Chapter 3 State-Variable Models

1

•State-Variable Models• State equations• State Variables of a Dynamic System• The concept of State• Form of the State Equations• The State Differential Equation• Transfer Function of a State Space Model• The State Transition Matrix• Characteristic Equation and Eigenvalues• Controllability & Observability

• The time-domain is the mathematical domain that incorporates the response and description of a system in terms of time, t.

State-Variable Models

Differential equations

Modeling

Nonlinear&Time Varying

LTI Transfer Function

Lapalce Transform

Inverse Lapalce Transform

Physicalsystem

State-VariableModels

State-VariableModels

(cause) (effect)

2

State-Variable Models

3

1 2

1 0 2 0 0

Consider this system shown in the above Figure. A set ofstate variables ( , , , ) for system is a set such that knowledge of the initial values of the state variables[ ( ), ( ), , ( )] at the

n

n

x x x

x t x t x t

…

… 0

1 2 0

initial time t and of the input signals ( ) and ( ) for will be sufficient to determine the future values of the outputs and state variables.

u t u t t t>

System (x)u(t)

Input

x(0) Initial conditions

y(t)

Output

The general form of a dynamic system is schematically shown as

State-Variable Models• State differential equations are an alternative way to describe

a dynamic system (i.e. time-domain method).

• The state-variable model, or state-space model is a particular differential equation model.

• Equations are expressed as n first-order coupled differential equations, but the choice of states is not unique

• All choices of state variables preserve the system’s input-output relationship (that of the transfer function)

4

State Equations

• Any nth order differential equation can easily be converted into a set of 1st order state equations for nonlinear or time-varying systems, the transfer function approach often breaks down.

• Most multivariable and many stochastic design methods are based on state equations

5

• For LTI systems, it is routine to move between state and transfer function representations i.e. between frequencyand time domains

The Concept of State

•Because the choice of states for a given system is not unique, we often choose states that represent physical measurements – voltages, velocity, position, etc.

• However, it may also be convenient to choose states that simplify the mathematical form of the state equations – for example the output and its derivatives – but are not easily measurable.

6

Form of the State EquationsThe state description always consists of two sets of equations, normally written in matrix form. The first set of equations describes the dynamics, and is a set of first order differential equations, each expressing the first derivative of a state as a function of all the (n) states and the (m) inputs (with no derivatives):

))(),(()(or

))(),(),()(),(()(

))(),(),()(),(()())(),(),()(),(()(

m121

m12122

m12111

tutxftx

tututxtxtxftx

tututxtxtxftxtututxtxtxftx

m

nnn

n

n

=

=

==

7

Form of the State Equations

))(),(()())(),(()(

tutxgtytutxftx

==

The second set of equations has no dynamics, and expresses the outputs as a function of the states and inputs:

))(),(()(or

))(),(),()(),(()(

))(),(),()(),(()())(),(),()(),(()(

m121

m12122

m12111

tutxgty

tututxtxtxgty

tututxtxtxgtytututxtxtxgty

npp

n

n

=

=

==

8

Linear Time Invariant Case

In this case, the functionsf and g become linearcombinations of x and ugiving the familiar form:

• If the are n states, m inputs, and p outputs, then A is square (nxn), B is (nxm), C is (pxn) and D is (pxm)

• For a single input, single output system, we have A square (nxn), B=b and is a (nx1) column vector, C=c is a (1xn) row vector, and D=d is a (1x1) scalar (often zero)

))()(()())()(()(

tu,txgtytu,txftx

==

DuCxyBuAxx

+=+=

9

1 11 1 12 2 1 11 1 1

2 21 1 22 2 2 21 1 2

1 1 2 2 1 1

n n m m

n n m m

n n n nn n n nm m

x a x a x a x b u b ux a x a x a x b u b u

x a x a x a x b u b u

= + +… + += + +… + +

= + +… + +

Form of the State Equations

1 2

The state of a linear time in-varying system is described by the set of firstorder differential equations written in terms of the state variables [ ... ]and can be written in general form as:

nx x x

1 11 12 1 111 1 1

2 21 22 2 2

11 2

nm

n

n nm mn n n nn n

x a a a xb b u

x a a a x

b b ux a a a x

= +

or in matrix form as follows:

Form of the State Equations1 11 12 1 1

11 1 12 21 22 2 2

11 2

nm

n

n nm mn n n nn n

x a a a xb b u

x a a a x

b b ux a a a x

= +

n: number of state variables, m: number of inputs.

The column matrix consisting of the state variables is called thestate vector and is written as

1

2

n

xx

x

x

=

Form of the State Equations1 11 12 1 1

11 1 12 21 22 2 2

11 2

nm

n

n nm mn n n nn n

x a a a xb b u

x a a a x

b b ux a a a x

= +

x A x Bu= +

1 11 12 111 1 1

2 21 22 2

11 2

, , ,

nm

n

n nm mn n n nn

x a a ab b u

x a a ax A B u

b b ux a a a

= = = =

Example:

State variables=?

)()()()(2

2tutky

dttdyb

dttydM =++

)()()()(12

2 tutkxtbxdt

tdxM =++

Thus the following two first-order DE’s

13

Mtutx

Mktx

Mb

dttdx

txdt

tdx

)()()()(

)()(

122

21

+−−=

=

1 1

2 2

0 1 0( )1

x xu tk bx x

M M M

= + − −

Example:how about ? )()()()()(

011

1

1 tutyadt

tdyadt

tydadt

tyda n

n

nn

n

n =++++ −

−

−

14

12

1

1 1 0 1

1 ( ( ))

nn

nn n

n

dx xdt

dx xdt

dx a x a x u tdt a

−

− −

=

=

= − + + +

111 2Let , , , .n

ndxdxx y x x

dt dt−= = =

1 1

2 2

0 1 1 1

0 1 00

0 0 0

1// / / n

n n n n n n

x xx xd u

dta

x a a a a a a x−

= +

The State Differential Equation

=

=

=

=

yy

xx

xyy

xx

x

2

1

2

1

)(2 2 tuyyyyy

+−−=

=

ωως

)(10

210

2 tuyy

yy

+

−−

=

ωςω

CxyBuAxx

=+=

)(2 2 tuyyy =++ ωως

[ ]01 10

210

2 =

=

−−

= CBAωςω

• Consider the second-order system:

Its state-space description is

With

15

Series RLC Circuit)()()( )( : 0v i tvtvtvtv CRL ++∑ ==

)()( )( tvRtidtdiLtv C++=

Cti

dtdv

dtdvCti CC )()( =⇔=

cvxix == 21 let

)()()( tvtvRtidtdiL C +−−=

Cti

dtdvC )( =

)(11 211 tv

Lx

Lx

LR

dtdx

+−−=

Cx

dtdx 12 =

ODE:

16

) : that(Notice 12 dt

dxx ≠

Series RLC Circuitcvxix == 21 let

11 2

2 1

1 1 ( ) dx R x x v tdt L L Ldx xdt C

= − − +

=

1 1

2 2

1 1( )

1 00 ux x

BA

Rx xL L v tLx x

C

− − = +

17

[ ] [ ] 12

2

If we let as the output, then = 0 1 , 0 1C C

xv C y x v

x

= = =

[ ] [ ] 11

2

If we let ( ) as the output, then = 1 0 , 1 0 ( ).x

i t C y x i tx

= = =

x Ax Bu= +

An RLC circuit. :0∑ =ii

RivdtdiL

itudt

dvCi

LcL

Lc

c

−=

−== )(

∑ = : 0v i

LcL

Lc

RiLRv

Ldtdi

tuC

iCdt

dv

−=

+−=

1

)(11

Lc ixvx == 21 let 21

2

21

1

)(11

xLRx

Ldtdx

tuC

xCdt

dx

−=

+−=

)( Lc iitu +=

RidtdiLvvv L

LoLc +=+=

18

An RLC circuit.Lc ixvx == 21 let

212

21

1

)(11

xLRx

Ldtdx

tuC

xCdt

dx

−=

+−=

19

1 1

2 2

1 10( )

1 0 ux x

BA

x xC u tCx xR

L L

− = + −

2( ) ( )oy t v t Rx= = ⇒

x Ax Bu= +

[ ]0c R=

[ ] 12 0

2

0 L

xy R Rx Ri v

x

= = = =

An RLC circuit.Lc ixvx == 21 let

212

21

1

)(11

xLRx

Ldtdx

tuC

xCdt

dx

−=

+−=

20

1 1

2 2

1 10( )

1 0 ux x

BA

x xC u tCx xR

L L

− = + −

2If we want ( ) ( )oy t v t Rx= = ⇒

x Ax Bu= +

[ ]0c R=

[ ] 12 0

2

0 L

xy R Rx Ri v

x

= = = =

What happen if is the output?Lv

An RLC circuit.Lc ixvx == 21 let

21

1 1

2 2

1 10( )

1 0 ux x

BA

x xC u tCx xR

L L

− = + −

What happen if is the output?Lv

1 1

2 1 1 2

let

c

L c o L

v x xv x v v x Ri x Rx

∗

∗

= =

= = − = − = −

*1 1

* * *2 1 2 1 2

1 1( ) ( )

x x

x x x x xR R

= ⇔

= + = +

[ ] 11 2 0

2

1 c L

xy R x Rx v v v

x

= − = − = − = ⇒

[ ]1c R= −

State Equation to SFG

LcL

Lc

RiLRv

Ldtdi

tuC

iCdt

dv

−=

+−=

1

)(11

1

2

let

c

L

x vx i

==

1 2

2 1 2

1 1 ( )

1

x x u tC C

Rx x xL L

= − +

= −

Output: .Rxtvty o 21 )()( ==

U(s) Vo(s)1x 2x

1x 2xs1

s1

RC1

C1

−

L1

LR

−

22

Inverted Pendulum on a Cart

State variables?

θMgLT =

23

Inverted Pendulum on a Cart

2 2

2 2( ) cosθ ( ) 0d y d θM m mL u tdt dt

+ + − =

The sum of the forces in the horizontal direction:

The sum of the torques about the pivot point:2 2

22 2 0d y d θmL mL mLg θ

dt dt+ − =

dttdxtx

dttdyxtyx )( )( )( )( 4321

θθ ====

mg

2

2

dtdlm θ

2 2

2 2( )d θ d ymLdt dt

← +

2 2

2 2( ) ( ) 0 for small d y d θM m m L u t θdt dt

+ + − =

24

Inverted Pendulum on a Cart

2 4( ) ( ) 0dx dxM m m l u tdt dt

+ + − =

The state variable for the second-order equations are:

The sum of the torques about the pivot point:

0 342 =−+ xg

dtdxl

dtdx

)()( 4321 dtd,,

dtdy,yx,x,x,x θθ=

25

Inverted Pendulum on a Cart2 4( ) ( ) 0dx dxM m m l u t

dt dt+ + − =

0342 =−+ xg

dtdxl

dtdx

)(32 tuxgm

dtdxM =+

23

1 ( )

1 ( ) (for )

dx m g x u tdt M M

u t M mM

= − +

≈ >>

dtdxgx

dtdxl 2

34 −=

26

(horizontal direction)

(about the pivot point)

2 4( ) ( ) 0dx dxM m m l u tdt dt

+ + − =

Inverted Pendulum on a Cart2 4( ) ( ) 0dx dxM m m l u t

dt dt+ + − =

0342 =−+ xg

dtdxl

dtdx

)(32 tuxgm

dtdxM =+

23

1 ( )

1 ( ) (for )

dx m g x u tdt M M

u t M mM

= − +

≈ >>

dtdxgx

dtdxl 2

34 −=

27

(horizontal direction)

(about the pivot point)

2 4( ) ( ) 0dx dxM m m l u tdt dt

+ + − =

(horizontal direction)

Inverted Pendulum on a Cart

0342 =−+ xg

dtdxl

dtdx

)(32 tuxgm

dtdxM =+

2 1 ( ) (for )dx u t M mdt M

≈ >>

28

(horizontal direction)

(about the pivot point)

43

1 ( ) 0dxu t l g xM dt

+ − =

0)(34 =+− tuxgM

dtdxlM

Inverted Pendulum on a Cart

)t(ulM

xlg

dtdx

xdt

dx

)t(uM

xM

gmdt

dx

xdtdx

1

1

34

43

32

21

−=

=

+−=

=

The four first-order differential equations:

CxyBuAxx

=+=

[ ]0100 =C29

0 1 0 00 0 00 0 0 10 0 0

mgM

gl

A−

=

1

1

0

0M

Ml

B

−

=

Transfer Function of a State Space Model

)()(A)( sBUsXssX +=

)()(C)()()(A)(

tDutxtytButxtx

+=+=

[ ] )()( sBUsXsI-A =

Taking the Laplace transform (assume zero initial conditions)

So,

[ ] 1( ) ( )X s sI A BU s−= −

)()(C)( sBUsXsY +=

30

)( ωσ js +=

Transfer Function of a State Space Model[ ] )()( 1 sBUsI-AsX −=

)()(C)( sBUsXsY +=Substituting into the output equation

Therefore, for single variable case, the transfer function of the system is

For multivariable case,

yields [ ] )()()(}{)( 1 sUsGsUDBsI-ACsY =+= −

)()()( sU/sYsG =

)()()( sU/sYsG jiij =

31

TF from State Space Model Example

3410

)()()( 2 ++

==sssU

sYsG

(a)

?)( =sG

A,B,C and D √

[ ] )()()(}{)( 1 sUsGsUDBsI-ACsY =+= −

[ ]xy

uxx

01010

4310

=

+

−−

=

32

TF from State Space Model Example

uxxx +−−= 212 43

110xy =

uyyy 1034 =++

10/10/

10/

12

12

1

yxxyxx

yx

====

=

3410

)()()( 2 ++

==sssU

sYsG

21 xx =

)(10)()34()(10)(3)(4)(

2

2

sUsYsssUsYssYsYs

=++

=++

(b)

?)( =sG[ ]xy

uxx

01010

4310

=

+

−−

=

Dynamics equation:

33

uxxx +−−= 212 43

TF from State Space Model Example

11

1

1)()()(

asasasUsYsG n

nn

n ++== −

+

11 1( ) ( ) ( )n n

n na s a s a Y s U s−+ + + =

L.F:

)()()()()(121

11 tutya

dttdya

dttyda

dttyda n

nnn

nn =++++ −

−

+

[ ]

+

==

− equation State

eqaution dyanmic:LF)()(

)(1 DBsI-ACsUsY

sG

34

TF from State Space Model Example

1

2 1

3 2

1

1

Let

n

n n

x yx x yx x y

d yx xdt

−

−

== == =

= =

1 2

1 1 2 21

1

1 ( );n n nn

x x

x a x a x a x ua

y x+

=

= − − − +

=

−−−

=

+++ 11211 ///100000010

nnnn aaaaaa

A

=

+1/1

00

na

B

[ ]001 =C D=?

? )()()()()(121

1

1 tutyadt

tdyadt

tydadt

tyda n

n

nn

n

n =++++ −

−

+

[ ] 1( )( )( )

Y sG s C sI - A B DU s

−= = + ⇒35

11 1

1 ( ) n nn n

G sa s a s a−

+

=+ + +

Realization of a Transfer Function,x Ax Bu y Cx Du= + = +• A state description is a

realization of G(s) if 1[ ] ( )C sI A B D G s−− + =

• A realization of G(s) is minimal if there exists no realization of G(s) of less order

• An LTI systems is observable if the initial state x(0) can be uniquely deduced from knowledge of u(t) and y(t) fort Є [0 T]

• An LTI system is controllable if for every x(t0) and every T >0, there exist u(t0+t), 0<t ≤ T such that x(t0+T) =0

36

From Transfer Function to State Space• For a given transfer function, there is no unique state

space realization

• Engineering dictates the use of a realization of leastorder, a minimal realization (A realization of G(s) is minimal if there exists no realization of G(s) of less order)

• A minimal realization is both controllable and observable

• All the possible A matrices for the different space realizations should have the same eigenvalues

37

[ ])()()( 1

sDsNDBsI-ACsG =+= −

0) Det( 0)( :Pole =⇒= sI-AsD

From Transfer Function to State Space

3)(1)( )()(1

)(63

6131

1

6)( ==⇔+

=+

=+

= sH,s

sGsGsH

sGs

s

ssT

2)(1)( )()(1

)(2

1121

1

)( ==⇔+

=+

=+

= sH,s

sGsGsH

sGs

s

ssT

Why & How?

1 2 1 2

15 5Δ 5( 1) 1 1( ) 5, 5 ,Δ Δ 1,Δ 1 51Δ 51 5

k kkP ssT s P P

s s ss

+ += = = = = = = = +

++

∑

1

1

5155

5)1(5)( −

−

+

+=

++

=ss

sssGc

38

From Transfer Function to State Space

1

1

5155

5)1(5)( −

−

+

+=

++

=ss

sssGc

39

1

1

12 1 2

ss s

−

−=+ +

1

1

6 63 1 3

ss s

−

−=+ +

From Transfer Function to State Space

1x

1x2x

2x

3x

3x

3211 063 xxxx ++−=

[ ]Xy

trxxx

xxx

X

001

)(150

5002020063

3

2

1

3

2

1

=

+

−−−

−=

=

State-variable differential equation:

]5)([5520 33212 xtrxxxx −++−=

)(500 3213 trxxxx +−−=

1xy =40

From Transfer Function to State Space

))()(()(

)3)(2)(5()1(30)(

)()(

321 sssssssq

sssssT

sRsY

−−−=

++++

==)3()2()5(

)()()( 321

++

++

+==

sk

sk

sksT

sRsY

1x

2x

3x

30 10 20 321 =−=−= kkk

State-variable differential equation:

)(111

300020005

3

2

1

3

2

1

trxxx

xxx

+

−−

−=

diagonal canonical form[ ]

=

3

2

1

3010-20- xxx

y41

The State Transition Matrix Φ(t)

How to compute this matrix?

nRtx ∈)(

42

The State Transition Matrix (u=0)

0)0()()(

)(

000

1

==−=∫=∫

==∈=⇒=

txetxttAx/xlnAdtx

dx

Adtx

dxAx

dt/dxRxifAxxAxx

At

• The state transition matrix satisfies the homogenous (i.e. zero-input) state equation.

• It represents the evolution of the system’s free response to non-zero initial conditions:

“zero input response”

x(0) is initial condition. Hence it is a constant.

Atet =)(φ

=?

)()( tAxtx =

)0((t))( xtx φ=

BuAxx += u=0

43

The State Transition Matrix (u=0)

• If , there is another way to find x(t) using Taylor series expansion

• Successive differentiation of gives:

0 At time

)(

32)3(

==

==

txAx

xAxAx

kk

nRtx ∈)(

Axx =

44

xAxAx 2==

)0()()( xttx φ=

The State Transition Matrix

+

+

++++=

++++=

kk

kk

txA!k

txA!

tAxx

tx!k

tx!

txxtx

)0(1)0(21)0()0(

)0(1)0(21)0()0()(

22

)(2

This series converges for all finite t. It is called the matrix exponential

)0()121( 22 xtA

!ktA

!AtI kk

+++++=

+++++= kkAt tA!k

tA!

AtIe 121 22

+++++= kkAt tA!k

tA!

AtIe 121 22

)0()( xetx At=45

The Matrix Exponential

122121 )( AtAtAtAtttA eeeee ==+

IeA =0

AtAt ee −− =1)( i.e. the matrix exponentialbehaves very much like thefamiliar scalar exponentialfunction. Note that A mustbe square.

T)( AttA eeT

=

AeAe AtAt =

AtAt Aeedtd

=

46

State Transition Matrix (u≠0) Φ(t)BuAxx +=

Buexedtd AtAt −− =)(

=∫ −t A dxe

dtd

0)( ττ

BueAxxe AtAt −− =− )(

∫ −+=∫+= −−tt tAAt dButxtBudexetx00

)( )()Φ()0()Φ()0()( τττττ

Atet =)Φ( 47

BuAxx =−

)0()( 0xetxe AAt −− − ∫= −t A dBue0

)( τττ

∫ −+=t

dButxttx0

)()Φ()0()Φ()( τττ

State transition Matrix (u≠0) –Φ(s)

Which compares with the time domain solution:

Let

BuAxx +=)()()0()( sBUsAXxssX +=−

)()()0()()( 11 sBUAsIxAsIsX -- −+−=

)()0()()( sBUxsXAsI +=−

1)()( -AsIs −=Φ)()()0()()( sBUsxssX ΦΦ +=

∫ −+=t

dτButxttx0

)()()0()()( ττΦΦ

Atet =)Φ( 48

)( ωσ js +=Can it exist?

State Transition Matrix

• Note that the system response has twocomponents:

• Natural response – “zero input response”due to initial conditions

• Forced response – “zero state response”due to input

• Overall response is the sum of the two components

BuAxx +=

∫ −+=t

dτButxttx0

)()()0()()( ττΦΦAtet =)Φ(

1)()( -AsIs −=Φ

49

Example

The time-domain state transition matrix can be obtained using the inverse Laplace transform

−−

=3120

A

+−

=−31

2 then

ss

AsI

−+=−=

ss

sAsIsΦ -

123

)Δ(1][ )( 1

)3)(1(232)3()Δ( with 2 ++=++=++= sssssss

50

Example

0,3 2211 == ααa=1,b=2

( ) ( 1)( 2)s s s∆ = + +

−+=−=

ss

sAsIsΦ -

123

)Δ(1][ )( 1

{ }

+−−+−−

== −−−−

−−−−

tttt

tttt-

eeeeeeeest 22

221

2222)( L)( Φφ

is response free the then11

)0( conditions inital Assuming

=x

)0()()( 2

2

== −

−

t

t

eexttx φ ttttt

ttttt

eeeeeeeeee

222

222

2222

−−−−−

−−−−−

=+−−

=+−−51

Example

Note that for otherinitial conditions,the phase planeplot will not be astraight line.

52

Characteristic Equation and Eigenvalues• Recall that, for a transfer function G(s)=N(s)/D(s),

the roots of the characteristic equation D(s)=0 are the poles of the system.

• Recall that the denominator of the transfer functionof a state-space representation is det(sI-A)

• The characteristic equation is then det(sI-A)=0• The roots of this equation are the eigenvalues of

the matrix A.

[ ] )()()(}{)( 1 sUsGsUDBsI-ACsY =+= −

For the stable system, the real parts of all the eigenvalues must be negative.

For the stable system, what must the egenvialues be ?

53

)( ωσ js +=

Controllability

DuCxyBuAxx

+=+= Theorem: This system is controllable if

and only if the following controllability matrix has rank n:

Note that for a single input system S will be an [n × n] square matrix and the rank test is that

][ 12 BABAABBS n−=

∆[S]≠0

54

Controllability Example

1 20 0

−?

=

−

−=

01

10

12BA

[ ]

:matrixility controllab

== ABBS

det S= 055

Observability

DuCxyBuAxx

+=+=

Theorem: This system is observable if and only if the following observability matrix has rank n:

TnCACACACV ][ 12 −=

56

Observability Example

1 02 0−

det V=

?

[ ]01 01

10

02=

=

−

−= CBA

:matrixity observabil

=

=CAC

V

057

Uncontrollable System

The state x2 cannot be affected by input u and henceis uncontrollable

58

Unobservable System

The state x2 does not affecte y and hence is unobservable

59

Controllability and Observability

60

Basic Idea of State Feedback

Consider the state feedback controller where is a constant feedback gain matrix

Then one can write

Whereas the poles of the open-loop system are given by the eigenvalues of A, the poles of the closed-loop systemare given by the eigenvalues of (A-BK).

BuAxx +=

rKxu +−=

BrxBKAr-KxBAxx

+−=++=

)( )(

61

State Feedback Design

-

Control system with state feedback

•The poles of the closed-loop system can be arbitrarily assigned if and only if the system is controllable

BrxBKAr-KxBAxx

+−=++=

)( )(

62

State Feedback Design•Often, states are not all measurable. Hence, it is

necessary to design an observer to construct them from the output vector.

• Such an observer can be designed if and only if the system is observable.

• The observed state is then used instead of the true state to generate the feedback.

63

State Variable Feedback

• Choice of feedback matrix gains allows the eigenvalues (or poles) to be assigned as we choose.

• Note that we have not addressed (yet) the issue of a reference input r.

• We shall be returning to state variable design later.

………and now back to transfer functions............

64