Chapter 6
Stability
Stability Definitions
• A system is stable if the natural response
approaches zero as time approaches infinity
• A system is unstable if the natural
response approaches infinity as time
approaches infinity
• A system is marginally stable if the
natural response neither decays nor grows
but remains constant or oscillates.
• A system is stable if every bounded input
yields a bounded output
• A system is unstable if any bounded input
yields an unbounded output
Common cause of problems in finding closed-loop poles:
a. original system;
b. equivalent system
Stable systems have closed-loop transfer functions with poles in the
left half-plane.
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
Marginally stable systems have closed-loop transfer functions with
only imaginary axis poles of multiplicity one and poles in the left halfplane.
Routh-Hurwitz Criterion
Using this method we can tell how many closed-loop poles
are in the left half-plane, in the right half-plane and on the
imaginary axis.
The method requires two steps: (1) Generate the data table
(Routh table) and (2) Interpret the table to determine the
number of poles in LHP and RHP.
Initial layout for Routh table
Completed Routh table
Feedback system for Example 6.1;
b. Equivalent closed-loop system
Completed Routh table for Example 6.1
Interpretation of Routh table
The number of roots of the polynomial that are in the
right half-plane is equal to the number of sign changes in
the first column.
R-H: Special case. Zero in the first column
 , the value of 
Replace the zero with
from –ive or +ive side.
is then allowed to approach zero
Problem Determine the stability of the closed-loop transfer function
10
T (s )  5
s  2s 4  3s 3  6s 2  5s  3
R-H: Special case. Zero in the first column
Problem Determine the stability of the closed-loop transfer function
T (s ) 
10
s 5  2s 4  3s 3  6s 2  5s  3
Solution Write a polynomial that has the reciprocal roots of the denominator
D (s )  3s 5  5s 4  6s 3  3s 2  2s  1
R-H: Special case. Entire row is zero
Problem Determine the number of right-half-plane poles in the closed
transfer function T (s ) 
10
s 5  7s 4  6s 3  42s 2  8s  56
Solution: Form an auxiliary polynomial, P(s) using the entries of row above
row of zeros as coefficient, then differentiate with respect to s finally use
coefficients to replace the rows of zeros and continue the RH procedure.
P (s )  s 4  6s 2  8
dP (s )
 4s 3  12s  0
ds
Root positions to generate even polynomials: A , B, C, or any combination
An entire row of zeros will appear when a
purely even or odd polynomial is a factor of
the original polynomial.
Even polynomials have roots that are
symmetrical about the origin.
If we don’t have a row of zeros, we cannot
possibly have roots on the imaginary axis.
Routh table for Example 6.5
Problem Determine the number of poles in the right-half-plane, left-halfplan and on the j  axis for the closed transfer function
T (s ) 
20
s 8  s 7  12s 6  22s 5  39s 4  59s 3  48s 2  38s  20
Summary of pole locations for Example 6.5
Feedback control system for Example 6.6
Problem Determine the number of poles in the right-half-plane, left-halfplan and on the j  axis for system
Solution: The closed loop transfer function is
200
T (s )  4
s  6s 3  11s 2  6s  200
Routh table for Example 6.6
2 poles in RHP, 2 poles in LHP no poles on j 
The system is unstable
axis
Feedback control system for Example 6.7
Problem Determine the number of poles in the right-half-plane, left-halfplan and on the j  axis for system
Solution: The closed loop transfer function is
1
T (s )  5
2s  3s 4  2s 3  3s 2  2s  1
Routh table for Example 6.7
Table for polynomial whose roots are
the reciprocal of the original
P (s )  s 5  2s 4  3s 3  2s 2  3s  2
2 sign changes, 2 poles in RHP system is unstable
Feedback control system for Example 6.8
Problem Determine the number of poles in the right-half-plane, left-halfplan and on the j  axis for system. Draw conclusions about stability of the
closed loop system.
Solution: The closed loop transfer function is
T (s ) 
128
s 8  3s 7  10s 6  24s 5  48s 4  96s 3  128s 2  192s  128
Routh table for Example 6.8
There is a row of zeros in the s5, so the s6 row forms an even
polynomial. 2 sign changes from the even polynomial so 2 poles in
the RHP and because of symmetry about origin 2 will be in the LHP
The 2 remaining poles are on the j  axis
Summary of pole locations for Example 6.5
Feedback control system for Example 6.9
Problem Find the range of gain K for the system that will cuase the
system to be stable, unstabel, and marginally stable. Assume K>0.
Solution: The closed loop transfer function is
K
T (s )  3
s  18s 2  77s  K
Routh table for Example 6.9
For K < 1386 the system is stable. For K > 1386 the system is unstable.
For K = 1386 we will have entire row of zeros (s row). We form the even
polynomial and differentiate and continue, no sign changes from the
even polynomial so the 2 roots are on the j  axis and the system is
marginally stable
Routh table for Example 6.10
Problem Factor the polynomial
s 4  3s 3  30s 2  30s  200
Solution: from the Routh table we see that the s1 row is a row of zeros.
So the even polynomial at the s2 row is P (s )  s 2  10
since this
polynomial is a factor of the original, dividing yields P (s )  s 2  3s  20
As the other factor so
2
2
s 4  3s 3  30s 2  30s  200  (s  10) (s  3s  20)
Stability is State Space Example 6.11
Problem Given the system
 0
X   2
 10
y  1 0
1
10 
8 1  X  0  u
0 
5 2 
0 X
3
Find out how many poles in the LHP, RHP and on the j  axis
Solution: 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) = s  6 s  7 s  52
3
2
Routh table for Example 6.11
One sign change, so 1 pole in the LHP and the
system is unstable