1

Chapter 3State Variable Models

The State Variables of a Dynamic SystemThe State Differential Equation

Signal-Flow Graph State VariablesThe Transfer Function from the State Equation

2

Introduction• In the previous chapter, we used Laplace transform to obtain the

transfer function models representing linear, time-invariant, physical systems utilizing block diagrams to interconnect systems.

• In Chapter 3, we turn to an alternative method of system modeling using time-domain methods.

• In Chapter 3, we will consider physical systems described by an nth-order ordinary differential equations.

• Utilizing a set of variables known as state variables, we can obtain a set of first-order differential equations.

• The time-domain state variable model lends itself easily to computer solution and analysis.

3

Time-Varying Control System• With the ready availability of digital computers, it is convenient to

consider the time-domain formulation of the equations representing control systems.

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

• The time-domain techniques can be utilized for nonlinear, time-varying, and multivariable systems (a system with several input and output signals).

• A time-varying control system is a system for which one or more of the parameters of the system may vary as a function of time.

• For example, the mass of a missile varies as a function of time as the fuel is expended during flight

4

Terms• State: The state of a dynamic system is the smallest set of variables

(called state variables) so that the knowledge of these variables at t= t0, together with the knowledge of the input for t ≥ t0, determines the behavior of the system for any time t ≥ t0.

• State Variables: The state variables of a dynamic system are the variables making up the smallest set of variables that determine the state of the dynamic system.

• State Vector: If n state variables are needed to describe the behavior of a given system, then the n state variables can be considered the n components of a vector x. Such vector is called a state vector.

• State Space: The n-dimensional space whose coordinates axes consist of the x1 axis, x2 axis, .., xn axis, where x1, x2, .., xn are state variables, is called a state space.

• State-Space Equations: In state-space analysis, we are concerned with three types of variables that are involved in the modeling of dynamic system: input variables, output variables, and state variables.

5

The State Variables of a Dynamic System• The state of a system is a set of variables such that the knowledge

of these variables and the input functions will, with the equations describing the dynamics, provide the future state and output of the system.

• For a dynamic system, the state of a system is described in terms of a set of state variables.

System

u1(t)

u2(t)

y1(t)

y2(t)

Input Signals Output Signals

6

State Variables of a Dynamic System

Dynamic SystemState x(t)

u(t) Input y(t) Output

x(0) initial condition

dynamics thedescribing equations theand inputs, excitation thestate,present given the

system, a of response future thedescribe variablesstate The

7

The State Differential EquationThe state of a system is described by the set of first-order differential

equations written in terms of the state variables (x1, x2, .., xn)

equation) aldifferenti (StateBu Axx.

+=

signals)output -equation(Output Du Cxy +=matrix nsmission direct tra :D matrix;Output :Cmatrixinput :B matrix; State :A

+

=

++++++=

++++++=

++++++=

mnmn

m

nnnnn

n

n

n

mnmnnnnnnn

mmnn

mmnn

u

u

bb

bb

x

xx

aaa

aaaaaa

x

xx

dtd

ububxaxaxax

ububxaxaxax

ububxaxaxax

.....

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

.

. . .

.

......

......

......

1

1

1112

1

21

22221

11211

2

1

112211

.212122221212

.111112121111

.

A x B u

dtdxx =&

8

Block Diagram of the Linear, Continuous Time Control System

)(x.

tB(t) ∫dt C(t)

D(t)

u(t) y(t)

A(t)

)(u )D()( x)C()()(u )B()x(t)A()(x

.

tttttytttt

+=+=

++

+

+x(t)

9

Mass Grounded, M (kg)Mechanical system described by the first-order differential equation

Fa(t)

x (t)

∫=

==

t

ta

a

dttFM

tv

dttxdM

dtdvMtF

txtvtT

0

)(1)(

)()(

(m) )(position Linear (m/sec) )(ocity Linear vel

m)-(N )( torqueAppied

2

2

a

v (t)

M

10

Mechanical Example: Mass-Spring DamperA set of state variables sufficient to describe this system includes the position and the velocity of the mass, therefore, we will define a set of

state variables as (x1, x2)

uM

xMkx

mb

dtdx

xdtdx

tukxbxdt

dxM

tukydtdyb

dtydM

dttdytx

tytx

1

;

)(

)(

)()(

)()(

122

21

122

2

2

2

1

+−−=

=

=++

=++

=

=

M

KbWall friction

y(t)

u(t)

constant Spring :k

11

Example 1: Consider the previous mechanical system. Assume that the system is linear. The external force u(t) is the input to the system, and the displacement y(t) of the mass is the output. The displacement y(t) is measured from the equilibrium position in the absence of the external force. This system is a single-input-single-output system.

[ ]

[ ] 0 ,0 1C ,10

B ,- -

1 0A

uCxy ;BAxx

:form standard in the areequation output theandequation state The

Equation)(Output 0 1

Equation) (State 10

21

mb-

mk-

1 0

have weform,matrix vector aIn :isequation output The

1 then );()();()(

)( and )( : variables twodefine usLet s.integrator twoinvolvesit meansIt system.order second a is This

.

2

1

2.

1.

1

212.

21..

21

21

...

==

=

=

+=+=

=

+

=

=

+−−=

===

=++

Dmm

bmk

Du

xx

y

um

xx

x

x

xy

um

xmbx

mkx

xxtytxtytx

txtx

ukyybym

12

Electrical and Mechanical Counterparts

ResistorRi2

Damper / Friction0.5 Bv2

Dissipative

Capacitor0.5 Cv2

Gravity: mghSpring: 0.5 kx2

Potential

Inductor0.5 Li2

Mass / Inertia0.5 mv2 / 0.5 jω2

Kinetic

ElectricalMechanicalEnergy

13

Resistance, R (ohm)

v(t) R

i(t)

)(1)(

)()()(Current

)( voltageAppied

tvR

ti

tRitvti

tv

=

=

14

Inductance, L (H)

v(t) L

i(t)

∫=

=

t

tdttv

Lti

dttdiLtv

titv

0

)(1)(

)()(

)(Current )( voltageAppied

15

Capacitance, C (F)

v(t) C

i(t)

dttdvCti

dttiC

tv

titv

t

t

)()(

)(1)(

)(Current )( voltageAppied

0

=

= ∫

16

Electrical Example: An RLC Circuit

C

L

Ru(t)

iL

iC

vC

( ) ( )

21

212

21

0201

2221

)()( : thenis signaloutput The

1

)(11 )( :by drepresente is system theofoutput The

)(

junction at the KCL USEnetwork theofenergy

initial total theis )( and )(2/12/1

)( );(

Rxtvty

xLRx

Ldtdx

tuC

xCdt

dxtRiv

vRidtdiL

itudt

dvCi

txtxCvLi

tixtvx

o

Lo

cLL

Lc

c

cL

LC

==

−+

+−=

=

+−=

−+==

+=

==

ξ

+Vo-

17

Example 2: Use Equations from the RLC circuit

[ ]

[ ]xy

ux

CLRxRy

tuCx

LR -

L

C -

x

3 0

02

3- 12- 0

x

have we1/2, 1, 3,When 0isoutput The

)( 0

1

1

10

.

.

=

+

=

====

+

=

18

Signal-Flow Graph ModelA signal-flow graph is a diagram consisting of nodes that are connected by several directed branches and is a graphical

representation of a set of linear relations. Signal-flow graphs are important for feedback systems because feedback theory is concerned

with the flow and processing of signals in system.

Vf(s) θ (s)G(s)

R1(s)

R2(s)

Y1(s)

Y2(s)

G11(s)

G22(s)

G12(s)

G21(s)

2.11-2.8 :Examples Read

19

Mason’s Gain Formula for Signal Flow GraphsIn many applications, we wish to determine the relationship between an input and output variable of the signal flow diagram. The transmittance

between an input node and output node is the overall gain between these two nodes.

.path h that toucloops theremovingby from obtained is cofactor theis, that removed,path forwardkth thetouching

loops thegraph with theoft determinanpath forwardkth theofcofactor

--1

..loops) gnontouchin theseof nscombinatio possible all of productsgain of (sum-loops) gnontouchin twoof nscombinatio possible all ofgain of (sum

gain) loop individual all of (sum-1graph oft determinan

path forward k ofgain path

1

th

k

k

k

a b,c d,e,ffcdcba

k

kkk

P

LLLLLL

P

PP

∆∆=∆

+=

+

+==∆=

∆∆

=

∑ ∑ ∑

∑

20

Signal-Flow Graph State Models

U(s)

1/C 1/s 1/L

1/s

R

x1x2 vo

-R/L

-1/C

( ) ( ) ( ) ( )LCsLRsLCR

LCsLsRLCsR

sUsVo

RxvxLRx

Lx

tuC

xC

x

sssUsV

sG

o

o

/1//;

/1/1/

)()(

;1

)(11)()(

)(

22

2

2212.

21.

2

++=

++=

=−=

+−=

++==

γβα

21

factor loopfeedback theof sum-1factorspath -forward theof Some

1)(

)()()(

....1....

)()()(

........

)()()(

1

)1(1

11

)1(1

)1(1

)(1

11

11

1

=−

=

∆∆

==

++++

++++==

++++

++++==

∑∑

∑

=

−−−−−

−−−+−−−

−−

−−

−−

Nq

k k

k k

no

nn

no

nmnm

mno

nn

no

mm

m

LqP

sG

kPsUsYsG

sasasasbsbsbs

sUsYsG

asasasbsbsbs

sUsYsG

22

Phase Variable Format: Let us initially consider the fourth-order transfer function. Four state variables (x1, x2, x3, x4); Number of

integrators equal the order of the system.

U(s)

1 1/s 1/s 1/s 1/s bo

Y(s)x1x2x3x4

U(s)1/s 1/s 1/s 1/s

x4x3 x2 x1

40

31

22

13

40

012

23

34

0

1

)()()(

−−−−

−

++++=

++++==

sasasasasb

asasasasb

sUsYsG

-a0-a1-a2-a3

Y(s)

23

40

31

22

13

40

31

22

13

012

23

34

012

23

3

1

)(

−−−−

−−−−

++++

+++=

++++

+++=

sasasasasbsbsbsb

asasasasbsbsbsb

sG

textbook theof 3.1 Examplev Read

U(s)

1 1/s 1/s 1/s 1/s bo

Y(s)x1x2x3x4

b1b2

b3

-a0-a1-a2

-a3

[ ]

==

+

=

+++=+−−−−=

===

4

3

2

1

3210

4

3

2

1

32104

3

2

1

43322110

433221104.

43.

32.

21.

Cx)(

)(

1 000

- - - -1 0 0 00 1 0 00 0 1 0

)(

;;

xxxx

bbbbty

tu

xxxx

aaaaxxxx

dtd

xbxbxbxbtyuxaxaxaxax

xxxxxx

24

Alternative Signal-Flow Graph State Models

)1()1(5)(

++

=sssGc )2(

1+s )3(

6+s

)2(1+s

R(s) Y(s)

I(s)U(s)

)5()1(5)(

++

=sssGc ( )3

6+s

Controller

Motor and Load

R (s)

1 1/s 5

U (s)

1 1/s 6 1/s 1

Y (s)

I (s)

5

-5 -2 -3

[ ]x0 0 1 )( 150

x5- 0 0

20- 2- 00 6 3-

x.

=

+

= ytr

25

The State Variable Differential Equations

[ ]x30 10- 20-)( );( 111

x 3- 0 0

0 2- 00 0 5-

x

30 and -10, ,20)3()2()5(

)()(

))()(()(

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

)(

.

321

321

321

=

+

=

==−=+

++

++

==

−−−=

++++

==

tytr

kkksk

sk

sksT

sRY(s)

sssssssq

sssssT

sRY(s)

1/s

1/s

1/s

1

1

1

-20

-20

30

-3

-5

form Canonicalor form Diagonal

26

The State Variable Differential Equations

[ ]x30 10- 20-)(

)( 111

x 3- 0 0

0 2- 00 0 5-

x

30 and -10, ,20)3()2()5(

)()(

))()(()(

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

)(

)( 150

x 5- 0 05- 2- 00 6 3-

x

.

321

321

321

.

=

+

=

==−=+

++

++

==

−−−=

++++

==

+

=

ty

tr

kkksk

sk

sksT

sRY(s)

sssssssq

sssssT

sRY(s)

tr

27

The Transfer Function from the State EquationGiven the transfer function G(s), we may obtain the state variable equations

using the signal-flow graph model. Recall the two basic equations

( )[ ]

B )( C)()()(

)( B )( C)( )( B )( )( X

)( A-I Since

)(B)(X AI)(CX)(

)( B)( XA )( XCx

BAxx

1

.

ssUsYsG

sUssYsUss

ss

sUssssY

sUsssy

u

Φ==

Φ=Φ=

Φ=

=−=

+==

+=

−

input. single theis andoutput single theis

uy

transformLaplace theTake

28

Exercises: E3.2 (DGD)A robot-arm drive system for one joint can be represented by the differential equation,

where v(t) = velocity, y(t) = position, and i(t) is the control-motor current. Put the equations in state variable form and set up the matrix form for k1=k2=1

)()()()(321 tiktyktvk

dttdv

+−−=

=

=

=+=

===

+

−

=

+−−=

=

vy

k

kkiu

ikv

ykv

ydtd

tiktyktvkdtdv

dtdyv

x,0

B ,1- 1-1 0

A Bu;Axx

1let and , Define

0k- 1 0

)()()(

3

.

21

312

321

29

E3.3: A system can be represented by the state vector differential equation of equation (3.16) of the textbook. Find the characteristic roots of the system (DGD).

−

=1- 11 0

A

( )

23

21;

23

21

0111

1)( 11-

Det A)-I(Det

21

2

jj −−=+−=

=++=++=

+

=

λλ

λλλλ

λλ

λ

BuAxx.

+=

30

E3.7: Consider the spring and mass shown in Figure 3.3 where M = 1 kg, k = 100 N/m, and b = 20 N/m/sec. (a) Find the state vector differential equation. (b) Find the roots of the characteristic equation for this system (DGD).

( ) -10 ;010

1002020 1001-

Det A)-I(Det

10

20- 100-1 0 20100

212

2

.

212.

21.

===+=

++=

+

=

+

=

+−−=

=

λλλ

λλλ

λλ

uxx

uxxx

xx

31

E3.8: The manual, low-altitude hovering task above a moving land deck of a small ship is very demanding, in particular, in adverse weather and sea conditions. The hovering condition is represented by the A matrix (DGD)

=

2- 5- 01 0 00 1 0

A

( )2--1 2;-1 ;0

0)52

2 5 01- 0

0 1- DetA)-IDet(

321

2

jj =+===++

=

==

λλλλλλ

λλ

λλ

32

E3.9: See the textbook (DGD)

[ ]

( )

[ ]z1.79-0.35-y ;1/2- 0

0 2z

02121

25s

x3/2-1y x, 3/2- 1

1/2 1-x

21

.

2

212.

121.

=

−=

=

++=++

=

=

−−=

−=

z

sss

xxx

xxx

33

P3.1 (DGD-ELG4152):

uL

CLL

xC

x

xL

xLRv

Lx

vxix

idtC

v

vdtdiLtRitv

c

c

01/

x 0 1/

1/- R/-x (c)

1

11:are equations state The (b)

and as variablesstate Select the (a)

1

)()(

KVLApply

.

12.

211.

c21

+

=

=

−−=

==

=

+++=

∫