Modern Control Theory

Modern Control Theory

10EE55

MODERN CONTROL THEORY Subject Code

: 10EE55

IA Marks

:

25

No. of Lecture Hrs./ Week

: 04

Exam Hours

:

03

Total No. of Lecture Hrs.

: 52

Exam Marks

: 100

PART - A UNIT - 1 & UNIT - 2

STATE VARIABLE ANALYSIS AND DESIGN: Introduction, concept of state, state variables and state model, state modeling of linear systems, linearization of state equations. State space representation using physical variables, phase variables & canonical variables 10 Hours UNIT - 3

Derivation of transfer function from state model, digitalization, Eigen values, Eigen vectors, generalized Eigen vectors. 6 Hours UNIT - 4 Solution of state equation, state transition matrix and its properties, computation using Laplace transformation, power series method, Cayley-Hamilton method, concept of controllability & observability, methods of determining the same 10 Hours PART - B UNIT - 5

POLE PLACEMENT TECHNIQUES: stability improvements by state feedback, necessary & sufficient conditions for arbitrary pole placement, state regulator design, and design of state observer, Controllers- P, PI, PID. 10 Hours

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UNIT - 6

Non-linear systems: Introduction, behavior of non-linear system, common physical non linearity-saturation, friction, backlash, dead zone, relay, multi variable non-linearity. 3 Hours

UNIT - 7 Phase plane method, singular points, stability of nonlinear system, limit cycles, construction of phase trajectories. 7 Hours UNIT - 8

Liapunov stability criteria, Liapunov functions, direct method of Liapunov & the linear system, Hurwitz criterion & Liapunov‟s direct method, construction of Liapunov functions for nonlinear system by Krasvskii‟s method. 6 Hours TEXT BOOKS: Digital control & state variable methods- M. Gopal - 2 1. 2003

nd

edition, THM Hill

2. Control system Engineering- I. J. Nagarath & M. Gopal, - 3 Age International (P) Ltd.

rd

edition, New

REFERENCE BOOKS:

1. Inc

State Space Analysis of Control Systems- Katsuhiko Ogata -Prentice Hall

Automatic Control Systems- Benjamin C. Kuo & Farid Golnaraghi, 8 2. edition, John Wiley & Sons 2003.

3.

th

Modern Control Engineering- Katsuhiko Ogata- PHI 2003

Control Control Engineering Engineering theory and practice- M. N. Bandyapadhyay PHI, 4. 2007

5.

Modern control systems- Dorf & Bishop- Pearson education, 1998

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UNIT - 6

Non-linear systems: Introduction, behavior of non-linear system, common physical non linearity-saturation, friction, backlash, dead zone, relay, multi variable non-linearity. 3 Hours

UNIT - 7 Phase plane method, singular points, stability of nonlinear system, limit cycles, construction of phase trajectories. 7 Hours UNIT - 8

Liapunov stability criteria, Liapunov functions, direct method of Liapunov & the linear system, Hurwitz criterion & Liapunov‟s direct method, construction of Liapunov functions for nonlinear system by Krasvskii‟s method. 6 Hours TEXT BOOKS: Digital control & state variable methods- M. Gopal - 2 1. 2003

nd

edition, THM Hill

2. Control system Engineering- I. J. Nagarath & M. Gopal, - 3 Age International (P) Ltd.

rd

edition, New

REFERENCE BOOKS:

1. Inc

State Space Analysis of Control Systems- Katsuhiko Ogata -Prentice Hall

Automatic Control Systems- Benjamin C. Kuo & Farid Golnaraghi, 8 2. edition, John Wiley & Sons 2003.

3.

th

Modern Control Engineering- Katsuhiko Ogata- PHI 2003

Control Control Engineering Engineering theory and practice- M. N. Bandyapadhyay PHI, 4. 2007

5.

Modern control systems- Dorf & Bishop- Pearson education, 1998

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10EE55 CONTENTS

Sl. No.

Titles

Page No.

1.

UNIT - 1 STATE VARIABLE ANALYSIS AND DESIGN

04

2.

UNIT - 2 STATE SPACE REPRESENTATION REPRESENTATION

24

3.

UNIT - 3 DERIVATION OF TRANSFER FUNCTION FROM STATE MODEL

42

4.

UNIT- 4 SOLUTION OF STATE EQUATIONS EQUATIONS

57

5.

UNIT - 5 POLE PLACEMENT TECHNIQUES TECHNIQUES

78

6.

UNIT -6 NON LINEAR SYSTEMS

99

7.

UNIT - 7 PHASE PLANE ANALYSIS

105

8.

UNIT -

121

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8 STABILITY ANALYSIS

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UNIT - 1 & UNIT - 2 STATE VARIABLE ANALYSIS AND DESIGN: Introduction, concept of state, state variables and state model, state modeling of linear systems, linearization of state equations. State space representation using physical variables, phase variables & canonical variables 10 Hours

State Variable Analysis & Design State Var i able Anal ysi s or State Space A nal ysi s : - The state variable approach is a powerful tool/technique for the analysis and design of control system. The state space analysis is a modern approach and also easier for analysis using digital computers. It's gives the total internal state of

the system considering all initial

conditions. Wh y do we need state space anal ysis?

The conventional approach used to study the behaviour of linear time invariant control systems, uses time domain or frequency domain methods. When performance specifications are given for single input, single output linear time invariant systems, then system can be designed by using Root locus. When time domain specifications are given, Root locus technique is employed in designing the system. If frequency domain specifications are given, frequency response plots like Bode plots are used in designing the system. In conventional methods, the systems are modelled using Transfer Function approach, which is the ratio of Laplace transform of output to input, neglecting all the initial conditions.

The drawbacks in the transfer function model and analysis are,

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1.

Transfer function is defined under zero initial conditions.

2.

transfer function is applicable to linear time invariant systems.

3.

It is restricted to single input and single output systems.

4.

Does not provides information regarding the internal state of the system.

5.

The classical methods like Root locus, Bode plot etc. are basically trial and

error procedures which fail to dive the optimal solution required State variables analysis can be applied for any type of systems like,  

Linear system Non- linear system



Time invariant system



Time varying system



Multiple input and multiple output system.

the analysis can be carried with initial conditions.

A dvantages of state var iabl e anal ysis

1.

Convenient tool for MIMO systems

2.

Uniform platform for representing time-invariant systems, time-varying

systems, linear systems as well as nonlinear systems 3.

Can describe the dynamics in almost all systems (mechanical systems,

electrical systems, biological systems, economic systems, social systems etc. 4.

It can be performed with initial conditions.

5.

Variables used to represent system can be any variables in the system.

6.

Using this analysis the internal states of the system at any time instant can be

predicted. 7.

As the method involves matrix algebra, can be conveniently adopted for

digital computers.

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Comparison: Classical vs. Modern Control Classical Control (Linear) 

Modern Control (Linear) 

Developed in 1920-

1950

1980





Frequency domain analysis & Design(Transfer function based) Based on SISO  models Deals with input and  output variables Well-developed  robustness concepts (gain/phase margins) No  Controllability/Observabilit y inference No optimality  concerns Well-developed  concepts and very much in use in industry

Developed in 1950-

Time domain analysis and design(Differential equation based) Based on MIMO  models Deals with input,  output and state variables Not well-developed  robustness concepts 

Controllability/Observabilit y can be inferred Optimality issues can  be incorporated Fairly well-developed  and slowly gaining popularity in industry

State:- The state is the condition of a system at any time instant 't'.

State vari able:- A set of variable which described the state of the system at any time instant are called state variables. OR The state of a dynamic system is the smallest number of variables (called state variables)

such that the knowledge of these variables at t=to, together with the

knowledge of the input for t=to, completely determine the behaviour of the system for any time t ≥ to. CITSTUDENTS.IN

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State space: - The set of all possible values which the state vector X(t) can have( or assume) at time t forms the state space of the system.

State vector : - It is a (nx1) column matrix whose elements are state variables of the system,(where n is order of system) it is denoted by X(t).

State Variable Selection 

Typically, the number of state variables (i.e. the order of the system)

is equal to the number of independent energy storage elements. However, there are exceptions! 

Is there a restriction on the selection of the state variables ?

YES! All state variables should be linearly independent and they must

collectively describe the system completely.

State Space F orm ul ati on

In the state variable formulation of a system, in general, a system consists of m-inputs, p-outputs and n-state variables. The state space representation of the system may be visualized as shown in figure1.1. Let,

State variables =

(t),

(t),

(t),...................

Input variables =

(t),

(t),

(t),...................

(t),

(t),

(t),

(t),...................

(t),

Output variables=

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(t),

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(t)

→

→

(t)

(t)

→

→

(t)

→

(t)

→

(t)

Control

(t) (t)

→

System

→ ↓

(t)

↓

↓

↓

(t)

(t) .....

(t)

Control

U→

System

→Y

↓

X The different variables may be represented by the vectors(column matrix) as shown below Input vector

Output vector

State variable vector

State variable representation can be arranged in the form of n number of first order differential equations.

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= ẋ = f 1 (x1, x2, x3,......x n; u1, u2, .......um) = ẋ2 = f 2 (x1, x2, x3,......xn; u1, u2, .......um) . . . = ẋn = f n (x1, x2, x3,......x n; u1, u2, .......u m) Any 'n' dimensional time invariant system has state equations in the function form as,

ẋ(t) = f( X, U) ...... State

equation ......(1)

While out puts of such system are dependent on the state of the system and instantaneous inputs. Functional output equation can be written as,

Y(t) = g(X, U) .......Output equation .....(2) State M odel Of L i near System 

State model of a system consist of the state equation & output equation.



The state equation of the system is a function of state variables and inputs as defined

by equation 1. For linear tine invariant systems the first derivatives of state variables can be expressed as linear combination of state variables and inputs.

ẋ1 = a11 x1 + a12 x2 + .............+ a 1n xn + b11 u1 +.............+ b 1m um ẋ2 = a21 x1 + a22 x2 + .............+ a 2n xn + b21 u1 +.............+ b 2m um CITSTUDENTS.IN

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.. .

ẋn = an1 x1 + an2 x2 + .............+ a nn xn + bn1 u1 +.............+ b nm um In matrix form the above equation can be expressed as

=

+

It can also be written as

ẋ(t) = A X(t) + B U(t) State space analysis Classical control theory Vs modern control theory The development of control system analysis and design can be divided into three eras. In the first era, we have classical control theory , which deals with techniques developed before 1950. classical control embodies such methods as root locus, Bode, Nyquist and Routh- Hurwitz. These methods have in common the use of transfer functions in the complex frequency(s) domain, emphasis on the use of

graphical

techniques, the use of feedback and the use of simplifying assumptions to approximate the time response . since computers were not available at that time , a great deal of emphasis was placed on developing methods that were amenable to manual computation and graphics. A major limitation of classical control methods was the use of single – input , single output (SISO) methods. Multivariable (i.e

multiple – input

multiple – output or MIMO) systems were analyzed and designed one loop at a time . Also the use of transfer functions and the frequency domain limited one to linear time – invariant systems. In the second era , we have modern control ( which is not so modern any longer ) CITSTUDENTS.IN

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which referees to state – space methods developed in the late 1950,s and early 1960s. In modern control, system models are directly written in the time domain. Analysis and design are done in time domain. It should denoted that before Laplace transforms and transfer functions became popular in the 1920s Engineering were studying s ystems in the time domain. Therefore the resurgence of time domain analysis was not unusual, but it was triggered by the development of computers and advances in numerical analysis. Because computers were available, it was no longer necessary to development analysis and design methods

that were strictly manual. An Engineer could use computers to

numerically solve or simulate large system that were nonlinear and time varying. State space methods removed the previously mentioned limitations of classical control. The period of the1960s was the heyday of modern control.

System representation in state variable form This chapter introduce the concept of state variable and the various means of representing control systems

in state variable form. Each method of state variable

representation results in a s ystem description interms of „n‟ first – order differential th

equations, as opposed to the usual n

– order equation. s A convenient tool for

this new system representations is matrix notation.

System state and state variable It is important to stress at out set that the concept of system state is first of all, a physical concept. However it is often convenient to interms of mathematical model. Here this mathematical model is assumed to consist of ordinary differential equations which have a unique solution for all inputs and initial conditions. It is in terms of this mathematical model that the „system state‟ or simply „state‟ is defined. Definition The state of system at any time „t0‟ is the minimum set of numbers

X1 (to) ,X2(to) -------- Xn(to) which along with the input to the system for t sufficient to determine the behaviour of the system for all t

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to

is

to.

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Modern Control Theory In

other

words,

the

state

10EE55 of

a

system

represents

the

minimum

amount

of

information that we need to know about a system at t o such that its future behaviour can be determined without reference to the input before to.

The idea of state is familiar from a knowledge of the physical world and the means of solving the differential equations used to model the physical world . Consider a ball flying through the air. Intuitively we feel that if we know the ball‟s position and velocity, we also know its future behaviour. It is on this basis that an outfielder positions himself to catch a ball. Exactly the same information is needed to solve a differential equation model of the problem. Consider for example the second order differential equation ·· · X +aX + bX = f(t) The solution to this equation be found as the sum of the forced response, due to f(t) and the natural or unforced response ie the solution of homogeneous equation ·· · X +aX + bX = 0 If X1

(t), X2(t) -------- Xn(t) are state variables of the system chosen then the

initial conditions of the state variables plus the u(t)‟s for t > 0 should be sufficient to decide the future behaviour i.e y(t)‟s for t > 0. Note that the state variables need not be physically measurable or observable quantities. Practically however it is convenient to chose easily measurable quantities. The number of state variable is then equal to the order of the differential equation which is normally equal to the number of energy storage elements in the system.

State equations for linear systems in matrix form

The state of a linear time – invariant nth order system is represented by the following set of „n‟ number of first order differential equations with constant coefficients in terms of n state variable X1,X2-------- Xn. CITSTUDENTS.IN

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X1 = a11 X1+a12 X2 + ------- --- +a1n Xn + b11U1 + ----- - +b1m U m

· X2 = a21 X1+a22 X2 + ------- --- +a21 Xn + b21U1 + ----- - +b2m U m

Xn = an1 X1+an2 X2 + ------- --- +ann Xn + bn1U1 + ----- - +bnm U In matrix form the above equations may be written as

X1

a11 a12 ----------- a1n

x1

b11 b12 ----------- b1m

u1

X2

a21 a22 ----------- a2n

x2

b21 b22 ----------- b2m

u2

bn1 bn2 ----------- bnm

um

=

an1 an2 ----------- ann

Xn

X

+

A

=

xn

X

+

B

U

X is called derivative of state vector whose size is (nx1) · X is called state vector whose size is (n x 1) A is called system matrix whose size is (n x n) B is called input matrix whose size is ( n x m) U is called input vector whose size is (m x 1)

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Output equation

The state variables X1(t) -------X n(t) represents the dynamic state of a system. The system output/ outputs may be used as some of the state variables themselves ordinarily, the output variables Y1 = C11 X1 + C12 X2 + ---------- C1n Xn Y2 = C21 X1 + C22 X2 + ---------- C2n X

Y p = C p1 X1 + C p2 X2 + ---------- CPn Xn In matrix form,

y1

c11c 12 ----------- c1n

x1

y2

c21 c22 ----------- c2n

x2

c p1 c p2 ----------- c pn

xn

=

y p or

Y

=

C

X

Y = output vector of size ( P x 1) C = Transmission matrix of size ( P x n) X = State vector of size ( n x 1) Sometimes the output is a function of both state variables and inputs . for this general case

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Modern Control Theory Y = CX +DU or

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Y1

c11 c12 ----------- c1n

x1

D11 D12 -------- D1m

u1

Y2

c21 c22 ----------- c2n

x2

D21 D22 -------- D2m

u2

DP1 DP2 --------- DPm

um

=

Y p

+

c pn -----------

cpn

xn

D matrix is of size ( p x m) State Model

The state equation of a system determines its dynamic state and the output equation gives its output at any time t > 0 provided the state at t = 0 and the control forces for t 0 are known. These two equations together form the state model of the system. The state model of a linear system is therefore given by · X = AX +BU Y = CX+ DU

(1 )

State Model of SISO linear and time invariant system.

If we let m =1 and p=1 in the state model of a multiple input multiple output linear time invariant system we obtain the following state model for SISO linear system. · X = AX +bu Y = CX+ du

(2 )

Where b and C are ( nx1) vectors

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State Model using phase – variables ( BUSH FORM)

Let us now consider how the state model defined by equation (2) may be obtained for an nth order SISO system whose describing differential equation relating output y with input u is given by n

d y n

dt

an-1

+

d

n-1

dt

where an-1 , an - 2

y

n-1

------------

dy (0), dt

y(0),

+ an-2

d

n-2

dt

y

n-2

+ ------------ a1

dy + a0 y dt

= b0 u

(3)

a1, a0 are constant coefficients d

n-1

------------

y (0),

are initial conditions

dt

To arrive at the state model of equation (3) it is rewritten in shorthand form as n

n-1

y +an-1 y

n-2

+ an-2 y

+--------------a1 y· + a0 y = b0u

(4)

We first define the state variables X1, X2, ----------- Xn which can be done in many possible w ays. A convenient way is define X1 = y X2 = ·y

· n-1

Xn = y With the above definition of state variables equation (4) is reduced to a set of „n‟ first order differential equations given below; · X1 = ·y = X2 · X2 = ·y· = X3

· n-1 Xn-1 = y = Xn · n X = y = -a0X1 – a1X2 – a2X3 -----a n-1Xn +b0u CITSTUDENTS.IN

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· the above equations result in the following state equations

It is to be noted that the matrix A has a special form. It has all 1‟s in the upper off – diagonal, its last row is comprised of the negative of the coefficients of the original

differential equation and all other elements are zero. This form of matrix A is known as the „BUSH FORM‟ . The set of state variables which yield „BUSH FORM‟ for the matrix A are called „Phase variables ‟

When „A‟ is in „BUSH FOR M‟, the vector b has the specialty that all its elements except the last are zero. In fact A and B and therefore the state equation can be written directly by inspection of the linear differential equation. The output being y = X1 , the output equation is given by

Where C = [ 1 0 ------0] Note: There is one more state model called canonica l state model . we shall consider this model after going through transfer function. CITSTUDENTS.IN

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Derivation of transfer function from a given state model

Having obtained the state model we next consider the problem of determining transfer function from a given state model of SISO / MIMO systems. 1) SISO SYSTEM y(s)

u(s) G(s)

G(s) is called transfer function defined as y(s) G(s) =

y(s) = u(s) G(s) or y (s) = G(s) u(s) -------- (1) u(s)

State Model is given by X (t) = AX (t) + Bu Y(t) = CX (t) + Du

----------- (2) ----------- (3)

Taking laplace transformation on both sides of equations ( 2) and (3) and neglect ing initial conditions We get sX(s) = AX (s) + Bu(s) -----(4) Y(s) = CX (s) + Du(s) ---- (5) From (4) (sI – A ) X(s) = Bu(s) -1

Or X(s) = --------- (sI – A ) Bu(s) --------(6)

Substituting ( 6) in (5) -1

Y(s) = C (sI – A ) Bu(s) + Du(s) -1

Y(s) = ( C (sI – A ) B+D) u(s) --------------- (7) Comparing ( 7 ) with (1)

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Modern Control Theory -1

G(s) = C( sI – A ) B + D

10EE55 ---------( 8)

An important observation that needs to be made here is that while the state model is not unique, the transfer function is unique. i.e. the transfer function of equation (8) must work out to be the same irrespective of which particular state model is used to describe the system. (ii) MIMO SYSTEM y1 (s)

u1(s)

G(s)

u2(s)

y2 (s)

m inputs

P out puts yn (s)

um(s)

G(s) = C( sI – A ) -1B + D Where y(s) =G(s) U(s)

G(s) matrix is called transfer matrix of size (p x m) y(s) matrix is of size (px1) u(s) matrix is of size (mx1) y1 (s) = G 11(s) u1 (s) + G 12(s) u2(s) + y1 (s) Transfer function 11(s)

=

-------------------+ G 1m(s)u m(s)

G

u1(s) u2(s) = u3(s) = ----- um(s) = 0

Similarly G 12 (s), -------- are defined CITSTUDENTS.IN

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Derivative of state models from transfer function

More often the system model is known in the transfer function form. It therefore becomes necessary to have methods available for converting the transfer function model to a state model. The process of going from the transfer function to the state equations is called decomposition of the transfer function. In general there are three basic ways of decomposing a transfer function in direct decomposition, parallel decomposition, and cascaded decomposition has its own advantage and is best suited for a particular situation. Decomposition of TF 1. Converting a TF with a constant term in numerator. Phase variablesvariables that are successive derivatives of each other.

3

2

s C + 9s C + 26sC + 24 c = 24 R

Take in LT

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2. Converting a TF with polynomial in numerator

Taking in LT

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2

C(S) = s X1 +7s X1 + 2 X1 .. . C(t) = X1 + 7 X1 + 2 X1 = X3 + 7 X2 + 2 X1

3. Cascading form

The denominator of TF is to be in factor form

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R (S)o.o-27-:-« r--s --31·.- ....--c;; ----....,

x3

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;»--s·-.1..-.---1---oo C(S)

x3x2 -2

X 1=X2- 4

X

1

X2= X3- 3

X

1

x2x1 -3

-4. .·

x 3=- 2X3+ 2 4 V

C =

[ 1

0

x1

0 ]

x2 x3

.

X1 x.2

X: l

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·- 4

1

0

X1

0

x2 +

0 R

0

-3

1

0

0

-2

x3

24

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UNIT-2 STATE-SPACE REPRESENTATION

1

Introduction

The classical control theory and methods (such as root locus) that we have been using in class to date are based on a simple input-output description of the plant, usually expressed as a transfer function. These methods do not use any knowledge of the interior structure of the plant, and limit us to single-input single-output (SISO) systems, and as we have seen allows only limited control of the closed-loop behavior when feedback control is used. Modern control theory solves many of the limitations by using a much “r icher” description of the plant dynamics. The so-called state-space description provide the dynamics as a set of coupled first-order diff erential equations in a set of internal variables known as state variables, together with a set of algebraic equations that combine the state variables into physical output variables. 1.1

Definition of System

State

The concept of the state of a dynamic system refers to a minimum set of variables, known as state variables, that fully describe the system and its response to any given set of inputs [1-3]. In particular a state-determined system model has the characteristic that: A mathematical description of the system in terms of a minimum set of variables xi (t ), i = 1 , . . . , n, together with knowledge of those variables at an initial time t 0 and the system inputs for time t ≥ t 0 , are sufficient to predict the future system state and outputs for all time t > t 0 . This defi nition asserts that the dynamic behavior of a state-determined system is completely characterized by the response of the set of n variables xi (t ), where the number n is defined to be the order of the system. The system shown in Fig. 1 has two inputs u1 (t ) and u2 (t ), and four output variables y1 (t ) , . . . , y4 (t ). If the system is state-determined, knowledge of its state variables ( x1 (t 0 ) , x2 (t 0 ) , . . . , xn (t 0 )) at some initial time t 0 , and the inputs u1 (t ) and u2 (t ) for t ≥ t 0 is sufficient to determine all future behavior of the s ystem. The state variables are an internal description of the system which completely characterize the system state at any time t , and from which any output variables yi (t ) may be computed. Large classes of engineering, biological, social and economic systems may be represented by state-determined system models. System models constructed withthe pure and ideal (linear) one-port elements (suchas mass, spring and damper elements) are state-determined CITSTUDENTS.IN

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Figure 1: System inputs and outputs. system models. For suchsystems the number of state variables, n , is equal to the number of independent energy storage elements in the system. The values of the state variables at any time t specify the energy of each energy storage element within the system and therefore the total system energy, and the time derivatives of the state variables determine the rate of change of the system energy. Furthermore, the values of the system state variables at any time t provide su fficient information to determine the values of all other variables in the system at that time.

There is no unique set of state variables that describe any given system; many diff erent sets of variables may be selected to yield a complete system description. However, for a given system the order n is unique, and is independent of the particular set of state variables chosen. State variable descriptions of systems may be formulated in terms of physical and measurable variables, or in terms of variables that are not directly measurable. It is possible to mathematically transform one set of state variables to another; the important point is that any set of state variables must provide a complete description of the system. In this note we concentrate on a particular set of state variables that are based on energy storage variables in physical systems.

1.2

The State Equations

A standard form for the state equations is used throughout system dynamics. In the standard form the mathematical description of the system is expressed as a set of n coupled first-order ordinary diff erential equations, known as the state equations , in which the time derivative of each state variable is expressed in terms of the state variables x1 (t ) , . . . , x n (t ) and the system inputs u1 (t ) , . . . , ur (t ). In the general case the form of the n state equations is: x˙1

=

f 1 (x , u , t )

x˙2

=

f 2 (x , u , t )

. x˙n CITSTUDENTS.IN

= =

( 1)

. f n (x , u , t Page 25


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and eachof the functions f i (x , u , t ), (i = 1 , . . . , n) may be a general where x˙i = dx /dt i nonlinear, time varying function of the state variables, the system inputs, and time. 1 It is common to express the state equations in a vector form, in which the set of n state T variables is written as a state vector x(t ) = [x 1 (t ) , x2 (t ) , . . . , x n (t )] , and the set of r inputs T is written as an input vector u(t ) = [u1 (t ) , u2 (t ) , . . . , ur (t )] . Each state variable is a time varying c omp onent of the column vector x(t ). This form of the state equations explicitly represents the basic elements contained in the de finition of a state determined system. Given a set of initial conditions (the values of the xi at some time t 0 ) and the inputs for t ≥ t 0 , the state equations explicitly specify the derivatives of all state variables. The value of each state variable at some time ∆ t later may then be found by direct integration. The system state at any instant may be interpreted as a point in an n-dimensional state space , and the dynamic state response x (t ) can be interpreted as a path or trajectory traced out in the state space. In vector notation the set of n equations in Eqs. (1) may be written: x ˙

=

f (x , u , t ) .

(2)

where f (x , u , t ) is a vector function with n components f i (x , u , t ). In this note we restrict attention primarily to a description of systems that are linear and time-invariant (LTI), that is systems described by linear di ff erential equations with constant coefficients. For an LTI system of order n , and with r inputs, Eqs. (1) become a set of n coupled first-order linear di ff erential equations with constant coefficients: x˙1 x˙2

= =

. x˙n =

a11 x1 a21 x1

. an 1 x1

+ +

a12 x2 a22 x2

+ an 2 x2

+ +

... ...

+ ...

+ +

+

a1 n x n a2 n x n

a nn xn

+ +

+

b11 u1 b21 u1 bn1 u1

+

... ...

+

+

...

+

+

+

b1r ur b2r ur

(3)

b nr ur

where the coefficients aij and bij are constants that describe the system. This set of n equations de fines the derivatives of the state variables to be a weighted sum of the state variables and the system inputs. Equations (8) may be written compactly in a matrix form: As given above in page 10 ˙ = Ax x

+

Bu (5)

In this note we use bold-faced typ e to denote vector quantities. Upper case letters are used to denote general matrices while lower case letters denote column vectors. See Appendix A for an introduction to matrix notation and o p e rations.

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where the state vector x is a column vector of length n , the input vector u is a column vector

of length r , A is an n × n square matrix of the constant coe ffi cients aij , and B is an n × r matrix of the coe fficients bij that weight the inputs.

1.3

Output Equations A system output is de fined to be any system variable of interest. A description of a

physical system in terms of a set of state variables does not necessarily include all of the variables of direct engineering interest. An important property of the linear state equation description is that all system variables may be represented by a linear combination of the

state variables xi and the system inputs ui . An arbitrary output variable in a system of order n with r inputs may be written: y(t ) = c1 x1

+

c2 x2

+

...

+

c n x n

+

d 1 u1

+

...

+

(6)

d r ur

where the ci and d i are constants. If a total of m system variables are de fined as outputs, the m suchequations may be written as: y1 y2

. y m

= =

c11 x1 c21 x1

. = c m1 x1

+

+

c12 x2 c22 x2

+

c m2 x2

+

+

+

c1 n x n c2 n x n

+

c mn x n

+

+

. .. . ..

+

. ..

+

+

... ...

+

d 1r ur d 2r ur

+

...

+

d mr ur

+

+

d 11 u1 d 21 u1

+

d m1 u1

(7)

or in matrix form: As given above in page 10

The output equations, Eqs. (8), are commonly written in the compa.ct form: y

=

Cx

+

(9)

Du

where y is a column vector of the output variables yi (t ), C is an m × n matrix of the constant coe fficients cij that weight the state variables, and D is an m × r matrix of the constant coe fficients d ij that weight the system inputs. For many physical systems the matrix D is the null matrix, and the output equation reduces to a simple weighted combination of the state variables: y = Cx. (10) CITSTUDENTS.IN

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State Equation Based Modeling P rocedure

The complete system model for a linear time-invariant system consists of (i) a set of n state equations, defined in terms of the matrices A and B , and (ii) a set of output equations that relate any output variables of interest to the state variables and inputs, and expressed in terms of the C and D matrices. The task of modeling the system is to derive the elements of the matrices, and to write the system model in the form: x ˙

=

Ax + Bu

y

=

Cx + Du.

(11) (12)

The matrices A and B are properties of the system and are determined by the system structure and elements. The output equation matrices C and D are determined by the particular choice of output variables. The overall modeling procedure developed in this chapter is based on the following steps: 1. Determination of the system order n and selection of a set of state variables from the linear graphsystem representation. 2. Generation of a set of state equations and the system A and B matrices using a well defined methodology. This step is also based on the linear graph system description. 3. Determination of a suitable set of output equations and derivation of the appropriate C and D matrices.

2

Block Diagram Representation Described by State Equations

of Linear Systems

The matrix-based state equations express the derivatives of the state-variables explicitly in terms of the states themselves and the inputs. In this form, the state vector is expressed as the direct result of a vector integration. The block diagram representation is shown in Fig. 2. This general block diagram shows the matrix operations from input to output in terms of the A , B , C , D matrices, but does not show the path of individual variables. In state-determined systems, the state variables may always be taken as the outputs of integrator blocks. A system of order n has n integrators in its block diagram. The derivatives of the state variables are the inputs to the integrator blocks, and each state equation expresses a derivative as a sum of weighted state variables and inputs. A detailed block diagram representing a system of order n may be constructed directly from the state and output equations as follows: Step 1: Draw n integrator (S −1 ) blocks, and assign a state variable to the output

of each block. CITSTUDENTS.IN

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Figure 2: Vector block diagram for a linear system described by state-space system dynamics. Step 2: At the input to each block (which represents the derivative of its state variable)

draw a summing element. Step 3: Use the state equations to connect the state variables and inputs to the

summing elements through scaling operator blocks. Step 4: Expand the output equations and sum the state variables and inputs through

a set of scaling operators to form the components of the output.

Example 1 Draw a block diagram for the general second-order, single-input single-output system: x˙1 x˙2 y (t )

=

a11 a12 a21 a22

=

c1 c2

x1 x2 x1 2

+

b1 b2

u(t )

+ du (t ).

(i)

Solution: The block diagram shown in Fig. 3 was drawn using the four steps

described above.

3 .Transformation From State-Space Equations to Classical Form The transfer function and the classical input-output di fferential equation for any system variable may be found directly from a state space representation through the Laplace transform. The following example illustrates the general method for a first order system.

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Figure 3: Block diagram for a state-equation based second-order system.

Example 2 Find the transfer function and a single first-order di ff erential equation relating the output y (t ) to the input u(t ) for a system described by the first-order linear state and output equations: dx

=

dt y(t ) =

ax(t ) + bu(t )

(i )

cx(t ) + du(t )

(ii)

Solution: The Laplace transform of the state equation is

sX ( s) = aX ( s) + bU ( s) ,

(iii)

which may be rewritten with the state variable X ( s) on the left-hand side:

( s − a ) X ( s))

=

bU ( s).

Then dividing by ( s

− a),

(iv) solve for the state variable: b U ( s) , X ( s) = s − a

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and substitute into the Laplace transform of the output equation Y ( s) = cX ( s) + dU ( s): Y ( s)

bc

=

+

s − a

d U ( s)

ds +U ( s) (bc − ad ) ( s − a)

(vi)

=

The transfer function is: H ( s) =

Y ( s) U ( s)

ds + (bc −

=

ad ) ( s − a )

(vii)

.

The d ifferential equation is found directly: ( s − a ) Y ( s) = (ds

+

( bc − ad )) U ( s) ,

(viii)

and rewriting as a d ifferential equation: dy − ay dt

=

d

du + ( bc − ad ) u(t ). dt

(ix)

Classical representations of higher-order systems may be derived in an analogous set of steps by using the Laplace transform and matrix algebra. A set of linear state and output equations written in standard form x ˙

=

Ax + Bu

y

=

Cx + Du

(13) (14)

may be rewritten in the Laplace domain. The system equations are then sX ( s) = A X ( s) + BU ( s) Y( s) =

(15)

CX( s) + DU ( s)

and the state equations may be rewritten: sx( s) − Ax( s) = [sI − A ] x( s) = Bu( s).

(16)

where the term sI creates an n × n matrix with s on the leading diagonal and zeros elsewhere. (This step is necessary because matrix addition and subtraction is only de fined for matrices of the same dimension.) The matrix [sI − A] appears frequently throughout linear system theory; it is a square n × n matrix withelements directly related to the A matrix:  

[ sI − A] =

CITSTUDENTS.IN







( s

− a11 )

··· − a 1n −a21 ( s − a 22 ) · · · −a 2n . . . . .. . −an1 −an2 · · · ( s − a nn ) −a12

   . 

(17)

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The state equations, written in the form of Eq. (16), are a set of n simultaneous operational expressions. The common methods of solving linear algebraic equations, for example Gaussian elimination, Cramer’s rule, the matrix inverse, elimination and substitution, may be directly applied to linear operational equations such as Eq. (16). For low-order single-input single-output systems the transformation to a classical formulation may be performed in the following steps: 1. Take the Laplace transform of the state equations. 2. Reorganize each state equation so that all terms in the state variables are on the left-hand side. 3. Treat the state equations as a set of simultaneous algebraic equations and solve for those state variables required to generate the output variable. 4. Substitute for the state variables in the output equation. 5. Write the output equation in operational form and identify the transfer function. 6. Use the transfer function to write a single di ff erential equation between the output variable and the system input. This method is illustrated in the following two examples.

Example 3 Use the Laplace transform method to derive a single d iffe rential equation for the capacitor voltage vC in the series R-L-C electric circuit shown in Fig. 4 Solution: The linear graph method of state equation generation selects the

Figure 4: A series RLC circuit.

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capacitor voltage vC (t ) and the inductor current i L (t ) as state variables, and generates the following pair of state equations: v˙ c i˙ L

1 /C

0 −1 /L

=

L R/

−

0

vc i L

+

+

0 V in

1 /L

V in .

(i)

The required output equation is: y (t ) =

vc i L

1 0

(ii)

Step 1: In Laplace transform form the state equations are: sV C ( s)

0V C ( s) + 1 /C I L ( s) + 0 V s ( s) /LV C ( s) − R/LI L ( s) + 1 /LV s ( s) −1

=

sI L ( s ) =

(iii)

Step 2: Reorganize the state equations: sV C ( s) − 1 /C I L ( s)

1 /LV C ( s) + [s + R/ L] I L ( s)

0V s ( s)

=

1 /LV s ( s)

=

(iv) (v )

Step 3: In this case we have two simultaneous operational equations in the state variables vC and i L . The output equation requires only vC . If Eq. L], and Eq. (v) is multiplied by 1 /C , and the (iv) is multiplied by [s + R/ equations added, I L ( s) is eliminated: [s ( s

R/L ) + 1 /LC ] V C ( s) = 1 /LCV s ( s)

+

(vi)

Step 4: The output equation is y = vC . Operate on both sides of Eq. (vi) by −1 [s2 + ( R/L) s + 1 /LC ] and write in quotient form: V C ( s) =

1 /LC

V s ( s) s2 + ( R/L) s + 1 /LC

Step 5: The transfer function H ( s) = V c ( s) /V s ( s) is: H ( s) = 1 /LC s2 + ( R/L) s + 1 /LC

(vii)

(viii)

Step 6: The di fferential equation relating vC to V s is: d 2 vC dt 2 CITSTUDENTS.IN

R dv C

+

L dt

+

1 LC

vC

=

1 LC

V s (t )

(i x )

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Cramer’ s Rule, for the solution of a set of linear algebraic equations, is a useful method to apply to the solution of these equations. In solving for the variable xi in a set of n linear algebraic equations, such as Ax = b the rule states: xi

det A(i) =

(18)

det [ A]

where A( i) is another n × n matrix formed by replacing the ith column of A with the vector b.

If (19)

[ sI − A ] X( s) = BU( s)

then the relationship b etween the ith state variable and the input is X i ( s ) =

det [sI − A](i) det [ sI − A]

(20)

U ( s)

(i )

where ( sI − A) is de fined to be the matrix formed by replacing the ith column of ( sI with the column vector B. The di fferential equation is det [ sI − A] xi

=

det ( sI

(i)

− A)

−

uk (t ).

(21)

Example 4 Use Cramer’ s Rule to solve for v L (t ) in the electrical system of Example 3. Solution: From Example 3 the state equations are:

v˙ c i˙ L

=

1 /C

0 /L −1

R/ L

−

vc i L

+

0 1 /L

V in (t )

(i )

and the output equation is: v L =

−vC − Ri L + V s (t ).

(ii)

In the Laplace domain the state equations are: s −1 /C 1 /L s + R/ L

CITSTUDENTS.IN

V c ( s) I L ( s)

=

0 1 /L

A)

V in ( s).

(iii)

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The voltage V C ( s) is given by: det

(1)

det ( sI − A )

V C ( s)

=

=

det [( sI − A)]

V in ( s) =

det

0

−

1 /C

1 /L ( s + R/ L) /C s −1 1 /L ( s + R/ L)

V in ( s)

1 /LC V in ( s). s2 + ( R/L) s + ( 1 /LC )

(iv)

The current I L (t ) is: det

(2)

det ( sI − A)

I L ( s)

=

det [( sI − A)]

=

V in ( s) =

det

s 0 1 /L 1 /L

/C s −1 1 /L ( s + R/ L)

V in ( s)

s/L V in ( s). s2 + ( R/L) s + (1 /LC )

(v)

The output equation may be written directly from the Laplace transform of Eq. (ii): V L ( s)

=

−V C ( s) − RI L ( s)

+ V s ( s)

1 /LC s2 + ( R/L) s + ( 1 /LC ) −

=

− =

=

1 /LC

( R/ L) s + 1 V s ( s) s2 + ( R/L) s + (1 /LC ) −

+

2

+ ( R/L) s + ( 1 /LC )) V s ( s) s2 + ( R/L) s + ( 1 /LC )

− ( R/L) s + ( s

s2 V s ( s) , s2 + ( R/L) s + ( 1 /LC )

(vi)

giving the diff erential equation d 2 v L dt 2

R dv L d 2 V s 1 . v L (t ) = + L dt LC dt 2

(vii)

+

For a single-input single-output (SISO) system the transfer function may be found directly by evaluating the inverse matrix X( s) = ( sI − A )

−

1

BU ( s).

(22)

Using the de finition of the matrix inverse: [sI − A ] CITSTUDENTS.IN

−1

=

adj [sI − A] , det [ sI − A]

(23) Page 35


10EE55

adj [ sI − A ] B U ( s ). det [sI − A] and su bstituting into the output equations gives: X( s) =

Y ( s) =

C [ sI − A ]

−1

B

+

C [sI − A ]

=

−1

(24)

BU ( s) + DU ( s)

D U ( s).

(25)

Expanding the inverse in terms of the determinant and the adjoint matrix yields: Y (S )

=

=

C adj ( sI − A) B + det [sI − A ] D U ( s) det [sI − A] H ( s)U ( s)

(26)

so that the required di ff erential equation may be found by expanding: det [sI − A] Y ( s) = [C adj ( sI

− A) B

+

det [sI − A] D] U ( s).

(27)

and taking the inverse Laplace transform of both sides.

Example 5 Use the matrix inverse method to find a di ff erential equation relating v L (t ) to V s (t ) in the system described in Example 3. Solution: The state vector, written in the Laplace domain,

X( s) = [ sI − A ]

−1

BU ( s)

(i)

from the previous example is: V c ( s) I L ( s)

=

s −1 /C 1 /L s + R/ L

−1

0 1 /L

V in ( s).

(ii)

The determinant of [sI − A] is det [sI − A] = s2 + ( R/L) s + ( 1 /LC ) ,

(iii)

and the adjoint of [ sI − A] is

adj

s −1 /C 1 /L s + R/ L

=

s

R/L 1 /C /L s −1

+

(iv)

.

From Example .5 and the previous example, the output equation v L (t ) = −vC − Ri L + V s (t ) speci fi es that C = [−1 − R] and D = [1]. The transfer function, Eq. (26) is: CITSTUDENTS.IN

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Since C adj ( sI − A) B

−1

=

=

s + R/L 1 /C /L s −1

− R

0 1 /L

R 1 s − , L LC

(vi)

−

the transfer function is H ( s)

=

=

−( R/L ) s − 1 / ( LC ) + ( s

2

+ ( R/L) s + (1 /LC ))

1

( s2 + ( R/L) s + (1 /LC )) s2 , (S 2 + ( R/L )S + ( 1 /LC ))

(vii)

which is the same result found by using Cramer ’s rule in Example 4.

4

Transformation from Classical Form to State-Space Representation

The block diagram provides a convenient method for deriving a set of state equations for a system that is specified in terms of a single input/output diff erential equation. A set of n state variables can be identified as the outputs of integrators in the diagram, and state equations can be written from the conditions at the inputs to the integrator blocks (the derivative of the state variables). There are many methods for doing this; we present here one convenient state equation formulation that is widely used in control system theory. Let the di ff erential equation representing the system be of order n , and without loss of generality assume that the order of the polynomial operators on both sides is the same: an sn + an−1 sn

−1

+ ···

+

a0 Y ( s) = b n sn

+

bn −1 s n

−1

+ ···

+

b0 U ( s).

(28)

We may multiply bothsides of the equation by s−n to ensure that all diff erential operators have been eliminated: an + an −1 s −1 + · · · bn

+

+

b n −1 s− 1

a1 s −( n −1) + a 0 s −n Y ( s) = +

···

+

b1 s − (n − 1) + · · ·

+

b0 s −n U ( s) ,

(29)

from which the output may be spec ifi ed in terms of a transfer function. If we define a dummy variable Z ( s), and split Eq. (29) into two parts

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Figure 5: Block diagram of a system represented by a classical differential equation.

Eq. (30) may be solved for U ( s)), U ( s) =

an + an−1 s −1 + · · ·

+

a1 s −( n −1) + a0 s −n X ( s)

(32)

and rearranged to generate a feedback structure that can be used as the basis for a block diagram: a0 1 1 a1 1 an 1 1 +··· + + Z ( s) = U ( s) − Z ( s). (33) n 1 −n −

an

an s

a n s

−

an s

The dummy variable Z ( s) is specifi ed in terms of the system input u(t ) and a weighted sum of successive integrations of itself. Figure 5 shows the overall structure of this direct-form block diagram. A string of n cascaded integrator (1 /s) blocks, with Z ( s) de fined at the input to the first block, is used to generate the feedback terms, Z ( s) /s −i , i = 1 , . . . n , in Eq. (33). Equation (31) serves to combine the outputs from the integrators into the output y (t ). A set of state equations may be found from the block diagram by assigning the state variables xi (t ) to the outputs of the n integrators. Because of the direct cascade connection of the integrators, the state equations take a very simple form. By inspection: x˙ 1

=

x2

x˙ 2

=

x3

.

.

x˙ n − 1 = xn x˙ n

CITSTUDENTS.IN

=

−

a1 a0 x1 − x2 − · · · an an

a 1 xn an n−

−

+

1 an

u(t ).

(34)

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In matrix form these equations are 





 

  

  

x˙ 1   x˙ 2

 

.

=

 

 

  ˙ n−1   x

 

x˙ n−2 x˙ n

0 0

1 0

.

.

··· ··· ..

.

0 0

0 0

−a0 /a n

−a1 /a n

−





.

 

  

  

0 1

      xn−2      xn− 1 

0 0

. 1 0

··· ··· ···



0 0

a n −2 /a n

−

x1   x2

..

a n 1 /a n

+

0 0

    

xn



.

   

0 0

    

u (t ) .

1 /an

(35) The A matrix has a very distinctive form. Each row, except the bottom one, is filled of zeroes except for a one in the position just above the leading diagonal. Equation (35) is a common form of the state equations, used in control system theory and known as the phase variable or companion form. This form leads to a set of state variables which may not correspond to any physical variables within the system. The corresp onding output relationship is spec ifi ed by Eq. (31) by noting that X i ( s) = −

Z ( s) /s ( n +1 −i) . y (t ) = b0 x1

+

b1 x2

+

b2 x3

+

···

+

bn −1 x n

+

b n z (t ) .

(36)

But z (t ) = dx n /dt , which is found from the nthstate equation in Eq. (34). When substituted into Eq. (36) the output equation is: 

Y ( s) =

b0 −

b n a0 b n a1 b1 − a

a n

···

bn −1 −

b n an −1 a

n



x1    x2 

+

 

u(t ).

a

.

n

bn



n

xn

Example 6 Draw a direct form realization of a block diagram, and write the state equations in phase variable form, for a system with the di fferential equation d 3 y d 2 y +7 2 dt 3 dt

+

19

dy + 13 y dt

=

13

du + 26u dt

(i)

Solution: The system order is 3, and using the structure shown in Fig. 5 the

block diagram is as shown in Fig. 6. The state and output equations are found directly from Eqs. (35) and ( 37): 

CITSTUDENTS.IN

x˙ 1



  ˙ x 2 



=

0

1

0

0

0

1

  



x1 x2



+



0 0





u(t ) ,

(ii) Page 39

(37)






x˙ 3

CITSTUDENTS.IN

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



−13

−19

−7



x3





1

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Figure 6: Block diagram of the transfer operator of a third-order system found by a direct realization. 

y (t ) =

5



x1   26 13 0  x2  + [0] u (t ) . x3

(iii)

The Matrix Transfer Function

For a multiple-input multiple-output system Eq. 22 is written in terms of the r component input vector U( s) −1 (38) X( s) = [ sI − A ] BU ( s) generating a set of n simultaneous linear equations, where the matrix is B is n × r . The m component system output vector Y( s) may be found by substituting this solution for X( s) into the output equation as in Eq. 25: Y ( s) = =

C [sI − A ]

−1

C [sI − A ]

B { U( s) } + D { U( s) }

−1

B

+

D { U ( s) }

(39)

and expanding the inverse in terms of the determinant and the adjoint matrix Y ( s)

=

C adj ( sI − A) B

+

det [sI − A] D

det [ sI − A]

U( s)

= H( s)U( s) ,

(40)

where H( s) is defi ned to be the matrix transfer function relating the output vector Y( s) to the input vector U( s):

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lOEESS

Modern Control Theory H(s)

=

(C adj (sl -A) B + det [si- A] D) det [si- A]

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(41)

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UNIT-3 DERIVATION OF TRANSFER FUNCTION FROM STATE MODEL UNIT - 3

Derivation of transfer function from state model, digitalization, Eigen values, Eigen vectors, generalized Eigen vectors. 6 Hours

Converting transfer-functions to state models using canonical forms

The state variables that produce a state model are not, in general, unique. However, there exist several common methods of producing state models from transfer functions. Most control theory texts contain canoni cal cal for m developments of a standard form called the control canoni , see, e.g., [1]. Another is the phase vari vari able canoni canoni cal cal f orm . Control Canonical Form

When the order of a transfer function‟s denominator is higher than the order of its numerator, the proper transfer function is called str ictl y proper . Consider the general, strictly proper proper third-order transfer function Y ( s) s ) U ( s) s )



b2 s  b1 s  b0 2

s 3  a2 s 2  a1 s  a0

.

(1)

Dividing each term by the highest order of s of s yields b2 Y ( s) s ) U ( s) s )





b1 2



b0 3

s s s a2 a1 a0

1

s



s

2



s

(2)

3

which is a function containing numerous 1/ s s terms, or integrators. There are various signal-flow graph configurations that will produce this function. One possibility is the control canonical form shown in Figure 1. The state model for the signal flow configuration in Figure 1 is C

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 x1   0  x    0  2   x3   a0 y   b0

0   x1 

0  1   x2   0  u      a2   x3  1 

1 0

a1

(3)

 x1  b2   x2    0 u    x3 

b1

b2 b1 U(s)

1/s



Y (s) (s)

1/s

X 3(s)



b0

1/s X 2(s)

X 1(s)

-a 2 - a1 -a0

Figure 1. Control canonical form block diagram.

The validity of (3) is tested by rearranging (2) to yield 1 1 1 1 1 1 s )  a 1 2 Y ( s) s )  a0 3 Y ( s) s )  b2 U ( s) s )  b1 2 U ( s) s )  b0 3 U ( s) s ) Y ( s) s )  a 2 Y ( s) s s s s s s

(4)

and then substituting the expression for Y ( s) s) from the state model output equation in (3) yields 1 1 1 b0 X 1 ( s) s )  b1 X 2 ( s) s )  b2 X 3 ( s) s )  b2 U ( s) s )  b1 2 U ( s) s )  b0 3 U ( s) s ) s s s 1 a s ) b X ( s) s )  b X ( s) s )  b X ( s) 2

a 1



s 1

2

0

1

1

2

2

3



s )  b0 X 1 ( s) s )  b1 X 2 ( s) s )  b2 X 3 ( s)



s 1  a0 3  b0 X 1 ( s) s ) . s )  b1 X 2 ( s) s )  b2 X 3 ( s) s (5) Equation (5) can be rewritten as CITSTUDENTS.IN

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 b0 X 1 ( s)  b1 X 2 ( s)  b2 X 3 ( s)  U ( s)

b2



Y ( s) U ( s)

s



1

b1



a2

s



s



2

b0 s

a1 s



2

3

,

a0 s

3

which is identical to equation (2) proving that the control canonical form is valid. As an example consider the transfer function

C( s)



b2

2 s  8 s  6 2

R( s) s  8 s  26 s  6 3

2



b1



2

b0

s s 2 s 3  a2 a 1 0a



1

s



s



2

s

3

s 1



8



6

s 2 s 3 . 8 26 6

s



s

2



s

3

The coefficients corresponding with the control canonical form are: b0  6,

b1  8,

b2  2,

a0  6,

a1  26,

and

a2  8 .

The state model based on control canonical form is 1 0   x1  0   x1   0  x    0 0 1   x2   0  r 2         x3   6 26 8  x3  1 

 x1  y   6 8 2  x2   [0]r .    x3  Examination of Figure 1 shows one potential benefit of manipulating systems to conform to control canonical form. If X 1(s) happened to carry units of inches (position), then X 2(s) might be inches/second (velocity), and X 3(s) might be inches/second/second (acceleration). Phase Variable Canonical Form

Development of the phase variable canonical form as presented in [2] begins with the general fourthorder, strictly proper transfer function

 b2s 2  b1s 3  b0s 4   U ( s) s 4  a3s 3  a2 s 2  a1s  a0 1  a3s 1  a2 s 2  a1s 3  a0 s 4 Y ( s)

b3s  b2 s  b1 s  b0 3

2

b3s

1

(6)

which can be rearranged to read 1

2

3

4

Y ( s )  b3s U ( s )  b2s U ( s )  b1 s U ( s )  b0 s U ( s )

 a3 s 1Y ( s )  a2 s 2Y ( s )  a1s 3Y ( s )  a0 s 4Y ( s ) CITSTUDENTS.IN

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(7)


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The denominator in (6) is fourth order and leads one to conclude that a block diagram consisting of four 1/s terms may be useful. Construction of the phase variable canonical form is initiated by setting the output to

Y ( s )  b0 s 4U ( s )

(8)

which may be represented using four integrators as shown in Figure 2.

U(s)

1/s

1/s

1/s

1/s

b0

Y (s)

Figure 2. Block diagram of the output equation.

After substituting Y(s) from (8) into (7), the expression becomes 1

2

3

4

Y ( s )  b3s U ( s )  b2s U ( s )  b1 s U ( s )  b0s U ( s )

a 3b0 s 5U ( s )  a2b0 s 6U ( s )  a1b0 s 7U ( s )  a0b0 s 8U ( s )

.

(9)

It is left for the reader to show that the additional terms when applied to the block diagram results in Figure 3.8 (b) in [2]. The reader may also wish to examine the similarities between control canonical form and phase variable form. It is important to note that a particular transfer function may be represented by block diagrams of many different canonical forms, yielding many different valid state models. The advances made in microprocessors, microcomputers, and digital signal processors have accelerated the growth of digital control systems theory. The discrete-time systems are dynamic systems in which the system-variables are defined only at discrete instants of time. The terms sampled-data control systems, discrete-time control systems and digital control systems have all been used interchangeably in control system literature. Strictly speaking, sampleddata are pulse-amplitude modulated signals and are obtained by some means of sampling an analog signal. Digital signals are generated by means of digital transducers or digital computers, often in digitally coded form. The discrete-time systems, in a broad sense, describe all systems having some form of digital or sampled signals. Discrete-time systems differ from continuous-time systems in that the signals for a discrete-time systems are in sampled-data form. In contrast to the continuous-time system, the operation of discrete-time systems are described by a set of difference-equations. The analysis and design of discrete-time systems may be effectively carried out by use of the z -transform, which was evolved from the Laplace transform as a special form.

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SAMPLED-DATA AND DIGITAL CONTROL SYSTEMS Figure 2.1 (a) shows the block-diagram of a sampled-data contr ol system. The continuous er r or - signal e( )t is sampled at an interval of time T by means of a sampler. The plant output is also a continuous-signal and it is due to the fact that a hold circuit preceeds the plant. (En" &iJnal) e{1) e'(1)

) ·t )

Hokl clrrot

(OJ!piJ) y)

T (

(a)

( OJip!J) )

{b)

Figur e 2.1: (a) Block-diagram of a sampled-data control system. (b) Block- diagram of a digital control system Figure 2.1 (b) shows the block-diagram of a digital control system. A digital contr o l system uses digital signals and a digital computer to contr ol a pr ocess. The continuous-time error -signals ar e converted from analog for m to digital form by means of an AID converter. In practice, the AID converter itself contains the sa m1)ler After pr o cessing the inputs, the digital computer pr ovides an output in digital form. This output is then convert ed to analog for m by means of a DIA converter. In pr a ctice, the Dl A converter itself contains the h old-device. .

The use of sampled data in control systems enables time-sharing among different input signals using the same contr ol equipment. Different input signals can be sampled periodically by staggering sampling time and thus a number of inputs can be handled over the same control equipment. Sampled data technique is most appr opr iate for control systems req uir ing long-distance data transmissi on. It is well k nown that pulses may be transmitt ed with little loss of accuracy. Data in analog form may suffer considerable distort ion in the I

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I ll l

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that pulses may be transmitted with little loss of accuracy. Data in analog form may suffer consider able distortion in the transmission channel. Using the sampled-data technique, a number of communication signals can be sequentially sampled and transmitted through a single transmission channel, thus decreasing the cost of transmission installation. This technique is referred to as time-multil,le xing. In some control applications , it is requir ed that the load be driven by a weak power input signal. The signal-sampling reduces the power demand made on the signal and is , therefor e , helpful for signals of weak power or igin.

Sampler A sampler is the basic e lement of a discrete-time contro l system, which converts a continuous-time signal into a train of pulses occuring at the sampling instants 0, T, 2T, ...... ,wher e T is the sampling period. Note that, in between the sampling instants, the sampler does not transmit any information. Figure 2.2 (a) shows a switch being used as a sa m1,ler. A continuous-t ime signal f{ t), as shown in Figur e 2.2 (b), is the input to the sampler . The switch is closed for a very short duration of time IJ.t, and then r emains open f or some duration of time. I)

'

_ _,1

I] + •

I

f '(l)

T lb)

.I

{C]

{

Figure 2.2: Uniform periodic sampling: (a)Sampler (b)Cont inuous signal (c)Pr actical sampling( Id eal sampling

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The process is r epeated with a time period T, called smn11ling time, and the signal at the output end of the sampler is availab le in the form of pulses of very short duration /J.I, each followed by a skip per iod when no output appears at the output end of the samp ler . The continuous-t ime signal f{l) is thus sampled at a regular interval of time T, as shown in Figure 2.2 (c). The sampled signal is denoted as f *{I ), which is obtained after sampling the continuous time input signal f{ l). Figure 2.2 (c) shows the signal-modulated 11ulse train having pulse-width /J.I and is a case of practical-sam11ling. However, in the case of idea l -sa m11ling, the pulse width /J.I approaches to zero and theref or e the output *f ( )I of an idea l sampler is the signal-modulated im1n1lse train as shown in Figure 2.2 (d). Figure 2.3 shows two samplers 011erating in synchronism i.e., their sampling time and instants of oper ation are the same. The continuous-time signal f{l) is thus being sampled twice. As the two samplers operate in synchronism, the output of the second sampler will obviously be f(l).

t(t)

---·

-- 1

1 11

T

')+--

.......

[r(t)J'

. r(t)

T Sampler·2

Sampler·1 Figure 2.3: Signal f(l) sampled twice Theref ore 'i'

*

[ f (t)] == ( t ) . Thus, starring a starred fun ctio n r esults in the starred function itself.

Sampling Process The sampler converts a continuous-time signa l f{ l ) into a sequence of pulses wher ein the magnitude of the pulse gives the strength of the input signal at the instant of sampling. The pulse considered is of very short time dur ation and thus may be approximated as an im1n1lse. Considering the sampler output to be a train of weighted impulses, we can r elate the continuous-time signal f(l} to the sampler outputr(t) as CITSTUDENTS.IN

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(2.1)f * (t) = f (t)or(t) where o.,{t ) represents a train of unit im11ulses. The sampler output is equal to the prcduct of the continuous-time input signal f(t) and the train of unit-impulses o.,{ t). In other words, the sampler may be considered as a modulator with the continuous-time input signal f(t) as the modulating signal and the train of unit-impulses 5r(t) as the carrier, as shown in Figur e 2.4. '•ill

·

·

"'"''"

-6t..,.t..tt4T -t o r 1TJt 4t n

Figure 2.4: Sampling process The unit-impulse tr ain may be expressed mathematically as

(2.2)orU)

= 2::

o( t - /{f)

k=-oo

where o(t - k 7) is the unit-impulse o:curing at t = k r As the continuous-time signal f(t) appears only at sampling instants in the output; the same is denoted by f(k 7) and the sampled-output signal can be represented as

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E f(kT)§(t kT )

(2.3)[ (t) =

k =-oo Equation (2.3) gives the sampler out1mt, which is a train of weighted impulses. Most of the standard time-f unctions are zero f or t < 0, then in Equation (2.3), the lower limit c an be put as zero inst e a as the function f( AT) is zero for t < 0. In this book, therefore, unless otherwise st ated, we sh all assume that this is th E

Thus, Equation (2.3) becomes

(2.4)[.(t) =

L f(kT)6(t kT ). k =O

2.2.2.1 laplace Transf orm of a Sampled (Starred) Function The unit-impulse o(t - AT), occur ing at sampling instants t = AT·, A = 0, 1, 2, ... ... , is the fir st time-derivative of the sampled unit- step function u( t - AT) occuring at t = AT ; A = 0, 1, 2,.......

Ther efore ,

£ o(t-kT)

£[ u( t-kT ) ] 1 skT

s . -e £ o(t kT)

e -skT .

Thus, the Laplace transform of Equation (2.4) is given b y

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(2.5)F (s) = 2: f(kT)e -sk T k =O

F *( s ) is known as the Laplace transform of a sampled (starred) function or simply a starred transf or m.

n

If f(t) is a unit-step function t hen f{k = 1 and the Laplace transform of the starr ed unit-step function f*(t ) is given by I

L I.e -sk1'

ill

F (1)

k =O

1 + e -sT + e -2s1'+ . , . F "' (1 )

l-

-sT'

Signal Reconstruction: Holding Devices The use of a sampler in control systems generates high-frequency components in the sampled ouput. The samp ler input F ourier-SI)ectm is shown in Figur e 2.5(b) wherein wm is the highest frequenc y contained in the input signal. The sa mpler output contains the original input f requenc y component wm and complementary high-frequency components in addition as I

shown in F igur e 2.5 (c) and Figure 2.5 (d) that correspond to w 8 > 2w m and w 8 < 2w m respectively; where w 8 is the sam1)ling fre
--1

1

T

F'w)

1

(a)

f{jtrl)

(b)

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F'(jM)

-2 (I (c)

----

-+

- 2 It)

----_. w •

- @$- COm

0

2

M

•

Figure 2.5: (a)Sampler (b)Sampler input Fourier spectra (c)Sampler output Fourier (0Sampler output Fourier spectr a: w s

< 2w m* spectra: w s > 2w m* It is immediately observed from Figure 2. 5 (c) and F igure 2.5 (d) that so long as w s :! 2 w m' the or i ginal spectr um is preserved in the sampled signal and can be extracted from it by loW-IHtss filtering (shown in the dotted line in Figure 2.5 (c)). This is the well-known Shanon's sam1,ling theorem accor ding to which the inf o rmation contained in a sign al is fully preserved in the sampled version so long as the sampling frequency (w J is at le a st twice the ma ximum frequenc y (w m) contained in the signal. That is,

The or igi nal signal is reconstructed from the sampled signal by means of various t ypes of hold-cir cuits (extr ap olat ors). The simplest hold-circuit is the zero-order-hold (ZOH) in which the reconstructed signa l acquir es the same value as the last received sample for the entir e sampling per iod. The zer o-order-hold (ZOH) circuit has the character i st ics of a low-pass filter and therefore, if a ZOH circuit is placed after the sampler , the high-frequency (complementar y) components present in the reconstr ucted signal are easily filtered out and at the output of ZOH the ori gina l signal appears. The ZOH device is also k no wn a s a "box-car" generator. The schematic diagram of sampler and ZOH is shown in Figure 2.6 (a) , while signal re constr uct ion is illustrated in Figures 2.6 (b) and 2.6 (c).

l' (l)

f{t) -----" .,_,

T

CITSTUDENTS.IN

ZO H

I

·

·

'1,(1) (Reconstructed signal)

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10EE55

Modern Control Theory f '(l)

0

.' (b)

.

I

'''

0 .... (cl Figure 2.6: (a)Sampler and zero-order-hold (ZOH). (b)Sampled signal. (c) Reconst r ucte d signal by ZOH _5

Note that the higher-order holds offer no particular advantage over the ZOH. The ZOH when used in conjunction with a high sampling rate provides satisfactor y performance. I

I

Transfer-Function of a ZOH Circuit In Figure 2. 7 a unit impulse input is given to a ZOH circuit that holds the input signal for a duration T and thus the output appears to be a unit-step function until duration T Therefore the output of ZOH can be given as I

I

h( t)

I)

0

0

T

Figure 2.7: Unit impulse input to ZOH

h( t )

= u( t) - u(t - T). CITSTUDENTS.IN

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Therefore, the Laplace transform of the output of ZOH is given as

1

fl(s )

1

s- s. e

H(s )

=

- sT

(1- e --s7)

s

Now, the transfer function of ZOH is given by

in place transf ormof ouiput of ZOH Laplace transjonnof input of ZOH

(2.7)or,

1 (1e" T )

or

s

2.2.3.2 Fre(1uency Response of ZOH We already know from Equation .7) that the transfer function of a zero-or der -hold circuit is

Put s = jw in the preceeding r elation and we have

Gho(Jw )

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7

(J-e -i"" )

.

J W

- jwT/ 2

jwT / 2

.e

.e

( e- j ..JT/ 2-e - j ..JT/2) jwT / 2

.e

.

J W

-e- j..,T/2 ]

- j.11TJ2

e

2}

[

· ( wT )

2e

sm 2 . Te -J"'T/2

T( )

Tsi n(1) )(

.

1 112 w ""

. sin( w[ ) - jwT / 2

.e

.

If the sampling frequency is wm. then the sampling time T is given by

Tsin(::}

_ j (

Gho( JW ) = -(---r 'r - w-)--"-'-- e •

rrw

w s

} •

W s

(2.8 A)IGh0 (jw)l =T CITSTUDENTS.IN

sin( rrw ) w s

(:) Page 55

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(2. 8B)

0,

sin( :) > 0

11' ,

sin( :) < 0

B

To view the pr evious der ivations, the amplitude and phase plots ar e shown in Figur e 2.8. IGt.,tiwll

T T

2'

0

Ol

w ,

2

3w4

(a)

LGoo(jw 3td

1

OJ

-180'

0000000000000 000000

0000000000000000 0000

0000000000000 0000000

00

(b)

Figm·e 2.8:Frequency response of the ZOH: (a)Amplitude plot (b)Phase plot

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UNIT-4 SOLUTION OF STATE EQUATIONS UNIT - 4

Solution of state equation, state transition matrix and its properties, computation using Laplace transformation, power series method, Cayley-Hamilton method, concept of controllability & observability, methods of determining the same 10 Hours

After obtaining

the various mathematical

models

such as physical, phase and canonical

forms of state models, the next step to analysis is to obtain the solution of state equation. From the solution of

the state equation, the transient response can then be obtained for specific input. This

completes the analysis of control system in state-space. The solution of state equation consists of two parts; homogenous solution and forced solution. The model with zero input is referred as homogenous system and with non-zero input is referred as forced system. The solution of state equation with zero input is called as zero input response (ZIR). The solution of state model with zero state (zero initial condition) is referred as zero state response (ZSR). Hence total response is sum of ZIR and ZSR. a)

Zero Input Response (ZIR):

Consider the n-th order LTI system with zero input. The state equation of such system with usual notations is given by X (t ) = AX (t) (1) with x(0) = x 0

∆

(2)

Let the solution of Eq.(1) be of the form (3)

at

x(t ) = e .k where eat is matrix exponential function defined as At

e

= I + At +

3

A 2 t 2

A t 2 +

2! 3! and k is a constant vector.

(4)

+ ......

Differentiating Eq. (3); at

X (t ) = Ae k At

−1

k = (e ) x(t 0) = e CITSTUDENTS.IN

− At

x(0) Page 57


10EE55 At

x(t ) = e e

− At

x(t 0)

A ( t − t 0 )

=e

x (t 0)

At t=0 (zero initial condition), at

x (t) = e x(0)

(8)

From Eq.8, it is observed that the initial state

x(0) = x 0 at t=0 is driven to a state x(t) at time „t‟ by matrix exponential function e

At

. This matrix

exponential function, which transfers the initial state of the system in „t‟ seconds, is called STATE TRANSITION MATRIX (STM) and is denoted by

Properties of State Transition Matrix

1).

(0)=I proof follows from the definition, 2

e

(t):

At

= I + At +

2

A 3 t 2

A t 2!

+

+ ......

3!

= φ (t ) Substituting t=0 results into

(0)=I .

−1

2). φ (t ) = φ ( −t ) : Proof: Consider φ (t ) = I + At +

φ ( −t ) = I − At +

A 2 t 2

A 2 t 2 2!

+ ....

3 3

−

A t

+ .....

2! 3! Multiplying φ (t ) and φ ( −t ) ,

φ (t )φ ( −t ) = I Pre-multiplying both sides with φ − 1

φ ( −t ) = φ −1 (t ) 3). φ (t 1 )φ (t 2 ) = φ (t 1 + t 2 ) : 2

Proof : φ (t 1 ) = I + At 1 + 2

t 1

2! t 2

2

2

+ 3

A3 t1

t 2

3!

3

+ ....

3

+ 3! + .... 2! A2(t1 +t2 )2 A3(t1 +t3)3 φ(t1)φ(t2 ) = I + A(t1 +t2) + + +....... 2! 3!

and φ (t 2 ) = I + At 2 +

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= φ (t 1 + t 2 ) 4). φ (t 2 − t 1 )φ (t 1 − t 0 ) = φ (t 2 − t 0 ) : (This property implies that state transition processes can be divided into a number of sequential transitions), i.e. the transition from t 0 to t 2, x(t 2 ) = φ (t 2 − t 0 ) x(t 0 ) ; is equal to transition from „t0‟ to „t1‟ and from „t1‟ to „t2‟ i.e., x(t 1 ) = φ (t 1 − t 0 ) x(t 0 ) x(t 2 ) = φ (t 2 − t 1 ) x(t 1 ) A (t 2 − t 1 ) + .... 2! 2 A (t 1 − t 0 ) + .... φ (t 1 − t 0 ) = I + A(t 1 − t 0 ) + 2! 2 3 A (t 2 +t0)2 A (t 2 +t0 )3 φ(t2 −t1)φ(t1 −t0) = I + A(t 2 +t 0) + + +....... 2! 3! = φ (t 2 − t 0 )

Proof : φ (t 2 − t 1 ) = I + A(t 2 − t 1 ) +

Evaluation of State Transition Matrix φ (t ) : Few method to evaluate the STM are;

1) Power series method 2) Inverse Laplace transform method 3) Cayley-Hamilton theorem 1). Power series method: It is infinite solution which can be trunkated to 2 or 3 terms. By the definition, 2

e

At

= I + At +

3 A t

2

A t

2

+ + ...... 2! 3! Example: Compute the STM by power series method given 0 a) A = −1

1 −2

A 2 t 2

a) φ (t) = I + At + =

2!

1 0

+

0 1 1−

= −t

CITSTUDENTS.IN

t

b) A =

,

2

2

+ 2

+t −

+

3! 1

−1

−2

+ ...... t+

1

−1

−2 2

+ ...

t −t +

t

te − t

3

2

e

− t

2

0

3

3

0 1

3 2

A t

0

t

1 1

2

t

t

2

2!

+

0

1

−1

−2

3

t

3

3!

+ ....

3

2

+ ...

+ ...

3

te

− t

+ Page 59

Modern Control Theory 1 − 2t +

=

− te

2t

3

+ ...

e −t − te

A 2 t 2

2!

1 0

+

A 3 t 2 3!

− t

+ ......

1 1 1 1 + t+ 0 1 0 1

0 1

1+ t + =

−

2

−t

b) φ (t ) = I + At +

=

3t

10EE55

t

2

2!

+

t

2

t2 2!

3

3!

+ ...

0

t+t +

+ ..... t

3

+ ... 2 t 2 t 3 + + + t 1 + ... 2! 3! 2

t

et te = t 0 e 2). Inverse Laplace Transform Method: Consider the state equation with zero input as x(t) = Ax(t ); ∆

x (t 0) = x(0) Taking Laplace Transform on both sides, Sx( s) − x(0) = Ax( s) [SI − A] x( s) = x(0) -1

Premultying both sides by [SI-A] , −1

x( s) = [SI − A] x(0) Taking Laplace Inverse on both sides, -1

x (t) =

−1

[ SI − A] x(0)

Comparing the above example with At

x (t) = e x(0) It shows that At e = STM = φ (t) =

φ ( t) =

-1 [SI

-1 [SI

− A]

−1

− A] −1

Example: Obtain the STM by Laplace Transform (Inverse Laplace Transform) method for given system matrices 1 0 1 1 0 = A a). A = b). , 1 1 CITSTUDENTS.IN

Page 60

Modern Control Theory −1

1

S −1

−2

a). A =

c). A =

,

0

1

−

−

2

−1

0 1

[SI − A] =

φ ( s) = [SI − A] −1 =

φ ( t) =

10EE55

-1

φ ( s) =

CITSTUDENTS.IN

e

t

0

0

S −1

−1

0

S −1

te e

3 S −1

=

S −1 1 0 S −1 [S − 1]

2

t t

Page 61


b). A =

0

1

−1

10EE55

−2

−1

S 1

[SI − A] =

(S + 2) S +2

S

φ ( t) = [SI − A] −1 =

-1

φ (t) =

φ (t) =

0

c). A =

1

−2

−3

S

[SI − A] −1 =

φ ( t) = =

−1

=

1 S+2

(1 + t ) e − te

−t

te

−t

−1

S +3

(S + 1)2 −1

(S + 1) S

(S + 1) 2

(S + 1)

=

2

− t − t

(1 − t)e

[SI − A] =

1

S

−1

2

S+3

S +3

1

−2

S

(S + 2)(S +1)

S +3 1 S −2

Hence φ (t) =

-1

2

−1

(S + 2)(S +1) S +3 1

φ (t) =

2e −t − 2e

−t

−e

-1

−2

S

(S + 2)(S +1)

−2 t

e

+ 2e − 2 t

−t

−e

−t

+ 2e − 2 t

−e

−2 t

3. STM by Calley-Hamilton Theorem: This method is useful for large systems. The theorem states that “Every nonsingular square matrix satisfies its own characteristic equation.” This theorem helps for evaluating the function of a matrix.

For nxn, non-singular matrix A the matrix poly function f(A) is given by f ( A) = α 0 I + α 1 A + α 2 A + .... + α n An −1 2

= φ (t ) where α 0 , α 1,.....α n constant coefficients which can be evaluated with eigen values of matrix A as described below. Step 1: For a given matrix, form its characteristic equation

λ I − A = 0 and find

eigen values as λ1 , λ2 ,.....λ n . CITSTUDENTS.IN

Page 62


10EE55

Step2 (Case 1):If all eigen values are distinct, then form „n‟ simultaneous equations as, e

e

λ1t

= f (λ1) = α 0 + α 1λ 1 + α2 λ1 + .....

λ2 t

2 = f (λ 2 ) = α 0 + α 1λ2 + α 2 λ2 + .....

λ nt

= f (λ n ) = α 0 + α 1λ n + α 2 λn + .....

e

2

2

Solve for α 0 , α 1 , α 2 ,....α n (Case 2): If some Eigen values are repeated then obtain one independent equation by using this Eigen value and find co-efficents α 0 , α 1 , α 2 ,....α n . Step 3: Substitute the co-efficients α 0 , α 1 , α 2 ,....α n in function f ( A) = α 0 I + α 1 A + .... + α n A n −1 = φ (t ) Examples: 10

1) Find f ( A) = A

for A =

0

1

−1

−2

Now characteristic equation is λ I − A = 0. λ

−1

1

λ+2

=0

( λ + 1)

2

= 0.

Hence, Eigen values are λ1 = −1, λ 2 = 0,λ2 = −1. Since „A‟ is of 2x2, the corresponding poly function is 10

f (λ ) = λ

=α 0 +

(1)

α 1λ 10

f (λ1 ) = λ 1 10

( −1)

α0

= α 0 + α 1λ1

= α0 − α 1

−α1

=1

Since it is case of repeated Eigen values, to obtain the second equation differentiating the expansion for f( ) on both sides (i.e. Eq 1), d d λ

[λ10 ]

10 λ

λ = −1

9 λ = −1

= α1

= α 1 = −10 Hence, α 0 = 1 + α 1 = 1 − 10 = −9

CITSTUDENTS.IN

Page 63

Modern Control Theory 10

f ( A) = A 1 0

= α0

0

α0

=

0

α0

0

= α 0 I + α 1 A

+ α1

1

+

10EE55

0

1

−1

−2

0

α1

− α1

− 2α 1

α0 α1 −9 = 10 − α 1 α 0 − 2α 1

=

− 10

11

2). Find STM by Cayley Hamlton method, given; a) A =

1 1

0 b) A= −1

0 1

1

c)

−2

A =

0

1

−2

−3

1 1

a) Consider A =

0 1

Characteristic equation is λ I − A = 0 =

λ −1

−1

0

λ −1

λ1 = 1and λ 2 = 1. Since A is of 2x2 second order system, hence,

λ

e t = α 0 + α 1 λ

(1)

e

λt

= α 0 + α 1λ at λ = λ1

will be e

λ1t

= α0 + α1 λ1

et = α 0 + α1

(2)

Differentiating Eq. (1) with respect to t e

λt

and substituting =

2

= α1at λ = λ2

t

t e = α 1

(3)

From Eqs. (2) and (3) α 0 = e t − α1 = e t − t e t t

= e (1 − t ) Hence STM is given by

φ (t) = e At = α 0 I + α 1 A =α0

1 0 0 1

t

=

+ α1

1 1 0 1

e (1 − t)

0

0

e t (1 − t)

CITSTUDENTS.IN

+

t

te

0

te

te

t

t

Page 64


=

e

t

0

te

t

e

t

1

0 b) A = −1

−2

10EE55

; characteristic equation is λ I − A = 0

λ

−1

1

λ+2

=0

λ1 = −1, λ 2 = −1.

For second order system,

e

λt

α

= α 0 + α1λ

e

λ1t

=

(1)

+α λ 0

e

−t

1

1

= α 0 − α 1 ; ( λ1 = −1 ) (2)

Differentiating Eq. (1) with respect to α1 = t e

and substituting = 2 =-1

− t

α 0 = e−t (1 + t) from Eq. (2).

hence

The STM is given by φ (t) = e On simplification,

φ (t) =

(1 + t )e − t − te

c). A =

− t

1

−2

−3

−1

2

λ+3

= α 0 I + A

− t

(1 − t)e

0

λ

te

At

− t

characteristic equation is λ I − A = 0.

λ1 = −1, λ2 = −2.

=0

Eigen values are distinct. The corresponding functions are,

e

λ1t

e

−t

= α 0 + α 1λ 1

(1)

= α 0 − α1

(2)

And e

e

λ2 t

=α 0 + α 1λ 2

− 2 t

(3)

= α0 − 2α1 From Eqs. (2) and (3) solving for α 0 and α 1 ,

α 0 = 2e −t α1 = e

− t

−e

−e

−2 t

−2 t

and

.

Hence the STM will be φ (t ) = α 0 I + α 1 A . On simplification CITSTUDENTS.IN

Page 65

Modern Control Theory α0

10EE55

α1

=

− 2α 1

α0

−t

− 3α 1

− t

−t

2e − e 2 = −t − 2e + 2e − 2 t

−2 t

e −e −t − e + 2e

− 2 t

as before.

Deterimine the STM of system matrix 2 1 A = 0

4

2 0 by Cayley-Hamilton method. 3 1

Now characteristic equation is, λ−2

−1

−4

0

λ−2

0

0

0

λ −1

λ I − A =

=0

Eigen values are λ1 = 2, λ1 = 2, λ2 = 1 Corresponding function will be e

λ1t

2

(1)

= α0 + 2α 1 + 4α 2

(2)

= α 0 + α 1λ 1 + α2λ 1

e

2t

Differentiating Eq. (1) with respect to On simplification t e

2t

1

and substituting

= α 1 + 4α 2

and e

λ 2t

1=2,

(3) 2

= α 0 + α1λ2 + α 2 λ 2

(4)

= α 0 + α1 + α 2 From Eqs. (2), (3) and (4), solving for α 0 , α 1 , α 2 , α0 = 2te

t

− 3e

α1 = −3te

2t

α 2 = te 2t

−e

2t

+ 4e

t

+ 4e ; 2t

2t

− 4e

t

t

+e Hence STM φ (t ) = α 0 I + α1 A + α 2 A 2 e =

2t

0 0

t

2t

12e − 12 e + 13te 2t e − 3e

CITSTUDENTS.IN

t

+ 3e

2t

2 t

− 4e

t

+ 4e 0

e

2 t

t

Page 66


10EE55

Solution of Non-homogeneous State Equation Consider the state equation with forced input u(t) as x(t) = Ax(t ) + Bu (t) ;

(1)

∆

x(t o ) = xo x(t) − Ax(t ) = Bu (t) Pre-multiplying both sides by e − At , − − e At [ x(t) − Ax(t )] = e At u(t ) Consider, d − At [e x(t)] dt − At

(2)

− At

x (t) + e x(t) − At e [ x 9T 0 − Ax (t)] hence Eq (2) can be written as d − At [e x (t)] = e − At u (t) dt Integrating both sides with time limits 0 and t − Ae

(3)

t

e

o

x(t ) | = e − Aτ Bu (τ ) d τ t

− At

o t

e

− At

x (t) − x(0) = e

− Aτ

Bu (τ ) d τ

o t

e

− At

x (t) = x(0) + e

− Aτ

Bu (τ ) d τ

o At

Pre-multiplying both sides by e , t

x (t) = e

x(0) + e A(t −τ ) Bu (τ ) d τ

Atx ( 0)

o

ZIR

ZSR t

x (t) = φ (t ) x(0) + φ (t − τ ) Bu (τ )d τ . o

If the initial time is t o instead of o then, t

x (t) = φ (t

− to

) x( t o ) + φ (t − τ ) Bu (τ ) d τ

(4)

t o

The Eq (4) is the solution of state equation with input as u(t) which represents the change of state from x(t o ) to x(t) with input u(t). Example 1: Obtain the response of the s ystem for step input of 10 units given state model as

CITSTUDENTS.IN

Page 67


−1

1

−1

− 10

x (t) =

10EE55

0

x (t) +

10

u (t )

y(t ) = [1 0] x(t); x(0) = 0 Solution: The state equation solution is At

x (t) = e x(0) + e

(1)

A(t −τ )

Bu(τ ) d τ

Since x(0)=0, t A(t −τ )

Bu (τ ) d τ

x(t) = e o

(2)

= φ (t − τ ) Bu (τ ) d τ Now, the state transition matrix φ (t) can be evaluated as −1 e At = φ (t) = -1 [SI − A]

1.0128 e

− a1t

− 0.0128 e

a 2t

=

− 0.114e

− a1t

)

φ (t

−τ )

=

+ 0.114e

1.0128 e

−

− a 2t

− a1 (t −τ )

− 0.114e

− a1 (t −τ )

0.114e

− a1t

− 0.0128e − 0.0128e

− a1t

− 0.114 e

− a 2 t

+1.012e

− a 2 (t −τ

+ 0.114e − a2 (t −τ )

− a 2t

0.114 e

− a1 ( t −τ )

− 0.0128e

− 0.114 e

− a1 (t −τ )

− a 2 (t −τ )

+ 1.012e − a2 ( t −τ )

Hence t

x(t) = φ (t − τ ) o

=

0 10

10 d τ

9.094 − 10.247 e − a1t + 1.153e − a 2 t − 0.132

+ 1.151e

[

∴ y(t ) = Cx(t ) = 1

− a1t

− 1.019e

− a 2t

0] x(t)

− a1t

−a t

= 9.094 − 10.247 e + 1.153e 2 ; where, a1=1.1125, a2=9.88 Example 2: Consider the system 0 1 0 x (t) = x (t) + u (t) −2 −3 1 y(t ) = [1 1] x (t); x(0) =

1

Solution: The −

1

1

CITSTUDENTS.IN

Page 68


2e

=

−t

− 2e

−e

−t

−2 t

+ 2e − 2t

10EE55

e

−t

−e

−e

−t

+ 2e

ZIR = φ ( t ) x(0) 2e −t − e −2 t = −t − 2t − 2e + 2e

2e − t − e −2 t

=

− 2e

=

−t

− 2 t − 2 t

e −t − e − 2 t −e

−e

−t

−t

+ 2e − 2 t

+e

1 −1

− 2 t

+ 2e − 2t + e −t − 2e − 2 t

e − t

−e

− t

t

ZSR = φ (t − τ ) Bu (τ ) d τ o

2e − ( t −τ )

t

= o

=

− 2e

− t (t −τ )

e − ( t −τ ) −e

− ( t −τ )

−e

−2 ( t −τ

e − ( t −τ ) − e −2 (t −τ )

+ 2e − 2 (t −τ ) −e

−e

− ( t −τ )

+ 2e − 2 ( t −τ )

0 1

1d τ

−2 (t −τ

d τ

+ 2e − 2 (t −τ )

1 1 − 2t − t −e + e = 2 2 −t −2 e − e t ∴ x (t) = ZIR + ZSR =

e −t −e

1 = 2

− t

1 += 2

(1 + e −e

−e

−t

e − t

+ −e

1 2

e

− 2t

− 2 t

− 2 t

− 2 t

y(t ) = Cx(t) = [1 1] x(t ) 1 (1 + e − 2 t = [1 1] = 2 − 2 t −e 1 − 2t − 2 t = (1 + e ) − e 2 1 = (1 − e − 2t ) ) 2 Example 3: x (t) =

2 0 x(t) + u ( t) 1 −2 −5 0

y(t ) = [2 1] x (t); x(0) = [1 CITSTUDENTS.IN

2]

T

Page 69


10EE55

Solution: State transition matrix 4 −t 1 − 4t 2 −t 2 − 4t e − e e − e 3 3 3 3 φ (t) = 2 − 4t 1 − t 4 − 2 −t e + e e + e − 4 t 3 3 3 3 t

x(t ) = φ (t ) x(0) + φ (t − τ ) Bu (τ ) d τ o

10 =

3 −5 3

e − t − e − 2t − e −t + e − 2t +

4

e

3 8 3

− 4 t

e − 4 t

Controllability and Observability Consider the typical state diagram of a system.

The system has two state variables. X1(t) and X2 (t). The control input u(t) effects the state variable X1(t) while it cannot effect the effect the state variable X2(t). Hence the state variable X2 (t) cannot be controlled by the input u(t). Hence the system is uncontrollable, th i.e., for n order, which has „n‟ state variables, if any one state variable is uncontrolled by the input u(t), the system is said to be UNCONTROLLABLE b y input u(t). Definition: For the linear system given by X (t ) = AX (t) + Bu (t) Y (t) = CX (t ) + Du (t) is said to be completely state controllable. If there exists an unconstrained input vector u(t), which transfers the initial state of the system x(t0) to its final state x(tf ) in finite time f(tf -t0) i.e. f f. It can be seen if all the initial states are controllable the system is completely controllable otherwise the system the system uncontrollable. Methods to determine the Controllability: 1) Gilbert‟s Approach 2) Kalman‟s Approach. Gilbert’s Approach: It determines not only the controllability but also the uncontrollable state variables, if the it uncontrollable. CITSTUDENTS.IN

Page 70


10EE55

Consider the solution of state equation X (t ) = AX (t) + Bu (t ); X (t o ) = X (0) as t A(t −τ )

At

Bu (τ ) d τ . Assuming without loss of generality that X(t)=0, the

X (t) = e X (0) + e 0

solution will be t

− X (0)

= e− A τ Bu (τ )d τ 0

The system is state controllable, iff the none of the elements of B is zero. If any element is zero, the corresponding initial state variable cannot be controlled by the input u(t). Hence the system is uncontrollable. The solution involves A, B matrices and is independent of C matrix, the controllability of the system is frequently referred to as the controllability of the pair [A, B]. Example: Check the controllability of the system. If it is uncontrollable, identify the state, which is uncontrollable. 1 − 1 −1 X (t) + u (t) X (t) = 0 0 −2 Solution: First, let us convert into diagonal form. λ +1 1 =0 λ I − A = 0 λ+2 [ λ + 1][λ + 2] = 0 2 λ + 3λ + 2 = 0 ( λ + 2)( λ + 1) = 0 λ = −2, λ = −1 Eigen vector associated with λ = −2 −2

0

1 0

v1 =

1 1

Eigen vector associated with λ = −1 −1 + 1

1

0 1

0

−1 + 2

0 1

v2 =

P −1 =

CITSTUDENTS.IN

1

P =

0 0

1

−1

1

1

1 1 1 0

0

1

−1

1 Page 71

Modern Control Theory −

1 P AP =

=

P −1 B =

0 −1

0 −1

10EE55

1 −1

−1

1 1

1

−2

1 0

0

0 −2 1 1 −2 = 1 −1 1 0 0 1 1 1 0

=

0 −1

0 −1

-1

As P B vector contains zero element, the system is uncontrolled and state X 1 (t) is uncontrolled.

2) Kalman’s approach:

This method determines the controllability of the system. Statement : The system described by x(t) = A x(t) + Bu(t ) y(t ) = Cx(t ) is said to completely state controllable iff the rank of controllability matrix Qc is equal to order of system matrix A where the controllability matrix Qc is given by, Qc = [ B |AB|A2 B|…… |An-1 B]. i.e., if system matrix A is of order nxn then for state controllability

ρ (Qc ) = n where ρ (Qc ) is rank of Q c. Example: Using Kalman‟ s Approach determine the state controllability of the system described by (1) y + 3 y + 2 y = u + u with x1 = y, x 2 = y − u x2 = −3 x 2 − 2 x1 − 2u x1 = x 2 + u x1 x 2

=

A =

0

1

x1

−2

−3

x 2

0

1

−2

−3

, B =

+

1 −2

u(t )

1 −2

0 CITSTUDENTS.IN

Page 72

1

1

Modern Control Theory Now AB =

−2

−3

Qc = [ B | AB ] =

Qc =

1

−2

−2

4

10EE55

−2

=

1

−2

−2

4

−2

4

=0

hence rank ρ (Qc ) is <2 and system matrix A is of 2x2. Therefore system is not state controllable. Example: Determine the state controllability of the system by Kalman‟s approach. 0 1 0 0 0 x(t) = 0 0 −2

CITSTUDENTS.IN

1 x(t ) + 0 u(t ) 1 −3

Page 73


10EE55

Verify the result with Gilbert‟s approach. Solution: Kalman‟s approach: Here, 0

1

0

A 0

0

1 , B = 0 ; AB =

0 −2

0

1

−3

0

1

1 ; A 2 B = − 3 −3

7

B|A2 B|AB

1

0

0

Q = 0 3 1 −c 1 −3 7

c

| Q |≠ 0.

ρ (Qc ) = 3 = Order of system. Therefore, system is state controllable. Verification by Gilbert’s approach:

Transferring the system model into canonical form with usual procedure.

1 2 Z(t ) = 0 − 1 0 Z (t ) + − 1 u (t) 1 0 0 −2 2 0

Here, B = 2

0

1 2

0

−1

1

has no zero element in any row, hence s ystem is controllable.

Observability: Concept: A system is completely observable, if every state variable of the system effects some of the outputs. In other words, it is often desirable to obtain information on state variables from the measurements of outputs and inputs. If an y one of the states cannot be observed from the measurements of the outpits and inputs, that state is unobservable and s ystem is not completely observable or simply unobservable.

Consider the state diagram of typical system with state variables as x 1 and x2 and y and u(t) as output and inputs respectively,

CITSTUDENTS.IN

Page 74


10EE55

On measurement of y(t), we can observe x 1(t) but we cannot observe x 2(t) by measuring out y(t), hence state is not completely observable. Definition: A system described by x(t) = Ax (t) + Bu (t) y(t ) = Cx (t) + Du (t) is said to be completely observable iff with measurement of output y(t) over finite time interval to t tf along with input u(t); t to the initial state x(to) can be estimated. Explanation: Consider the output equation y(t ) = Cx (t) + Du (t) Assuming D=0, y(t)=Cx(t). All the state variables x(t) can be estimated w ith measurement of y(t) provided C matrix do not contain the „zero‟ element. If any element of C matrix is zero, the corresponding state cannot be estimated and hence system is unobservable. Kalman’s Approach: The system described by the model x(t) = Ax (t) + Bu (t) y(t ) = Cx (t) + Du (t)

is said to be completely observable iff the rank of observability matrix φ o is equal to the order of system matrix A where, 2

T T φ o = [ C T AT C A T C

AT

n −1

CT ]

th

For n order system, rank of φ o must be n. i.e. ρ (φ o ) = n It involves A and C matrices. The pair A, C is observable. Example: Consider the system 3 0 −2 x (t) = x (t) + u(t ) 1 0 −1 y(t ) = [1

0] x(t )

Test the observability using Kalman‟s approach. Solution: Here A =

−2

0

; C = 1 0]

−

0

1

A T C T = T

CITSTUDENTS.IN

0

1

−1

0

−2

0

−2

=

0

φ Page 75


o

[

= C

10EE55

T

]

T

A C =

−2

1 00

ρ (φ o ) < 2

φo = 0

Order of system is 2. So, system is unobservable. Example: Show that the system 0 x(t) =

1

0

0

0 1 x(t ) + 0 u(t ) 1 − 11 − 6

0 −6

y(t ) = [4 5 1] x(t ) is not completely observable by 1) Kalman‟s approach 2) Gilbert‟s approach Solution: Kalman’s approach 0

1

0

0 0

−6

T

A =

0 −6

0

1

− 11

−6

A

= 1 0 0 1

− 11 −6

4 C = [4 5 1]

C

T

= 5

1 −6

6 T

2

T

A C = − 7 ; A T C T =

5

−1

−1

CT o

T

4 =

T

T

A C

2

A T C T −6

6

5

−7

5

1

−1

−1

o

φ =0

ρ (φ o ) < 3 and order of system is 3. Therefore system is unobservable. Gilbert’s approach: Transferring the system model into observable canonical form with usual procedure, the output equation will be y(t ) = [0 − 2− 2] z (t

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10EESS

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UNIT-5 POLE PLACEMENT PLACEMENT TECHNIQUES UNIT - 5

POLE PLACEMENT TECHNIQUES: stability improvements by state feedback, necessary & sufficient conditions for arbitrary pole placement, state regulator design, and design of state observer, Controllers- P, PI, PID. 10 Hours

Stability improvement by state Feedback: The main goal of a feedback design is is to stabilize a system, if it is initially unstable or to improve its stability if transient phenomena don‟t die out sufficiently fast. Consider single input single out put system with n-th order described by

X(t) = AX(t) + Bu(t) Let us assume that all the state variables are measured accurately, and then it is possible to implement a linear control law of the form u(t) = −KX(t) where, K = [k 1k 2 ..........k n ] is (1xn) feedback gain matrix. Substitute Eq (2) in Eq (1) to obtain closed loop system model as

X(t) = AX(t) − BKX(t) = [A − BK]X(t) The characteristic characteristic Eqn Eq n of closed loop system will be SI − (A − BK) = 0 th

When Eq (4) is evaluated, this yields an n order polynomial in S containing the n-gains K 1 , K 2,…………K n. The control Law design then consists of picking the gains so that the roots of Eq (4) are in desired locations. In the next section it can be seen that if the state model is in controllable canonical form, then the roots of characteristic Eqn (4) can be chosen arbitrarily. If all the given values of (A-BK) are placed in left- half of s-plane, the closed loop s ystem is asymptotically stable. And the state X(t) decays to Zero irrespective of initial state X(0)- the initial perturbation in the state . The system state can be maintained Zero value in spite of disturbances that act upon the system. S ystem with this property are called REGULATOR SYSTEMS. Control configuration of such s ystem as shown

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Algorithm Algorithm for f or Design: a) Let the the desired Locations of closed closed loop Poles be λ1 , λ 2 ........λ n b) The correspondry correspondry cha Eqn will w ill be

(S − 1 )(S − 2 )......(S − n ) = 0 Sn + 1S n −1 + 2S n − 2 + ....... + n = 0 c) Let Let the controller gain matrix be K=[K1 K2----------K n]. n n −1 n−2 d) SI − (A − BK) = S + 1S + 2 S + ...... e) Corresponding equation is n S + 1Sn −1 + 2S n − 2 + ......

n

(1)

n

= 0 -------(2)

f) Comparing the co-efficient of like powers of s form Eqn (1) & (2) K (2) K 1 ,K 2 ,--------K n can be Evaluated and this completes design of state back controller K. Example: Design a controller K for the state model 0 1 0 X(t) = X(t X(t ) + u(t) 0 0 1 Such that the system shall have damping ratio as 0.707 & settling time as 1 sec. Solution: a) Converting the desired specifications as locations of of closed loop loop poles at − 4 ± j4 j 4 b) The corresponding cha c ha Eqn will be (S+4+J4) (S+4-J4)=0 (1) S 2 + 8S + 32 = 0 c) Let the controller be of the form K=[K1 K2]. d) From given state model

A − BK ] = [ A

=

0

1

0 0

−

0

0

K1 K 2

0

1

−K1

−K 2

2 SI − (A − BK) = S + K1 S + K 2

e) The corresponding characteristic equation is S2 + K2S + K 1 = 0 Comparing the co-efficient of like powers of S from Equ (1) and (2) K2=8, K1=32. =32 .

(2)

Hence controller K=[32 8]. Controller gain matrix K=[32 8] in feed back will have desired specifications. CITSTUDENTS.IN

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Necessary and sufficient conditions: conditions : The sufficient conditions for arbitrarily pole-placement is, the system must be completely controllable. Consider the state model

(1)

X(t) = AX(t AX( t ) + Bu(t) Y(t)=CX(t)

(2)

The above model can be converted into controller canonical form through the transformation

X(t) = PX(t) Where P is nxn non singular transformable matrix, and assumed as P1n P11 P12 P=

P21 P22

(3)

P 2n

Pn2

P

Pn1

nn

P1 Pi = [Pi1 Pi2

= P2 ;

Pin ]

Pn Taking the derivative of Eqn (3) and substituting for X(t) from Equ(1)

X(t) = AX + B u(t) Where A = PAP −1 ,

=

(4) (5)

B = PB

0

1

0......

0

0

0

1......

0

.

.

.

.

−

−

n

n −1

......

−

(6) n

0 B = PB =

. 0

(7)

1 From Equ.(3),

x1 = P11x 1 + P12 x 2 + ......... + P1n x n = P1x Taking the derivative x1 = P1x(t) = P1[ Ax(t Ax( t) + Bu(t)] = P1 Ax(t) + P1Bu(t)

(8) (9)

Comparing (8) and (9), P1B = 0 CITSTUDENTS.IN

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10EE55

& x 2 = P1 Ax(t)

(10)

Taking again derivative of Equ(10) Hence the corresponding characteristic poly will be

SI − (A − B K = Sn + (1

+ K n )Sn −1 + ....... + n

and characteristic equation will be n S + ( 1 + K n )Sn −1 + ( 2 + K n )Sn − 2 ....... +

n

+ K1

(14)

+ K1 = 0

(15)

Where K1, K 2 ,.....K n can be chosen real numbers arbitrarily. Let the desired characteristic equation be n S + a1Sn −1 + ....... + a n = 0 Equating the coefficients of like powers of S from equation (15) and (16)

a1 =

+ Kn

1

a2 = . an =

2

+ K n −1

K n =a1 −

(16)

1

K n −1 =a 2 −

2

(17) n

+ K1

K1 =a n −

n

Hence the controller K will be (18) K = [an − n a n −1 − n −1 a1 − 1 ] Transferring to original system K = K P = [an − n a n −1 − n −1 a1 − 1 ]P The necessary condition is the system should be completely controllable. If the system is not completely controllable then there are certain states that cannot be controlled by stable feedback. Thus the necessary and sufficient condition for arbitrarily pole placement is the system should be completely controllable in s-plane. Steps to design the controller K: 1) For given state model, test the controllability of the system by determining the rank of controllable matrix Qc. 2) If it is controllable , transfer the model into controllable canonical form through similarly transformation matrix P ; where P can be -1 obtained as. a) P1=[00---------0 1]Qc P2=P1 A 2 P3=P1 A Pn=P1 An-1 P1 b)

P =P1 P A A 1

n −1

3) Compute A & B as

A = PAB PAB = PB 4) Assume the controller K CITSTUDENTS.IN

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[

10EE55

]

K = K1K 2 ....Kn and

]

[

Determine A A − B K = −

0

1 0

.

0

0 .

0 1 0 .

1

.

.

.

.

.

.

.

.

.

− K1

n

.

-

i

− Kn

5) Determine the Cha equation as

SI − (A − B K = S n + (

+ K 1 )Sn −1 + ....... +

n

1

−Kn

=0

6) Determine the desired cha equation as Sn + a 1Sn −1 + a S n − 2 + ....... + a n = 0 7) Determine the elements of gain matrix K , K 1 , K 2 ,..... K K n as K1 = an −

n

K 2 = a n −1 − K n = a1 −

[

n −1

and hence

1

]

K = K1K 2 ....Kn 8) The The feedback controller K for original s ystem is K = KP This completes the design procedure Design of controller by Ackermann’ s formula: We have

P1

[ ]

& P1 = Q−c1 ⋅ [00

P = P1 A

01]

P1 An −1 K = [an

−

n

an − 1 −

a1

n −1

−

1

]P −

1 00 01] (Q c ) [00

= [a n −

n

a n −1 −

a1 −

n −1

1

−1

]

(Qc ) A[00 −1

(Qc ) An −1 [00

= [00

01]Q c (a1

−

1

)A

(a 2

−

2

)A

01] 01]

+ ...... + (a n −

n)I

] (1)

Characteristics poly of A matrix is −

−2

SI − A = Sn + 1S n 1 + 2Sn + ....... + n According to Cayley-Hammilton theorem, n

A + 1 A n −1 + ....... + n

CITSTUDENTS.IN

(2)

I= 0

n

n −1

Page 82


K = [00

01]Q c

−1

10EE55

(A) -------------(4)

Where

(A) = A n + a1A n −1 + a 2 A n − 2 + a 3 An − 3 + ....... + a nI

[

& Q c = B | AB | A 2B | ....... A n −1B

(5)

]

(6)

Design procedure: 1. For given model, determine Qc and test the controllability. If controllable, 2. Compute the desired characteristics equation as Sn + a1Sn −1 + ......... + a n = 0 and determine coefficients a1 , a 2 ,

an

3. Compute the characteristics poly of system matrix A as

(A) = A n + a1 A n −1 + a2 A n − 2 + ....... + an −1

−1

4. Determine Q c and then controller k will be K=[0 0 ….0 1] . Φ ( A) c Q This completes the design procedure. Ex. Design controller k which places the closed loop poles at − 4 ± j4 for system

X(A) =

0 1 1 0

X(t) +

0 1

u(t) using Acermann‟s formula.

Answer: Check for controllability I

Q c = [B | AB ] = Qc−1 =

0 1

Its controllable as f(Qc)=2.

1 0

0 1

1 0 II. Desired characteristics equation is (S+4+j4)(S+4-j4) S 2 + 8S + 32 = 0 a1 = 8,a 2 = 32. III.

(A) = A

2

+ a 1 A + a 2I =

1 0 0 1 302

32 0 0

8

0

IV. K = [0 1] State regulator design: The state Regulator system is a system whose non zero initial state can be transferred to zero initial state by the application of the input u(t). Design of such system requires a control law U(t)=-KX(t) which transfers the non-zero initial state to zero initial state (Rejection of disturbances) by properly designing controller K. The following steps give design of such controller. For a system given by

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X(t) = Ax(t) + Bu(t) Y(t) = CX(t)with u(t) = −KX(t)where K = [K 1K 2 ......Kn ] From above relations, the closed loop system is X(t) = [A − BK]X(t) 1. From the characteristics poly of matrix A SI − A = Sn +

1

S

n −1

+

2

S

n −2

+ ....... +

(1)

n

Determine α 1 , α 2 ,.......α n 2. Determine the transformation matrix P, which transfers the given state model into controllable caunomical form.

P1 P=

P1

A

1

; P = [0..0…0c 1 ] Q − 1

P1 A n −1 c

[

−

2 n 1 Q = B | AB | A B | ....... A B

]

3. Using the desired location of closed loop poles (desired Egen values) obtain the characteristics poly

(S − 1 )(S − 2 ).....(S − determine a1 , a 2 ,.......a n

n n

) = S + a1 Sn −1 + a2 Sn − 2 + .....a n

4. The required state feed back controller K is K = [a n − na n −1 − n −1.....a 1 − 1 ]P Example: A regulator system has the plant

X(t) =

0

1

0 −6

0 − 11

0

0

1 X(t) + 0 u(t) 1 −6

Y = [1 0 0]X(t) Problems 1) Design a state feed back controller which place the closed loop poles at − 2 ± j 3.464 and-5. Give the block diagram of the control configuration Solution: By observing the structure of A and B matrices, the model is in controllable conmical form. Hence controller K can be designed.

S1.

SI − A = 0 6

−1

S

0 −1

S 3 + 6S 2 + 11S + 17

11 S + 6 CITSTUDENTS.IN

2

Page 84

Modern Control Theory 3

Comparing with S +

10EE55 S +

1

2

S+

3

, we obtain

1

= 6,

2

= 11,

3

= 17

2. Desired characteristics poly is (S+2+j3.464)(S+-j3.464)(S+5) 3

2

+ 9S + 20S + 60 3 Comparing with s2 + a 1s + a 2s + a 3 we obtain ≅S

a1 =9, a 2 =20, a3 =60 3. Hence the required controller K is K=[a3- 3 a2- 2 a1- 1] = [ K1 K2 K3] = [43 9 3] Block Diagram:

K 1=43 K 2=9 K 3=3

Design of state observer: Design of observer: The device that estimates the state variables with measurement of output along with input u(t) is called state observer.

If all the state variables are estimated, it is called Full-order state observer. If only few statevariable are estimated (which can‟t be measured accurately), the observer is called Reduced-order state observer. Design of Full-order observer: In this case all the state variables are to be estimated. a) Open loop observer: Here it is assumed that all the state variables are estimated correctly and initial state are also estimated correctly. i.e For State model

X(t) = Ax(t) + Bu(t) CITSTUDENTS.IN

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Y(t) = Cx(t); initial state X(0) are known and X (t ) is estimated states and X (0) is estimated initial states the resulting observer in open loop observer. Draw backs of open loop observer. 1. Precise Lack of information of x(0) 2. If X (0) X(0), the estimated state X (t ) obtain from open loop estimator will have continuously growing error and hence the output. 3. Small errors in A,B, and disturbances that enters the system but not in estimator will cause slow estimate process. Hence better to go for closed loop estimator. Design of closed loop Full-order state observer : Figure shows closed loop Full order state observer (Luen berger state observer)

State model is

Y(t) = Cx(t)

X(t) = Ax(t) + Bu(t)

Estimator output is Y(t) = CX(t)

CITSTUDENTS.IN

(1) (2) (2a)

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[

Y(t) − Y(t) = c X(t) − X(t) Error in output is Error in state vector ~ X(t) = X(t) − X(t)

]

From figure X(t) = AX(t) + Bu(t) + m(Y(t) − Y(t)

CITSTUDENTS.IN

(2b)

(3) (4)

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m is nx1 constant gain matrix. Differentiating equation (3) ~ X = X(t) − X(t) ~ = [ A − mc ]X(t) …….. by equations (2 to 3)

(5)

The characteristics equation of the error is given by (6) SI − (A − mc) = 0 ~ If „m‟ is chosen properly, (A-mc) have reasonably fast roots, then X (t ) will decay to ~ zero irrespective of (0) i.e X (t ) ~ − X (t ) , regardless of X (0) . Further it can be seen that equation (5) is independent of control input u(t). The observer gain matrix „m‟ can be obtained in exactly the same fashion as controller gain matrix „K ‟ Design procedure: 1. Let the desired pole locations of observer error

β 1 , β 2 ,.......β n then the

roots be corresponding poly equation will be (S − 1 )(S − 2 ).....(S − n ) = 0

(1)

Sn + a1Sn −1 + a 2Sn − 2 + .....a n = 0 determine co-effients a1 , a 2 ,.......a n

(2)

2. Let the observer gain matrix be m m = [m1,m 2 ,.......mn ]T obtain the characteristics poly equation of the closed loop system with observer as

SI − (A − mc) = 0 Sn + ( 1 + mn )Sn −1 + ( 2 + mn −1 )Sn − 2 + ....... + ( n + m1 ) = 0 (3) 3. Comparing the co-efficient of like powers of S from equations (2) and (3) a1 = 1 + mn m n = a1 − 1 an = n + m1 m1 = a n − n where the co-efficients α 1 , α 2 ,.......α n are obtained from poly equation SI − A = 0

Sn + 1Sn −1 + ....... + n =0. 4. The observer gain matrix „m‟ will be then m = [m1,m 2 ,.......mn ]T Example : Consider the state model

X(t) =

0

1

0 −6

0 − 11

Y(t) = [0 CITSTUDENTS.IN

0

0

1 X(t) + 0 u 1 −6

0 1]X(t) Page 88


10EE55

2) Design an observer; where observer-error poles are required to be located at − 2 ± j3.64 and -5 Solution: Model in oberserver conamical form will be

0 1 X(t) = 0 0

−6 − 11

1 Y(t) = [0

0 X(t) + 0

−6

1

0 1]X(t)

1) Desired characteristic equation will be (S+2+j 3 .64) (S+2-j3.64) (S+5)=0. 3

2

=> S + 9S + 20S + 60 = 0 hence comparing with S 3 + a 1s 2 + a 2s + a 3 = 0 a1 =9, a 2 =20, a3 =60

(I)

2) Let the observer be m = [m1 , m2 ,.......m n ]

0 0

[ A − mc ] =

−6

1 0

0 −6

1

m2

0 m3

− (6 + m1 )

0 0

[ A − mc ] =

0 0 m1

− (11 + m 2 )

1 0

− (6 + m 3 )

1

SI − (A − mc ) = 0 S 3 + (6 + m 3 )S2 + (11 + m 2 )S + (6 + m1 ) = 0 Comparing with s3 + 1s2 + 2 s + 1

= (6 + m3 ),

2

3

=0

= (11 + m 2 ),

3

= (6 + m1

(II)

) 3) From I and II,

9 = 6 + m3

m3 = 3

20 = 11 + m 2 m2 = 9 60 = 6 + m1 m1 = 54 4) Hence m = [54

CITSTUDENTS.IN

9 3] T

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Reduced-order state observer Required when few of state variables cannot be measured accurately. Design of Reducedorder state observer :

In this design procedure, let us assume one measurement Y(t). Which measures one state X1(t). The output equation is given by Y(t)=CX(t) (1) Where C=[1 0 -----0] Partitioning the states X(t) into states directly measured X1(t) and states directly non measurable Xe(t), which are to be estimated. X (t) 1 i.e X(t) = (2) X e(t ) Partionionly the matrixes accordly & state equation can be written as

X1(t)

=

Xe (t)

a11

a1e

X1(t)

a e1 a ee

X e (t)

and Y(t) = [1 0 ]

+

b1 be

u(t)

(3)

X1 (t)

(4)

X e (t)

from equation (3) (5)

Xe (t) = a ee X e(t) + a X (t) + ebu e1 Known inpu t

from (4) Y(t) = X1(t) => X1(t) = Y(t) Also from Equation (3) X1(t) = a11 X1(t) + a1e Xe (t) + b1u(t). Collecting the known terms. (6)

Y(t) − a 11 X1 (t) - b1u(t) = a1e X e (t) Known terms

Comparing equation (5) & (6) with

X(t) = AX(t) + bu(t) and Y(t) = CX(t) we can have similarity as X(t) → Xe (t) A(t) → aee (7)

bu → a e1 X1 (t) + b eu(t) c = a1e Y = Y(t) − a11 X1 (t) − b1u(t) Using Equation (7), the state model for reduced order state observer can be obtained as ê (t) = aee X ˆ e (t) + a e1 y + b e u(t)X 1(t) + m (Y(t) + (Y X CITSTUDENTS.IN

ˆ e (t)) − a 11y − b 1u − a 1e X Page 90

(8)


10EE55

ˆ (t) (same as X ˆ (t) = AX ˆ (t) + bu(t) + m(y(t) − y ˆ (t) = a X ˆ (t)) where y e 1e ~ ˆ Defining Xe (t) = X e − X e

~ ˆ (t) Xe (t) = Xe (t) − X e ˆ (t) in above equation & on simplification, Substituting for Xe (t) & X e

(9)

X(t) = (a ee − ma1e )Xe Corresponding characteristic equation is SI − (a ee − ma1e ) = 0

(10)

Comparing with desired characteristic equation „m‟ can be evaluated same as m in fullorder observer. To implement the reduced-order observer, Equation (8) can be written as ˆ e (t) = (a ee X

− ma 1e

ˆ e (t) + (a e1 − ma 11)Y(t) + (be − mb 1)u + m y(t) )X

(11)

As Y(t) is difficult to realize, Define ′ ˆ (t) − my X e (t) = X e

ˆ e (t) − m y(t) X e′ (t) = X ˆ (t) + (a − ma )y(t) + (b − mb )u(t) = (a ee − ma 1e )X e e1 11 e 1

(12)

Equation (12) can be represented by block diagram as shown below.

Example: System is described by

0 0 0 Y(t) = [0

0 − 24

1 −10 0 1]X(t)

50 X(t) + 2 u(t) 1

To design the reduced-order observer for un measurable states such that t he given values will be at – 10, -10. CITSTUDENTS.IN

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Solution: From output Equation it follows that states X1(t) & X2(t) cannot be measured as corresponding elements of C matrix are zero. Let us write the state equation as

X1

0

X2 = 1 0 0

X

3

0

0

•

1

X1(t)

50

− 24

X 2 (t) + 2 u(t)

− 10

X3 (t)

1

, aee =

0 0 1

0

m=

a1e = [0 1]

− 24

50

a 11 = [− 10 ]; be =

Let

0

, ae1 =

2

b1 = [1]

m1 m2

Characteristic equation of an observer is

SI − (a ee

− ma1e )

=0

S 2 + m 2S + m 1 = 0 Desirable characteristic equation is (S + 10)(S + 10) = 0 S 2 + 20S + 100 = 0 From two characteristic equations, equating the co-efficient of like powers of S 20=m2 & 100=m1 . 100 m= 20 Hence state model of observer will be X e ′ (t) = {

0 0 1 0

−

100

[

20

ˆ (t) + { 0 1]}X e

0 − 24

−

100 20

× { −10}]}Y(t) + {

50 2

−

100 20

Design of compensator by the principle of separation: Consider the completely controllable and observable system described by

He nc e the sta te fee db ac k co

X(t) = AX(t) + Bu(t) Y(t) = CX(t) Let the control law be u(t) = −kX(t) Assume the full-order observer model be ˆ (t) = AX ˆ (t) + Bu + M(y − cx ˆ (t)) X CITSTUDENTS.IN

1}u(t)

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10EE55

ntrol law using the estimated states will be Using equation (4) in (1a)

(5)

X(t) = AX(t) − BKXˆ (t) (6)

ˆ (t)) = (A − BK)X(t) + BK(X − X Defining observer state error vector as ~ ˆ (t) X(t) = X(t) − X using (6) in (5) ~ X(t) = (A − BK)X(t) + BKX(t) WKT observer error-vector model is ~ ~ X(t) = (A − mC)X(t) Equation (7) & (8) can be obtained in vector model as

X(t) (A − BK) = ~ 0 X(t)

(7)

(8)

BK

X(t) ~ (A − mC) X(t)

(9)

Equation (9) represents the model of 2n dimensional system. The corresponding characteristic equation is SI − (A − BK) SI − (A − mC) = 0 (10) Equation (10) represents the poles (given values) of the system, which is union of poles of controller & observers. The design of controller & observer can be carried by separately, & when used together, the system is unchanged. This special case of separation principle controller can be designed using

SI − (A − BK) = 0 and observer by SI − (A − mC) = 0 . The corresponding block diagram is as shown

Transfer f unction of compensator: ˆ (t) = AX ˆ (t) + Bu(t) + m(y − cx ˆ (t)) Now X CITSTUDENTS.IN

(1) Page 93


From

10EE55

ˆ (t) U(t) = −KX

(2)

(1) & (2)

(3)

ˆ (t) = (A − BK − mC)X ˆ (t) + mY(t) X Taking L.T. ˆ (s) = (A − BK − mC)X ˆ (S) + mY(S) SX ˆ (S) mY(s) = [SI − (A − BK − mC)]X Taking L.T. of equation (2)

(4)

ˆ (S) U(S) = −K.X From equations (4) & (5) U(S) −1 = K[SI − (A − BK − mC) ]m − Y(S) = D(S)

(5)

Example: A System is described by

X(t) =

0

1

0

0 −5

0

1

0

X(t) + 0 u(t) 1 − 6 − 11

Y(t) = [1 0 0]X(t) Design controller to take place closed loop poles at – 1 ±j1, -5. Also design an observer such that observer poles are at – 6, -6, -6. Derive the Un compensated & comp ensated T.F. Solution: System is observable & Controllable. Using the procedure of controller design & observer design, the controller will be K=[4 1 1] and observer will be T M=[12 25 -72] Transfer function of uncompensated system will be -1 Mu(S)= C[SI-A] B. 1 Mu(S) = 3 2 S + 6S + 11S + 6 Transfer function of compensated system will be U(S) MC(s) = = K[SI − (A − BK − mC)] −1 m Y(S) CITSTUDENTS.IN

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CONTROLLERS The controller is an element which accepts the error in some form and decides the proper corrective action.

The output of the controller is then applied to the system or process. The accuracy of the entire

system depends on how sensitive is the controller to the error detected and how it is manipulating such an error. The controller has its own logic to handle the error. the response and

mode

of operation.

Continuous and Discontinuous controllers.

The controllers are classified based on

On the basis of mode of operation, they are classified into The discontinuous mode controllers are further classified as

ON-OFF controllers and multi position controllers.

Continuous

mode controllers, depending on the input-output relationship, are classified into three

basic types named as Proportional controller, Integral controller and Derivative controller.

In

many

practical cases, these controllers are used in combinations. The

examples

of

such

composite controllers are Proportional – Integral (PI) controllers,

Proportional – Derivative (PD) controllers and Proportional - Integral – Derivative (PID) controllers. The block diagram of a basic control system with controller is shown in Figure. The error detector compares the feedback signal b(t) with t he reference input r(t) to generate an error. e(t) = r(t) – b(t).

Proportional Controller: In the proportional control mode, the output of the controller is proportional to the error e(t). The relation between the error and the controller output is determined by a constant called proportional gain constant denoted as K P. i.e. p(t) = K P e(t). Though there exists linear relation between controller output and the error, for zero error the controller output s hould not be zero as this will lead to zero input to the system or process. there exists some controller output P o for the zero error. CITSTUDENTS.IN

Hence,

Therefore mathematically the proportional

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control mode is expressed as P(t) = K P.e(t) + P o. The performance of proportional controller depends on the proper design of the gain K P. As the proportional gain K P increases, the system gain will increase and hence the steady state error will decrease.

But due to high gain, peak overshoot and settling time increases and this may lead t o

instability of the system. acceptable limits. make it zero.

So, compromise is made to keep steady state error and overshoot within

Hence, when

the

proportional controller is used, error reduces but can not

The proportional controller is suitable where manual reset of the operating point is

possible and the load changes are small.

Integral Controller: We have seen that proportional controller can not adapt with the changing load conditions. To overcome this situation, integral mode or reset action controller is used. In this controller, the controller output P(t) is changed at a rate which is proportional to actuating error signal e(t).

Mathematically the integral controller mode is expressed as The constant K i is called integral constant.

The output from the controller at any instant is the area

under the actuating error curve up to that instant. If the error is zero, the controller output will not change. The integral controller is relatively slow co ntroller. It changes its output at a rate which is dependent on the integrating time constant, until the error signal is cancelled.

Compared

to

the proportional controller, the integral control requires time to build up an appreciable output. However it continues to act till the error si gnal disappears. steady state error can be made to zero.

Hence, with the integral controller the

The reciprocal of integral constant is known as integral time

constant Ti. i.e., Ti = 1/K i.

Derivative Controller: In this mode, the output of the controller depends on the rate of change of error with respect to time. Hence it is also known as rate action mode or anticipatory action mode. The mathematical equation for derivative controller is

.

Where K dis the derivative gain constant. The derivative gain constant indicates by how much percentage the controller output must change for every percentage per second rate of change of the error. The advantage of the derivative control action is that it responds to the rate of change of error and can produce the significant correction before the magnitude of the actuating error becomes too large. Derivative control thus anticipates the actuating error, initiates an early corrective action and tends to increase stability of the system by improving the transient response. When the error is zero or constant, the derivative controller output is zero. Hence it is never used alone.

PI Controller: CITSTUDENTS.IN

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Modern Control Theory This is a composite control mode obtained by combining the proportional mode integral mode. The mathematical expression for PI controller is

10EE55 and

The transfer function is given by

The advantage of this controller is that the one to one correspondence of proportional controller and the elimination of steady state error due to integral controller. Basically integral controller is a low-pass circuit. The PI Controller has following effects on the system. 1. 2. 3. 4. 5. 6.

It increases the order of the system It increases the type of the system It improves the steady state accuracy. It increases the rise time so response become slow. It filters out the high frequency noise It makes the response more oscillatory.

PD Controller: This is a composite control mode obtained by combining the proportional mode and derivative mode. The mathematical expression for PI controller is

The transfer function is given by

PID Controller: This is a composite control mode obtained by combining the proportional mode, integral mode and derivative mode. The mathematical expression for PI controller is

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The PD Controller has following effects on the system.

1. It increases the damping ratio and reduces overshoot. 2. It reduces the rise time and makes response fast. 3. It reduces the settling time 4. The type of the system remains unchanged. 5. Steady state error remains unchanged. In general it improves transient part without affecting steady state. The transfer function is given by

This is to note that derivative control is effective in the transient part of the response as error is varying, whereas in the steady state, usually if any error is there, it is constant and does not vary with time. In this aspect, derivative control is not effective in the steady state part of the response.

In the steady state part, if any error is there,

integral control will be effective to give proper correction to minimize the steady state error. An integral controller is basically a low pass circuit and hence will not be effective in transient part of the response where error is fast changing.

Hence for the whole range of

time response both derivative and integral control actions should be provided in addition to the inbuilt proportional control action for negative feedback control systems.

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UNIT-6 NON LINEAR SYSTEMS UNIT - 6

Non-linear systems: Introduction, behavior of non-linear system, common physical non linearitysaturation, friction, backlash, dead zone, relay, multi variable non-linearity. 3 Hours

Many practical systems are sufficiently nonlinear so that the important features of their performance may be completely overlooked if they are analyzed and designed through linear techniques. The mathematical models of the nonlinear systems are represented b y nonlinear differential equations. Hence, there are no general methods for the analysis and synthesis of

nonlinear

control systems.

superposition principle does not apply to nonlinear

The fact that

systems makes generalisation

difficult and study of many nonlinear systems has to be specific for typical situations.

6.2 Behaviour of Nonlinear Systems

The most important feature of nonlinear systems is that nonlinear systems do not obey the principle of superposition. Due to this reason, in contrast to the linear case, the response of nonlinear systems to a particular test signal is no guide to their behaviour to other inputs. The nonlinear system response may be highly sensitive to input amplitude. For example, a nonlinear system giving best response for a certain step input may exhibit highly unsatisfactory behaviour when the input amplitude is changed.

Hence, in a nonlinear system, the stability is very much dependent

on the input and also the initial state. Further,

the

nonlinear

systems

may

exhibit

limit

sustained oscillations of fixed frequency and amplitude.

cycles

which

are

self-

Once the system trajectories

converge to a limit cycle, it will continue to remain in the closed trajectory in the state space identified as limit cycles.

In many systems the limit cycles are undesirable

particularly when the amplitude is not small and result in some unwanted phenomena.

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Figure 6.1

A nonlinear system, when excited by a sinusoidal input, may generate several harmonics in addition to the fundamental corresponding to the input frequency.

The

amplitude of the fundamental is usually the largest, but the harmonics may be of significant amplitude in many situations.

Another

peculiar

jump phenomenon.

characteristic

exhibited

by

nonlinear

systems

is

called

For example, let us consider the frequency response curve of

spring-mass- damper system.

The frequency responses of the system with a linear

spring, hard spring and soft spring are as shown in Fig. 6.2(a), Fig. 6.2(b) and Fig. 6.2(c) respectively.

For a hard

spring,

as

the

input

frequency

is

gradually

increased from zero, the measured response follows the curve through the A, B and C, but at C an increment in frequency results in discontinuous jump down to the point D, after which with further increase in frequency, the response curve follows through DE. If the frequency is now decreased, the response follows the curve EDF with a jump up to B from the point F and then the response curve

moves

towards

A. This

phenomenon which is peculiar to nonlinear s ystems is known as jump resonance. For a soft spring, jump phenomenon will happen as shown in fig. 6.2(c).

Fig. 6.2(a) CITSTUDENTS.IN

Fig. 6.2(b)

Fig. 6.2 (c)

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When excited by a sinusoidal input of constant frequency and the amplitude is increased from low values, the output frequency at some point exactly matches with the input frequency and continue to remain as such thereafter.

This phenomenon which results in a

synchronisation or matching of the output frequency with the input frequency is called fr equency entrainm ent or s ynchronisation.

6.3 Methods of Analysis

Nonlinear systems are difficult to analyse and arriving at general conclusions are tedious. However,

starting with the classical techniques for the solution of standard nonlinear

differential equations, several techniques have been evolved which suit different types of analysis. It should be emphasised that very often the conclusions arrived at will be useful for the system under specified conditions and do not always lead to generalisations.

The

commonly used methods are listed below.

Linearization Techniques: In reality all systems are nonlinear and linear systems are only

approximations of the nonlinear systems.

In some cases, the linearization yields useful

information whereas in some other cases, linearised model has to be modified when the operating point moves from one to another. Many techniques like perturbation method, series approximation techniques, quasi-linearization techniques etc. are used for linearise a nonlinear system.

Phase Plane Analysis: This method is applicable to second order linear or nonlinear systems

for the study of the nature of phase trajectories near the equilibrium points.

The system

behaviour is qualitatively analysed along with design of system parameters so as to get the desired response from the system.

The

periodic

oscillations

in

nonlinear systems

called limit cycle can be identified with this method which helps in investigating the stability of the system.

Describing

Function

Analysis:

This method is based on the principle of harmonic

linearization in which for certain class of nonlinear systems with low pass characteristic. This method is useful for the study of existence of limit cycles and determination of the amplitude, frequency and stability of these limit cycles.

Accuracy is better for higher order systems as they

have better low pass characteristic. CITSTUDENTS.IN

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Liapunov’s Method for Stability: The analytical solution of a nonlinear system is rarely possible.

If a numerical solution is attempted, the question of stability behaviour can not be fully answered as solutions to an infinite set of initial conditions are needed.

The Russian mathematician A.M.

Liapunov introduced and formalised a method which allows one to conclude about the stability without solving the system equations. 6.4 Classification of Nonlinearities

The nonlinearities are classified into i) Inherent nonlinearities and ii) Intentional nonlinearities. The nonlinearities which are present in the components used in system due to the inherent imperfections or properties of the system are known as inherent nonlinearities. gears etc.

Examples are saturation in magnetic circuits, dead zone, back lash in However in some cases introduction of nonlinearity may improve the

performance of the system, make the system more economical consuming less space and more reliable than the linear system designed to achieve the

same objective.

Such

nonlinearities introduced intentionally to improve the system performance are known as intentional nonlinearities.

Examples are different types of relays which are very

frequently used to perform various tasks. But it should be noted that the improvement in system performance due to nonlinearity is possible only under specific operating conditions.r

other conditions, generally nonlinearity degrades the performance of the

system.

6.5

Common

Physical

Nonlinearities:

The

common

examples

of

physical

nonlinearities are saturation, dead zone, coulomb friction, stiction, backlash, different types of springs, different types of relays etc.

Saturation: This is the most common of all nonlinearities.

All practical systems, when

driven by sufficiently large signals, exhibit the phenomenon of saturation due to limitations of physical

capabilities of their components. Saturation is a common

phenomenon in magnetic circuits and amplifiers.

Dead zone: Some systems do not respond to very small input signals.

range of input, the output is zero.

For a particular

This is called dead zone existing in a system.

The

input-output curve is shown in figure. CITSTUDENTS.IN

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Figure 6.3

Backlash: Another important nonlinearity commonly occurring in physical systems is

hysteresis in mechanical transmission such as gear trains and linkages. This nonlinearity is somewhat different from magnetic hysteresis and is commonly referred to as backlash. In servo systems, the gear backlash may cause sustained oscillations or chattering phenomenon and the system may even turn unstable for large backlash.

Figure 6.4

Relay:

A

relay

is

a

nonlinear

power

amplifier

which

can

provide

large

power

amplification inexpensively and is therefore deliberately introduced in control systems. A relay controlled system can be switched abruptly between several discrete states which are usually off, full forward and full reverse.

Relay

controlled

systems

find

wide applications in the

control field. The characteristic of an ideal relay is as shown in figure. In practice a relay has a definite amount of dead zone as shown. This dead zone is caused by the facts that relay coil requires a finite amount of current to actuate the relay. Further, since a larger coil current is CITSTUDENTS.IN

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needed to close the relay than the current at which the relay drops out, the characteristic always exhibits hysteresis.

Multivariable Nonlinearity: Some nonlinearities such as the torque-speed characteristics of a

servomotor, transistor characteristics etc., are functions of more than one variable. Such nonlinearities are called multivariable nonlinearities.

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UNIT-7 PHASE PLANE ANALYSIS UNIT - 7 Phase plane method, singular points, stability of nonlinear system, limit cycles, construction of phase trajectories. 7 Hours

Phase plane analysis is one of the earliest techniques developed for the study of second order nonlinear system.

It may be noted that in the state space formulation, the state variables

chosen are usually the output and its derivatives. The phase plane is thus a state plane where the two state variables x 1 derivative

.

and x2

are analysed which may be the output variable y and its

The method was first introduced by Poincare, a French mathematician.

The

method is used for obtaining graphically a solution of the following two simultaneous equations of an autonomous system.

Where and are either linear or nonlinear functions of the state variables x1 and x2 respectively. The state plane with coordinate axes x1 and x2 is called the phase plain. In many cases, particularly in the phase variable representation of systems, take the form

The plot of the state trajectories or phase trajectories of above said equation thus gives an idea of the solution of the state as time t evolves without explicitly solving for the state. The phase plane analysis is particularly suited to second order nonlinear systems with no input or constant inputs.

It can be extended to cover other inputs as well such as ramp inputs, pulse

inputs and impulse inputs.

7.2 Phase Portraits From the fundamental theorem of uniqueness of solutions of the state equations or differential equations, it can be seen that the solution of the state equation starting from an initial state in the state space is unique. This will be true if

and

are analytic. For

such a system, consider the points in the state space at which the derivatives of all the state CITSTUDENTS.IN

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variables are zero. These points are called singul ar points . These are in fact equil ibri um points of the system. undisturbed.

If the system is placed at such a point, it will continue to lie there if left A family of phase trajectories starting from different initial states is called a

phase portrait.

As time t increases, the phase portrait graphically shows how the system moves

in the entire state plane from the initial states in the different regions.

Since the solutions from

each of the initial conditions are unique, the phase trajectories do not cross one another.

If the

system has nonlinear elements which are piece-wise linear, the complete state space can be divided into different regions and phase plane trajectories constructed for each of the regions separately.

7.3 Phase Plane Method Consider the homogenous second order system with differential equations

This equation may be written in the standard form

where

and

n

are the damping factor and undamped natural frequency of the system. Defining the state

variables as x = x 1 and variable form as

= x2, we get the state equation in the state

These equations may then be solved for phase variables x 1 and x2. The time response plots of x1, x 2 for various values of damping with initial conditions can be plotted. When the different ial equations describing the dynamics of the system are nonlinear, it is in general not possible to obtain a closed form solution of x 1, x2. For example, if the spring force is nonlinear say (k 1x + k 2x3) the state equation takes the form

Solving these equations by integration is no more an easy task.

In such situations, a graphical

method known as the phase-plane method is found to be very helpful.

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The coordinate plane

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with axes that correspond to the dependent variable x 1 and x2 is called phase-plane. The curve described by the state point (x1,x2) in the phase-plane with respect to time is called a phase trajectory. A phase trajectory can be easily constructed by graphical techniques. 7.3.1 Isoclines Method: Let the state equations for a nonlinear system be in the form

Where both

and

are analytic.

From the above equation, the slope of the trajectory is given by

Therefore, the locus of constant slope of the trajectory is given by f 2(x1,x2) = Mf 1(x 1,x2)

The above equation gives the equation to the famil y of isoclines. For different values of M, the slope of the trajectory, different isoclines can be drawn in the phase plane. Knowing the value of M on a given isoclines, it is easy to draw line segments on each of these isoclines. Consider a simple linear system with state equations

Dividing the above equations we get the slope of the state trajectory in the x 1-x2 plane as

For a constant value of this slope say M, we get a set of equations

which is a straight line in the x1-x2 plane. We can draw different lines in the x1-x2 plane for different values of M; called isoclines. If draw sufficiently large number of isoclines to cover the complete state space as shown, we can see how the state trajectories are moving in the state plane. Different trajectories can be drawn from different initial conditions.

A large number of

such trajectories together form a phase portrait. A few typical trajectories are shown in figure given below.

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Figure 7.1 The Procedure for construction of the phase trajectories can be summarised as below:

1. For the given nonlinear differential equati on, define the state variables as x 1 and x2 and obtain the state equations as

2. Determine the equation to the isoclines as

3. For typical values of M, draw a large number of isoclines in x 1-x 2 plane 4. On each of the isoclines, draw small line segments with a slope M. 5. From an initial condition point, draw a trajectory following the line segments with slopes M on each of the isoclines. Example 7.1: For the system having the transfer function

and a relay with

dead zone as nonlinear element, draw the phase trajectory originating from the initial condition (3,0).

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The differential equation for the system is given by

where u is

Since the input is zero, e = r – c = -c and the differential equation in terms of error will be

Defining the state variables as x 1 = e and x2 = equations as

we

get

the

state

The slope of the trajectory is given by

Equation to the isoclines is given by

We can identify three regions in the state plane depending on the values of e = x 1. Re gio n 1:

Here u = 1, so that the isoclines are given by

or

For different values of M, these are a number of straight lines parallel to the x-axis. M

0

1/3

1/2

2

3

-4

-3

-2

-1

-1

-1.5

-2

1

0.5

-0.2

-0.25

-0.33

-0.5

Here u = 0, so that the isoclines are given by

or M = -1, which are

parallel lines having constant slope of -1. Trajectories are lines of constant slope -1. CITSTUDENTS.IN

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Region 3:

Here u = -1 so that on substitution we get

or

These are also lines parallel to x – axis at a constant distance decided by the value of the slope of the trajectory M. M

0

1/3

1

2

3

-5

-4

-3

-2

1

0.75

0.5

1/3

0.25

-0.25

-0.33

-0.5

-1

Figure 7.2: Phase portrait of Example 1

Isoclines drawn for all three regions are as shown in figure. It is seen that trajectories from either region 1 or 2 approach the boundary between the regions and approach t he origin along a line at -1 slope. The state can settle at any value of x1 between -1 and +1 as this being a dead zone and no further movement to the origin along the x1-axis will be possible. This will result in a steady state error, the maximum value of which is equal to half the dead zone. However, the presence of dead zone can reduce the oscillations and result in a faster response. In order that the steady state error in the output is reduced, it is better to have as small a dead zone as possible. Example 7.2: For the system having a closed loop transfer function

,

plot the phase trajectory originating from the initial condition (-1,0). The differential equation for the system is given by

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The slope of the trajectory is given by

When x2 = 0,

ie., all the isoclines will pass through the point x 1 = 2, x2 = 0. When M = 0,

When M = 2, When M = 4, When M = 8,

When M = -2, When M = -4,

When M = -6, When M = -10,

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Figure 7.3 The isoclines are drawn as shown in figure. 1,0). At (-1,0), the slope is , ie., next isoclines, the average slope is (8+4)/2 = o angle 80.5 ).The same procedure is repeated as shown in figure.

The starting point of the trajectory is marked at (o the trajectory will start at an angle 90 . From the 6, ie., a line segment with a slope 6 is drawn (at an and the complete phase trajectory will be obtained

7.3.2 Delta Method: The delta method of constructing phase trajectories is applied to systems of the form

Where may be linear or nonlinear and may even be time varying but must be continuous and single valued. With the help of this method, phase trajectory for any system with step or ramp or any time varying input can be conveniently drawn. The method results in considerable time saving when a single or a few phase trajectories are required rather than a complete phase portrait. While applying the delta method, the above equation is first converted to the form

In general, depends upon the variables x, and t, but for short intervals the changes in these variables are negligible. Thus over a short interval, we have , where

is a constant.

Let us choose the state variables as x 1 = x;

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, then

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Therefore, the slope equation over a short interval is given by

With known at any point P on the trajectory and assumed constant for a short interval, we can draw a short segment of the trajectory by using the trajectory slope dx 2/dx1 given in the above equation. A simple geometrical construction given below can be used for this purpose. 1. From the initial point, calculate the value of . 2. Draw a short arc segment through the initial point with (- , 0) as centre, thereby determining a new point on the trajectory. 3. Repeat the process at the new point and continue.

Example 7.3: For the system described by the equation given below, construct the trajectory starting at the initial point (1, 0) using delta method.

Let

then

The above equation can be rearranged as

So that At initial point is calculated as = 0+1-1 = 0. Therefore, the initial arc is centred at point (0, 0). The mean value of the coordinates of the two ends of the arc is used to calculate the next value of and the procedure is continued. By constructing the small arcs in this way the complete trajectory will be obtained as shown in figure.

Figure 7.4 CITSTUDENTS.IN

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7.4 Limit Cycles:

Limit cycles have a distinct geometric configuration in the phase plane portrait, namely, that of an isolated closed path in the phase plane. A given system may have more than one limit cycle. A limit cycle represents a steady state oscillation, to which or from which all trajectories nearby will converge or diverge. In a nonlinear system, limit cycles describes the amplitude and period of a self sustained oscillation. It should be pointed out that not all closed curves in the phase plane are limit cycles. A phase-plane portrait of a conservative s ystem, in which there is no damping to dissipate energy, is a continuous family of closed curves. Closed curves of this kind are not limit cycles because none of these curves are isolated from one another. Such trajectories always occur as a continuous family, so that there are closed curves in any neighbourhood of any particular closed curve. On the other hand, limit cycles are periodic motions exhibited only by nonlinear non conservative systems. As an example, let us consider the well known Vander Pol ‟s differential equation

which describes physical situations in many nonlinear systems. In terms of the state variables , we obtain

The figure shows the phase trajectories of the system for > 0 and < 0. In case of > 0 we observe that for large values of x 1(0), the system response is damped and the amplitude of x1(t) decreases till the s ystem state enters the limit cycle as shown by the outer trajectory. On the other hand, if initially x 1(0) is small, the damping is negative, and hence the amplitude of x 1(t) increases till the system state enters the limit cycle as shown by the inner trajectory. When < 0, the trajectories moves in the opposite directions as shown in figure.

Figure 7.5 A limit cycle is called stable if trajectories near the limit cycle, originating from outside or inside, converge to that limit cycle. In this case, the system exhibits a sustained CITSTUDENTS.IN

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oscillation with constant amplitude. This is shown in figure (i). The inside of the limit cycle is an unstable region in the sense that trajectories diverge to the limit cycle, and the outside is a stable region in the sense that trajectories converge to the limit cycle. A limit cycle is called an unstable one if trajectories near it diverge from this limit cycle. In this case, an unstable region surrounds a stable region. If a trajectory starts within the stable region, it converges t o a singular point within the limit cycle. If a trajectory starts in the unstable region, it diverges with time to infinity as shown in figure (ii). The inside of an unstable limit cycle is the stable region, and the outside the unstable region.

7.5 Analysis and Classification of Singular Points: Singular points are points in the state plane where . At these points the slope of the trajectory dx2/dx1 is indeterminate. These points can also be the equilibrium points of the nonlinear system depending whether the state trajectories can reach these or not. Consider a linearised second order system represented by

Using linear transformation x = Mz, the equation can be transformed to canonical form

Where,

1

and

2

are the roots of the characteristic equation of the system.

The transformation given simply transforms the coordinate axes from x 1-x2 plane to z1-z2 plane having the sa me origin, but does not affect the nature of the roots of the characteristic equation. The phase trajectories obtained by using this transformed state equation still carry the same information except that the trajectories may be skewed or stretched along the coordinate axes.In general, the new coordinate axes will not be rectangular. The solution to the state equation being given by

The slope of the trajectory in the z 1 – z2 plane is given by

Based on the nature of these eigen values and the trajectory in z 1 – z2 plane, the singular points are classified as follows. Nodal Point:

Consider eigen values are real, distinct and negative as shown in figure (a). For this case the equation of the phase trajectory follows as

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Consider now a system with eigen values are real, distinct one positive and one negative. Here, one of the states corresponding to the negative eigen value converges and the one corresponding to positive eigen value diverges so that the -k k trajectories are given by z 2 = c(z1) or (z1) z2 = c which is an equation to a rectangular hyperbola for positive values of k. The location of the eigen values, the phase portrait in z1 – z2 plane and in the x1 – x2 plane are as shown in figure. The equilibrium point around which the trajectories are of this type is called a saddle point.

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Focus Point: Figure 7.7

Consider a system with complex conjugate eigen values. The canonical form of the state equation can be written as

Using linear transformation, the equation becomes

The slope

We get,

This is an equation for a spiral in the polar coordinates. A plot of this equation for negative values of real part is a family of equiangular spirals. The origin which is a singular point in this case is called a stable focus. When the eigen values are complex conjugate with positive real parts, the phase portrait consists of expanding spirals as shown in figure and the singular point is an unstable focus. When transformed into the x 1-x2 plane, the phase portrait in the above two cases is essentially spiralling in nature, except that the spirals are now somewhat twisted in shape.

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Centre or Vortex Point: Figure 7.8

Consider now the case of complex conjugate eigen values with zero real parts. ie.,

1,

2

= ±j

Integrating the above equation, we get which is an equation to a circle of radius R. The radius R can be evaluated from the initial conditions. The trajectories are thus concentric circles in y 1-y2 plane and ellipses in the x1-x 2 plane as shown in figure. Such a singular points, around which the state trajectories are concentric circles or ellipses, are called a centre or vortex.

Figure 7.9 Example 7.4:

Determine the kind of singularity for the following differential equation.

Let the state variables be The corresponding state model is

At singular points,

and

Therefore, the singular point is at (0,0)

The characteristic equation is i.e, 1,

2

I-A = 0 or

2

+3 +2=0

= -2, -1. Since the roots are real and negative, the type of singular point is stable

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Example 7.5: For the nonlinear system having differential equation:

find all singularities. Defining the state variables as

, the state equations are

At singular points, So that the singular points

The singularities are thus at (0,0) and (-1,0). Linearization about the singularities:

The Jacobean matrix J =

Linearization around (0,0), i.e., substituting x 1 = 0 and x 2 = 0

The characteristic equation is

i.e., ( -0.1)+1 = 0 or

2

I-A = 0

– 0.1 + 1 = 0

The eigen values are complex with positive real part. The singular point is an unstable focus. Linearization around (-1,0)

The characteristic equation will be

Therefore

2

– 0.1 - 1 = 0

= 1.05 and -0.98. Since the roots are real and one negative and another positive, the singular point is a saddle point. 1,

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Example 7.6:

Determine the kind of singularity for the following differential equation.

Let the state variables be The corresponding state model is

At singular points, So that the singular points

The singularities are thus at (0,0) and (2,0). Linearization about the singularities:

The Jacobean matrix J =

Linearization around (0,0), i.e., substituting x 1 = 0 and x2 = 0

The characteristic equation is

i.e., ( + 0.5) + 2 = 0 or

2

I-A = 0

+ 0.5 + 2 = 0

The eigen values are complex with negative real parts. The singular point is a stable focus. Linearization around (-2,0)

The characteristic equation will be

Therefore

2

+ 0.5 - 2 = 0

= 1.19 and -1.69. Since the roots are real and one negative and another positi ve, the singular point is a saddle point. 1,

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UNIT-8: STABILITY ANALYSIS UNIT - 8 Liapunov stability criteria, Liapunov functions, direct method of Liapunov & the linear system,

Hurwitz criterion & Liapunov‟s direct method, construction of Liapunov functions for nonlinear system by Krasvskii‟s method. 6 Hours

For linear time invariant (LTI) systems, the concept of stability is simple and can be formalised as per the following two notions:

a) A system is stable with zero input and arbitrary initial conditions if the resulting trajectory tends towards the equilibrium state. b) A system is stable if with bounded input, the system output is bounded.

In nonlinear systems, unfortunately, there is no definite correspondence between the two notions. The linear autonomous systems have only one equilibrium state and their behaviour about the equilibrium state completely determines the qualitative behaviour in the entire state plane.

In nonlinear systems, on the other hand, system behaviour for small

deviations about the equilibrium

point may be different from that for large deviations.

Therefore, local stability does not imply stability in the overall state plane and the two concepts should be considered separately.

Secondly, in a nonlinear system with multiple

equilibrium states, the system trajectories ma y move away from one equilibrium point and tend to other as time progresses. Thus it appears that in case of nonlinear systems, there is no point in talking about system stability. More meaningful will be to talk about the stability of an equilibrium point.

Stability in the region close to the equilibrium point or in the

neighbourhood of equilibrium point is called stability in the small. For a larger region around the equilibrium point, the stability may be referred to as stability in the large.

In the

extreme case, we can talk about the stability of a trajectory starting from anywhere in the complete state space, this being called global stability. A simple physical illustration of different types of stability is shown in Fig. 8.1

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Fig. 8.1 Global and local stability Equilibrium State: In the system of equation , a state x e where f (x e,t) = 0 for all „t‟ is called an equilibrium state of the system. If the system is linear time invariant, namely f ( x ,t) = Ax , then there exists only one equilibrium state if „A‟ is non-singular and there exist infinitely many equilibrium states if „A‟ is singular. For nonlinear systems, there may be one or more equil ibrium states. Any isolated equilibrium state can be shifted to the origin of the coordinates, or f(0,t) = 0, by a translation or coordinates.

8.2 Stability Definitions: The Russian mathematician A.M. Liapunov has clearly defined the different types of stability. These are discussed below. Stability: An equilibrium state x e of the number > 0 there is a real number for all t t0. The real number depends on t0, the equilibrium state is said to be

system is said to be stable if for each real ( ,t 0 ) > 0 such that implies on and in general, also depends on t 0. If does not depend uniformly stable.

An equilibrium state x e of the system of equation is said to be stable in the sense of Liapunov if, corresponding to each S( ), there is an S( ) such that trajectories starting in S( ) do not leave S( ) as t increases indefinitel y. is said to be Asymptotic Stability: An equilibrium state x e of the system asymptotically stable, i f it is stable in the sense of Liapunov and every solution starting within S( ) converges, without leaving S( ), to x e as „t‟ increases indefinitely. Asymptotic Stability in the Large: If asymptotic stability holds for all states from which trajectories originate, the equilibrium state is said to be asymptotically stable in the large. An equilibrium state x e of the system is said to be asymptotically stable in the large, if it is stable and if every solution converges to x e as „t‟ increases indefinitely. Obviously a necessary condition for asymptotic stability in the large is that there is only one equilibrium state in the whole state space.

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The above said definitions are represented graphically in Fig. 8.2

Fig. 8.2 Liapunov‟s Stability

> 0 and any Instability: An equilibrium state x e is said to be unstable, if for some real number real number > 0, no matter how small, there is always a state x 0 in S( ) such that the trajectory starting at this state leaves S( ).

8.3 Stability by the Method of Liapunov Russian mathematician A.M. Liapunov has proposed a few theorems for the study of stability of the system. The most popular among this is called the “Second Method of Liapunov” or This method is very general in its “Direct Method of Liapunov”. formulation and can be used to study of stability of linear or nonlinear systems. The method is called „direct‟ method as it does not involve the solution of the system differential equations and stability information is available without solving the equations which is definitely an advantage for nonlinear systems. The stability information obtained by this method is precise and involved no approximation. First Method of Liapunov: The first method of Liapunov, though rarely talked about, is essentially a theorem stating the conditions under which system stability information can be inferred by examining the simplified equations obtained through local linearization. This theorem is applicable only to autonomous systems.

8.4 Sign Definiteness Let V(x1, x2, x3, ……. Xn) be a scalar function of the state variables x 1, x2, x3, …….., xn. Then the following definitions are useful for the discussion of Liapunov‟s second method. 8.4.1 Scalar Functions:

A scalar function V(x) is said to be positive definite in a region states x in the region and V(0) = 0.

if V(x) > 0 for all nonzero

A scalar function V(x) is said to be negative definite in a region states x in the region and V(0) = 0.

if V(x) < 0 for all nonzero

A scalar function V(x) is said to be positive semi-definite in a region if it is positive for all states in the region and except at the origin and at certain other states, where it is zero. CITSTUDENTS.IN

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A scalar function V(x) is said to be negative semi-definite in a region if it is negative for all states in the region and except at the origin and at certain other states, where it is zero. A scalar function V(x) is said to be indefinite if in the region negative values, no matter how small the region is.

it assumes both positive and

Examples:

8.4.2 Sylvester’s Criteria for Definiteness: T

A necessary and sufficient condition in order that the quadratic form x Ax , where A is an nxn real symmetric matrix, be positive definite is that the determinant of A be positive and the successive principal minors of the determinant of A be positive.

T

A necessary and sufficient condition in order that the quadratic form x Ax , where A is an nxn real symmetric matrix, be negative definite is that the determinant of A be positive if n is even and negative if n is odd, and the successive principal minors of even order be positive and the successive principal minors of odd order be negative. i . e . ,

T

A necessary and sufficient condition in order that the quadratic form x Ax , where A is an nxn real symmetric matrix, be positive semi-definite is that the determinant of A be singular and the successive principal minors of the determinant of A be nonnegative. i.e.,

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T

A necessary and sufficient condition in order that the quadratic form x Ax , where A is an nxn real symmetric matrix, be negative semi-definite is that the determinant of A be singular and all the principal minors of even order be nonnegative and those of odd orders be non positive.

Example 8.1: Using Sylvester ‟s criteria, determine the sign definiteness of the following quadratic forms

Successive minors are

principal

Hence the given quadratic form is positive definite.

Example 8.3:

Successive principal minors are

Hence the quadratic form is negative definite.

8.5 Second Method of Liapunov The second method of Liapunov is based on a generalization of the idea that if the system has an asymptotically stable equilibrium state, then the stored energy of the system displaced within the domain of attraction decays with increasing time until it finally assumes its minimum value at the equilibrium state. The second method of Liapunov consists of determination of a fictitious energy function called a Liapunov function. The idea of the Liapunov function is more general than that of energy and is more widely applicable. Liapunov functions are functions of x 1, x2, x3, …….., xn. and t. We denote Liapunov functions V(x 1, x2, x3, …….., xn, t) or V(x ,t) or V(x ) if functions do not include t explicitly. In the second method of Liapunov the sign behaviours of V(x ,t) and its time derivative give information on stability, asymptotic stability or instability of the equilibrium state under CITSTUDENTS.IN

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consideration at the origin of the state space. Theorem 1: Suppose that a system is described by , where f(0,t) = 0 for all t. If there exists a scalar function V( x ,t) having continuous first partial derivatives and satisfying the following conditions. 1. V( x ,t) is positive definite 2. Then the equilibrium state at the origin is uniformly asymptotically stable. If in additi on, then the equilibrium state at the origin is uniformly asymptotically stable in the large. A visual analogy may be obtained by considering the surface

. This is a

cup shaped surface as shown. The constant V loci are ellipses on the surface of the cup. Let be the initial condition. If one plots trajectory on the surface shown, the representative point x(t) crosses the constant V curves and moves towards the lowest point of t he cup which is the equilibrium point.

Fig. 8.3 Energy function and movement of states Theorem 2: Suppose that a system is described by , where f (0,t) = 0 for all t 0. If there exists a scalar function V( x ,t) having continuous first partial derivatives and satisfying the following conditions. 1. V( x ,t) is positive definite CITSTUDENTS.IN

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2. 3. 0, where

does not vanish identically in t t0 for any t 0 and any x0 denotes the trajectory or solution starting from x0 at t 0.

Then the equilibrium state at the origin of the system is uniformly asymptotically stable in the large. If however, there exists a positive definite scalar function V( x ,t) such that is identically zero, then the system can remain in a limit cycle. The equilibrium state at the origin, in this case, is said to be stable in the sense of Liapunov. Theorem 3: Suppose that a system is described by where f (0,t) = 0 for all t t0. If there exists a scalar function W ( x ,t) having continuous first partial derivatives and satisfying the following conditions 1. W (x ,t) is positive definite in some region about the origin. 2. is positive definite in the same region, then the equilibrium state at the origin is unstable.

Example 8.4: Determine the stability of following system using Liapunov‟s method.

Let us choose Then

This is negative definite. Hence the system is asymptotically st able. Example 8.5: Determine the stability of following system using Liapunov‟s method.

Let us choose Then This is a negative semi definite function. If x2 must be zero for all t t0.

is to be vanish identically for t

t0, then

This means that vanishes identically only at the origin. Hence the equilibrium state at the origin is asymptotically stable in the large.

Example 8.6: Determine the stability of following system using Liapunov‟s method.

Let us choose Then This is an indefinite function. CITSTUDENTS.IN

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Let us choose another Then This is negative semi definite function. If is to be vanish identically for t t0, then x2 must be zero for all t t0. This means that vanishes identically only at the origin. Hence the equilibrium state at the origin is asymptotically stable in the large. Example 8.7: Consider a nonlinear system governed by the state equations

Let us choose Then

Therefore, for asymptotic stability we require that the above condition is satisfied.

The

region of state space where this condition is not satisfied is possibly the region of instability. Let us concentrate on the region of state space where this condition is satisfied.

The limiting condition for such a region is.The dividing lines lie in the first and

third quadrants and are rectangular hyperbolas as shown in Figure. 8.4. In the second and fourth quadrants, the inequality is satisfied for all values of x1 region of stabilit y and possible instability.

and x2. Figure 8.4 shows the

Since the choice of Liapunov function is not

unique, it may be possible to choose another Liapunov function for the system under consideration which yields a larger region of stability.

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Conclusions:

1. Failure in finding a „V‟ function to show stability or asymptotic stability or instability of the equilibrium state under consideration can give no information on stability. 2. Although a particular V function may prove that the equilibrium state under consideration is stable or asymptotically stable in the region , which includes this equilibrium state, it does not necessarily mean that the motions are unstable outside the region . 3. For a stable or asymptotically stable equilibrium state, a V function with the required properties always exists.

8.6 Stability Analysis of Linear Systems Theorem: The equilibrium state x = 0 of the system given by equation is asymptotically stable if and only if given any positive definite Hermitian matrix Q (or positive definite real symmetric matrix), there exists a positive definite Hermitian matrix P (or positive definite real symmetric matrix) such that . The scalar function is a Liapunov function for the system.

Hence, for the asymptotic stability of the system of , it is sufficient that Q be positive definite. Instead of first specifying a positive definite matrix P and examining whether or not Q is positive definite, it is convenient to specify a positive definite matrix Q first and then examine whether or not P determined from is also positive definite. Note that P being positive definite is a necessary condition.

Note:

1. If does not vanish identically along any trajectory, then Q may be chosen to be positive semi definite. 2. In determining whether or not there exists a positive definite Hermitian or real symmetric matrix P, it is convenient to choose Q = I, where I is the identity matrix. Then the elements of P are determined from and the matrix P is tested for positive definiteness.

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Example 8.8: Determine the stability of the system described by

Assume a Liapunov function V(x) = T X PX

-2P12 = -1 P11 – P12 – P22 = 0 2P12 – 2P22 = -1 Solving the above equations, P11 = 1.5;

P12 = 0.5; P22 = 1

and

1.5 > 0 and det(P) > 0. Therefore, P is positive define. Hence, the equilibrium state at origin is asymptotically stable in the large. The Liapunov function V(x) = T X PX

Example 8.9: Determine the stability of the equilibrium state of the following system.

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4P11 + (1 – j)P12 + (1+j)P21 = 1 (1 – j)P11 + 5P21 + (1 – j)P22 = 0 (1 + j)P11 + 5P12 + (1 +j)P22 = 0 (1 – j)P12 + (1 + j)P21 + 6P22 = 1 Solving the above equations, P11 = 3/8;

P12 = - (1 + j)/8;

P12 = - (1 - j)/8;

P22 = 1/4

P is positive definite. Hence the origin of the system is asymptotically stable.

Example 8.10: Determine the stability of the system described by

T

Assume a Liapunov function V(x) = X PX

-2P11 + 2P12 = -1 -2P11 – 5P12 + P22 = 0 -4P12 – 8P22 = -1 Solving the above equations, P11 = 23/60;

P12 = -7/60;

and

P22 = 11/60

23 > 0 and det(P) > 0. Therefore, P is positive define. Hence, the equilibrium state at origin is asymptotically stable in the large.

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Example 8.11: Determine the stability range for the gain K of the system given below.

In determining the stability range of k, we assume u = 0.

Let us choose

a positive semi definite real symmetric matrix

This choice is permissible since except at the origin. To verify this, note that

cannot be identically equal to zero

being identically zero implies that x3 is identically zero. If x3 is identically zero, then x1 must be must be zero since we have 0 = - kx 1 – 0. If x1 is identically zero, then x 2 must also be identically zero since 0 = x 2. Thus is identically zero only at the origin. Hence we may use the Q matrix defined by a psd matrix. Let us solve,

i.e.,

-k p13 – k p13 = 0 -k p23 + p11 – 2p12 = 0 -k p33 + p12 – p13 = 0 P 12 – 2p22 + p12 – 2p22 = 0 P 13 – 2p23 + p22 – p23 = 0 P 23 – p33 + p23 – p33 = -1

Solving the above equations, we get

For P to be positive definite, it is necessary and sufficient that 12 – 2k > 0 and k > 0 or 0 < k <6. Thus for 0 < k <6, the system is stable; that is, the origin is asymptotically stable in the large.

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8.7 Stability Analysis of Nonlinear Systems In a linear free dynamic system if the equilibrium state is locally asymptotically stable, then it is asymptotically stable in the large. In a nonlinear free dynamic system, however, an equilibri um state can be locally asymptotically stable without being as ymptotically stable in the large. Hence i mplications of asymptotic stability of equilibrium states of linear systems and those of nonlinear systems are quite different. If we are to examine asymptotic stability of equilibrium states of non linear systems, stability analysis of linearised models of nonlinear systems is completely inadequate. We must investigate nonlinear systems without linearization. Several methods based on the second method of Liapunov are available for this purpose. They include Krasovskii‟s method for testing sufficient conditions for asymptotic stability of nonlinear systems, Schultz – Gibson‟s variable gradient method for generating Liapunov functions etc. Krasovsk ii’s Method:

Consider the system defined by , where x is an n-dimensional vector. Assume that f(0) = 0 and that f(x) is differentiable with respect to x i where, I = 1,2,3,…..,n. The Jacobian matrix F(x) for the system is

Define is the conjugate transpose of F(x). If the Hermitian matrix is negative definite, then the equilibrium state x = 0 is . If asymptoticall y stable. A Liapunov function for this system is in addition , then the equilibrium state is asymptotically stable in the large. Proof: If is negative definite for all x 0, the determinant of is nonzero everywhere except at x = 0. There is no other equilibrium state than x = 0 in the entire state space. Since f(0) = 0, f(x) 0 for x 0, and , is positive definite. Note that We can obtain

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If is negative definite, we see that is negative definite. Hence V(x)is a Liapunov function. Therefore, the origin is asymptotically stable. If tends to infinity as , then the equilibrium state is asymptotically stable in the large. Example 8.12: Using Krasovskii‟s theorem, examine the stability of the equilibrium state x = 0 of the system given by

The Jacobian matrix is given by

This is a negative definite matrix and hence the equilibrium state is asymptotically stable. . Therefore the equilibrium state is asymptotically stable in the large. Example 8.13: Using Krasovskii‟s theorem, examine the stability of the equilibrium state

x = 0 of the system given by

This is a negative definite matrix and hence the equilibrium state is asymptotically stable. CITSTUDENTS.IN

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f 7 _ x ) f ( ::; J - :::;:'!: - x::! ) :- (_ ·

·

_

, , ,

1

- x. - xY

-4 f"£" •

equilibrium state is asymptotically stable in the large.

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Modern Control Theory

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