Stability Analysis

1
Stability Analysis

• Chapter 6 Basic Concept
• Chapter 7 Stability Analysis in Time Domain
• (Chapter 8 -- Frequency Response)
• Chapter 9 Stability Analysis in Frequency
• Domain

2
Example of an Unstable System Tacoma Narrows
Bridge (Figure 6.3, p. 358)

• Good web site www.youtube.com/watch?vHxTZ446tbzE

3
CH. 6 Stability of Linear Feedback Systems

• Chapter 6 introduces the concept of stability of
  control systems. Topics include
• - The Concept of Stability (6.1)
• Absolute Stability
• Relative Stability
• BIBO stability
• - Stability Criterion (6.2)
• Routh-Hurwitz Criterion
• - Stability of State Variable Systems (6.4)
• - Design Example (6.5)
• - Stability Analysis using Matlab

4
Concept of Stability(Figure 6.1, p. 357)
5
Concept of Stability

• Absolute Stability
• A system is either stable or unstable. This type
  of stable/unstable is referred as absolute
  stability.

6
Concept of Stability

• Absolute vs. Relative Stability
• A system is either stable or unstable. This type
  of stable/unstable is referred as absolute
  stability.
• If the system is stable, we can further
  investigate the degree of stability. This is
  referred as relative stability.

7
Definition of System Stability

• BIBO Stability
• A stable system is a dynamic system with a
  bounded response to a bounded input.
• This is called a BIBO stability.

8
Definition of System Stability

• 1. A stable system is a dynamic system with a
  bounded response to a bounded input.
• This is called a BIBO stability.
• Ex. Step response of a 2nd-order system

BIBO Stability
9
Ex. BIBO-Stable Inverted Pendulum (Fig. 2.6, p.
50)
10
Unstable Inverted Pendulum System (Fig. 3.22, p.
169)
11
CH. 6 Stability of Linear Feedback Systems

• Chapter 6 introduces the concept of stability of
  control systems. Topics include
• - The Concept of Stability (6.1)
• BIBO stability
• Absolute Stability
• Relative Stability
• - Stability Criterion (6.2) ?
• - Stability of State Variable Systems (6.4)
• - Design Example (6.5)
• - Stability Analysis using Matlab

12
Stability depends on the pole locations in the
s-plane
Poles
13
General Stability Condition

• A necessary and sufficient condition for a
  feedback system to be stable is that all the
  poles of the system transfer function have
  negative real parts. (p. 359)

14
Effect of Pole Locations of 2nd-Order Systems
(review)

• Standard transfer function
• G(s) ?n2/(s22? ?ns ?n2)
• Poles of G(s) are
• p1,2 - ? ?n j?n -?2

?lt1
15
Effect of Pole Locations of 2nd-Order Systems
(review)

• Standard transfer function
• G(s) ?n2/(s22? ?ns ?n2)
• gt Poles are p1,2 - ? ?n j?n -?2
• Unit Step Responce
• Y(s)G(s)R(s), R(s)1/s
• ?n2/s(s22? ?ns ?n2) (5.8)
• gt
• y(t) 1 (1/ß)e - ? ?nt sin(?n ß t ?)
  (5.9)
• gt System is stable if all poles have negative
  real parts.

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Effect of Pole Locations on Impulse Responses
(Figure 5.17)
Marginally stable
Unstable
Stable
17
Example

• Let G(s)2/(s1)(s2)
• p(s)/q(s)
• Then,
• poles are -1,-2
• gt
• System impulse (or natural) response is
• y(t)k1e-t k2e-2t
• gt
• system is stable.

18
Exercise

• Show that
• 1. G(s)s/(s21) is stable
• 2. G(s)10(s2)/(s1)(s-3)(s4) is unstable.

19
CH. 6 Stability of Linear Feedback Systems

• Chapter 6 introduces the concept of stability of
  control systems. Topics include
• - The Concept of Stability (6.1)
• BIBO stability
• Absolute Stability
• Relative Stability
• - Stability Test (6.2) ?
• Routh-Hurwitz Stability Test
• - Stability of State Variable Systems (6.4)
• - Design Example (6.5)
• - Stability Analysis using Matlab

20
System Stability Criterion

• Recall Let the system transfer function be
  G(s)p(s)/q(s).
• Then,
• The (System) Characteristic Eqn. is
  q(s)0, and
• Poles are the roots of q(s)o.
• Necessary condition It is necessary for a stable
  system to have all the coefficients of the
  characteristic equation have the same sign.
• Proof Page 361, Equations (6.3)-(6.6)

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System Stability Criterion

• Necessary condition It is necessary for a stable
  system to have all the coefficients of the
  characteristic equation have the same sign.
• Ex.
• 1. q(s)s3s22s8
• 2. q(s)s44
• necessary condition is satisfied for both.
• Are they stable?

22
System Stability Criterion

• Necessary condition It is necessary for a stable
  system to have all the coefficients of the
  characteristic equation have the same sign.
• Ex.
• 1. q(s)s3s22s8
• 2. q(s)s44
• necessary condition is satisfied for both.
• Are they stable?
• 1. q(s)(s2)(s2-s4) gt poles -2, ½j /2 gt
  stable
• 2. q(s)(s22s2)(s2-2s2) gt poles -1j1, 1j1
• 
  gt unstable

23
Routh-Hurwitz Stability Test

• Necessary and Sufficient condition
• R-H test
• The Routh-Hurwitz stability criterion
  states that the number of roots of q(s) with
  positive real parts is equal to the number of
  changes in sign of the first column of the Routh
  array.
• Proof See References 6,7 on page 999-1000.
• John Routh, English mathematician
• Adolf Hurwitz, German mathmatician

24
6.2 The Routh-Hurwitz Stability Testfor SISO-LTI
Systems

• A necessary condition for a feedback system to be
  stable is that all the poles of the system
  transfer function have negative real parts. How
  about a sufficient condition?
• gt Use the Routh-Hurwitz test. (Note Matlab may
  not be used when non-numerical parameters are
  involved.)
• Step 1. In order to test system stability, first
  construct the Routh array from the system
  characteristic equation q(s)0.
• Step2. The number of changes in sign of the first
  column of the Routh array is equal to the number
  of poles in the right half-plane.

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Routh Array
26
Routh Array
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Examples

• Example 6.1
• Example 6.2

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The Routh-Hurwitz Stability Test

• Exercise.

29
Example of The Routh Array constructed from q(s)0

• gt Thus, the system is unstable.

30
Special Case

• Zeros in the first column (p.364)
• gt Replace the zero with a small positive number
  e.
• Ex.
• q(s)s52s42s34s211s10 (6.10)
• gt
• This system is unstable. (Exercise)

31
ExampleAntenna Position Control
32
ExampleAntenna Position Control

• System Block Diagram

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ExampleAntenna Position Control

• System Block Diagram

34
Stability Analysis

• q(s)s3101.7s2171s6.63K
• gt
• Routh array
• s3 1 171
• s2 101.7 6.63K
• s 17392.4-6.63K 0
• s0 6.63K
• gt
• 17392.4-6.63K0 when K2623
• gt Stable for 0ltKlt2623.

35
CH. 6 Stability of Linear Feedback Systems

• Chapter 6 introduces the concept of stability of
  control systems. Topics include
• - The Concept of Stability (6.1)
• BIBO stability
• Absolute Stability
• Relative Stability
• - Stability Criterion (6.2)
• Routh-Hurwitz Criterion
• - Stability of State Variable Systems (6.4) ?
• - Design Example (6.5)
• - Stability Analysis using Matlab

36
6.4 Stability of State Variable Systems

• Ex. 6.8 Consider the following 2nd-order system
• dx1/dt -3x1 x2
• dx1/dt -Kx1 x2 Ku
• gt
• dx/dt A x Bu
• where
• A-1 1-K -1
• and
• B0 KT ?

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Any Questions on System Stability so far?
38
Solution of State Equation(review)

• dx/dt Ax Bu and yCx
• gt
• x(s) F(s)x0 F(s)BU(s), x0x(0)
• where F(s) (sI-A)-1 State
  Transition Matrix
• and det(sI-A) 0 System
  Characteristic Equation
• So if x00, then x(s) F(s)BU(s)
• Thus the system transfer function is
• T(s)Y(s)/U(s)CF(s)U(s)
• C(sI-A)-1B
  (3.79)

39
System Poles

• dx/dt Ax Bu and yCx
• x(s) F(s)x0 F(s)BU(s), x0x(0)
• where F(s) (sI-A)-1 State
  Transition Matrix
• and det(sI-A) 0 System
  Characteristic Equation
• gt
• det(sI-A) 0 gt system poles

40
Note System Poles and Eigenvalues

• dx/dt Ax Bu and yCx
• x(s) F(s)x0 F(s)BU(s), x0x(0)
• where F(s) (sI-A)-1 State
  Transition Matrix
• and det(sI-A) 0 System
  Characteristic Equation
• det(sI-A) 0 gt system poles
• Note Eigenvalue of a matrix A is defined as
• the solution of
• det(?I-A) 0
• ?
• System poles Eigenvalues of A

41
Eigenvector and Rotation

• In linear algebra, every linear transformation
  between finite-dimensional vector spaces can be
  expressed as a matrix. Using eigenvalue ? and the
  correspondng eigenvector v (solution of Av ?v),
  you can show that the two pictures of Mona Lisa
  can be mapped from one to the other, i.e. they
  represent the same image.

42
Note System Poles and Eigenvalues

• dx/dt Ax Bu and yCx
• ?
• System poles Eigenvalues of A
• Ex.
• dx/dt Ax , where A0 -12 3
• Then, det(sI-A)s 1-2 s-3
• s2-3s2(s-1)(s-2)
• ?
• eigenvalues of A 1 and 2 ? poles

43
6.4 Stability of State Variable SystemsExamples

• Ex. 1
• dx(t)/dt0 -12 3x(t)
• Ex. 2
• dx(t)/dt0 3 12 8 1-10 -5 -2x(t)
  1000u(t)

44
6.4 Stability of State Variable Systems

• Ex. 6.8 Consider the following 2nd-order system
• dx1/dt -3x1 x2
• dx2/dt -Kx1 x2 Ku
• gt
• dx/dt A x Bu
• and
• the system characteristic equation is given
  by
• q(s) det(sI A ) 0

45
Stability of State Variable Systems

• Ex. 6.8 Consider the following 2nd-order system
• dx1/dt -3x1 x2
• dx2/dt -Kx1 x2 Ku
• gt
• dx/dt A x Bu
• and
• the system characteristic equation is given
  by
• q(s) det(sI A ) 0
• gt
• q(s) s2 2s (K-3) 0.
• gt
• From the Routh array, the system is stable
  if
• Kgt3

46
6.5 Design ExamplesEx. 6.10 Automatic vehicle
turning control

• Figure 6.8 (Note A vehicles powertrain consists
  of all the components that generate power and
  deliver it. This includes the engine,
  transmission, drivershafts and wheels.)

47
6.5 Design ExamplesEx. 6.10 Automatic vehicle
turning control

• Figure 6.8 (Note A vehicles powertrain consists
  of all the components that generate power and
  deliver it. This includes the engine,
  transmission, drivershafts and wheels.)

48
Stability Region

• System Ch. Eqn. is
• q(s)s(s1)(s2)(s5)K(sa)
• s48s317s2(K10)sKa0 (6.28)
• The Routh array ?
• b3(126-K)/8, c3b3(K10)-8Ka/b3
• gt
• Klt126, Kagt0, (K10)(126-K)-64Kagt0
• gt
• Region of Stability is shown in Figure 6.9
  (next slide)

49
Stability Region on the K-a space

• Figure 6.9 for Kgt0 and agt0 (Exercise)

50
CH. 6 Stability of Linear Feedback Systems

• Chapter 6 introduces the concept of stability of
  control systems. Topics include
• - The Concept of Stability (6.1)
• Absolute Stability
• Relative Stability
• BIBO stability
• - Stability Criterion (6.2)
• Routh-Hurwitz Criterion
• - Stability of State Variable Systems (6.4)
• - Design Example (6.5)
• - Stability Analysis using Matlab ?

51
Pole locations using pole(sys)
52
Ex. Stability Region as a function of K(Fig.
6.19)
53
Plot of Root Locations (Fig. 6.20)

• Matlab commands
• gtgtK00.520
• for i1length(K)
• q1 2 4 K(i)
• proots(q)
• plot(real(p),imag(p),x)
• hold on
• end
• grid on

54
Ex. 6.13 Aircraft control

• where zgto and pgto.

55
Ex. 6.11 Aircraft control

• Step 1. Ch. Eq. for the closed-loop system is
• q(s) s3(p-1)s2 (K-p)sKz 0.
• Step 2. From the Routh array, the stability
  condition is
• Kgtp(p-1)/(p-1)-z (6.36)
• Step 3. Use Matlab to plot the stability region
  on p-z coordinates using K as a parameter
• gt
• Figure 6.21

56
Matlab command meshgrid

• The following example shows how to use meshgrid
  to create a surface plot of a function
  ZXexp(-X2-Y2)
• gtgt X,Ymeshgrid(-2.22,-2.22)
• ZX.exp(-X.2-Y.2)
• surf(X,Y,Z)

57
Matlab command meshgrid

• The following example shows how to use meshgrid
  to create a surface plot of a function
  ZXexp(-X2-Y2)
• gtgt X,Ymeshgrid(-2.22,-2.22)
• ZX.exp(-X.2-Y.2)
• surf(X,Y,Z)
• Note
• surf will plot a 3-D shaded surface.
• mesh will plot a non-shaded surface.
• . is the array multiplication in Matlab.

58
Matlab commands for the stability region of the
control system for Ex. 6.13

• p,zmeshgrid(1.20.210,0.1.210)
• kp.(p-1)./(p-1-z)
• mesh(k)

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Stability surface and region(Fig. 6.27)
60
?

• Any Questions?