Stability of Feedback Systems

1
Stability of Feedback Systems

• SYSTEMS AND CONTROL I ECE 09.321
• 10/24/07 Lecture 12
• ROWAN UNIVERSITY
• College of Engineering
• Prof. John Colton
• DEPARTMENT OF ELECTRICAL COMPUTER ENGINEERING
• Fall 2007 - Semester One

2
Some Administrative Items

• Quiz 3 will be on Wednesday 10/31 950 AM to
  1040 AM.
• Material through todays lecture
• Through section 6.2 of textbook
• Through problem set 8
• Problem Set 7 is due today.
• Problem Set 8 is available on the course
  website
• Monday lab is on the course website

3
Stability of Linear Feedback Control Systems

• Stability Defined
• Routh Hurwitz Stability Criterion
• Root Locus
• Frequency Response Methods
• Gain and phase margin
• Nyquist Criterion

4
So Why Is Stability So Important?
5
So Why Is Stability So Important?
6
So Why Is Stability So Important?
Tahoma Narrows Bridge across Puget Sound 07/01/40
Tahoma Narrows Bridge across Puget Sound 11/07/40
Cause of failure oscillations due to wind
loading
7
Some Stability Definitions

• A stable system is a system with a bounded
  response to every bounded input.
• This is sometimes referred to as BIBO or NBSC
• A closed loop feedback system is either stable
  or unstable there is
• no middle ground. This is absolute stability.
  We can now drop the adjective
• absolute since stability and absolute stability
  are synonyms
• A necessary and sufficient condition for
  stability is that all of the poles
• of the system transfer function have negative
  real parts
• A necessary and sufficient condition for
  stability is that the system
• impulse response h(t) is absolutely
  integrable,
• 8
• i.e. ?h(t)dt lt 8
• -8
• Given that a closed loop system is stable, we
  can further characterize
• the degree of stability. This is relative
  stability.
• If all transfer function poles are in the left
  half plane, we can

8
Some Stability Definitions

• A stable system is a system with a bounded
  response to every bounded input.
• This is sometimes referred to as BIBO or NBSC
• A closed loop feedback system is either stable
  or unstable there is
• no middle ground. This is absolute stability.
  We can now drop the adjective
• absolute since stability and absolute stability
  are synonyms
• A necessary and sufficient condition for
  stability is that all of the poles
• of the system transfer function have negative
  real parts
• A necessary and sufficient condition for
  stability is that the system
• impulse response h(t) is absolutely
  integrable,
• 8
• i.e. ?h(t)dt lt 8
• -8
• Given that a closed loop system is stable, we
  can further characterize
• the degree of stability. This is relative
  stability.
• If all transfer function poles are in the left
  half plane, we can

9
Some Stability Definitions

• A stable system is a system with a bounded
  response to every bounded input.
• This is sometimes referred to as BIBO or NBSC
• A closed loop feedback system is either stable
  or unstable there is
• no middle ground. This is absolute stability.
  We can now drop the adjective
• absolute since stability and absolute stability
  are synonyms
• A necessary and sufficient condition for
  stability is that all of the poles
• of the system transfer function have negative
  real parts
• A necessary and sufficient condition for
  stability is that the system
• impulse response h(t) is absolutely
  integrable,
• 8
• i.e. ?h(t)dt lt 8
• -8
• Given that a closed loop system is stable, we
  can further characterize
• the degree of stability. This is relative
  stability.
• If all transfer function poles are in the left
  half plane, we can

10
Stability Example
MicrophoneAmplifiers Speakers
Acoustic Energy from speakers
Speech into microphone
Error
Reflected acoustic energy back to mike
11
Stability Considerations in the s plane
H(s) Y(s)/R(s) K(s z1) (s z2)(s zM)/s
N(s p1) (s p1) (s pM) N(s)/?(s)
Q
NM-Q H(s) Y(s)/R(s) ? Ai /(s
si) ? (Bks Ck ) /( s 2 2aks ak2 ?k2)
i 1
k 1 Q
NM-Q h(t) ? Ai e sit ? Dk e
akt sin (?kt ?k ) u(t) i 1
k 1 where the coefficients Ai
and Dk depend on system parameters si , zi , ak,
K , and ?k
Poles are the roots of the characteristic
equation ?(s) 0 For a stable system, all poles
will have negative real parts For an unstable
system, at least one pole must be in the RHP or
on the j? axis How do we determine if (how many)
poles are in the RHP without having to factor
the characteristic equation?
12
Stability Considerations in the s plane
H(s) Y(s)/R(s) K(s z1) (s z2)(s zM)/s
N(s p1) (s p1) (s pM) N(s)/?(s)
Q
NM-Q H(s) Y(s)/R(s) ? Ai /(s
si) ? (Bks Ck ) /( s 2 2aks ak2 ?k2)
i 1
k 1 Q
NM-Q h(t) ? Ai e sit ? Dk e
akt sin (?kt ?k ) u(t) i 1
k 1 where the coefficients Ai
and Dk depend on system parameters si , zi , ak,
K , and ?k
Poles are the roots of the characteristic
equation ?(s) 0 For a stable system, all poles
will have negative real parts For an unstable
system, at least one pole must be in the RHP or
on the j? axis How do we determine if (how many)
poles are in the RHP without having to factor
the characteristic equation?
13
Stability Considerations in the s plane
H(s) Y(s)/R(s) K(s z1) (s z2)(s zM)/s
N(s p1) (s p1) (s pM) N(s)/?(s)
Q
NM-Q H(s) Y(s)/R(s) ? Ai /(s
si) ? (Bks Ck ) /( s 2 2aks ak2 ?k2)
i 1
k 1 Q
NM-Q h(t) ? Ai e sit ? Dk e
akt sin (?kt ?k ) u(t) i 1
k 1 where the coefficients Ai
and Dk depend on system parameters si , zi , ak,
K , and ?k
Poles are the roots of the characteristic
equation ?(s) 0 For a stable system, all poles
will have negative real parts For an unstable
system, at least one pole must be in the RHP or
on the j? axis How do we determine if (how many)
poles are in the RHP without having to factor
the characteristic equation?
14
Polynomial Roots
Characteristic equation ?(s) an s n an-1 s
n -1 a1 s a0 0 How would we determine
whether any of the roots of this equation lies in
the LHP? ?(s) an s n an-1 s n -1 a1 s
a0 0 Factored form ?(s) an ( s r1) (s
r2 ) (s rn ) 0 where the ri are
complex Multiplying out, ?(s) an s n an (r1
r2 rn ) s n -1
an (r1 r2 r2 r3 r1 r3 )
s n -2
an (r1 r2 r3 r1 r2 r4 ) s n -3

an ( 1 ) n (r1 r2 r3 rn ) 0
?(s) an s n an (sum of all roots)
s n -1
an (sum of products of roots taken 2 at a
time) s n -2
an (sum of products of roots taken 3 at a
time) s n -3
an ( 1 ) n (product of all n roots) 0 All
polynomial coefficients ak must have the same
sign if all the roots are in the LHP All
polynomial coefficients ak must be nonzero if
all the roots are in the LHP
15
Polynomial Roots
Characteristic equation ?(s) an s n an-1 s
n -1 a1 s a0 0 How would we determine
whether any of the roots of this equation lies in
the LHP? ?(s) an s n an-1 s n -1 a1 s
a0 0 Factored form ?(s) an ( s r1) (s
r2 ) (s rn ) 0 where the ri are
complex Multiplying out, ?(s) an s n an (r1
r2 rn ) s n -1
an (r1 r2 r2 r3 r1 r3 )
s n -2
an (r1 r2 r3 r1 r2 r4 ) s n -3

an ( 1 ) n (r1 r2 r3 rn ) 0
?(s) an s n an (sum of all roots)
s n -1
an (sum of products of roots taken 2 at a
time) s n -2
an (sum of products of roots taken 3 at a
time) s n -3
an ( 1 ) n (product of all n roots) 0 All
polynomial coefficients ak must have the same
sign if all the roots are in the LHP All
polynomial coefficients ak must be nonzero if
all the roots are in the LHP
16
Polynomial Roots
Characteristic equation ?(s) an s n an-1 s
n -1 a1 s a0 0 How would we determine
whether any of the roots of this equation lies in
the LHP? ?(s) an s n an-1 s n -1 a1 s
a0 0 Factored form ?(s) an ( s r1) (s
r2 ) (s rn ) 0 where the ri are
complex Multiplying out, ?(s) an s n an (r1
r2 rn ) s n -1
an (r1 r2 r2 r3 r1 r3 )
s n -2
an (r1 r2 r3 r1 r2 r4 ) s n -3

an ( 1 ) n (r1 r2 r3 rn ) 0
?(s) an s n an (sum of all roots)
s n -1
an (sum of products of roots taken 2 at a
time) s n -2
an (sum of products of roots taken 3 at a
time) s n -3
an ( 1 ) n (product of all n roots) 0 All
polynomial coefficients ak must have the same
sign if all the roots are in the LHP All
polynomial coefficients ak must be nonzero if
all the roots are in the LHP
17
Examples Polynomial roots

• Necessary conditions that all roots have negative
  real parts If conditions, then poles NRP
• All coefficients have same sign
• All coefficients must be nonzero
• ?(s) (s 1)(s 2) s 2 3s 2 all
  roots in LHP
• ?(s) (s 10)(s) s 2 10s
  missing coefficient ? not all roots are in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 all roots in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2
• ?(s) (s j)(s j) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1)(s 1) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1) 2 s 2 2s 1 different signs
  ? not all roots are in LHP
• ?(s) (s 1)(s 2)
• s 2 s 1 different signs ? not all
  roots are in LHP
• Are these also sufficient conditions? If all
  poles NRP then conditions
• ?(s) (s 2)(s 2 s 4) complex conjugate
  roots with positive real part

18
Examples Polynomial roots

• Necessary conditions that all roots have negative
  real parts If conditions, then poles NRP
• All coefficients have same sign
• All coefficients must be nonzero
• ?(s) (s 1)(s 2) s 2 3s 2 all
  roots in LHP
• ?(s) (s 10)(s) s 2 10s
  missing coefficient ? not all roots are in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 all roots in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2
• ?(s) (s j)(s j) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1)(s 1) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1) 2 s 2 2s 1 different signs
  ? not all roots are in LHP
• ?(s) (s 1)(s 2)
• s 2 s 1 different signs ? not all
  roots are in LHP
• Are these also sufficient conditions? If all
  poles NRP then conditions
• ?(s) (s 2)(s 2 s 4) complex conjugate
  roots with positive real part

19
Examples Polynomial roots

• Necessary conditions that all roots have negative
  real parts If conditions, then poles NRP
• All coefficients have same sign
• All coefficients must be nonzero
• ?(s) (s 1)(s 2) s 2 3s 2 all
  roots in LHP
• ?(s) (s 10)(s) s 2 10s
  missing coefficient ? not all roots are in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 all roots in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2
• ?(s) (s j)(s j) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1)(s 1) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1) 2 s 2 2s 1 different signs
  ? not all roots are in LHP
• ?(s) (s 1)(s 2)
• s 2 s 1 different signs ? not all
  roots are in LHP
• Are these also sufficient conditions? If all
  poles NRP then conditions
• ?(s) (s 2)(s 2 s 4) complex conjugate
  roots with positive real part

20
Examples Polynomial roots

• Necessary conditions that all roots have negative
  real parts If conditions, then poles NRP
• All coefficients have same sign
• All coefficients must be nonzero
• ?(s) (s 1)(s 2) s 2 3s 2 all
  roots in LHP
• ?(s) (s 10)(s) s 2 10s
  missing coefficient ? not all roots are in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 all roots in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 different signs ? not
  all roots are in LHP
• ?(s) (s j)(s j) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1)(s 1) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1) 2 s 2 2s 1 different signs
  ? not all roots are in LHP
• ?(s) (s 1)(s 2)
• s 2 s 1 different signs ? not all
  roots are in LHP
• Are these also sufficient conditions? If all
  poles NRP then conditions
• ?(s) (s 2)(s 2 s 4) complex conjugate
  roots with positive real part

21
Examples Polynomial roots

• Necessary conditions that all roots have negative
  real parts If conditions, then poles NRP
• All coefficients have same sign
• All coefficients must be nonzero
• ?(s) (s 1)(s 2) s 2 3s 2 all
  roots in LHP
• ?(s) (s 10)(s) s 2 10s
  missing coefficient ? not all roots are in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 all roots in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 different signs ? not
  all roots are in LHP
• ?(s) (s j)(s j) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1)(s 1) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1) 2 s 2 2s 1 different signs
  ? not all roots are in LHP
• ?(s) (s 1)(s 2)
• s 2 s 1 different signs ? not all
  roots are in LHP
• Are these also sufficient conditions? If all
  poles NRP then conditions
• ?(s) (s 2)(s 2 s 4) complex conjugate
  roots with positive real part

22
Examples Polynomial roots

• Necessary conditions that all roots have negative
  real parts If conditions, then poles NRP
• All coefficients have same sign
• All coefficients must be nonzero
• ?(s) (s 1)(s 2) s 2 3s 2 all
  roots in LHP
• ?(s) (s 10)(s) s 2 10s
  missing coefficient ? not all roots are in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 all roots in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 different signs ? not
  all roots are in LHP
• ?(s) (s j)(s j) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1)(s 1) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1) 2 s 2 2s 1 different signs
  ? not all roots are in LHP
• ?(s) (s 1)(s 2)
• s 2 s 1 different signs ? not all
  roots are in LHP
• Are these also sufficient conditions? If all
  poles NRP then conditions
• ?(s) (s 2)(s 2 s 4) complex conjugate
  roots with positive real part

23
Examples Polynomial roots

• Necessary conditions that all roots have negative
  real parts If conditions, then poles NRP
• All coefficients have same sign
• All coefficients must be nonzero
• ?(s) (s 1)(s 2) s 2 3s 2 all
  roots in LHP
• ?(s) (s 10)(s) s 2 10s
  missing coefficient ? not all roots are in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 all roots in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 different signs ? not
  all roots are in LHP
• ?(s) (s j)(s j) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1)(s 1) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1) 2 s 2 2s 1 different signs
  ? not all roots are in LHP
• ?(s) (s 1)(s 2)
• s 2 s 1 different signs ? not all
  roots are in LHP
• Are these also sufficient conditions? If all
  poles NRP then conditions
• ?(s) (s 2)(s 2 s 4) complex conjugate
  roots with positive real part

24
Examples Polynomial roots

• Necessary conditions that all roots have negative
  real parts If conditions, then poles NRP
• All coefficients have same sign
• All coefficients must be nonzero
• ?(s) (s 1)(s 2) s 2 3s 2 all
  roots in LHP
• ?(s) (s 10)(s) s 2 10s
  missing coefficient ? not all roots are in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 all roots in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2
• ?(s) (s j)(s j) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1)(s 1) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1) 2 s 2 2s 1 different signs
  ? not all roots are in LHP
• ?(s) (s 1)(s 2)
• s 2 s 1 different signs ? not all
  roots are in LHP
• Are these also sufficient conditions? If all
  poles NRP then conditions
• ?(s) (s 2)(s 2 s 4) complex conjugate
  roots with positive real part

25
Examples Polynomial roots

• Necessary conditions that all roots have negative
  real parts If conditions, then poles NRP
• All coefficients have same sign
• All coefficients must be nonzero
• ?(s) (s 1)(s 2) s 2 3s 2 all
  roots in LHP
• ?(s) (s 10)(s) s 2 10s
  missing coefficient ? not all roots are in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 all roots in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2
• ?(s) (s j)(s j) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1)(s 1) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1) 2 s 2 2s 1 different signs
  ? not all roots are in LHP
• ?(s) (s 1)(s 2)
• s 2 s 1 different signs ? not all
  roots are in LHP
• Are these also sufficient conditions? If all
  poles NRP then conditions?
• ?(s) (s 2)(s 2 s 4) complex conjugate
  roots with positive real part

26
Examples Polynomial roots

• Necessary conditions that all roots have negative
  real parts If conditions, then poles NRP
• All coefficients have same sign
• All coefficients must be nonzero
• ?(s) (s 1)(s 2) s 2 3s 2 all
  roots in LHP
• ?(s) (s 10)(s) s 2 10s
  missing coefficient ? not all roots are in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2 all roots in LHP
• ?(s) (s 1 j)(s 1 j)
• s 2 2s 2
• ?(s) (s j)(s j) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1)(s 1) s 2 1 missing
  coefficient ? not all roots are in LHP
• ?(s) (s 1) 2 s 2 2s 1 different signs
  ? not all roots are in LHP
• ?(s) (s 1)(s 2)
• s 2 s 1 different signs ? not all
  roots are in LHP
• Are these also sufficient conditions? If all
  poles NRP then conditions
• ?(s) (s 2)(s 2 s 4) complex conjugate
  roots with positive real part

27
Routh Hurwitz Stability Criterion
The Routh Hurwitz criterion is a necessary and
sufficient criterion for the stability of linear
time invariant systems
General procedure Characteristic equation ?(s)
an s n an-1 s n -1 a1 s a0 0 Form
Routh array s n an an-2 an-4 The
Routh - Hurwitz criterion says s n 1 an-1
an-3 an-5 the number of roots of
?(s) with
positive real
parts is equal to the s n 2 bn-1 bn-3 bn-5
number of changes of sign in the s
n 3 cn-1 cn-3 cn-5 first
column of the Routh array . . . s 0
hn-1
an an-2 bn-1 (1/
an-1 ) an-1 an-3 (an-1 an-2 an
an-3 )/ an-1
an an-4 bn-3 (1/
an-1 ) an-1 an-5 (an an-5 an-1
an-5 )/ an-1
an-1 an-3 cn-1 (1/
bn-1 ) bn-1 bn-3 (an-3 bn-1
an-1 bn-3 )/ an-1
28
Routh Hurwitz Stability Criterion
The Routh Hurwitz criterion is a necessary and
sufficient criterion for the stability of linear
time invariant systems
General procedure Characteristic equation ?(s)
an s n an-1 s n -1 a1 s a0 0 Form
Routh array s n an an-2 an-4 The
Routh - Hurwitz criterion says s n 1 an-1
an-3 an-5 the number of roots of
?(s) with
positive real
parts is equal to the s n 2 bn-1 bn-3 bn-5
number of changes of sign in the s
n 3 cn-1 cn-3 cn-5 first
column of the Routh array . . . s 0
hn-1
an an-2 bn-1 (1/
an-1 ) an-1 an-3 (an-1 an-2 an
an-3 )/ an-1
an an-4 bn-3 (1/
an-1 ) an-1 an-5 (an an-5 an-1
an-5 )/ an-1
an-1 an-3 cn-1 (1/
bn-1 ) bn-1 bn-3 (an-3 bn-1
an-1 bn-3 )/ an-1
29
Routh Hurwitz Stability Criterion
The Routh Hurwitz criterion is a necessary and
sufficient criterion for the stability of linear
time invariant systems
General procedure Characteristic equation ?(s)
an s n an-1 s n -1 a1 s a0 0 Form
Routh array s n an an-2 an-4 The
Routh - Hurwitz criterion says s n 1 an-1
an-3 an-5 the number of roots of
?(s) with
positive real
parts is equal to the s n 2 bn-1 bn-3 bn-5
number of changes of sign in the s
n 3 cn-1 cn-3 cn-5 first
column of the Routh array . . . s 0
hn-1
an an-2 bn-1 (1/
an-1 ) an-1 an-3 (an-1 an-2 an
an-3 )/ an-1
an an-4 bn-3 (1/
an-1 ) an-1 an-5 (an-1an-4 an
an-5 )/ an-1
an-1 an-3 cn-1 (1/
bn-1 ) bn-1 bn-3 (an-3 bn-1
an-1 bn-3 )/ an-1
30
Routh Hurwitz Stability Criterion
The Routh Hurwitz criterion is a necessary and
sufficient criterion for the stability of linear
time invariant systems
General procedure Characteristic equation ?(s)
an s n an-1 s n -1 a1 s a0 0 Form
Routh array s n an an-2 an-4 The
Routh - Hurwitz criterion says s n 1 an-1
an-3 an-5 the number of roots of
?(s) with
positive real
parts is equal to the s n 2 bn-1 bn-3 bn-5
number of changes of sign in the s
n 3 cn-1 cn-3 cn-5 first
column of the Routh array . . . s 0
hn-1
an an-2 bn-1 (1/
an-1 ) an-1 an-3 (an-1 an-2 an
an-3 )/ an-1
an an-4 bn-3 (1/
an-1 ) an-1 an-5 (an-1 an-4 an
an-5 )/ an-1
an-1 an-3 cn-1 (1/
bn-1 ) bn-1 bn-3 (an-3 bn-1
an-1 bn-3 )/ an-1
31
Routh Hurwitz Stability Criterion
The Routh Hurwitz criterion is a necessary and
sufficient criterion for the stability of linear
time invariant systems
General procedure Characteristic equation ?(s)
an s n an-1 s n -1 a1 s a0 0 Form
Routh array s n an an-2 an-4 The
Routh - Hurwitz criterion says s n 1 an-1
an-3 an-5 the number of roots of
?(s) with
positive real
parts is equal to the s n 2 bn-1 bn-3 bn-5
number of changes of sign in the s
n 3 cn-1 cn-3 cn-5 first
column of the Routh array . . . s 0
hn-1
an an-2 bn-1 (1/
an-1 ) an-1 an-3 (an-1 an-2 an
an-3 )/ an-1
an an-4 bn-3 (1/
an-1 ) an-1 an-5 (an-1 an-4 an
an-5 )/ an-1
an-1 an-3 cn-1 (1/
bn-1 ) bn-1 bn-3 (an-3 bn-1
an-1 bn-3 )/ an-1
32
Example Case 1 no 0s in first column
?(s) a2 s 2 a1 s a0 second order
system Routh array s 2 a2 a0 s 1
a1 0 s 0 b1 0 where b1
1/ a1 ( 0 a2 a0 a1) a0 Therefore, the
necessary and sufficient conditions for stability
of a second order system is that the
coefficients ak be all positive or all
negative.
33
Example Case 1 no 0s in first column
?(s) a3 s 3 a2 s 2 a1 s a0 third
order system Routh array s 3 a3
a1 s 2 a2 a0 s 1 b1
0 s 0 c1 0 where b1 1/ a2 (a2 a1
a0 a3 ) and c1 (1/b1 ) (b1 a0 ) a0
Therefore, the necessary and sufficient
conditions for stability of a third order system
is that the coefficients ak be all positive
and a2 a1 gt a0 a3 (the condition a2 a1 a0 a3
results in a 0 in the first column and will be
treated separately as a case 3 example)
34
Example Case 1 no 0s in first column
?(s) s 3 s 2 2 s 24 (s 1 jv7)(s 1
jv7 )(s 3) third order system Routh
array s 3 1 2 s 2 1 24
s 1 -22 0 s 0 24 0 Two sign
changes mean that two roots of ?(s) lie in the
RHP, so the system is unstable
35
Example Case 2 0 in first column, non-zero in
associated row
If only one element in the array is zero, it may
be replaced with a small positive number e that
is allowed to approach zero after completing the
array. ?(s) s 5 2s 4 2s 3 4s 2 11 s
10 0 Routh array s 5 1 2
11 s 4 2 4 10 s 3 0 6
0 s 2 c1 10 0 s 1 d1 0
0 s 0 10 0 0
36
Example Case 2 0 in first column, non-zero in
associated row
If only one element in the array is zero, it may
be replaced with a small positive number e that
is allowed to approach zero after completing the
array. ?(s) s 5 2s 4 2s 3 4s 2 11 s
10 0 Routh array s 5 1 2
11 s 4 2 4 10 s 3 0 6
0 s 2 c1 10 0 s 1 d1 0
0 s 0 10 0 0
37
Example Case 2 0 in first column, non-zero in
associated row
If only one element in the array is zero, it may
be replaced with a small positive number e that
is allowed to approach zero after completing the
array. ?(s) s 5 2s 4 2s 3 4s 2 11 s
10 0 Routh array s 5 1 2
11 s 4 2 4 10 s 3 e 6
0 s 2 c1 10 0 s 1 d1 0
0 s 0 10 0 0 c1 (4 e 12)/ e
12/ e and d1 (6c1 10 e )/ c1 ? 6 Two
sign changes in the first column say that the
system is unstable with two roots in the RHP
38
Example Case 2 0 in first column, non-zero in
associated row
For what range of K is the system stable? ?(s)
s 4 s 3 s 2 s K 0 Routh array s 4
1 1 K s 3 1 1 0 s 2
0 K 0 s 1 c1 0 0 s 0
K 0 0 c1 (4 e 12)/ e 12/ e and
d1 (6c1 10 e )/ c1 ? 6 Two sign changes in
the first column say that the system is unstable
with two roots in the RHP
39
Example Case 2 0 in first column, non-zero in
associated row
For what range of K is the system stable? ?(s)
s 4 s 3 s 2 s K 0 Routh array s 4
1 1 K s 3 1 1 0 s 2
e K 0 s 1 c1 0 0 s 0
K 0 0 c1 (e K)/ e ? K/
e Therefore for any value of K gt 0, the system
is unstable. Is the system stable for K lt 0? for
K 0?