Lecture "Fundamentals of control systems - Chapter 2: Mathematical models of continuous control systems" presentation of content: Presentation of content, transfer function, block diagram algebra, signal flow diagram, state space equation, linearized models of nonlinear systems.
Lecture Notes Lecture Notes
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The concept of mathematical model The concept of mathematical model Transfer function Block diagram algebra Block diagram algebra Signal flow diagram State space equation State space equation Linearized models of nonlinear systems
Nonlinear state equation Nonlinear state equation Linearized equation of state
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If you design a control system, what do you need to know If you design a control system what do you need to know
Question Question
What are the advantages of mathematical models?
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about the plant or the process to be controlled?
Practical control systems are diverse and different in nature Practical control systems are diverse and different in nature. It is necessary to have a common method for analysis and design of different type of control systems Mathematics The relationship between input and output of a LTI system of can be described by linear constant coefficient equations:
Why mathematical model? Why mathematical model?
u(t) y(t)
n
1
d
a
a
n
tya )( n
0
a 1
1
)( ty n 1
)( tdy dt
dt
n )( tyd n dt
m
d
b 0
b 1
b m
tub )( m
1
1 1 tu )( 1 m
m tud )( m dt
dt
tdu )( dt
Linear Time- Invariant System
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n: system order, for proper systems: nm. n: system order for proper systems: nm ai, bi: parameter of the system
M M
tBv )( )( tBv
f f
t )( )( t
tdv )( dt
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M: mass of the car, B friction coefficient: system parameters M: mass of the car B friction coefficient: system parameters f(t): engine driving force: input v(t): car speed: output v(t): car speed: output
M
B
tKy )(
f
t )(
tdy )( dt
2 tyd )( 2 dt
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M: equivalent mass B friction constant, K spring stiffness f(t): external force: input y(t): travel of the car body: output (t) t t b d f th t l
Example: Elevator Example: Elevator
t b l
g
MB MB Counter- balance
ML Cabin & load
M
B
)(
gMtKgM
L
T
B
2 tyd )( 2 dt dt
tdy )( dt dt
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ML: mass of cabin and load, MB: counterbalance M B friction constant, K gear box constant (t): driving moment of the motor y(t): position of the cabin
Difficult to solve differential equation order n (n>2) Difficult to solve differential equation order n (n>2)
n
1
d
a
a
0
a 1
n
tya )( n
1
ty )( n 1
n tyd )( n dt dt
dt dt
tdy )( dt dt
m
d
b 1
b 0
b m
tub )( m
1
1 tu )( 1 m
m tud )( m dt
dt
tdu )( dt
System analysis based on differential equation model is
System design based on differential equations is almost
difficult.
It is necessary to have another mathematical model that makes It is necessary to have another mathematical model that makes
impossible in general cases.
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the analysis and design of control systems easier: transfer function state space equation
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st
f
sF )(
f
et ( ).
dt
t )(
The Laplace transform of a function f(t), defined for all real The Laplace transform of a function f(t) defined for all real numbers t ≥ 0, is the function F(s), defined by:
0 0
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where: s : complex variable (Laplace variable) L : Laplace operator F(s) Laplace transform of f(t). The Laplace transform exists if the integral of ƒ(t) in the interval [0,+) is convergence. interval [0 +) is convergence
sG sG )( )(
tg )( )( tg
Given the functions f(t) and g(t), and their respective Laplace Given the functions f(t) and g(t) and their respective Laplace transforms F(s) and G(s):
f f
sF sF )( )(
t )( )( t
Linearity
t )(
sFa )( .
sGb . )(
fa .
tgb )(.
Time shifting
f
.
sF )(
Tt (
Ts e
)
Differentiation
sF
s )(
f
)0(
tdf )( )( dt
t t
Integration
f
)( sF sF )( s
0
f f
t )( )( t
sF sF
s )( )( s
Final value theorem Final value theorem
df lim lim t
)( d lim lim s 0
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Unit step function:
Laplace transform of basic functions Laplace transform of basic functions
tf tf
u(t)
tu )(
)( tu
1
1 1 s
tf
0 0 0
1 1 i i 0 i
Dirac function:
t t 0 0
t )(
tf tf tf
0 0 0
0 i i i
(t)
1
tL )(
)( dt )( dt
t t
1 1
1
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t t 0 0
Ramp function: f
)( tr
tu
)( t
tf tf
0 0
t i 0 i
R ti
r(t)r(t)
)(. tut
1
1 2 s
0 0
at
Exponential function
f
t )(
tu )(.
e
0
f t f t f t i
ate e i i 0
t 1 0
f(t)
e at
)(. tu
1
1 as
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t t 0 0
Sinusoidal function id l f
Laplace transform of basic functions (cont’) Laplace transform of basic functions (cont’)
Si ti
tf tf
f
t )(
(sin
tut ). )(
t t 0
0 0 0
i i tf i
sin sin
f( ) f(t)
(sin
)() tut
t t 0 0
2
s
2
Table of Laplace transform: Appendix A, Feedback control
p
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pp of dynamic systems, Franklin et. al.
Consider a system described by the differential equation: Consider a system described by the differential equation:
u(t) y(t)
n
1
d
a a
a a
0 0
a a 1 1
n
tya )( )( tya n
1 1
ty )( 1 1 n
n tyd )( n dt
dt
tdy )( dt
m
d
b 0 0
)( )( tub m m
b m m
b 1 1
1 1
1 tu )( 1 m m 1
dt d
m tud )( m m dt d Taking the Laplace transform the two sides of
Linear time invariant system invariant system
tdu )( dt d the above equation, using differentiation property and assuming that the g p p initial condition are zeros, we have:
n sYsa )(
a
sY
s )(
0
n 1 sYsa )( 1
s )(
n 1 m sUsb )( 0
sYa )( n 1 m sUsb )( 1
sUb )( m
sUb 1 m
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q g y ,
Transfer function: Transfer function:
m
m
1
)( sG )(
n n
n n
1 1
sY )( )( sU U )(
sb 0 sa 0
sb 1 sa 1
n
n
bs b m m 1 asa 1
Definition: Transfer
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y function of a system is the ratio between the Laplace transform of the output signal and the Laplace transform of the input signal assuming that initial conditions are zeros. conditions are zeros
Step 1: Establish the differential equation describing the
Procedure to find the transfer function of a component Procedure to find the transfer function of a component
p p
y input-output relationship of the components by: p p Applying Kirchhoff's law, current-voltage relationship of
Applying Newton's law, the relationship between friction
resistors, capacitors, inductors,... for the electrical components. components
h i th ti f f t l i l
and velocity, the relationship between force and deformation of springs ... for the mechanical elements. d f Apply heat transfer law, law of conservation of energy, for
...
Step 2: Taking the Laplace transform of the two sides of the differential equation established in step 1, we find the transfer differential equation established in step 1 we find the transfer function of the component.
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the thermal process. p
R
First order integrator:
Passive compensators Passive compensators
C C
sG )(sG )(
1 RCs
1
C
First order differentiator:
R
sG )( )(
RCs RC
1 1
RCs
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g
C
Phase lead compensator:
R1 R
sG )(
K
C
R2
Ts 1 1Ts Ts 1
R R 1
R R 2
1
T
KC
R R 2
R 1
R 2
R 2
CRR 12 R R 1 2
R2
Phase lag compensator :
R1
)( )( sG
K
C C
Ts 1 Ts 1
C
1
T
(
R 1
) CR 2
1CK
R 1R 1 RR 2
1
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ass e co pe sato s Passive compensators
Proportional Controller (P)
sG )( G )(
K P K P
PK K 2R 1R
Active controllers Active controllers
I
sG
K
)(
P P
K s
K I I
K P P
1 CR CR 1
2R 1R
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) Proportional Integral controller (PI) g p (
Transfer function of some type of controllers (cont’) Transfer function of some type of controllers (cont’)
Proportional Derivative controller (PD)
sG
K
)(
P
sK D
CR CR
K D K
2
K P K
2R 2 R 1
Active controllers Active controllers
I
sG
K
)(
P
sK D
Proportional Integral Derivative controller (PID) ) K s s
11
2
2
K I
K P
1 CR 1
2
CRCR CR 21
K D
2CR 1
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g p (
ia
Ra
La
Ua
Ea a
ML ,B, J ML B J
Equivalent diagram of a DC motor Equivalent diagram of a DC motor
: motor speed ML : load inertia ti B : friction constant
La : armature induction Ra : armature resistance t Ua : armature voltage Ea : back electromotive force J : moment of inertia of the
a rotor
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R M l d i i t
Applying Kirchhoff's law for the armature circuit:
Transfer function of DC motors Transfer function of DC motors
ff f
(
tU )( a
i a
Rt ). a
L a
tE )( a
di t )( a dt dt
(1)
(2) where: t )( K tE )( a
Applying Newton’s law for the rotating part of the motor:
)( )( t
K : electromotive force constant K : electromotive force constant : excitation magnetic flux
)( )(
J J
tMtM )( tMtM )(
)( )( tB tB
L
d dt
(3) (3)
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(4) where: tM )( t )( iK a
Taking the Laplace transform of (1), (2), (3), (4) leads to:
f (1) (2) (3) (4) l d t th L l T ki t f
a
a
(5) I ( s )( sU )( a Rs ). a sIL a sE )( a
(6) s )( K sE )( a
t
(7) )( Js s )( sMsM )( )( sB
Denote:
(8) sM )( s )( iK a
T a a
L a R a
Electromagnetic time constant Electromagnetic time constant
Tc
J B B
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Mechanical time constant
From (5) and (7), we have: From (5) and (7) we have:
I
s )(
a
sU )( a 1( R R 1( a
sE )( a sT sT ) ) a
s )(
)( B B 1( 1(
sMsM )( L sT sT ) ) c
From (5’), (6), (7’) and (8) we can develop the block diagram
(5’)
)(sM L
)(sU a )(a
/1 R/1 R a 1 asT
)(sEa
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of the DC motor as follow: of the DC motor as follow:
Transfer function of a thermal process Transfer function of a thermal process
Temperature of the oven th
Electric power supplying to the oven 100% 100% t
th
y(t) u(t)
y(t) (t) y(t) (t)
“Exact” characteristic of the oven
Approximate characteristic of the oven
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sG )(
The approximate transfer function of the thermal The approximate transfer function of the thermal process can be calculated by using the equation:
sY )(sY )( sU )(
The input is the unit step signal, then
sU
)(
1 1 s
y ty )( )(
f f
( (
) )
ate output s The approximate output is:
e app o
where:
f
t )(
K
1(
The Laplace transform of f (t) is: The Laplace transform of f (t) is:
sF F )( )(
1(
s
)
Applying the time delay theorem: Applying the time delay theorem:
sY sY )( )(
s
)
1Tt 1 2/ Tte ) K 2sT sT 1Ke 1(
2sT
sG )( )( G
)( sY )( sU )(
sT 1Ke sT 1 2
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Transfer function of a car Transfer function of a car
M
tBv )(
f
t )(
Differential equation:
tdv )( dt dt
Transfer function
)( sG
sG
)(
)( sV )( )( sF F
1 BM BMs
K 1Ts T 1
M: car mass B: friction constant f(t): driving force v(t): car speed
K K
T T
1 B
M B
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where
q
Differential equation:
tKy )(
f
t )(
M
B
)( tdy dt dt
2 )( tyd 2 dt dt
Transfer function: Transfer function:
)( sG sG )(
2
)( sY )( sF
1 Bs
Ms
K
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M: equivalent car mass B: friction constant K: spring stiffness f(t): external force f(t): external force y(t): travel of car body
y(t) y(t) yfb(t) yfb(t)
y( )
g
Feedback signal yfb(t) is proportional to y(t), so transfer functions of
yfb( )
p p sensors are usually constant:
sH
)(
fbK
Ex: Suppose that temperature of a furnace changing in the range y(t) = 05000C, if a sensor converts the temperature to a voltage in the range yfb(t) 05V, then the transfer function of the sensor is: the range y (t) 05V then the transfer function of the sensor is:
0
sH )(
K
(5
V
/)
(500
C
)
(01.0
0 CV /
)
fb
If the sensor has a delay time, then the transfer function of the
sensor is:
)( sH )( sH
1
K fb sT fb
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Sensor
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Block diagram is a diagram of a system in which the principal Block diagram is a diagram of a system, in which the principal parts or functions are represented by blocks connected by lines, that show the relationships of the blocks.
A block diagram composes of 3 components:
Function block Summing point Summing point Pickoff point
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Summing point Function block Pickoff point
Un (s)
Yn (s)
( ) U(s)
( ) Y(s)
1 ( ) U1 (s)
1 ( ) Y1 (s)
G1 G
G2 G
Gn G
U2(s)
Y2 (s)
Transfer function of systems in series Transfer function of systems in series
Y (s)
U(s)
n
)( sG s
)( sG i
i 1 1
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Gs
U1 (s) U (s)
Y1 (s) Y (s)
G1
Y(s) Y(s)
U(s) U(s)
Y2 (s) Y2 (s)
U2(s) U (s)
G2
Transfer function of systems in parallel Transfer function of systems in parallel
Y (s)
U(s)
Yn (s)
Un (s)
Gn
n
)( sG )( p p
)( sG )( i i
1 i
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Gp
Block diagram algebra (cont’) Block diagram algebra (cont’)
Negative feedback
Unity negative feedback
Y(s)
Y(s)
R(s)
R(s)
E(s)
E(s)
G(s)
G(s)
+
+
Yfb(s)
Yfb(s)
H(s)
sGcl sG )( )(
sGcl )( )( sG
1
)( sG sG )(
sG )( sHsG ( ). )(
1
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Transfer function of feedback systems Transfer function of feedback systems
Positive feedback
Unity positive feedback
Y(s)
Y(s)
R(s)
R(s)
E(s)
E(s)
G(s)
G(s)
+ +
+ +
Yfb(s) Y (s)
Yfb(s) Y (s)
H(s)
sGcl )(
sGcl )(
1 1
sG )( sG )( )( sG
1 1
sG )( sHsG )( ( ). )( )( sHsG
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Transfer function of feedback systems Transfer function of feedback systems
For a complex system consisting of multi feedback loops, we
Transfer function of multi-loop systems Transfer function of multi loop systems
i l f t bl th t k di ti t f l
perform equivalent block diagram transformation so that simple i connecting blocks appears, and then we simplify the block diagram from the inner loops to the outer loops. diagram from the inner loops to the outer loops
g p p
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q Two block diagrams are equivalent if their input-output relationship are the same.
Moving a pickoff point behind a block
x1 x3 x3 x1
G(s) G(s)
x
x
x 3
x
x
2 3
1 Gx G 1
2
Gx 1 xGx / G / 3 1
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x2 x2 1/G(s)
Moving a pickoff point ahead a block ff
x3 x1 x1 x3
G(s) G(s)
x
x
2 2
x 3 3
Gx 1 1
x
2 3
Gx 1 Gx G 1
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x2 x2 G(s)
Moving a summing point behide a block
x3 x1 x1 x3
+
+
G(s) G(s)
x
(
x
x
(
x
3
GxGx 1
2
) Gx 2
1
3
) Gx 2
1
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x2 x2 G(s)
Moving a summing point ahead a block
x3 x1 x1 x3
+
+
G(s) G(s)
x
(
x
)
/
xGxGGx
x
3
1
2
1
2
3
xGx 1
2
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x2 x2 1/G(s)
= - +
= + -
x
x
(
)
x
x
)
(
x 4
1
2
x 3
x 4
1
3
x 2
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Interchanging the positions of the two consecutive summing points
= - + = +
x x
x x
x x
x x
( (
) )
x x 4
= - + = + 1
2
x x 3
x x 4
1
2
x x 3
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Splitting a summing point
Block diagram algebra (cont’) Block diagram algebra (cont’)
Do not
NoteNote
Do Do
interchange the positions of a pickoff point and a g p summing point :
not not interchange interchange
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the the positions of two summing points there exists a pickoff point if between them: b t th
Find the equivalent transfer function of the following system:
Y(s)
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Interchanging the summing points and ,
Y(s)
)(
)(
sGsGsGA )( 3
4
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Eliminating GA(s)=[G3(s)//G4(s)]
GB(s)=[G1(s) // unity block] , G (s)=[G (s) // unity block]
Y(s) Y( )
1)( 1)(
sGB sG
1 sG sG )( )(
sGC )(
sG )( 2 sGsG sGsG ( )( ). )( )(
).[ ) [
( (
)] )]
1 1
1 1
A
2
sG )( 2 sGsGsG sGsGsG )( )( ( ( 3
2
4
Equivalent transfer function of the system:
)(
).
(
sGsGsG )( B
eq
C
(
sGeq )( sG )(
(
)]
1
sGsG 1[ )]. )( 1 2 sGsGsG ).[ )( ( 3
4
2
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GC (s)= feedback loop[G2(s),GA(s)]:
Find the equivalent transfer function of the following system:
Y(s)
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Interchanging the positions of the summing points and
Y(s)
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Moving the pickoff point behind the block G2(s)
GB(s) = feedback loop [G2(s), H2(s)] [GA(s)// unity block] GC(s) GC(s) = [GA(s)// unity block]
Y(s)Y(s)
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GD(s) = cascade(GB (s), GC(s), G3(s)) G ( ) d (G ( ) G ( ) G ( ))
Y(s)
GE(s) = feedback loop(GD(s), H3(s))
Y(s)Y(s)
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Detailed calculation: Detailed calculation:
1
*
GA GA
H G 2
*
G GB
1
G 2 HG 2
2
1
2
1
*
G
1
1
G C
A
H H G 2
HG HG G 2
1
2
1
*
.
G
D
C
B
. GGG 3
1
2 1
3
HGGG 3 HG 2
2
G 2 HG 2
2
HG G 2
G 3
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Example 2 (cont’) algebra –– Example 2 (cont’) Block diagram Block
1
2 1
3
3
3
G
*
E
1
1
1
G D HG D D
3 3
HGGG 1 2 HHGHGGHG 3 3
2 2
3 3
3 3
1 1
2 2
2 2
3 3
1 1
H H
3
2 1
3
HGGG HGGG 3 HG 2 2 HGGG 3 HG 2
2
Equivalent transfer function of the system: f
3
3
.
G G 1
1
2
2
3
2
3
3
1
G
*
eq
3
3
1
E
GG E 1 GG 1
1
.
G 1
1 1
HGGG 1 2 HHGHGGHG 3 HGGG 1 2 HHGHGGHG HHGHGGHG 3
2
2
2
3
3
1
3
1
3
1
1
Geq eq
1 1
2
HGGGGG 3 HGGGGGHHGHGGHG HGGGGGHHGHGGHG 1
3
2
2
3
2
2
3
3
1
3
1
3
1
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E i f th t t ti t f l
Find the equivalent transfer function of the following system:
Y(s)
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56
Move the summing point ahead the block G1(s),
Hint to solve example 3 Hint to solve example 3 Hint to solve example 3 Hint to solve example 3
Y(s)Y(s)
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57
then interchange the position of the summing points and then interchange the position of the summing points and Move the pickoff point behind the block G2(s)
Students calculate the equivalent transfer function themselves
Solution to Example 3 Solution to Example 3 Solution to Example 3 Solution to Example 3
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58
using the hints in the previous slide. using the hints in the previous slide
Block diagram algebra is a relatively simple method to Block diagram algebra is a relatively simple method to calculate the equivalent transfer function of a control system. The main disadvantage of block diagram algebra is its lack of systematic diagram transformation; each particular block diagram can be transformed by different heuristic ways. transformed by different heuristic ways.
When calculating the equivalent
perform the procedure block to
function, transfer it
is necessary to manipulate many calculations on algebraic the fractions. This could be a potential source of error if the fractions This could be a potential source of error if system is complex enough.
Block diagram algebra is only appropriate for
finding
transfer
function of complex systems, flow graph method (to be discussed later) is more flow graph method (to be discussed later) is more
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59
equivalent transfer function of simple systems. To find equivalent signal signal effective.
Y(s) Y(s)
Y(s) Y(s)
Block diagram
Signal flow graph
l di
th
h
d
ti
f
t
i
i
Source node: a node from which there are only out-going
hi h th
d
t
f
l
i
Sink node: a node to which there are only in-going branches. Hybrid node: a node which both has in-going branches and out-
going branches.
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60
p
q
g
Y(s) Y(s)
Y(s)
Forward path
Loop
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G
kk P
1 1
k
The equivalent transfer function from a source The equivalent transfer function from a source node to a sink node of a system can be found by using the Mason’s formula:
Pk: is the gain of kth forward path from the considered source node
to the considered sink node.
1
LL i
L i
j
LLL mj i
: is the determinant of the signal flow graph. : is the determinant of the signal flow graph
i
, ji nontouchin nontouchin g g
, , mji nontouchin nontouchin g g
iL : is the gain of the ith loop k: is the cofactor of the kth path .
k is inferred from by removing all the gain(s) of the loop(s) touching the forward path Pk
Note: Nontouching loops do not have any common nodes. A loop and Note: Nontouching loops do not have any common nodes A loop and
a path touch together if they have at least one common node.
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Find the equivalent transfer function of the system described Find the equivalent transfer function of the system described
Example 1 Signal flow graph –– Example 1 Signal flow graph
R(s)
Y(s)
Solution Solution
1 1
3
2
2
6
6
2
Forward paths: GGGGGP GGGGGP 1 2 5 1 4 GGGGP 2 1 5 4 GGGP GGGP 3 3 1 1 7 7
2 2
Loop: L L 1 1 L 2 L 3 L 4
HG HG 4 4 HGG 7 HGGG 4 5 HGGGG 4
5
2
3
2
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63
by the following signal flow graph:
)
The determinant of the SFG: The determinant of the SFG: L (1 4
L 2
L 3
L 1
LL 21
The cofactors of the paths
11 12 3 1 L 1
The equivalent transfer function of the system:
( (
) )
Geq G
P P 1 2
P P 2
P P 3
1
3
1
1(
)
1
2
1
4
2
6
1
3
4
Geq G
1
GGGGGGGGGGGG 1 7
HG 4
5
5 HGGHGHGGGGHGGGHGGHG 2
1
6
2
4
5
7
2
3
5
2
4
2
1
4
7
4
2
2
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Find the equivalent transfer function of the system described Find the equivalent transfer function of the system described
R(s)
Y(s)
Solution:
R(s)
Y(s)
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65
by the following block diagram:
R(s)
Y(s)
Forward paths:
2
3
GGGP 1 1 3 GHGP 2 1 3
1
1
3 3
1 1
3 3
Loop L 1 L L 2 L 3 L L 4 4 L 5
HG 2 2 HGG HGG 3 2 GGG 3 2 HHG HHG HGG 3
1
1
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The determinant of the SFG: f th SFG
(1
)
L 1
L 2
L 3
L 4
L 5
The cofactors of the paths
11 1 12
The equivalent transfer function of the system: f
Th d t t i
(
)
P 1
P 2
1
2
f th Th t t t l i f
1
1
3
1
Geq
1 1
2
3
HGGGGG HGGHHGGGGHGGHG HGGHHGGGGHGGHG 2
1
2
3
2
3
3
3
3
2
1
3
1
1
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ti 1 Geq
Find the equivalent transfer function of the system described
q y
Y(s)
Solution:
Y(s)
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by the following block diagram:
Y(s)
Forward path
2
1
GGGP 1 1 3 2 GP 4
1
3 3
Loop L 1 L L 2 L 3 L L 4 4 L 5
HG 1 2 HGG HGG 2 1 GGG 3 2 HGG HGG 3 3 2 2 G 4
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Determinant of the SFG: Determinant of the SFG:
(1
)
(
)
L 1
L 2
L 3
L 4
L 5
LL 41
LL 51
LL 52
LL 54
LLL 541
The cofactor:
(1
)
(
)
11
2
L 1
L 2
L 4
LL 41
The equivalent transer function of the system: The equivalent transer function of the system:
G
(
)
P 1
P 2
1
2
1
Num Den
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State: The state of a system is a set of variables whose State: The state of a system is a set of variables whose values, together with the equations described the system dynamics, will provide future state and output of the system. A nth order system has n state variables. The state variables can be physical variables, but not necessary.
State vector: n state variables form a column vector called
T
nx
x 1x
2 x
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72
the state vector.
By using state variables, we can transform the n-order o de differential equation describing the system dynamics into a set of n first order differential equations (called state equations) of the form: equations) of the form:
tu )(
a ab es, y us g s a e a s o e ca e
Ax B
t )( )( t
x t )( ty )( Cx
n
c
where where
c 1C
2
nc
a 11 a a 21 21
a 12 a a 22 22
a 1 n a a 2 2
a
a 1 n
n
2
a nn
b 1 b b 2 2 nb
Note: Depending on how we chose the state variables, a y system can be described by many different state equations.
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q y y
B
M
y tKy )( )(
f f
)( )( t
)( tdy dt
Differential equation: 2 tyd )( 2 2 dt
tx )( 1
f
t )(
tx )( 2
tx )( 1
tx )( 2
Denote: ty y )(1 )( )( )( tx 1 tx ty )( )( 2
1 1 M
f
t )(
tx )( 2 K K M )( tx )( 1 1 tx )( 2
)( )( tx 1 1 )( tx 2
B B M
.
1 B B M
0 1 1 M
ty )(
01
0 K K M tx )(1 tx )( 1 )( tx 2
)( )( t
)( )( t
)( )( t
A suspension system A suspension system A suspension system A suspension system
01C
t )(
0 1 M
0 K M
1 B M
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: motor speed Mt : load inertia M : load inertia B : friction constant J : moment of inertia of the rotor
La : armature induction Ra : armature resistance R : armature resistance Ua : armature voltage Ea : back electromotive force
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DC motor DC motor
Applying Kirchhoff s law for the armature circuit: Applying Kirchhoff's law for the armature circuit:
(
)( tU a
i a
). Rt a
L a
tE )( a
)( di t a dt dt
(1)
t )(
K
tE )( a
(2) where:
Applying Newton s law for the rotating part of the motor: Applying Newton’s law for the rotating part of the motor:
K : electromotive force constant : excitation magnetic flux
(for simplicity, assuming that load torque is zero)
)( tM
J
)( tB
d )( t d t )( dt
(3) (3)
tM )( )( tM
t )( )( t
iK iK a
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(4) (4) where: h
t )(
Example 2 (cont’) State equations –– Example 2 (cont’) State equations
(1) & (2)
t )(
t )(
i ö
tU )( ö
di ö dt
R ö L ö
K L ö
1 L ö
t )(
(5)
t )(
t )(
(3) & (4)
i ö
d dt
K J
B J
t )(
Denote:
t )( )( t
i ö
tx )(1 tx )( )(2 tx
tx tx )( )( 1
tx tx )( )( 1
tx )( )( tx 2
tU )( )( tU ö
1 L ö
(5) & (6)
)( )( tx 2 2
)( )( tx 1 1
)( )( tx 2 2
R ö ö L ö K J
K L ö B J
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(6)
ö
)( tx 1 )( tx 2
tx )( 1 )( tx 2
1 tUL )( tUL )( ö 0
R ö L ö ö K J
K L ö ö B J
t )(
tx )( )( 1 tx )( 2
10
Example 2 (cont’) State equations –– Example 2 (cont’) State equations
tU )( u
t )( )( t
t )( )( t
where:
10C
1 öL 0 0
R ö L ö K J
K L ö B J
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The differential equation describing the system dynamics is:
n
1
d
a
a
0
a 1
n
)( tya n
)( tub 0
1
)( ty 1 n
n )( tyd n dt
dt
)( tdy dt
Define the state variables as follow:
The first state is the system output: The first state is the system output: The i th state (i=2..n) is chosen to be the first derivative of the (i1)th state :
tx tx )( )(1 1 tx )( 2 )( )( tx 3 3
ty )( )( ty tx )( 1 )( )( tx 2 2
t )(
tx )( n
n
1
x
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q g y y
t )(
tu )(
t )(
State equation:
t )( )( t
0 0
1 1
0 0
0 0
0
0
1
0
where:
)( tx 1 )( tx 2
)( )( t
2
1
)( x t n 1 )(txn )( tx
0 a n a 0
0 a n a 0
1 a 1 a 0
0 0 0 0 b 0 0a
C
01
0 a n a 0 00
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Write the state equations describing the following system:
)(6)(5)(2 ty ty ty
10
ty )(
tu )(
Define the state variables as:
ty )( tx )( 1 tx )( 2
tr )(
State equation:
t )( t )( )(
0
0
where
0 0
0
1
0
0
1
0
5.0
0
0
1
0 0 b 0 a
0
3 3
2 2
5 5
3 3
5.2 52
0 a a
0 a a
1 a 1 1 a
0
0
0
001C
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Consider a system described by the differential equation:
n
1
d d
a
a
n
)( )( tya n
0
a 1
1
)( )( ty ty 1 n
n )( )( tyd tyd n dt
dt
n
n
d
d
b b n
b b 0 0
b b 1 1
2 2
)(1 )( tub tub 1 n
1 )( tu 1 n
)( )( tdy tdy dt 2 )( tu 1 n
dt
dt
)( tdu dt
Define the state variables as follow:
tu )(
tx )( 1 tx )( 2 2 tx )( 3
ty )( tx )( 1 1 tx )( 2
tu )( 1 1 2
The first state is the system output: The i th state (i=2 n) is equal to the state (i 2..n) is equal to the The i first derivative of the (i1)th state minus a quantity proportional to the input: th i t
x
t )(
tx )( n
n
n
tu )( 1
1
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82
t )(
tu )(
t )(
State equation:
t )( )( t
0 0
1 1
0 0
0 0
0
0
1
0
tx )( 1 tx )( 2
t )(
where:
2
1
x t )( 1 n t )(txn )(
1 2 1 n n
0 a n a 0
0 a n a 0
0 a n a 0
1 a 1 a 0
01
00
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Case #2 (cont ) Case #2 (cont’) Case #2 (cont ) Case #2 (cont’)
1
b b 0 a 0 b b 1 1
2
a
b 2 2
12 12
3
a a 11 11 a 0 a 21 21 a 0
2
b n
a 1 n
a n
1
1
11
n
a 2 n a 0
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84
The coefficients in the vector B are calculated as follow:
Write the state equations describing the following system: )(6)(5)(2 ty ty ty
tu )(
tu )(
ty )(
10
10
20
Define the state variables:
tu )(
)( tx 1 )( tx 2 tx )( 3
)( ty )( tx 1 tx )( 2
)( tu 1 2
)( )( tu
The state equation: The state equation:
)( )( t t )(
0 0
0 0
1 1
0 0
1 1
0 0
0
0
1
3 3
2 2
5 5
3 3
5.2 52
1 B 2 3
001C
0 a a
0 a a
1 a 1 1 a
0
0
0
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The elements of vector B are calculated as follow:
0
10 10
b 0 a 0 b b 1
5
a
20
0655
b 2 2
05 05 2 2 2 1 1
2
5 2
0 2 a a 11 a 0 a 1 1 2 2 a
0
1 2 3
0
Case #2: Example (cont )) Case #2: Example (cont’)) Case #2: Example (cont )) Case #2: Example (cont’))
5 5 5 2
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86
Consider a system described by the differential equation: Consider a system described by the differential equation:
n
1
d
a
a
0
a 1
n
tya )( n
1
)( ty n 1
n )( tyd n dt dt
dt dt
)( tdy dt dt m d
b 0
b 1
b m
tub )( m
1
1 tu )( 1 m
m tud )( m dt
dt
tdu )( dt
space equations in controllable canonical form StateState--space equations in controllable canonical form
m
m
1
sG )( )( G
n
n
1
... ...
sb 0 0 sa 0
sb 1 1 sa 1
n
n
bsb 1 1 m m m m asa 1
The controllable canonical state equations of the system is Th t t
or equivalently by the transfer function:
f th ti t i t i
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87
l ll bl presented in the next slide.
tr )( )( t
State equations:
t )( )( t t )(
1
0
0
0
0
0
0
1
Where:
t )(
2
1
tx )( )(t 1 )( tx 2 )(txn
0 0 0 1
0 a n a 0 0
0 a n a 0 0
0 a n a 0 0
1 a 1 a 0 0
1
0 0
0 0
b m a 0
b m a 0
b 0 a 0
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88
)(2 ty
Write the controllable canonical state equations of the following Write the controllable canonical state equations of the following ty ty )(4)(5)( ty
tu )(3)( tu
tr )(
system:
Solution:
t )( )( t
0
1
0
0
1
0
0
0
1
2
5.2
5.0
0 0 1
0 a 3 a 0
0 a 2 a 0
1 a 1 a 0
where:
5.005.1 50051
b 2 a 0
b 1 a 0
b 0 a 0
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89
Establishing state equations from block diagrams Establishing state equations from block diagrams
Establish the state equations describing the system below:
R(s)
Y(s)
Example Example Example Example
10 )(1
ss (
s
)3
Define the state variables as in the block diagram:
( ) R(s)
Y(s) Y(s)
2(s) X2(s)
X3(s) X3(s)
X1(s) X1(s)
+
(
)1
(
)3
1 1 s
10 10 s
1 1 s
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90
+
Establishing state equations from block diagrams Establishing state equations from block diagrams
From the block diagram, we have:
sX
10
3)( s
sX )( 1
sX )( 2
1
sX )( 1
sX )( 2
10 10 s 3
Example (cont ) Example (cont’) Example (cont ) Example (cont’)
10
tx )( 1
tx )(3 1
tx )( 2
sX
)( )( s
)( )( sX 2 2
)( )( sX 3 3
2 2
)( )( sX 2 2
)( )( sX 3 3
1 1
1 s
(1)
tx )( 2
tx )( 2
tx )( 3
sR )(
sY
)(
sX
s )(
sR )(
sX )(3
3
sX )( 1
1 s
tr )( )(
(2)
)( tx )( 3
tx )( )( 1
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91
(3) (3)
Establishing state equations from block diagrams Establishing state equations from block diagrams
Combining (1), (2), and (3) leads to the state equations:
0
3
10
0
11 0 0
0 1
)( )( t
)( )( t
Output equation:
ty )(
tx )( 1
tx )(tx )( 1 tx )( 2 )(3 tx )( t
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92
Given a system described by the state equations:
tu )(
ib d b th t t Gi d ti t
t )( t )( )( t
Then the transfer function of the system is:
1
)( sG
BAIC
s
)( sY )( sU
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93
Calculate the transfer function of the system described by the Calculate the transfer function of the system described by the
Example State equation to transfer function –– Example State equation to transfer function
tu )(
state equation:
t )( t )(
0
1
where
01C
2
3 1
3
Solution: The transfer function of the system is:
1
)( sG
BAIC
s
)( sY )( sU
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94
0
1
s
1
s
s
Example (cont ) Example (cont’) Example (cont ) Example (cont’)
10 10
2 2
2 2
s s
01
3 3
3 3
1
1AI s
s 2
s
( ss
)1.(2)3
1
1 3
13 s 2 s
13
1
01 01
s s
13 13
2
2
s
1 s 3
s
2
1 s 3
s
2
s 2
1
s s
s s
2
1 s 3
s
2
(3 2 s
1)3 s s 3 2
3 13 13 1
sG sG )( )(
s
2
s 3 10 2 s 3
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95
tu
t )(
t )(
Solution to the state equation ?)(
t
)( t t )(
( t t (
u u
) ) B B
d )( )( d
0
1
[ [
( (
s s
)] )]
t )( )( t
where where transient matrix transient matrix
(
1)
s )(
System response?
)( ty
)( t
AIs
Example:
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Diff. equation
Define x
1
sG )(
s
BAIC
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Transfer function State equation
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98
Nonlinear systems do not satisfy the superposition
principle and cannot be described by a linear differential
equation.
Most of the practical systems are nonlinear: Most of the practical systems are nonlinear:
Fluid system (Ex: liquid tank,…)
Thermal system (Ex: furnace,…) Thermal system (Ex: furnace )
Mechanical system (Ex: robot arm,….)
Electro-magnetic system (TD: motor,…) Electro magnetic system (TD: motor )
Hybrid system ,…
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99
Input – output relationship of a continuous nonlinear system Input output relationship of a continuous nonlinear system can be expressed in the form of a nonlinear differential equations.
n
1
d
,
,
,
),
,
,
tu )(,
g
( ty
)( ty n n 1 1
)( tdy dt dt
)( tdu dt dt
n )( tyd n n dt dt
dt dt
m )( tud m m dt dt
where: u(t): input signal, where: u(t): input signal,
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100
y(t): output signal, g(.): nonlinear function
qin
t
u(t)
a: cross area of the dischage valve a: cross area of the dischage valve A: cross area of the tank g: gravity acceleration k: constant t k CD: discharge constant
Balance equation:
tyA )(
t )(
tq )( in
q out
y(t) (t) qout
tku )( )( tku
t )(
aC
2
tgy )(
tqin tq )( )( q out
D
ty )( )( ty
aC aC
2 2
tgy tgy
where: where:
D D
1 A
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(first order nonlinear system ) ) li t
l
m m
u
J: moment inertia of the robot arm J: moment inertia of the robot arm M: mass of the robot arm m: object mass l: length of robot arm l: length of robot arm lC : distance from center of gravity to rotary axis B: friction constant g: gravitational acceleration g: gravitational acceleration u(t): input torque (t): robot arm angle
J
(
(
ml
)
cos
tu )(
)() t
According to Newton’s Law 2 tB ml )(
gMl C
t )(
g
cos
tu )(
)( t
2
2
)
(
J
ml ( ( J
Ml ) C 2 ml )
)
(
J
B ml
1 ml
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(second order nonlinear system)
Moving direction
(t)
: steering angle : ship angle k: constant i: constant
(t) ( )
The differential equation describing the steering dynamic of a
g y g q
t )( )( t
t )( )( t
t )( )( t
t )( )( t
)( )( t t
ship:
t )( )( t 3
k 21
1 1 2
1
1 21
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(third order nonlinear system) (third order nonlinear system)
A continuous nonlinear system can be described by the state A continuous nonlinear system can be described by the state
Describing nonlinear systems by state equations Describing nonlinear systems by state equations
equation:
), ), ( ( )) )) tut tut ), ( ))
where: u(t): input,
y(t): output,
x(t): state vector,
x(t) = [x1(t), x2(t),…,xn(t)]T
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f(.), h(.): nonlinear functions
StateState--space model of nonlinear system Example 1 space model of nonlinear system –– Example 1
ty )(
aC
2
tgy
qin
Differential equation: Differential equation: tku )(
u(t)
D
1 A A
ty )( )(
Define the state variable: tx )(1 )(
State equation:
y(t) (t) qout
St t ti
tut ), ( )) tut ), ( )) ), tut ( ))
aC
)( t
gx 1
D
where
tu )(
),( u
k A
h
tut ), (
))
2 A 1 tx )(
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StateState--space model of nonlinear system Example 2 space model of nonlinear system –– Example 2
t )(
g
cos
tu )(
)( t
2
2
l
J
(
)
Ml ) C 2 ) ml
)
(
J
1 ml
Differential equation: Differential equation: B ml ( ml ( J
m
Define the state variable:
t )( )( t
t )(
tx )( 1 tx )( )(2 tx 2
t )(
u u
tut ), (
))
State equation:
q
h h
)) ))
where
u ),(
tu )(
cos
tx )( 1
tx )( 2
2
2
J
)
(
J
)(2 tx ml ( J (
gMl C 2 ml
) )
)
(
B ml
1 ml
h
tut ), (
))
1 tx )(
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Consider a nonlinear system described by the diff. equation: Consider a nonlinear system described by the diff. equation:
tut ), ( )) tut ), ), ( ( )) )) tut
The state is called the equilibrium point of the nonlinear
u
,( ux ( ux
) )
If If
system if the system is at the state system if the system is at the state and the control signal is and the control signal is fixed at then the system will stay at state forever.
is equilibrium point of the nonlinear system then: is equilibrium point of the nonlinear system then:
tut ), (
))
0
uu ,
The equilibrium point is also called the stationary point of the
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nonlinear system.
)( tx 1 )( )( tx 2 2
Consider a nonlinear system described by the state equation: Consider a nonlinear system described by the state equation:
1
tu )(
( ). )( txtx u 2 1 )( )(2)( )( tx tx 2 1 1 2 u Find the equilibrium point when
Solution:
tut ( ),
Example 1 Equilibrium point of nonlinear system –– Example 1 Equilibrium point of nonlinear system
))
0
The equilibrium point(s) are the solution to the equation:
01
0
xx . 1 2 x x 2 1 1 2 2
2
1x
2x
2x
1x
2 2 2
2 2
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or
Consider a nonlinear system described by the state equation: d
Example 2 Equilibrium point of nonlinear system –– Example 2 Equilibrium point of nonlinear system
)
1 x
3
3
x 1 1 x 2 x
3
2 2 x x u 2 3 2 3 sin( x x 1 2 x u 3
y
1x
ib d b th t t id C ti li t
tu )( )( tu
0 0
u u
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Find the equilibrium point when Find the equilibrium point when
Consider a nonlinear system described by the diff. equation: Consider a nonlinear system described by the diff equation:
(1)
Expanding Taylor series for f(x,u) and h(x,u) around the
,( ux
)
y g ) ( )
( equilibrium point , we can approximate the nonlinear system (1) by the following linearized state equation:
(2)
t )(
where:
)( )( tu ty )(
u y
(
y
h
u
))
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110
The matrix of the linearized state equation are calculated as
i d t t f th li l t d t i ti l
Th follow:
f 1 u f f 2 2 u f f n u
)
f 1 x 1 f f 2 x 1 f n x 1
f f 1 1 x x 2 n f f f f 2 2 x nx 2 f f n n x x 2 n
(
)
h u
h x 1 1
h x 2 2
h nx n
) )
) )
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Linearized state--space model Linearized state Example 1 space model –– Example 1
2
2
a
cm 1
,
A
100
cm
The parameter of the tank: The parameter of the tank:
3 3
qin
k k
150 150
cm
/ /
sec
CV CV . ,
8.0 80
D
u(t) u(t)
2
g
981
cm
/
sec
t )(
y(t) qout
tut ), (
))
Nonlinear state equation:
h h
)) ))
)( )( t
aCD D
tu )( )(
3544 3544
.0 0
9465 9465
tu )( )(
where
.0 0
tx )( )( 1
g gx 2 1 1 A
k k A
h
), ), ( ( tut
)) ))
)( )( 1 tx 1
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Linearized state--space model Linearized state Example 1 (cont’) space model –– Example 1 (cont’)
The equilibrium point:
2020
9465
.0u
1 x uxf ,(
)
.0
3544
u 5.1
0
x 1
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Linearize the system around y = 20cm: Linearize the system around y = 20cm:
The matrix of the linearized state-space model: The matrix of the linearized state space model:
g
Linearized state--space model Linearized state Example 1 (cont’) space model –– Example 1 (cont’)
.0
0396
1
aC 2
2 D xA 1
f 1 x (1
)
h x (1
)
(
)
0
5.1
h u
k A
f 1 u
(
)
(
)
(
)
The linearized state equation describing the system around
)( t
gx 1
tu )(
0396
2 A
k A
h
the equilibrium point y 20cm is: the equilibrium point y=20cm is:
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Linearized state--space model Linearized state Example 2 space model –– Example 2
The parameters of the robot: The parameters of the robot:
l
,2.0
mm
1.0
kg
lm ,5.0 C
l
2
M M
kg kg ,5.0 50
J J
02.0 020
. mkg mkg
m
2
B
.0
,005
g
81.9
m
/
sec
t )(
u u
tut ), (
))
Nonlinear state equation :
h h
)) ))
where:
),( u
cos
tu )(
tx )( 1
)( tx 2
2
2
)(2 tx ( ml J (
gMl C 2 ml
) )
)
(
J
)
(
J
B ml
1 ml
h
tut ), (
))
1 tx )(
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Linearize the system around the equilibrium point y = /6 (rad): Linearize the system around the equilibrium point y = /6 (rad):
Calculating the equilibrium point:
6/x 6/ 1 x
x 2 (
u
0
Linearized state--space model Linearized state Example 2 (cont’) space model –– Example 2 (cont’)
, ,
) )
cos cos
u u
x x 1
x x 2
2
2
ml J (
gMl C C 2 ml
) )
)
(
J
)
(
J
1 ml
B ml
0
2744
2x .1u
Then the equilibrium point is: Th ilib i i t i th
x x 1 x 2
6/ 6/ 0
2744 2744
.1u 1
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The system matrix around the equilibrium point: The system matrix around the equilibrium point:
Linearized state--space model Linearized state Example 2 (cont’) space model –– Example 2 (cont’)
a a 11 12 a 21 a a a 22
1
0
a 12
a 11
f 1 x x (2 (2
) )
f 1 x x (1 1
)
a
sin
21 21
)( )( tx 1 1
ml ( J (
) Ml C 2 2 ml )
(
)
f 2 x (1
)
a a 22
2
)
(
)
f 2 2 x (2
)
cos
tu )(
tx )( 1
tx )( 2
2
2
gMl gMl C C 2 ml
J
) ) )
)
(
)
(
J
1 1 ml
B B ml
B J ml ( )( tx 2 ml ( ( ml J (
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The input matrix around the equilibrium point: The input matrix around the equilibrium point:
Linearized state--space model Linearized state Example 2 (cont’) space model –– Example 2 (cont’)
b 1 2b b
0 0
b b 1
f 1f 1 u
(
)
b 2
2
f f 2 u
J
1 1 ml
(
)
cos
tu )( )(
tx )( )( 1
tx )( )( 2
2
2
gMl gMl C C 2 ml
)
(
J
)
(
tx )( 2 ml ( ( ml J (
) ) )
J
1 1 ml
B B ml
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The output matrix around the equilibrium point: The output matrix around the equilibrium point:
0
c 2
1
c 1
cC 1
2 c
h x (2 (2
u,x
) )
h x x (1 1
Linearized state--space model Linearized state Example 2 (cont’) space model –– Example 2 (cont’)
)
0
d 1
1dD
h u ,x u ( ( u
) )
Then the linearized state equation is:
0D 0D
01C 01C
0 1 a 21 a
22
0 2b
h
1 tx )(
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Drive the nonlinear system to the neighbor of the equilibrium Drive the nonlinear system to the neighbor of the equilibrium
Around the equilibrium point, use a linear controller to maintain Around the equilibrium point, use a linear controller to maintain
point (the simplest way is to use an ON-OFF controller)
the system around the equilibrium point.
Linear Linear control r(t) e(t) u(t) y(t)
Nonlinear system system +
ON-OFF
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