s
0
2
3 
s0
2
s3
2
s
The first sign change:
The second sign change:

1

Noting ε→0+, there are two sign changes of the coefficients in the first
column of the array. This agrees with the correct result indicated by the
factored form of the polynomial equation.
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Stability analysis - Routh’s stability criterion


Another method that can be used when a zero appears only in the first
column of a row is derived from the fact that a polynomial that has the
reciprocal roots of the original polynomial has its roots distributed the same
because taking the reciprocal of the root value does not move it to another
region. This method is usually computationally easier.
We now show that the polynomial we are looking for, the one with the
reciprocal roots, is simply the original polynomial with its coefficients written
in reverse order. Assume the equation:
s n  an 1s n 1    a1s  a0  0

If s is replaced by 1/d, then d will have roots which are the reciprocal of s.
Making this substitution in the last equation,
n
1
1
   an 1  
d 
d 

n 1
1
   a1    a0  0
d 
Factoring out (1/d)n,
n
1
(1 n )
n
1 
1
1
1 
 a0   
  1  an 1      a1  
 d  
d 
d 
 d  
n
1
   1  an 1d    a1d ( n 1)  a0 d n   0
d 
Kwang-Hyun Cho
Stability analysis - Routh’s stability criterion

Thus, the polynomial with reciprocal roots is a polynomial with the
coefficients written in reverse order.
Example 6.3


Determine the stability of the closed-loop transfer function
10
T ( s)  5
s  2 s 4  3s 3  6 s 2  5s  3
First, write a polynomial that has the reciprocal roots of the denominator of
the last equation. From our discussion this polynomial is formed by writing
the denominator in reverse order. Hence,
D( s )  3s 5  5s 4  6 s 3  3s 2  2 s  1


We form the Routh table as shown in Table
6.6 using the last equation.
Since there are two sign changes, the
system is unstable and has two right-halfplane poles.
Table 6.6
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Stability analysis - Routh’s stability criterion
Special cases (2) – Entire row is zero


If all of the coefficients in any derived row are zero, then there are roots of
equal magnitude lying radially opposite each other in the s-plane; that is,
there are two real roots of equal magnitude and opposite sign and/or two
conjugate imaginary roots. (Figure 6.5 from Control Systems Engineering
book)
In such a case, the evaluation of the rest of the array can be continued by
forming an auxiliary polynomial with the coefficients of the last row and by
using the coefficients of the derivative of this polynomial in the next row.

Figure 6.5 (from Control
Systems Engineering book)

Such roots of equal magnitude and
lying radially opposite each other in
the s-plane can be found by solving
the auxiliary polynomial, which is
always even.
For a 2n-degree auxiliary
polynomial, there are n pairs of
equal and opposite roots.
Kwang-Hyun Cho
Stability analysis - Routh’s stability criterion

For example, consider the following equation:
s 5  2 s 4  24 s 3  48s 2  25s  50  0

The array of coefficients is:
s5
s4
s3

1 24 25
2 48 50
0 0
Auxiliary polynomial P(s)
The terms in the s3 row are all zero. The auxiliary polynomial is then formed
from the coefficients of the s4 row. The auxiliary polynomial P(s) is:
P ( s )  2 s 4  48s 2  50


which indicates that there are two pairs of roots of equal magnitude and
opposite sign. These pairs are obtained by solving the auxiliary polynomial
equation P(s) = 0.
The derivative of P(s) with respect to s is:
dP( s)
 8s 3  96 s
ds
The terms in the s3 row are replaced by the coefficients of the last equation,
that is, 8 and 96.
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Stability analysis - Routh’s stability criterion

The array of coefficients then becomes:
s5
s4
1
2
25
50
24
48
Coefficients of dP(s)/ds
8
96
s3
50
24
s2
s1 112.7 0
s 0 50

We see that there is one change in sign in the first column of the new array.
Thus, the original equation has one root with a positive real part. By solving
for roots of the auxiliary polynomial equation,
2 s 4  48s 2  50  0
we obtain:
or

s 2  1,
s 2  25
s  1,
s   j5
These two pairs of roots are a part of the roots of the original equation. As
a matter of fact, the original equation can be written in factored form as
follows:
( s  1)( s  1)( s  j 5)( s  j 5)( s  2)  0
Kwang-Hyun Cho
Stability analysis - Routh’s stability criterion
Relative stability analysis





Routh’s stability criterion provides the answer to the question of absolute
stability. In many practical cases, however, this answer is not sufficient; we
usually require information about the relative stability of the system.
A useful approach to examining relative stability is to shift the s-plane axis
and apply Routh’s stability criterion. That is, we substitute:
s  sˆ  
(  positive constant)
into the characteristic equation of the system, write the polynomial in terms
of ŝ , and apply Routh’s stability criterion to the new polynomial in ŝ.
The number of changes of sign in the first column of the array developed
for the polynomial in ŝ is equal to the number of roots that are located to
the right of the vertical line s   .
Thus, this test reveals the number of roots that lie to the right of the
vertical line s   .
Therefore, the maximum value of σ that makes the system stable is
referred as to stability margin.
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Stability analysis - Routh’s stability criterion
Application of Routh’s stability criterion to control systems
analysis



Routh’s stability criterion is of limited usefulness in linear control systems
analysis, mainly because it does not suggest how to improve relative
stability or how to stabilize an unstable system.
It is possible, however, to determine the effects of changing one or two
parameters of a system by examining the values that cause instability.
Consider the system shown in Figure 10-39. Let us determine the range of
K for stability. The closed-loop transfer function is
C (s)
K

R ( s ) s ( s  1)( s  2)  K

The characteristic equation is:
s ( s  1)( s  2)  K  0

Figure 10-39
The array of coefficients becomes:
1
2
s3
2
s
K
3
6 K
1
s
3
s0
K
Kwang-Hyun Cho
Stability analysis – Stability in state space
(from Control Systems Engineering book)

For stability, K must be positive, and all coefficients in the first column of
the array must be positive. Therefore,
6K 0

When K = 6, the system becomes oscillatory, and, mathematically, the
oscillation is sustained at constant amplitude.
Stability in state space



Up to this point we have examined stability from input and output
description viewpoint. Now we look at stability from the perspective of state
space description.
Note that the values of the system’s poles are equal to the eigenvalues of
the system matrix, A. We also state that the eigenvalues of the matrix A
were solutions of the equation det (sI-A) = 0, which also yielded the poles
of the transfer function.
Review : The eigenvalues of a matrix, A, are values λ of that permit a
nontrivial solution (other than 0) for x in the equation
Ax   x
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Stability analysis – Stability in state space
(from Control Systems Engineering book)

In order to solve for the values of λ that do indeed permit a solution for x,
we rearrange the last equation as follows:
or

 x  Ax  0
( I  A ) x  0
Solving for x yields:
or
x  ( I  A) 1 0
x


adj ( I  A)
0
det( I  A )
We see that all solutions will be the null vector except for the occurrence of
zero in the denominator. Since this is the only condition where elements of
x will be 0/0, or indeterminate, it is the only case where a nonzero solution
is possible.
The values of λ are calculated by forcing the denominator to zero:
det( I  A)  0
Eq. 6.31
Kwang-Hyun Cho
Stability analysis – Stability in state space
(from Control Systems Engineering book)



This equation determines the values λ of for which a nonzero solution for x.
We defined x as eigenvectors and the values of λ as the eigenvalues of the
matrix A.
Let us now relate the eigenvalues of the system matrix, A, to the system’s
poles. In Topic 6, we derived the equation of the system transfer function
(from the state description) as:
G ( s)  C( sI  A)1 B  D

The system transfer function has det (sI-A) in the denominator because
the presence of (sI-A)-1. Thus,
det( sI  A)  0
Eq. 6.32
is the characteristic equation for the system from which the system poles
can be found.


Since Eqs. 6.31 and 6.32 are identical apart from a change in variable
name, we conclude that the eiganvalues of the matrix A are identical to the
system’s poles.
Thus, we can determine the stability of a system represented in state space
by finding the eigenvalues of the system matrix, A, and determining their
locations on the s-plane.
Kwang-Hyun Cho
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Stability analysis – Stability in state space
(from Control Systems Engineering book)
Example 6.11

Given the system
3 1
 0
10 
x   2
8 1  x   0  u
 10 5 2 
 0 


y  1 0 0 x
Find out how many poles are in the left half-plane, in the right half-plane,
and on the jω-axis.
First form (sI-A):
3 1 s
3
1 
 s 0 0  0
( sI  A)  0 s 0    2
8 1    2 s  8 1 
0 0 s   10 5 2  10
5
s  2 

Now find the det(sI-A):
det( sI  A )  s 3  6s 2  7 s  52


Using this polynomial, form the Routh
table, as shown in Table 6.18.
Since there is one sign change, the system
has one pole on right-half-plane .
Table 6.18
Kwang-Hyun Cho
15