phase_plane_analysis for analyzing non linear system.pdf

Stability and Phase Plane
Analysis
Advanced Control (Mehdi Keshmiri, Winter 95)
1

Objectives
Objectives of the section:
 Introducing the Phase Plane Analysis
 Introducing the Concept of stability
 Stability Analysis of Linear Time Invariant Systems
 Lyapunov Indirect Method in Stability Analysis of
Nonlinear Systems
2

Introducing
the Phase Plane Analysis
3

Phase Plane Analysis
Phase Space form of a Dynamical System:
𝑥1 = 𝑓1 𝑥1, 𝑥2, … , 𝑥𝑛, 𝑢1, 𝑢2, … , 𝑢𝑚, 𝑡
𝑥2 = 𝑓2(𝑥1, 𝑥2, … , 𝑥𝑛, 𝑢1, 𝑢2, … , 𝑢𝑚, 𝑡)
⋮
𝑥𝑛 = 𝑓𝑛 𝑥1, 𝑥2, … , 𝑥𝑛, 𝑢1, 𝑢2, … , 𝑢𝑚, 𝑡
𝑋 = 𝐹 𝑋, 𝑈, 𝑡
𝑋 ∈ ℝ𝑛 𝑈 ∈ ℝ𝑚
𝑋 = 𝐹 𝑋, 𝑈, 𝑡
𝑼 𝑿
𝑋 = 𝐹 𝑋, 𝑈
𝑼 𝑿
Time-Varying System Time-Invariant System
4

Phase Space form of a Linear Time Invariant (LTI) System:
𝑿 = 𝑨𝑿 + 𝑩𝑼
𝑋 ∈ ℝ𝑛
𝑈 ∈ ℝ𝑚
 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 is a graphical method for studying
second-order systems respect 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
𝑥1 = 𝑓1(𝑥1, 𝑥2)
𝑥2 = 𝑓2(𝑥1, 𝑥2)
6

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

Disadvantage of Phase Plane Method:
 It is restricted to at most second-order
 graphical study of higher-order is computationally and
geometrically complex.
8

Example: First Order LTI System
𝒙 = 𝐬𝐢𝐧(𝒙)
𝑑𝑥
sin(𝑥)
= 𝑑𝑡
Analytical Solution
𝑥0
𝑥
𝑑𝑥
sin(𝑥)
=
0
𝑡
𝑑𝑡
𝑑𝑥
𝑑𝑡
= sin(𝑥)
𝑡 = ln
cos 𝑥0 + cot(𝑥0)
cos 𝑥 + cot(𝑥)
Graphical Solution
-6 -4 -2 0 2 4 6
-1
-0.5
0
0.5
1
x
sin(x)
9

Concept of Phase Plane Analysis:
 Phase plane method is applied to Autonomous Second Order System
 System response 𝑋 𝑡 = (𝑥1 𝑡 , 𝑥2(𝑡)) to initial condition 𝑋0 = 𝑥1 0 , 𝑥2 0
is a mapping from ℝ(Time) to ℝ2(𝑥1, 𝑥2)
 The solution can be plotted in the 𝑥1 − 𝑥2 plane called State Plane or
Phase Plane
 The locus in the 𝑥1 − 𝑥2 plane is a curved named Trajectory that pass through
point 𝑋0
 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 (𝑥, 𝑥) plane
𝑥1 = 𝑓1(𝑥1, 𝑥2) 𝑥2 = 𝑓2(𝑥1, 𝑥2)
10

Example: Van der Pol Oscillator Phase Portrait
x





11

Plotting Phase Plane Diagram:
Analytical Method
Numerical Solution Method
Isocline Method
Vector Field Diagram Method
Delta Method
Lienard’s Method
Pell’s Method
12

Analytical Method
 Dynamic equations of the system is solved, then time parameter is omitted
to obtain relation between two states for various initial conditions
𝑥1 = 𝑓1(𝑥1, 𝑥2)
𝑥2 = 𝑓2(𝑥1, 𝑥2)
Solve 𝑥1 𝑡, 𝑋0 = 𝑔1(𝑡, 𝑋0)
𝑥2 𝑡, 𝑋0 = 𝑔2(𝑡, 𝑋0)
𝐹 𝑥1, 𝑥2 = 0
 For linear or partially linear systems
13

Example: Mass Spring System
𝑚𝑥 + 𝑘𝑥 = 0
For 𝑚 = 𝑘 = 1 : 𝑥 + 𝑥 = 0
𝑥 𝑡 = 𝑥0 cos(𝑡) + 𝑥0 sin(𝑡)
𝑥 𝑡 = −𝑥0 sin 𝑡 + 𝑥0 cos(𝑡)
𝑥2
+ 𝑥2
= 𝑥0
2
+ 𝑥0
2
x
y
x2
+ y2
- 4
-2 -1 0 1 2
-2
-1
0
1
2
14

Analytical Method
 Time differential is omitted from dynamic equations of the system, then
partial differential equation is solved
𝑥1 = 𝑓1(𝑥1, 𝑥2)
𝑥2 = 𝑓2(𝑥1, 𝑥2)
Solve
𝐹 𝑥1, 𝑥2 = 0
𝑑𝑥2
𝑑𝑥1
=
𝑓2(𝑥1, 𝑥2)
𝑓1(𝑥1, 𝑥2)
 For linear or partially linear systems
15

𝑚𝑥 + 𝑘𝑥 = 0
For 𝑚 = 𝑘 = 1 : 𝑥 + 𝑥 = 0
x
y
x2
+ y2
- 4
-2 -1 0 1 2
-2
-1
0
1
2
𝑥1 = 𝑥2
𝑥2 = −𝑥1
𝑑𝑥2
𝑑𝑥1
=
−𝑥1
𝑥2
𝑥20
𝑥2
𝑥2𝑑𝑥2 =
𝑥10
𝑥1
−𝑥1𝑑𝑥1
𝑥2 + 𝑥2 = 𝑥0
2
+ 𝑥0
2
16

Numerical Solution Method
Dynamic equations of the system is solved numerically (e.g. Ode45) for
various initial conditions and time response is obtained, then two states are
plotted in each time.
Example: Pendulum 𝜃 + sin 𝜃 = 0
0 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
x1
x
2
0 2 4 6 8 10
-1
-0.5
0
0.5
1
Time(sec)
x
1
(t)
17

Isocline Method
Isocline: The set of all points which have same trajectory slope
First various isoclines are plotted, then trajectories are drawn.
𝑥1 = 𝑓1(𝑥1, 𝑥2)
𝑥2 = 𝑓2(𝑥1, 𝑥2)
𝑑𝑥2
𝑑𝑥1
=
𝑓2(𝑥1, 𝑥2)
𝑓1(𝑥1, 𝑥2)
= 𝛼 𝑓2 𝑥1, 𝑥2 = 𝛼𝑓1(𝑥1, 𝑥2)
18

𝑚𝑥 + 𝑘𝑥 = 0
For 𝑚 = 𝑘 = 1 : 𝑥 + 𝑥 = 0
𝑥1 = 𝑥2
𝑥2 = −𝑥1
𝑑𝑥2
𝑑𝑥1
=
−𝑥1
𝑥2
= 𝛼
𝑥1 + 𝛼𝑥2 = 0
x
y
x2
+ y2
- 4
-5 0 5
-5
0
5
Slope=1
Slope=infinite
Slope=-1
19

Vector Field Diagram Method
Vector Field: A set of vectors that is tangent to the trajectory
 At each point (𝑥1, 𝑥2) vector
𝑓1(𝑥1, 𝑥2)
𝑓2(𝑥1, 𝑥2)
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
1
2
sin( ) 0
sin( )
x x
x
x
 


 
   

 
  

   

f
20

 Singular point is an important concept which reveals great info about
properties of system such as stability.
 Singular points are only points which several trajectories pass/approach
them (i.e. trajectories intersect).
Singular Points in the Phase Plane Diagram:
Equilibrium points are in fact singular points in the phase plane
diagram
21
𝑓1 𝑥1, 𝑥2 = 0
𝑓1 𝑥1, 𝑥2 = 0
Slope of the trajectories at
equilibrium points
𝑑𝑥2
𝑑𝑥1
=
0
0

22
Example: Using Matlab
x ' = y
y ' = - 0.6 y - 3 x + x2
-6 -4 -2 0 2 4 6
-10
-5
0
5
10
x
y
𝑥 + 0.6𝑥 + 3𝑥 + 𝑥2 = 0

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);
𝑥 + 0.6𝑥 + 3𝑥 + 𝑥2
= 0

24
Example: Using Maple Tools
𝑥 + 0.6𝑥 + 3𝑥 + 𝑥2
= 0

25
Phase Plane Analysis for Single DOF Mechanical System
In the case of single DOF mechanical system
𝑥 + 𝑔 𝑥, 𝑥 = 0
𝑥1 = 𝑥
𝑥2 = 𝑥
𝑥1 = 𝑥2
𝑥2 = −𝑔(𝑥1, 𝑥2)
 The phase plane is in fact (𝑥 − 𝑥) plane and every point shows the
position and velocity of the system.
 Trajectories are always clockwise. This is not true in the general phase
plane (𝑥1 − 𝑥2)

26
Introducing
the Concept of Stability

Stability, Definitions and Examples
27
 Stability analysis of a dynamic system is normally introduced in the state
space form of the equations.
𝑋 = 𝐹 𝑋, 𝑈, 𝑡
𝑋 ∈ ℝ𝑛 𝑈 ∈ ℝ𝑚
 Most of the concepts in this chapter are introduced for autonomous
systems
u x
x = f(x,u)
Autonomous Dynamic System
u x
,t
x = f(x,u )
Dynamic System

28
 Our main concern is the first type analysis. Some preliminary issues of the
second type analysis will be also discussed.
Stability analysis of a dynamic system is divided in three
categories:
1. Stability analysis of the equilibrium points of the systems. We study the
behavior (dynamics) of the free (unforced, 𝑢 = 0) system when it is perturbed
from its equilibrium point.
2. Input-output stability analysis. We study the system (forced system 𝑢 ≠ 0)
output behavior in response to bounded inputs.
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.

29
Reminder:
𝑋𝑒 is said an equilibrium point of the system if once the system reaches this
position it stays there for ever, i.e. 𝑓 𝑋𝑒 = 0
Definition (Lyapunov Stability):
The equilibrium point 𝑋𝑒 is said to be stable (in the sense of Lyapunov
stability) or motion of the system about its equilibrium point is said to be
stable if the system states (𝑋) is perturbed away from 𝑋𝑒 then it stays close to
𝑋𝑒. Mathematically 𝑋𝑒 is stable if
0
)
(
,
0 


 


 (0) ( ) 0
e e
t t
 
      
x x x x

30
 Without loss of generality we can present our analysis about equilibrium
point 𝑋𝑒 = 0, since the system equation can be transferred to a new form
with zero as the equilibrium point of the system.
𝑋 = 𝑌 − 𝑌𝑒
𝑌 = 𝐹 𝑌
𝑌𝑒 ≠ 0
𝑋 = 𝐹 𝑋
𝑋𝑒 ≠ 0
The equilibrium point 𝑋𝑒 is said to be stable (in the sense of Lyapunov
stability) or motion of the system about its equilibrium point is said to be stable
if for any 𝑅 > 0, there exists 0 < 𝑟 < 𝑅 such that
A more precise definition:
(0) ( ) 0
e e
r t R t
      
x x x x

31
The equilibrium point 𝑋𝑒 = 0 is said to be
Definition (Lyapunov Stability):
Stable if
∀𝑅 > 0 ∃0 < 𝑟 < 𝑅 𝑠. 𝑡.
𝑋 0 < 𝑟 𝑋 𝑡 < 𝑅 ∀𝑡 > 0
Unstable if it is not stable.
Asymptotically stable if it is stable and
∀𝑟 > 0 𝑠. 𝑡.
𝑋 0 < 𝑟 lim
𝑡→∞
𝑋(𝑡) = 0
Marginally stable if it is stable and not
asymptotically stable
Exponentially stable if it is asymptotically stable with an exponential rate
𝑋(0) < 𝑟 𝑋(𝑡) < 𝛼𝑒−𝛽𝑡 𝑋(0) 𝛼, 𝛽 > 0

32
Example: Undamped Pendulum
𝜃 +
𝑔
𝑙
sin(𝜃) = 0 𝜃𝑒1 = 0 , 𝜃𝑒2 = 𝜋

 
R
B
r
B
 𝜃𝑒1 is a marginally stable point and 𝜃𝑒2 is an unstable point
0 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)

33
Example: damped Pendulum
𝜃 + 𝐶𝜃 +
𝑔
𝑙
sin(𝜃) = 0 𝜃𝑒1 = 0 , 𝜃𝑒2 = 𝜋
 𝜃𝑒1is an exponentially stable point and 𝜃𝑒2 is an unstable point
0 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)

34
stateform
1 2
2
1 2
2
2 1 1 2
(1 ) 0 0
(1 ) e e
x x
x x x x x x
x x x x


      

   


Example: Van Der Pol Oscillator
 𝑥𝑒 = 0 is an unstable point

35
Definition:
if the equil. point 𝑋𝑒 is asymptotically stable, then the set of all
points that trajectories initiated at these point eventually converge
to the origin is called domain of attraction.
Definition:
if the equil. point 𝑋𝑒 is asymptotically/exponentially stable, then
the equil. point is called globally stable if the whole space is
domain of attraction. Otherwise it is called locally stable.

36
The origin in the first order system of 𝑥 = −𝑥 is globally exponentially stable.
Example 1:
0 0
( ) lim ( ) 0 0
t
t
x x x t x e x t x


       
The origin in the first order system 𝑥 = −𝑥3
is globally asymptotically but
not exponentially stable.
Example 2:
3 0
0
2
0
( ) lim ( ) 0
1 2 t
x
x x x t x t x
tx 
      


37
The origin in the first order system 𝑥 = −𝑥2 is semi-asymptotically but not
exponentially stable.
Example 3:
0
0
2 0
0
0 1/
lim ( ) 0 if 0
( )
lim ( ) if 0
1
t
t x
x t x
x
x x x t
x t x
tx


 


     
  
 

Domain of attraction is 𝑥0 > 0.

38
Example 4:
𝑥 + 𝑥 + 𝑥 = 0

39
Example 5:
𝑥 + 0.6𝑥 + 3𝑥 + 𝑥2
= 0

40
Example 6:
𝑥 + 𝑥 + 𝑥3 − 𝑥 = 0

41
Stability Analysis of
Linear Time Invariant Systems

Phase Plane Analysis of LTI Systems
42
It is the best tool for study of the linear system graphically
This analysis gives a very good insight of linear systems
behavior
The analysis can be extended for higher order linear system
Local behavior of the nonlinear systems can be understood from
this analysis
The analysis is performed based on the system eigenvalues and
eigenvectors.

43
Consider a second order linear system:
 If the A matrix is nonsingular, origin is the only equilibrium
point of the system
 If the A matrix is singular then the system has infinite number
of equilibrium points. In fact all of the points belonging to the
null space of A are the equilibrium point of the system.
isnon-singular e
 
A x 0
 
* *
issingular | ( )
e Null
  
A x x x A
𝑥 = 𝐴𝑥 𝐴 ∈ ℝ2×2
, 𝑥 ∈ ℝ2

44
Consider a second order linear system:
𝑥 = 𝐴𝑥
𝐴 ∈ ℝ2×2 , 𝑥 ∈ ℝ2
The analytical solution can be obtained based on eigenvalues (𝜆1, 𝜆2):
If 𝜆1, 𝜆2 are real and distinct
If 𝜆1, 𝜆2 are real and similar
If 𝜆1, 𝜆2 are complex conjugate
𝑥 𝑡 = 𝐴𝑒𝜆1𝑡 + 𝐵𝑒𝜆2𝑡
𝑥 𝑡 = (𝐴 + 𝐵𝑡)𝑒𝜆𝑡
𝑥 𝑡 = 𝐴𝑒𝜆1𝑡
+ 𝐵𝑒𝜆2𝑡
= 𝑒𝛼𝑡(𝐴 sin 𝛽𝑡 + 𝐵 cos(𝛽𝑡))

45
Jordan Form (almost diagonal form)
This representation has the system eigenvalues on the leading diagonal, and
either 0 or 1 on the super diagonal.
𝑥 = 𝐴𝑥 𝑦 = 𝐽𝑦
𝐽 =
𝐽1
⋱
𝐽𝑛
𝐽𝑖 =
𝜆𝑖 1
⋱ 1
𝜆𝑖
Obtaining Jordan form:
𝑦 = 𝑃−1𝑥
𝑃 = 𝑣1 … 𝑣𝑛
𝐽 = 𝑃−1
𝐴𝑃

46
The A matrix has 2 eigenvalues (either two real, or two complex
conjugates) and can have either two eigenvectors or one
eigenvectors. Four categories can be realized
1. Two distinct real eigenvalues and two real eigenvectors
2. Two complex conjugate eigenvalues and two complex
eigenvectors
3. Two similar (real) eigenvalues and two eigenvectors
4. Two similar (real) eigenvalues and one eigenvectors
A is non-singular:

47
1. Two distinct real eigenvalues and two real eigenvectors
𝑦 = 𝐽𝑦 𝐽 =
𝜆1 0
0 𝜆2
𝑦1 = 𝜆1𝑦1
𝑦2 = 𝜆2𝑦2
𝑦1 = 𝑦10𝑒𝜆1𝑡
𝑦2 = 𝑦20𝑒𝜆2𝑡
ln
𝑦1
𝑦10
= 𝜆1𝑡
ln
𝑦2
𝑦20
= 𝜆2𝑡
𝜆2
𝜆1
ln
𝑦1
𝑦10
= ln
𝑦2
𝑦20
𝑦1
𝑦10
𝜆2
𝜆1
=
𝑦2
𝑦20
𝑦2 =
𝑦20
𝑦10
𝜆2/𝜆1
𝑦1
𝜆2/𝜆1
𝑦2 = 𝐾𝑦1
𝜆2/𝜆1

48
1. A ) 𝝀𝟐 < 𝝀𝟏 < 𝟎 𝑦2 = 𝐾𝑦1
𝜆2/𝜆1
 System has two eigenvectors 𝑣1, 𝑣2 the phase plane portrait is as the
following
 Trajectories are:
 tangent to the slow eigenvector (𝑣1) for near the origin
 parallel to the fast eigenvector (𝑣2) for far from the origin
 The equilibrium point 𝑋𝑒 = 0 is called stable node
x ' = - 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

49
Example 7: 𝑥 + 4𝑥 + 2𝑥 = 0
𝑋 =
0 1
−2 −4
𝑋
det 𝜆𝐼 −
0 1
−2 −4
= 𝜆2 + 4𝜆 + 2 = 0 𝜆1 = −0.59 , 𝜆2 = −3.41
−0.59 −1
2 −0.59 + 4
𝑣1
1
𝑣1
2 = 0
𝑣1 =
0.86
−0.51
−3.41 −1
2 −3.41 + 4
𝑣1
1
𝑣1
2 = 0
𝑣2 =
−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]

50
1. B ) 𝝀𝟐 > 𝝀𝟏 > 𝟎 𝑦2 = 𝐾𝑦1
𝜆2/𝜆1
 System has two eigenvectors 𝑣1 and 𝑣2 the phase plane portrait is
opposite as the previous one
 Trajectories are:
 tangent to the slow eigenvector 𝑣1 for near origin
 parallel to the fast eigenvector 𝑣2 for far from origin
 The equilibrium point 𝑋𝑒 = 0 is called unstable node
x ' = - 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

51
Example 8: 𝑥 − 3𝑥 + 2𝑥 = 0
𝑋 =
0 1
−2 3
𝑋
det 𝜆𝐼 −
0 1
−2 3
= 𝜆2 − 3𝜆 + 2 = 0 𝜆1 = 1 , 𝜆2 = 2
1 −1
2 1 − 3
𝑣1
1
𝑣1
2 = 0
𝑣1 =
−0.71
−0.71
2 −1
2 2 − 3
𝑣1
1
𝑣1
2 = 0
𝑣2 =
−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]

52
1. C ) 𝝀𝟐 < 𝟎 < 𝝀𝟏 𝑦2 = 𝐾𝑦1
𝜆2/𝜆1
 System has two eigenvectors 𝑣1 and 𝑣2, the phase plane portrait is as the
following
 Only trajectories along 𝑣2 are stable trajectories
 All other trajectories at start are tangent to 𝑣2 and at the end are tangent
to 𝑣1
 This equilibrium point is unstable and is called saddle point
x ' = 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

53
Example 9: 𝑥 − 𝑥 − 2𝑥 = 0
𝑋 =
0 1
2 1
𝑋
det 𝜆𝐼 −
0 1
2 1
= 𝜆2 − 𝜆 − 2 = 0 𝜆1 = −1 , 𝜆2 = 2
−1 −1
−2 −1 − 1
𝑣1
1
𝑣1
2 = 0
𝑣1 =
−0.71
0.71
2 −1
−2 2 − 1
𝑣1
1
𝑣1
2 = 0
𝑣2 =
−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]

54
2. Two complex conjugate eigenvalues and two complex eigenvectors
𝛼 −𝛽
𝛽 𝛼
𝑟 ≡ 𝑦1
2
+ 𝑦2
2
𝜃 ≡ tan−1(
𝑦2
𝑦1
)
𝑟𝑟 = 𝑦1𝑦1 + 𝑦2𝑦2 = 𝑦1 𝛼𝑦1 − 𝛽𝑦2 + 𝑦2 𝛽𝑦1 + 𝛼𝑦2 = 𝛼𝑟2
𝜃 1 + tan 𝜃2
=
𝑦1𝑦2 − 𝑦2𝑦1
𝑦1
2 =
𝑦1 𝛽𝑦1 + 𝛼𝑦2 − 𝑦2 𝛼𝑦1 − 𝛽𝑦2
𝑦1
2 = 𝛽 1 + tan 𝜃2
𝑟 = 𝛼𝑟
𝜃 = 𝛽
𝑟 𝑡 = 𝑟0𝑒𝛼𝑡 𝜃 𝑡 = 𝜃0 + 𝛽𝑡

55
2. A ) 𝝀𝟐, 𝝀𝟏 = 𝜶 ± 𝜷𝒊 , 𝜶 < 𝟎 , 𝜷 ≠ 𝟎
 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.
𝑟 𝑡 = 𝑟0𝑒𝛼𝑡
𝜃 𝑡 = 𝜃0 + 𝛽𝑡
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

56
Example 10: 𝑥 + 𝑥 + 𝑥 = 0
𝑋 =
0 1
−1 −1
𝑋
det 𝜆𝐼 −
0 1
−1 −1
= 𝜆2 + 𝜆 + 1 = 0 𝜆1, 𝜆2 = −0.5 ± 0.866𝑖
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]

57
2. B ) 𝝀𝟐, 𝝀𝟏 = 𝜶 ± 𝜷𝒊 , 𝜶 > 𝟎 , 𝜷 ≠ 𝟎
 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.
𝜃 𝑡 = 𝜃0 + 𝛽𝑡
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

58
Example 11: 𝑥 − 𝑥 + 𝑥 = 0
𝑋 =
0 1
−1 1
𝑋
det 𝜆𝐼 −
0 1
−1 1
= 𝜆2 − 𝜆 + 1 = 0 𝜆1, 𝜆2 = 0.5 ± 0.866𝑖
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]

59
2. C ) 𝝀𝟐, 𝝀𝟏 = ±𝜷𝒊 , 𝜶 = 𝟎 , 𝜷 ≠ 𝟎
 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.
𝜃 𝑡 = 𝜃0 + 𝛽𝑡
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

60
Example 12: 𝑥 + 3𝑥 = 0
𝑋 =
0 1
−3 0
𝑋
det 𝜆𝐼 −
0 1
−3 0
= 𝜆2 + 3 = 0 𝜆1, 𝜆2 = ±1.732𝑖
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]

61
3. Two similar (real) eigenvalues and two eigenvectors
𝜆 0
0 𝜆
𝑦1 = 𝜆𝑦1
𝑦2 = 𝜆𝑦2
𝑦1 = 𝑦10𝑒𝜆𝑡
𝑦1
𝑦2
=
𝑦10
𝑦20
𝑦2 = 𝐾𝑦1

62
x ' = 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
𝜆 > 0
x ' = - 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
𝜆 < 0
3 ) 𝝀𝟐 = 𝝀𝟏 = 𝝀 ≠ 𝟎
 System has two similar eigenvalues and two different eigenvectors. The
phase plane portrait is as the following, depending to the sign of 𝜆
 The trajectories are all along the initial conditions and they are 𝜆 < 0
toward 𝜆 > 0 or outward the origin

63
4. Two similar (real) eigenvalues and One eigenvectors
𝜆 1
0 𝜆
𝑦1 = 𝜆𝑦1 + 𝑦2
𝑦2 = 𝜆𝑦2
𝑦1 = 𝑦10𝑒𝜆𝑡 + 𝑦20𝑡𝑒𝜆𝑡
𝑦1 = 𝑦10
𝑦2
𝑦20
+ 𝑦2
1
𝜆
ln(
𝑦2
𝑦20
)
𝑦1 = 𝑦2(
𝑦10
𝑦20
+
1
𝜆
ln
𝑦2
𝑦20
)

64
𝜆 > 0
𝜆 < 0
4 ) 𝝀𝟐 = 𝝀𝟏 = 𝝀 ≠ 𝟎
 System has two similar eigenvalues and only one eigenvector. The phase
plane portrait is as the following, depending to the sign of 𝜆
 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

65
Example 13: 𝑥 + 2𝑥 + 𝑥 = 0
𝑋 =
0 1
−1 −2
𝑋
det 𝜆𝐼 −
0 1
−1 −2
= 𝜆2 + 2𝜆 + 1 = 0 𝜆1, 𝜆2 = −1
−1 −1
1 −1 + 2
𝑣1
1
𝑣1
2 = 0
𝑣1 =
−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]

66
A is singular (𝒅𝒆𝒕 𝑨 = 𝟎):
 System has at least one eigenvalue equal to zero and therefore infinite
number of equilibrium points. Three different categories can be
specified
 𝜆1 = 0 , 𝜆2 ≠ 0
 𝜆1, 𝜆1 = 0 , 𝑅𝑎𝑛𝑘 𝐴 = 1
 𝜆1, 𝜆1 = 0 , 𝑅𝑎𝑛𝑘 𝐴 = 0

67
1) 𝜆1 = 0 , 𝜆2 ≠ 0 𝑦 = 𝐽𝑦
𝑦1 = 0
𝑦2 = 𝜆𝑦2
𝑦1 = 𝑦10
𝐽 =
0 0
0 𝜆
 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 𝑣1 along 𝑣2
x ' = 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

68
Example 14: 𝑥 + 𝑥 = 0
𝑋 =
0 1
0 −1
𝑋
det 𝜆𝐼 −
0 1
0 −1
= 𝜆2 + 𝜆 = 0 𝜆1 = 0 , 𝜆2 = −1
0 −1
0 1
𝑣1
1
𝑣1
2 = 0
𝑣1 =
1
0
−1 −1
0 −1 + 1
𝑣1
1
𝑣1
2 = 0
𝑣1 =
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]

69
2) 𝜆1, 𝜆1 = 0 , 𝑅𝑎𝑛𝑘 𝐴 = 1 𝑦 = 𝐽𝑦
𝑦1 = 𝑦2
𝑦2 = 0
𝑦1 = 𝑦20𝑡 + 𝑦10
𝑦2 = 𝑦20
𝐽 =
0 1
0 0
 System has infinite number of non-isolated equilibrium points along a line
 System has only one eigenvector, and it is loci of the equilibrium points
 All the trajectories move toward infinity along the system eigenvector
(unstable system).

70
2) 𝜆1, 𝜆1 = 0 , 𝑅𝑎𝑛𝑘 𝐴 = 0
𝑦 = 𝐽𝑦
𝑦1 = 0
𝑦2 = 0
𝑦1 = 𝑦10
𝑦2 = 𝑦20
𝐽 =
0 0
0 0
 System is a static system. All the points are equilibrium points

71
Summary
Six different type of isolated equilibrium points can be identified
Stable/unstable node
Saddle point
Stable/ unstable focus
Center

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 A matrix.

73
Consider a linear time invariant (LTI) system
Origin is the only equilibrium point of the system if A is non-
singular
Otherwise the system has infinite number of equilibrium points,
all the points on null-space of A are in fact equil. points of the
system.

 
x = Ax Bu
y Cx Du
det( ) 0
e



A
x = Ax x 0
 
det( ) 0
* *
| Nullspace( )
e

 

A
x = Ax x x x A

74
Details for Case of Non-Singular A
 Origin is the only equilibrium point of the system
 This equilibrium point (system) is
 Exponentially stable if all eigenvalues of A are either real
negative or complex with negative real part.
 Marginally stable if eigenvalues of A have non-positive real
part and 𝑟𝑎𝑛𝑘 𝐴 − 𝜆𝐼 = 𝑛 − 𝑟 for all repeated imaginary
eigenvalues, 𝜆 with multiplicity of r
 Unstable, otherwise.

75
A is Non-Singular
 Classification of the equilibrium point of higher order system
into node, focus, and saddle point is not as easy as second order
system. However some points can be emphasized:
 The equilibrium point is stable/unstable node if all
eigenvalues are real and have the negative/positive sign.
 The equilibrium point is center if a pair of eigenvalues are
pure imaginary complex conjugate and all other eigenvalues
have negative real
 In the case of different sign in the real part of the
eigenvalues trajectories have the saddle type behavior near
the equilibrium point

76
Trajectories are along the eigenvector with minimum
absolute real part near the equilibrium point and along the
eigenvector with maximum absolute real part.
Trajectories have spiral behavior if there exist some complex
(obviously conjugates) eigenvalues.
Spiral behavior is toward/outward depending on the sign
(negative and positive) of real part of the complex conjugate
eigenvalues.
These concepts can be visualized and better understood in three
dimensional case

77
Lyapunov Indirect Method in
Stability Analysis of
Nonlinear Systems

78
 There are two conventional approaches in the stability analysis
of nonlinear systems:
 Lyapunov direct method
 Lyapunov indirect method or linearization approach
 The direct method analyzes stability of the system (equilibrium
point) using the nonlinear equations of the system
 The indirect method analyzes the system stability using the
linearized equations about the equilibrium point.

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 𝑥𝑒 :
Motivation:
( )

x f x
( ) 0 e
 
f x x
ˆ
e
 
x x x
ˆ ˆ
( ) ( ) ( )
e e
    
x f x x f x x f x ˆ H.O.T ˆ ˆ
e

  


x
x
x x
x A
f
if
ˆ ˆ ˆ
( )
e 
  

x 0
x f x x Ax x x

80
This means stability of the equilibrium point may be studied
through the stability analysis of the linearized system.
This is the base of the Lyapunov Indirect Method
Example: in the nonlinear second order system
origin is the equilibrium point and the linearized system is given
by
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
1 1
2
0
2 2 1 2
ˆ ˆ
ˆ ˆ
ˆ
ˆ ˆ ˆ ˆ
e
x x
x x
x
x x x x


  
 

 
  
 
    
 

   
 
  
  
x
f
x

81
Theorem (Lyapunov Linearization Method):
 If the linearized system is strictly stable (i.e. all eigenvalues of A
are strictly in the left half complex plane ) then the equilibrium
point in the original nonlinear system is asymptotically stable.
 If the linearized system is unstable (i.e. in the case of right half
plane eigenvalue(s) or repeated eigenvalues on the imaginary axis
with geometrical deficiency (𝑟 > 𝑛 − 𝑟𝑎𝑛𝑘(𝜆𝐼 − 𝐴)), then the
equilibrium point in the original nonlinear system is unstable.
ˆ ˆ isstrictlystable ( ) isasymptoticallystable
  
x Ax x f x
ˆ ˆ isunstable ( ) isunstable
  
x Ax x f x

82
Theorem (Lyapunov Linearization Method):
 If the linearized system is marginally stable (i.e. all eigenvalues of A are in the
left half complex plane and eigenvalues on the imaginary axis have no
geometrical deficiency) then one cannot conclude anything from the linear
approximation. The equilibrium point in the original nonlinear system may
be stable, asymptotically stable, or unstable.
 The Lyapunov linearized approximation method only talks about the local
stability of the nonlinear system, if anything can be concluded.
( ) isasymptoticallystable
ˆ ˆ ismarginallystable ( ) ismarginallystable
( ) isunstable



  

 

x f x
x Ax x f x
x f x

83
 The nonlinear system 𝑥 = 𝑎𝑥 + 𝑏𝑥5 is
 Asymptotically stable if 𝑎 < 0
 Unstable if 𝑎 > 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 ˆ
1 0
ˆ
1 1
ˆ
x x
x
x
   
 
 

   
 
   
 
 

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