Introduction to Linear Systems and State Space

7/30/2019 Introduction to Linear Systems and State Space

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1. Introduction and Motivation

In this week, we will review classical control, highlight some of the shortcoming of the classical approach,

which motivates the study of state-space models. To prepare for further study, we will also review

knowledge of vectors and matrices.

1.1 Introduction

Linear systems are usually mathematically described in one of two domains: time-domain orfrequency domain. The classical frequency-domain approach results in a system representation inthe form of a transfer function where the initial conditions are assumed to be zero. In the time-domain approach, the systems representation is in the form of a differential equation. However,rather than working with the high-order differential equation, we will use a set of 1st order

differential equations to describe the equivalent high-order differential equation. Of course if theorginal differential equation is of 4

thorder, then we will need four 1

storder differential equations

in order to describe the same system. Using this 1st

order representation (essentially stateequations), we will see that it provide a convenient time-domain representation that is the samefor systems of any order. Furthermore, state space descriptions do not assume zero initialconditions, and allow for the analysis and design of system characteristics that are not possiblewith frequency domain representations.

A Control System is a system in which some physical quantities are controlled by regulatingcertain energy inputs.

A system is a group of physical components assemblied to perform a specific function. It may beelectrical, mechanical, thermal, biomedical, chemical, pneumatic or any combination of the theabove.

A physical quantity may be temperature, pressure, electrical voltage, mechanical position orvelocity, liquid level, etc.

Some applications of automatic control

Space vehicle systems

Missile Guidance Systems Aircraft autopilot systems Robotic systems Weapon control systems

Power systems

Home appliances: aircons, refrigerators, dryers, microwave ovens etc. Automatic assembly line

Chemical processescontrolling pressures, temperature,humidity, pH values, etc.

Manufacturing systemsCNC machines, machine tool control, EDM control, etc

Quality control and Inspection of manufactured parts

Classical Control VS Modern Control:

The following are some of the features of Classical Control Theory:


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1. It uses extensively the Laplace operator in the description of a system.2. It is based on input/output relationship called transfer function. In computing the transferfunction of a system, all initial conditions are set to zero.3. It was primarily developed for single-input-single-output systems, although extension to multi-input and multi-output is now possible.4. It can only describe linear time-invariant differential equation systems.5. Based on methods like frequency response, root locus etc.6. Its formulation is not very well suited for digital computer simulation.

The following are some of the features of Modern Control Theory:

1. Uses the time domain representation of the system known as the state space representation.2. Can be used to describe multi-input-multi-output systems. Extensions to linear time varying,nonlinear and distributed systems are easy and straight forward.3. The formulation incorporates initial conditions into descriptions.4. It is based on time domain approach and is therefore, very well suited for computersimulations.

5. Provides a complete description of the system.

Open-loop control systemsTypically represented in block diagram form

Output has no effect on the control action. Examples: washing machine, traffic lights, microwave oven, dryers,toasters, etc.

Closed-loop control systemsTypically represented as:

Actuating signal, and hence output, depends on the input and the currentstate of the output.


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Examples are servo controlled motor, missile, robots etc.

Laplace transformation:

Given a functionf(t) where t 0.

Let sbe the complex variable defined as s = +j where , are variables.Then the Laplace transform of f(t), denoted as F(s), is

0( ) [ ( )] ( ) stF s L f t f t e dt

Transfer Function:

Consider the system given by

where1

0 1 11

1

0 1 11

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

n n

n nn n

m m

m mm m

d d da c t a c t a c t a c t

dt dt dt

d d db r t b r t b r t b r t

dt dt dt

with . n m

The transfer function of a linear, time-invariant, differential equationsystem is defined

1 2

0 1 2 1

1 2

0 1 2 1

( ) ( )( ) .

( ) ( )

m m m

m m

n n n

n n

b s b s b s b s bC S L output G s

R s L input a s a s a s a s a

Poles of a system: The roots of the denominator of the transfer function G(s) arecalled the poles of the system.

e.g.,2

2

2 2( )

( 3 2)

s sG s

s s s

poles of the system are s=0, s=-2. s=-1.

When is a SISO system stable?Ans: depends on the definition of stability.

Simplest answer: all poles have negative real parts.However, this leads to some problems, which we will see below.

1.2 Definitions

Dynamic System: A dynamic system (or a system with memory) is one whose output depends

on itself from an earlier point in time. A system whose output depends only on the current timeand the current input is memoryless.

Dynamic Systems most often occur as differential equation (continuous-time), or as differenceequations (discrete-time) because closed-form solutions of such systems require integration or


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summation, of a quantity over past time. The system equation will be algebraic if the system ismemoryless.

Causality: A system is said to be causal if the value of the output at time to depends on the values

of the input and output for all time t up to and including to (i.e for tto).

Systems that are not causal are called anticipatory, because they violate the seemingly impossibleconditoin that they depend on some future values for use at the current time. Physical systemsobviously must be causal but non-causal systems are often used in data filtering and imageprocessing applications. In those cases, an entire data set is first acquired so that at any giventime, you have data available for processing at a time ahead of the present time.

If a system is causal, then its transfer function will be proper, i.e the degree of its numeratorpolynomial must be no greater than its denominator polynomial.

Time-invariance: A time-invariant system is one whose output depends only on the differencebetween the initial time and the present time, i.e. y = y(t-t0) where t is the present time and t0 is the

initial time. Otherwise the system is time-varying.

Time-varying systems are typically systems in which time appears as an explicit variable in thedifferential, difference or algebraic equation that describe the system. A time-invariantdifferential equation must, by necessity, be one with constant coefficients.

Example: For a system with output y and input u,

uuuyyy 6286 is time-invariant

uuducbyyay is time-invariant provided a, b, c and d are constants

uuducytbytay )()( is time-varying

Linearity: A linear system is one that satisfies the homogeneity and additivity conditions. Ahomogeneous system is one for which, given y(t) = S(u(t)), then S(au(t)) = aS(u(t)) for all a and u,and an additive system is one for which S(u1 + u2)= S(u1)+ S(u2).

Linear systems are thus systems for which the principle of superposition holds, i.e the effect ofeach input can be considered independently of one another. In systems with memory, the systemis linear if the nth derivative depends in a linear way on each of the lower derivatives, and also ina linear way on the input, if any. If the coefficients of the differential equation are functions of

the dependent variable then a nonlinear differential equation results.

Example:

y + ay +by = cu+ du+ u is linear time-invariant (LTI)

uuducytbytay )()( is linear time-varying

uuducytbyyy )(2 is nonlinear, time-varying

1.3 Motivation for State-Space model


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Consider an unstable plant G ss

( )

1

1. In order to design a compensator for such a plant,

one might use a compensator1

1)(

s

ssH as shown in the block diagram Figure 2.1.

V(s) U(s) Y(s)

Y(s)V(s)

G(s)H(s)

GH(s)

1

1)()(

ssHsG

Now, as far as the input/output is concerned the transfer function from V(s) to Y(s) is now stable.In fact, the solution is:

)(*

)()(

0

)(

tve

dvety

t

tt

which looks completely stable.

But this controller will not work because the system will tend to saturate or burn out. To see this,instead of just looking at the input/output alone, we really need to analyze the complete

representation of the cascaded system and not only look at the input and output. In fact, theactual system consists of the two cascaded systems as shown below:

Y ss

U s

sY s Y s U s

sY S Y s U s

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

1

1

Let x t y t2 ( ) ( ) , therefore we get ( ) ( ) ( )x t x t u t2 2 (2.1)

Consider


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or

U ss

V s( ) ( )

12

1

(2.2)

Multiply throughout by et, we get

e x t e v d x

x t e e v d e x

x t e v d e x

x t e v x e

tt

tt

t

tt

t

t t

1

0

1

1

0

1

1

0

1

1 1

2 0

2 0

2 0

2 0

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) * ( )

( )

(2.3)

From (2.1), ( ) ( ) ( )x t x t u t2 2

The solution is given by (multiply byte )

2 2 2( ) ( ) ( )t t t

de x x e x e u t

dt

2 2

0

2 2

0

( )

2 1 2

0

( ) ( ) (0)

( ) ( ) (0)

( ) [ ( ) ( )] (0)

t

t

t

t t

t

t t

e x t e u d x

x t e e u d e x

x t e v x d e x

If you worked this through, youll get

x t e x e e x e vt t t t 2 212 1

0 0( ) ( ) ( ) ( ) * (2.4)

U

V

s

s

1

1

Let so thatX ss

V s

U s V s X s

Now sX s X s V s

1

1

1 1

2

1

2

( ) ( )

( ) ( ) ( )

, ( ) ( ) ( )

( ) ( ) ( )x t x t v t1 1 2

e x x e x e vt d

dt

t t( ) ( )1 1 1 2


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Therefore, because of the term et, unless the initial conditions can be guaranteed to be zero or

x x1 20 2 0( ) ( ) , the output will grow without bounds! In practice, it is impossible to guaranteesuch initial conditions as noise exists invariably in all physical systems and any small noise will

cause the system to go unstable.

If you had used Laplace to solve the equation, things would be much easier!!From (2.2),

1 1( ) ( ) 2 ( )x t x t v t

Taking the Laplace Transform, we get,

sX s x X s V s1 1 10 2( ) ( ) ( ) ( )

i.e.

X sx

s

V s

s1

1 0

1

2

1( )

( ) ( )

(2.5)

Also, ( ) ( ) ( )x t x t u t2 2

so that ( ) ( ) ( ) ( )x t x t x t v t2 1 2

Taking Laplace,

sX s X s X s V s x2 1 2 2 0( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )s X s X s V s x

1 02 1 2

)1(

)(

)1)(1(

)0(

)1(

)0()(

)0()(1

)(2

1

)0()()1(

122

21

2

s

sV

ss

x

s

xsX

xsVs

sV

s

xsXs

(2.6)

From Inverse Laplace Table, the solution is

x t e x e e x e vt t t t

2 212 1

0 0( ) ( ) ( ) ( ) *

Q: What went wrong in the previous analysis which concluded that the system is stable?Ans: Only the input/output was considered, not the whole system (i.e., internal behavior).

We are able to see why the solution is unbounded because we write the equations as

( ) ( )t fx x, v

This is called a state-space equation which is a system of 1st order differential equation. Notethat the variables x and v are written as bold to signify that they are vectors, ie.


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x

x

x

xn

1

2

wherex1,x2 are state variables.

In the example above, we see that we cannot just look at the external (input/output) behavior ofthe system alone. We need to know the internal behavior as well. Only by knowing the internalbehavior will we able to understand why the system goes unstable. This is the advantage of thestate-space approach because it gives us the internal behavior of the system. But, compared to thetransfer function approach, it is more difficult to design and implement. Note that any controllerthat is obtained using state space can also be found using the transfer function approach, andusually easier too. However, state-space gives us the insight into why a controller is designed assuch, which we may not know if we only use the transfer function approach.

1.4 Review of vector and matrix properties

Linear Independence of vectors : A set of vectors nRm21 ,...xx,x is said to be linearly

dependentif and only if there exist scalars ic not all zero, such that

0...21 m21 xxx mccc

If the only set ofci for which the above equation hold is 021 mccc , then the

set of vectorsnRm21 ,...xx,x is said to be linearly independent.

Alternatively, a set of vectors are linearly dependent if and only ifxi can be expressed as a linear

combination ofxj (j=1,2,....m,ji).

Examples:

0

0

1

2

2

121 cc

021 cc

We introduce the notation

01

21

m

m

c

c

ccc m1m21 xxxxx


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Hence, if m1 xx are linearly independent, then 021 mccc

Matrices:

Transpose of a matrixA.

IfnmRA is given by

mnmm

n

aaa

a

aaa

21

21

11211

then the transpose ofA, denoted byAT

nmR , is given by

mnnn

m

aaa

a

aaa

21

12

12111

Note that (A+B)T

=AT

+BTand (AB)

T=B

TA

T

Symmetric Matrix:

A matrixA is said to be symmetric if

TAA

Skew-symmetric Matrix.

A matrix A is said to be skew-symmetric if


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TAA

Remarks:

(i) Given any square matrixA, thenTAA is symmetric and TAA is skew-symmetric.

(ii) For any matrixA, AAB T is a symmetric matrix.

Properties ofDeterminant of a square matrix.

21212121 abbabb

aaA

21

213

31

312

32

321

321

321

321

cc

bba

cc

bba

cc

bba

ccc

bbb

aaa

A

1.1.1.If any two row or any two columns of a square matrix A are linearly dependent, thendet(A)=0.

1.1.2.If det(A)=0,A is asingular matrix ( inverse does not exist). Otherwise, it is nonsingular.1.1.3.det(A)=det(AT).1.1.4.det(AB)=det(A)det(B) ifA andB are both square matrices.1.1.5.det(A)= n ....21 where s are the eigenvalues of A.1.1.6.If a matrix is singular, at least one of its eigenvalues is zero. (why?)

1.1.7.det(A-1) =)det(

1

A.

1.1.8.If two rows ( or two columns ) of the determinant are interchanged, only the sign of thedeterminant is changed.

1.1.9.If a row (or a column ) is multiplied by a scalark, then the determinant is multiplied by k.1.1.10. If nmmnnn RCRBRA ,, and mmRD , then


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)det()det(0

det0

det DADC

A

D

BA

Rank of a matrix.

The rank ofA is the maximum number of linearly independent columns or rows inA. IfnmRA , then

nmArank ,min)( .

)(),(min)( BrankArankABrank .

Remarks:

(i) If rank(A) is equal to the number of columns or the number of rows ofA, thenA is said

to be full rank.

(ii) IfA is square and full rank, thenA is nonsingular.

Derivatives:

1 Ifx Rn

, then

d

d tx(t)

d

dtx1(t)

.

.d

dtxn (t)

2 dAdt daij

dt

and Adt aijdt

.

3 IfJ(x) is a scalar function of x Rn , then


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2J

x2

2J

x12

. .

2J

x1xn. . . .

. . . .

2J

xnx1. .

2J

xn2

Trace:

IfARnn , then Tr(A)= Trace of A= aiii1

n

.

Tr(A+B)=Tr(A)+Tr(B).

Tr(AB)=Tr(BTAT)=Tr(BA)=Tr(ATBT).

Eigenvalues and Eigenvectors:

A scalar ( which can be complex ) is called an eigenvalue ofA if there exists a nonzero

vectorx ( which can be complex ) such that Ax x . Any nonzero vectorx satisfyingAx x is called an eigenvector ofA associated with the eigenvalue .

If ARnn , the determinant

det(I-A)

is called the characteristic polynomial ofA. It is an nth-degree polynomial in . The

characteristic equation is given by

det(I-A)=0

If the determinant det(I-A) is expanded, the characteristic equation becomes

det(I-A)=n a1

n1 ... an 0

The n roots of the characteristic equation are called the eigenvalues ofA.

Note : Eigenvectors are unique up to a non-zero scalar multiple

Cayley-Hamilton Theorem:


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Suppose ARnn and det( I-A)= n a1n1 ... an 0 is the characteristic

polynomial ofA. Then

An a1A

n1 ... anI 0

Positive and Nonnegative definite matrices.

A square matrix ARnn is said to be positive definite if for all x Rn , x 0 ,x

TAx>0.

A square matrix ARnn is said to be non-negative definite ( or positive semi-definite )if for all x Rn , x 0 , x

TAx 0 .

Remarks:

(i) ForA to be positive definite, all leading minors must be positive; i.e.,

a11 0 ,a11 a12

a21 a22 0 ,

a11 a12 a13

a21 a22 a23

a31 a32 a33

0 ,....

(ii) A symmetric matrix A is positive definite if and only if its eigenvalues are

positive. It is positive semi-definite if all eigenvalues are non-negative.

(iii) If DRnm , then DDT A is always positive semi-definite. Furthermore,it is positive definite if and only ifD has full rankn.

Negative and Non-positive definite matrices.

A square matrix ARnn is said to be negative definite if for all x Rn , x 0 ,x TAx


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2 System Modelling

In this chapter, we will look at physical modeling of systems, i.e. using known laws of physics to derive the

underlying state equations that described the system. Whether the system is mechanical, electrical,

chemical or biological, a mathematical description should be written in a unified manner so that a single

theory of stability, control or analysis can be applied to the model. This mathematical description that we

will use is a set of equations known as the state equations.

2.1 Modeling of dynamical system

Example 2.1 Mechanical System Equations

Derive the equations of motion for the spring-mass-damper system shown in Figure 2.1. In thesystem, the mass is M, the spring constant K and the damping constant B. An externally appliedforcefis applied to the mass.

Solution:

Applying Newtons law

Fma

we get

Md z t

dt

f t Bdz t

dt

kz t

2

2

( )( )

( )( )

This is a second order ODE. If we want to write this as a set of 1st order ODE, we can

choosex z

x z

1

2

then

( ( ) )

x z x

x z M f t Bx kx

1 2

2

1

2 1

i.e.

x x

xk

Mx

B

Mx

Mf

1 2

2 1 21

or, in matrix form

x

xk

M

B

M

x

xM

f1

2

1

2

0 1 01

(3.1)

If we are interested in finding the position of the mass, then the output isy(t), is then y(t)=x1(t),or

forcefM

B

K

z


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yx

x

1 0

1

2

(Q: what is the physical meaning of 2x ?)

More compactly we can write,

)()()(

)()()(

tDtCt

tBtAt

uxy

uxx

(3.2)

where

0

01

10

10

D

CM

B

M

B

M

kA

Equation (3.2) is known as a state-space equation or simply state equations and the variables x1andx2 are known as state variables. In the equation,A is known as the state matrix,B the inputmatrix, Cthe output matrix andD the feedthrough matrix.

Definition: State Equation: The state equations of a system are the set of n first-orderdifferential equations, where n is the number of independent states.

If we are interested in the force acting on the spring, then the outputy(t) will become

y t kx t kx

xf( ) ( ) .

1

1

20 0

If we are interested in both the force acting on the dashpot as well as the force acting on thespring, then our output becomes

y t Bx t Bx

xf

y t kx t k

x

x f

1 21

2

2 1

1

2

0 0

0 0

( ) ( ) .

( ) ( ) .

or

y( )( )

( ).t

y t

y t

B

k

x

xf

1

2

1

2

0

00

or if we want to know about the total force acting on the mass, then the output is

1

2

( ) 1x

y t k B fx


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Example 2.2: Electrical Systems Equations: Series RLC circuit

vi(t)

R L

Cv(t)

i(t)

It will be shown later that the choice of state variables is not unique. In general, then number ofstate variables will depend on the number of energy storage elements in the system. Onlyindependent state variables can be chosen. In this example, there are 2 energy storage elementsin the circuit, the inductor and the capacitor. Therefore we can choose i(t) andVo(t) as our state

variables.

idt

dvC

vvRidt

diL i

or, we can write as

ivL

tv

ti

C

LL

R

tv

ti

0

1

)(

)(

01

1

)(

)(

If we now letxi=i, x2=v andu=vi, then we can write the state equation as

x

x

R

L L

C

x

xL u

1

2

1

2

1

10

1

0

or in a more compact form

uA bxx

where x is a nx1 state vectorA is an nxn matrixb is a nx1 vector

What about the output quantity? If, in this circuit example, we are interested in the output voltageacross the capacitor, then we let the output,y(t) to be

y tx

x x v( )

0 1

1

22


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again, in a more compact form, we get

y=cTx

Example 2.3 DC Motor

On the electrical side, by KVL,

e-Ri=vb (KVL)

=kii (Torque-current relation)

vb=k2 (back emf)

On the mechanical side, using Newtons law

Jd

dt

Hence, we have,

Jd

dtk i k

Re v

Jd

dt

k

Re k

i i b

i

1

2

( )

( )

Jk k

R

k

R

e

k k

JR

k

JRe

i i

i i

2

2

Since

d

dt, therefore we have

k k

JR

k

JRe

i i2

or

+

-

e

J

+

-

vb


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0 1

0

02k k

JR

k

JR

ei i

The above 2 examples illustrate the form of the final expressions of the system, ie. the state-space

equations. Note that the general form is

( ) ( ) ( )

( ) ( ) ( )

x Ax Bu

y Cx Du

t t t

t t t

where

A B C D R R R Rn n n r m n m r , , , andx u R Rn r1 1,

Both x andu are function of time, i.e. we should write x(t), andu(t). However, for the sake ofnotational simplicity, we normally write it as x and u. In the above cases,

A B C D R R R Rn n n r m n m r , , , are constant matrices and the state equation is alinear-time-invariant (LTI) system.

IfA B C D R R R Rn n n r m n m r , , , are themselves time-varying, then the systemis known as linear time-varying system, and so the state equations are given by

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

x A x B u

y C x D u

t t t t t

t t t t t

A more general form the above is

, )

( , )

( , )

( , , )

x f(x,u

x,u

x,u

y g x u

t

f t

f t

t

n

1

This represents a class of nonlinear time varying state equations.

2.2 State Concepts

State: The state of a system is a mathematical structure containing a set of n variablesx1,x2

...xn(t), called the state variables such that knowledge of these variables at t=t0, togetherwith the input fort>t0 completely determines the behavior of the sytem for any time t>t0.

There is a minimum set of state variables which is required to represent the systemaccurately. The initial starting time is normally taken as zero.

Note: state variables need not be physically observable and measurable quantities (infact, in most cases they are not).

Example:

dx

dt ax

au t x t x

x t e x

a

e u dat

at t

1 1

1

0 0

0

10

( ) ( )

( ) ( )( )

( )

(2.3)


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Herex(t) is the state variable, and given the input u(t) and the initial value ofx(t)at t= t0, x(t0 ) Eq(2.3) completely define the trajectory ofx(t) for all time t>t0.

State Variable: The state variables of a dynamical system are the variables making up thesmallest set of variables that determine the state of the system. If at least n variablesx1, , xn areneeded to completely describe the behavior of a dynamical system (so that once the input is given

for t t 0 and the initial state at t=t0 is specified, the future state of the system is completelydetermined), then such n variables are a set of state variables.As noted above, the state variables need not be physically measurable or observable quantities.Variables that do not represent physical quantities and those that are neither measurable norobservable can be chosen as state variables.

State Vector: If n state variablexi(t) , i=1,...n, are needed to completely describe the behavior

of a given system, then these n state variables can be considered as the elements orcomponents of the n-dimensional vectorx(t). Such a vectorx(t) is called a state vector.

State Space: State space is defined as the n-dimensional space in which the components of the

state vector (i.e., state variables) represent its' coordinate axes. Eg. phase plane.

State Trajectory: The path produced in the state space by the state vectorx(t) as itchanges with the passage of time. Eg. phase trajectory

The first step in applying these definitions is the selection of the state variable, given a physicalsystem. NOTE that the state variables are NOT unique.

What is the relationship of a same system represented using different sets of state variables?

Now, the state of a system at any given time can be seen as a point in the state space. Suppose we

set up three coordinate systems (U, I, II). Let a ,a ,a1 2 3 R3

be the unit vector of the axes of

coordinate system I andb ,b ,b1 2 3 R3

be the unit vector of coordinate system II. Then the

point P can be described using coordinate I as

O1

U

O

III

x1

x2

x3

x1

_

x2_

x3

_ P


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3

2

1

3211

x

x

x

O

xxxOPO

321

321

aaa

aaa

Alternatively, P can also be described by using coordinate II as

3

2

1

3211

x

x

x

O

xxxOPO

321

321

bbb

bbb

The above equations show that

a1

a2

a3

x1

x2

x3

b1 b2 b3 x1

x2

x3

or

x

x

x

x

x

x

T

x

x

x

1

2

3

11

2

3

1

2

3

a a a b b b1 2 3 1 2 3

provided that the inverse exists. The equation shows that the coordinate of a point can always beobtained from another coordinate system through a nonsingular square matrix T! Extending theargument to n dimension,

x Tx

relates the coordinate of a point in one coordinate system to another. We now look at the effect ofa coordinate change when applied to the state-space model.

( ) ( ) ( )

( ) ( ) ( )

x Ax Bu

y Cx Du

t t t

t t t

Since x T x1

,

)()()()()(

)()()()()(

ttttt

ttttt

DuxCTDuCxy

BuxATBuAxxTx

or

)()()(

)()( 11

ttt

tt

DuxCTy

BuTxATTx

Note that these two equations are the state space equations of the same system! In general, one

can define as many different coordinate system as can be by specifying a different T (as long as itis nonsingular), it is obvious that there are many different state space representations of the same


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system. In summary, the state-space representation of a system is non-unique and we have aninfinite choice of state vectors!

Example:

x

x

x

x

x

x

1

2

1

2

1

2

1 2

0 2

0

1

1 0

u

y

Let

x

x

x

x

1

2

1

2

2 0

0 5

then,

.

x

x

x

x

x

x

1

2

1

2

1

2

1

0 2

0

5

05 0

45

u

y

Since there are infinite representations of the same physical system, a natural question is whetherthere are some representations that are more useful or more insightful than others. It turns out thatthere are: eg controllable canonical forms, observable canonical forms and diagonal forms, as weshall see later.

2.3 Selection of state variables

Consider the following SISO nth order system,

Now, if we know the values of all the initial conditions for y and its derivatives, then we can findthe future behaviour of the system completely. Recall the definition of states! Hence, we cantakey and all its' differentials as the state variables

We therefore, now define state variables as

then, we can rewrite the equation as:

y a y a y a y un nn( ) ( )

1 2 1

x yx y

x ynn

1

2

1

( )


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in matrix form, we get

( ) ( ) ( )x Ax But t t where

A

0 1 0 0

0 0 1 0

0 0 0 1

1 2 3

a a a an

Output is

Example: (from Ogata)

Choose

then,

x x

x x

x x

x a x a x u

n n

n n n

1 2

2 3

1

1 1

y 1 0 0 0 x

or

xy =cT

y y y y u 6 11 6 6

x y

x y

x y

1

2

3

.

x x

x x

x x x x u

ie

1 2

2 3

3 1 2 36 11 6 6


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x

x

x

x

x

x

u

y

x

x

x

1

2

3

1

2

3

1

2

3

0 1 0

0 0 1

6 11 6

0

0

6

1 0 0

(3.4)

Since state variables are not unique, someone else using a different representation can come upwith a different state variables that can describe the same system. Of course the solution thoughwill be the same.

Example: For the same system above, we can choose

so that

3

2

1

3

2

1

3

2

1

001

6

00

6100

101010

x

x

x

y

u

x

xx

x

xx

(3.5)

Although equations (3.4) and (3.5) do not look the same, the solution is the same.

For the general differential equation:

)()()(

)()()()(

01011

1

1 tubdt

tdub

dt

tudbtya

dt

tdya

dt

tyda

dt

tyd

n

n

nn

n

nn

n

if we follow the same method to get the state equations, we will find that we would need thederivatives ofu(t) in the state equations. However, this will then not be in the standard format of(3.2). Instead, there are other commonly used formulations for state equations. In such cases, thestate variables are chosen so that the matrices A, B, C, and D have particular forms. We call suchstandard forms canonical forms. We illustrate one such canonical form using an example.

Example:

x y

x y

x y y

1

2

3

x x

x x x

x x x u

1 2

2 3 1

3 3 26 10 6


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Given a system described by the following differential equation, find an appropriate state spaceformulation.

)(2)(

3)()(

2)(

2

2

tudt

tduty

dt

tdy

dt

tyd (3.6)

If we take the Laplace Transform of (3.6) and assuming zero initial conditions, we get

2 ( ) 2 ( ) ( ) 3 ( ) 2 ( )s Y s sY s Y s sU s U s

so that the transfer function from U(s) to Y(S) is

G ss

s s( )

3 2

2 12

Y s

U s

( )

( )

We could rewrite this as:

Y ss

s sU s( ) ( )

3 2

2 12

(this gives us back the nice form of the previous example)

Let

then we get,

and the state equation is

Also, the output is given by

Let X ss s

U s( ) ( ) 1

2 12

s X sX X U2

2

x x

x x x

1

2 1

x x x u2 2 12

x x

x x x U

1 2

2 2 12

y t x t x t

x x

( ) ( ) ( )

3 2

3 22 1


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x

x

x

xu

yx

x

1

2

1

2

1

2

0 1

1 2

0

1

2 3

In general, for the general differential equation given by

)()()(

)()()()(

01011

1

1 tubdt

tdub

dt

tudbtya

dt

tdya

dt

tyda

dt

tyd

n

n

nn

n

nn

n

(Q: why the right hand side is at most ordern?)

the transfer function is

1

1 1 0

1

1 1 0

1

1 1 1 1 0 0

1

1 1 0

( )

( )

( ) ( ) ( )

n n

n n

n n

n

n

n n n n nn n n

n

b s b s b s bY s

U s s a s a s a

b b a s b b a s b b ab

s a s a s a

A simulation diagram for the above realization is

u(t)

an-1

an-2

a0

bn

bn-1

b1

b0

xn xn-1 x1x2

1nx

y(t)

-- -

and a representation of the state equation is


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u

x

x

x

aaaax

x

x

nnn

n

1

0

0

100

000

010

2

1

1210

2

1

x110 ncccy + bnu

wherec0 = b0 a0bnc1 = b1 a1bn

.and cn-1 = bn-1 - an-1bn

This is known as the controller canonical form. In this form, the feedback coefficients ai onlyand appears in the final state equation. The significance of the controller canonical form will bemake known later when you study controllability of a system. A simulation diagram of thecontroller canonical form is given in Figure 2.1. (Sometimes this diagram is preferable to the onegiven earlier because in this case, the output is taken directly from the states while earlier there is

a output from nx which has to be calculated.)

Figure 2.1 Simulation Diagram for controllable canonical form

u(t)

an-1

an-2

a0

bn

cn-1

c1

c0

xn xn-1 x1x2

1nx

y(t)

- - -


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If we define

C B

A A

B C

D D

T

T

T

T

, then, we get another realization given by

( ) ( ) ( )

( ) ( ) ( )

x Ax Bu

y Cx Du

t t t

t t t

then we get the observable canonical form below:

ub

x

x

x

y

u

c

c

c

x

x

x

a

a

a

a

x

x

x

n

n

nn

n

n

n

2

1

1

1

0

2

1

1

2

1

0

2

1

1000

1000

10

001

000

The simulation diagram for the observable canonical form is shown in Figure 2.6.

Figure 3.6 Simulation Diagram for observable canonical form

c0 1

c1 c2 bn

-a0 -a1 -a2 -an-1

cn-1

yu


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An important realization is the diagonal form. It can be shown that the diagonal form is best fromthe viewpoint of sensitivity, ie it will be not as sensitive to roundoff errors, truncation errors or thefinite word length in a computer.

The analog simulation for the diagonal realization is

For Y sB s

A s

U s

g

sU s g b ci

i

i i i

i

n

( )( )

( )

( ),

( ) :

1

c1b1

1

cnbn

n

1/(s-2) c2b2

y

u


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Diagonal forms are useful from the viewpoint of sensitivity and also in solving the equations.This is because with state equations, the problem in solving is the coupling between the statevariables. Now in diagonal form, we only have a set of independent 1st order differentialequations and that can be solved easily. But it is not always possible to obtain a diagonal form. Itis only possible if and only if A has n linearly independent eigenvectors.

2.4 Relation between Transfer Funct ion and State Equation representations

It would be reasonable to ask how the state space description, in time domain, is related to thetransfer function representation of a system. In this section, we will show the relationship

between the system matrices (A,B, C, D) and the transfer function. Since the transfer functiontypically refers to a single-input-single output system, we will consider state-space representations

that are single-input-single-output, ie. for the case where B R C R D Rn n 1 1 1 1, ,

Given a system with transfer function given by

and the state-space description given by

( ) ( ) ( )

( ) ( ) ( )

x t Ax t B t

Cx

u

y t t Du t (11.1)

From the state equation (11.1), taking the Laplace Transform, we get

(11.2)

Since transfer function is obtained with zero initial conditions, therefore, setting x(0)=0 andrearranging, (11.2) becomes

(11.3)

Comparing (11.3) with Y(s)=G(s)U(s), we get

DBAsICsG 1)()( (11.4)

Note that since B R C R A Rn n n n 1 1, , , therefore the product C sI A B R( ) 1 1 1

is a scalar.

We now take a closer look at equation (11.4). Since, in general, the inverse of a matrix H can be

expressed as

G s Y sU s

( ) ( )( )

sX s x AX s BU s

Y s CX s DU s

( ) ( ) ( ) ( )

( ) ( ) ( )

0

X s sI A BU s

and

Y s C sI A B D U s

( ) ( ) ( )

( ) ( ) ( )

1

1


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)det(

)(adj1

H

HH

where adj(H) refers to the adjoint of A, therefore the inverse of (sI-A) is given by

)det(

)(adj)( 1

AsI

AsIAsI

From (11.4), we get

])det()(adj[)det(

1)( DAsIBAsIC

AsIsG

In the case of a SISO system, the expression ])det()(adj[ DAsIBAsIC is a polynomial ins. Also, det(sI-A) is a polynomial and so we would get

)()(

)det()()(

sDSN

AsISNsG

From classical control, the characteristic roots (or poles) of the system are the values ofs suchthat D(s)=0. Clearly this is also the same as the values of s such det(sI-A)=0. From this, we canclearly see that the eigenvalues ofA are the poles of the system G(s)! (Q: what does that imply interms of stability of the system?)

Example:From the earlier example, we have the state equations:

x

xk

M

B

M

x

xM

f

yx

x

1

2

1

2

1

2

0 1 01

1 0

Using G s C sI A B D( ) ( ) 1 , we get

M

B

sM

ks

M

B

M

k

s

sAsI

110

0

0

( )( )

sI As s

sB

Mk

Ms

BM

kM

1 11


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Example:Given the state equations

( ) ( ) ( )

,

x xt A t Bu t

A B

where

0 1

2 3

0

1

and

( ) 3 1 ( )y t t x

then using , we get

3 1

0( 1)( 2) ( 1)( 2)( ) 3 1

2 1

( 1)( 2) ( 1)( 2)

3

( 1)( 2)

s

s s s sG s

s

s s s s

s

s s

As an exercise, check that the state equation from this transfer function is the same as above.Note that the above is in controller canonical form.Write out the observable canonical form of the transfer function, and compute the related transfer

function. It should be the same.

2.5 Notes:The matrix (sI-A) is sometimes called the characteristic matrix of A. Its determinant,

is known as the characteristic polynomial ofA, and its roots [i] are the eigenvalues or

characteristic values ofA (recall that they are the poles of the system). An importantproperty ofa(s) is that

This is called the Cayley-Hamilton theorem, which we will need to use later.

Example:

Given anA matrix A=[ 1 2; 0 3], show that it satisfies the Cayley-Hamilton theorem.

30

21A

and

G s C sI A B D( ) ( ) 1

a s sI A s a s a s a

s s s

n

n

n

n

( ) det( )

( )( ) ( )

12 1

1 2

a A A a A a In nn( ) 1

10


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30

21

30

21

10

01)(

s

ssAsI

34)3)(1()det()( 2 ssssAsIsa

and so

10

013

30

214

30

21

30

2134)( 2 IAAAa

010

013

30

214

90

81

We have stated earlier that state-space equations are not unique and so there are infinite state-

space representations of a physical system. However, there is only one transfer functionrepresentation of the same system. Hence in the next section, we will now show that no matterwhat state-space forms describing a system are found, the eigenvalues of the state space equationswill be the same. We call this the invariant of eigenvalues undersimilarity transform.

2.6 Invariant of eigenvalues under similarity transform

Given ( ) ( ) ( )t t t x Ax Bu

Let x Px (Similarity Transform)

Now for P non-singular, we have

-1 -1P x = AP x +Bu

i.e., -1x = PAP x + PBu

To show invariant of eigenvalues (ie for different state space representation of the same system,

the eigenvalues are the same), we need to show that are identical.

To do that, we consider

I A and I PA P -1

I PA P PA

P I A

P I A

P I A

P I A

I A

P P P

P

P

P

P

-1 -1 -1

-1

-1

-1

-1

( )


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Therefore, the eigenvalues are invariant under a similarity transform

Appendix: Inverse of Matrix

(the following taken from www.mathwords.com)

Cofactor

The determinant obtained by deleting the row andcolumn of a given element of a matrix

or determinant. The cofactor is preceded by a + or sign depending whether the element

is in a + or position.

Cofactor Matrix - Matrix of Cofactors

A matrix with elements that are the cofactors, term-by-term, of a given square matrix.


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Adjoint

The matrix formed by taking the transpose of the cofactor matrix of a given original

matrix. The adjoint of matrix A is often written adj A.

Example: Find the adjoint of the following matrix:

Solution: First find the cofactorof each element.

As a result the cofactor matrix of A is

Finally the adjoint of A is the transpose of the cofactor matrix:

Inverse of matrix

A-1 = (adjoint of A) or A-1 = (cofactor matrix of A)T


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Example: The following steps result in A-1 for .

The cofactor matrix for A is , so the adjoint is

. Since det A = 22, we get


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3 Solutions to State Space Models

In this week we will discuss the methods to solve linear systems represented using state-spacemodels. We will also address the issue of handling non-linearity in modeling.

3.1 Linearization

The utility of the linear-time-invariant (LTI) model goes beyond systems that are described bylinear model. Although most physical systems encountered in the real world are nonlinear, lineartechniques can still be used to analyze the local behaviour of a nonlinear system about smalldeviations about an operating point. While this may appear restrictive, in practice it is not too badas most control systems are designed to maintain variables close to a particular operating point

anyway. In order to obtain a linear model from a nonlinear system, we introduce a linearizationtechnique based on Taylor series expansion of a function.

Specifically, if a nonlinear time invariant system is given by

( , )x f x u

We can linearize about any given operating point using Taylors series expansion about aparticular operatin point. Suppose we wish to find the linear equation about the operating point(x0, u0 ), then at the operating point, we get

( , )x f x u0 0 0 (3.1.1)

Considerx x x

u u u

0

0

then

( , )

x x x

f x x u u

0

0

Expanding using Taylors series expansion, we get,

( , )

, ,

x x f x uf

xx

f

uu0 0

x u x u

0

0 0 0 0

and so

, ,

xf

xx

f

uu

x u x u

0 0 0 0

(3.1.2)

Equation (3.1.2) is the state space equation with A corresponding to

f

xx u0 0,

. i.e.


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A =

f

xx ,u

x ,u

0 0

0 0

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

n

n

n n n

n

1

1

1

2

1

2

1

2

1 2

and

B =

00

00

u,x

u,xu

f

n

nnn

n

n

uf

uf

uf

u

f

u

f

u

f

u

f

u

f

21

2

1

2

1

2

1

1

1

Example: Inverted Pendulum

As an example, consider the classical inverted pendulum problem. The system consists of a cartwith an inverted pendulum attached. This system is also commonly known as pole-balancerbecause the problem of applying a force u(t) to keep the pendulum upright is similar to that ofbalancing a pole on one's hand. It has more practical interpretation in terms of launching of arocket but we will look at this system because it can illustrate certain control concepts very

clearly.

Figure Pole Balancing cart system

Define the center of gravity of the mass as (xG, yG), then,

xG = x + l sin

yG = l cos

Applying Newtons law in the x -direction, we get

Md x

dt

md x

dt

uG

2

2

2

2

ie.

Force u(t)

M

m

l

P

x

y


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Md x

dtm

d x l

dtu

2

2

2

2

( sin )(3.1.3)

Since

d

dtd

dt

d

dt

d

dt

sin (cos )

sin (sin ) (cos )

cos (sin )

cos (cos ) (sin )

2

22

2

22

Eq. (3.1.3) can be written as

( ) (sin ) (cos ) M m x ml ml u 2 (3.14)

The equation of motion of the mass m in the y direction will also contain terms related to the

motion of the mass in the x direction. Instead, we consider the rotational motion of the mass maround point P. Applying Newtons second law to the rotational motion, we get

md x

dtl m

d y

dtl mgl

G G

2

2

2

2cos sin sin

(3.15)

Substituting forxGand yG, we get

md x l

dtl m

d l

dtl mgl

2

2

2

2

( sin )cos

( cos )sin sin

m x l l l m l l l mgl [ (sin ) (cos ) ] cos [ (cos ) (sin ) ] sin sin 2 2

Simplifying by using cos2

+ sin2

= 1, we get

mxl ml ml mgl cos (cos ) (sin ) sin 2 2 2 2 or

mx ml mg cos sin (3.16)

The nonlinear equations of motion are then given by Eqs. (3.14) and (3.16), i.e.

( ) (sin ) (cos ) M m x ml ml u 2

mx ml mg cos sin


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u

Ml

Mx

Ml

gmM

M

mgx

1

10

0

00)(0

000

1000

0100


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3.2 Solution of time-invariant state equation

Consider the scalarhomogeneous differential equation

(7.1)

Assume the solution of the form

Substitute in (7.1) to get

Now, this must hold for all t, so equating the coefficients, we get

At t=0, x(0)=b0, so that we get

As an exercise, substitute back this into (7.1) to check it indeed is the solution.

Now, we will use a similar method to try and solve the state equation. Consider

)()( tAt xx where x: n-dimensional vectorA: n by n matrix

By analogy with the scalar case, we can assume the solution in the form of a vector power series

in t, or

ktttt k210 bbbbx2)(

By substituting for )(tx in the state equation, we get

)(32 2123 kk tttAtktt k210k21 bbbbbbbb

This must hold for all t, so

x ax

x t b b t b tkk( ) 0 1

b b t kb t

a b b t b t

k

k

k

k

1 2

1

0 1

2

( )

b ab

b ab a b

b a bk kk

1 0

212 1

12

2

0

10

!

x t at a t a t x

x t e x

kk k

at

( ) ( ) ( )

( ) ( )

! !

1 0

0

12

2 2 1


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10

0k

012

01

bb

bbb

bb

kAk

AA

A

!1

2

1

2

1 2

At t=0, 0bx )0( ,

)0()!

1

!2

1()( 22 xx kktA

ktAAtIt

We call the power series in the parenthesis the matrix exponential because of its similarity to theinfinite power series for a scalar exponential. ie.

Atkk etAk

tAAtI !

1

!2

1 22

Therefore, the solution to the state equation is

)0()( xx Atet

whereAte is also known as thestate transition matrix

3.3 Properties of State Transition Matrix

1.

The convergence of the series can be proven for any finite t.That is,Ate always

exists.

2.Proof:

eA t

jI AtAt

n

j j

A tn

lim!

!

2 2

2

0

d

dte Ae e AAt At At


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11

At

At

Ae

tA

tAAtIA

tA

tAtAA

tA

tAAtI

dt

de

dt

d

]!3!2

[

]!3!2

0[

]!3!2

[][

33

22

34

232

33

22

3.AsAtstA eee )(

Proof:

00

!!k

kk

k

kk

AsAt

k

sA

k

tAee

By direct expansion, well get

)(

222

22

22

2

33

22

33

22

)!2

2()(

!2!2)(

)!3!2

)(!3!2

(

stA

AsAt

e

ststAstAI

tsAs

At

AstAI

sA

sAAsI

tA

tAAtIee

More rigorous derivation:

0 0

0 0

0

( )

! !

!( )!

( )

!

k k k k At As

k k

i k ik

k i

kk

k

A t s

A t A se e

k k

t sA

i k i

t sA

k

e

4. expAtis nonsingular and [ expAt]-1

= exp ( -At)

Proof:

Since2121 )( AtAtttA eee , therefore if we let tt 1 , and tt 2 , then


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AtAt

AtAt

AtAtttA

eeei

eeI

eee

1

)(

][..

SinceAte always exists, therefore inverse ofeAtalways exists and so eAt is nonsingular.

5. BtAttBA eee )( ifAB=BABtAttBA eee )( ifABBA

Proof:

!3

)(!2

)()(3

32

2)( tBAt

BAtBAIe tBA (4.3.1)

!3!2!2!3

!2!2)(

)!3!2

)(!3!2

(

33

32

32

33

222

22

33

22

33

22

tB

tAB

tBA

tA

tBABt

tAtBAI

tB

tBBtI

tA

tAAtIee BtAt

(4.3.2)

Subtracting (4.3.2) from (4.3.1), we get

!3

)22(!2

)(3

22222

)( tABBABABABABABAt

ABBAeee BtAttBA

so that ifAB=BA (this is called A and B commute), then the right hand terms vanish and so

BtAttBA eee )(

3.4 Solution ofinhomogenousstate equation (ie. with input)

Consider the scalar case,

Integrating w.r.t t, we get

[ ]

. . [ ( )]

x ax bu

x ax b u

e x ax e bu

i ed

dte x t e b u

at at

at at


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taat dbuextxe

0

)()0()(

ttaat dbuexetx

0

)( )()0()(

The former term is thefree response (zero input response) and the latter term known as theforcedresponse (zero state response).

We will again extend the result to the matrix case. Consider the state equation,

Following the scalar case, we get

Integrating from 0 to t,

0

( ) (0) ( )t

At Ae x t x e Bu d

ttAAt duBexetx

0

)( )()0()(

(4.4.1)

3.5 How to find eAt?

Computing eAtby summing up infinite terms is impossible. So we discuss how to compute itmore conveniently. In this section, we illustrate the method using inverse Laplace transforem.

Again we will look at the scalar case to get an idea of what to do. Consider the scalar 1st orderdifferential equation,

( ) ( ) ( )

:

:: tan

: tan

x t Ax t Bu t

where

x n vector

u m vector A n n cons t matrix

B n m cons t m atrix

1

1

[ ]

. . [ ( )]

x Ax Bu

x Ax Bu

e x Ax e Bu

i e ddt

e x t e Bu

At At

At At

x ax


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Taking the Laplace Transform, we get

1

( ) (0) ( )

(0)( )

( ) (0)

sX s x aX s

xX s

s a

s a x

Taking the Inverse Laplace Transform, we get

Compare with the solution we obtained earlier, ie

Hence, we get

Similarly, for the matrix case,

It can be shown that,

Hence, we have just shown that

The importance of this equation is that it provides a convenient means to compute the matrix

exponential: compute1( )sI A first, then take inverse Laplace transform (on each of its entries).

For the case of the state equation with input,we get

x t L s a x( ) {( ) } ( ) 1 1 0

x t e xat( ) ( ) 0

e L s aat 1 1{( ) }

X s sI A x

x t L sI A x

( ) ( ) ( )

( ) {( ) } ( )

1

1 1

0

0

( )

{( ) }!

sI AI

s

A

s

A

s

L sI A I At A t

eAt

1

2

2

3

1 12 2

2

L sI A eAt 1 1{( ) }


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From the previous result foreAt, we get

X s L e x L e BU sAt At( ) { } ( ) { } ( ) 0

Taking the inverse Laplace Transform,

x(t) eAtx(0) eA(t)Bu()d0

t

(4.4.2)

If the initial time is not zero but at t0, then the solution becomes

x(t) eA(tt0 )x(t0) eA(t)Bu()d

t0

t

(4.4.3)

We can see this because from (4.4.2), the solution at time t0 is given by

x(t0

) eAt0 x(0) eA(t0 )Bu()d0

t0

so that

x(0) eAt0 x(t0) eAt0 eA(t0 )Bu()d

0

t0

But the solution at time tfor zero initial condition is given by (4.4.1)


t

Therefore, substituting forx(0), we get

x(t) eAt(eAt0 x(t0) eABu()d

0

t0

) eA(t)Bu()d0

t

x(t) eA(tt0 )x(t0) eA(t)Bu()d

t0

t

Another way to think of it is that if the intial time is t0, consider some one (observer 2) else whohas a watch pointed at zero. From the view point of observer 2, the intial value is x(t0), and thesystem will evolve for a time t-t0. Solving the equation of observer 2, one get (4.4.3). Notice thatthe system is independent to the observer.

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

x t Ax t Bu t

sX s x AX s BU s

sI A X s x BU s

X s sI A x sI A BU s

0

0

01 1


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Example:

Solve ux

x

x

x

1

0

21

10

2

1

2

1

Solution:

21

10

0

0

s

sAsI

21

1

s

s

Hence we get

s

s

ssAsI

1

12

12

1)(

2

1

22

22

)1()1(

1

)1(

1

)1(

2

s

s

s

ss

s

Therefore, using the Inverse Laplace Table,

tt

ttAt

ette

teetAsILe

)1(

)1(}){( 11

and so

duet

etx

ette

teettx

t

t

t

tt

tt

)()](1[

)()0(

)1(

)1()(

0)(

)(

3.6 Other methods to computeAte

It was shown previously thatAte can be obtained using the Inverse Laplace Transform. This is

the most common way to findAte in closed-form, i.e.

}{ 11 AsILeAt

Other methods are as follows:

3.6.1 Series expansion


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SinceAte was defined as an infinite series, that can be used to compute Ate to any degree of

accuracy required. This is used in most computer code (MATLAB for example). Recall that

Since the series converges,Ate can be approximated by evaluating up toNterms only.

3.6.2 Using similarity transform

If we can transform the matrixA into either a diagonal or Jordan (discuss later) canonical form,

thenAte can be easily evaluated. We shall first consider the case whereA has only distinct

eigenvalues. In that case,A can be transformed into a diagonal matrix using similarity transform.IfA has repeated eigenvalues, then it may not always be able to be diagonalized, but it may be

transformed into a form called the Jordan canonical form using similarity transform.

Case 1: Matrix A is diagonalizable

Consider a diagonal matrix where

n

00

0

0

00

2

1

We will now show that it is possible to obtain from matrixA by a similarity transform, i.e. we

will show that it is possible to get 1PAP .

Similarity transform:

Let 1PAP , then premultipy byP-1 (assuming that the inverse exists) on both side to get

11 PAP

or

1

2

0 00

0

0 0 n

A

1 2 n 1 2 nc c c c c c

where cIare the columns ofP-1

then

1 2 nA A A 1 2 n 1 2 nc c c c c c

It is easy to see that this is equivalent to the following set of equations

Ac1

1c1

e

A t

j I AtAt

n

j j

A tn

lim !!

2 2

20


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Ac2

2c2 (4.7.2.1)

and so on.

But, recall the definition of eigenvector: Any nonzero vectorx satisfying Ax x is called an

eigenvector ofA associated with the eigenvalue . Hence, from the equation above, we can seethat c1 is the eigenvector associated with eigenvalue 1 , c2is the eigenvector associated with

eigenvalue 2 and so ciis the eigenvector associated with eigenvalue i . This shows that byselecting the eigenvectors as columns of theP-1 matrix, and if the n eigenvectors are independent(which will be the case if the eigenvalues are unique), then it is possible to diagonalize the matrixA.

We can use similarity transform to findAte as we will now show.

Consider the state space equation

A Bu

y C Du

x x

x

where the solution is given by


t

(4.7.2.2)

Using theP defined earlier, if we let

x Px , then

1

1

PAP PBu

y CP Du

x x

x

We have shown earlier that 1PAP , and so we get

Bu

y C Du

x x

x

The solution to this state equation would be

x(t) etx(0) e(t)B u()d0

t

(4.7.2.3)and using

x Px , we get

x(t) P1etPx(0) P1e(t)B u()d0

t

(4.7.2.4)

Comparing (4.7.2.4) with (4.7.2.2), we can see that

PePe tAt 1


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Since is diagonal, i.e.

n

0

01

and

t

t

t

t

ne

e

e

e

00

0

0

2

1

Ate can be easily evaluated as

P

e

e

e

Pe

t

t

t

At

n

00

0

0

2

1

1

Note: From (4.7.2.4), if we assume that the input u(t) is zero, then we get

x(t) P1etPx(0)or

x(t) P1etx(0)

so that

1

2

1

2

(0)0

(0)( )

0

(0)0 0 n

t

t

t n

xe

xet

xe

1 2 nx c c c

i.e. 1 21 2( ) (0) (0)t tt e x e x

1 2x c c

This shows that the solutionx(t) is a combinations of the responses )0(it

i xeci and hence is a

combination (superposition) of the eigenmodestie

. We will discuss this later in this chapter.

Example:


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Given

20

10A , find Ate using the diagonalization method.

Solution:

The first thing we need to do is to find the eigenvalues. This can be found by solving thecharacteristic polynomial det(sI-A)=0.

0)2(20

1)det(

ss

s

sAsI

hence the eigenvalues ares=0, -2.

Next, we need to find the eigenvectors associated with the eigenvalues. Considering s=0, we needto seek solution ofx such that

(0I-A)x=0;

i.e. 020

10

2

1

x

x

and sox1=k, where kis a constant andx2=0.Normalizing, we get the eigenvector associated withs=0 as x=[1 0]T.

Similarly for the other eigenvalue,

(-2I-A)x=0

i.e. 000

12

2

1

x

x

so that 02 21 xx or 12 2xx . Hence the eigenvector associated withs=2 is given byx=[1 -2]

T, or if we normalized it, we get

5

251

x

Hence, using the eigenvectors as columns of the matrixP-1, we get

520

511

1P

and

250

211

P


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therefore, since PePe tAt 1 , we get

2

50

211

0

0

520

511

2

0

t

tAt

e

ee

t

t

t

tAt

e

e

e

ee

2

2

2

2

0

)1(2

11

250

211

520

511

end

Example 2:

Given Ax x where

31

20A , find x(t) for the intial conditions [1 1]T and [2 1]T

respectively.

Solution:

The eigenvalues of theA matrix ares=-2,-1 and the associated eigenvectors are (not normalized)

c1 1

1

and c2

2

1

(notice conincide with the initial conditions)

Hence we form

11

211P

and so

11

21P

Since )0()( 1 xePtx t or

x(t) c1e1tx1(0) c2e

2tx2(0), i.e. we get

)0(1

2)0(

1

1)( 21

2 xexet tt

x


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If

1

1)0(x , then

0

1

1

1

11

21)0(x and so

tet 2

1

1)(

x

If

1

2)0(x , then

1

0

1

2

11

21)0(x , and so

tet

1

2)(x

Hence we can conclude that if the initial point lies on the eigenvector, it will remain on that line

forever. A phase plane plot (x2 vs x1 ) for the two initial condistions is shown in the figure below.

For arbitrary initial points, a series of phase plots can be found as shown below:

x2

x1

(2,1)

(1,1)

x2


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Example 3:

Given

32

12A ,find the phase portrait.

Solution (as an exercise, please fill out the details):

The eigenvalues of theA matrix ares=-1,-4 and the associated eigenvectors are (not normalized)

2

1

1

121 candc

and the corresponding phase-plane plot is

Case 2: Matrix A is not diagonalizable

We will not be going into the details of this case for this course. Suffice to say that in case where

A has repeated eigenvalues, thenA may or may not be diagonalizable. In such cases, it can be

shown that thebest one can do is to reduce the matrixA to a form known as the Jordan canonicalform which has the following structure:

2

1


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1

1

1

1

001

00

10

001

J

The above shows the Jordan form for the case where the eigenvalues are repeated n times. The

matrixJte is then given by

t

tn

tt

tn

ttt

Jt

e

en

ttee

en

te

ttee

e

1

111

1111

00

0)!2(

0

)!1(!22

12

{ Review:

Examples of eigenvalues and eigenvectors

Find the eigenvalues and eigenvectors of the matrix

41

23A

Solution:

Using the determinant rule,

107

2)4)(3(

41

23

2

ss

ss

s

sAsI

so that 0 AsI gives the eigenvalues ass1=2, ands2= 5.

To find the eigenvectors associated with the eigenvalues, first consider the eigenvalues1=2. Todetermine the eigenvector, we must find x such that

xx 1sA

ie. 021

21

)( 1

xx

AIs


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and therefore

1

2x is a solution for any scalar. We can choose the solution =1 or the

normalized solution

51

52

x

Similarly, the eigenvector associated withs2=5 is given by

1

1x for any scalar. We can

choose the solution =1 or the normalized solution

21

21

x

It is understood that any scalar multiples of such normalized eigenvectors will also be aneigenvector.

}

Example:

Consider the matrix

A

1 3 2

0 4 2

0 3 1

Using the determinant rule,

sI A

s 1 3 2

0 s 4 2

0 3 s 1

(s 1)[(s 4)(s 1) 6]

(s 1)(s2 3s 2)

(s 1)2(s 2)

so that 0 AsI gives the eigenvalues ass1=1, ands2= 2.

To find the eigenvectors associated with the eigenvalues, first consider the eigenvalues1=1. Todetermine the eigenvector, we must find x such that

Ax x

ie.

sI A x 0 3 2

0 3 2

0 3 2

x 0


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i.e.

3x2 2x3 0We get 1 equation (i.e. 1 constraint) with 3 unknowns and so we can get 2 linearly independenteigenvectors such as:

1 0 0 T

and

1 2 3

For eigenvalues2= 2, we get the eigenvector

1 1 1 T and so the matrix can be diagonalized.(Prove this!)

Example:

Consider the matrix

A

2 1 2

0 2 1

0 0 1

The eigenvalues are:s1=1, ands2= 2 (solving2( 2) ( 1) 0s s ).

Eigenvector associated withs1=1 is given by:

sI A x 1 1 2

0 1 1

0 0 0

x 0

so that the eigenvector is

1 1 1 T

Eigenvector associated withs1=2 is given by:

sI A x 0 1 2

0 0 1

0 0 1

x 0

so that the eigenvector is

1 0 0 T

For this example, there are only one linearly independent eigenvector of the repeated eigenvalues1=2, hence it is not diagonalizable. Instead, we can only convert it into Jordan canonical form

2 1 0

0 2 0

0 0 1


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Example:

Consider the matrix

A 3 2

4 1

Characteristics equation is given by:

sI A s 3 2

4 s 1

s2 2s 5

so that 0 AsI gives the eigenvalues as

s1,2 2 i 16

2

1 2i

.

Eigenvectors associated with the eigenvalues1 = 1+2i is given by:

(s1I A)x 2 2i 2

4 2 2i

x 0

or

(2 2i)x1 2x2 0

4x1 (2 2i)x2 0

Note that the 2 equations are the same since

(2 2i) (22i) 8

So that the eigenvector is given by

x 1

1 i

.

The other eigenvector is

x 1

1 i

3.7 Modes of the System

Consider the time invariant linear system

0)0(, xxAxx

where nnn RARx , .

For simplicity, let use assume that all the eigenvalues of A are distinct. Let the

eigenvectors ofA be npp ,...,1 associated with the eigenvalues n ,...,1 . As these

eigenvectors are linearly independent, they form a basis.

Let us define the nonsingular matrix

PRnn as


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1 1 2 nP p p p

where the columns ofP-1 are the eigenvector ofA. Let us also denote

PRn n as

1

2

n

q

q

P

q

where nqq ,...,1 are the row vectors of

P. Therefore,

qipj 1 i j

0 i j

We know that

0)( xetx At for 0t , and

x(t) P1etPx0 , for 0t ,

where ).,...,,( 21 ndiag Hence, the solution ofx(t) can be rewritten as

0

1

)( xqpetxn

i

ii

ti

, for 0t ,

or

n

i

ii

tpetx i

1

)( , (5.1)

where 0xqii , i=1,..,n.

(1) The above (5.1) shows that the response of the system is a composition of motions along theeigenvectors of A. We call such a motion a mode or eigenmode of the system. A particular

mode of the system can be excited by choosing the initial state0

x as a component along the

corresponding eigenvector.

(2) Each mode of the system varies in time according to the exponential function oftie

.

(3) An arbitrary initial condition will in general excite all n modes of the system. The amount of

excitation of each mode depends on the value of 0xqii .

(4) For example, suppose 10 kpx where kis a constant, then1

1)( pketx t .This implies that the solution x(t) consists of only the response of only the first mode. In state

space, this response can be seen to move along the eigen direction of1

p . The other modes

are suppressed.

Example:

Given a system xx

54

10


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The eigenvalues of A are 1 and 4 respectively. The eigenvectors are

1

1and

4

1.

Therefore,

P1 1 1

1 4

and

Pq1

q2

4

3

1

31

3

1

3

Therefore, 221121)( pepetx tt where

20

10

11x

xq and

20

10

22x

xq and

20101

3

1

3

4xx and 20102

3

1

3

1xx . Thus

tt exxexxtx 4201020101 )3

1

3

1()

3

1

3

4()( and

tt exxexxtx 4201020102 )3

1

3

1(4)

3

1

3

4()(


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4 Controllability and Observability

In this week, we are interested in qualitative features of state-space representation of

system. Specifically, we will be discussing two ideas known as controllability and

observability of a system.

4.1 Introduction and Motivation

While many features in state-space analysis of system have their parallels in frequency

domain analysis or classical analysis, Controllability and Observability are unique

features of state-space analysis. These ideas were first introduced by E.G.Gilbert and R.F.

Kalman in the early 1960s. They give a clear explanation as to why cancellation of

unstable poles are undesirable even if perfect cancellation is possible. ( c.f. the very first

example given in the introduction to state-space system ). Using these ideas, it can be

shown that the overall system is unstable although the transfer function of the system isstable. Hence, the idea of controllability and observability is an important concept in the

state-space analysis of the system. Before we proceed with the definitions of

controllability and observability, let us look at some motivating examples.

Example 1 : Consider

ux

ux

2

1

,

If )0()0( 21 xx , then for all time t and all control u. So, there are no

possible u and tto make say 1 2( ) ( ) 1x t x t .

Example 2 : Consider

2

2

21

0xy

x

xx

The above shows thaty(t) will always be a constant. Hence, observingy(t) does not tell us

what is doing.

A more interesting example is one given by the following diagram. It consists of two carts

coupled by a passive spring. In addition to the spring force, an active control force fis to

be provided by some means within the system, so thatfacts on both cart 1 and 2.

Example 3.

x 1

x 2

m 1 m 2f f

k Figure: Example of two-cart system

x t x t1 2( ) ( )

x1


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The equations of motion of the system can be shown to be

The state equation using the state vector is given by

00//

00//

1000

0100

22

11

mkmk

mkmkA ,

2

1

/1

/1

0

0

m

mB

It is well known from the law of physics that the force can change the relative distance

between the two carts. i.e., but it cannot change the variables or

independently. However, if only the matricesA andB are given, based on what we have

learnt so far, we have no way of knowing that such a constraint exists. The problem is

even more pronounce since state-space representation of system is not unique. The

question is: is it possible to identify such constraints bsed on the matricesA andB?

Example 4

Consider the system given below:

1R 2R0 0R

1C 2C

3R

1v 2vV

Figure: An uncontrollable system

By Physics, we know

1 2 11 1

1 3

2 1 2

2 22 3

;V v v v

C vR R

V v v v

C v R R

dx

dtx1 1

dxdt

x2 2

dx

dt

k

mx x

f

m( )1

1

1 2

1

dx

dt

k

mx x

f

m( )2

2

2 1

2

x x x x xT

1 2 1 2

x x2 1 x1 x2


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Hence, the differential equation can be described by

VRCvRRCvRCv 222

3221

322

11111

Consider the voltage across 3R , 21 vvv then

VRRCC

CRCRv

RRCRCv

RCRRCv

2121

11222

32231

1

32311

11111111

If 2211 CRCR , then the coefficient ofVvanishes and

1 2 3

1 1 3

R R Rv v

C R R

This implies that v is not influenced by Vand the voltage v decays from whatever initial

voltage to zero, i.e., v is not controllable.

The above examples illustrate the concept of controllability of system.

Practical examples of unobservability are also easily available. Typically examples of

unobservability of system occur when two systems are connected in tandem. If the output

consists of the output from the first system only, then the states of the second system are

unobservable.

Controllability and Observability are concerned with providing an analysis

tool to the above situations. For a given system of the form

x = Ax + B u

y = Cx + Du(1)

where , and matrices. Concept of Controllability

and Observabilty are useful in answering the following questions.

Can we drivex(t) wherever we want using some u?

Can we causex(t) to follow a given path?

Can the measurement ofy(t) tell us whatx(t0) was?

Can we trackx(t) by observingy(t)?

The concepts of controllability and observability are defined under the assumption that

we have complete knowledge of a dynamical equation; i.e., matrices A, B, Cand D are

known.

4.2 Controllability

A R B Rn n n r , C Rm n D Rm r

VRC

vRC

vRRC

v11

2

31

1

311

1

11111


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Definition : A Linear Time-invariant system is said to be controllable if there exists an

input, u(t), 10 tt which drives the system from any initial state 0)0( xx to any

other state 11)( xtx in a finite time 1t .

The key to the above definition are the words "any" and "finite". If the

input is only possible to make the system go from some states to some other states, the

system is not controllable. Moreover, if it takes an infinite amount of time to go from the

arbitrary initial state to the arbitrary final state, the system is likewise uncontrollable.

It is possible to define the concept of controllability to accommodate time-

varying systems, in which it may happen that the possibility of reaching 1x may depend

on the initial time 0t (in that case 0t may not be 0 anymore). For our case, we will

confine our studies on time-invariant systems. In that case, there is no loss of generality in

taking the initial time 0t to be zero.

Theorem 1 - ControllabilityThe n-dimensional linear time-invariant state equation in (1) is controllable if and only if

the following condition is satisfied:

1) The nrn controllability matrix BAABBU n 1

is full row rank (i.e., rank(U)=n).

Proof:

and .There are various ways of showing this. The following is a more direct proof showing

the connection to the Cayley-Hamilton principle.

For the system

x = Ax + B u

y = Cx + Du(2)

The time solution of the system given in (2) at time1t is

1

11

0

)(

01 )(

t

tAAtdBuexex

(3)

or

1

1

0

01 )(

t

AAt dBuexxe (4)

Note that ( From Cayley-Hamilton Theorem and infinite series expansion of )

(5)

e A

e AA kk

nk

( )0

1


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Substituting (5) into (4) gives

1

1

0

1

0

01 )()(

t

k

n

k

kAtduBAxxe (6)

If we denote

1

0

)()(

t

kk du

then

1

1

0

11

0

011

n

n

k

n

k

kAt BAABBBAxxe

(7)

For the system to be controllable, it means that

Introduction to Linear Systems and State Space

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Transcript of Introduction to Linear Systems and State Space