Fundamentals of Control Systems - Chapter 2: Mathematical Models of Continuous Control Systems

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Nonlinear systems
 Nonlinear systems do not satisfy the superposition
principle and cannot be described by a linear differential
equation.
 Most of the practical systems are nonlinear:
Fluid system (Ex: liquid tank,)
Thermal system (Ex: furnace ),
Mechanical system (Ex: robot arm,.)
Electro magnetic system (TD: motor )- ,
Hybrid system ,
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Mathematical model of nonlinear systems
 Input output relationship of a continuous nonlinear system–
can be expressed in the form of a nonlinear differential
equations.



 

)(,)(,,)(),(,)(,,)()( 1
1
tu
dt
tdu
dt
tudty
dt
tdy
dt
tydg
dt
tyd
m
m
n
n
n
n

where: u(t): input signal,
y(t): output signal,
g(.): nonlinear function
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Nonlinear system – Example 1
a: cross area of the dischage valve
A: cross area of the tank
g: gravity acceleration
k t t(t)
u(t)
qin
: cons an
CD: discharge constant
y qout
 Balance equation: )()()( tqtqtyA outin 
)()( tkutq where: in
)(2)( tgyaCtq Dout 
 (first order li t ) )(2)(1)( tgyaCtkuty D
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non near sys em A
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Nonlinear system – Example 2
J: moment inertia of the robot arm 
M: mass of the robot arm
m: object mass
l: length of robot armm
l
lC : distance from center of gravity to rotary axis 
B: friction constant
g: gravitational acceleration
u 
u(t): input torque
(t): robot arm angle
 According to Newton’s Law
)(cos)()()()( 2 tugMlmltBtmlJ C   
 )(
)(
1cos
)(
)()(
)(
)( 222 tumlJ
g
mlJ
Mlmlt
mlJ
Bt C 
  
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(second order nonlinear system)
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Nonlinear system – Example 3
: steering angle
: ship angle
k: constant
i: constant(t)
(t)
Moving direction 
 The differential equation describing the steering dynamic of a
ship:
   )()()()(1)(11)( 3 ttktttt    
(third order nonlinear system)
3
212121
 
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Describing nonlinear systems by state equations
 A continuous nonlinear system can be described by the state
equation:
  ))(),(()( tutt xfx  ))(),(()( tuthty x
where: u(t): input,
y(t): output,
x(t): state vector,
x(t) = [x1(t), x2(t),,xn(t)]T
f(.), h(.): nonlinear functions
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State-space model of nonlinear system – Example 1
 Differential equation:
(t)
u(t)
qin  )(2)(1)( tgyaCtku
A
ty D
 Define the state variable:
)()(
y qout
1 tytx 
St t ti   ))(),(()( tutt xfx a e equa on:   ))(),(()( tuthty x
)(
)(2
),( 1 tu
A
k
A
tgxaC
u D xfwhere
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)())(),(( 1 txtuth x
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State-space model of nonlinear system – Example 2
 Differential equation:
m
u
l )(
)(
1cos
)(
)()(
)(
)( 222 tumlJ
g
mlJ
Mlmlt
mlJ
Bt C 
  
 Define the state variable:




)()(
)()(
2
1
ttx
ttx




 State equation:

 
))()(()(
))(),(()(
h
tutt xfx
 , tutty x

)(2 tx
where
 

 )(
)(
1)(
)(
)(cos
)(
)(),(
22212 tumlJ
tx
mlJ
Btx
mlJ
gMlmlu Cxf
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)())(),(( 1 txtuth x
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Equilibrium points of a nonlinear system
 Consider a nonlinear system described by the diff. equation:




))(),(()(
))(),(()(
tuthty
tutt
x
xfx
 The state is called the equilibrium point of the nonlinear 
system if the system is at the state and the control signal is
x
x
 If is equilibrium point of the nonlinear system then:)( ux
fixed at then the system will stay at state forever.u x
,
0))(),(( ,  uutut xxxf
 The equilibrium point is also called the stationary point of the
nonlinear system.
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Equilibrium point of nonlinear system – Example 1
 Consider a nonlinear system described by the state equation:







)(2)(
)().(
)(
)(
21
21
2
1
txtx
utxtx
tx
tx


Find the equilibrium point when 1)(  utu
 Solution:
0))(),(( ,  uutut xxxf
The equilibrium point(s) are the solution to the equation:




02
01.
21
21
xx
xx


 
2
21x


 
2
21x or
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  22x   22x
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Equilibrium point of nonlinear system – Example 2
C id li t d ib d b th t t ti ons er a non near sys em escr e y e s a e equa on:
  uxxx
2
3
2
21 1
 


 ux
xxx
x
x
2
3
313
3
2 )sin(

Find the equilibrium point when 0)(  utu
1xy 
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Linearized model of a nonlinear system around an equilibrium point
 Consider a nonlinear system described by the diff equation:




))()(()(
))(),(()(
tuthty
tutt
x
xfx
.
(1) 
,
 Expanding Taylor series for f(x,u) and h(x,u) around the 
equilibrium point , we can approximate the nonlinear 
system (1) by the following linearized state equation: 
),( ux




)(~)(~)(~
)(~)(~)(~
tutty
tutt
DxC
BxAx (2) 
where:
ututu
tt


)()(~
)()(~ xxx
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ytyty  )()(~ )),(( uhy x
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Linearized model of a nonlinear system around an equilibrium point
Th t i f th li i d t t ti l l t d e ma r x o e near ze s a e equa on are ca cu a e as
follow:
1
2
1
1
1
n
fff
x
f
x
f
x
f










 
2
1
f
u
f






2
2
2
1
2
nxxxA



 


f
uB







)(21 un
nnn
x
f
x
f
x
f
,x
 




  )( u
n
u ,x 
)(21 nx
h
x
h
x
hC 







 
)( uu
h
x
D 




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u,x ,
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Linearized state-space model – Example 1
The parameter of the tank:
u(t)
qin
3
22
80/150
100 ,1
CVk
cmAcma 
y(t) qout 2sec/981
. ,.sec
cmg
cm D


 Nonlinear state equation:




))()(()(
))(),(()(
tuthty
tutt
x
xfx
,
)(94650)(35440)(
)(2
)( 1 k
tgxaCDf
where
.., 1 tutxtuAA
u x
)())(),(( 1 txtuth x
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Linearized state-space model – Example 1 (cont’)
Linearize the system around y = 20cm:
 The equilibrium point:
201 x
05.13544.0),( 1  uxuxf 9465.0u
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Linearized state-space model – Example 1 (cont’)
 The matrix of the linearized state space model:
0396.021  D gaCfA
-
1 hC
2
)(1)(1

uu
xAx
,x,x )(1
 ux ,x
5.1
)()(
1 

uu A
k
u
f
,x,x
B 0
)(


uu
h
,x
D
 The linearized state equation describing the system around
the equilibrium point y=20cm is:
)(
)(2
),( 1 tu
A
k
A
tgxaC
u D xf

 
)(~)(~
)(~5.1)(~0396.0)(~ tutt xx
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)())(),(( 1 txtuth x tty x
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Linearized state-space model – Example 2
The parameters of the robot:
2
C
02050
1.0,2.0 ,5.0
mkgJkgM
kgmmlml 
m
u
l
2sec/81.9 ,005.0
..,.
mgB 

 Nonlinear state equation :




))()(()(
))(),(()(
tuthty
tutt
x
xfx
,

)(2 tx
where:
 

 )(
)(
1)(
)(
)(cos
)(
)(),(
22212 tumlJ
tx
mlJ
Btx
mlJ
gMlmlu Cxf
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)())(),(( 1 txtuth x
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Linearized state-space model – Example 2 (cont’)
Linearize the system around the equilibrium point y = /6 (rad):
 Calculating the equilibrium point:
6/x1
01cos)(),(
2




 uxBxgMlml
x
u Cxf
)()()( 22212  
 mlJmlJmlJ
   02x  2744.1u
Th th ilib i i t i  6/xen e equ r um po n s:  02
1
x
x
27441
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.u
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Linearized state-space model – Example 2 (cont’)
 The system matrix around the equilibrium point:


 1211
aa
aa
A
0
1
1
11 

x
fa 1
)(2
1
12 

x
fa
2221
)( u,x
12
2
21 )(sin
)( C txMlmlfa 

u,x
)()(1 )( uu mlJx ,x,x 
2 Bfa 
  1)(
)(
)(
2
BgMlml
tx
u Cxf
)(
2
)(2
22 )( uu mlJx ,x,x 

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 
 )()()()()(cos)(
,
22212 tumlJ
tx
mlJ
tx
mlJ
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Linearized state-space model – Example 2 (cont’)
 The input matrix around the equilibrium point:


 1
b
b
B
01fb
2
)(
1  uu ,x
1f
2
)(
2
2 mlJu
b
u 
 ,x




 )(1)()()(
)(
)(
2
BgMlml
tx
u Cxf
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 
 )()(cos)(
,
22212 tumlJ
tx
mlJ
tx
mlJ
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Linearized state-space model – Example 2 (cont’)
 The output matrix around the equilibrium point:
1
1
1 

x
hc 21 ccC 0
)(2
2 

x
hc
)( u,x u,x
1dD 0
)(
1 

u
hd
u,x
 Then the linearized state equation is: 



)(~)(~)(~
)(~)(~)(~
ttt
tutt
DxC
BxAx
 uy

10
A 
0
B  01C 0D 2221 aa  2b
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)(),( 1 txuh x
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Regulating nonlinear system around equilibrium point
 Drive the nonlinear system to the neighbor of the equilibrium
point (the simplest way is to use an ON-OFF controller)
 Around the equilibrium point, use a linear controller to maintain
the system around the equilibrium point.
Linear
r(t) Nonlinear 
system+
y(t)
control u(t)e(t)
ON-OFF
Mode 
select
6 November 2012 120© H. T. Hoang - www4.hcmut.edu.vn/~hthoang/