phase_plane_analysis for analyzing non linear system.pdf
ronald551060
39 views
83 slides
Jul 08, 2024
Slide 1 of 83
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
About This Presentation
Analysis of non-linear systems
Slide Content
Stability and Phase Plane
Analysis
Advanced Control (Mehdi Keshmiri, Winter 95)
1
Analysis
Advanced Control (Mehdi Keshmiri, Winter 95)
1
Advanced Control (Mehdi Keshmiri, Winter 95)
Objectives
Objectives of the section:
Introducing the Phase Plane Analysis
Introducing the Concept of stability
Stability Analysis of Linear Time Invariant Systems
LyapunovIndirect Method in Stability Analysis of
Nonlinear Systems
2
Objectives
Objectives of the section:
Introducing the Phase Plane Analysis
Introducing the Concept of stability
Stability Analysis of Linear Time Invariant Systems
LyapunovIndirect Method in Stability Analysis of
Nonlinear Systems
2
Advanced Control (Mehdi Keshmiri, Winter 95)
Introducing
the Phase Plane Analysis
3
Introducing
the Phase Plane Analysis
3
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Phase Space form of a Dynamical System:
�
1=�
1�
1,�
2,…,�
�,�
1,�
2,…,�
�,�
�
2=�
2(�
1,�
2,…,�
�,�
1,�
2,…,�
�,�)
⋮
�
�=�
��
1,�
2,…,�
�,�
1,�
2,…,�
�,�
�=??????�,??????,�
�∈ℝ
�
??????∈ℝ
�
�=??????�,??????,�
?????? ??????
�=??????�,??????
?????? ??????
Time-Varying System Time-Invariant System
4
Phase Plane Analysis
Phase Space form of a Dynamical System:
�
1=�
1�
1,�
2,…,�
�,�
1,�
2,…,�
�,�
�
2=�
2(�
1,�
2,…,�
�,�
1,�
2,…,�
�,�)
⋮
�
�=�
��
1,�
2,…,�
�,�
1,�
2,…,�
�,�
�=??????�,??????,�
�∈ℝ
�
??????∈ℝ
�
�=??????�,??????,�
?????? ??????
�=??????�,??????
?????? ??????
Time-Varying System Time-Invariant System
4
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Phase Space form of a Linear Time Invariant (LTI) System:
??????=�??????+�??????
�∈ℝ
�
??????∈ℝ
�
Multiple isolated equilibria
Limit Cycle
Finite escape time
Harmonic, sub-harmonic and almost periodic Oscillation
Chaos
Multiple modes of behavior
Special Properties of Nonlinear Systems:
5
Phase Plane Analysis
Phase Space form of a Linear Time Invariant (LTI) System:
??????=�??????+�??????
�∈ℝ
�
??????∈ℝ
�
Multiple isolated equilibria
Limit Cycle
Finite escape time
Harmonic, sub-harmonic and almost periodic Oscillation
Chaos
Multiple modes of behavior
Special Properties of Nonlinear Systems:
5
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Phase Plane Analysisis a graphical method for studying
second-order systemsrespect to initial conditions by:
providing motion trajectories corresponding to various initial
conditions.
examining the qualitative features of the trajectories
obtaining information regarding the stability of the equilibrium
points
�
1=�
1(�
1,�
2)
�
2=�
2(�
1,�
2)
6
Phase Plane Analysis
Phase Plane Analysisis a graphical method for studying
second-order systemsrespect to initial conditions by:
providing motion trajectories corresponding to various initial
conditions.
examining the qualitative features of the trajectories
obtaining information regarding the stability of the equilibrium
points
�
1=�
1(�
1,�
2)
�
2=�
2(�
1,�
2)
6
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Advantages of Phase Plane Analysis:
It is graphical analysis and the solution trajectories can be
represented by curves in a plane
Provides easy visualization of the system qualitative
Without solving the nonlinear equations analytically, one can study
the behavior of the nonlinear system from various initial
conditions.
It is not restricted to small or smooth nonlinearities and applies
equally well to strong and hard nonlinearities.
There are lots of practical systems which can be approximated by
second-order systems, and apply phase plane analysis.
7
Phase Plane Analysis
Advantages of Phase Plane Analysis:
It is graphical analysis and the solution trajectories can be
represented by curves in a plane
Provides easy visualization of the system qualitative
Without solving the nonlinear equations analytically, one can study
the behavior of the nonlinear system from various initial
conditions.
It is not restricted to small or smooth nonlinearities and applies
equally well to strong and hard nonlinearities.
There are lots of practical systems which can be approximated by
second-order systems, and apply phase plane analysis.
7
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Disadvantage of Phase Plane Method:
It is restricted to at most second-order
graphical study of higher-order is computationally and
geometrically complex.
8
Phase Plane Analysis
Disadvantage of Phase Plane Method:
It is restricted to at most second-order
graphical study of higher-order is computationally and
geometrically complex.
8
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Example: First Order LTI System
??????=??????????????????(??????)
��
sin(�)
=��
Analytical Solution
??????
0
??????
��
sin(�)
=
0
??????
��
��
��
=sin(�)
�=ln
cos�
0+cot(�
0)
cos�+cot(�)
Graphical Solution-6 -4 -2 0 2 4 6
-1
-0.5
0
0.5
1
x
sin(x)
9
Phase Plane Analysis
Example: First Order LTI System
??????=??????????????????(??????)
��
sin(�)
=��
Analytical Solution
??????
0
??????
��
sin(�)
=
0
??????
��
��
��
=sin(�)
�=ln
cos�
0+cot(�
0)
cos�+cot(�)
Graphical Solution-6 -4 -2 0 2 4 6
-1
-0.5
0
0.5
1
x
sin(x)
9
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Concept of Phase Plane Analysis:
Phase plane method is applied to Autonomous Second Order System
System response ��=(�
1�,�
2(�))to initial condition �
0=�
10,�
20
is a mapping from ℝ(Time) to ℝ
2
(�
1,�
2)
The solution can be plotted in the �
1−�
2plane called State Plane or
Phase Plane
The locus in the �
1−�
2plane is a curved named Trajectorythat pass through
point �
0
The family of the phase plane trajectories corresponding to various initial
conditions is called Phase portrait of the system.
For a single DOF mechanical system, the phase plane is in fact is (�, �)plane
�
1=�
1(�
1,�
2) �
2=�
2(�
1,�
2)
10
Phase Plane Analysis
Concept of Phase Plane Analysis:
Phase plane method is applied to Autonomous Second Order System
System response ��=(�
1�,�
2(�))to initial condition �
0=�
10,�
20
is a mapping from ℝ(Time) to ℝ
2
(�
1,�
2)
The solution can be plotted in the �
1−�
2plane called State Plane or
Phase Plane
The locus in the �
1−�
2plane is a curved named Trajectorythat pass through
point �
0
The family of the phase plane trajectories corresponding to various initial
conditions is called Phase portrait of the system.
For a single DOF mechanical system, the phase plane is in fact is (�, �)plane
�
1=�
1(�
1,�
2) �
2=�
2(�
1,�
2)
10
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Example: Van der Pol Oscillator Phase Portraitx
11
Phase Plane Analysis
Example: Van der Pol Oscillator Phase Portraitx
11
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Plotting Phase Plane Diagram:
Analytical Method
Numerical Solution Method
Isocline Method
Vector Field Diagram Method
Delta Method
Lienard’sMethod
Pell’s Method
12
Phase Plane Analysis
Plotting Phase Plane Diagram:
Analytical Method
Numerical Solution Method
Isocline Method
Vector Field Diagram Method
Delta Method
Lienard’sMethod
Pell’s Method
12
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Analytical Method
Dynamic equations of the system is solved, then time parameter is omitted
to obtain relation between two states for various initial conditions
�
1=�
1(�
1,�
2)
�
2=�
2(�
1,�
2)
Solve�
1�,�
0=�
1(�,�
0)
�
2�,�
0=�
2(�,�
0)
??????�
1,�
2=0
For linear or partially linear systems
13
Phase Plane Analysis
Analytical Method
Dynamic equations of the system is solved, then time parameter is omitted
to obtain relation between two states for various initial conditions
�
1=�
1(�
1,�
2)
�
2=�
2(�
1,�
2)
Solve�
1�,�
0=�
1(�,�
0)
�
2�,�
0=�
2(�,�
0)
??????�
1,�
2=0
For linear or partially linear systems
13
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Example: Mass Spring System
� �+��=0
For �=�=1: �+�=0
��=�
0cos(�)+ �
0sin(�)
��=−�
0sin�+ �
0cos(�)
�
2
+ �
2
=�
0
2
+ �
0
2x
y
x
2
+ y
2
- 4
-2 -1 0 1 2
-2
-1
0
1
2
14
Phase Plane Analysis
Example: Mass Spring System
� �+��=0
For �=�=1: �+�=0
��=�
0cos(�)+ �
0sin(�)
��=−�
0sin�+ �
0cos(�)
�
2
+ �
2
=�
0
2
+ �
0
2x
y
x
2
+ y
2
- 4
-2 -1 0 1 2
-2
-1
0
1
2
14
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Analytical Method
Time differential is omitted from dynamic equations of the system, then
partial differential equation is solved
�
1=�
1(�
1,�
2)
�
2=�
2(�
1,�
2)
Solve
??????�
1,�
2=0
��
2
��
1
=
�
2(�
1,�
2)
�
1(�
1,�
2)
For linear or partially linear systems
15
Phase Plane Analysis
Analytical Method
Time differential is omitted from dynamic equations of the system, then
partial differential equation is solved
�
1=�
1(�
1,�
2)
�
2=�
2(�
1,�
2)
Solve
??????�
1,�
2=0
��
2
��
1
=
�
2(�
1,�
2)
�
1(�
1,�
2)
For linear or partially linear systems
15
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Example: Mass Spring System
� �+��=0
For �=�=1: �+�=0x
y
x
2
+ y
2
- 4
-2 -1 0 1 2
-2
-1
0
1
2
�
1=�
2
�
2=−�
1
��
2
��
1
=
−�
1
�
2
??????
20
??????
2
�
2��
2=
??????
10
??????
1
−�
1��
1
�
2
+ �
2
=�
0
2
+ �
0
2
16
Phase Plane Analysis
Example: Mass Spring System
� �+��=0
For �=�=1: �+�=0x
y
x
2
+ y
2
- 4
-2 -1 0 1 2
-2
-1
0
1
2
�
1=�
2
�
2=−�
1
��
2
��
1
=
−�
1
�
2
??????
20
??????
2
�
2��
2=
??????
10
??????
1
−�
1��
1
�
2
+ �
2
=�
0
2
+ �
0
2
16
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Numerical Solution Method
Dynamicequationsofthesystemissolvednumerically(e.g.Ode45)for
variousinitialconditionsandtimeresponseisobtained,thentwostatesare
plottedineachtime.
Example: Pendulum ??????+sin??????=00 2 4 6 8 10
-1
-0.5
0
0.5
1
Time(sec)
x
2
(t) -1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
x
1
x
2 0 2 4 6 8 10
-1
-0.5
0
0.5
1
Time(sec)
x
1
(t)
17
Phase Plane Analysis
Numerical Solution Method
Dynamicequationsofthesystemissolvednumerically(e.g.Ode45)for
variousinitialconditionsandtimeresponseisobtained,thentwostatesare
plottedineachtime.
Example: Pendulum ??????+sin??????=00 2 4 6 8 10
-1
-0.5
0
0.5
1
Time(sec)
x
2
(t) -1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
x
1
x
2 0 2 4 6 8 10
-1
-0.5
0
0.5
1
Time(sec)
x
1
(t)
17
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Isocline Method
Isocline: The set of all points which have same trajectory slope
First various isoclines are plotted, then trajectories are drawn.
�
1=�
1(�
1,�
2)
�
2=�
2(�
1,�
2)
��
2
��
1
=
�
2(�
1,�
2)
�
1(�
1,�
2)
=��
2�
1,�
2=��
1(�
1,�
2)
18
Phase Plane Analysis
Isocline Method
Isocline: The set of all points which have same trajectory slope
First various isoclines are plotted, then trajectories are drawn.
�
1=�
1(�
1,�
2)
�
2=�
2(�
1,�
2)
��
2
��
1
=
�
2(�
1,�
2)
�
1(�
1,�
2)
=��
2�
1,�
2=��
1(�
1,�
2)
18
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Example: Mass Spring System
� �+��=0
For �=�=1: �+�=0
�
1=�
2
�
2=−�
1
��
2
��
1
=
−�
1
�
2
=�
�
1+��
2=0x
y
x
2
+ y
2
- 4
-5 0 5
-5
0
5
Slope=1
Slope=infinite
Slope=-1
19
Phase Plane Analysis
Example: Mass Spring System
� �+��=0
For �=�=1: �+�=0
�
1=�
2
�
2=−�
1
��
2
��
1
=
−�
1
�
2
=�
�
1+��
2=0x
y
x
2
+ y
2
- 4
-5 0 5
-5
0
5
Slope=1
Slope=infinite
Slope=-1
19
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Vector Field Diagram Method
Vector Field: A set of vectors that is tangent to the trajectory
At each point (�
1,�
2)vector
�
1(�
1,�
2)
�
2(�
1,�
2)
is tangent to the trajectories
Hence vector field can be constructed in the phase plane and direction of
the trajectories can be easily realized with that 1 2
12
sin()0
sin()
x x
xx
f
20
Phase Plane Analysis
Vector Field Diagram Method
Vector Field: A set of vectors that is tangent to the trajectory
At each point (�
1,�
2)vector
�
1(�
1,�
2)
�
2(�
1,�
2)
is tangent to the trajectories
Hence vector field can be constructed in the phase plane and direction of
the trajectories can be easily realized with that 1 2
12
sin()0
sin()
x x
xx
f
20
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
Singularpointisanimportantconceptwhichrevealsgreatinfoabout
propertiesofsystemsuchasstability.
Singularpointsareonlypointswhichseveraltrajectoriespass/approach
them(i.e.trajectoriesintersect).
Singular Points in the Phase Plane Diagram:
Equilibrium points are in fact singular points in the phase plane
diagram
21
�
1�
1,�
2=0
�
1�
1,�
2=0
Slope of the trajectories at
equilibrium points
��
2
��
1
=
0
0
Phase Plane Analysis
Singularpointisanimportantconceptwhichrevealsgreatinfoabout
propertiesofsystemsuchasstability.
Singularpointsareonlypointswhichseveraltrajectoriespass/approach
them(i.e.trajectoriesintersect).
Singular Points in the Phase Plane Diagram:
Equilibrium points are in fact singular points in the phase plane
diagram
21
�
1�
1,�
2=0
�
1�
1,�
2=0
Slope of the trajectories at
equilibrium points
��
2
��
1
=
0
0
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
22
Example: Using Matlabx ' = y
y ' = - 0.6 y - 3 x + x
2
-6 -4 -2 0 2 4 6
-10
-5
0
5
10
x
y
�+0.6 �+3�+�
2
=0
Phase Plane Analysis
22
Example: Using Matlabx ' = y
y ' = - 0.6 y - 3 x + x
2
-6 -4 -2 0 2 4 6
-10
-5
0
5
10
x
y
�+0.6 �+3�+�
2
=0
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
23
Example: Using Maple Code
with(DEtools):
xx:=x(t): yy:=y(t):
dx:=diff(xx,t): dy:=diff(yy,t):
e0:=diff(dx,t)+.6*dx+3*xx+xx^2:
e1:=dx-yy=0:
e2:=dy+0.6*yy+3*xx+xx^2=0:
eqn:=[e1,e2]: depvar:=[x,y]:
rang:=t=-1..5: stpsz:=stepsize=0.005:
IC1:=[x(0)=0,y(0)=1]:
IC2:=[x(0)=0,y(0)=5]:
IC3:=[x(0)=0,y(0)=7]:
IC4:=[x(0)=0,y(0)=7]:
IC5:=[x(0)=-3.01,y(0)=0]:
IC6:=[x(0)=-4,y(0)=2]:
IC7:=[x(0)=1,y(0)=0]:
IC8:=[x(0)=4,y(0)=0]:
IC9:=[x(0)=-6,y(0)=3]:
IC10:=[x(0)=-6,y(0)=6]:
ICs:=[IC||(1..10)]:
lincl:=linecolour=sin((1/2)*t*Pi):
mtd:=method=classical[foreuler]:
phaseportrait(eqn,depvar,rang,ICs,stpsz,lincl,mtd);
�+0.6 �+3�+�
2
=0
Phase Plane Analysis
23
Example: Using Maple Code
with(DEtools):
xx:=x(t): yy:=y(t):
dx:=diff(xx,t): dy:=diff(yy,t):
e0:=diff(dx,t)+.6*dx+3*xx+xx^2:
e1:=dx-yy=0:
e2:=dy+0.6*yy+3*xx+xx^2=0:
eqn:=[e1,e2]: depvar:=[x,y]:
rang:=t=-1..5: stpsz:=stepsize=0.005:
IC1:=[x(0)=0,y(0)=1]:
IC2:=[x(0)=0,y(0)=5]:
IC3:=[x(0)=0,y(0)=7]:
IC4:=[x(0)=0,y(0)=7]:
IC5:=[x(0)=-3.01,y(0)=0]:
IC6:=[x(0)=-4,y(0)=2]:
IC7:=[x(0)=1,y(0)=0]:
IC8:=[x(0)=4,y(0)=0]:
IC9:=[x(0)=-6,y(0)=3]:
IC10:=[x(0)=-6,y(0)=6]:
ICs:=[IC||(1..10)]:
lincl:=linecolour=sin((1/2)*t*Pi):
mtd:=method=classical[foreuler]:
phaseportrait(eqn,depvar,rang,ICs,stpsz,lincl,mtd);
�+0.6 �+3�+�
2
=0
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
24
Example: Using Maple Tools
�+0.6 �+3�+�
2
=0
Phase Plane Analysis
24
Example: Using Maple Tools
�+0.6 �+3�+�
2
=0
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis
25
Phase Plane Analysis for Single DOF Mechanical System
In the case of single DOF mechanical system
�+��, �=0
�
1=�
�
2= �
�
1=�
2
�
2=−�(�
1,�
2)
Thephaseplaneisinfact(�− �)planeandeverypointshowsthe
positionandvelocityofthesystem.
Trajectoriesarealwaysclockwise.Thisisnottrueinthegeneralphase
plane(�
1−�
2)
Phase Plane Analysis
25
Phase Plane Analysis for Single DOF Mechanical System
In the case of single DOF mechanical system
�+��, �=0
�
1=�
�
2= �
�
1=�
2
�
2=−�(�
1,�
2)
Thephaseplaneisinfact(�− �)planeandeverypointshowsthe
positionandvelocityofthesystem.
Trajectoriesarealwaysclockwise.Thisisnottrueinthegeneralphase
plane(�
1−�
2)
Advanced Control (Mehdi Keshmiri, Winter 95)
26
Introducing
the Concept of Stability
26
Introducing
the Concept of Stability
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
27
Stabilityanalysisofadynamicsystemisnormallyintroducedinthestate
spaceformoftheequations.
�=??????�,??????,�
�∈ℝ
�
??????∈ℝ
�
Mostoftheconceptsinthischapterareintroducedforautonomous
systemsu x x = f(x,u)
Autonomous Dynamic Systemu x ,tx = f(x,u )
Dynamic System
Stability, Definitions and Examples
27
Stabilityanalysisofadynamicsystemisnormallyintroducedinthestate
spaceformoftheequations.
�=??????�,??????,�
�∈ℝ
�
??????∈ℝ
�
Mostoftheconceptsinthischapterareintroducedforautonomous
systemsu x x = f(x,u)
Autonomous Dynamic Systemu x ,tx = f(x,u )
Dynamic System
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
28
Ourmainconcernisthefirsttypeanalysis.Somepreliminaryissuesofthe
secondtypeanalysiswillbealsodiscussed.
Stability analysis of a dynamic system is divided in three
categories:
1.Stabilityanalysisoftheequilibriumpointsofthesystems.Westudythe
behavior(dynamics)ofthefree(unforced,�=0)systemwhenitisperturbed
fromitsequilibriumpoint.
2.Input-outputstabilityanalysis.Westudythesystem(forcedsystem�≠0)
outputbehaviorinresponsetoboundedinputs.
3. Stability analysis of periodic orbits. This analysis is for those systems which
perform a periodic or cyclic motion like walking of a biped or orbital motion
of a space object.
Stability, Definitions and Examples
28
Ourmainconcernisthefirsttypeanalysis.Somepreliminaryissuesofthe
secondtypeanalysiswillbealsodiscussed.
Stability analysis of a dynamic system is divided in three
categories:
1.Stabilityanalysisoftheequilibriumpointsofthesystems.Westudythe
behavior(dynamics)ofthefree(unforced,�=0)systemwhenitisperturbed
fromitsequilibriumpoint.
2.Input-outputstabilityanalysis.Westudythesystem(forcedsystem�≠0)
outputbehaviorinresponsetoboundedinputs.
3. Stability analysis of periodic orbits. This analysis is for those systems which
perform a periodic or cyclic motion like walking of a biped or orbital motion
of a space object.
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
29
Reminder:
�
??????issaidanequilibriumpointofthesystemifoncethesystemreachesthis
positionitstaysthereforever,i.e.��
??????=0
Definition (LyapunovStability):
Theequilibriumpoint�
??????issaidtobestable(inthesenseofLyapunov
stability)ormotionofthesystemaboutitsequilibriumpointissaidtobe
stableifthesystemstates(�)isperturbedawayfrom�
??????thenitstayscloseto
�
??????.Mathematically�
??????isstableif0)(,0 (0) ( ) 0
ee
tt x x x x
Stability, Definitions and Examples
29
Reminder:
�
??????issaidanequilibriumpointofthesystemifoncethesystemreachesthis
positionitstaysthereforever,i.e.��
??????=0
Definition (LyapunovStability):
Theequilibriumpoint�
??????issaidtobestable(inthesenseofLyapunov
stability)ormotionofthesystemaboutitsequilibriumpointissaidtobe
stableifthesystemstates(�)isperturbedawayfrom�
??????thenitstayscloseto
�
??????.Mathematically�
??????isstableif0)(,0 (0) ( ) 0
ee
tt x x x x
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
30
Withoutlossofgeneralitywecanpresentouranalysisaboutequilibrium
point�
??????=0,sincethesystemequationcanbetransferredtoanewform
withzeroastheequilibriumpointofthesystem.
�=�−�
??????
�=??????�
�
??????≠0
�=??????�
�
??????≠0
Theequilibriumpoint�
??????issaidtobestable(inthesenseofLyapunov
stability)ormotionofthesystemaboutitsequilibriumpointissaidtobestable
ifforany??????>0,thereexists0<�<??????suchthat
A more precise definition: (0) ( ) 0
ee
r t R t x x x x
Stability, Definitions and Examples
30
Withoutlossofgeneralitywecanpresentouranalysisaboutequilibrium
point�
??????=0,sincethesystemequationcanbetransferredtoanewform
withzeroastheequilibriumpointofthesystem.
�=�−�
??????
�=??????�
�
??????≠0
�=??????�
�
??????≠0
Theequilibriumpoint�
??????issaidtobestable(inthesenseofLyapunov
stability)ormotionofthesystemaboutitsequilibriumpointissaidtobestable
ifforany??????>0,thereexists0<�<??????suchthat
A more precise definition: (0) ( ) 0
ee
r t R t x x x x
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
31
The equilibrium point �
??????=0is said to be
Definition (LyapunovStability):
Stableif
∀??????>0∃0<�<??????�.�.
�0<� ��<??????∀�>0
Unstableif it is not stable.
Asymptotically stable if it is stable and
∀�>0�.�.
�0<� lim
??????→∞
�(�)=0
Marginallystableifitisstableandnot
asymptoticallystable
Exponentiallystableifitisasymptoticallystablewithanexponentialrate
�(0)<� �(�)<��
−�??????
�(0)�,�>0
Stability, Definitions and Examples
31
The equilibrium point �
??????=0is said to be
Definition (LyapunovStability):
Stableif
∀??????>0∃0<�<??????�.�.
�0<� ��<??????∀�>0
Unstableif it is not stable.
Asymptotically stable if it is stable and
∀�>0�.�.
�0<� lim
??????→∞
�(�)=0
Marginallystableifitisstableandnot
asymptoticallystable
Exponentiallystableifitisasymptoticallystablewithanexponentialrate
�(0)<� �(�)<��
−�??????
�(0)�,�>0
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
32
Example: UndampedPendulum
??????+
�
�
sin(??????)=0 ??????
??????1=0,??????
??????2=?????? R
B r
B
??????
??????1is a marginally stablepoint and ??????
??????2is an unstablepoint0 2 4 6 8 10
-1.5
-1
-0.5
0
0.5
1
1.5
Time(sec)
Teta (rad)
X(0) = (0,1) 0 2 4 6 8 10
0
10
20
30
40
Time(sec)
Teta (rad)
X(0) = (0,4)
Stability, Definitions and Examples
32
Example: UndampedPendulum
??????+
�
�
sin(??????)=0 ??????
??????1=0,??????
??????2=?????? R
B r
B
??????
??????1is a marginally stablepoint and ??????
??????2is an unstablepoint0 2 4 6 8 10
-1.5
-1
-0.5
0
0.5
1
1.5
Time(sec)
Teta (rad)
X(0) = (0,1) 0 2 4 6 8 10
0
10
20
30
40
Time(sec)
Teta (rad)
X(0) = (0,4)
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
33
Example: damped Pendulum
??????+� ??????+
�
�
sin(??????)=0 ??????
??????1=0,??????
??????2=??????
??????
??????1is an exponentially stablepoint and ??????
??????2is an unstablepoint0 5 10 15
-2.5
-2
-1.5
-1
-0.5
0
0.5
Time(sec)
Teta (rad)
X(0) = (-2.5,0) 0 5 10 15
-7
-6
-5
-4
-3
Time (sec)
Teta (rad)
X(0) = (-3.5,0) 0 5 10 15
-4
-2
0
2
4
6
8
Time(sec)
Teta (rad)
X(0) = (-3,10)
Stability, Definitions and Examples
33
Example: damped Pendulum
??????+� ??????+
�
�
sin(??????)=0 ??????
??????1=0,??????
??????2=??????
??????
??????1is an exponentially stablepoint and ??????
??????2is an unstablepoint0 5 10 15
-2.5
-2
-1.5
-1
-0.5
0
0.5
Time(sec)
Teta (rad)
X(0) = (-2.5,0) 0 5 10 15
-7
-6
-5
-4
-3
Time (sec)
Teta (rad)
X(0) = (-3.5,0) 0 5 10 15
-4
-2
0
2
4
6
8
Time(sec)
Teta (rad)
X(0) = (-3,10)
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
34stateform
122
122
2 1 1 2
(1 ) 0 0
(1 )
ee
xx
x x x x x x
x x x x
Example: Van Der Pol Oscillator
�
??????=0is an unstablepoint
Stability, Definitions and Examples
34stateform
122
122
2 1 1 2
(1 ) 0 0
(1 )
ee
xx
x x x x x x
x x x x
Example: Van Der Pol Oscillator
�
??????=0is an unstablepoint
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
35
Definition:
iftheequil.point�
??????isasymptoticallystable,thenthesetofall
pointsthattrajectoriesinitiatedatthesepointeventuallyconverge
totheoriginiscalleddomainofattraction.
Definition:
iftheequil.point�
??????isasymptotically/exponentiallystable,then
theequil.pointiscalledgloballystableifthewholespaceis
domainofattraction.Otherwiseitiscalledlocallystable.
Stability, Definitions and Examples
35
Definition:
iftheequil.point�
??????isasymptoticallystable,thenthesetofall
pointsthattrajectoriesinitiatedatthesepointeventuallyconverge
totheoriginiscalleddomainofattraction.
Definition:
iftheequil.point�
??????isasymptotically/exponentiallystable,then
theequil.pointiscalledgloballystableifthewholespaceis
domainofattraction.Otherwiseitiscalledlocallystable.
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
36
The origin in the first order system of �=−�is globally exponentially stable.
Example 1:00
( ) lim ( ) 0 0
t
t
x x x t x e x t x
Theorigininthefirstordersystem �=−�
3
isgloballyasymptoticallybut
notexponentiallystable.
Example 2:3 0
0
2
0
( ) lim ( ) 0
12
t
x
x x x t x t x
tx
Stability, Definitions and Examples
36
The origin in the first order system of �=−�is globally exponentially stable.
Example 1:00
( ) lim ( ) 0 0
t
t
x x x t x e x t x
Theorigininthefirstordersystem �=−�
3
isgloballyasymptoticallybut
notexponentiallystable.
Example 2:3 0
0
2
0
( ) lim ( ) 0
12
t
x
x x x t x t x
tx
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
37
The origin in the first order system �=−�
2
is semi-asymptotically but not
exponentially stable.
Example 3:0
0
2 0
00
1/
lim ( ) 0 if 0
()
lim ( ) if 01
t
tx
x t x
x
x x x t
x t xtx
Domain of attraction is �
0>0.
Stability, Definitions and Examples
37
The origin in the first order system �=−�
2
is semi-asymptotically but not
exponentially stable.
Example 3:0
0
2 0
00
1/
lim ( ) 0 if 0
()
lim ( ) if 01
t
tx
x t x
x
x x x t
x t xtx
Domain of attraction is �
0>0.
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
38
Example 4:
�+ �+�=0
Stability, Definitions and Examples
38
Example 4:
�+ �+�=0
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
39
Example 5:
�+0.6 �+3�+�
2
=0
Stability, Definitions and Examples
39
Example 5:
�+0.6 �+3�+�
2
=0
Advanced Control (Mehdi Keshmiri, Winter 95)
Stability, Definitions and Examples
40
Example 6:
�+ �+�
3
−�=0
Stability, Definitions and Examples
40
Example 6:
�+ �+�
3
−�=0
Advanced Control (Mehdi Keshmiri, Winter 95)
41
Stability Analysis of
Linear Time Invariant Systems
41
Stability Analysis of
Linear Time Invariant Systems
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
42
Itisthebesttoolforstudyofthelinearsystemgraphically
Thisanalysisgivesaverygoodinsightoflinearsystems
behavior
Theanalysiscanbeextendedforhigherorderlinearsystem
Localbehaviorofthenonlinearsystemscanbeunderstoodfrom
thisanalysis
Theanalysisisperformedbasedonthesystemeigenvaluesand
eigenvectors.
Phase Plane Analysis of LTI Systems
42
Itisthebesttoolforstudyofthelinearsystemgraphically
Thisanalysisgivesaverygoodinsightoflinearsystems
behavior
Theanalysiscanbeextendedforhigherorderlinearsystem
Localbehaviorofthenonlinearsystemscanbeunderstoodfrom
thisanalysis
Theanalysisisperformedbasedonthesystemeigenvaluesand
eigenvectors.
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
43
Considerasecondorderlinearsystem:
IftheAmatrixisnonsingular,originistheonlyequilibrium
pointofthesystem
IftheAmatrixissingularthenthesystemhasinfinitenumber
ofequilibriumpoints.Infactallofthepointsbelongingtothe
nullspaceofAaretheequilibriumpointofthesystem.isnon-singular
e
A x 0
**
issingular | ( )
e
Null A x x x A
�=���∈ℝ
2×2
,�∈ℝ
2
Phase Plane Analysis of LTI Systems
43
Considerasecondorderlinearsystem:
IftheAmatrixisnonsingular,originistheonlyequilibrium
pointofthesystem
IftheAmatrixissingularthenthesystemhasinfinitenumber
ofequilibriumpoints.Infactallofthepointsbelongingtothe
nullspaceofAaretheequilibriumpointofthesystem.isnon-singular
e
A x 0
**
issingular | ( )
e
Null A x x x A
�=���∈ℝ
2×2
,�∈ℝ
2
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
44
Considerasecondorderlinearsystem:
�=��
�∈ℝ
2×2
,�∈ℝ
2
The analytical solution can be obtained based on eigenvalues (??????
1,??????
2):
If ??????
1,??????
2are realand distinct
If ??????
1,??????
2are realand similar
If ??????
1,??????
2are complex conjugate
��=��
??????1??????
+��
??????2??????
��=(�+��)�
????????????
��=��
??????
1??????
+��
??????
2??????
=�
�??????
(�sin��+�cos(��))
Phase Plane Analysis of LTI Systems
44
Considerasecondorderlinearsystem:
�=��
�∈ℝ
2×2
,�∈ℝ
2
The analytical solution can be obtained based on eigenvalues (??????
1,??????
2):
If ??????
1,??????
2are realand distinct
If ??????
1,??????
2are realand similar
If ??????
1,??????
2are complex conjugate
��=��
??????1??????
+��
??????2??????
��=(�+��)�
????????????
��=��
??????
1??????
+��
??????
2??????
=�
�??????
(�sin��+�cos(��))
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
45
Jordan Form (almost diagonal form)
Thisrepresentationhasthesystemeigenvaluesontheleadingdiagonal,and
either0or1onthesuperdiagonal.
�=�� �=��
�=
�
1
⋱
�
�
�
??????=
??????
??????1
⋱1
??????
??????
Obtaining Jordan form:
�=??????
−1
�
??????=�
1…�
�
�=??????
−1
�??????
Phase Plane Analysis of LTI Systems
45
Jordan Form (almost diagonal form)
Thisrepresentationhasthesystemeigenvaluesontheleadingdiagonal,and
either0or1onthesuperdiagonal.
�=�� �=��
�=
�
1
⋱
�
�
�
??????=
??????
??????1
⋱1
??????
??????
Obtaining Jordan form:
�=??????
−1
�
??????=�
1…�
�
�=??????
−1
�??????
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
46
TheAmatrixhas2eigenvalues(eithertworeal,ortwocomplex
conjugates)andcanhaveeithertwoeigenvectorsorone
eigenvectors.Fourcategoriescanberealized
1.Twodistinctrealeigenvaluesandtworealeigenvectors
2.Twocomplexconjugateeigenvaluesandtwocomplex
eigenvectors
3.Twosimilar(real)eigenvaluesandtwoeigenvectors
4.Twosimilar(real)eigenvaluesandoneeigenvectors
A is non-singular:
Phase Plane Analysis of LTI Systems
46
TheAmatrixhas2eigenvalues(eithertworeal,ortwocomplex
conjugates)andcanhaveeithertwoeigenvectorsorone
eigenvectors.Fourcategoriescanberealized
1.Twodistinctrealeigenvaluesandtworealeigenvectors
2.Twocomplexconjugateeigenvaluesandtwocomplex
eigenvectors
3.Twosimilar(real)eigenvaluesandtwoeigenvectors
4.Twosimilar(real)eigenvaluesandoneeigenvectors
A is non-singular:
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
47
1.Two distinct real eigenvalues and two real eigenvectors
�=���=
??????
10
0??????
2
�
1=??????
1�
1
�
2=??????
2�
2
�
1=�
10�
??????
1??????
�
2=�
20�
??????
2??????
ln
�
1
�
10
=??????
1�
ln
�
2
�
20
=??????
2�
??????
2
??????
1
ln
�
1
�
10
=ln
�
2
�
20
�
1
�
10
??????
2
??????1
=
�
2
�
20
�
2=
�
20
�
10
??????2/??????1
�
1
??????
2/??????
1
�
2=��
1
??????
2/??????
1
Phase Plane Analysis of LTI Systems
47
1.Two distinct real eigenvalues and two real eigenvectors
�=���=
??????
10
0??????
2
�
1=??????
1�
1
�
2=??????
2�
2
�
1=�
10�
??????
1??????
�
2=�
20�
??????
2??????
ln
�
1
�
10
=??????
1�
ln
�
2
�
20
=??????
2�
??????
2
??????
1
ln
�
1
�
10
=ln
�
2
�
20
�
1
�
10
??????
2
??????1
=
�
2
�
20
�
2=
�
20
�
10
??????2/??????1
�
1
??????
2/??????
1
�
2=��
1
??????
2/??????
1
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
48
1. A ) ??????
�<??????
�<��
2=��
1
??????2/??????1
Systemhastwoeigenvectors�
1,�
2thephaseplaneportraitisasthe
following
Trajectories are:
tangent to the slow eigenvector (�
1)for near the origin
parallel to the fast eigenvector (�
2)for far from the origin
The equilibrium point �
??????=0is called stable nodex ' = - 2 x
y ' = - 12 y
-10 -8 -6 -4 -2 0 2 4 6 8 10
-10
-8
-6
-4
-2
0
2
4
6
8
10
x
y x ' = y
y ' = - 2 x - 3 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Phase Plane Analysis of LTI Systems
48
1. A ) ??????
�<??????
�<��
2=��
1
??????2/??????1
Systemhastwoeigenvectors�
1,�
2thephaseplaneportraitisasthe
following
Trajectories are:
tangent to the slow eigenvector (�
1)for near the origin
parallel to the fast eigenvector (�
2)for far from the origin
The equilibrium point �
??????=0is called stable nodex ' = - 2 x
y ' = - 12 y
-10 -8 -6 -4 -2 0 2 4 6 8 10
-10
-8
-6
-4
-2
0
2
4
6
8
10
x
y x ' = y
y ' = - 2 x - 3 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
49
Example 7: �+4 �+2�=0
�=
01
−2−4
�
det??????�−
01
−2−4
=??????
2
+4??????+2=0 ??????
1=−0.59,??????
2=−3.41
−0.59 −1
2 −0.59+4
�
1
1
�
1
2
=0
�
1=
0.86
−0.51
−3.41 −1
2 −3.41+4
�
1
1
�
1
2
=0
�
2=
−0.28
0.96
-4 -2 0 2 4
-4
-2
0
2
4
x
y 0 2 4 6 8 10
-1
-0.5
0
0.5
1
1.5
X(0) = [-1,10]
X(0) = [-1,1]
Phase Plane Analysis of LTI Systems
49
Example 7: �+4 �+2�=0
�=
01
−2−4
�
det??????�−
01
−2−4
=??????
2
+4??????+2=0 ??????
1=−0.59,??????
2=−3.41
−0.59 −1
2 −0.59+4
�
1
1
�
1
2
=0
�
1=
0.86
−0.51
−3.41 −1
2 −3.41+4
�
1
1
�
1
2
=0
�
2=
−0.28
0.96
-4 -2 0 2 4
-4
-2
0
2
4
x
y 0 2 4 6 8 10
-1
-0.5
0
0.5
1
1.5
X(0) = [-1,10]
X(0) = [-1,1]
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
50
1. B ) ??????
�>??????
�>��
2=��
1
??????2/??????1
Systemhastwoeigenvectors�
1and�
2thephaseplaneportraitis
oppositeasthepreviousone
Trajectories are:
tangent to the slow eigenvector �
1for near origin
parallel to the fast eigenvector �
2for far from origin
The equilibrium point �
??????=0is called unstable nodex ' = - 2 x
y ' = - 12 y
-10 -8 -6 -4 -2 0 2 4 6 8 10
-10
-8
-6
-4
-2
0
2
4
6
8
10
x
y x ' = y
y ' = - 2 x - 3 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Phase Plane Analysis of LTI Systems
50
1. B ) ??????
�>??????
�>��
2=��
1
??????2/??????1
Systemhastwoeigenvectors�
1and�
2thephaseplaneportraitis
oppositeasthepreviousone
Trajectories are:
tangent to the slow eigenvector �
1for near origin
parallel to the fast eigenvector �
2for far from origin
The equilibrium point �
??????=0is called unstable nodex ' = - 2 x
y ' = - 12 y
-10 -8 -6 -4 -2 0 2 4 6 8 10
-10
-8
-6
-4
-2
0
2
4
6
8
10
x
y x ' = y
y ' = - 2 x - 3 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
51
Example 8: �−3 �+2�=0
�=
01
−23
�
det??????�−
01
−23
=??????
2
−3??????+2=0 ??????
1=1,??????
2=2
1−1
21−3
�
1
1
�
1
2
=0
�
1=
−0.71
−0.71
2−1
22−3
�
1
1
�
1
2
=0
�
2=
−0.45
−0.89
-4 -2 0 2 4
-4
-2
0
2
4
x
y 0 0.2 0.4 0.6 0.8 1
-4
-2
0
2
4
6
X(0) = [-2,2]
X(0) = [-2,0.1]
Phase Plane Analysis of LTI Systems
51
Example 8: �−3 �+2�=0
�=
01
−23
�
det??????�−
01
−23
=??????
2
−3??????+2=0 ??????
1=1,??????
2=2
1−1
21−3
�
1
1
�
1
2
=0
�
1=
−0.71
−0.71
2−1
22−3
�
1
1
�
1
2
=0
�
2=
−0.45
−0.89
-4 -2 0 2 4
-4
-2
0
2
4
x
y 0 0.2 0.4 0.6 0.8 1
-4
-2
0
2
4
6
X(0) = [-2,2]
X(0) = [-2,0.1]
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
52
1. C ) ??????
�<�<??????
�
�
2=��
1
??????2/??????1
Systemhastwoeigenvectors�
1and�
2,thephaseplaneportraitisasthe
following
Only trajectories along �
2are stable trajectories
All other trajectories at start are tangent to �
2and at the end are tangent
to �
1
This equilibrium point is unstable and is called saddle pointx ' = 3 x
y ' = - 3 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = y
y ' = y + 2 x
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Phase Plane Analysis of LTI Systems
52
1. C ) ??????
�<�<??????
�
�
2=��
1
??????2/??????1
Systemhastwoeigenvectors�
1and�
2,thephaseplaneportraitisasthe
following
Only trajectories along �
2are stable trajectories
All other trajectories at start are tangent to �
2and at the end are tangent
to �
1
This equilibrium point is unstable and is called saddle pointx ' = 3 x
y ' = - 3 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = y
y ' = y + 2 x
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
53
Example 9: �− �−2�=0
�=
01
21
�
det??????�−
01
21
=??????
2
−??????−2=0 ??????
1=−1,??????
2=2
−1−1
−2−1−1
�
1
1
�
1
2
=0
�
1=
−0.71
0.71
2−1
−22−1
�
1
1
�
1
2
=0
�
2=
−0.45
−0.89
-4 -2 0 2 4
-4
-2
0
2
4
x
y 0 2 4 6 8 10
-1
0
1
2
3
x 10
8
X(0) = [1,0.5]
X(0) = [1,-1.5]
Phase Plane Analysis of LTI Systems
53
Example 9: �− �−2�=0
�=
01
21
�
det??????�−
01
21
=??????
2
−??????−2=0 ??????
1=−1,??????
2=2
−1−1
−2−1−1
�
1
1
�
1
2
=0
�
1=
−0.71
0.71
2−1
−22−1
�
1
1
�
1
2
=0
�
2=
−0.45
−0.89
-4 -2 0 2 4
-4
-2
0
2
4
x
y 0 2 4 6 8 10
-1
0
1
2
3
x 10
8
X(0) = [1,0.5]
X(0) = [1,-1.5]
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
54
2. Two complex conjugate eigenvalues and two complex eigenvectors
�=���=
�−�
��
�≡�
1
2
+�
2
2
??????≡tan
−1
(
�
2
�
1
)
� �=�
1 �
1+�
2 �
2=�
1��
1−��
2+�
2��
1+��
2=��
2
??????1+tan??????
2
=
�
1 �
2−�
2 �
1
�
1
2
=
�
1��
1+��
2−�
2��
1−��
2
�
1
2
=�1+tan??????
2
�=��
??????=�
��=�
0�
�??????
??????�=??????
0+��
Phase Plane Analysis of LTI Systems
54
2. Two complex conjugate eigenvalues and two complex eigenvectors
�=���=
�−�
��
�≡�
1
2
+�
2
2
??????≡tan
−1
(
�
2
�
1
)
� �=�
1 �
1+�
2 �
2=�
1��
1−��
2+�
2��
1+��
2=��
2
??????1+tan??????
2
=
�
1 �
2−�
2 �
1
�
1
2
=
�
1��
1+��
2−�
2��
1−��
2
�
1
2
=�1+tan??????
2
�=��
??????=�
��=�
0�
�??????
??????�=??????
0+��
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
55
2. A ) ??????
�,??????
�=�±�??????, �<�,�≠�
System has no real eigenvectors the phase plane portrait is as the following
The trajectories are spiral around the origin and toward the origin.
This equilibrium point is called stable focus.
��=�
0�
�??????
??????�=??????
0+��x ' = - 2 x - 3 y
y ' = 3 x - 2 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = y
y ' = - x - y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Phase Plane Analysis of LTI Systems
55
2. A ) ??????
�,??????
�=�±�??????, �<�,�≠�
System has no real eigenvectors the phase plane portrait is as the following
The trajectories are spiral around the origin and toward the origin.
This equilibrium point is called stable focus.
��=�
0�
�??????
??????�=??????
0+��x ' = - 2 x - 3 y
y ' = 3 x - 2 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = y
y ' = - x - y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
56
Example 10: �+ �+�=0
�=
01
−1−1
�
det??????�−
01
−1−1
=??????
2
+??????+1=0 ??????
1,??????
2=−0.5±0.866??????x ' = y
y ' = - x - y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y 0 2 4 6 8 10
-1
-0.5
0
0.5
1
1.5
2
x(0) = [1,-2]
X(0) = [1,2]
Phase Plane Analysis of LTI Systems
56
Example 10: �+ �+�=0
�=
01
−1−1
�
det??????�−
01
−1−1
=??????
2
+??????+1=0 ??????
1,??????
2=−0.5±0.866??????x ' = y
y ' = - x - y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y 0 2 4 6 8 10
-1
-0.5
0
0.5
1
1.5
2
x(0) = [1,-2]
X(0) = [1,2]
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
57
2. B ) ??????
�,??????
�=�±�??????, �>�,�≠�
System has no real eigenvectors the phase plane portrait is as the following
The trajectories are spiral around the origin and diverge from the origin.
This equilibrium point is called unstable focus.
��=�
0�
�??????
??????�=??????
0+��x ' = - 2 x - 3 y
y ' = 3 x - 2 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = y
y ' = - x - y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Phase Plane Analysis of LTI Systems
57
2. B ) ??????
�,??????
�=�±�??????, �>�,�≠�
System has no real eigenvectors the phase plane portrait is as the following
The trajectories are spiral around the origin and diverge from the origin.
This equilibrium point is called unstable focus.
��=�
0�
�??????
??????�=??????
0+��x ' = - 2 x - 3 y
y ' = 3 x - 2 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = y
y ' = - x - y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
58
Example 11: �− �+�=0
�=
01
−11
�
det??????�−
01
−11
=??????
2
−??????+1=0 ??????
1,??????
2=0.5±0.866??????x ' = y
y ' = - x + y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y 0 2 4 6 8 10
-600
-400
-200
0
200
X(0) = [1,2]
X(0) = [1,-2]
Phase Plane Analysis of LTI Systems
58
Example 11: �− �+�=0
�=
01
−11
�
det??????�−
01
−11
=??????
2
−??????+1=0 ??????
1,??????
2=0.5±0.866??????x ' = y
y ' = - x + y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y 0 2 4 6 8 10
-600
-400
-200
0
200
X(0) = [1,2]
X(0) = [1,-2]
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
59
2. C ) ??????
�,??????
�=±�??????, �=�,�≠�
System has two imaginary eigenvalues and no real eigenvectors the phase
plane portrait is as the following
The trajectories are closed trajectories around the origin.
This equilibrium point is marginally stable and is called center.
��=�
0�
�??????
??????�=??????
0+��x ' = 3 y
y ' = - 3 x
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = y
y ' = - 5 x
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Phase Plane Analysis of LTI Systems
59
2. C ) ??????
�,??????
�=±�??????, �=�,�≠�
System has two imaginary eigenvalues and no real eigenvectors the phase
plane portrait is as the following
The trajectories are closed trajectories around the origin.
This equilibrium point is marginally stable and is called center.
��=�
0�
�??????
??????�=??????
0+��x ' = 3 y
y ' = - 3 x
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = y
y ' = - 5 x
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
60
Example 12: �+3�=0
�=
01
−30
�
det??????�−
01
−30
=??????
2
+3=0 ??????
1,??????
2=±1.732??????x ' = y
y ' = - 3 x
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y 0 2 4 6 8 10
-2
-1
0
1
2
X(0) = [1,-2]
X(0) = [1,2]
Phase Plane Analysis of LTI Systems
60
Example 12: �+3�=0
�=
01
−30
�
det??????�−
01
−30
=??????
2
+3=0 ??????
1,??????
2=±1.732??????x ' = y
y ' = - 3 x
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y 0 2 4 6 8 10
-2
-1
0
1
2
X(0) = [1,-2]
X(0) = [1,2]
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
61
3. Two similar (real) eigenvalues and two eigenvectors
�=���=
??????0
0??????
�
1=??????�
1
�
2=??????�
2
�
1=�
10�
????????????
�
2=�
20�
????????????
�
1
�
2
=
�
10
�
20
�
2=��
1
Phase Plane Analysis of LTI Systems
61
3. Two similar (real) eigenvalues and two eigenvectors
�=���=
??????0
0??????
�
1=??????�
1
�
2=??????�
2
�
1=�
10�
????????????
�
2=�
20�
????????????
�
1
�
2
=
�
10
�
20
�
2=��
1
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
62x ' = 2 x
y ' = 2 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
??????>0x ' = - 2 x
y ' = - 2 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
??????<0
3 ) ??????
�=??????
�=??????≠�
System has two similar eigenvalues and two different eigenvectors. The
phase plane portrait is as the following, depending to the sign of ??????
The trajectories are all along the initial conditions and they are ??????<0
toward ??????>0or outward the origin
Phase Plane Analysis of LTI Systems
62x ' = 2 x
y ' = 2 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
??????>0x ' = - 2 x
y ' = - 2 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
??????<0
3 ) ??????
�=??????
�=??????≠�
System has two similar eigenvalues and two different eigenvectors. The
phase plane portrait is as the following, depending to the sign of ??????
The trajectories are all along the initial conditions and they are ??????<0
toward ??????>0or outward the origin
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
63
4. Two similar (real) eigenvalues and One eigenvectors
�=���=
??????1
0??????
�
1=??????�
1+�
2
�
2=??????�
2
�
1=�
10�
????????????
+�
20��
????????????
�
2=�
20�
????????????
�
1=�
10
�
2
�
20
+�
2
1
??????
ln(
�
2
�
20
)
�
1=�
2(
�
10
�
20
+
1
??????
ln
�
2
�
20
)
Phase Plane Analysis of LTI Systems
63
4. Two similar (real) eigenvalues and One eigenvectors
�=���=
??????1
0??????
�
1=??????�
1+�
2
�
2=??????�
2
�
1=�
10�
????????????
+�
20��
????????????
�
2=�
20�
????????????
�
1=�
10
�
2
�
20
+�
2
1
??????
ln(
�
2
�
20
)
�
1=�
2(
�
10
�
20
+
1
??????
ln
�
2
�
20
)
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
64
??????>0
??????<0
4 ) ??????
�=??????
�=??????≠�
System has two similar eigenvalues and only one eigenvector. The phase
plane portrait is as the following, depending to the sign of ??????
The trajectories converge to zero or diverge to infinity along the system
eigenvector. x ' = 0.5 x + y
y ' = 0.5 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = - 0.5 x + y
y ' = - 0.5 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Phase Plane Analysis of LTI Systems
64
??????>0
??????<0
4 ) ??????
�=??????
�=??????≠�
System has two similar eigenvalues and only one eigenvector. The phase
plane portrait is as the following, depending to the sign of ??????
The trajectories converge to zero or diverge to infinity along the system
eigenvector. x ' = 0.5 x + y
y ' = 0.5 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = - 0.5 x + y
y ' = - 0.5 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
65
Example 13: �+2 �+�=0
�=
01
−1−2
�
det??????�−
01
−1−2
=??????
2
+2??????+1=0 ??????
1,??????
2=−1
−1−1
1−1+2
�
1
1
�
1
2
=0
�
1=
−0.71
0.71
-4 -2 0 2 4
-4
-2
0
2
4
x
y 0 2 4 6 8 10
-1
0
1
2
3
X(0) = [2,3]
X(0) = [2,-4]
Phase Plane Analysis of LTI Systems
65
Example 13: �+2 �+�=0
�=
01
−1−2
�
det??????�−
01
−1−2
=??????
2
+2??????+1=0 ??????
1,??????
2=−1
−1−1
1−1+2
�
1
1
�
1
2
=0
�
1=
−0.71
0.71
-4 -2 0 2 4
-4
-2
0
2
4
x
y 0 2 4 6 8 10
-1
0
1
2
3
X(0) = [2,3]
X(0) = [2,-4]
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
66
A is singular (��??????�=�):
Systemhasatleastoneeigenvalueequaltozeroandthereforeinfinite
numberofequilibriumpoints.Threedifferentcategoriescanbe
specified
??????
1=0, ??????
2≠0
??????
1,??????
1=0, ??????����=1
??????
1,??????
1=0, ??????����=0
Phase Plane Analysis of LTI Systems
66
A is singular (��??????�=�):
Systemhasatleastoneeigenvalueequaltozeroandthereforeinfinite
numberofequilibriumpoints.Threedifferentcategoriescanbe
specified
??????
1=0, ??????
2≠0
??????
1,??????
1=0, ??????����=1
??????
1,??????
1=0, ??????����=0
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
67
1) ??????
1=0, ??????
2≠0 �=��
�
1=0
�
2=??????�
2
�
1=�
10
�
2=�
20�
????????????
�=
00
0??????
System has infinite number of non-isolated equilibrium points along a line
System has two eigenvectors. Eigenvector corresponding to zero eigenvalue
is in fact loci of the equilibrium points
Depending on the sign of the second eigenvalue, all the trajectories move
inward or outward to �
1along �
2x ' = 0
y ' = 2 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = - x + y
y ' = - 3 x + 3 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Phase Plane Analysis of LTI Systems
67
1) ??????
1=0, ??????
2≠0 �=��
�
1=0
�
2=??????�
2
�
1=�
10
�
2=�
20�
????????????
�=
00
0??????
System has infinite number of non-isolated equilibrium points along a line
System has two eigenvectors. Eigenvector corresponding to zero eigenvalue
is in fact loci of the equilibrium points
Depending on the sign of the second eigenvalue, all the trajectories move
inward or outward to �
1along �
2x ' = 0
y ' = 2 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y x ' = - x + y
y ' = - 3 x + 3 y
-4 -3 -2 -1 0 1 2 3 4
-4
-3
-2
-1
0
1
2
3
4
x
y
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
68
Example 14: �+ �=0
�=
01
0−1
�
det??????�−
01
0−1
=??????
2
+??????=0 ??????
1=0,??????
2=−1
0−1
01
�
1
1
�
1
2
=0
�
1=
1
0
−1−1
0−1+1
�
1
1
�
1
2
=0
�
1=
1
−1
-4 -2 0 2 4
-4
-2
0
2
4
x
y 0 2 4 6 8 10
-2
-1
0
1
2
3
X(0) = [2,-4]
X(0) = [3,-4]
Phase Plane Analysis of LTI Systems
68
Example 14: �+ �=0
�=
01
0−1
�
det??????�−
01
0−1
=??????
2
+??????=0 ??????
1=0,??????
2=−1
0−1
01
�
1
1
�
1
2
=0
�
1=
1
0
−1−1
0−1+1
�
1
1
�
1
2
=0
�
1=
1
−1
-4 -2 0 2 4
-4
-2
0
2
4
x
y 0 2 4 6 8 10
-2
-1
0
1
2
3
X(0) = [2,-4]
X(0) = [3,-4]
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
69
2) ??????
1,??????
1=0, ??????����=1 �=��
�
1=�
2
�
2=0
�
1=�
20�+�
10
�
2=�
20
�=
01
00
Systemhasinfinitenumberofnon-isolatedequilibriumpointsalongaline
Systemhasonlyoneeigenvector,anditislocioftheequilibriumpoints
Allthetrajectoriesmovetowardinfinityalongthesystemeigenvector
(unstablesystem).
Phase Plane Analysis of LTI Systems
69
2) ??????
1,??????
1=0, ??????����=1 �=��
�
1=�
2
�
2=0
�
1=�
20�+�
10
�
2=�
20
�=
01
00
Systemhasinfinitenumberofnon-isolatedequilibriumpointsalongaline
Systemhasonlyoneeigenvector,anditislocioftheequilibriumpoints
Allthetrajectoriesmovetowardinfinityalongthesystemeigenvector
(unstablesystem).
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
70
2) ??????
1,??????
1=0, ??????����=0
�=��
�
1=0
�
2=0
�
1=�
10
�
2=�
20
�=
00
00
Systemisastaticsystem.Allthepointsareequilibriumpoints
Phase Plane Analysis of LTI Systems
70
2) ??????
1,??????
1=0, ??????����=0
�=��
�
1=0
�
2=0
�
1=�
10
�
2=�
20
�=
00
00
Systemisastaticsystem.Allthepointsareequilibriumpoints
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
71
Summary
Six different type of isolated equilibrium points can be identified
Stable/unstable node
Saddle point
Stable/ unstable focus
Center
Phase Plane Analysis of LTI Systems
71
Summary
Six different type of isolated equilibrium points can be identified
Stable/unstable node
Saddle point
Stable/ unstable focus
Center
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
72
Stability Analysis of Higher Order Systems:
Analysis and results for the second order LTI system can be
extended to higher order LTI system
Graphical tool is not useful for higher order LTI system except
for third order systems.
This means stability analysis of mechanical system with more
than one DOF can not be materialized graphically
Stability analysis is performed through the eigenvalue analysis of
the Amatrix.
Phase Plane Analysis of LTI Systems
72
Stability Analysis of Higher Order Systems:
Analysis and results for the second order LTI system can be
extended to higher order LTI system
Graphical tool is not useful for higher order LTI system except
for third order systems.
This means stability analysis of mechanical system with more
than one DOF can not be materialized graphically
Stability analysis is performed through the eigenvalue analysis of
the Amatrix.
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
73
Consideralineartimeinvariant(LTI)system
OriginistheonlyequilibriumpointofthesystemifAisnon-
singular
Otherwisethesystemhasinfinitenumberofequilibriumpoints,
allthepointsonnull-spaceofAareinfactequil.pointsofthe
system.
x=AxBu
yCxDu det( ) 0
e
A
x = Ax x 0
det( ) 0
**
| Nullspace( )
e
A
x = Ax x x x A
Phase Plane Analysis of LTI Systems
73
Consideralineartimeinvariant(LTI)system
OriginistheonlyequilibriumpointofthesystemifAisnon-
singular
Otherwisethesystemhasinfinitenumberofequilibriumpoints,
allthepointsonnull-spaceofAareinfactequil.pointsofthe
system.
x=AxBu
yCxDu det( ) 0
e
A
x = Ax x 0
det( ) 0
**
| Nullspace( )
e
A
x = Ax x x x A
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
74
Details for Case of Non-Singular A
Origin is the only equilibrium point of the system
This equilibrium point (system) is
Exponentially stableif all eigenvalues of Aare either real
negative or complex with negative real part.
Marginally stable if eigenvalues of Ahave non-positive real
part and �����−??????�=�−�for all repeated imaginary
eigenvalues, ??????with multiplicity of r
Unstable, otherwise.
Phase Plane Analysis of LTI Systems
74
Details for Case of Non-Singular A
Origin is the only equilibrium point of the system
This equilibrium point (system) is
Exponentially stableif all eigenvalues of Aare either real
negative or complex with negative real part.
Marginally stable if eigenvalues of Ahave non-positive real
part and �����−??????�=�−�for all repeated imaginary
eigenvalues, ??????with multiplicity of r
Unstable, otherwise.
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
75
AisNon-Singular
Classificationoftheequilibriumpointofhigherordersystem
intonode,focus,andsaddlepointisnotaseasyassecondorder
system.Howeversomepointscanbeemphasized:
Theequilibriumpointisstable/unstablenodeifall
eigenvaluesarerealandhavethenegative/positivesign.
Theequilibriumpointiscenterifapairofeigenvaluesare
pureimaginarycomplexconjugateandallothereigenvalues
havenegativereal
Inthecaseofdifferentsignintherealpartofthe
eigenvaluestrajectorieshavethesaddletypebehaviornear
theequilibriumpoint
Phase Plane Analysis of LTI Systems
75
AisNon-Singular
Classificationoftheequilibriumpointofhigherordersystem
intonode,focus,andsaddlepointisnotaseasyassecondorder
system.Howeversomepointscanbeemphasized:
Theequilibriumpointisstable/unstablenodeifall
eigenvaluesarerealandhavethenegative/positivesign.
Theequilibriumpointiscenterifapairofeigenvaluesare
pureimaginarycomplexconjugateandallothereigenvalues
havenegativereal
Inthecaseofdifferentsignintherealpartofthe
eigenvaluestrajectorieshavethesaddletypebehaviornear
theequilibriumpoint
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
76
Trajectoriesarealongtheeigenvectorwithminimum
absoluterealpartneartheequilibriumpointandalongthe
eigenvectorwithmaximumabsoluterealpart.
Trajectorieshavespiralbehaviorifthereexistsomecomplex
(obviouslyconjugates)eigenvalues.
Spiralbehavioristoward/outwarddependingonthesign
(negativeandpositive)ofrealpartofthecomplexconjugate
eigenvalues.
Theseconceptscanbevisualizedandbetterunderstoodinthree
dimensionalcase
Phase Plane Analysis of LTI Systems
76
Trajectoriesarealongtheeigenvectorwithminimum
absoluterealpartneartheequilibriumpointandalongthe
eigenvectorwithmaximumabsoluterealpart.
Trajectorieshavespiralbehaviorifthereexistsomecomplex
(obviouslyconjugates)eigenvalues.
Spiralbehavioristoward/outwarddependingonthesign
(negativeandpositive)ofrealpartofthecomplexconjugate
eigenvalues.
Theseconceptscanbevisualizedandbetterunderstoodinthree
dimensionalcase
Advanced Control (Mehdi Keshmiri, Winter 95)
77
LyapunovIndirect Method in
Stability Analysis of
Nonlinear Systems
77
LyapunovIndirect Method in
Stability Analysis of
Nonlinear Systems
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
78
Therearetwoconventionalapproachesinthestabilityanalysis
ofnonlinearsystems:
Lyapunovdirectmethod
Lyapunovindirectmethodorlinearizationapproach
Thedirectmethodanalyzesstabilityofthesystem(equilibrium
point)usingthenonlinearequationsofthesystem
Theindirectmethodanalyzesthesystemstabilityusingthe
linearizedequationsabouttheequilibriumpoint.
Phase Plane Analysis of LTI Systems
78
Therearetwoconventionalapproachesinthestabilityanalysis
ofnonlinearsystems:
Lyapunovdirectmethod
Lyapunovindirectmethodorlinearizationapproach
Thedirectmethodanalyzesstabilityofthesystem(equilibrium
point)usingthenonlinearequationsofthesystem
Theindirectmethodanalyzesthesystemstabilityusingthe
linearizedequationsabouttheequilibriumpoint.
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
79
A nonlinear system near its equilibrium point behaves like a linear:
•Nonlinear system:
•Equilibrium point:
•Motion about equilibrium point:
•Linearized motion:
•It means near �
??????:
Motivation: ()x f x ( ) 0
e
f x x ˆ
e
x x x ˆˆ( ) ( ) ( )
ee
x f x x f x x f x ˆH.O.Tˆˆ
e
x
x
xxx A
f if
ˆ ˆ ˆ()
e
x0
x f x x Ax x x
Phase Plane Analysis of LTI Systems
79
A nonlinear system near its equilibrium point behaves like a linear:
•Nonlinear system:
•Equilibrium point:
•Motion about equilibrium point:
•Linearized motion:
•It means near �
??????:
Motivation: ()x f x ( ) 0
e
f x x ˆ
e
x x x ˆˆ( ) ( ) ( )
ee
x f x x f x x f x ˆH.O.Tˆˆ
e
x
x
xxx A
f if
ˆ ˆ ˆ()
e
x0
x f x x Ax x x
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
80
Thismeansstabilityoftheequilibriumpointmaybestudied
throughthestabilityanalysisofthelinearizedsystem.
ThisisthebaseoftheLyapunovIndirectMethod
Example:inthenonlinearsecondordersystem
originistheequilibriumpointandthelinearizedsystemisgiven
by2
1 2 1 2
2 2 1 1 1 2
cos
(1 ) sin
x x x x
x x x x x x
111 1
20
2 2 1 2
ˆˆˆ ˆ
ˆˆ ˆ ˆ ˆ
e
xxx x
xx x x x
x
f
x
Phase Plane Analysis of LTI Systems
80
Thismeansstabilityoftheequilibriumpointmaybestudied
throughthestabilityanalysisofthelinearizedsystem.
ThisisthebaseoftheLyapunovIndirectMethod
Example:inthenonlinearsecondordersystem
originistheequilibriumpointandthelinearizedsystemisgiven
by2
1 2 1 2
2 2 1 1 1 2
cos
(1 ) sin
x x x x
x x x x x x
111 1
20
2 2 1 2
ˆˆˆ ˆ
ˆˆ ˆ ˆ ˆ
e
xxx x
xx x x x
x
f
x
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
81
Theorem(LyapunovLinearizationMethod):
Ifthelinearizedsystemisstrictlystable(i.e.alleigenvaluesofA
arestrictlyinthelefthalfcomplexplane)thentheequilibrium
pointintheoriginalnonlinearsystemisasymptoticallystable.
Ifthelinearizedsystemisunstable(i.e.inthecaseofrighthalf
planeeigenvalue(s)orrepeatedeigenvaluesontheimaginaryaxis
withgeometricaldeficiency(�>�−����(??????�−�)),thenthe
equilibriumpointintheoriginalnonlinearsystemisunstable.ˆˆ isstrictlystable ( ) isasymptoticallystable x Ax x f x ˆˆ isunstable ( ) isunstable x Ax x f x
Phase Plane Analysis of LTI Systems
81
Theorem(LyapunovLinearizationMethod):
Ifthelinearizedsystemisstrictlystable(i.e.alleigenvaluesofA
arestrictlyinthelefthalfcomplexplane)thentheequilibrium
pointintheoriginalnonlinearsystemisasymptoticallystable.
Ifthelinearizedsystemisunstable(i.e.inthecaseofrighthalf
planeeigenvalue(s)orrepeatedeigenvaluesontheimaginaryaxis
withgeometricaldeficiency(�>�−����(??????�−�)),thenthe
equilibriumpointintheoriginalnonlinearsystemisunstable.ˆˆ isstrictlystable ( ) isasymptoticallystable x Ax x f x ˆˆ isunstable ( ) isunstable x Ax x f x
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
82
Theorem (Lyapunov Linearization Method):
Ifthelinearizedsystemismarginallystable(i.e.alleigenvaluesofAareinthe
lefthalfcomplexplaneandeigenvaluesontheimaginaryaxishaveno
geometricaldeficiency)thenonecannotconcludeanythingfromthelinear
approximation.Theequilibriumpointintheoriginalnonlinearsystemmay
bestable,asymptoticallystable,orunstable.
TheLyapunovlinearizedapproximationmethodonlytalksaboutthelocal
stabilityofthenonlinearsystem,ifanythingcanbeconcluded.( ) isasymptoticallystable
ˆˆ is marginallystable ( ) is marginallystable
( ) is unstable
x f x
x Ax x f x
x f x
Phase Plane Analysis of LTI Systems
82
Theorem (Lyapunov Linearization Method):
Ifthelinearizedsystemismarginallystable(i.e.alleigenvaluesofAareinthe
lefthalfcomplexplaneandeigenvaluesontheimaginaryaxishaveno
geometricaldeficiency)thenonecannotconcludeanythingfromthelinear
approximation.Theequilibriumpointintheoriginalnonlinearsystemmay
bestable,asymptoticallystable,orunstable.
TheLyapunovlinearizedapproximationmethodonlytalksaboutthelocal
stabilityofthenonlinearsystem,ifanythingcanbeconcluded.( ) isasymptoticallystable
ˆˆ is marginallystable ( ) is marginallystable
( ) is unstable
x f x
x Ax x f x
x f x
Advanced Control (Mehdi Keshmiri, Winter 95)
Phase Plane Analysis of LTI Systems
83
The nonlinear system �=��+��
5
is
Asymptotically stable if �<0
Unstable if �>0
No conclusion from linear approximation can be drawn if
The origin in the nonlinear second order system
is unstable because the linearized system is
unstable
Example 15: 2
1 2 1 2
2 2 1 1 1 2
cos
(1 ) sin
x x x x
x x x x x x
1 1
2
2
ˆ ˆ10
ˆ11ˆ
x x
xx
Phase Plane Analysis of LTI Systems
83
The nonlinear system �=��+��
5
is
Asymptotically stable if �<0
Unstable if �>0
No conclusion from linear approximation can be drawn if
The origin in the nonlinear second order system
is unstable because the linearized system is
unstable
Example 15: 2
1 2 1 2
2 2 1 1 1 2
cos
(1 ) sin
x x x x
x x x x x x
1 1
2
2
ˆ ˆ10
ˆ11ˆ
x x
xx
Tags
Categories
GeneralDownload
Download Slideshow Get the original presentation fileQuick Actions
Statistics
- Views 39
- Slides 83
- Age 515 days
Related Slideshows
22
Pray For The Peace Of Jerusalem and You Will Prosper
RodolfoMoralesMarcuc
32 views
26
Don_t_Waste_Your_Life_God.....powerpoint
chalobrido8
35 views
31
VILLASUR_FACTORS_TO_CONSIDER_IN_PLATING_SALAD_10-13.pdf
JaiJai148317
32 views
14
Fertility awareness methods for women in the society
Isaiah47
30 views
35
Chapter 5 Arithmetic Functions Computer Organisation and Architecture
RitikSharma297999
29 views
5
syakira bhasa inggris (1) (1).pptx.......
ourcommunity56
30 views