Notes on Control Principles

Chapter 3 State-Variable Models

•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

(cause) (effect)

State-Variable Models

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

x t x t x t

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)

x(0) Initial conditions

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)

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

• 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.

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

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

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

m12122

m12111

tutxftx

tututxtxtxftx

tututxtxtxftxtututxtxtxftx

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

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

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

m12122

m12111

tutxgty

tututxtxtxgty

tututxtxtxgtytututxtxtxgty

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

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 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

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

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

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

Form of the State Equations1 11 12 1 1

11 1 12 21 22 2 2

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

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

)()()()(

+−−=

0 1 0( )1

x xu tk bx x

= + − −

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

1 tutyadt

tdyadt

tydadt

tyda n

n =++++ −

1 1 0 1

1 ( ( ))

dx xdt

dx a x a x u tdt a

− −

= − + + +

111 2Let , , , .n

ndxdxx y x x

dt dt−= = =

0 1 1 1

0 1 00

1// / / n

n n n n n n

x xx xd u

x a a a a a a x−

The State Differential Equation

)(2 2 tuyyyyy

+−−=

ωως

2 tuyy

−−

ωςω

CxyBuAxx

)(2 2 tuyyy =++ ωως

[ ]01 10

−−

= CBAωςω

• Consider the second-order system:

Its state-space description is

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

)()( )( tvRtidtdiLtv C++=

dtdvCti CC )()( =⇔=

cvxix == 21 let

)()()( tvtvRtidtdiL C +−−=

dtdvC )( =

)(11 211 tv

+−−=

dtdx 12 =

) : that(Notice 12 dt

dxx ≠

Series RLC Circuitcvxix == 21 let

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

= − − +

1 1( )

1 00 ux x

Rx xL L v tLx x

− − = +

[ ] [ ] 12

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

xv C y x v

[ ] [ ] 11

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

−== )(

∑ = : 0v i

Lc ixvx == 21 let 21

)( Lc iitu +=

RidtdiLvvv L

LoLc +=+=

An RLC circuit.Lc ixvx == 21 let

1 10( )

1 0 ux x

x xC u tCx xR

− = + −

2( ) ( )oy t v t Rx= = ⇒

x Ax Bu= +

[ ]0c R=

[ ] 12 0

xy R Rx Ri v

= = = =

1 10( )

1 0 ux x

x xC u tCx xR

− = + −

2If we want ( ) ( )oy t v t Rx= = ⇒

x Ax Bu= +

[ ]0c R=

[ ] 12 0

xy R Rx Ri v

= = = =

What happen if is the output?Lv

1 10( )

1 0 ux x

x xC u tCx xR

− = + −

What happen if is the output?Lv

2 1 1 2

L c o L

v x xv x v v x Ri x Rx

= = − = − = −

* * *2 1 2 1 2

1 1( ) ( )

x x x x xR R

= + = +

[ ] 11 2 0

xy R x Rx v v v

= − = − = − = ⇒

[ ]1c R= −

State Equation to SFG

x vx i

1 1 ( )

x x u tC C

Rx x xL L

= − +

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

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

1x 2xs1

Inverted Pendulum on a Cart

State variables?

θMgLT =

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 ====

dtdlm θ

2 2( )d θ d ymLdt dt

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

+ + − =

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

)()( 4321 dtd,,

dtdy,yx,x,x,x θθ=

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

dt dt+ + − =

0342 =−+ xg

)(32 tuxgm

dtdxM =+

1 ( ) (for )

dx m g x u tdt M M

u t M mM

= − +

≈ >>

dtdxgx

dtdxl 2

34 −=

(horizontal direction)

(about the pivot point)

+ + − =

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

dt dt+ + − =

0342 =−+ xg

)(32 tuxgm

dtdxM =+

1 ( ) (for )

dx m g x u tdt M M

u t M mM

= − +

≈ >>

dtdxgx

dtdxl 2

34 −=

+ + − =

0342 =−+ xg

)(32 tuxgm

dtdxM =+

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

≈ >>

1 ( ) 0dxu t l g xM dt

+ − =

0)(34 =+− tuxgM

dtdxlM

)t(ulM

The four first-order differential equations:

CxyBuAxx

[ ]0100 =C29

0 1 0 00 0 00 0 0 10 0 0

Transfer Function of a State Space Model

)()(A)( sBUsXssX +=

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

tDutxtytButxtx

[ ] )()( sBUsXsI-A =

Taking the Laplace transform (assume zero initial conditions)

[ ] 1( ) ( )X s sI A BU s−= −

)()(C)( sBUsXsY +=

)( 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 =

TF from State Space Model Example

)()()( 2 ++

==sssU

?)( =sG

A,B,C and D √

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

−−

uxxx +−−= 212 43

110xy =

uyyy 1034 =++

10/10/

yxxyxx

)()()( 2 ++

==sssU

21 xx =

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

sUsYsssUsYssYsYs

?)( =sG[ ]xy

−−

Dynamics equation:

uxxx +−−= 212 43

1)()()(

asasasUsYsG n

n ++== −

11 1( ) ( ) ( )n n

n na s a s a Y s U s−+ + + =

)()()()()(121

11 tutya

dttdya

dttyda

dttyda n

nn =++++ −

− equation State

eqaution dyanmic:LF)()(

)(1 DBsI-ACsUsY

x yx x yx x y

d yx xdt

== == =

1 1 2 21

1 ( );n n nn

x a x a x a x ua

= − − − +

−−−

+++ 11211 ///100000010

nnnn aaaaaa

[ ]001 =C D=?

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

1 tutyadt

tdyadt

tydadt

tyda n

n =++++ −

[ ] 1( )( )( )

Y sG s C sI - A B DU s

−= = + ⇒35

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

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

[ ])()()( 1

sDsNDBsI-ACsG =+= −

0) Det( 0)( :Pole =⇒= sI-AsD

From Transfer Function to State Space

3)(1)( )()(1

6)( ==⇔+

= sH,s

sGsGsH

2)(1)( )()(1

)( ==⇔+

= sH,s

sGsGsH

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

+ += = = = = = = = +

5)1(5)( −

12 1 2

−=+ +

6 63 1 3

−=+ +

3211 063 xxxx ++−=

5002020063

−−−

State-variable differential equation:

]5)([5520 33212 xtrxxxx −++−=

)(500 3213 trxxxx +−−=

1xy =40

))()(()(

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

321 sssssssq

sssssT

−−−=

==)3()2()5(

)()()( 321

30 10 20 321 =−=−= kkk

State-variable differential equation:

300020005

−−

diagonal canonical form[ ]

3010-20- xxx

The State Transition Matrix Φ(t)

How to compute this matrix?

nRtx ∈)(

The State Transition Matrix (u=0)

0)0()()(

==−=∫=∫

==∈=⇒=

txetxttAx/xlnAdtx

dt/dxRxifAxxAxx

• 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

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

nRtx ∈)(

xAxAx 2==

)0()()( xttx φ=

The State Transition Matrix

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

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

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

)0()121( 22 xtA

!AtI kk

+++++=

+++++= kkAt tA!k

AtIe 121 22

+++++= kkAt tA!k

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

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

Buexedtd AtAt −− =)(

=∫ −t A dxe

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:

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

Example

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

−−

=−31

2 then

−+=−=

sAsIsΦ -

)Δ(1][ )( 1

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

Example

0,3 2211 == ααa=1,b=2

( ) ( 1)( 2)s s s∆ = + +

−+=−=

sAsIsΦ -

)Δ(1][ )( 1

+−−+−−

== −−−−

−−−−

eeeeeeeest 22

2222)( L)( Φφ

is response free the then11

)0( conditions inital Assuming

)0()()( 2

== −

eexttx φ ttttt

eeeeeeeeee

−−−−−

=+−−

=+−−51

Example

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

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 ?

)( 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

Controllability Example

1 20 0

: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 −=

Observability Example

1 02 0−

det V=

[ ]01 01

−= CBA

:matrixity observabil

Uncontrollable System

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

Unobservable System

The state x2 does not affecte y and hence is unobservable

Controllability and Observability

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

+−=++=

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

+−=++=

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.

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

Notes on Control Principles

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