1. Modern Control Systems (MCS)
Lecture-11
Phase Plane Analysis
Dr. Imtiaz Hussain
Assistant Professor
URL :

2. Introduction
The definition for stability for LTI systems is an easy concept to understand (eigenvalues).
It is necessary to extend the concept of stability to nonlinear systems.

3. Basic Idea
To generate motion trajectories corresponding to various initial conditions in the phase plane.
To examine the qualitative features of the trajectories.
In such a way, information concerning stability and other motion patterns of the system can be obtained.

4. State Space Trajectories
The unforced response of a system released from any initial point x(to) traces a curve or trajectory in state space, with time t as an implicit function along the trajectory.
When state variables are represented as phase variables, the state space is called phase space or phase plane.
The family of all trajectories (which started by different initial points) is called phase portrait. 

5. Phase Plane Analysis
A graphical method: to visualize what goes on in a nonlinear system without solving the nonlinear equations analytically.
2. Limitation: limited for second-order (or first –order) dynamic system; however, some practical control systems can be approximated as second-order systems.

6. How to Plot?
There are a number of methods for constructing phase plane trajectories for linear or nonlinear system, such as
Solution of state equations
analytical method
The method of isoclines
The delta method
Lienard’s method
Pell’s method
Software Programs (based on above six methods)

7. Phase Plane Analysis of Linear Systems

8. Consider the second-order linear system
is the general solution
Where the eigenvalues and are the solutions of the characteristic equation

9. There is only one singular point (assuming
), namely the origin.
and are both real and have the same sign (positive or negative)
and are both real and have opposite signs (saddle point)
and are complex conjugate with non-zero real parts
and are complex conjugates with real parts equal to zero (center point)

15. Phase Portrait of LTI Systems
The system that is studied has no forcing function and is represented by the state equation
Let us consider a system with the given initial condition
The system’s response is overdamped with eigenvalues at -1 and -3.
Because one state equation is 𝑥 1 =3 𝑥 2 , then 𝑥 2 is a phase variable, therefore resulting state space is called phase plane.
𝑥 = 𝑎 11 𝑎 12 𝑎 21 𝑎 22 𝑥
𝑥 = 0 3 −1 −4 𝑥
𝑥 0 = 0 2

16. Phase Portrait (Solution of state equations)
The state transition matrix ∅(𝑡) is used to obtain the expressions for state space trajectories.
State transition matrix for LTI system can be obtained as
Then the solution of state equation is obtained as
𝑥 = 0 3 −1 −4 𝑥
𝑥 0 = 0 2
∅ 𝑡 = L −1 𝑆𝐼−𝐴 −1
𝑥 𝑡 =∅ 𝑡 𝑥(0)
𝑥 𝑡 = 3 𝑒 −𝑡 −3 𝑒 −3𝑡 − 𝑒 −𝑡 +3 𝑒 −3𝑡

17. Phase Portrait (Solution of state equations)
x1=3*exp(-t)-3*exp(-3*t);
x2=-exp(-t)+3*exp(-3*t);
plot(x1,x2)
𝑥 𝑡 = 3 𝑒 −𝑡 −3 𝑒 −3𝑡 − 𝑒 −𝑡 +3 𝑒 −3𝑡

18. Example-RLC Circuit
For the RLC circuit draw the state space trajectory with following initial conditions.
Solution Vc + -

19. Example-RLC Circuit
Following trajectory is obtained

20. Example-RLC Circuit

21. Equilibrium Point and Singular Points
Equilibrium point is defined as a point where the system states can stay forever
This implies that
A singular point is an equilibrium point in the phase plane.

22. Equilibrium Point
The equilibrium or stationary state of the system is when

23. Examples of LTI systems
𝑥 𝑥 2 = 0 1 −1 −4 𝑥 1 𝑥 (overdamped Stable System, [stable node])
𝑥 𝑥 2 = −5 𝑥 1 𝑥 (overdamped unstable [saddle point]
𝑥 𝑥 2 = 0 1 −4 −1 𝑥 1 𝑥 (underdamped stable System, stable focus)
𝑥 𝑥 2 = 0 1 − 𝑥 1 𝑥 (underdamped Unstable System, unstable focus)
𝑥 𝑥 2 = 0 1 − 𝑥 1 𝑥 (overdamped Unstable System, [unstable node])

24. Software Solution 𝑥 1 𝑥 2 = 0 1 −1 −4 𝑥 1 𝑥 2
𝑥 𝑥 2 = 0 1 −1 −4 𝑥 1 𝑥 2
x1dom = linspace(-5,5,51);
x2dom = linspace(-5,5,51);
[x1,x2] = meshgrid(xdom,ydom); % generate mesh of domain
x1dot = x1; % dx1/dt
X2dot= = -x1-4*x2; % dx2/dt
quiver(x1,x2,x1dot,x2dot) % velocity Vectors

25. Enter Equations in PPlane Applet

26. Phase Plane Analysis of Nonlinear Systems

27. Introduction
In discussing the phase plane analysis of nonlinear systems, two points should be kept in mind:
Phase plane analysis of nonlinear systems is related to that of liner systems, because the local behavior of nonlinear systems can be approximated by the behavior of a linear system.
Nonlinear systems can display much more complicated patterns in the phase plane, such as multiple equilibrium points and limit cycles.

28. Example 𝑟 =𝑟(1−𝑟) 𝜃 = sin2 ( 𝜃 2 ) 𝑟= 𝑥 1 2 + 𝑥 2 2
Consider the second-order system with state variables 𝑥 1 and 𝑥 2 whose dynamics are most easily described in polar coordinates via the equations
Where the radius r is given as
And the angle θ is
𝑟 =𝑟(1−𝑟)
𝜃 = sin2 ( 𝜃 2 )
𝑟= 𝑥 𝑥 2 2
0≤𝜃= tan −1 𝑥 2 𝑥 1 <2𝜋

29. Equilibrium States 𝑟= 𝑥 1 2 + 𝑥 2 2 𝑟 =𝑟(1−𝑟) 𝜃 = sin2 ( 𝜃 2 )
It is easy to see that there are precisely two equilibrium points: one at the origin, and the other at r = 1, 𝜃 = 0.
𝑟= 𝑥 𝑥 2 2
𝑟 =𝑟(1−𝑟)
𝜃 = sin2 ( 𝜃 2 )
0≤𝜃= tan −1 𝑥 2 𝑥 1 <2𝜋

30. Example: Simple Pendulum
Consider the pendulum shown in figure.
𝑀 𝐿 2 𝜃 +𝑏 𝜃 +𝑀𝑔𝐿 sin 𝜃 =0
Where,
L: the pendulum’s length M: its mass b: the friction coefficient at the hinge g: the gravity constant.

31. Example: Simple Pendulum
Letting 𝑥 1 =𝜃 and 𝑥 2 = 𝜃
𝑀 𝐿 2 𝜃 +𝑏 𝜃 +𝑀𝑔𝐿 sin 𝜃 =0
State equation of the system are given as

32. The equilibrium points:
x2 = 0, sin x1 = 0
(0, 0) and (π, 0)
The pendulum resting exactly at the vertical up and down positions.

33. Example
Consider a nonlinear system described by following state equations determine the number of equilibrium point(s) of the system.
𝑥 =𝑥−𝑦
𝑦 = 𝑥 2 + 𝑦 2 −2
Answer
(1, 1) and (−1, −1).

34. Example
Consider a nonlinear system described by following state equations determine the number of equilibrium point(s) of the system.
𝑥 =𝑥−𝑥𝑦
𝑦 =𝑦+2𝑥𝑦
Answer
(0, 0) and (−1/2, 1).

35. Local Behavior of Nonlinear Systems
If the singular point of interest is not at the origin, by defining the difference between the original state and the singular point as a new set of state variables, we can shift the singular point to the origin.
Using Taylor expansion a second order nonlinear system can be rewritten in the form
𝑥 1 =𝑎 𝑥 1 + 𝑏𝑥 2 +𝑔( 𝑥 1 , 𝑥 2 )
𝑥 2 =𝑐 𝑥 1 + 𝑑𝑥 2 +ℎ( 𝑥 1 , 𝑥 2 )

36. Local Behavior of Nonlinear Systems
In the vicinity of the origin, the higher order terms can be neglected, and therefore, the nonlinear system trajectories essentially satisfy the linearized equation
As a result, the local behavior of the nonlinear system can be approximated by the patterns shown by linear systems
𝑥 1 =𝑎 𝑥 1 + 𝑏𝑥 2
𝑥 2 =𝑐 𝑥 1 + 𝑑𝑥 2

37. Limit Cycle
In the phase plane, a limit cycle is defied as an isolated closed curve.
The trajectory has to be both closed, indicating the periodic nature of the motion, and isolated, indicating the limiting nature of the cycle (with near by trajectories converging or diverging from it).

38. Limit Cycle
Depending on the motion patterns of the trajectories in the vicinity of the limit cycle, we can distinguish three kinds of limit cycles.
Stable Limit Cycles: all trajectories in the vicinity of the limit cycle converge to it as t →∞
Unstable Limit Cycles: all trajectories in the vicinity of the limit cycle diverge to it as t →∞

39. Limit Cycle
Semi-Stable Limit Cycles: some of the trajectories in the vicinity of the limit cycle converge to it as t →∞

40. Example
Find the equilibrium point(s) of the nonlinear system. Then determine the type and stability of each equilibrium point.
If system exhibits limit cycles then determine the nature of limit cycle as well.
𝑥 =𝑦−𝑥( 𝑥 2 + 𝑦 2 −1)
𝑦 =−𝑥−𝑦( 𝑥 2 + 𝑦 2 −1)

41. Example
Find the equilibrium point(s) of each nonlinear system given below. Then determine the type and stability of each equilibrium point.
If system exhibits limit cycles then determine the nature of limit cycle as well.
𝑥 =𝑦+𝑥( 𝑥 2 + 𝑦 2 −1)
𝑦 =−𝑥+𝑦( 𝑥 2 + 𝑦 2 −1)
𝑥 =𝑦−𝑥 ( 𝑥 2 + 𝑦 2 −1) 2
𝑦 =−𝑥−𝑦 ( 𝑥 2 + 𝑦 2 −1) 2
1).
2).

42. Exercise
Find the equilibrium point(s) of each nonlinear system given below. Then determine the type and stability of each equilibrium point. If system exhibits limit cycles then determine the nature of limit cycle as well.
𝑥 = 𝑥 2 𝑦+3𝑥𝑦−10𝑦
𝑦 =𝑥𝑦−4𝑥
4).
𝑥 =𝑥𝑦−3𝑦
𝑦 =𝑥𝑦−3𝑥
1).
𝑥 =2− 𝑥 2 − 𝑦 2
𝑦 = 𝑥 2 − 𝑦 2
5).
𝑥 = 𝑥 2 −3𝑥𝑦+2𝑥
𝑦 =𝑥+𝑦−1
2).
𝑥 = 𝑥 2 + 𝑦 2 −13
𝑦 =𝑥𝑦−2𝑥−2𝑦+4
3).

43. Solution Equilibrium points are (0, 0) and (−3, 3).
𝑥 =𝑥𝑦−3𝑦
𝑦 =𝑥𝑦−3𝑥
1).
Equilibrium points are (0, 0) and (−3, 3).
(0, 0) is a stable center,
(−3, 3) is an unstable saddle point.

44. Solution Equilibrium points are (0, 1) and (1/4, 3/4)
𝑥 = 𝑥 2 −3𝑥𝑦+2𝑥
𝑦 =𝑥+𝑦−1
2).
Equilibrium points are (0, 1) and (1/4, 3/4)
(0, 1) is an unstable saddle point,
(1/4, 3/4) is an unstable spiral point.

45. Solution Equilibrium points are (2, 3), (2, −3), (3, 2) and (−3, 2).
𝑥 = 𝑥 2 + 𝑦 2 −13
𝑦 =𝑥𝑦−2𝑥−2𝑦+4
3).
Equilibrium points are (2, 3), (2, −3), (3, 2) and (−3, 2).
(2, 3) is an unstable saddle point
(2, −3) is an unstable saddle point
(3, 2) is an unstable node,
(−3, 2) is an asymptotically stable node.

46. Solution
𝑥 = 𝑥 2 𝑦+3𝑥𝑦−10𝑦
𝑦 =𝑥𝑦−4𝑥
4).
Equilibrium points are (1, 1), (1, −1), (−1, 1), and (−1, −1).
(1, 1) is an asymptotically stable spiral point,
(1, −1) and (−1, 1) both are unstable saddle points,
(−1, −1) is an unstable spiral point.

47. Solution Equilibrium points are (0, 0), (2, 4), and (−5, 4).
𝑥 =2− 𝑥 2 − 𝑦 2
𝑦 = 𝑥 2 − 𝑦 2
5).
Equilibrium points are (0, 0), (2, 4), and (−5, 4).
(0, 0) is an unstable saddle point,
(2, 4) is an unstable node,
(−5, 4) is an asymptotically stable node.

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