Vision
System Lab
Chapter 6. Stability
Youngjoon, Han
[email protected]
VISION SYSTEM
Vision
System Lab
Introduction
 C(t)=Cforced(t)+Cnatural(t)
 Depending upon natural response, as
time approaches infinity a linear, timeinvariant system is
– Stable if the natural response approaches
zero
– Unstable if the natural response grows
without bound
– Marginally stable if the natural response
neither decays nor grows
VISION SYSTEM
Vision
System Lab
Introduction
 C(t)=Cforced(t)+Cnatural(t)
 Depending upon natural response, the
definition of stability,
– A system is stable if every bounded input
yields a bounded output (BIBO)
– If input is bounded but the total response is
not bounded, the system is unstable
VISION SYSTEM
Vision
Stability System Lab
Marginally stable
Note that the
roots are on the
imaginary axis
VISION SYSTEM
Closed-loop poles
and
Vision
System Lab
response
VISION SYSTEM
Vision
System Lab
Routh-Hurwitz Criterion
 Routh-Hurwitz Criterion
– 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
– The method requires two step
• Generate a data table(Routh table)
• Interpret the Routh table to tell how many closedloop system poles are in the left half-plane, the
right half-plane, and on the jw-axis
VISION SYSTEM
Vision
System Lab
Routh-Hurwitz Criterion
 Generating a Basic Routh Table
Equivalent Closed-loop
Transfer function
If no sign changes, stable
VISION SYSTEM
Vision
System Lab
Routh-Hurwitz Criterion
 Example 6.1
1000
G ( s)
( s  2)( s  3)( s  5)

1000
1  G ( s) 1 
( s  2)( s  3)( s  5)
1000
1000

 3
( s  2)( s  3)( s  5)  1000 s  10 s 2  31s  1030
VISION SYSTEM
Vision
System Lab
Routh-Hurwitz Criterion
 Example 6.1
Two sign changes, two nstable poles
VISION SYSTEM
Routh-HurwithVision
Criterion:
System Lab
Special Case
 Zero Only in the first Column
10
T ( s)  5
s  2s 4  3s 3  6s 2  5s  3
VISION SYSTEM
Two sign changes; two unstable poles
Routh-HurwithVision
Criterion:
System Lab
Special Case
 Zero Only in the first Column  A polynomial
that has the reciprocal root of the original
polynomial has its roots distributed the same
T (s) 
10
s 5  2 s 4  3s 3  6 s 2  5 s  3
1
, criteria can be converted to
d
D ( s )  3s 5  5 s 4  6 s 3  3s 2  2 s  1
s
VISION SYSTEM
Vision
Routh-Hurwith Criterion:
System Lab
Special Case
 Entire Row is zero
10
T ( s)  5
4
3
2
s  7 s  6s  42s  8s  56
P(s)
All
zero
P’(s)
VISION SYSTEM
P(s)=s4+6s2+8,
P’(s)=4s3+12s
Pole distribution Vision
via Routh table
System
Lab
with row of zeros
 An entire row of zeros
will appear in the Routh
table when a purely even
or purely odd polynomial
is a factor of the original
polynomial.
 Even polynomials only
have roots that are
symmetrical about the
origin.
VISION SYSTEM
Pole distribution Vision
via Routh table
System
Lab
with row of zeros
 Since jω roots are symmetric about the
origin, if we do not have a row of zeros, we
cannot possibly have jω roots.
 Everything from the row containing the
even polynomial down to the end of the
Routh table is a test of only the even
polynomial.
VISION SYSTEM
Pole distribution
Vision
via Routh table with
row of
System
Lab
zeros
T (s) 
All
zero
VISION SYSTEM
20
s 8  s 7  12s 6  22s 5  39s 4  59s 3  48s 2  38s  20
Pole distribution Vision
via Routh table
System
Lab
with row of zeros
 Total 8 roots exist because it is 8th order
P(s)=s4+3s2+3, no sign change after this
means no real pair  4 roots on j-axis
VISION SYSTEM
Pole distribution Vision
via Routh table
System
Lab
with row of zeros
 Example 6.8
128
G(s)
s ( s 7  3s 6  10 s 5  24 s 4  48s 3  96 s 2  128s  192)
T (s) 

128
1  G(s) 1 
s ( s 7  3s 6  10 s 5  24 s 4  48s 3  96 s 2  128s  192)
1
 8
s  3s 7  10 s 6  24 s 5  48s 4  96 s 3  128s 2  192 s  128
VISION SYSTEM
Pole distribution
Vision
via Routh table with
row of
System
Lab
zeros
 Example 6.8
P(s)
P(s)=s6+8s4+32s2+64; two rhs (two sign changes),
VISION SYSTEM
so two lhs, the remaining two on j-axis
Stability designVision
via RouthSystem Lab
Hurwitz
 Example 6.9
VISION SYSTEM
K
G ( s)
s ( s  7)( s  11)
T (s) 

K
1  G (s) 1 
s ( s  7)( s  11)
K
 3
s  18s 2  77 s  K
Stability designVision
via RouthSystem Lab
Hurwitz
 If no sign change at the first column;
unstable
 If there is a zero row, jw is possible
Stable
0<K<1386
VISION SYSTEM
Stability designVision
via RouthSystem Lab
Hurwitz
 Routh table for
Example 6.9 with K = 1386
P(s)=18s2+1386
P’(s)=36s
As there is no
sign change
below, there are
two jw poles
VISION SYSTEM
Marginally stable
Vision
System Lab
Stability in State Space
 System poles are equal to the eigen
values of the system matrix A
 Ax=lx
 (lI-A)x=0
 x= (lI-A)-10
 x= [adj(lI-A) /det (lI-A)]0
 det (lI-A)=0
 Use det (sI-A)=0
VISION SYSTEM
Vision
System Lab
Stability in State Space
 Example 6.11
s(s-8)(s+2)+30+10
+10(s-8)+5s-6(s+2)
=s3-6s2-7s-52
VISION SYSTEM
 0
x   2
 10
y  1 0
3
8
5
0x
1 
10
1  x   0 u
 0 
 2
3
 s 0 0  0
( sI  A)  0 s 0   2
8
0 0 s   10  5
3
1 
 s
  2 s  8
 1 
 10
5
s  2
det( sI  A)  s 3  6 s 2  7 s  52
1 
1 
 2
Vision
System Lab
Stability in State Space
 Example 6.11
One rhs, two lhs; unstable
VISION SYSTEM