Chapter 3 State Variables

8/3/2019 Chapter 3 State Variables

1/55

Chapter 3: Modeling in the Time Domain1

Chapter 3

Modeling in the Time Domain


2/55


Cramers rule (1704-1752 Gabriel Cramer)

Ax+ By=f

Cx+Dy=g how do you solve it?

Lets eliminate y ADx+BDy=fD

- BCx+BDy=gB

(AD-BC)x=fD-gB x= (fD-gB)/(AD-BC)

Determinant=AD-BC


3/55


Cramers rule

BCAD

fCgAy

BCAD

gBfDx

BCAD

gC

fA

yBCAD

Dg

Bf

x

BCADDet

g

f

y

x

DC

BA

!

!

-

!

-

!

!

!

-

Ax+By=f

Cx+Dy=g


4/55


Examples; Cramers rule

1

5

491

5

38

1*12*3

31

43

1*12*3

23

14

3

4

21

13

32

43

!

!!

!

-

!

-

!

!

!

-

!

!

yx

yx

BCADDet

y

x

yx

yx


5/55


Review of Matrices

? A ? A

-

!

-

!

-

-

!

-

!

-

-

!

-

-

963

852

741

987

654

321

,2

121

42

31

43

21

1210

86

87

65

43

21

T

T

T

T

yxy

x


6/55


Review of Matrices

_ a

_ a j

j

T

Xyx

yxyx

Xyxy

x

AA

BABA

cwcz

cycx

wz

yxc

1i1ij

11j1

AC222

1

AC221

matrixsymmetric

11

113

33

33

43

21

76

54

43

21

76

54

)(

,86

42

43

212

!

-

!

!!

!

-

!

-

!

-

-

!

-

-

!

-

!

-

-

!

-


7/55


Review of Matrices

!

-

{

-

!

-

!

-

-

-

!

-

!

-

-

zyx

zyx

zyx

z

y

x

ABBA

987

654

32

987

654

321

**

4631

3423

4*82*73*81*7

4*62*53*61*5

43

21*

87

65

5043

2219

8*46*37*45*3

8*26*17*25*1

87

65*

43

21


8/55


Introduction

Frequency-domain technique rapidly providing stability and transient

response information Immediate can see the effect of varyingsystem parameters

State-space approach (modern, ortime-domain approach)can be used fornon-linear system with backlash,saturation and dead zones


9/55


State-space Representation

Select a particular subset of allpossible system variables; state

variables For an nth order system, write n

simultaneous, first-order differentialequations in terms of the statevariables; simultaneous differentialequationsstate equations


10/55


State-space Representation

Solve the simultaneous equations withthe known initial conditions of all thestate variables at t

0

as well as thesystem input

Algebraically combine the statevariables with the systems input and

find all other system variablesoutputequations State equation +output equation

state space representation


11/55


State-space representation

x=Ax+Bu

y=Cx+Du

x=state vector Y=output vectorx=time derivative of the state vector

u=input vector

A=system matrixB=input matrix

C=output matrix

D=feedforward matrix


12/55


Example

fz

x

z

x

fxzfxxzx

xz

xz

fxxx

-

-

-

!

-

!!!

!

!

!

1

0

23

10

3232

32


13/55


Example II

f

y

zx

y

zx

fxzy

fxxxxzyxzy

xz

fxxxx

-

-

-

!

-

!

!!!

!!

!

!

1

00

234

100010

432

432

432


14/55


Next Move?

-

!

-

-

!

-

!!

!

!

!!

!

v

xy

v

x

v

x

xxxv

xv

Solution

xx

xxx

]01[

23

10

32

5at t

0)0(,1)0(

032


15/55


-

!

-

-!

-

-

!

-

-

!

-

-

!

-

!(

((!(

3.011.0

30

01

)1.0()1.0(

3

0

0

1

23

10

)0(

)0(

23

10

)0(

)0(1.0

2

)()()()( 2

vx

v

x

v

xtLet

tty

ttytytty


16/55


-

!

-

-

!

-

-

!

-

-

!

-

-

!

-

!

24.0

97.01.0

4.2

3.0

3.0

1

)2.0(

)2.0(

4.2

3.0

3.0

1

23

10

)1.0(

)1.0(

23

10

)1.0(

)1.0(

2.0

v

x

v

x

v

x

t


17/55


iii

iii

iii

ii

vtvv

xtxx

vxv

vx

Scheme

*

*

23

1

1

(!

(!

!

!

Simple Euler Scheme


18/55


Exact Solution of the Example

)2sin2

12(cos)(

2)1(

1)1()(

)()32(0)0(,1)0(

032

2

2

ttetx

s

ssX

ssXssxx

xxx

t !

!

!!!

!


19/55


Simple Euler Scheme

Time x(I) v(I) x dot v dot x(I+1) v(I+1) Exact

0 1 0 0 -3 1 -0.03 1

0.01 1 -0.03 -0.03 -2.94 0.9997 -0.0594 0.999851

0.02 0.9997 -0.0594 -0.0594 -2.8803 0.999106 -0.0882 0.999408

0.03 0.999106 -0.0882 -0.0882 -2.82091 0.998224 -0.11641 0.998677

0.04 0.998224 -0.11641 -0.11641 -2.76185 0.99706 -0.14403 0.9976640.05 0.99706 -0.14403 -0.14403 -2.70312 0.99562 -0.17106 0.996374

0.06 0.99562 -0.17106 -0.17106 -2.64474 0.993909 -0.19751 0.994814

0.07 0.993909 -0.19751 -0.19751 -2.58671 0.991934 -0.22338 0.99299

0.08 0.991934 -0.22338 -0.22338 -2.52905 0.9897 -0.24867 0.990907

0.09 0.9897 -0.24867 -0.24867 -2.47177 0.987213 -0.27338 0.98857

0.1 0.987213 -0.27338 -0.27338 -2.41487 0.98448 -0.29753 0.985987

0.11 0.98448 -0.29753 -0.29753 -2.35837 0.981504 -0.32112 0.983161

0.12 0.981504 -0.32112 -0.32112 -2.30228 0.978293 -0.34414 0.98010.13 0.978293 -0.34414 -0.34414 -2.2466 0.974852 -0.36661 0.976808

0.14 0.974852 -0.36661 -0.36661 -2.19134 0.971186 -0.38852 0.973291

0.15 0.971186 -0.38852 -0.38852 -2.13652 0.9673 -0.40988 0.969555

0.16 0.9673 -0.40988 -0.40988 -2.08213 0.963202 -0.43071 0.965604

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6

Euler Scheme

exact


20/55


)22(6

1

),(*

)

2

,

2

(*

)2

,2

(*

),(*

),(

1

1

1

nnnnii

niin

n

iin

n

iin

iin

iiii

DCBAyy

CytfhD

By

htfhC

Ay

htfhB

ytfhA

ythfyy

!

!

!

!

!

!

Runge- Kutta Scheme; General


21/55

Chapter 3: Modeling in the Time Domain21Figure 3.1RLnetwork

Initial condition i(0)

Ldi/dt+Ri=v(t) state equation

i(t) is state variable; but you

can pick up something else;e.g. vR


22/55


RLnetwork

Laplace transform

L[sI(s)-i(o)]+RI(s)=V(s)

For unit step v(t)=1, V(s)=1/s I(s)=1/{s(Ls+R)}+ i(o)L/(Ls+R)

1/{s(Ls+R)}=(1/L)[A/s+B/(s+R/L)]

A=L/R, B=-L/R I(s)=(1/R)[1/s-1/(s+R/L)]+ i(o)/(s+R/L)

i(t)=(1/R)[1-e-(R/L)t]+ i(o) e-(R/L)t


23/55


RLnetwork

i(t) state variable Ldi/dt+Ri=v(t) state equation If initial condition i(0), input v(t) are

known, we know i(t) Then we know other network variables;

output equations

VR(

t)=R

i(t

) VL(t)=V(t)-Ri(t) As VL(t)=Ldi/dt di/dt= VL(t)/L= [V(t)-Ri(t)]/L


24/55


Figure 3.2RLC network

Second order system; twostate variables needed; e.g.i(t), q(t) charge on thecapacitor


25/55


RLC network

Ldi/dt+Ri+(1/c)+idt=v(t) i(t)=dq/dt, i=q

Lq+Rq+(1/c)q=v(t) To make two first-order equations dq/dt=i

di/dt=-(1/LC)q-R

/Li+v(t)/L State equations If we know initial conditions, and

input v(t), we can solve the problem


26/55


RLC network

Output equation

vL(t)=-(1/C)q(t)-Ri(t)+v(t) [note that

vL+vR+vC=v(t)] Both system equations and output

equationstate-space representation

You can pick up other state variables,e.g. vR, vC i(t) must be continuous, i=vR/R


27/55


RLC network

vL=Ldi/dt=(L/R) dvR/dt

As vL+vR+vC=v(t)

(L/R) dvR/dt+vR+vC= v(t) dvR/dt=-(R/L) vR-(R/L) vC+(R/L)v(t)

vC=(1/c)(vR/R)dt

Differentiate dvC/dt=(1/RC)vR New state equations


28/55



State variables are linearlyindependent; no state variable can be

written as a linear combination of theother state variables

Summary of the RLC state-spacerepresentation


29/55


1D,R-C

1-C),(

v(t)u,10

B,

110

A,/

/

)(1LC

1

!

-

!!

!

!

-

!

-

!

-

!

-

!

!!

!

tvy

DuCxy

equationsoutput

Li

qx

L

R

LC

dtdi

dtdqx

tvL

iLRq

dtdii

dtdq

BuAxx

L


30/55


Figure 3.3Graphicrepresentationof state space

and a statevector

For RLC network;

vR, vC wereselected as statevariables Function of time t


31/55



x=Ax+Bu

y=Cx+Du

x=state vector Y=output vectorx=time derivative of the state vector

u=input vector

A=system matrix

B=input matrix

C=output matrix

D=feedforward matrix


32/55


Applying the State-space representation

A minimum number of state variablesmust be selected as components ofthe state vector (sufficient todescribe completely the state of thesystem

State variables must be linearly

independent Usually number of energy-storageelements becomes number of statevariables


33/55


Figure 3.4Block diagram of amass and damper

Mdv/dt+Dv=f(t);(Ms+D)V(s)=F(s)

One energy-storageelement (mass)

First order equation, so one state-variable may be enough; but massmust have relative position use two

state-variables, v(t), x(t)


34/55


Example 3.1 Electrical network for representation in

state space

Output iR(t)


35/55


Solution to the Example 3.1

Write equation for all energy-storageelements; inductor & capacitor

CdvC/dt=iC LdiL/dt=vL Select vC& iLas state variables

Next, change iC& vLin terms of statevariables and input v(t)

iC= iL- iR= iL- vC/R

vL= v(t)-vC


36/55


-

-

!

-

-

-

!

-

!

!

!

!

!

L

C

L

C

L

C

C

CL

LCC

CL

LCC

i

v

R

tv

Li

v

L

CRCi

v

vR

output

tvLvLdt

di

iC

vRCdt

dv

tvvdtdiL

ivRdt

dvC

01

i

)(1

0

01

11

1i

)(11

11

)(

1

R

R


37/55


Figure 3.6

Electrical network forExample 3.2


38/55


Figure 3.7

Translationalmechanical system


39/55


Figure 3.8

Electric circuitfor Skill-AssessmentExercise 3.1


40/55


Figure 3.9Translationalmechanical systemfor Skill-AssessmentExercise 3.2


41/55


Figure 3.10a. Transfer function; b. equivalent block diagramshowing phase-variables. Note: y(t) = c(t)


42/55


24926-24-

2424269

)(24)(24269

24269

24

)(

)(

1

3213

32

21

3

2

1

23

23

xcy

rxxxx

xx

xx

cx

cxcx

rcccc

sRsCsss

ssssR

sC

!!

!

!

!

!

!

!

!

!

!

State variables

Output equation

System

equations


43/55


? A

001

1

0

0

92624

100

010

24926-24-

3

2

1

3

2

1

2

2

1

1

3212

32

21

-

!

-

-

-

!

-

!!

!

!

!

x

xx

y

r

x

x

x

x

x

x

xcy

rxxxx

xx

xx


44/55


Can you explain?


45/55


Figure 3.11

Decomposing atransfer function

Output equation

y(t)=b0x1+b1x2+b2x3


46/55


Figure 3.12a. Transfer function;b. decomposedtransfer function;c. equivalent block

diagram. Note:y(t) = c(t)


47/55


? A

172

27

)()27()(

1

00

92624

100010

3

2

1

123

12

3

2

1

2

2

1

-

!

!!

-

-

-

!

-

x

x

x

y

xxxy

sXsssC

r

x

xx

x

xx


48/55


Figure 3.13Walking robots,such as Hannibalshown here, canbe used to explore

hostileenvironments andrough terrain,such as that foundon other planetsor inside

volcanoes.

Bruce Frisch/S.S./Photo Researchers


49/55


Converting to a Transfer Function

)()(

)()(

)(])([

)()()()(

)()()(

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

)()()(

1

1

1

1

DBAsICsU

sYsT

sUDBAsIC

sDUsBUAsICsY

sBUAsIsX

sBUsXAsIsDUsCXsY

sBUsAXssX

DuCxy

BuAxx

!!

!

!

!

!!

!

!!


50/55


Example 3.6

? A

12s3ss

)1(2

)3(1

1323

)(

321

10

01

321

100

010

00

00

00

)(

001

0

0

10

321

100

010

23

2

2

1

3

2

1

3

2

1

2

2

1

-

!

-

!

-

-

!

-

!

-

-

-

!

-

sss

sss

sss

AsI

s

s

s

s

s

s

AsI

x

x

x

y

u

x

x

x

x

x

x


51/55


? A

? A

? A

12s3ss

)23(10

10

10

)23(10

00112s3ss

1

0

0

10

12s3ss

)1(2

)3(1

1323

001

)()(

0

001

0

0

10

23

2

2

23

23

2

2

1

!

-

!

-

-

!

!

!

!

-

!

ss

s

ss

sss

sss

sssDBAsICsT

D

C

B

Example 3.6


52/55


Figure 3.14

a.Simple pendulum;b. force components

of Mg;c. free-body diagram


53/55


Figure 3.15

Nonlinear translationalmechanical systemfor Skill-AssessmentExercise 3.5


54/55


Figure 3.16

Pharmaceutical drug-levelconcentrations in a human


55/55


Figure 3.17

Aquifer systemmodel

Chapter 3 State Variables

Documents