KS40602 State Space Modeling

8/9/2019 KS40602 State Space Modeling

1/25

KS40602/kent

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KS40602/kent

1.2

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State M del

1.2.1 Stat The stat(knownvariables

0t t , c

0t t > . Insystem at

In the staconsistsspace repFig. 1.1.

States var

Input vari

Output v

The diff (column

The statenumber o

xdt

dx1

1 = &

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M

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e Space F

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te space re

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(1.1)

2

2


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State Space Modeling 3

State Space Modeling 3 KS40602/kenteo/0809(2)

The n numbers of differential equations may be written in vectornotation as

( ))(),()( t t f t UXX =& (1.2)

The set of all possible values which the input vector )(t U canhave (assume) at time forms the input space of the system.

Similarly, the set of all possible values which the output vector)(t Y can assume at time t forms the output space of the system

and the set of al possible values which the state vector )(t X canassume at time t forms the state space of the system.

1.2.2 State Space Model of Linear System

The state model of a system consists of the state equation andoutput equation. The state equation of a system is a function of state variables and inputs as defined by Eqn. (1.2).

For linear time invariant systems the first derivatives of statevariables can be expresses as a linear combination of statevariables and inputs.

mmnn ububub xa xa xa x 121211112121111 .................. +++++=&

mmnn ububub xa xa xa x 222212122221212 .................. +++++=& M

mnmnnnnnnnn ububub xa xa xa x .................. 22112211 +++++=&

(1.3)where the coefficients ija and ijb constants.

In the matrix form the above equations can be expressed as

+

=

mnmnn

m

m

m

nnnnn

n

n

n

n u

u

u

u

bbb

bbb

bbb

bbb

x

x

x

x

aaa

aaa

aaa

aaa

x

x

x

x

M

LL

MLLMM

LL

LL

LL

M

LL

MLLMM

LL

LL

LL

&

M

&

&

&

3

2

1

21

33231

22221

11211

3

2

1

21

33231

22221

11211

3

2

1

(1.4)

The matrix Eqn. (1.4) can also be written as)()()( t Bt At UXX +=& (1.5)

The equation )()()( t Bt At UXX +=& is called the state equation of Linear Time Invariant (LTI) system.


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The output at any time, )(t Y is the function of state variables andinputs.

( ))(),()( t t f t UXY = (1.6)

Hence the output variables can be expressed as a linearcombination of state variables and inputs.

mmnn ud ud ud xc xc xc y 121211112121111 .................. +++++=

mmnn ud ud ud xc xc xc y 222212122221212 .................. +++++= M

mnmnnnnnnnn ud ud ud xc xc xc y .................. 22112211 +++++=(1.7)

where the coefficients ijc and ijd constants.

In the matrix form the above equations can be expressed as

+

=

m pm p p

m

m

m

n pn p p

n

n

n

n u

u

u

u

d d d

d d d

d d d

d d d

x

x

x

x

ccc

ccc

ccc

cac

y

y

y

y

M

LL

MLLMM

LL

LL

LL

M

LL

MLLMM

LL

LL

LL

M3

2

1

21

33231

22221

11211

3

2

1

21

33231

22221

11211

3

2

1

(1.8)

The matrix Eqn. (1.8) can also be written as)()()( t Dt C t UXY += (1.9)

The equation )()()( t Dt C t UXY += is called the output equationof Linear Time Invariant (LTI) system.

The state model of a system consists of state equation and outputequation. The state equation and output equation together calledas state model of the system.

Hence the state model of a linear time invariant system (LTI)system is given by the following equations.

)()()( t Bt At UXX +=& State equation)()()( t Dt C t UXY += Output equation


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State Diag1. Block D2. Signal F

1.3

eo/0809(2)

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iagram arelements o

5

5


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KS40602/kent

The state

invariant sequations.

)( At XX =&

State equat

)( C t XY =Output equ

eo/0809(2)

odel of li

stem is gi

)()( t Bt U+ion

)()( t Dt U+ation

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Fig. 1.5:

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tput of the

6

6


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If initial conditions are given, then they can be added at theoutput of integrators using adders.

In each state equation, the first derivative of state variable isexpressed as a function of state variables and inputs.

Therefore from the knowledge of a state equation, the statevariables and inputs are multiplied by appropriate scalars andthen added to get the first derivative of a state variable.

Now, the first derivative of the state variable is given as input tothe corresponding integrator. Similarly the input of all otherintegrators is obtained by considering the state equations one byone.

Each output equation is a function of state variables and inputs.Therefore from the knowledge of an output equation, the statevariables and inputs are multiplied by appropriate scalars andthen added to get an output. Similar procedure is followed togenerate all other outputs.


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Advantage1. Stateutilized f feedback.

2. Designfeedback forward.

3. Solutiogives tivariables.

Disadvant1. Solutiomay beco

1.4

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ments R,

ystem

8

8


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KS40602/kent

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A minimstate modelectricalelements.

The enerphysicalstate varidifferenticonstitute The inputsources.currents idissipatinvariables

ExampleObtain th1.1.1 by c

SolutionLet the th

quantitiesLet the in At node

21++ ii

1 + x x

3 x =&

l numberel of the ssystem ar

y storage evariables iables andl equatio

s the state

s to the syhe output

n energy dg elementcan be any

1.1e state mohoosing mi

Fig. 1.1.1

1.1ree state v

as 11 i x = ;put variabl

, apply Ki

0=dt

dv c

03 =+ xC &

111

xC

xC

f state varistem. The

e currents

lements arthe diffe

the equatins. Thesequation of

stem are ein electric

issipatingin electricvoltage or

del of thenimal num

Electrical

riables 1 x ,

22 i x = ;)(t eu = , i

chhoffs C

2

ables are cbest choicand volta

inductancential equns are rea

set of the system.

citing voltal systemlement. Tl network.urrent in t

electrical nber of state

Network E

2 x and 3 x

cv x =3 .nput to the

rrent Law

State Sp

State Sp

osen for os of statees in ene

and capactions arerranged asirst order

age sourcere usuallye resistancIn generale network.

etwork shvariables.

xample 1.1

be related

system.

ce Modeling

ce Modeling

taining theariables ingy storage

itance. Theeplaced byfirst ordeequation

or currentvoltages oe is energythe output

wn in Fig.

to physical

(1.1.1)

(1.1.2)

9

9


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KS40602/kent

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At 1 st loo

Rit e + 11)(

1 xu +

x1 =&

At 2 nd lo

L Ri + 222

22 R x

2 x =&

Arrangin

x

x x

=

3

2

1

&

&&

Let choosvariables

1 x y =

Arrangin

=

2

1

y

y

, apply Ki

dt di

L =+ 11

111 x L =+ &

L

x

L

R

1

1

1

1 1+

p, apply Ki

cvdt

di =2

322 x x L =&

22

2

2 1 L

x L

R +

the state e

C C

L R

L R

11

0

0

2

21

1

e the voltawhich den

11 R ; 2 y =

the state e

2

1

2

1 0

x

x

R

chhoffs V

cv

3 x

u

L

x1

31

rchhoffs

3 x

quations in

x

x x

L

L

0

1

1

3

2

1

2

1

es across tted by 1 y

22 R

quations in

oltage Law

oltage La

matrix for

[u L

+0

0

1

1

e resistanc

11 Ri ; 2 y

matrix for

State Spa

State Spa

,

,

,

es as the o

22 Ri= .

,

e Modeling 1

e Modeling 1

(1.1.3)

(1.1.4)

(1.1.5)

(1.1.6)

(1.1.7)

tput

(1.1.8)

0

0


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KS40602/kent

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ExampleObtain th1.2.1 by c

SolutionLet the stLet the in

Let the o Connect

Convert t

At node 1

21 + R

vv

1 R

x x

1 x =&

1.2e state mohoosing 1v

Fig. 1.2.1

1.2te variable

put variabl

tput variab

voltage so

he voltage

, apply Kir

01 =dt dv

C

01 =+ xC &

1

11C

xCR

+

del of the)(t and (2v

Electrical

s )(11 t v x = )(t vu = .

le 1 )(t v y =

urce at the

ource to th

hhoffs C

2 x

electrical n)t as state

Network E

; 22 v x =

1 x= .

input.

e current s

rrent Law,

State Spa

State Spa

etwork shariables.

xample 1.2

)t .

urce.

e Modeling 1

e Modeling 1

wn in Fig.

(1.2.1)

(1.2.2)

1

1


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KS40602/kent

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At node

R

vv 12 +

R

x x 2

x2 =&

Arrangin

x

x

=

2

1

&

&

The outp

Arrangin

[= 01 y

, apply Kir

dt

dvC

R

v 22 +

C R

x ++ 21 &

C x

CR1

1

the state e

C CR

CRCR

21

11

t 1 )(t v y =

the state e

]

2

1

x

x

hhoffs C

Rt v )(=

Ru=

2

CR x 12 +

quations in

x

x

+

2

1

1 x= .

quations in

rrent Law,

matrix for

[ ]u R

10

matrix for

State Spa

State Spa

,

,

e Modeling 1

e Modeling 1

(1.2.3)

(1.2.4)

(1.2.5)

(1.2.6)

2

2


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KS40602/kent

Advantage1. The statdirectly f from thegoverning

2. The pha link bfunction dtime doma

Disadvant1. The phphysical vand theref for measupurpose.

1.5

eo/0809(2)

:space mo

rmed byifferentialhe system.

se variabltween theesign apprin design a

ge:se variablriables of tre are not

rement an

State Sp

el can beinspectionequations

s providetransfer

oach andproach.

s are nothe systemavailable

control

ace Rep The phaswhich arderivativ Usually tstate vari The statethe systetransfer f There arvariables

1.5.1 Pha Considerrelating t

ya ynn

+&&

)1(

1

By choos

)1(

3

2

1

=

===

n

n y x

y x

y x y x

&

M

&&

&

Substitutigovernin

xa xnn

+1

&

xn =& The state

a x

x x

x x

x x

n

nn

==

==

1

1

32

21

&

&

M

&

&

esentati

variablesobtained

s.

e variablebles are th

model usinmodel is

nction for

three mand they ar

se Variabl

the folloe output y

yan

++&

)2(

2

ing the out

ng the stthe syste

xan

++12

xa xa n 21

equations

xa nn

12

on usin

are definedfrom one

used is then derivati

g phase vaalready kn

.

thods of e explained

es Method

ing nth o)(t to the i

a n+ &L 2

ut y and t

n

n y x && =

te variabl, Eqn. (1.1

xan

+2

L

1 LL

f the syste

a n

LL

Phase V as those pof the sys

system oues of the o

iables canwn in the

odelingin the foll

1 (Bush-C

der linearput )(t u o

ya n ++ && 1

heir derivat

es in the),

xan

++ 213

a xa nn 32

are

xa x n

132

ariables State Spa

State Spa

rticular statem variab

put and thtput.

e easily differential

system uwing secti

ompanion)

differentia system.

bu yn =

ives as stat

differenti

bu xan

=1

xa x n 121

b xa n+

1

e Modeling 1

e Modeling 1

te variableles and it

remaining

termined iequation o

sing phasens.

l equation

(1.10)

variables,

l equation

bu

u

3

3


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Arranging the state equations in the matrix form,

[ ]u

b x

x

x

x

x

aaaaa x

x

x

x

x

n

n

nnnnn

n

+

=

0

0

0

0

10000

01000

00100

00010

1

3

2

1

1321

1

3

2

1

MM

L

L

MMMMML

L

L

&

&

M&

&

&

(1.11) UXX B A +=&

Matrix A (system matrix) has a very special form. It has all 1s inthe upper off-diagonal; its last row is comprised of the negativecoefficients of the original differential equation and all otherelements are zero. This form of matrix A is known as Bush formor Companion Form .

Matrix B has the specialty that all its elements except the lastelement are zero. The output being 1 x y = , and the outputequation is given by

[ ]

=

n x

x

x

x

y

M

LL3

2

1

0001

(1.12) XY C =

The advantage in using phase variables for state space modelingis that the system state model can be written directly byinspection from the differential equation governing the system.

Example 1.3

Obtain the state model of the system with the transfer function

12410

)()(

23 +++=

ssssU sY

using Phase Variables Method 1 (Bush-Companion).

Solution 1.3

Given that124

10)()(

23 +++=

ssssU sY

(1.3.1)

( ) )(10124)( 23 sU ssssY =+++ )(10)()(2)(4)( 23 sU sY ssY sY ssY s =+++ (1.3.2)

Inverse Laplace Transform of Eqn. (1.3.2)u y y y y 1024 =+++ &&&&&& (1.3.3)

UXY DC +=

0=

DQ

XY C =


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Define state variables as follows, y x =1 ; y x &=2 ; y x &&=3

Substitute 3 x y &&&& = ; 3 x y =&& ; 2 x y =& ; 1 x y = into Eqn. (1.3.3),

u x x x x 1024 1233 =+++& u x x x x 1024 1233 +=&

Hence, the state equations are

21 x x =& ; 32 x x =& ; u x x x x 1024 1233 +=&

Output equation is

1 x y =

The state model in the matrix form is

[ ]u x

x

x

x

x

x

+

=

10

00

421

100010

3

2

1

3

2

1

&

&

&

[ ]

=

3

2

1

001

x

x

x

y

1.5.2 Phase Variables Method 2 (Signal Flow Graph)

Consider the following nth order linear differential equationrelating the output )(t y to the input )(t u of a system.

ubububub

ya ya ya ya ya y

mm

mm

nnn

nnn

++++=

++++++

&LL&&

&&&LL&&&

1

)1(

10

12

)2(

2

)1(

1 (1.13)

Let n = m = 3, ubububub ya ya ya y 3210321 +++=+++ &&&&&&&&&&&& (1.14)

Taking Laplace Transform of Eqn. (1.14) with zero initialcondition

)()()()(

)()()()(

322

13

0

322

13

sU bssU bsU sbsU sb

sY assY asY sasY s

+++

=+++

( ) ( ) )()( 32213032213 sU bsbsbsbsY asasas +++=+++

322

13

322

13

0

)()(

asasas

bsbsbsbsU sY

++++++

=

+++=

+++

+++=

33

221

3

3

2

210

33

2213

33

221

03

11s

a

s

a

s

as

b

s

b

s

bb

s

a

s

a

s

as

s

b

s

b

s

bbs

(1.15)


16/25

KS40602/kent

eo/0809(2)

From thesystem is

=sT 1)(

where P

The transwith thre

1)(sT =

Compari

01 bP = ;

sa

P 111 =

Hence, a1.7. Theloops are

Let assigsignal floderivativequationsnodes, wvariables.

Fig. 1

Mansonsgiven by

K

K K P

= path ga= 1 (su

+ (sumof two

K = forK th for

fer functiofeedback l

( 1211321

PPPP

+++

g Eqn. (1.1

s

bP 12 = ;

; 12P =

signal flosignal flowtouching l

state variw graph. H

of the sare formehose valu

.7 Signal Fl

gain form

in of K th foof loop gaof gain pr

non-touchihat part of ward path

n of a systoops (touc

)134

PP+

5) and (1.1

22

3 s

bP = ;

22

s

a; 13P =

graph canis construops.

ables at thence at thetate variab

by summis corresp

ow Graph

ula, the tr

ward pathin of all inducts of alg loops)

the graph

em with f ing each o

7),

33

4 s

bP = ;

33

s

a

be constrted such t

output of input of e

le will beng all the inds to fir

f the Syste

State Spa

State Spa

nsfer func

ividual lool possible chich is not

ur forwarher) is giv

cted as shhat all K

each integch integratavailable.

ncoming sist derivati

m with Eq

e Modeling 1

e Modeling 1

tion of the

(1.16)

ps)ombination

touching

paths andn by

(1.17)

wn in Fig.1= and all

rator in theor, the first

The statenals to thee of state

. (1.15)

6

6


17/25

KS40602/kent

eo/0809(2)

Summing

a x 11 (=& x1 =& Summing

a x 22 =& x2 =& Summing

a x 33 (=& x2 =&

The outpoutput no x y 1= Arrangin

x

x

x

=

3

2

1

&

&

&

[ y 01=

up the inc

ub x 01 ) ++ x x 211 ++

up the inc

ub x 01 ) ++ x x 312 ++

up the inc

ub x 01 ) ++b x ( 313 +

t equationde.

ub0

the state e

a

a

a

3

2

1

00

10

01

] x

x

x

3

2

1

0 +

ming sign

ub x 12 + ubab )011

ming sign

ub x 23 + bab )( 022

ming sign

u3 uba )03

is given by

quations a

b

b

b

x

x

x

+

3

2

1

3

2

1

u0

ls at node

ls at node

ls at node

the sum of

d output e

[ ]uba

ba

ba

03

02

01

State Spa

State Spa

1 x&

2 x&

3 x&

incoming s

uation in

e Modeling 1

e Modeling 1

(1.18)

(1.19)

(1.20)

ignals to

(1.21)

atrix form,

(1.22)

(1.23)

7

7


18/25

KS40602/kent

eo/0809(2)

Consider= n is giv

x

x

x

x

n

n

=

1

2

1

&

&

M

&

&

[ y 01=

ExampleObtain th

)()( =

ssU sY

using Pha Solution

Given tha

)()(

sU sY

The signconstructconsistingraph wi

22 s an

Fig. 1.

nth order dn below.

a

a

a

a

n

n

1

2

1

0

0

0

1

MM

0 LL

1.4state mod

241023 ++ s

se Variabl

1.4

t)()( =

ssU sY

+3 41

1

ss

=4

1

1

s

l flow grd as shoof three

l have thr

d 31 s .

.1 Signal F

ifferential e

0

0

1

0

LL

LL

M

LL

LL

] b

x

x

x

x

n

3

2

1

0 +

M

el of the sy

1+

s Method

241023 ++ s

+

32

120

ss

32

3

12ss

s

ph for then in Fig.integratorse individu

low Graph

quation, th

x

x

x

x

n

n

+

1

2

1

0

1

0

0

MM

u0

stem with t

(Signal Fl

1+

above tra1.3.1 withand withal loops w

of the Syst

State Spa

State Spa

e general

bab

ab

bab

bab

nn

nn

0

11

022

011

M

he transfer

w Graph).

nsfer functa single fopath gainith loop g

m with Eq

e Modeling 1

e Modeling 1

odel for m

[ ]u

0

(1.24)

(1.25)

unction

(1.4.1)

(1.4.2)

ion can berward path

310 s . Theins s4

n. (1.4.2)

8

8


19/25

KS40602/kent

eo/0809(2)

Assign ststate equthe inpucorrespon

Hence, th

114 x x =&

12 2 x x =&

x x 13 =&

Output eq

1 x y =

The state

x

x

x

=

3

2

1

&

&

&

[= 01 y

1.5.3 Pha Considerrelating t

bub

ya ym

nn

+=

+

&

&&

10

)1(

1

Let n = m a y 1+ &&&& Takingcondition

)(

)(3

0

3

sU sb

sY s +

ate variabltions are ot of theding first d

e state equ

2 x+

31 x+ u10

uation is

model in t

001

102

014

]

3

2

1

0

x

x

x

se Variabl

the folloe output y

u

yam

n

+

++

LL&

&

)1(

)2(

2

= 3,a ya 32 ++ &&

aplace Tr

)(

)(2

1

21

sU sb

sY s

+

+

s at the obtained by

integratorerivative o

tions are

e matrix fo

[ x

x

x

+

10

0

0

3

2

1

es Method

ing nth o)(t to the i

bub

a

m

n

++

+

&

&L

1

2

bub y 0 += &&&

ansform o

)(

)(

2

2

ssU b

assY

+

+

tput of thsumming t

and eqthe state v

rm is

]u

3 (Pole-Ze

der linearput )(t u o

u

ya n ++ && 1

bubu 21 ++ &&&

Eqn. (1.

)(

)(

3 sU b

sY =

State Spa

State Spa

integratoe incomin

uating theariable.

ro)

differentia system.

yn

u3

27) with

e Modeling 1

e Modeling 1

(1/s). Thesignals t

m to the

l equation

(1.26)

(1.27)

ero initial

9

9


20/25



( ) ( ) )()( 32213032213 sU bsbsbsbsY asasas +++=+++

322

13

322

13

0

)()(

asasas

bsbsbsb

sU sY

++++++

=

Let)(

)()()(

)()(

1

1

s X sY

sU s X

sU sY =

where32

21

31 1

)()(

asasassU

s X +++

= (1.28)

and 322

13

01 )(

)(bsbsbsb

s X sY +++= (1.29)

Rearranging Eqn. (1.28),[ ] )()( 322131 sU asasass X =+++

)()()()()( 131212113 sU s X assX as X sas X s =+++ (1.30)

Inverse Laplace Transform of Eqn. (1.30)u xa xa xa x =+++ 1312111 &&&&&& (1.31)

Let the state variables be 1 x , 2 x and 3 x ,

where 12 x x &= ; 123 x x x &&& == ; 123 x x x &&&&&& ==

Substitute into Eqn. (1.31)

u xa xa xa x =+++ 1322313& u xa xa xa x += 1322313&

Hence, the state equations are21 x x =& ; 32 x x =& ; u xa xa xa x += 1322313&

Rearranging Eqn. (1.29),)()()()()( 13121

211

30 s X bssX bs X sbs X sbsY +++= (1.32)

Inverse Laplace Transform of Eqn. (1.32)13121110 xb xb xb xb y +++= &&&&&& (1.33)

Substitute the state variables into Eqn. (1.33)13223130 xb xb xb xb y +++= & (1.34)

Substitute u xa xa xa x += 1322313& into Eqn. (1.34)( ) 1322311322310 xb xb xbu xa xa xab y ++++=

( ) ( ) ( ) ub xbab xbab xbab y 0301120221033 +++= (1.35)

Hence, the output equation is( ) ( ) ( ) ub xbab xbab xbab y 0301120221033 +++=


21/25



Arranging the state equations and output equation in matrix form,

[ ]u

x

x

x

aaa x

x

x

+

=

1

0

0

100

010

3

2

1

1233

2

1

&

&

&

(1.36)

[ ] ub x

x

x

babbabbab y 0

3

2

1

011022033+

= (1.37)

Consider n th order differential equation, the general model for m = n is given below.

[ ]u

x

x

x x

aaaa x

x

x x

n

n

nnn

n

+

=

1

0

00

1000

00000010

1

2

1

121

1

2

1

MM

LL

LL

MMMM

LL

LL

&

&

M

&

&

(1.38)

[ ] ub

x

x

x

x

babbabbabbab y

n

n

nnnn 0

1

2

1

0110220110 +

=

ML

(1.39)

Example 1.5Obtain the state model of the system with the transfer function

)3)(1()4(10

)()(

+++=sss

ssU sY

using Phase Variables Method 3 (Pole-Zero).

Solution 1.5

Given that)3)(1(

)4(10)()(

+++=sss

ssU sY

(1.5.1)

Let)(

)()()(

)()(

1

1

s X sY

sU

s X

sU sY

=

where)3)(1(

1)()(1

++=

ssssU s X

(1.5.2)

and )4(10)(

)(1

+= ss X

sY (1.5.3)


22/25



Rearranging Eqn. (1.5.2),)(34)( 231 sU ssss X =++

)()(3)(4)( 112

13 sU ssX s X ss X s =++ (1.5.4)

Inverse Laplace Transform of Eqn. (1.5.4)u x x x =++ 111 34 &&&&&& (1.5.5)

Let the state variables be 1 x , 2 x and 3 x ,

where 12 x x &= ; 123 x x x &&& == ; 123 x x x &&&&&& ==

Substitute into Eqn. (1.5.5)u x x x =++ 233 34&

u x x x += 233 34&

Hence, the state equations are21 x x =& ; 32 x x =& ; u x x x += 233 34&

Rearranging Eqn. (1.5.3),)(40)(10)( 11 s X ssX sY += (1.5.6)

Inverse Laplace Transform of Eqn. (1.5.6)

11 4010 x x y += & (1.5.7)

Substitute the state variables into Eqn. (1.5.7)

12 4010 x x y += (1.5.8)

Hence, the output equation is

12 4010 x x y +=

Arranging the state equations and output equation in matrix form,

[ ]u x

x

x

x

x

x

+

=

1

0

0

430

100

010

3

2

1

3

2

1

&

&

&

[ ]

=

3

2

1

01040

x

x

x

y


23/25

KS40602/kent

1.6

eo/0809(2)

State Sp ace Rep 1.6.1 Ca In canonidiagonalthe transf By Partiathe n th or

bsU sY =

)()(

where C 1denomin

The Eqn.

= bsU sY

)()(

b=

Hence,

)( 0bsY =

L

The bloc

Fi

esentati

onical Fo

cal form omatrix. Thr function

l fraction eer system

s

C

++

+1

1

C C LL,, 2tor polyno

(1.40) can

+

+

ss

C

1 11

s

sC

++

1 11

0

(111

1)(s

++

++

L

diagram o

g. 1.7: Blo

on usin

m

state modelements

of the syste

pansion, tan be expr

s

C

++

L2

2

n are residuial (or pol

be rearrang

+

+

s

C

1

2

s

sC

++1 2

2

( )

)

11

11

C ss

C ss

nn

f Eqn. (1.4

k Diagram

Canoni

l, the syston the diam.

e transferssed as sh

n

n

s

C +

+L

es and ,1es of the sy

d as belo

+ 2

LL

s

sC

n

n

+

+1

L

)(

1)(1

sU

sU

+

) is shown

of Canoni

al VariaState Spa

State Spa

m matrixonal are t

function Y wn in Eqn

n LL,2

stem).

.

+

ss

C

n

n

1

( )11

2ss

+

in Fig. 1.7.

al State M

les e Modeling 2

e Modeling 2

will bee poles o

)() sU s o

. (1.40).

(1.40)

are roots o

)(2 sU C +

(1.41)

del

3

3


24/25



Assign state variables at the output of integrator. The input of theintegrator will be the first derivative of state variable.

The state equations are formed by adding the incoming signals tothe integrator and equating to first derivative of state variable.

The state equations are,

u x x

u x x

u x x

nnn+=

+=+=

&

M

&

&

222

111

The output equation is,

ub xC xC xC y nn 02211 ++++= LL

The canonical form of the state model in the matrix form is givenbelow.

[ ]u

x

x

x

x

x

x

x

x

n

n

n

n

n

n

+

=

1

1

1

1

000

000

000

000

1

2

1

1

2

1

1

2

1

MM

LL

LL

MMMM

LL

LL

&

&

M

&

&

(1.42)

[ ] ub

x

x

x

x

C C C C y

n

n

nn 0

1

2

1

121+

=

MLL (1.43)

The advantage of canonical form is that the state equations areindependent of each other. The disadvantage is that the canonicalvariables are not physical variables and so they are not availablefor measurement and control.

Advantage:

1. State equations independentto each other.

Disadvantage:1. Not measurable


25/25

n

ExampleObtain th

)()( =

ssU sY

using Ca Solution

Given tha

By partial

4)()( =

sU sY

)(sY =

Fig. 1. Assign stFig. 1.6.1state variby addincorrespon The state

The outp

The state

x

x

x

=

3

2

1

&

&

&

1.6state mod

3)(1()4(10

+++ss

s

onical Var

1.6

t)()( =

ssU sY

fraction e

11530+

ss

3401 U

s

.1: Block

ate variabl. At the inbles willincoming

ding first d

equations

t equation

model in t

30

01

00

el of the sy

) iables.

3)(1()4(10

+++ss

s

pansion Y

335

+s

11

1

)

s

ss

+

iagram of

s at the ouut of the ine availablesignals toerivative o

re u x =1 ;&

is3

40 x y =

e matrix fo

[ x

x

x

+

1

1

1

3

2

1

stem with t

)

)() sU s ca

(151

U

Eqn. (1.6.2

put of thetegrator, th. The statehe integratthe state v

x x += 22&

21 15 x +

rm is show

];

= y

State Spa

he transfer

n be expres

11

1

)

s

s

++

) in Canoni

integratorfirst deriv

equations aor and equariable.

xu =3 ; &

3 x

n as below.

1530

e Modeling 2

unction

(1.6.1)

sed as

)(35

3sU

(1.6.2)

cal Form

s shown inative of there obtainedating to the

u x +3

3

2

1

35

x

x

x

(1.6.3)

5

KS40602 State Space Modeling

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Transcript of KS40602 State Space Modeling