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Chapter 2 (page#143)
2.5 Stability Testing
2.5.1 Coefficient Tests
2.5.4 Case 1
Introduction
2.5.2 Routh-Hurwitze Stability Testing
2.5.5 Case 2
2.5.6 Case 3
2.6 Parameter Shifting (Page 159)
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System should be
Stable: BIBO stable if, for every bounded input, the
output is bounded for all time
LTI system must have all poles in the left-half of the s-
plane (negative real parts)
Transfer function poles are same as roots of the
characteristic polynomial are the same as system
Eigen- values
All Eigen-values must have negative real parts for BIBOstability
Poles on imaginary axis are not stable by this definition
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Stability
Example:1)s)(s()s(R
)s(C)s(G)s(T
21
2
Stable or Unstable? Why Characteristic equation (s+1)(s+2)=0
Hence s+1=0 or s=-1 and s+2=0 or s=-2
system to be stable is that all roots (-1 & -2)
of the characteristic equation (poles of the
closed-loop transfer function) lie in the left
half of the s-plane.
x x
-2 -1
S-plane
The natural-response terms for the system are k1e-t and k2e
-2t
22
2
2232
2
nn
n
sss
Over Damped System? Critically Damped System?
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Example 2:12112
241023
sss
s)s(T
)s)s)(s(
s
431
2410
Hence roots are s= -1, s= 3, & s= -4
The natural-response terms for the system are k1e-t ,k2e
3t,and k3e-4t
Stable or Unstable? Unstable due to s=3
Unstable due to term k2e3t
MATLAB Program:
>> p = [1 2 -11 -12];
>> r = roots(p)
Note: Numerator 10s+24 has NO role in the stability of the system
x x
-4 -1
x
3
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Example:312
s
s)s(T
)js)(js(
s
Hence roots are s= -j, s=j
The natural-response terms for the system are ksin(t+)
Stable or Unstable? Marginally stable
Marginally stable because no exponential term
MATLAB Program:
>> p = [1 0 1];
>> r = roots(p)
x
x
j
-j
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Example:4 (Example 3)
12
s
s)s(T
)s()s(
s)s(R)s(T)s(C
1
1
1 22
The response of the system c(t) = tsint
Marginally stable
Let the input r(t)=sint1
12
s
)s(R
)js)(js(
s
Hence roots are s= -j, s=j
x
x
j
-j
x
x
j
-j
Stable or Unstable? Unstable due to t
sint is marginally stable but the multiplication of time (t) it makes it unstable
NOTE: The natural-response of the system are ksin(t+)
bounded (marginally stable) but for unbounded output
(unstable) for certain bounded input
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Necessary condition
If any coefficientai of Q(s) is zero or negative
then not all roots lie in the left half of the s-plane
Otherwise: set up Routharray and use Routh-
Hurwitz criterion:
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It requires that all poles of the closed-loop TF lie in the
LHP.
HenceAbsoluteStability analysis requires determiningif any poles are in the RHP or on the jwaxis.
Routh-Hurwitz Criterion (Doesnt actually calculate
roots)
Routh-Hurwitz Stability Testing (Criterion)
Gives the number of roots with positive real parts.
01
1
1)( asasasasQn
n
n
n
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First step make the Routh Array:
s
n
an an-2 an-4 an-6
sn-1 an-1 an-3 an-5 an-7
sn-2 b1 b2 b3 b4
sn-3 c1 c2 c3 c4
. .
. .
. .s2 k1 k2
s1 l1
s0 m1
014
43
32
21
1 asasasasasasa)s(Qn
nn
nn
nn
nn
n
31
2
1
1 1
nn
nn
naaaa
ab
51
4
1
2
1
nn
nn
naa
aa
ab
21
31
1
1
1
bb
aa
b
cnn
31
51
1
2
1
bb
aa
bc
nn
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014
43
32
21
1 asasasasasasa)s(Qn
nn
nn
nn
nn
n
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Example: (Case: 1)
Number of sign changes in the first column = number of
unstable poles
Note: Case 1
(No zero element in first column)
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Example: (Case: 2) Un-stable system
Case 2 (Steps)
1. First Element of a row is Zero (0)
2. Replace 0 by small number. (=0.00000000..01)3. Continue the array
Q(s) = s5 + 2s4 + 2s3 + 4s2 + 11s + 10
s5 1 2 11s4 2 4 10
6
s2 -12/ 10s1 6
s0
10Two sign changes whether was
assumed +ve orve.
0)2241(21
4221
21
1 xxb
0b
1s
3
21
31
1
10
1
bb
aa
bcthen
nn
12412
1
6
4211 )(cthen
)valuesmallveryvery(blet 1
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Example: (Case: 3)Case 3: (Steps)
1. All Element of a row are Zero. (Premature Termination )
2. Auxiliary Polynomial
3. Aux Poly is differentiated with respect to s4. Coefficients of the Polynomial replaces the zero row
Ex. Q(s) = s2 + 1
s2 1 1
s1 0
s0
No sign change in the first column
implies that system is STABLE.
Aux poly indicates roots on jw axis, which implies
MARGINALLY STABLE
Aux Pol = s2 + 1
d/ds(s2 + 2)
2s
2
1
S2=-1; S=j
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What is Auxiliary Polynomial
Q(s) contains an even
polynomial as factor. An evenpolynomial is one in which the
exponents of s are even
integers or zero
This even polynomial is called
AuxiliaryPolynomial
In Routh array, coefficients of
Aux Poly are those directly
above the zero row. (See
examples above)
s2 1 1
s1 0s0
Aux Pol =s2 + 1
Aux Pol =s2
+ 2
s2 1 2
s1 0s0
s3 1 2
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Example: (Case: 3)Case 3: (Steps)
1. All Element of a row are Zero. (Premature Termination )
2. Auxiliary Polynomial
3. Aux Poly is differentiated with respect to s4. Coefficients of the Polynomial replaces the zero row
Ex. Q(s) = (s+1) (s2 + 2) =s3 + s2 + 2s + 2
s3 1 2
s2 1 2
s1 0
s0
Aux Pol = s2 + 2
d/ds(s2 + 2)
2s2
1
RHP=2
LHP=2
IA = 0
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Example: (Case: 3)
Ex. Q(s) = s4 + s3 + 3s2 + 2s + 2
s4 1 3 2
s3 1 2
s2 1 2
s1 0
s0
Aux Pol = s2 + 2
d/ds(s2 + 2)
2s2
2
RHP=0
LHP=2
IA = 2
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Example_Case_1 from Book (Page 146)
P(s) = 2s4 + 3s3 + 5s2 + 2s + 6 (2.14)
s4 2 5 6
s3 3 2
s2 +11/3 6
s1 -32/11
s0 + 6
Number of sign changes in the first column = number of
unstable poles
Two signchanges
+
-
+
Twounstable
poles
RHP=2
LHP=2
IA = 0
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Case 2 (Steps)
1. First Element of a row is Zero (0)
2. Replace 0 by small number. (=0.00000000..01)3. Continue the array
Q(s) = 3s4 + 6s3 + 2s2 + 4s +5 (2.16)
s4 3 2 5s3 6 4
2
s1 -12/s0 2
Two sign changes whether was
assumed +ve orve.
0b1
s2
Example_Case_2 from Book (Page 149)
RHP=2
LHP=2
IA = 0
E l C 2 f B k (P 149)
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Q(s) = s5 + s4 + 2s3 + 3s2 + s + 4 (2.20)
s5 1 2 1
s4 1 3 4
s3 -1 -3
s2 4
s1 4/
s0 4
0
Example_Case_2 from Book (Page 149)
Two sign changes whether was
assumed +ve orve.
RHP=2
LHP=3
IA = 0
E l C 3 f B k (P 150)
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Q(s) = s5 +2s4 + 8s3 + 11s2 + 16s + 12 (2.24)
s5 1 8 16
s4 2 11 12
s3 2.5 10
s2 3 12
s1 0
s0 12
Example_Case_3 from Book (Page 150)
Aux Pol = 3s2 + 12
d/ds(3s2 + 12)
6s
6
No sign change in the first column
MARGINALLY STABLE.
Aux poly indicates roots on jwaxis
3s2=-12 s2=-4 s1,2=2
E l E l C 3 f B k (P 152)
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s6 + s5 + 5s4 + s3 + 2s2 - 2s 8 (2.26)
s6 1 5 2 -8
s5 1 1 -2
s4 4 4 -8
s
3
0 0
s2 2 -8
s1 72
s0 -8
Aux Pol = 4s4 +4s2 + 8
16
Example Example_Case_3 from Book (Page 152)
d/ds(4s4 +4s2 + 8)
16s3 +8s2
8
RHP=1
LHP=3
IA = 2
E l f B k (P 160)
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S3 2 2
S2 3 K
S1 (6-2K)/3
S0 K
(6-2K)/3 > 0
K> 0
Stable
Stable
For what value of K the system will be marginally stable?
S4 1 4 K
Stability range 0 < K < 3
Example_ from Book (Page 160)
P(s) = S4 + 2s3 + 4s2 + 2s + K
E l f B k (P 160)
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S3 2 K
S
2
4-K/2 6S1
S0 6
> 0
K < 8
Stable
Stable
S4 1 4 6
Stability range can not be satisfied for any value of K
(complex roots). Polynomial has RHP roots for all K
Example_ from Book (Page 160)
complexroots).(
K
502
241640124
2
2
KK
P(s) = S4 + 2s3 + 4s2 + Ks + 6
24
12
/KK
Example (Important used in designing)
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Example (Important used in designing)
Design specification that ess must be less than 2% of the constant
unit step input. Find value of K that will produce error > 2%. Using
Routh Hurwitze criteria, verify the values of K for stability.
S3 1 5
S2 4 2+2K
S1 1/4(18-2K)
S0 2+2K
K -1
Stable
Stable
pcs
p GGlim0
K Ksss
K
sp
254
2lim
230K
50
1
1
1
K
ess
254
223
sss
KKGp
49K
For ess49system to be stable -1 < K
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_
K
2
s3 + 4s2 + 5s + 2
Q(s) = s4 + 4s3 + 5s2 + (2 + 2Kp) s + 2 Ki
s4 1 5 2Ki
s3 4 2+2Kp
s0 2Ki
Kp 0
Stable
Stable
s
ksKGwithreplacedisKtheexampleprevioustheIn
ipc
s
ksK ip
ksKK
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Q(s) = s4 + 4s3 + 5s2 + (2 + 2Kp) s + 2 Ki
s4 1 5 2Ki
s3 4 2+2Kp
s0 2Ki
Kp 0
Stable
Stable
s
ksK
s
KKGwithreplacedisKexampleprevioustheIn
ipppc
Choose Kp &Ki = 3; using simulink simulate the problem
_
K
2
s3 + 4s2 + 5s + 2
_
K
2
s3 + 4s2 + 5s + 2
_
K
2
s3 + 4s2 + 5s + 2
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