control lecture notes 01

8/3/2019 Control Lecture Notes 01

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Control Systems

IMAD M. JAIMOUKHA

EE2/ISE2

Room 1111C

Extension 46279

[email protected]

www3.imperial.ac.uk/controlandpowerwww3.imperial.ac.uk/people/i.jaimouka

1

Organisation

18 lectures and 9 tutorials/study groups.

Notes to be handed out.

The notes are only notes, use books for background.

Supplementary notes (NOT EXAMINABLE) oc-casionally handed out.

Problem sheets (together with answers) will behanded out.

Past exam papers (from 3 and 4 years ago) will b ehanded out.

The problems will be of type you would expect toanswer in the exam, with some exceptions:

There are book-work type questions which makesense in an exam but which do not on a problemsheet.

Exam questions are generally easier.

Computationally intensive questions are not suit-able for examination.

2

Useful Books

(Use latest editions)

Modern Control System Theory and Design, S.M.Shinners, Wiley. Good for the state-space approach.

Modern Control Systems, R.C. Dorf and R.H.Bishop, Addison-Wesley. Good for the transferfunction approach.

Feedback Control of Dynamic Systems, G.F. Franklin,J.D. Powell and A. Emami-Naeini, Addison-Wesley.Good for practical design issues.

Feedback Control Systems, C.L. Phillips and R.DHarbor, Prentice-Hall. Follows the lecture courseclosely, and is useful for third year course too.

3

Course OutlinePage

1. Introduction 6 Aims of control Examples of Control Systems The feedback design dilemma Control design tasks Feedback for static systems

2. Mathematical Models of Dynamic Systems 19 Laplace transform Transfer functions

Electric circuit models Mechanical systems Control System Modelling

3. Block Diagram Algebra and Signal Flow Graphs(Not examinable) 33 Prototype Feedback System Block Diagram Algebra Signal Flow Graphs and Masons Rule

4. State-Variable Models (Not examinable) 55 General state-variable form Simulation diagrams Solution of state-variable equations Relation to transfer functions Stability and the matrix exponential function

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5. System Responses 71 Test signals First order system Steady-state response Second order system Time response specifications Frequency response of systems Steady-state Accuracy

6. Stability Analysis 93 The problem of stability in control Routh-Hurwitz stability criterion

7. Root-locus Analysis and Design Techniques 113 Stability of feedback systems The root-locus Rules for plotting the root-locus Root-locus design tools 138

8. Frequency Response Methods 158 Conformal mapping

The Nyquist stability criterion Phase lead - phase lag compensation PID control

5

Aims of Control

Use of feedback to force a given plant toexhibit desired characteristics

PlantControllerE E

T

cE EE

-

controlerror outputreference disturbance

Examples of control systems include:

Domestic hot water heater - temperature feed-back used to control the water temperature.

Control of the concentration of a fluid - opacityused to control concentration.

Lateral and longitudinal control of a Boeing747.

Active control of a segmented optical system.

6

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9 10

The Control Problem

Typical aims of feedback:

disturbance rejection

tracking improvement

transient response shaping

sensitivity reduction

reduction in effect of nonlinearity

closed-loop stability

The feedback design dilemma:Almost all the benefits of feedback can be achieved,

provided that the loop gain is sufficiently high(over a certain frequency range). Unfortunately,

for most plants, high loop gain tends to drivethe system into instability.

11

Control Design Tasks

Most control systems contain some of the following:

actuatorcontrollercontroller

sensorcontroller

plantE E E E E E

''

T

reference output

-

Output and sensor selection

Input and actuator selection

Model building

Controller design

Simulation testing

Hardware testing

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17 18


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Mathematical Models of Dynamic Systems

Introduction

Laplace Transform

Inverse Laplace Transform

Applications to Differential Equations

Models for Electric Circuits

Transfer Functions

Models for Mechanical Systems

19

Introduction

There are many ways of describing linear dynamicalsystems, but the following three occur most frequently:

A system of ordinary differential equa-tions often the initial description of a system,in terms of its physics and interconnections. Thevariables are just the inputs and outputs.

A transfer function model. A single inputsinge output process is described by its transferfunction: the ratio of the Laplace transform of out-put and input. Particularly convenient for the de-sign of controllers for dynamical systems.

A state variable model: a system of first orderordinary (linear) differential equations. Descrip-tions of the first form can be reduced to this formby the introduction of the intermediate variablescalled states. This form is convenient for simula-

tion, computation and certain forms of analysis.

20

Laplace Transforms

Used to solve differential equations via algebraicequations.

Given f(t), 0 t , with f(t) = 0 for t 0,we define the Laplace transform off(t) via:

F(s) = L{f(t)} =0

estf(t)dt.

Example: f(t) = 1 for t 0:

F(s) = 0

estdt = 1

sest|0 =

1

s

Example: f(t) = eat, for t 0:F(s) =

0

esteatdt =0

e(sa)tdt =1

s a

The Laplace transform is a linear operator:L{k1f1(t) + k2f2(t)} = k1L{f1(t)} + k2L{f2(t)}

Example: f(t) = cos t = 0.5(ejt + ejt):

F(s) = 0.5s j + 0.5s + j = ss2 + 2

21

There is a table of standard transforms and ma-nipulative rules-properties.

Example: Suppose L{f(t)} = F(s). Then

L{etf(t)} = F(s + ).

Proof: L{etf(t)} = 0

estetf(t)dt

=0

et(s+)f(t)dt = F(s + ).

Example:

L{et cos t

}:

L{cos t} = ss2 + 2

L{et cos t} = s + (s + )2 + 2

Example: L{x(t)} = sL{x(t)} x(0).Proof : L{x(t)} =

0

estx(t)dt

= estx(t)|0 +s0

estx(t)dt

= sL{x(t)} x(0)

Example: L{x(t)} = s2L{x(t)}sx(0) x(0).

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23 24

Inverse Laplace Transform

Let F(s) = L{f(t)}. Then

f(t) =1

2j

+jj

F(s)estds, .

In general, this expression is difficult to evaluate.However, we can usually use partial fraction ex-pansion and tables of Laplace transforms.

Example: Using the covering rule:F(s) =

s+1

s(s+2)=

0.5

s+

0.5

s+2 f(t) = 1+ e

2t

2

Example:F(s) =

s + 3/5

(s + 1)2 + 22=

(s + 1) 2/5(s + 1)2 + 22

=s + 1

(s + 1)2 + 22 1

5

2

(s + 1)2 + 22

f(t) = et cos2t 0.2et sin2t Example:

F(s) =1

(s+1)2(s+2)=

1

s+2 1

s+1+

1

(s+1)2

f(t) = e2t+ettet

The inverse Laplace transform of any strictly

proper rational function of s is a combinationof exponentials, sinusoids and polynomials in t.

25

Applications to Differential Equations

Consider the differential equation:y(t) + 4y(t) + 5y(t) = u(t), y(0) = y0, y(0) = 0

Taking Laplace transforms:s2Y(s) sy0 + 4(sY(s) y0) + 5Y(s) = U(s)

Y(s) = U(s)s2 + 4s + 5

+(s + 4)y0

(s2 + 4s + 5)

Suppose that U(s) = 1/s:Y(s) =

1

s(s2 + 4s + 5)+

(s + 4)y0(s + 2)2 + 1

=1

5{1

s s + 4

(s + 2)2 + 1} + (s + 4)y0

(s + 2)2 + 1

=1

5{1

s s + 2

(s + 2)2 + 1 2

(s + 2)2 + 1}

+(s + 2)y0

(s + 2)2 + 1+

2y0(s + 2)2 + 1

y(t) = 15{1 e2t cos t 2e2t sin t}

+y0e2t(cos t + 2 sin t)

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Models for Electric Circuits

We know that:

v(t) = Ri(t) V(s) = RI(s)

is a model for a resistor.

For a capacitor (assuming zero initial conditions):

i(t) = Cv(t) V(s) = 1sC

I(s)

For an inductor (assuming zero initial conditions):

v(t) = Ldi(t)

dt V(s) = sL I(s)

The impedance Z(s) of an electrical componentis the ratio of the Laplace transform of the voltage(across the component) to the Laplace transform

of the current (through the component), with allinitial conditions assumed to be zero:

For a resistor: Z(s) = R

For an inductor: Z(s) = sL

For a capacitor: Z(s) = 1sC

27

A more complex example is provided by the fol-lowing RC circuit:

e

ee

ee

ee

e..................... .......................................... i

i

.....................$$

..................... T

c

R1 i(t)

R2

Cv1(t) v2(t)

V1(s) = I(s)(R1 + R2 +1

sC)

V2(s) = I(s)(R2 +1

sC)

V2(s)V1(s)

=1 + sCR2

1 + sC(R1 + R2)

28

Another example is a circuit containing an op-amp:

e

ee

ee

ee

e..................... ..........................................

e

ee

ee

ee

e..................... ..........................................

h

+

vi(t)i(t)

vo(t)-

R2 C

R1

Making the virtual earth assumption:

Vi(s) = I(s) R1

Vo(s) = I(s)(R2 + 1sC

)

Vo(s)Vi(s)

= R2 +1

sC

R1

= 1 + sCR2sCR1

29

Transfer Functions

The transfer function G(s) of a linear systemis the ratio of the Laplace transform of the outputto the Laplace transform of the input, with all ini-tial conditions assumed to be zero:

For the first example:

G(s) =V(s)

I(s)= R

For the second:

G(s) =V2(s)

V1(s)=

1 + sCR21 + sC(R1 + R2)

For the third:

G(s) =Vo(s)

Vi(s)= 1 + sCR2

sCR1

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Models for Mechanical Systems

Example: Assume x(0) = x(0) = 0.

..........................r

rr

r..........................

....................................................................

c

c

K B

f(t)

x(t)

M

Balancing forces on M:

f(t) = Kx(t) + Bx(t) + Mx(t)

Taking Laplace transform:

G(s) =X(s)

F(s)=

1

M s2 + Bs + K

31

Example: Balancing forces on M2 and M1:

f(t) = M2x2 + K2(x2 x1)0 = M1x1 + K2(x1 x2) + K1x1 + Bx1

Taking Laplace transforms (assume xi(0)= xi(0)=0)

(s2M2 + K2)X2(s) = F(s) + K2X1(s)

K2X2(s) = X1(s)(s2M1 + Bs + K1 + K2)

X1(s)K2

= F(s)+ K2X1(s)(s2M2+K2)(s2M1+Bs +K1+K2)

X1(s)F(s)

=K2

(s2M2+K2)(s2M1+Bs +K1+K2)K22

..........................r

rr

r..........................

..........................r

rr

r..........................

....................................................................

M1

c

M2

cc

K1B

x1(t)

x2(t)f(t)

K2

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SUPPLEMENTARY NOTES

(NOT EXAMINABLE)

Block Diagram Algebra and Signal FlowGraphs

Introduction

Prototype Feedback System

Block Diagram Algebra

Block Diagram Transformations

Examples

Signal Flow Graphs

Masons Rule

Examples

33

Introduction

Many practical control systems consist of a com-plicated interconnection of smaller subsystems.

Before tackling a control system design for suchsystems, it is usually helpful to simplify the com-plex interconnection of subsystems.

Essentially, we seek a systematic way of eliminat-ing variables (signals) we do not want to control ormeasure.

We discuss two methods of performing these sim-plifications:

Block diagram algebra.

Signal flow graphs.

34

Prototype Feedback System

Usually, we will want to reduce the interconnectionto a diagram of the form:

n P(s)G(s)C(s)

F(s)

E E E E

'

E

T

r(s)e(s)

-

y(s)u(s)

In this diagram:

G(s) = system or plant

P(s) = post compensatorC(s) = pre compensatorF(s) = feedback compensator

r(s) = reference signal

e(s) = tracking error signal

u(s) = input (or actuator) signal

y(s) = output signal

35

We will often make use of the following definitionswhich refer to this diagram:

Loop gain function: C(s)G(s)P(s)F(s)

Closed loop transfer function: y(s)/r(s)

Tracking error ratio (or sensitivity function):

e(s)/r(s)

From the diagram

e(s) = r(s) F(s)P(s)G(s)C(s)e(s)

So, the sensitivity function is given by

e(s)

r(s)=

1

1 + F(s)P(s)G(s)C(s)

It follows that the closed loop transfer function is

y(s)

r(s)

=C(s)G(s)P(s)

1 + C(s)G(s)P(s)F(s)

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Block Diagram Algebra

ET

E

'

BlockE

u yu

y

Block diagrams consist of:

1. Blocks: these give a description of subsystemdynamics.

2. Summers: add or subtract two or more sig-nals.

3. Arrows: these give the direction of signal prop-agation.

4. Take off points.

37

Block Diagram Transformations

The reduction of complex block diagrams is facili-tated by a series of easily derivable transformationswhich are summarised in diagrams 1 and 2 below.

The following steps may be used to simplify com-

plicated block diagrams:

Combine cascaded blocks using 1.

Combine parallel blocks using 2.

Eliminate minor loops using 4.

Shift summers left using 7 and shift take offpoints right using 10 and 12.

Go around the block again if necessary.

General methods best learned from examples.

38

Signal Flow Graphs

A signal flow graph is a pictorial representa-tion of a set of simultaneous equations describinga system.

Equations are represented with the aid ofbranchesand nodes:

nodes represent variables.

branches relate variables.

Example: The signal flow graph below is a pic-torial representation of:

y2 = ay1 + by2 + cy4y3 = dy2y4 = ey1 + f y3y5 = hy4 + gy3

y1y2 y3 y4 y5

a

c g

h

e

fd

node

branch

b

39

Before proceeding further, we will define some termswhich we will need later:

A source is a node which has outgoing branchesonly (y1).

A sink only has incoming branches (y5).

A path is a set of branches having the samesense of direction (adf h, eh and adf c).

A forward path originates from a source and

terminates in a sink. No node may be encoun-tered more than once (eh, ecdg and adf h).

The path gain is the product of the coeffi-cients associated with the branches of the path.

A feedback loop is a path that begins andends at the same node; additionally no nodemay be encountered more than once (b anddfc).

The loop gain is the path gain of a feedbackloop.

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45 46

Masons Rule

The general formula for the path gain between the sin-gle source and the single sink is:

G =kPkk

where:

= 1 L1 + L2 L3 + ... + (1)mLm

and

L1 = gain of each closed loop in the graph

L2 = product of loop gains of any two nontouching

loops. (loops are called nontouching if they have

no node on common)...

Lm = product of loop gains of any m nontouching loops

Pk = gain of the kth forward path

k = the value of remaining with the loops touching

the path Pk are removed

47

Example

g

E

'

T

Er y

y = r + gyy

r=

1

1 g

r

1 1 y

g

= 1 gP1 = 1

1 = 1y

r=

1

1 g

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Example

f

g1

g2

E

E

ET T

E

'

E

cr y

yr

= g1 + g21 f(g1 + g2)

49

r 1

g1

g2

1 y

f

P1 = g1, 1 = 1

P2 = g2, 2 = 1

= 1 g1f g2f yr

=g1 + g2

1 f(g1 + g2)

50

Example

u yG1 G2 G3 G4

H3-

- H2

- H1

1 1

Forward paths:

P1 = G1G2G3G4.

Closed loops:

G2G3H2, G3G4H3, G1G2G3G4H1

L1 = (G2G3H2 + G3G4H3 + G1G2G3G4H1)

Non-touch loops: None

51

= 1 + G2G3H2 + G3G4H3 + G1G2G3G4H1

1 = 1

G = yu

=G1G2G3G4

1 + G2G3H2 + G3G4H3 + G1G2G3G4H1

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Example

u1G2G1G41 1

y

F2

G3

F1

-

Forward paths: P1 = G1G2G4, P2 = G1G3G4

Closed loops: G1G4F1, G1G2G4F2, G1G3G4F2

L1 = G1G4F1 G1G4F2(G2 + G3)

= 1 G1G4F1 + G1G4F2(G2 + G3)

1 = 2 = 1

G = yu

=G1G4(G2 + G3)

1 G1G4F1 + G1G4F2(G2 + G3)

53

Example

a b c d e f g h

i j k

m

= 1 (bi + dj + f k + bcdefgm) +(bidj + bifk + djfk) (bidjfk)

P1 = abcdefgh

1 = 1

G =abcdefgh

1bidjf kbcdefgm+bidj +bifk +djfkbidjfk

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State-variable Models

Introduction

Standard State-variable Format

Advantages of State-variable Models

Simulation Diagrams

State-variable Models from Transfer Functions

Transfer Functions from State-variable Models

The Matrix Exponential Function

Solving the State-variable Equations

Stability and the Matrix Exponential Transfer Functions from State-variable Models

Computing the Matrix Exponential Function

55

Introduction

We have already come across:

linear differential equation models

transfer function models

We now consider the state-variable model, alsocalled the state-space model.

This is a differential equation model, but the equa-tion is written in a specific format.

The idea is that an nth order differential equationis decomposed into a set of n 1st order equationswritten in matrix-vector form.

This decomposition is achieved by defining inter-nal variables, called states, whose time evolutioncompletely characterises the behaviour of the sys-tem.

56

Example

..........................r

rr

r..........................

....................................................................

M

cc

KB

u(t) y(t)

My(t) + By(t) + Ky (t) = u(t)

Define x1(t) = y(t), x2(t) = y(t). x1(t) = x2(t)

x2(t) = KM

x1(t) BM

x2(t) +u(t)

Mor in matrix form:

x1(t)

x2(t)

=

0 1KM BM

x1(t)

x2(t)

+

01

M

u(t)

y(t) =

1 0 x1(t)

x2(t)

This is an example of a single input/singleoutput state-variable model.

57

Standard State-variable Format

The state of a system at time t1 is theamount of information at t1 that, togetherwith all inputs for t t1, uniquely deter-mines the system behaviour for all t t1.

State-variable models have the form:x(t) = Ax(t) + Bu(t) (state equation)

y(t) = Cx(t) + Du(t) (output equation)

x(0) = x0 (initial condition)

where

A : n

n (system matrix)

B : n r (input matrix)C : p n (output matrix)D : p r (direct feedthrough matrix)

x(t) : n 1 (state vector)u(t) : r 1 (input vector)y(t) : p 1 (output vector)

j

nB C

D

A

E E E

TE

T

'

E

E

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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.......................

........................

.......................

.........

.......................

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........................

.......................

........................

.......................

.......................

........................

.......................

.......................

........................

.......................

........................

.......................

.........

c

x x

y

u

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Example

Consider the coupled differential equations:y1 + k1y1 + k2y1 = u1 + k3u2y2 + k4y2 + k5y1 = k6u1

Define: x1 = y1, x2 = y1, x3 = y2

This gives the state equations:x1 = x2x2 = k2x1 k1x2 + u1 + k3u2x3 = k5x2 k4x3 + k6u1

and the output equations:

y1 = x1, y2 = x3

Collecting these into matrix form gives the 2-input2-output state-variable model

x

x1

x2x3

=

A

0 1 0

k2 k1 00 k5 k4

x

x1

x2x3

+

B

0 0

1 k3k6 0

u

u1u2

y1

y2

y

=

1 0 0

0 0 1

C

x1x2x3

x

59

Advantages of State-variable Models

State-variable models give knowledge about theinternal structure as well as the input-outputcharacteristics of the system.

There are computational advantages: time-domain matrix methods lend themselves

naturally to computer solution, especially forhigh order models.

matrix methods enable us to easily determinethe transient response and evaluate the sys-tem performance.

A good unified framework for several advancedcontrol theories, such as optimal control designmethods.

A natural framework for system simulation.

Extensions to: multivariable systems

nonlinear systems

time-varying systems

are (relatively) straightforward.

60

Simulation Diagrams

These are block diagrams (or signal flow graphs)that are constructed to have a given transfer func-tion or to model a set of differential equations.

There are three elements in a simulation diagram: an integrator (whose transfer function is 1s)

a pure gain

a summer

All these components can be easily constructed us-ing simple electronic devices.

They are useful in constructing computer simula-tions (digital or analogue) of a given system.

For the mass-spring system we have:x1(t) = x2(t)

x2(t) = KM

x1(t) BM

x2(t) +u(t)

M

"!#

1M

1s

1s

BM

K

M

E E E

ee

ee

eeeuT

'

E

'

...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...............................................

....................

....................

....................

...................

....................

....................

....................

...................

....................

....................

....................

....................

....................

...................

....................

.......

...............................................

....................

....................

....................

...................

....................

....................

....................

....................

...................

....................

....................

....................

...................

....................

....................

.......

Ex2 x1

yu- -

61

State-variable Models from TransferFunctions

There are simulation diagrams which can be de-rived from general transfer functions of the form:

y(s)

u(s)= g(s) =

bn1sn1 + bn2sn2 + + b0sn + an1sn1 + an2sn2 + + a0

Divide numerator and denominator by sn:

y(s) =bn1s1 + bn2s2 + + b0sn

1 + an1

s

1 + an2

s

2 +

+ a0s

nu(s)

Sete(s) :=

u(s)

1 + an1s1 + an2s2 + + a0sn

so that

y(s) = [bn1s1 + bn2s2 + + b0sn] e(s)

This givese(s) = u(s)[an1s1 + an2s2 + + a0sn] e(s)

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This has the following simulation diagram:

1

s1s

E E

ee

ee

eeuT

1s

an1

an2

a1

a0

E................................................... EU&&b

'

...........................................

E E

'

'

'

--

xn1 x2 x1xn

--

ue

Complete the diagram by settingy(s) = [bn1s1 + bn2s2 + + b0sn] e(s)

nE E E................................................... E E E E.....................1

s1s

1s b0

b1

bn2

bn1

E

E

E

................................

...........

c

x1xn1xn x2e y

63

The state variables satisfy:x1 = x2x2 = x3

...

xn1 = xnxn = a0x1 a1x2 an2xn1

an1xn + u

while the output is

y = b0x1 + b1x2 + + bn2xn1 + bn1xn

In matrix form this yields the state-variable model:

x =

0 1 0 00 0 1 0... ... . . . . . . 00 0 0 1

a0 a1 an2 an1

x +

00...01

u

y =

b0 b1 bn2 bn1

x

Note the direct connection with the coefficients ofthe transfer function:

g(s) =bn1sn1 + bn2sn2 + + b0

sn + an1sn1 + an2sn2 + + a0

64

The Matrix Exponential Function

Suppose that A is a square matrix and let t denotethe time variable and I denote the identity ma-trix.

Define the matrix exponential function eAtvia the power series

eAt = I+ At +A2t2

2!+

A3t3

3!+

+

Aktk

k!+

It can be shown this series converges for any square

matrix A and any scalar t.

Some, but not all, of the properties of the scalarexponential function are inherited by the matrixexponential function.

65

A most useful property concerns the derivative:

d

dteAt = A + A2t +

A3t2

2!+

= A(I + At +A2t2

2!+ )

= AeAt = eAtA

The most useful properties are summarised below:

P1: e0 = I

P2: eTT1

= T eT1 for any nonsingular T

P3: e(+)A = eAeA

P4: eA = (eA)1

P5: ddteAt = AeAt = eAtA

P6: L(eAt) = (sI A)1

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Solving the State-variable Equations

Consider the state-variable model:x(t) = Ax(t) + Bu(t), x(0) = x0y(t) = Cx(t) + Du(t)

Suppose first that u(t) = 0 for all t:x(t) = Ax(t), x(0) = x0

Trial solution: x(t) = eAtx0. Check:

x(t) = A eAtx0 x(t)

= Ax(t), x(0) = e0I

x0 = x0

When u(t) = 0 we proceed as follows:eAtx(t) = eAtAx(t) + eAtBu(t)

(P5) ddt

[eAtx(t)] = eAtBu(t)

t0d

dt[eAtx(t)]dt =

t0 e

ABu()d

(P1) eAtx(t) x0 =t0 e

ABu()d

(P4, P3)

x(t) = eAtx0 + t

0

eA(t)Bu()d

Finally

y(t) = CeAtx0 initial cond.response

+

input response t0 Ce

A(t)Bu()d convolution integral

+ Du(t)

67

Stability and the Matrix Exponential

Consider the eigenvalue decomposition of A:A = TT1

= diag(1, 2, , n) eigenvalues of A

Using property P2:

e

At

= e

TT1t

= I+ TT1t +TT1TT1t2

2!+

= T[I+ t +2t2

2!+ ]T1 = T etT1

= T diag(e1t, e2t, , ent) T1since is a diagonal matrix.

Consider the response of the unforced systemx(t) = Ax(t), x(0) = x0

x(t) = eAtx0 = T etT1x0

It follows thatlim

t x(t) = 0, x0 Re[i(A)] < 0, i

The unforced system is stable if andonly if all the eigenvalues of A lie in the

left half of the complex plane

68

Transfer Functions from State-variableModels

Consider a state-variable model initially at rest:x(t) = Ax(t) + Bu(t), x(0) = 0

y(t) = Cx(t) + Du(t)

Taking Laplace transforms:sx(s) = Ax(s) + Bu(s)

(sI A)x(s) = Bu(s) x(s) = (sI A)1Bu(s)

and so

y(s) = Cx(s) + Du(s)

y(s) = [D + C(sI A)1B]u(s)

The transfer function from u(s) to y(s) is theng(s) = [D + C(sI A)1B]

Note that (sI A)1 will exist for all values of swhich are not eigenvalues ofA. We call i(A) thepoles of g(s)

69

Computing the Matrix ExponentialFunction

1. Truncation of power series expansion:

eAt I + At + A2t2

2!+

A3t3

3!+ + A

ktk

k!

Usually, only a few terms are needed

2. Eigenvalue-eigenvector decomposition:

LetA = TT1 = Tdiag(1, 2, , n)T1

eAt = T etT1 = Tdiag(e1t, e2t, , ent)T1 The eigenvalue matrix is not always diagonal Method not always accurate

3. Inverse Laplace transform:

Since L{eAt} = (sI A)1 eAt = L1{(sI A)1}

Feasible only for matrices of small dimension Almost never used in practice

4. Matlab: Just type E = expm(A

t)

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System Responses

Introduction

Test Signals

Response of First Order Systems

Steady-state Response

Step Response of Second Order Systems

Time Response Specifications

Interpretation of Pole Locations

Step Response for Higher Order Systems

Introduction to Frequency Response Steady-state Accuracy of Feedback Systems

71

Introduction

For control system design, we need a basis ofcom-parison of performance of different designs.

One way of setting up this basis is to specify par-

ticular test signals and compare the response ofdifferent designs to these signals.

The use of test signals is justified since there is aclose correlation between the capability of systemsto respond to actual input signals and the responsecharacteristics to test signals.

We analyse the transient response as well asthe steady-state response of simple systems.

Although practical control systems are generallyofhigh order, we will be mainly concerned withfirst & second order systems, since these cap-ture the main features of practical control systems.

72

Test Signals

All the following test signals are defined for t 0. The unit impulse function is based on a rec-

tangular function (t) such that

(t) = 1, 0 t , >0

As 0, (t) approaches the unit impulse (t):0

(t)dt = 1,0

(ta)g(t)dt = g(a), L{(t)}= 1

It is useful for modeling shock inputs.

The unit step function:

r(t) = 1, t 0 L{r(t)} = 1s

is useful for modeling sudden disturbances.

The unit ramp function:r(t) = t, t 0 L{r(t)} = 1

s2

is useful for modeling gradually changing inputs.

The sinusoidal functionscos t = Re ejt L{cos t} = s

s2 + 2

sin t = Im ejt L{sin t} = s2 + 2

are important in frequency response techniques.

73

Response of First Order Systems

Consider the first order transfer function

g(s) =y(s)

r(s)=

K

s + 1

The differential equation representation is

y(t) + y(t) = Kr(t)

When r(t) is a unit step

y(s) =K

s(s + 1)= K(

1

s 1

s + 1/)

y(t) = K(1 et/) = K Ket/

If > 0, the second term tends to 0 as t .

The first term is called the steady-state re-sponse and K is called the steady-state value.

The second term is called the

transient responseand is called the time constant of the system.

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The following figure shows the step response.

t

y(t)

K

K ( 1 - e )- t /

The exponential function decays to less than 2%of its initial value within 4 time constants.

The output y(t) reaches about 63% of its finalvalue when t = .

An example of a first order system is provided bythe following RC circuit.

e

ee

ee

ee

e..................... .......................................... g

g

T

c

i(t)

CVi(t) Vo(t)

R

V0(s)

Vi(s)=

1

1 + sCR(K = 1, = RC)

75

Steady-state Response

The concept of the steady-state response can begeneralized to transfer functions of any order.

Suppose thaty(s) = g(s)r(s)

where g(s) is a given transfer function.

Suppose that the limitlim

t y(t)

exists.

Using the final value theorem:lim

t y(t) = lims 0 sy(s) = lims 0 sg(s)r(s)

If r(t) is a unit step, this becomes:

limt y(t) = lims0 sg(s)

1s

= lims0 g(s)

= g(0)

g(0) is often called the d.c. gain of the system.

76

Response of Second Order Systems

Consider the damped mass spring system again

..........................r

rr

r..........................

....................................................................

M

c

c

k B

y(t)r(t)

The transfer function can be written asg(s) =

y(s)

r(s)=

1

Ms2 + Bs + k

=1

M

s2 + BMs +k

M

=1

k

kM

s2 + BMs +k

M

= K2n

s2 + 2ns + 2nwhere

K =1

k, n =

kM

, =B

2

kM

77

The standard form for second order systems is:

g(s) = K2n

s2 + 2ns + 2n

where

: damping ratio

n : undamped natural frequency

K : d.c. gain

Since K only determines the d.c. magnitude of theresponse, we will set

K = 1

and so

g(s) =2n

s2 + 2ns + 2n

=2n

(s p1)(s p2)

where

p1, p2 = n n

2 1

are the poles ofg(s).

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Pole Locations

The system characteristics depend on and n(equivalently p1, p2), since they are the only para-meters that appear in the transfer function.

There are three cases of practical interest:1. > 1 (overdamped system):

p1, p2 = n n2 1are real, negative and unequal.

2. = 1 (critically damped system):

p1, p2 = nare real, negative and equal.

3. 0 < 1 (underdamped system):p1, p2 = n jn

1 2

are complex conjugate and have negative realparts (zero real part if = 0).

Another case is < 0 (unstable system):p1, p2 = n n

2 1

have positive real parts will be covered later.

79

The three cases are illustrated in the figure below.

T

E

T

T

E

E

5

55

5

5555

55

x x

x

unequal real LHP poles - overdamped

equal real LHP poles - critical damping

complex conjugate LHP poles - underdamped

x

x

= tan1

1 2

80

Step Response of Second Order Systems

If r(t) is a unit step

y(s) =2n

s(s2 + 2ns + 2n)

Expanding in partial fractions and completing thesquare

y(s) =1

s s + 2n

s2 + 2ns + 2n

=1

s s + n

(s + n)2 + 2d n

(s + n)2 + 2d

where

d =

1 2nis called the damped natural frequency.

Taking inverse Laplace transform

y(t) = 1ent(cos dt+ 12 sin dt)

= 1 112e

nt sin(

12nt+)

where

= tan1 1 2

81

The step responses are shown for several values of as a function of nt.

wn t

y(t)

Step response for second order system for various damping ratios

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2From: U(1)

To:Y(1)

2

1

0.7

0.4

0.2

0.1

0

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If=0, the frequency of the sinusoid is n - henceundamped natural frequency.

If 1 > > 0, d =

1 2n is the frequency ofthe sinusoid - hence damped natural frequency.

The time constant of the exponential envelopeis = 1

n

For 0 < < 1 the response is a damped sinusoid -hence damped (or underdamped) system.

As increases from 0 to 1, the response becomesless oscillatory - hence damping ratio.

For 1, the oscillations have ceased - henceoverdamped system( > 1) and critically dampedsystem( = 1).

When < 0, the response grows without limit -hence unstable system.

83

Time Response Specifications

Some system design specifications can be describedin terms of the step response.

For underdamped second order systems, we definethe following specifications:

Tr : rise time (10% 90%)Mp : peak value

Tp : time to first peakyss : steady state value

% overshoot :Mp yss

yss 100

Ts : settling time

0 2 4 6 8 10 12 14 16 180

0.2

0.4

0.6

0.8

1

1.2

1.4Typical step response

y(t)

Mpt

TpTr

1+d

1d

Ts

0.1

0.9

yss

84

Ts is the time required to enter a yss d tube.Common values are 2%, 4% and 5%.

The time constant is = 1/n so Ts 4= 4/n.

For overdamped (or critically damped) systems,Tr, Ts and yss (but not Mp or Tp) are well-defined.

With = tan1 1 2/, the step response foran underdamped 2nd order system is given by:

y(t) = 1 11

2

e nt sin(

1 2nt + )

Differentiating y(t) and equating to 0 gives

Tp =

1 2n , Mp = 1 + e

12

Since yss = 1,

% overshoot =Mp yss

yss 100 = 100e

12

As increases from 0 to 1, Mp, Ts and the percent-age overshoot decrease, while Tp and Tr increase.

A compromise must be found between the speed

of response (measured by Tp & Tr) and track-ing properties (measured by Ts & % overshoot).

85 86


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Interpretation of Pole Locations

Suppose thatg(s) =

y(s)

r(s)=

n(s)

d(s)=

(s z1)(s z2) (s zl)(s p1)(s p2) (s pm)

where n(s) is the zero polynomial and d(s) isthe pole polynomial.

The roots ofn(s) (z1, . . . , zl) are the zeros ofg(s)while the roots ofd(s) (p1, . . . , pm) are the polesofg(s).

Assuming no repeated poles and expanding in par-tial fractions the response to r(s) has the form

y(s) =r1

s p1 +r2

s p2 + +rm

s pm + a(s)where a(s) contains terms due to r(s).

y(t) = r1ep1t + r2ep2t + + rmepmt + a(t)

where a(t) is the forced response.

Since n(s) and d(s) are real, poles and zeros areeither real or occur in complex conjugate pairs.

Each real pole contributes a pure exponential, whilea pair of complex conjugate poles contribute an ex-ponentially decaying (or expanding) sinusoid.

87 88

Step Response of Higher Order Systems

Good test signal since it is easy to interpret.

Useful if system is predominantly second order.

A system g(s) is predominantly second order if

g(s) = a(s) +s +

s2 + bs + c

where the poles ofa(s) are located far into the lefthalf plane compared with the roots of s2 + bs + c,called the dominant poles.

The response associated with the poles ofa(s) willdecay fast, and the response will be largely deter-mined by the dominant poles.

E

T

k

xx

xroot of

poles ofa(s) s2 + bs + c

x

x

x

89

Introduction to Frequency Response

Let g(s) be stable and suppose that

u(t) = ejt = u(s) = 1s j

Expanding in partial fractions

y(s) = g(s)1

s j =g(j)

s j + a(s)

where a(s) includes all the poles of g(s).

Taking inverse Laplace transform

y(t) = g(j)ejt + a(t).

Since a(s) is stable the steady-state response is

yss(t) = g(j)ejt = |g(j)|ej[t+()]

where

g(j) =

gain |g(j)| ej

phase ()

Since ejt = cos t + j sin t

Im {u(t)} = sin tIm {yss(t)} = |g(j)| sin[t + ()]

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Steady-state Accuracy of Feedback Systems

Consider the feedback system shown below. Here,Q(s) represents the loop gain. Assume the feed-back loop is stable.

0)1(

Q(s)E ET

E

-

r e y

The Laplace transform of the error signal is

e(s) =r(s)

1 + Q(s)

ess = limt e(t) = lims 0sr(s)

1 + Q(s)

The system type N is defined to be the numberoffree integrators in the loop.

For a unit step

ess =1

1 + Kp, Kp = lim

s 0 Q(s)

For a type 0 system (N = 0), ess = 1/[1 +Q(0)],while ess =0 for type 1 and type 2 systems.

91

The following table gives ess for type 0, 1, and 2systems for step, ramp and parabolic inputs.

Type r(s) = 1s r(s) =1

s2 r(s) =1

s3 Error constants

0 11+Kp Kp = lims0 Q(s)

1 0 1Kv Kv = lims0 sQ(s)

2 0 0 1Ka Ka = lims0 s2Q(s)

Steady-state error constants

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Stability Analysis - The Routh-HurwitzStability Criterion

Introduction

Bounded-input Bounded-output Stability

General Properties of Polynomials

The Routh-Hurwitz Stability Criterion

The Routh Array

Stability for First & Second Order Systems

Applications in Feedback Design

93

Introduction

One of the fundamental design objectives in con-trol is the maintenance of system stability.

Even if a system is open-loop stable, the intro-duction of feedback may destabilise it.

Furthermore, most of the benefits of feedback areobtained by increasing the loop gain. Unfortu-nately, for most feedback systems, this tends todestabilise the closed-loop.

It is therefore important to investigate the stabilityproperties of control systems.

For a general (possibly nonlinear or time-varying)system, the problem of stability is usually one of

the most difficult to analyse.

Fortunately, for linear, time-invariant (LTI)systems, the question of stability is much easier todetermine.

94

Bounded-input Bounded-output Stability

A system is bounded-input bounded-outputstable, if, for every bounded input, the output re-mains bounded for all time.

Let g(s) be a transfer function representation ofan LTI system. Then, g(s) can be written as

g(s) =ms

m + m1sm1 + + 1s + 0nsn + n1sn1 + + 1s + 0

where the denominator polynomial is called the

characteristic polynomial ofg(s).

The equation

nsn + n1sn1 + + 1s + 0 = 0

is called the characteristic equation for g(s).

The roots of the characteristic equation are calledthe poles ofg(s).

The order of the system g(s) is defined to be thedegree of the characteristic polynomial.

95

We already know that Fact: An LTI system is bounded-input bounded-

output stable provided all its poles lie in the

left half-plane.

Fact: An LTI system is marginally stableif all its poles lie in the left half-plane, and

at least one pole on the imaginary axis.

Fact: An LTI system is unstable if any ofits poles lie in the right half-plane.

One way of determining stability is to calculate theroots of the characteristic equation.

The disadvantage of this is that system parametersmust be assigned numerical values, which makes itdifficult to find the range of values of a parameterthat ensures stability.

Furthermore, in a feedback control system, even ifwe know the open-loop system p oles, we still needto determine the closed-loop poles.

An alternative procedure which determines the lo-cation of the poles is presented next.

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General Properties of Polynomials

A left half-plane root of a characteristic polynomial(pole) is called a stable root (pole).

A right half-plane root is called an unstable root.

An imaginary axis root is called a marginallystable root.

Assume all polynomial coefficients are real.

Consider the second order polynomialP(s)= s2+1s+0 = (sp1)(sp2)

=s2(p1+p2)s+p1p2 1=(p1+p2), 0 =p1p2

p1 andp2 are either real, or else complex conjugate.If p1 & p2 are stable, we have 1 > 0 & 0 > 0.

97

Consider next the third order polynomial

P(s)=s3+2s2+1s+0 = (sp1)(sp2)(sp3)

=s3(p1+p2+p3)s2+(p1p2+p2p3+p1p3)sp1p2p3

2=(p1+p2+p3)1=(p1p2+p2p3+p1p3),

0=p1p2p3

Again, if all the roots are stable, all the coefficientswill have the same sign.

NOT THE OTHER WAY

98

Consider a general nth order polynomial:

P(s) = sn + n1sn1 + ... + 1s + 0

= (s p1)(s p2) (s pn)

By analogy with the previous examples, we have:

n1 = ni=1

pi

n2 =

product of roots taken 2 at a time

n3 = product of roots taken 3 at a time...

0 = (1)n ni=1

pi

We conclude:

1. If all the roots are stable, all the polynomial co-efficient will be positive.

2. If any coefficient is negative, at least one rootis unstable.

3. If any coefficient is zero, not all roots are stable.

99

The Routh-Hurwitz Stability Criterion

The stability of an LTI system requires that all thepoles (roots of the characteristic equation) must liein the left half-plane.

The Routh-Hurwitz stability criterion is an an-alytical procedure for determining if all roots of apolynomial lie in the left half-plane, without eval-uating these roots.

Consider the nth order polynomial

P(s) = nsn + n1sn1 + ... + 1s + 0

in which we can always assume 0 = 0.

If 0 = 0, we can write :

P(s) = s (nsn1 + n1sn2 + + 1)

P(s)

and work with P(s) instead.

Note that in the case that 0 = 0, P(s) will have

at least one root at the origin and we conclude thatthe LTI system is marginally stable or unstable.

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The Routh Array

The first 2 rows of the Routh array are formedfrom the characteristic polynomial coefficients

P(s) = nsn+n1sn1+n2sn2+n3sn3+n4sn4

+ +1s+0sn n n2 n4 n6 ...

sn

1

n1 n3 n5 n7 ...sn2 b1 b2 b3 b4 ...sn3 c1 c2 c3 c4 ...

... ... ... ... ... ...s2 k1 k2s1 l1s0 m1

The remaining rows are formed as follows:

b1 = 1n1

n n2

n1 n3

, b2 = 1

n1

n n4

n1 n5

c1 = 1b1

n1 n3b1 b2

, c2 = 1

b1

n1 n5b1 b3

...

m1 = 1l1

k1 k2l1 0

101

Once the Routh array is completed, we have the

Routh-Hurwitz Criterion

The number of unstable roots is equal to the

number of sign changes in the first column of

the array

102

Example: Consider the third order polynomial:

P(s) = s3 + s2 + 2s + 8

roots : 2.0, 0.5 j1.9365

The Routh array is:

s3 1 2s2 1 8s 61 8

The two sign changes (from +1 to 6 and from6 to +8) indicate two unstable roots.

Not all arrays can be completed in thismanner.

103

Case 2 Problems If the first element of a rowis zero, with at least one nonzero element in

the same row, the procedure is modified by re-

placing that first element with a small number such that || 1 and proceeding as before.

Example: Consider the polynomialP(s) = s5 + 2s4 + 2s3 + 4s2 + 11s + 10

roots : 1.31, 1.24 j1.04, 0.89 j1.46

The modified Routh array iss

5

1 2 11s4 2 4 10s3 0 6 0s2 12/ 10s 61 10

There are two sign changes (irrespective of the signof) indicating two unstable roots.

The 1st element of the 4th row is calculated as1

2 4 6

= 1

(12 4) = 4 12

12

A similar procedure is used to calculate the remain-ing rows.

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Case 3 Problems If every entry in a row iszero, the last modification will not give useful

information and another modification is needed.

An even polynomial is one in which the powersofs are even integers or zero only: s0, s2, s4, . . .

The roots of an even polynomial are symmetric

with respect to both the real and imaginary axes.

Example: Consider the even polynomialp(s) = s2 + 1

roots : js2 1 1s 0 01 ?

Using the previous modification would gives2 1 1

s 01 1

If is chosen positive - there are no sign changes,while a negative would give two sign changes.

The previous test is inconclusive.105

The modification is illustrated with an example.

Example: P(s) = (s+1)(s2+2) = s3+s2+2s+2 roots : 1, j

2

s3 1 2 auxiliary polynomial

s2 1 2 Pa(s) = s

2 + 2

s 0 2 0 2s1 2

dPa(s)ds

dPa(s)ds = 2s allows the array to b e completed.

The absence of a sign change indicates no unstableroots, but there are marginally stable roots (dueto the auxiliary p olynomial).

Applying the modified criterion to Pa(s) = s2 + 2:

s2 1 2

Pa(s) = s2 + 2

s 0 2 dPa(s)ds = 2s1 2

The absence of a sign change indicates no unstableroots, so all the roots are on the imaginary axis.We conclude the system is marginally stable.

106

Example: This example has a combination ofcase 2 and 3 problems:

P(s) = s4 + 4

roots : 1 j, 1 j

The modified Routh array is:s4 1 0 4 Pa(s) = s4 + 4s3 0 4 0 0 dPa(s)ds = 4s3s2 0 4s

16/

1 4

Summary of procedure: All elements of row 2 are zero (case 3 problem)

Row 1 defines the auxiliary polynomial Pa(s)

Row 2 is replaced by coefficients ofdPa(s)

ds

Only element 1 of row 3 is zero (case 2 problem)

The zero element is replaced by with

|

| 1

The two sign changes indicate two unstable roots.107

Stability for First & Second Order Systems

For a 1st order system with characteristic polyno-mial

P(s) = 1s + 0

it is clear that stability requires both coefficientsto have the same sign.

A 2nd order system with characteristic polynomialP(s) = 2s

2 + 1s + 0

has a Routh array

s2 2 0s 11 0

A necessary and sufficient conditionfor stabilityis that all coefficients have the same sign.

Thus, for 1st and 2nd order systems, stability prop-erties can be determined by inspection.

For higher order systems we can only conclude thata necessary condition for stability is that all thecoefficients have the same sign.

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Applications in Feedback Design

Consider the feedback system involving a plantG(s) and compensator K(s).

K(s) G(s)EET

EEr y

-

The closed-loop transfer function H(s) is

H(s) =G(s)K(s)

1 + G(s)K(s)

The closed-loop poles are given by the roots of thecharacteristic equation

1 + G(s)K(s) = 0

Suppose we write

G(s) =Ng(s)

Dg(s), K(s) =

Nk(s)

Dk(s)

where Ng(s), Dg(s), Nk(s) and Dk(s) are polyno-mials.

109

Then, the closed-loop transfer function is given by

H(s) =

Ng(s)Nk(s)

Dg(s)Dk(s)

1 +Ng(s)Nk(s)

Dg(s)Dk(s)

=Ng(s)Nk(s)

Ng(s)Nk(s) + Dg(s)Dk(s)

The poles of the closed-loop system are also givenby the roots of the characteristic polynomial

Ng(s)Nk(s) + Dg(s)Dk(s) = 0

The poles of the open-loop system G(s) are theroots of its characteristic polynomial

Dg(s) = 0

which are generally different from closed-loop poles.

A fundamental design objective in control is thestabilisation of unstable systems.

The Routh-Hurwitz stability criterion can be usedas an aid in feedback design.

110

Example: Let G(s) be given by

G(s) =1

s3+5s2+2s8 =1

(s1)(s+2)(s+4) poles : +1, 2, 4 unstable

Let K(s) = K be a constant controller. Find therange of values of K for which the closed-loop isstable.

The closed-loop poles are the roots of the charac-teristic equation

1 + KG(s) = 1 +K

s3+5s2+2s8 = 0

s3+5s2+2s+(K 8) = 0

The Routh array is:s3 1 2s2 5 K 8s 0.2(18 K)1 K 8

For stability we need8 < K < 18

111

Example: Let G(s) be given by

G(s) =1

s3 s2 10s 8

Since the coefficients of the characteristic polyno-mial do not have the same sign, we conclude thatG(s) is unstable.

Let K(s) = K be a constant controller. Find therange of values of K for which the closed-loop isstable.

The characteristic equation is given by

1 + KG(s) = 1 +K

s3 s2 10s 8 = 0

s3 s2 10s + (K 8) = 0

It follows that no choice of K can ensure all coef-ficients have the same sign.

We conclude that G(s) cannot be stabilised by aconstant controller.

We need dynamic compensation.112


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Root-Locus Analysis

Introduction

The Root-Locus

Criteria for Plotting the Root-Locus

Rules for Plotting the Root-Locus

Derivation of Root-Locus Plotting Rules

The Magnitude Criterion

Closed-Loop Stability Range

Other Feedback Configurations

Root-Locus Plotting Rules (Negative K)

113

Introduction

Consider the following feedback system:

K(s) G(s)EET

EEr y

-

The closed-loop transfer function is

H(s) =G(s)K(s)

1 + G(s)K(s)

The closed-loop poles are given by the roots of thecharacteristic equation

1 + G(s)K(s) = 0

We have seen that the response of an LTI systemis largely determined by the location of its poles.

In control system analysis, we are interestedin how the location of the closed-loop poles is de-termined by K(s).

In control system design, we are concernedwith finding a compensator K(s) that places theclosed-loop poles in some desired region of the com-plex plane.

114

The Root-Locus

Suppose first that the feedback compensator is aconstant, K(s) = K.

K G(s)EET

EEr y

-

The system characteristic equation becomes1 + KG(s) = 0

The root-locus (RL) of a system is a plotof the roots of the system characteristicequation (closed-loop poles) as K variesfrom 0 to .

For an nth order system, the root-locus is a familyof n curves traced out by the n closed-loop polesas K varies over its range.

An inspection of the root-locus allows us to findthe possible location of the closed-loop poles as afunction of K.

This enables us to determine the nature of the re-sponse of the closed-loop system for all possiblevalues of K.

115

Example

Suppose that

G(s) =1

s(s + 2)

so that the characteristic equation becomes

1 +K

s(s + 2)= 0 = s2 + 2s + K = 0

The closed-loop poles are then given by

s = 1 1 K

1. For 0 K < 1, the poles are real and unequal

2. For K = 1, the poles are real and equal

3. For K > 1, the poles are complex conjugateand lie on the s = 1 line in the complex plane.

The RL is shown on the following page.

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ImagAxis

Root-Locus Plot for G(s) = 1s(s+2)

117

Criteria for Plotting the Root-Locus

A point s lies on the RL if1 + KG(s) = 0 K = 1

G(s)

G(s) = 1K

Since K is positive, this is equivalent to:

1. The magnitude criterion:

K =1

|G(s)|

2. The angle criterion:

arg G(s) = r, r = 1, 3, 5, . . .

Since the angle criterion is independent of K, wecan use it to:1. determine whether a point s lies on the RL,

2. plot the RL.

The magnitude criterion can be used to find Kcorresponding to a point on the RL.

118

The figure illustrates the angle criterion for

G(s) =s z1

(s p1)(s p2)

A point s lies on the RL if

args z1

(s p1)(s p2) =

That is, ifarg (s

z1)

1 [arg (s

p1)

1

+arg (s

p2)

2

] =

h

T

E

U

}

.

x x

s

z1

1

s-plane

p1p2

12

In general, a point s lies on the RL ifi (angles from zi)i (angles from pi)=(2l+1)

119

Rules for Plotting the Root-Locus

1. The RL is symmetric w.r.t. the real axis.

2. The RL originates on the poles of G(s) (K= 0)and terminates on the zeros of G(s) (K).

3. IfG(s) has zeros at infinity, the RL will ap-proach asymptotes as K approaches infinity.The angles of the asymptotes are

A = (2l + 1)

, l = 0, 1, . . .

These asymptotes intersect the real axis at

A =

poles zerosnumber of poles number of zeros

4. The RL includes all points on the real axis to theleft of an odd number of poles and zeros.

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5. The breakaway points b are roots of

dG(s)

ds= 0

6. The RL departs from a pole pj at an angle

d =i

zi i=j

pi + (2l + 1)

and arrives at a zero zj at an angle

a =i

pi i=j

zi + (2l + 1)

where pi is the angle from pi to pj and zi is theangle from zi to zj.

121

6 5 4 3 2 1 0 1 25

4

3

2

1

0

1

2

3

4

5

Real Axis

ImagAxis

>bc

A

Root-Locus Plot for G(s) = s+2s(s+1)(s+3)(s+4), = 3

x: pole ofG(s)

: zero ofG(s)b: breakaway point

c: centre of asymptotes

A: angles of asymptotes

122

Derivation of Root-Locus Plotting Rules


This follows from the assumption that G(s) is aratio of two polynomials with real coefficients.So, the characteristic polynomial roots are ei-ther real or occur in complex conjugate pairs.

2. The RL originates on the poles (for K= 0)and terminates on the zeros (as K),of G(s).

Suppose that

G(s) =b

m(s

z

1)(s

z

2)

(s

zm

)

(s p1)(s p2) (s pn)

Assumption: bm > 0. (If bm < 0, replacewith bm > 0 and apply rules for negative K.)

Then the characteristic equation becomes1 + KG(s) = 0

or

(sp1) (spn)+ Kbm(sz1) (szm) = 0

If K = 0 then p1, . . . , pn must lie on the RL.

As K approaches infinity but s remains finite,the root-loci approach z1, . . . , zm.

123

3. If G(s) has zeros at infinity, the RL willapproach asymptotes as K . Theangles of the asymptotes are

A = (2l + 1)

, l = 0, 1, . . .

and they intersect the real axis at

A = poles zeros

number of poles number of zeros

Write the open-loop transfer function G(s) as

G(s) =bms

m+bm1sm1+

sn

+an1sn

1

+ =

bmsm +

sm+

+ where

= nm > 0 (no asymptotes if = 0)so that G(s) has zeros at infinity.

Then the RL for large s satisfieslim

s [1+ KG(s)]=1+Kbm

s= 0 s +Kbm = 0

which has roots s =Kbm and angles A.

In the previous example, = 3 and soA =

3,

A =(134)(2)

3=2

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4. The RL includes all points on the realaxis to the left of an odd number of polesand zerosThis follows from the following observations:

x

x

x

.

s

A point s lies on the RL ifi(angles from zi)

i(angles frompi)=(2l+1)

The point s is shown off the real axis for clarity.

The angle contribution from a pair of complexconjugate poles or zeros cancels out.

The angle contribution from a pole or zero tothe left ofs is = 0.

The angle contribution from a pole or zero tothe right ofs is =

125

5. The breakaway points b are among thereal roots of

dG(s)

ds= 0

This follows from the following observations:

The breakaway point b must be real.

The characteristic equation has at least two rootsat the breakaway point.

If a rational function R(s) has at least two rootsat b, then

dR(s)ds has at least one root at b:

R(s) = (s b)2C(s)dR(s)

ds= 2(s b)C(s) + (s b)2dC(s)

ds

= (s b)[2C(s) + (s b)dC(s)ds

]

Since

R(s) = 1 + KG(s)

has at least two roots at b,

dR(s)

ds= K

dG(s)

ds

has at least one root at b.

126

Example: LetG(s) =

s + 2

s(s + 1)=

s + 2

s2 + s

dG(s)

ds=

s2 + s (2s + 1)(s + 2)s2(s + 1)2

= s2 + 4s + 2

s2(s + 1)2

The derivative of G(s) vanishes atb = 2

2 = 3.4142, 0.5858

The root-locus is shown below.

5 4 3 2 1 0 1 21.5

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0

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Real Axis

ImagAxis

>

127

6. The RL departs from a complex pole pj(arrives at a complex zero zj) at an angled (a) given by

d =i

zi i=j

pi (2l + 1)

a =i

pi i=j

zi (2l + 1)

where pi (zi) is the angle from pi (zi) topj (zj)

This follows from the following observations:

A point s lies on the RL ifi(angles from zi)

i(angles frompi)=(2l+1)

For departure from pj (arrival to zj), s can betaken to be very close to pj (zj).

The angles to s from the other poles pi, i = j(other zeros zi, i = j) can be approximated bythe angles from these poles (zeros) to pj (zj).

The rule is illustrated in the following example.

128


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Example: Let

G(s) =1

(s + 2)(s2 + 2s + 2)

The root-locus is shown below.

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>


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Other Feedback Configurations

The RL technique can be used to analyse moregeneral feedback configurations.

Example: Consider the feedback loop in the fig-ure with

G(s) =1

s(s + k)and it is required to plot the roots of the charac-teristic equation (closed-loop poles)

1 + G(s) = 1 +1

s(s + k)= 0

as k is varied.

The characteristic equation can be rearranged ass2 + 1 + ks = 0 1 + k s

s2 + 1= 0

We can now plot the RL ofG(s) = s

s2 + 1

E G(s) ET

E

-

133

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0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

Real Axis

ImagAxis

< >

Root-Locus Plot for G(s) = ss2+1

134

Root-Locus Plotting Rules (Negative K)


2. The RL originates on the poles (for K = 0) andterminates on the zeros (as K), ofG(s).

3. IfG(s) has zeros at infinity, the RL will ap-proach asymptotes as K approaches infinity.The angles of the asymptotes are

A = 2l

, l = 0, 1, . . .

and these asymptotes intersect the real axis at

A

=

poles zerosnumber of poles number of zeros

4. The RL includes all points on the real axis to theleft of an even number of poles and zeros.

5. The breakaway points b are roots of

dG(s)

ds= 0

6. The RL departs from a pole pj at an angle

d =i

zi i=j

pi + 2l

and arrives at a zero zj at an angle

a =i

pi i=j

zi + 2l

where pi is the angle from pi to pj and zi is theangle from zi to zj.

135

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> >


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137


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Root-Locus Design

Introduction

Proportional (P) Controllers

Phase-Lead Compensation

Rate Feedback

Proportional-Plus-Derivative (PD) Controllers

Phase-Lag Compensation

Proportional-Plus-Integral (PI) Controllers

Proportional-Plus-Integral-Plus-Derivative (PID) Con-trollers

Compensator Realisation

138

Introduction

Consider the feedback control system involving aplant G(s) and compensator Kc(s).

Kc(s) G(s)EET

EE

-

er y

The closed-loop transfer function is

H(s) =G(s)Kc(s)

1 + G(s)Kc(s)

and the characteristic equation is given by

1 + G(s)Kc(s) = 0

Suppose that the open-loop response is deemed un-satisfactory and it is required to use feedbackcompensation to improve the response.

Root-locus principles can be used for:

1. closed-loop pole placement

2. improving stability and transient response

3. improving steady-state response.

139

Proportional Controllers

It may be possible to satisfy the design specifica-tions using a constant compensator Kc(s) = Kp,called a proportional (P) controller.

Example: Let G(s) = 1(s + 1)(s + 2)(s + 3)

. The

design specifications on the closed-loop poles are:

1. the dominant poles have damping ratio = .707

2. the third p ole has a decay rate faster than e3t.

The figure shows the RL with the = 0.707 damp-

ing ratio line. It is clear that the specifications canbe satisfied using a proportional controller.

4 3 2 1 0 1 22

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140

Using the rlocfind command in MATLAB givesKp = 2.4450 with the resulting closed-loop polesat 3.5923, 1.2038 j0.9495.

The following figure shows the step response of theresulting closed-loop system.

Time (sec.)

Amplitude

Step Response

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Note the small overshoot and the nonzero steady-state error (since the system is type 0).

141


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Phase-Lead Compensation

A phase-lead compensator has transfer function

Kc(s) = Ks zs p

where z and p are real and negative and

|z| < |p|.

A phase-lead compensator has a positive con-tribution to the angle criterion and can be usedto shift the RL towards the left in the complexplane.

e E

T

x

.

s

p z > 0

Hence a phase-lead compensator will improve thetransient response by improving closed-loop stabil-ity.

142

Example: LetG(s) =

1

s(s + 1)

(type 1 system). The design specifications are:

1. The two dominant poles have damping ratio = .707 and settling time of 2s.

2. The third pole decays at least as fast as e2t.

The figure shows the uncompensated system RL.

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> >

control lecture notes 01

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