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state variable theory
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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
+
=
=
−−=
==
=
+++=
∫