Chap7_student.pdf

Chapter 7. Phase Plane Analysis

Modern Control Theory (Course Code: 10213403)

Professor Jun WANG

(�� Ç)

Department of Control Science & Engineering

School of Electronic & Information Engineering

Tongji University

Spring semester, 2012

Outline

1 Introduction

2 Phase plane portraits

3 Properties of phase plane

4 Construction of phase plane portraits

5 Singular points and limit cycles

6 Phase plane analysis of linear systems

7 Phase plane analysis of nonlinear systems

8 Simulations with MATLAB

9 Summary

Introduction

Introduction

Limitations of describing function method

◮ Approximated method

◮ Used for simple nonlinearities

◮ Time response not available

◮ Not suitable for aperiodical inputs

Phase Plane of x + f (x, x) = 0

◮ Phase variables x and x

◮ Phase plane as shown right x

x

Prof J Wang (Tongji Uni) Chap 7. Phase plane analysis Spring 2012 4 / 58

Introduction

Let x1 = x, x2 = x. Equation x + f (x, x) = 0 becomes

dx1dt = x2

dx2dt = −f (x, x) = −f (x1, x2)

Re-Define−−−−−−→

dx1dt = f1(x1, x2)

dx2dt = f2(x1, x2)

Then we get a trajectory in the phase plane

dx2

dx1

=f2(x1, x2)

f1(x1, x2)with (x10, x20) ⇒ x2 = φ(x1)

and at a specific point (x1, x2),dx2dx1

is the direction of the motion along

the trajectory.

Hence, we have phase plane portraits

A set of (x10, x20) ⇒ A set of trajectories.


Outline

1 Introduction








9 Summary

Phase plane portraits


Ordinary point dx2dx1

= α for a point (x1, x2)

◮ Direction at this point is unique.

◮ The trajectory started from

an ordinary point is unique.

Singular point dx2dx1

= 00

◮ The slope of the trajectory ( i.e. the direction of the motion ) is an

uncertain value.

◮ Infinite number of trajectories may depart from or arrive at this

point.

◮ The singular point is an equilibrium point. (Why?)

◮ Isolated singular point: a singular point in the neighborhood where

no other singular points exist.

x1

x2

b

dx2

dx1= α



Example

Find the singular points of a system x + x = 0.

Solutions

Let x1 = x, x2 = x, then

dx1dt = x = x2

dx2dt = x = −x = −x1

⇒(x1 = 0, x2 = 0)

is the equilibrium point.

dx2

dx1

= −x1

x2⇒

(x1 = 0, x2 = 0) is a singular point and

the only singular point.

NB: This is an undamped oscillation with x2 + x2 = R2.



Example

Find the singular points of a system x + x = 0.

Solutions


dx1dt = x = x2

dx2dt = x = −x = −x2

x1

x2 A

A ′

B

B ′

dx2dx1

= − x2x2

⇒ All point with x2 = 0 are singular points.

i.e. all points on the x1 -axis are singular points

NB: This is a mass-damper system with motion

x(t) =[

x(0) + x(0)]

− x(0)e−t


Outline

1 Introduction








9 Summary

Properties of phase plane

Motion along a trajectory

Upper half plane

x > 0 ⇒ x ↑

Lower half plane

x < 0 ⇒ x ↓

x

x



Symmetry in phase plane portraits

x =dx

dt=

dx

dx· dx

dt=

dx

dxx

∴ x + f (x, x) = 0 can be written as dxdx = x

x = −f(x,x)

x .

Case 1: Symmetry about the x-axis

A :dx

dx= −

f (x, x)

x= a

B :dx

dx= −

f (x,−x)

−x= −a

⇒ f (x, x) = f (x,−x)

i.e. f (x, x) is even function of x.

x

xbA

a

bB

−a



Case 2: Symmetry about the x-axis

A : dxdx = −

f(x,x)x = a

B : dxdx = −

f(−x,x)−x = −a

⇒ f (x, x) = −f (−x, x)

i.e. f (x, x) is odd function of x.

x

xbA

abB−a

Case 3: Symmetry about the origin

A : dxdx = −

f(x,x)x = a

B : dxdx = −

f(−x,−x)−x = a

⇒ f (x, x) = −f (−x,−x)

x

xbA

a

bB

a


Outline

1 Introduction








9 Summary

Construction of phase plane portraits

Analytical methods for phase plane portraits

Example

To draw the phase plane portrait of x +ω2x = 0.

Solutions

Using x = x dxdx , we have

xdx

dx+ω2x = 0

⇒ x dx +ω2x dx = 0

x

x

A

ωA

Integrating the above equation gives x2

ω2 + x2 = A2.



Example

Draw the phase plane portrait of x = −M with x(0) = x0 and x(0) = 0.

Solutions

From x = −M, we have

x = −Mt+C1 ⇒ x = −1

2Mt2+C1t+C2

From the initial conditions, we have

C1 = 0

C2 = x0

⇒

x = −Mt

x = − 12Mt2 + x0

x2 = 2M(x0 − x) = −2M(x − x0)

x

x

x0

M = 1

x

x

x0

M = −1



Isocline methods for phase plane portraits

An isocline

dx2

dx1

=f2(x1, x2)

f1(x1, x2)= α ⇒ f2(x1, x2) = αf1(x1, x2) ⇒ x2 = ϕ(x1,α)

All trajectories passing through any point on the curve

x2 = ϕ(x1,α) have the same slope α, i.e. the same direction of

motion.

x2 = ϕ(x1,α) is an isocline .

A set of different values of α ⇒ A set of different isoclines

All these isoclines give the field of direction for the tangents to

trajectories.



Example

To draw the phase plane portrait by the isocline method for the system

x + 2ζωx +ω2x = 0

Solutions

The system can be rewritten as xdxdx + 2ζωx +ω2x = 0.

(i) Draw the isoclines

Let dxdx = α, then αx + 2ζωx +ω2x = 0, i.e. dx

dx = −ω2

2ζω+α.

Assume ζ = 0.5 and ω = 1, i.e. x + x + x = 0

⇒ x =−1

1 + αx



(ii) Draw short line segments on the isoclines to represent the direction

of field.

Since x = −11+α

x, we have

α = 0 ⇒ x = −x

α = ∞ ⇒ x = 0

α = −1 ⇒ xx = −1

0 ,

x = 0

α = −2 ⇒ x = x

α = −4 ⇒ x = 13x

−0.4 −0.2 0 0.2 0.4 0.6

−0.4

−0.2

0

0.2

0.4

0.6

x

x

−0.4 −0.2 0 0.2 0.4 0.6

−0.4

−0.2

0

0.2

0.4

0.6

x

x

−0.4 −0.2 0 0.2 0.4 0.6

−0.4

−0.2

0

0.2

0.4

0.6

x

x

−0.4 −0.2 0 0.2 0.4 0.6

−0.4

−0.2

0

0.2

0.4

0.6

x

x

−0.4 −0.2 0 0.2 0.4 0.6

−0.4

−0.2

0

0.2

0.4

0.6

x

x



(iii) Draw the phase trajectory from an ordinary point.

−0.4 −0.2 0 0.2 0.4 0.6

−0.4

−0.2

0

0.2

0.4

0.6

x

x

−0.4 −0.2 0 0.2 0.4 0.6

−0.4

−0.2

0

0.2

0.4

0.6

x

x


Outline

1 Introduction








9 Summary

Singular points and limit cycles

Singular points

Definition

A singular point is the point, which satisfies

dx1

dt= f1(x1, x2) = 0

dx2

dt= f2(x1, x2) = 0

i.e. dx2dx1

= 00 , and motion cannot be determined with dx2

dx1.

NB: Special care should be paid to singular points.



Properties of singular points

Let’s linearize the motion equation in the neighborhood of the origin

dx1

dt= a1x1 + b1x2,

dx2

dt= a2x1 + b2x2

Let x = x1, then

x = x1 = a1x1 + b1x2 = a1x + b1x2

x = x1 = a1x + b1x2 = a1x + b1(a2x1 + b2x2) = a1x + b1a2x + b1b2x2

Since x = a1x + b1x2 −→ b1x2 = x − a1x,

x = a1x + b1a2x + b2(x − a1x) = (a1 + b2)x + (b1a2 − b2a1)x



That is x + ax + bx = 0, where a = −(a1 + b2) and b = a1b2 − a2b1.

The roots of the char. eqn λ2 + aλ + b = 0 are

λ1,2 =−a ±

√a2 − 4b

2

(Suppose λ1 , 0 and λ2 , 0)



Classification of singular points

(1) Stable focus

σ

jωλ1

λ2

×

×

x

x

(2) Unstable focus

σ

jωλ1

λ2

×

×

x

x



(3) Stable node

σ

jω

λ1 λ2

× × x

x

λ1

λ2

x(t) = C1eλ1t + C2eλ2t

The lines with slopes λ1 and λ2 are phase trajectories.

The lines with slopes λ1 and λ2 are separatrices as well.

All trajectories approach the separatrice of λ2 when λ1 < λ2.



(4) Unstable node

σ

jω

λ2λ1

×× x

xλ2

λ1

If λ1 < λ2, Ceλ2t will dominate the motion as t increases.



(5) Center

σ

jω

λ2

λ1×

×x

x

(6) Saddle point

σ

jω

λ1λ2×× x

xλ1

λ2



Limit cycles

The limit cycle is an isolated closed path in the phase plane.

x

x

Stable limit cycle

x

x

Unstable limit cycle

x

x

Semi-stable limit cycle



Example

Obtain the phase plane portrait of x + 0.5x + 2x + x2 = 0

Solutions

(i) Standard form

Let x1 = x and x2 = x, then

x1 = x = x2 , f1(x1, x2)

x2 = x = −0.5x − 2x − x2 = −2x1 − 0.5x2 − x21 , f2(x1, x2)

(ii) Singular points

x1 = x2 = 0, x2 = −2x1 − 0.5x2 − x21 = 0

Hence, singular points are

x1 = 0

x2 = 0

x1 = −2

x2 = 0, or equivalently

x = 0

x = 0

x = −2

x = 0Prof J Wang (Tongji Uni) Chap 7. Phase plane analysis Spring 2012 30 / 58


(iii) Properties of the singular points

• Singular point (0, 0)

∂f1∂x1

∣

∣

∣

(0,0)= 0,

∂f1∂x2

∣

∣

∣

(0,0)= 1

∂f2∂x1

∣

∣

∣

(0,0)= (−2 − 2x1)

∣

∣

x1=0= −2,

∂f2∂x2

∣

∣

∣

(0,0)= −0.5

The linearized equation

x1 = x2

x2 = −2x1 − 0.5x2

i.e.

dxdt = xdxdt = −2x − 0.5x

a1 = 0, b1 = 1, a2 = −2, b2 = −0.5 ⇒ x + 0.5x + 2x = 0

Thus the char. eqn. is λ2 + 0.5λ + 2 = 0, and λ1,2 = −0.25 ± 1.987.

Therefore, (0, 0) is a stable focus.Prof J Wang (Tongji Uni) Chap 7. Phase plane analysis Spring 2012 31 / 58


• Singular point (−2, 0)

Let y = x + 2, we obtain y + 0.5y − 2y + y2 = 0

In the similar way, we have

y + 0.5y − 2y = 0

λ2 + 0.5λ − 2 = 0

λ1,2 = 1.186, −1.686

0.5

1.0

1.5

−0.5

−1.0

−1.51−1−2−3

x

x

Therefore, (−2, 0) is a saddle point.



(iv) Actual phase plane portrait

1

2

3

−1

−2

−3

1 2−1−2−3−4x(s)

x



NB: The isocline method may be used for sketching the phase plane

portraits in this example.

The isocline is given by

−0.5x − 2x − x2

x= α ⇒ x = −

(x + 1)2

α+ 0.5+

1

α+ 0.5

Parabolas passing through (−2, 0) and (0, 0).



An isocline for α = 0

1

2

3

−1

−2

−3

1 2−1−2−3−4x(s)

x


Outline

1 Introduction








9 Summary

Phase plane analysis of linear systems


ExampleObtain the phase plane

trajectories of the step response

of the given 2nd-order system.

R(s)K

s(Ts+1) C(s)E(s)

−

Solutions

Basic equations:

C(s)

R(s)=

K

Ts2 + s + K⇒ Tc + c + Kc = Kr

E(s)

R(s)=

Ts2 + s

Ts2 + s + K⇒ Te + e + Ke = Tr + r

r(t) = R · 1(t) ⇒ r = r = 0Prof J Wang (Tongji Uni) Chap 7. Phase plane analysis Spring 2012 37 / 58


(i) For error equation

Te + e + Ke = 0

e(0) = R, e(0) = 0

Let x1 = e, x2 = e, then

dx1dt = e = x2

dx2dt = e = − e

T − KT e = − x2

T − KT x1

Thus, the singular point is

x1 = 0

x2 = 0i.e.

e = 0

e = 0



• Properties of (0, 0)

The char. eqn. is Tλ2 + λ + K = 0.When 1 − 4KT < 0

⇒ stable focus

e

e

When 1 − 4KT > 0

⇒ stable node

e

e



(ii) For output equation

Tc + c + Kc = KR

c(0) = 0, c(0) = 0

Singular point

x1 = R

x2 = 0i.e.

c = R

c = 0

• Properties of (R, 0)

When 1 − 4KT < 0

⇒ stable focus

e

e

R

When 1 − 4KT > 0

⇒ stable node

e

e

R



Example

Obtain the phase plane portrait of the given system Te + e = P.

NB: No item e !

Solutions

(i) When P = 0

Te + e = 0 ⇒ Tede

de+ e = 0 ⇒ de

de= −

e

Te

For e = 0 Continuous singular points:

e = 0.

For e , 0 Isoclines: d ede = − 1

T

e

e



(ii) When P , 0, Te + e = P

No singular points

Isoclines:de

de=

P − e

Te= α ⇒ e =

P

1 + αT

When α = 0, e = P.

Therefore, e = P is a phase plane trajectory.

• P > 0

e

ɺe

P α = 0

α > 0

01 <<− αT

T

1−<α

α = ∞O

• P < 0

e

ɺe

P α = 0

α > 0

α = ∞O

01 <<− αT

T

1−<α


Outline

1 Introduction








9 Summary

Phase plane analysis of nonlinear systems

Piecewise analysis

A simple example

f (x, x, x) = u =

M x > 0

−M x < 0

f (x, x, x) = −M f (x, x, x) = M

II I

x

x



Actual and virtual singular points

Let f (x, x, x) = M for Area I,

and PI1 and PI2 be singular points of f (x, x, x) = M.

I

x

x

PI1PI2PI1: Actual singular point

PI2: Virtual singular point



Example

Obtain the phase plane portrait for the following system

R(s) GNK

s(Ts+1) C(s)E(s) M(s)

−

where GN is defined as follows

m =

e |e| > e0

ke |e| < e0

k < 1

e

m

−e0

e0

k



Solutions

System equations:

R(s) GNK

s(Ts+1) C(s)E(s) M(s)

−

E(s) = R(s) − C(s) = R(s) −K

s(Ts + 1)M(s)

Te + e = Tr + r − Km

In e − e plane

3 areas

2 different equations

I II III

e

m

−e0

e0

k



Step response

r(t) = 1(t), r = 0, r = 0

e(0) = E0, e(0) = 0⇒

Te + e + Ke = 0 Areas I and III

Te + e + kKe = 0 Area II

Singular point: e = 0 and e = 0

◮ Actual singular point for system in II

◮ Virtual singular point for system in I, IIII II III

−e0

e0

Properties of the singular point:

Suppose 1 − 4kKT = 0. Since k < 1 , so 1 − 4KT < 0◮ Small input: |e| < e0

Te + e + kKe = 0 ⇒ (0,0) is the stable node

◮ Large input: |e| > e0

Te + e + Ke = 0 ⇒ (0,0) is the stable focus



Phase trajectory

Let A be the initial point

◮ For A, (0, 0) is the stable focus

◮ For B, (0, 0) is the stable node

◮ For C, (0, 0) is the stable focus

◮ For D, (0, 0) is the stable node

e0

I II III

O A

B

C

D

Features: Speeds up the regulation

◮ When large signal in the loop

• The origin is the stable focus (underdamped)

• Error decreases fast

◮ When small signal in the loop

• The origin is the stable node (critically damped)

• Oscillation avoided


Outline

1 Introduction








9 Summary

Simulations with MATLAB


Problems to solve

◮ How to solve differential equations?

◮ How to plot phase plane portraits?

◮ How to simulate nonlinear systems?

Tools to solve the problems

◮ “ode45” command

◮ Simulink nonlinear blocks



“ode45” Command

ODE45 solves ordinary differential equations with medium-order

method. It integrates the system of differential equations y = f (t, y)

from time t0 to tf with initial conditions y0.

[tout,yout] = ode45(function,[t0,tf],y0)

The solution array “yout” corresponding to a time vector “tout”

function Function handle of differential equations programmed in

a separate “.m” file

t0, tf Initial and final time of integral

y0 Initial conditions of the DEs



Example

An example of Van de Pol equation x − (1 − x2)x + x = 0.


x1 = x2

x2 = (1 − x21)x2 − x1

.

% This is script, saved as /VandePol.m0 in a separate file,

% defines the ODEs for the Van de Pol equations.

function [sys,x0]= VandePol(t,x)

sys=[x(2);(1-x(1)*x(1))*x(2)-x(1)];

% This script solves the DEs defined in VandePol.m

% with the ODE45 method.

[t, x] = ode45(-VandePol’, [0, 20], [3 ;4] );plot(x(:,1),x(:,2),’-b’);



The simulation result is shown below

1

2

3

4

−1

−2

−31 2 3 4−1−2−3

x

x


Outline

1 Introduction








9 Summary

Summary

Summary

How to analyze a nonlinear system?

◮ Analysis by linearization

• Linearization around neighborhood of an equilibrium

• Linearization by nonlinear feedback

• Linearization by inverse nonlinearity

◮ Extended linear analysis method

• Equivalent gain analysis by the root locus method

• Equivalent gain analysis by the describing function method

◮ Nonlinear analysis method

• Phase plane method


Summary

Summary of DF and phase-plane methods

DF method Phase-plane analysis

Plant complexity X 1st & 2nd order system

NL complexity × Piecewise linearity

Time response × X

Stability analysis X X

Limit cycle analysis X X

Accuracy × X

Method used Equi. linearization Graphical solution


Chapter 7. Phase Plane Analysis

Modern Control Theory (Course Code: 10213403)

Professor Jun WANG

(�� Ç)

Department of Control Science & Engineering

School of Electronic & Information Engineering

Tongji University

Spring semester, 2012

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