IntroCtrlSys_Chapter2.pdf

Lecture Notes
Lecture Notes
Introduction to Control Systems
Introduction to Control Systems
Instructor: Dr. Huynh Thai Hoang
Department of Automatic Control
Faculty of Electrical & Electronics Engineering
Ho Chi Minh City University of Technology
Email: hthoang@hcmut.edu.vn
huynhthaihoang@yahoo.com
Homepage: www4.hcmut.edu.vn/~hthoang/
20 September 2011 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1

Chapter 2
Chapter 2
MATHEMATICAL MODELS OF
MATHEMATICAL MODELS OF
CONTINUOUS CONTROL SYSTEMS
CONTINUOUS CONTROL SYSTEMS
20 September 2011 2
© H. T. Hoang - www4.hcmut.edu.vn/~hthoang/

‘ The concept of mathematical model
Content
Content
‘ The concept of mathematical model
‘ Transfer function
‘ Block diagram algebra
‘ Block diagram algebra
‘ Signal flow graph
‘ State space equation
‘ State space equation
‘ Linearized models of nonlinear systems
Ž Nonlinear state equation
Ž Nonlinear state equation
Ž Linearized state equation
20 September 2011 3

The concept of mathematical models
The concept of mathematical models
20 September 2011 4

‘ If you design a control system what do you need to know about the
Question
Question
‘ If you design a control system, what do you need to know about the
plant or the process to be control?
‘ What are the advantages of mathematical models?
20 September 2011 5

‘ Practical control systems are diverse and different in nature
Why mathematical model?
Why mathematical model?
‘ 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
d e e t type o co t o syste s ⇒ at e at cs
‘ The relationship between input and output of a LTI system of can be
described by linear constant coefficient equations:
y q
Linear Time-
I i t S t
u(t) y(t)
=
+
+
+
+
−
)
(
)
(
)
(
)
(
1
1
1
1
0 t
y
a
t
dy
a
t
y
d
a
t
y
d
a
n
n
L )
(
)
(
)
(
)
(
1
1
1
1
0 t
u
b
t
du
b
t
u
d
b
t
u
d
b
m
m
+
+
+
+
−
L
Invariant System
+
+
+
+ −
−
)
(
1
1
1
0 t
y
a
dt
a
dt
a
dt
a n
n
n
n
)
(
1
1
1
0 t
u
b
dt
b
dt
b
dt
b m
m
m
m
+
+
+
+ −
−
n: system order, for proper systems: n≥m.
ai, bi: parameter of the system
20 September 2011 6
ai, bi: parameter of the system

Example: Car dynamics
Example: Car dynamics
)
(
)
(
)
(
t
f
t
Bv
t
dv
M =
+ )
(
)
( t
f
t
Bv
dt
M +
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
20 September 2011 7

Example: Car suspension
Example: Car suspension
)
(
)
(
)
(
)
(
2
2
t
f
t
Ky
dt
t
dy
B
dt
t
y
d
M =
+
+
dt
dt
M: equivalent mass
B friction constant, K spring stiffness
f(t): external force: input
20 September 2011 8
f(t): external force: input
y(t): travel of the car body: output

Example: Elevator
Example: Elevator
M
t
K
M
t
dy
B
t
y
d
M +
+
+ )
(
)
(
)
(
2
g
M
t
K
g
M
dt
y
B
dt
y
M T
L B
+
=
+
+ )
(
)
(
)
(
2
τ
ML
Cabin &
MB
Counter-
balance
M f bi d l d M b l
Cabin &
load
ML: mass of cabin and load, MB: counterbalance
B friction constant, K gear box constant
τ(t): driving moment of the motor: input
20 September 2011 9
( ) g p
y(t): position of the cabin: output

‘ Difficult to solve differential equation order n (n>2)
Disadvantages of differential equation model
Disadvantages of differential equation model
‘ Difficult to solve differential equation order n (n>2)
=
+
+
+
+ −
−
−
)
(
)
(
)
(
)
(
1
1
1
1
0 t
y
a
dt
t
dy
a
dt
t
y
d
a
dt
t
y
d
a n
n
n
n
n
n
L
dt
dt
dt
)
(
)
(
)
(
)
(
1
1
1
1
0 t
u
b
dt
t
du
b
dt
t
u
d
b
dt
t
u
d
b m
m
m
m
m
m
+
+
+
+ −
−
−
L
‘ System analysis based on differential equation model is difficult.
‘ System design based on differential equation model is almost
impossible in general cases.
‘ It is necessary to have another mathematical model that makes the
‘ It is necessary to have another mathematical model that makes the
analysis and design of control systems easier:
Ž transfer function
20 September 2011 10
Ž state space equation

T f f ti
T f f ti
Transfer functions
Transfer functions

The Laplace transform of a function f(t) defined for all real numbers
Definition of Laplace transform
Definition of Laplace transform
The Laplace transform of a function f(t), defined for all real numbers
t ≥ 0, is the function F(s), defined by:
+∞
where:
{ } ∫
−
=
=
0
).
(
)
(
)
( dt
e
t
f
s
F
t
f st
L
where:
− s : complex variable (Laplace variable)
− L : Laplace operator
p p
− F(s) Laplace transform of f(t).
The Laplace transform exists if the integral of ƒ(t) in the interval
[0,+∞) is convergence.

Gi th f ti f(t) d (t) d th i ti L l
Properties of Laplace transform
Properties of Laplace transform
Given the functions f(t) and g(t), and their respective Laplace
transforms F(s) and G(s):
{ } )
(
)
( s
F
t
f =
L { } )
(
)
( s
G
t
g =
L
{ } )
(
)
( s
F
t
f =
L { } )
(
)
( s
G
t
g =
L
‘ Linearity { } )
(
.
)
(
.
)
(
.
)
(
. s
G
b
s
F
a
t
g
b
t
f
a +
=
+
L
‘ Time shifting { } )
(
.
)
( s
F
e
T
t
f Ts
−
=
−
L
)
( ⎫
⎧df
‘ Differentiation )
0
(
)
(
)
( +
−
=
⎭
⎬
⎫
⎩
⎨
⎧
f
s
sF
dt
t
df
L
s
F
t
)
(
⎫
⎧
‘ Integration
s
s
F
d
f
t
)
(
)
(
0
=
⎭
⎬
⎫
⎩
⎨
⎧
∫ τ
τ
L
‘ Final value theorem )
(
lim
)
(
lim s
sF
t
f =
‘ Final value theorem )
(
lim
)
(
lim
0
s
sF
t
f
s
t →
∞
→
=

Laplace transform of basic functions
Laplace transform of basic functions
‘ Unit step function:
1
⎧ ≥ 0
f
1 i
u(t)
{ }
s
t
u
1
)
( =
L
⎩
⎨
⎧
<
≥
=
0
t
f
0
0
t
f
1
)
(
i
i
t
u
t
0
1
‘ Dirac function:
t
0
⎩
⎨
⎧
=
∞
≠
=
0
t
f
0
t
f
0
)
(
i
i
t
δ δ(t)
⎩ =
∞ 0
t
f
i
∫
+∞
=1
)
( dt
t
δ
{ } 1
)
( =
t
δ
L
t
0
1
∫∞
−
)
( t
0

Laplace transform of basic functions (cont’)
R f i
‘ Ramp function:
⎧ ≥ 0
t
f
i
t
r(t)
1
⎩
⎨
⎧
<
≥
=
=
0
t
f
0
0
t
f
)
(
)
(
i
i
t
t
tu
t
r
t
0
1
1
{ } 2
1
)
(
.
s
t
u
t =
L
‘ Exponential function
t
0 1
⎧ ≥
−
0
f t
i
at
e
f(t)
{ }
at 1
)
(
L
⎩
⎨
⎧
<
≥
=
= −
0
f
0
0
f
)
(
.
)
(
t
i
t
i
at e
t
u
e
t
f
t
0
1 { } a
s
t
u
e at
+
=
− 1
)
(
.
L
t
0

Si id l f i
‘ Sinusoidal function
⎧ ≥ 0
t
f
sin i
t
ω
f(t)
⎩
⎨
⎧
<
≥
=
=
0
t
f
0
0
t
f
sin
)
(
).
(sin
)
(
i
i
t
t
u
t
t
f
ω
ω
f( )
t
0 t
0
{ } 2
2
)
(
)
(sin
ω
ω
ω
+
=
s
t
u
t
L
‘ Table of Laplace transform: Appendix A, Feedback control of
f p f pp f
dynamic systems, Franklin et. al.

‘ Consider a system described by the differential equation:
Definition of transfer function
Definition of transfer function
Linear time
i i t t
u(t) y(t)
=
+
+
+
+
−
)
(
)
(
)
(
)
(
1
1
1
1
0 t
y
a
t
dy
a
t
y
d
a
t
y
d
a
n
n
L
invariant system
+
+
+
+ −
−
)
(
1
1
1
0 t
y
a
dt
a
dt
a
dt
a n
n
n
n
)
(
)
(
)
(
)
(
1
1
1
1
0 t
u
b
d
t
du
b
d
t
u
d
b
d
t
u
d
b m
m
m
m
m
m
+
+
+
+ −
−
L
‘ Taking the Laplace transform the two sides of the above equation,
using differentiation property and assuming that the initial condition
)
(
1
1
1
0
dt
dt
dt
m
m
m
m −
g p p y g
are zeros, we have:
=
+
+
+
+ −
−
)
(
)
(
)
(
)
( 1
1
1
0 s
Y
a
s
sY
a
s
Y
s
a
s
Y
s
a n
n
n
n
L
)
(
)
(
)
(
)
( 1
1
1
0 s
U
b
s
sU
b
s
U
s
b
s
U
s
b m
m
m
m
+
+
+
+ −
−
L

‘ Transfer function:
Definition of transfer function (cont’)
Definition of transfer function (cont’)
‘ Transfer function:
m
m
m
m
b
s
b
s
b
s
b
s
Y
s
G
+
+
+
+
=
= −
−
1
1
1
1
0
)
(
)
(
L
‘ Definition: Transfer function of a system is the ratio between the
n
n
n
n
a
s
a
s
a
s
a
s
U +
+
+
+ −
−
1
1
1
0
)
(
)
(
L
y
Laplace transform of the output signal and the Laplace transform
of the input signal assuming that initial conditions are zeros.

Transfer function of components
Transfer function of components
Procedure to find the transfer function of a component
Procedure to find the transfer function of a component
‘ Step 1: Establish the differential equation describing the input-
output relationship of the components by:
p p p y
¾ Applying Kirchhoff's law, current-voltage relationship of
resistors, capacitors, inductors,... for the electrical components.
¾ Applying Newton's law the relationship between friction and
¾ Applying Newton s law, the relationship between friction and
velocity, the relationship between force and deformation of
springs ... for the mechanical components.
A l h f l l f i f f h
¾ Apply heat transfer law, law of conservation of energy,… for the
thermal processes.
¾ ...
‘ Step 2: Taking the Laplace transform of the two sides of the
differential equation established in step 1, we find the transfer
function of the component
function of the component.

Transfer function of some type of controllers
Transfer function of some type of controllers
Passive compensators
‘ First order integrator:
R
C
1
)
(s
G
g C
1
)
(
+
=
RCs
s
G
R
C
‘ First order differentiator:
1
)
( =
RC
RCs
s
G
1
)
(
+
RCs

Transfer function of some type of controllers (cont’)
‘ Phase lead compensator:
C
R 1
+
Ts
α
R1
R2
1
1
)
(
+
+
=
Ts
Ts
K
s
G C
α
R R
R
2
1
2
R
R
R
KC
+
=
2
1
1
2
R
R
C
R
R
T
+
= 1
2
2
1
>
+
=
R
R
R
α
‘ Phase lag compensator : R1
R2 1
)
(
+
=
Ts
K
s
G C
α
C
1
)
(
+
Ts
C
1
1
<
R
α
1
=
C
K C
R
R
T )
( 2
1 +
=
1
2
1
<
+
=
R
R
α

Active controllers
Active controllers
K
G )
(
‘ Proportional Controller (P)
P
K
s
G =
)
(
2
R
KP −
=
1
R
KP
‘ Proportional Integral controller (PI)
p g ( )
s
K
K
s
G I
P +
=
)
(
s
2
R
KP −
=
C
R
KI
1
−
=
1
R
P
C
R
I
1

Active controllers
Active controllers
‘ Proportional Derivative controller (PD)
s
K
K
s
G D
P +
=
)
(
2
R
K C
R
K
‘ Proportional Integral Derivative controller (PID)
1
2
R
KP −
= C
R
KD 2
−
=
p g ( )
s
K
s
K
K
s
G D
I
P +
+
=
)
(
2
1
2
2
1
1
C
R
C
R
C
R
KP
+
−
=
2
1
1
C
R
KI −
=
1
2C
R
KD −
=

Transfer function of DC motors
Equivalent diagram of a DC motor
Equivalent diagram of a DC motor
− La : armature induction − ω : motor speed
R t i t M l d i ti
− Ra : armature resistance − Mt : load inertia
− Ua : armature voltage − B : friction constant
− Ea : back electromotive force − J : moment of inertia of the rotor

‘ Applying Kirchhoff's law for the armature circuit:
)
(
)
(
).
(
)
( t
E
dt
t
di
L
R
t
i
t
U a
a
a
a
a
a +
+
= (1)
dt
)
(
)
( t
K
t
E ω
Φ
=
a
where:
K : electromotive force constant
(2)
Φ : excitation magnetic flux
‘ Applying Newton’s law for the rotating part of the motor:
t
d
J
t
B
t
M
t
M
)
(
)
(
)
(
)
(
ω (3)
dt
J
t
B
t
M
t
M L
)
(
)
(
)
(
)
( ω +
+
=
where: )
(
)
( t
i
K
t
M a
Φ
=
(3)
(4)

T ki h L l f f (1) (2) (3) (4) l d
‘ Taking the Laplace transform of (1), (2), (3), (4) leads to:
(5)
)
(
)
(
).
(
)
( s
E
s
sI
L
R
s
I
s
U a
a
a
a
a
a +
+
=
(6)
(7)
)
(
)
( s
K
s
E ω
Φ
=
a
)
(
)
(
)
(
)
( s
Js
s
B
s
M
s
M L ω
ω +
+
=
(8)
)
(
)
( s
i
K
s
M a
Φ
=
‘ Denote:
a
a
L
T = Electromagnetic time constant
a
a
R
B
J
Tc = Mechanical time constant
B

‘ From (5) and (7) we have:
‘ From (5) and (7), we have:
)
1
(
)
(
)
(
)
(
s
T
R
s
E
s
U
s
I a
a
a
+
−
= (5’)
)
1
( s
T
R a
a +
)
1
(
)
(
)
(
)
(
s
T
B
s
M
s
M
s L
+
−
=
ω
)
1
( s
T
B c
+
‘ From (5’), (6), (7’) and (8) we can develop the block diagram of
the DC motor as follow:
the DC motor as follow:
)
(s
Ua
)
(s
ML
R
/
1
)
(
a
)
(s
Ea
a
sT
R
+
1
/
1 a

Transfer function of a thermal process
Transfer function of a thermal process
Temperature of
th
u(t) y(t)
Electric power supplying
t th 100% the oven
to the oven 100%
(t) (t)
y(t) y(t)
“Exact” characteristic of the oven Approximate characteristic of the oven

Transfer function of a thermal process (cont’)
Transfer function of a thermal process (cont’)
‘ The approximate transfer function of the thermal )
(s
Y
‘ The approximate transfer function of the thermal
process can be calculated by using the equation: )
(
)
(
)
(
s
U
s
Y
s
G =
1
‘ The input is the unit step signal, then
s
s
U
1
)
( =
‘ The approximate output is: )
(
)
( 1
T
t
f
t
y −
=
‘ The approximate output is: )
(
)
( 1
T
t
f
t
y
where: )
1
(
)
( 2
/T
t
e
K
t
f −
−
=
The Laplace transform of f (t) is: )
(
K
F
The Laplace transform of f (t) is:
)
1
(
)
(
2s
T
s
s
F
+
=
)
(
1
Ke
s
Y
s
T
=
−
Applying the time delay theorem:
)
1
(
)
(
2s
T
s
s
Y
+
=
Applying the time delay theorem:
)
(
)
(
1
−
Ke
s
Y
G
s
T
1
)
(
)
(
)
(
2 +
=
=
s
T
s
U
s
G
⇒

Transfer function of a car
Transfer function of a car
M: car mass
B: friction constant
f(t): driving force
v(t): car speed
)
(
)
(
)
(
t
f
t
Bv
dt
t
dv
M =
+
‘ Differential equation:
dt
B
M
F
s
V
s
G =
=
1
)
(
)
(
)
( ⇔
1
)
( =
T
K
s
G
B
Ms
s
F +
)
( 1
+
Ts
where K
1
=
M
T =
B
K
B
T

Transfer function of an suspension system
Transfer function of an suspension system
M: equivalent car mass
q
K: spring stiffness
f(t): external force
f(t): external force
y(t): travel of car body
‘ Differential equation: )
(
)
(
)
(
)
(
2
2
t
f
t
Ky
dt
t
dy
B
dt
t
y
d
M =
+
+
dt
dt
‘ Transfer function s
Y
s
G =
=
1
)
(
)
(
K
Bs
Ms
s
F
s
G
+
+
=
= 2
)
(
)
(

Transfer functions of sensors
Transfer functions of sensors
y(t) yfb(t)
Sensor
y(t) yfb(t)
‘ Feedback signal yfb(t) is proportional to y(t), so transfer functions of
g yfb( ) p p y( ),
sensors are usually constant:
fb
K
s
H =
)
(
‘ Ex: Suppose that temperature of a furnace changing in the range y(t)
= 0÷5000C, if a sensor converts the temperature to a voltage in the
( ) 0 5V h h f f i f h i
range yfb(t) 0÷5V, then the transfer function of the sensor is:
)
/
(
01
.
0
)
(
500
/
)
(
5
)
( 0
0
C
V
C
V
K
s
H fb =
=
=
‘ If the sensor has a delay time, then the transfer function of the sensor
is: K
s
H fb
=
)
(
s
T
s
H
fb
+
=
1
)
(

Transfer functions of control systems
Transfer functions of control systems

Block diagram
Block diagram
‘ Block diagram is a diagram of a system in which the principal parts
‘ 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:
Ž Function block
Ž Summing point
Ž Summing point
Ž Pickoff point
Function block Pickoff point
Summing point

Block diagram algebra
Transfer function of systems in series
Transfer function of systems in series
G
U1 (s) Y1 (s)
G G
Un (s) Yn (s)
U(s) Y(s)
G1
1 ( )
G2
U2(s) Y2 (s)
Gn
…
( ) ( )
Gs
U(s) Y (s)
∏
=
n
i
s s
G
s
G )
(
)
(
∏
=
i 1

Block diagram algebra (cont’)
Transfer function of systems in parallel
Transfer function of systems in parallel
U (s) Y ( )
G1
U1 (s) Y1 (s)
U (s) Y2 (s)
U(s) Y(s)
G2
U2(s) Y2 (s)
…
U(s) ( )
Gp
U(s) Y (s)
Gn
Un (s) Yn (s)
∑
=
n
i s
G
s
G )
(
)
(
∑
=
i
i
p s
G
s
G
1
)
(
)
(

Transfer function of feedback systems
‘ Negative feedback ‘ Unity negative feedback
Y(s)
R(s)
− G(s)
E(s)
+
Y(s)
R(s)
− G(s)
E(s)
+
H(s)
Yht(s) Yht(s)
)
(
)
(
s
G
s
G l =
)
(
)
(
s
G
s
G l =
)
(
).
(
1
)
(
s
H
s
G
s
Gcl
+ )
(
1
)
(
s
G
s
Gcl
+

‘ Positive feedback ‘ Unity positive feedback
Y(s)
R(s)
+ G(s)
E(s)
Y ( )
+
Y(s)
R(s)
+ G(s)
E(s)
Y ( )
+
H(s)
Yfb(s) Yfb(s)
)
(
)
(
1
)
(
)
(
s
H
s
G
s
G
s
Gcl
−
=
)
(
1
)
(
)
(
s
G
s
G
s
Gcl
−
=
)
(
).
(
1 s
H
s
G )
(
1 s
G

Transfer function of multi loop systems
Transfer function of multi-loop systems
‘ For a complex system consisting of multi feedback loops, we perform
equivalent block diagram transformation so that simple connecting
q g p g
blocks appears, and then we simplify the block diagram from the inner
loops to the outer loops.
‘ Two block diagrams are equivalent if their input-output relationship
are the same

Moving a pickoff point behind a block
Moving a pickoff point behind a block

Moving a pickoff point ahead a block
Moving a pickoff point ahead a block

Moving a summing point behind a block

M i i i t h d bl k
Moving a summing point ahead a block

Interchanging the positions of the
two consecutive summing points

Splitting a summing point

Note
Note
‘ Do not interchange the positions of a pickoff point and a summing
point :
p
‘ Do not interchange the positions of
‘ Do not interchange the positions of
two summing points if there exists
a pickoff point between them:

Block diagram algebra –
– Example 1
Example 1
‘ Find the equivalent transfer function of the following system:
Y(s)

– Example 1 (cont’)
Example 1 (cont’)
I h i h i i c d d
‘ Interchanging the summing points c and d,
Eliminating GA(s)=[G3(s)//G4(s)]
Y(s)
)
(
)
(
)
( 4
3 s
G
s
G
s
GA −
=

Example 1 (cont’)
‘ G ( ) [G ( ) // unity block]
‘ GB(s)=[G1(s) // unity block] ,
GC (s)= feedback loop[G2(s),GA(s)]:
Y( )
)
(
1
)
( s
G
s
G +
=
Y(s)
)
(
1
)
( 1 s
G
s
GB +
=
)]
(
)
(
) [
(
1
)
(
)
(
)
(
1
)
(
)
( 2
2
s
G
s
G
s
G
s
G
s
G
s
G
s
G
s
GC
−
+
=
+
=
)]
(
)
(
).[
(
1
)
(
).
(
1 4
3
2
2 s
G
s
G
s
G
s
G
s
G A +
+
‘ Equivalent transfer function of the system:
)
(
).
(
)
( s
G
s
G
s
G C
B
eq =
)
(
)].
(
1
[
)
( 2
1 s
G
s
G
s
G
+
=
)]
(
)
(
).[
(
1
)
(
4
3
2 s
G
s
G
s
G
s
Geq
−
+
=

– Example 2
Example 2
Y(s)

Example 2 (cont’)
‘ Interchanging the positions of the summing points d and e
Moving the pickoff point f behind the block G2(s)
Y(s)

Example 2 (cont’)
‘ GB(s) = feedback loop [G2(s), H2(s)]
GC(s) = [GA(s)// unity block]
GC(s) [GA(s)// unity block]
Y(s)
Y(s)

Example 2 (cont’)
‘ G ( ) d (G ( ) G ( ) G ( ))
‘ GD(s) = cascade(GB (s), GC(s), G3(s))
Y(s)
‘ GE(s) = feedback loop(GD(s), H3(s))
Y(s)

Example 2 (cont’)
‘ Detailed calculation:
‘ Detailed calculation:
1
*
H
GA =
2
G
GA
2
*
G
G =
2
2
1 H
G
GB
+
=
H
G
H +
2
1
2
2
1
1
1
*
G
H
G
G
H
G
G A
C
+
=
+
=
+
=
2
2
1
3
3
2
3
2
1
2
2
2
2
3
1
1
.
.
*
H
G
H
G
G
G
G
G
H
G
H
G
G
G
G
G
G C
B
D
+
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
=

Example 2 (cont’)
H
G
G
G +
3
1
3
3
3
2
2
2
1
3
3
2
1
3
3
2
2
2
1
3
3
2
3 1
1
1
1
*
H
H
G
H
G
G
H
G
H
G
G
G
H
H
G
G
G
H
G
H
G
G
G
H
G
G
G
D
D
E
+
+
+
+
=
+
+
+
+
=
+
=
3
1
3
3
3
2
2
2
3
2
2
3
1
1 H
H
G
D
+
+
‘ E i l t t f f ti f th t
1
3
3
2 H
G
G
G
G
+
‘ Equivalent transfer function of the system:
1
3
3
2
1
3
1
3
3
3
2
2
2
1
1
1
1
.
1
1
.
1
*
H
H
G
H
G
G
H
G
H
G
G
G
G
H
H
G
H
G
G
H
G
G
G
G
G
G
G
E
E
eq
+
+
+
+
+
+
+
+
=
+
=
3
1
3
3
3
2
2
2
1 H
H
G
H
G
G
H
G +
+
+
1
3
1
3
2
1
1 H
G
G
G
G
G
H
H
G
H
G
G
H
G
H
G
G
G
G
G
Geq
+
=
⇒
1
3
1
3
2
1
3
1
3
3
3
2
2
2
1 H
G
G
G
G
G
H
H
G
H
G
G
H
G
eq
+
+
+
+
+

– Example 3
Example 3
Y(s)

Example 3 (cont’)
Hint to solve example 3
‘ Move the summing point e ahead the block G1(s),
then interchange the position of the summing points d ande
g p g p
Move the pickoff point f behind the block G2(s)
Y(s)

Example 3 (cont’)
S l ti t E l 3
S l ti t E l 3
Solution to Example 3
Solution to Example 3
‘ Students calculate the equivalent transfer function themselves using
the hints in the previous slide.
the hints in the previous slide.

Remarks on block diagram algebra
Remarks on block diagram algebra
‘ Bl k di l b i l ti l i l th d t l l t th
‘ 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 procedure to perform the block diagram transformation;
each particular block diagram can be transformed by different
heuristic ways.
heuristic ways.
‘ When calculating the equivalent transfer function, it is necessary to
manipulate many calculations on algebraic fractions. This could be a
potential source of error if the system is complex enough
potential source of error if the system is complex enough.
⇒ Block diagram algebra is only appropriate for finding equivalent
transfer function of simple systems.
To find equivalent transfer function of complex systems, signal flow
graph method (to be discussed later) is more effective.

Definition of signal flow graph
Definition of signal flow graph
Y(s)
Y(s) Y(s)
Y(s)
‘ Si l fl h t k i ti f d d b h
Block diagram Signal flow graph
‘ Signal flow graph: a networks consisting of nodes and branches.
‘ Node: a point representing a signal or a variable in the system.
‘ Branch: a line directly connecting two nodes, each branch has an
arrow showing the signal direction and a transfer function
representing the relationship between the signal at the two nodes of
the branch
the branch
‘ Source node: a node from which there are only out-going branches.
‘ Sink node: a node to which there are only in-going branches.
‘ Hybrid node: a node which both has in going branches and out
‘ Hybrid node: a node which both has in-going branches and out-
going branches.

Definition of signal flow graph (cont’)
Definition of signal flow graph (cont’)
‘ Forward path: is a path consisting of continuous sequence of
‘ Forward path: is a path consisting of continuous sequence of
branches that goes in the same direction from a source node to a sink
node without passing any single node more than once.
‘ Path gain is the product of all transfer functions of the branches
‘ Path gain is the product of all transfer functions of the branches
belonged to the path.
‘ Loop: is a closed path consisting of continuous sequence of
branches that goes in the same direction without passing any single
node more than once.
Loop gain is the product of all transfer functions of the branches
Loop gain is the product of all transfer functions of the branches
belonged to the loop.
Y(s) Y(s)
Y(s)
Loop
Forward path

Mason’s formula
Mason’s formula
‘ The equivalent transfer function from a source 1
‘ The equivalent transfer function from a source
node to a sink node of a system can be found
by using the Mason’s formula:
∑Δ
Δ
=
k
k
k P
G
1
ƒ Pk: is the gain of kth forward path from the considered source node
to the considered drain node.
ƒ Δ: is the determinant of the signal flow graph
K
+
−
+
−
=
Δ ∑
∑
∑
g
nontouchin
m
j
i
m
j
i
g
nontouchin
j
i
j
i
i
i L
L
L
L
L
L
,
,
,
1
ƒ Δ: is the determinant of the signal flow graph.
g
nontouchin
g
nontouchin
i
L : is the gain of the ith loop
ƒ Δk: is the cofactor of the kth path .
Δk is inferred from Δ 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.

Signal flow graph
Signal flow graph –
– Example 1
Example 1
‘ Find the equivalent transfer function of the system described by the
following signal flow graph:
Y(s)
R(s)
‘ Solution
‘ Solution
Ž Forward paths: Ž Loop:
G
G
G
G
G
P = 1
4
1 H
G
L −
=
5
4
3
2
1
1 G
G
G
G
G
P =
5
4
6
1
2 G
G
G
G
P =
7
2
1
3 G
G
G
P =
1
4
1 H
G
L
2
7
2
2 H
G
G
L −
=
2
5
4
6
3 H
G
G
G
L −
=
7
2
1
3 G
G
G
P
2
5
4
3
2
4 H
G
G
G
G
L −
=

Signal flow graph
Example 1 (cont’)
‘ The determinant of the SFG:
2
1
4
3
2
1 )
(
1 L
L
L
L
L
L +
+
+
+
−
=
Δ
‘ The cofactors of the paths
1
1 =
Δ
1
2 =
Δ
1
3 1 L
−
=
Δ
‘ The equivalent transfer function of the system:
)
(
1
Δ
+
Δ
+
Δ P
P
P
G )
( 3
3
2
2
1
1 Δ
+
Δ
+
Δ
Δ
= P
P
P
Geq
1
4
7
2
1
5
4
6
1
5
4
3
2
1 )
1
( H
G
G
G
G
G
G
G
G
G
G
G
G
G
G
+
+
+
=
2
7
2
1
4
2
5
4
3
2
2
5
4
6
2
7
2
1
4
1 H
G
G
H
G
H
G
G
G
G
H
G
G
G
H
G
G
H
G
Geq
+
+
+
+
+
=

Signal flow graph
– Example 2
Example 2
following block diagram:
Y(s)
R(s)
‘ Solution:
Y(s)
R(s)

Signal flow graph
Example 2 (cont’)
Y(s)
R(s)
Ž Forward paths: Ž Loop
3
2
1
1 G
G
G
P = 2
2
1 H
G
L −
=
H
G
G
L
3
1
1
2 G
H
G
P = 3
3
2
2 H
G
G
L −
=
3
2
1
3 G
G
G
L −
=
3
1
3
4 H
H
G
L −
=
3
1
3
4 H
H
G
L
1
3
1
5 H
G
G
L −
=

Signal flow graph
Example 2 (cont’)
Th d i f h SFG
)
(
1 5
4
3
2
1 L
L
L
L
L +
+
+
+
−
=
Δ
‘ The cofactors of the paths:
1
1 =
Δ1
1
2 =
Δ
Th i l f f i f h
‘ The equivalent transfer function of the system:
)
(
1
2
2
1
1 Δ
+
Δ
Δ
= P
P
Geq
Δ
1
3
1
3
1
3
3
2
1
3
3
2
2
2
1
3
1
3
2
1
1 H
G
G
H
H
G
G
G
G
H
G
G
H
G
H
G
G
G
G
G
Geq
+
+
+
+
+
+
=
1
3
1
3
1
3
3
2
1
3
3
2
2
2
1 H
G
G
H
H
G
G
G
G
H
G
G
H
G +
+
+
+
+

Signal flow graph
– Example 3
Example 3
q y y
following block diagram:
Y(s)
‘ Solution:
Y(s)

Signal flow graph
Example 3 (cont’)
Y(s)
3
2
1
1 G
G
G
P = 2
1
1 H
G
L −
=
H
G
G
L
Ž Forward path Ž Loop
4
2 G
P = 1
2
1
2 H
G
G
L −
=
3
2
1
3 G
G
G
L −
=
3
3
2
4 H
G
G
L −
=
3
3
2
4 H
G
G
L
4
5 G
L −
=

Signal flow graph
Example 3 (cont’)
‘ Determinant of the SFG:
‘ Determinant of the SFG:
5
4
1
5
4
5
2
5
1
4
1
5
4
3
2
1 )
(
)
(
1 L
L
L
L
L
L
L
L
L
L
L
L
L
L
L
L −
+
+
+
+
+
+
+
+
−
=
Δ
‘ The cofactor:
1
1 =
Δ
‘ The equivalent transer function of the system:
)
(
)
(
1 4
1
4
2
1
2 L
L
L
L
L +
+
+
−
=
Δ
‘ The equivalent transer function of the system:
)
(
1
2
2
1
1 Δ
+
Δ
Δ
= P
P
G
Num
Den

State space equations
State space equations

‘ State: The state of a system is a set of variables whose values
State of a system
State of a system
‘ 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 the
state vector.
[ ]T
n
x
x
x K
2
1
=
x

State equations
State equations
‘ By using state variables, we can transform the n-order differential
y g ,
equation describing the system dynamics into a set of n first order
differential equations (called state equations) of the form:
⎧
⎩
⎨
⎧
=
+
=
)
(
)
(
)
(
)
(
)
(
t
t
y
t
u
t
t
Cx
B
Ax
x
&
where
⎥
⎥
⎤
⎢
⎢
⎡ n
a
a
a
a
a
a
K
K
2
22
21
1
12
11
⎥
⎥
⎤
⎢
⎢
⎡
b
b
2
1
[ ]
where
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣
=
nn
n
n
n
a
a
a
a
a
a
K
M
M
M
K
2
1
2
22
21
A
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣
=
n
b
b
M
2
B [ ]
n
c
c
c K
2
1
=
C
‘ Note: Depending on how we chose the state variables, a system can
be described by many different state equations.
y y q

State equations
State equations –
– Example 1
Example 1
A suspension system
A suspension system
A suspension system
A suspension system
)
(
)
(
)
(
)
(
2
t
f
t
Ky
t
dy
B
t
y
d
M =
+
+
(*)
)
(
)
(
2
t
f
t
Ky
dt
B
dt
M =
+
+ ( )
⎪
⎨
⎧ =
1
)
(
)
( 2
1
B
K
t
x
t
x
&
‘ Denote:
⎨
⎧ = )
(
)
(
1 t
y
t
x
⎪
⎩
⎪
⎨ +
−
−
= )
(
1
)
(
)
(
)
( 2
1
2 t
f
M
t
x
M
B
t
x
M
K
t
x
&
⇒
⎩
⎨
= )
(
)
(
)
(
)
(
2
1
t
y
t
x
y
&
1
0
)
(
1
0
)
( 1
1 t
x
B
K
t
x
⎥
⎤
⎢
⎡
⎤
⎡
⎥
⎤
⎢
⎡
⎤
⎡ &
)
(
1
)
(
)
(
.
)
(
)
(
2
1
2
1
t
f
M
t
x
M
B
M
K
t
x ⎥
⎥
⎦
⎢
⎢
⎣
+
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎥
⎦
⎢
⎢
⎣
−
−
=
⎥
⎦
⎤
⎢
⎣
⎡
&
[ ] ⎤
⎡ )
(
1 t
x
⇔
[ ] ⎥
⎦
⎤
⎢
⎣
⎡
=
)
(
)
(
0
1
)
(
2
1
t
x
t
x
t
y
⎨
⎧ +
= )
(
)
(
)
( t
f
t
t B
Ax
x
&
⇔ ⎥
⎤
⎢
⎡
= B
K
1
0
A ⎥
⎤
⎢
⎡
= 1
0
B [ ]
0
1
=
C
⎩
⎨
= )
(
)
(
)
(
)
(
)
(
t
t
y
f
Cx
⇔ ⎥
⎥
⎦
⎢
⎢
⎣
−
−
M
M
A
⎥
⎥
⎦
⎢
⎢
⎣M
B [ ]

State equations
State equations –
– Example 2
Example 2
DC motor
DC motor
− La : armature induction − ω : motor speed
R t i t M l d i ti
− Ra : armature resistance − Mt : load inertia
− Ua : armature voltage − B : friction constant
− Ea : back electromotive force − J : moment of inertia of the rotor

State equations
State equations –
Example 2 (cont’)
‘ Applying Kirchhoff's law for the armature circuit:
)
(
)
(
).
(
)
( t
E
dt
t
di
L
R
t
i
t
U a
a
a
a
a
a +
+
= (1)
dt
)
(
)
( t
K
t
E ω
Φ
=
a
where:
K : electromotive force constant
(2)
Φ : excitation magnetic flux
‘ Applying Newton’s law for the rotating part of the motor:
‘ Applying Newton s law for the rotating part of the motor:
(for simplicity, assuming that load torque is zero)
t
d
J
t
B
t
M
)
(
)
(
)
(
ω (3)
dt
J
t
B
t
M
)
(
)
(
)
( ω +
=
where: )
(
)
( t
i
K
t
M a
Φ
=
(3)
(4)

State equations
State equations –
Example 2 (cont’)
‘ (1) & (2) ⇒ )
(
1
)
(
)
(
)
(
t
U
L
t
L
K
t
i
L
R
dt
t
di
ö
ö
ö
ö
ö
ö
ö
+
Φ
−
−
= ω (5)
L
L
L
dt ö
ö
ö
‘ (3) & (4) ⇒ )
(
)
(
)
(
t
J
B
t
i
J
K
dt
t
d
ω
ω
−
Φ
= ö
(6)
‘ Denote:
⎩
⎨
⎧
=
=
)
(
)
(
)
(
)
(
2
1
t
t
x
t
i
t
x
ω
ö
⎩
⎪
⎪
⎧
+
Φ
−
−
= )
(
1
)
(
)
(
)
( 2
1
1 t
U
L
t
x
L
K
t
x
L
R
t
x
& ö
ö
‘ (5) & (6) ⇒
⎪
⎪
⎩
⎪
⎪
⎨
−
Φ
= )
(
)
(
)
( 2
1
2 t
x
J
B
t
x
J
K
t
x
L
L
L
&
ö
ö
ö

State equations
State equations –
Example 2 (cont’)
)
(
1
)
(
)
( 1
1
t
U
L
t
x
L
K
L
R
t
x
ö
ö
ö
⎥
⎤
⎢
⎡
+
⎥
⎤
⎢
⎡
⎥
⎥
⎤
⎢
⎢
⎡ Φ
−
−
=
⎥
⎤
⎢
⎡ &
⎤
⎡ )
(
⇔
)
(
0
)
(
)
( 2
2
t
U
L
t
x
J
B
J
K
t
x
ö
ö
ö
ö
⎥
⎥
⎦
⎢
⎢
⎣
+
⎥
⎦
⎢
⎣
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣
−
Φ
=
⎥
⎦
⎢
⎣ &
[ ] ⎥
⎦
⎤
⎢
⎣
⎡
=
)
(
)
(
1
0
)
(
2
1
t
x
t
x
t
ω
⎩
⎨
⎧
=
+
=
)
(
)
(
)
(
)
(
)
(
t
t
t
U
t
t
Cx
B
Ax
x
ω
u
&
⇔
⎤
⎡
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
Φ
Φ
−
−
=
B
K
L
K
L
R
ö
ö
ö
A [ ]
1
0
=
C
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
=
0
1
ö
L
B
where:
⎥
⎥
⎦
⎢
⎢
⎣
−
J
J
⎥
⎦
⎢
⎣ 0

Method for establishing state equations from differential equations
Case #1: The differential equation
do not involve the input derivatives
do not involve the input derivatives
‘ The differential equation describing the system dynamics is:
)
(
)
(
)
(
)
(
)
(
0
1
1
1
1
0 t
u
b
t
y
a
dt
t
dy
a
dt
t
y
d
a
dt
t
y
d
a n
n
n
n
n
n
=
+
+
+
+ −
−
−
L
q g y y
)
(
)
(
1 t
y
t
x =
‘ Define the state variables as follow:
Ž The first state is the system output:
)
(
)
(
)
(
)
(
)
(
)
(
2
3
1
2
1
t
x
t
x
t
x
t
x
t
y
t
x
=
=
&
&
Ž The i th state (i=2..n) is chosen to be the
first derivative of the (i−1)th state :
)
(
)
(
)
(
)
(
1
2
3
t
x
t
x n
n −
= &
M

Case #1 (cont’)
Case #1 (cont’)
Case #1 (cont )
Case #1 (cont )
‘ State equation:
⎩
⎨
⎧
=
+
=
)
(
)
(
)
(
)
(
)
(
t
t
y
t
u
t
t
Cx
B
Ax
x
&
⎩ )
(
)
( t
t
y Cx
where:
⎤
⎡ ⎤
⎡ 0
0
1
0 ⎤
⎡ 0
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
)
(
)
(
)
(
2
1
t
x
t
x
M ⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
0
1
0
0
0
0
1
0
M
M
M
M
K
K
A ⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
0
0
M
B
⎥
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎢
⎣
=
−
)
(
)
(
)
(
1
t
x
t
x
t
n
M
x
⎥
⎥
⎥
⎥
⎥
⎢
⎢
⎢
⎢
⎢
−
−
−
−
=
−
− 1
2
1
1
0
0
0
a
a
a
a n
n
n
K
K
M
M
M
M
A
⎥
⎥
⎥
⎥
⎥
⎢
⎢
⎢
⎢
⎢
=
0
0
b
M
B
⎥
⎦
⎢
⎣ )
(t
xn ⎥
⎦
⎢
⎣ 0
0
0
0 a
a
a
a ⎥
⎦
⎢
⎣ 0
a
[ ]
0
0
0
1 K
=
C

Case #1: Example
Case #1: Example
Case #1: Example
Case #1: Example
‘ Write the state equations describing the following system:
)
(
)
(
10
)
(
6
)
(
5
)
(
2 t
u
t
y
t
y
t
y
t
y =
+
+
+ &
&
&
&
&
&
⎪
⎪
⎨
⎧
=
=
)
(
)
(
)
(
)
(
1
2
1
t
x
t
x
t
y
t
x
&
‘ Define the state variables as:
⎪
⎩ = )
(
)
( 2
3 t
x
t
x &
‘ State equation:
⎩
⎨
⎧ +
=
)
(
)
(
)
(
)
(
)
( t
r
t
t
C
B
Ax
x
&
⎤
⎡
⎤
⎡
⎥
⎥
⎤
⎢
⎢
⎡
=
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
= 0
0
0
0
B
⎩
⎨
= )
(
)
( t
t
y Cx
where
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎢
⎡
=
5
2
3
5
1
0
0
0
1
0
1
0
0
0
1
0
1
2
3 a
a
a
A
⎥
⎥
⎦
⎢
⎢
⎣
=
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣
=
5
.
0
0
0
0
0
a
b
B
⎥
⎦
⎢
⎣ −
−
−
⎥
⎥
⎦
⎢
⎢
⎣
−
−
− 5
.
2
3
5
0
1
0
2
0
3
a
a
a [ ]
0
0
1
=
C

Case #2: The differential equation involve the input derivatives
−
)
(
)
(
)
(
)
( 1
t
dy
t
y
d
t
y
d n
n
=
+
+
+
+ −
−
)
(
)
(
)
(
)
(
1
1
1
0 t
y
a
dt
t
dy
a
dt
t
y
d
a
dt
t
y
d
a n
n
n
n
L
)
(
)
(
)
(
)
(
1
2
2
1
1
0 t
u
b
t
du
b
t
u
d
b
t
u
d
b
n
n −
−
+
+
+
+ L )
(
1
2
1
1
1
0 t
u
b
dt
b
dt
b
dt
b n
n
n
n −
−
−
−
+
+
+
+
‘ Define the state variables as follow:
Ž The i th state (i=2 n) is equal to the )
(
)
(
)
(
)
(
)
(
1
1
2
1
t
r
t
x
t
x
t
y
t
x
−
=
=
β
&
Ž The i state (i 2..n) is equal to the
first derivative of the (i−1)th state
minus a quantity proportional to the
i t
)
(
)
(
)
( 2
2
3
1
1
2
t
r
t
x
t
x −
= β
M
&
input:
)
(
)
(
)
( 1
1 t
r
t
x
t
x n
n
n −
− −
= β
&

Case #2 (cont’)
Case #2 (cont’)
⎩
⎨
⎧
=
+
=
)
(
)
(
)
(
)
(
)
(
t
t
y
t
r
t
t
Cx
B
Ax
x
&
Case #2 (cont )
Case #2 (cont )
‘ State equation:
⎩ = )
(
)
( t
t
y Cx
where:
⎤
⎡ 0
0
1
0
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
)
(
)
(
2
1
t
x
t
x
M ⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
0
1
0
0
0
0
1
0
M
M
M
M
K
K
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
β
β
2
1
⎥
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎢
⎣
=
−
)
(
)
(
)
(
1
t
t
x
t
n
M
x
⎥
⎥
⎥
⎥
⎢
⎢
⎢
⎢
−
−
−
−
=
−
− 1
2
1
1
0
0
0
a
a
a
a n
n
n
K
M
M
M
M
A
⎥
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎢
⎣
=
−
n
β
β 1
M
B
⎥
⎦
⎢
⎣ )
(t
xn ⎥
⎥
⎦
⎢
⎢
⎣ 0
0
0
0 a
a
a
a
K
[ ]
C
⎥
⎦
⎢
⎣ n
β
[ ]
0
0
0
1 K
=
C

Case #2 (cont’)
Case #2 (cont’)
Case #2 (cont )
Case #2 (cont )
The coefficients β in the vector B are calculated as follow:
b
1
1
1
0
0
1
a
b
a
b
β
β
−
=
1
2
2
1
2
0
1
1
1
2
a
a
b
a
a
b
β
β
β
β
β
−
−
=
0
1
2
2
1
2
3
a
β
β
β =
M
0
1
1
2
2
1
1
1
a
a
a
a
b n
n
n
n
n
β
β
β
β −
−
−
− −
−
−
−
=
K

Case #2: Example
Case #2: Example
Case #2: Example
Case #2: Example
‘ Write the state equations describing the following system:
)
(
20
)
(
10
)
(
10
)
(
6
)
(
5
)
(
2 t
u
t
u
t
y
t
y
t
y
t
y +
=
+
+
+ &
&
&
&
&
&
&
‘ Define the state variables:
⎪
⎩
⎪
⎨
⎧
−
=
=
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
1
1
2
1
t
r
t
x
t
x
t
y
t
x
β
β
&
&
‘ The state equation:
⎩
⎨
⎧ +
=
)
(
)
(
)
(
)
(
)
( t
r
t
t
C
B
Ax
x
&
⎪
⎩ −
= )
(
)
(
)
( 2
2
3 t
r
t
x
t
x β
&
⎩
⎨
= )
(
)
( t
t
y Cx
⎤
⎡
⎤
⎡
where:
⎥
⎤
⎢
⎡ 1
β
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎢
⎡
=
5
2
3
5
1
0
0
0
1
0
1
0
0
0
1
0
1
2
3 a
a
a
A ⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣
=
3
2
β
β
B
[ ]
0
0
1
=
C
⎥
⎦
⎢
⎣ −
−
−
⎥
⎥
⎦
⎢
⎢
⎣
−
−
− 5
.
2
3
5
0
1
0
2
0
3
a
a
a

Case #2: Example (cont’))
Case #2: Example (cont’))
Case #2: Example (cont ))
Case #2: Example (cont ))
‘ The elements of vector B are calculated as follow:
⎪
⎪
⎪
⎧
×
=
=
=
0
5
10
0
2
0
0
0
1
b
a
b
β
β
⎪
⎪
⎪
⎪
⎨
−
=
×
−
×
−
=
−
−
=
=
×
−
=
−
=
15
0
6
10
5
20
5
2
0
5
10
1
2
2
1
2
0
1
1
1
2
a
a
b
a
a
b
β
β
β
β
β
⎪
⎩
=
=
= 15
2
0
3
a
β
⎤
⎡ 0
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
=
15
5
0
B
⇒
⎦
⎣

State
State-
-space equations in controllable canonical form
space equations in controllable canonical form
=
+
+
+
+ −
−
−
)
(
)
(
)
(
)
(
1
1
1
1
0 t
y
a
dt
t
dy
a
dt
t
y
d
a
dt
t
y
d
a n
n
n
n
n
n
L
dt
dt
dt
)
(
)
(
)
(
)
(
1
1
1
1
0 t
u
b
dt
t
du
b
dt
t
u
d
b
dt
t
u
d
b m
m
m
m
m
m
+
+
+
+ −
−
−
L
or equivalently by the transfer function:
m
m
m
m
b
s
b
s
b
s
b
G
+
+
+
+ −
1
1
1
0 ...
)
(
n
n
n
n
m
m
a
s
a
s
a
s
a
s
G
+
+
+
+
=
−
−
−
1
1
1
0
1
1
0
...
)
(
‘ Th t ll bl i l t t ti f th t i t d
‘ The controllable canonical state equations of the system is presented
in the next slide.

State
State-
-space equations in controllable canonical form (cont’)
space equations in controllable canonical form (cont’)
⎧ )
(
)
(
)
( t
t
t B
A
&
⎩
⎨
⎧
=
+
=
)
(
)
(
)
(
)
(
)
(
t
t
y
t
r
t
t
Cx
B
Ax
x
‘ State equations:
Where:
⎥
⎤
⎢
⎡ 0
0
1
0 K
⎥
⎤
⎢
⎡0
⎤
⎡ )
(t
⎥
⎥
⎥
⎥
⎢
⎢
⎢
⎢
=
0
1
0
0
M
M
M
M
K
A
⎥
⎥
⎥
⎥
⎢
⎢
⎢
⎢
=
0
M
B
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
=
)
(
)
(
)
( 2
1
t
x
t
x
t
M
x
⎥
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎢
⎣
−
−
−
− −
−
0
1
0
2
0
1
0
1
0
0
0
a
a
a
a
a
a
a
a n
n
n
K
K
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣1
0
⎥
⎥
⎦
⎢
⎢
⎣ )
(t
xn
M
⎦
⎣ 0
0
0
0
⎥
⎤
⎢
⎡
= −
0
0
0
1 b
b
b m
m
C
⎥
⎦
⎢
⎣
0
0
0
0
0
K
K
a
a
a
C

State
State-
-space equations in controllable canonical form (cont’)
space equations in controllable canonical form (cont’)
‘ Write the controllable canonical state equations of the following system:
‘ Write the controllable canonical state equations of the following system:
)
(
3
)
(
)
(
4
)
(
5
)
(
)
(
2 t
u
t
u
t
y
t
y
t
y
t
y +
=
+
+
+ &
&
&
&
&
&
&
&
S l i
‘ Solution:
⎩
⎨
⎧
=
+
=
)
(
)
(
)
(
)
(
)
(
t
t
y
t
r
t
t
Cx
B
Ax
x
⎤
⎡
⎥
⎤
⎢
⎡
0
1
0
0
1
0 ⎤
⎡0
where:
⎩ )
(
)
(
y
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
=
⎥
⎥
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎢
⎢
⎣
−
−
−
=
5
.
0
5
.
2
2
1
0
0
0
1
0
1
0
0
0
1
0
1
2
3 a
a
a
A
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
1
0
0
B
⎦
⎣
⎥
⎦
⎢
⎣ 0
0
0 a
a
a
⎦
⎣
[ ]
5
.
0
0
5
.
1
0
1
2
=
⎥
⎦
⎤
⎢
⎣
⎡
=
b
b
b
C
[ ]
0
0
0
⎥
⎦
⎢
⎣ a
a
a

Method for establishing state equations from block diagrams
Example
Example
Example
Example
‘ Establish the state equations describing the system below:
R(s)
+
−
Y(s)
)
3
)(
1
(
10
+
+ s
s
s
‘ Define the state variables as in the block diagram:
R(s) Y(s)
10
1
1 X1(s)
X2(s)
X3(s)
( )
+
−
Y(s)
)
3
(
10
+
s
)
1
(
1
+
s
s
1 X1(s)
2(s)
X3(s)

Example (cont’)
Example (cont’)
Example (cont )
Example (cont )
‘ From the block diagram, we have:
10
)
(
3
10
)
( 2
1 s
X
s
s
X
+
=
• )
(
10
)
(
3
)
( 2
1
1 s
X
s
X
s
sX =
+
⇒
)
(
10
)
(
3
)
( 2
1
1 t
x
t
x
t
x +
−
=
⇒ & (1)
)
(
1
1
)
( 3
2 s
X
s
X =
• )
(
)
(
)
( 3
2
2 s
X
s
X
s
sX =
+
⇒
)
(
1
)
( 3
2
s +
)
(
)
(
)
( 3
2
2
)
(
)
(
)
( 3
2
2 t
x
t
x
t
x +
−
=
⇒ & (2)
( )
)
(
)
(
1
)
(
3 s
Y
s
R
s
s
X −
=
• )
(
)
(
)
( 1
3 s
X
s
R
s
sX −
=
⇒
)
(
)
(
)
(
& (3)
)
(
)
(
)
( 1
3 t
r
t
x
t
x +
−
=
⇒ & (3)

Example (cont’)
Example (cont’)
Example (cont )
Example (cont )
‘ Combining (1), (2), and (3) leads to the state equations:
)
(
0
0
)
(
)
(
1
1
0
0
10
3
)
(
)
(
2
1
2
1
t
r
t
x
t
x
t
x
t
x
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
+
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
−
−
=
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
&
&
{
1
)
(
)
(
0
0
1
)
(
)
( 3
3
t
t
x
t
t
x
B
x
A
x
⎥
⎦
⎢
⎣
⎥
⎦
⎢
⎣
⎥
⎦
⎢
⎣−
⎥
⎦
⎢
⎣ 3
2
1
4
4 3
4
4 2
1
3
2
1
&
&
⎤
⎡ )
(t
x
‘ Output equation:
[ ]
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
=
)
(
)
(
)
(
0
0
1
)
(
)
(
3
2
1
1
t
x
t
x
t
x
t
x
t
y
4
3
4
2
1
C
⎥
⎦
⎢
⎣ )
(
3 t
x
C

State equation to transfer function
Gi d ib d b h i
‘ Given a system described by the state equations:
⎩
⎨
⎧ +
=
)
(
)
(
)
(
)
(
)
(
t
t
t
u
t
t
C
B
Ax
x
&
⎩
⎨
= )
(
)
( t
t
y Cx
‘ Then the transfer function of the system is:
( ) B
A
I
C
1
)
(
)
(
)
(
−
−
=
= s
s
U
s
Y
s
G

State equation to transfer function –
– Example
Example
‘ Calculate the transfer function of the system described by the state
‘ Calculate the transfer function of the system described by the state
equation:
⎨
⎧ +
= )
(
)
(
)
( t
u
t
t B
Ax
x
&
⎩
⎨
= )
(
)
( t
t
y Cx
⎤
⎡
where
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
3
2
1
0
A ⎥
⎦
⎤
⎢
⎣
⎡
=
1
3
B [ ]
0
1
=
C
‘ Solution: The transfer function of the system is:
( ) B
A
I
C
1
)
(
)
(
)
(
−
−
=
= s
s
U
s
Y
s
G

Calculate transfer functions from state equations
Calculate transfer functions from state equations
Example (cont’)
Example (cont’)
Example (cont )
Example (cont )
( ) ⎥
⎦
⎤
⎢
⎣
⎡
+
−
=
⎥
⎦
⎤
⎢
⎣
⎡
−
⎥
⎦
⎤
⎢
⎣
⎡
=
−
3
2
1
3
2
1
0
1
0
0
1
s
s
s
s A
I
⎦
⎣ +
⎦
⎣ −
−
⎦
⎣ 3
2
3
2
1
0 s
( ) ⎥
⎤
⎢
⎡ +
=
⎥
⎤
⎢
⎡ −
=
−
−
− s
s
s
1
3
1
1
1
1
A
I
( ) ⎥
⎦
⎢
⎣ −
−
−
+
⎥
⎦
⎢
⎣ + s
s
s
s 2
)
1
.(
2
)
3
(
3
2
( ) [ ] [ ]
1
3
1
1
3
0
1
1
1
+
=
⎥
⎤
⎢
⎡ +
=
−
−
s
s
s A
I
C( ) [ ] [ ]
1
3
2
3
2
0
1
2
3 2
2
+
+
+
=
⎥
⎦
⎢
⎣ −
+
+
= s
s
s
s
s
s
s A
I
C
( ) [ ] 1
)
3
(
3
3
1
3
1
1 +
+
=
⎥
⎤
⎢
⎡
+
=
−
− s
s
s B
A
I
C( ) [ ]
2
3
1
1
3
2
3 2
2
+
+
=
⎥
⎦
⎢
⎣
+
+
+
=
−
s
s
s
s
s
s B
A
I
C
10
3
)
(
+
s
s
G
⇒
2
3
)
( 2
+
+
=
s
s
s
G
⇒

Solution to state equations
Solution to state equations
‘ Solution to the state equation ?
)
(
)
(
)
( t
u
t
t B
Ax
x +
=
&
∫ −
Φ
+
Φ
= +
t
d
u
t
t
t )
(
)
(
)
0
(
)
(
)
( τ
τ
τ B
x
x ∫ −
Φ
+
Φ
= d
u
t
t
t
0
)
(
)
(
)
0
(
)
(
)
( τ
τ
τ B
x
x
)]
(
[
)
( 1
s
t Φ
=
Φ −
L
where transient matrix
)]
(
[
)
( s
t Φ
=
Φ L
1
)
(
)
( −
−
=
Φ A
I
s
s
where transient matrix
‘ System response?
)
(
)
( t
t
y Cx
=
‘ Example:

Relationship between the mathematical models
Relationship between the mathematical models
Diff. equation
L L -1 Define x
Transfer function State equation
( ) B
A
I
C
1
)
(
−
−
= s
s
G

Linearized models of nonlinear systems
Linearized models of nonlinear systems

‘ N li t d t ti f th iti i i l d
Nonlinear systems
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:
‘ Most of the practical systems are nonlinear:
Ž Fluid system (Ex: liquid tank,…)
Ž Thermal system (Ex: furnace,…)
Ž Mechanical system (Ex: robot arm,….)
Ž Electro-magnetic system (TD: motor,…)
Ž Hybrid system ,…
y y ,

Mathematical model of nonlinear systems:
Mathematical model of nonlinear systems:
‘ Input output relationship of a continuous nonlinear system can be
‘ Input – output relationship of a continuous nonlinear system can be
expressed in the form of a nonlinear differential equations.
⎞
⎛ −
)
(
)
(
)
(
)
(
)
( 1
t
d
t
d
t
d
t
d
t
d m
n
n
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
= −
)
(
,
)
(
,
,
)
(
),
(
,
)
(
,
,
)
(
)
(
1
1
t
u
dt
t
du
dt
t
u
d
t
y
dt
t
dy
dt
t
y
d
g
dt
t
y
d
m
m
n
n
n
n
L
L
where: u(t): input signal,
y(t): output signal,
g(.): nonlinear function
g(.): nonlinear function

Nonlinear system
Nonlinear system –
– Example 1
Example 1
f th di h l
a: cross area of the dischage valve.
A: cross area of the tank
g: gravity acceleration
( )
u(t)
qin
k: constant
CD: discharge constant
y(t) qout
‘ Balance equation: )
(
)
(
)
( t
q
t
q
t
y
A out
in −
=
&
)
(
)
( t
ku
t
qi =
where: )
(
)
( t
ku
t
qin
)
(
2
)
( t
gy
aC
t
q D
out =
where:
⇒
(first order
li )
( )
)
(
2
)
(
1
)
( t
gy
aC
t
ku
t
y D
−
=
&
⇒
nonlinear system )
( )
)
(
2
)
(
)
( t
gy
aC
t
ku
A
t
y D
=

Nonlinear system
– Example 2
Example 2
J: moment inertia of the robot arm
J: moment inertia of the robot arm
M: mass of the robot arm
m: object mass
l l h f b
l
l: length of robot arm
lC : distance from center of gravity to rotary axis
m
u θ
g: gravitational acceleration
u(t): input torque
θ(t): robot arm angle
θ(t): robot arm angle
‘ According to Newton’s Law
)
(
cos
)
(
)
(
)
(
)
( 2
t
u
g
Ml
ml
t
B
t
ml
J C =
+
+
+
+ θ
θ
θ &
&
&
C
⇒ )
(
)
(
1
cos
)
(
)
(
)
(
)
(
)
( 2
2
2
t
u
ml
J
g
ml
J
Ml
ml
t
ml
J
B
t C
+
+
+
+
−
+
−
= θ
θ
θ &
&
&
(second order nonlinear system)

Nonlinear system
– Example 3
Example 3
δ: steering angle
ψ: ship angle
k: constant
τi: constant
ψ(t)
δ(t)
Moving direction
ψ( )
‘ The differential equation describing the steering dynamic of a ship:
( ) ( )
)
(
)
(
)
(
)
(
1
)
(
1
1
)
( 3
3
t
t
k
t
t
t
t δ
δ
τ
ψ
ψ
ψ
ψ +
⎟
⎟
⎞
⎜
⎜
⎛
+
+
⎟
⎟
⎞
⎜
⎜
⎛
−
⎟
⎟
⎞
⎜
⎜
⎛
+
−
= &
&
&
&
&
&
&
&
(third order nonlinear system)
( ) ( )
)
(
)
(
)
(
)
(
)
(
)
( 3
2
1
2
1
2
1
t
t
t
t
t
t δ
δ
τ
τ
τ
ψ
ψ
τ
τ
ψ
τ
τ
ψ +
⎟
⎟
⎠
⎜
⎜
⎝
+
+
⎟
⎟
⎠
⎜
⎜
⎝
⎟
⎟
⎠
⎜
⎜
⎝
+

Describing nonlinear system by state equations
Describing nonlinear system by state equations
‘ A continuous nonlinear system can be described by the state
equation:
⎧ = ))
(
)
(
(
)
( t
u
t
t x
f
x
&
⎩
⎨
⎧
=
=
))
(
),
(
(
)
(
))
(
),
(
(
)
(
t
u
t
h
t
y
t
u
t
t
x
x
f
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

State
State-
-space model of nonlinear system
space model of nonlinear system –
– Example 1
Example 1
( )
u(t)
qin
( )
)
(
2
)
(
1
)
( t
gy
aC
t
ku
A
t
y D
−
=
&
‘ Define the state variable:
)
(
)
(
y(t) qout
A
)
(
)
(
1 t
y
t
x =
S i ⎨
⎧ = ))
(
),
(
(
)
( t
u
t
t x
f
x
&
‘ State equation:
⎩
⎨
⎧
= ))
(
),
(
(
)
(
))
(
),
(
(
)
(
t
u
t
h
t
y x
f
)
(
2 k
t
gx
aC
)
(
)
(
2
)
,
( 1
t
u
A
k
A
t
gx
aC
u D
+
−
=
x
f
)
(
))
(
),
(
( 1 t
x
t
u
t
h =
x
trong ñoù:

State
State-
-space model of nonlinear system
space model of nonlinear system –
– Example 2
Example 2
m
l )
(
)
(
1
cos
)
(
)
(
)
(
)
(
)
( 2
2
2
t
u
ml
J
g
ml
J
Ml
ml
t
ml
J
B
t C
+
+
+
+
−
+
−
= θ
θ
θ &
&
&
‘ Define the state variable:
⎩
⎨
⎧
=
=
)
(
)
(
)
(
)
(
2
1
t
t
x
t
t
x
θ
θ
&
u θ
⎩ = )
(
)
(
2 t
t
x θ
‘ State equation:
⎩
⎨
⎧ =
))
(
)
(
(
)
(
))
(
),
(
(
)
(
h
t
u
t
t x
f
x
&
q
⎩
⎨
= ))
(
),
(
(
)
( t
u
t
h
t
y x
⎥
⎤
⎢
⎡ )
(
2 t
x
where
⎥
⎥
⎦
⎢
⎢
⎣ +
+
+
−
+
+
−
=
)
(
)
(
1
)
(
)
(
)
(
cos
)
(
)
(
)
,
(
2
2
2
1
2
t
u
ml
J
t
x
ml
J
B
t
x
ml
J
g
Ml
ml
u C
x
f
)
(
))
(
),
(
( 1 t
x
t
u
t
h =
x

Equilibrium points of a nonlinear system
Equilibrium points of a nonlinear system
‘ Consider a nonlinear system described by the diff. equation:
⎩
⎨
⎧
=
=
))
(
),
(
(
)
(
))
(
),
(
(
)
(
t
u
t
h
t
y
t
u
t
t
x
x
f
x
&
‘ Consider a nonlinear system described by the diff. equation:
⎩ ))
(
),
(
(
)
( t
u
t
h
t
y x
‘ The state is called the equilibrium point of the nonlinear system if
the system is at the state and the control signal is fixed at then
x
x u
‘ If is equilibrium point of the nonlinear system then:
)
( u
x
the system is at the state and the control signal is fixed at then
the system will stay at state forever.
x u
x
‘ If is equilibrium point of the nonlinear system then:
)
,
( u
x
0
))
(
),
(
( ,
=
=
= u
u
t
u
t x
x
x
f
‘ The equilibrium point is also called the stationary point of the
nonlinear system.

Equilibrium point of nonlinear system
Equilibrium point of nonlinear system –
– Example 1
Example 1
‘ Consider a nonlinear system described by the state equation:
⎥
⎦
⎤
⎢
⎣
⎡
+
+
=
⎥
⎦
⎤
⎢
⎣
⎡
)
(
2
)
(
)
(
).
(
)
(
)
(
2
1
2
1
2
1
t
x
t
x
u
t
x
t
x
t
x
t
x
&
&
⎦
⎣
⎦
⎣ )
(
)
(
)
( 2
1
2
Find the equilibrium point when 1
)
( =
= u
t
u
‘ Solution:
0
))
(
),
(
( ,
=
=
= u
u
t
u
t x
x
x
f
The equilibrium point(s) are the solution to the equation:
, u
u
x
x
⎩
⎨
⎧
=
+
=
+
0
2
0
1
.
2
1
2
1
x
x
x
x
⇔
⎩ 2
1
⎪
⎪
⎨
⎧ =
2
2
1
x
⎪
⎪
⎨
⎧ −
=
2
2
1
x
⇔ or
⎪
⎩
⎨
−
=
2
2
x ⎪
⎩
⎨
+
=
2
2
x

Equilibrium point of nonlinear system
Equilibrium point of nonlinear system –
– Example 2
Example 2
⎥
⎤
⎢
⎡ +
+
+
⎤
⎡ u
x
x
x 2
3
2
2
1 1
&
⎥
⎥
⎦
⎢
⎢
⎣ +
−
+
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
u
x
x
x
x
x
x
x
2
3
3
1
3
3
2
3
2
1
)
sin(
&
&
Find the equilibrium point when 0
)
( =
= u
t
u
1
x
y =

Linearized model of a nonlinear system around an equilibrium point
‘ Consider a nonlinear system described by the differential equation:
⎩
⎨
⎧
=
=
))
(
)
(
(
)
(
))
(
),
(
(
)
(
t
u
t
h
t
y
t
u
t
t
x
x
f
x
&
‘ Consider a nonlinear system described by the differential equation:
(1)
⎩ = ))
(
),
(
(
)
( t
u
t
h
t
y x
‘ Expanding Taylor series for f(x,u) and h(x,u) around the equilibrium
p g y f( ) ( ) q
point , we can approximate the nonlinear system (1) by the
following linearized state equation:
⎧
)
,
( u
x
⎩
⎨
⎧
+
=
+
=
)
(
~
)
(
~
)
(
~
)
(
~
)
(
~
)
(
~
t
u
t
t
y
t
u
t
t
D
x
C
B
x
A
x
&
(2)
where:
u
t
u
t
u
t
t
−
=
−
=
)
(
)
(
~
)
(
)
(
~ x
x
x
y
t
y
t
y −
= )
(
)
(
~
)
(
)
(
))
,
(
( u
h
y x
=

Th i f h li i d i l l d f ll
1
1
1 f
f
f
⎥
⎤
⎢
⎡ ∂
∂
∂ 1
f ⎤
⎡ ∂
‘ The matrix of the linearized state equation are calculated as follow:
2
2
2
1
2
1
1
1
n
x
f
x
f
x
f
x
f
x
f
x
f
A ⎥
⎥
⎥
⎥
⎢
⎢
⎢
⎢
∂
∂
∂
∂
∂
∂
∂
∂
∂
=
L
L
2
1
u
f
u
f
B
⎥
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎢
⎡
∂
∂
∂
=
2
1
n
n
n
n
x
f
x
f
x
f
x
x
x
A
⎥
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎢
⎣ ∂
∂
∂
∂
∂
∂
∂
∂
∂
K
M
O
M
M
)
( u
n
u
f
u
,
x
⎥
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎢
⎣ ∂
∂
∂
M
)
(
2
1 u
n
x
x
x ,
x
⎦
⎣ ∂
∂
∂
⎤
⎡ h⎤
⎡∂
)
(
2
1 u
n
x
h
x
h
x
h
,
x
C ⎥
⎦
⎤
⎢
⎣
⎡
∂
∂
∂
∂
∂
∂
= K
)
( u
u
h
,
x
D ⎥
⎦
⎤
⎢
⎣
⎡
∂
∂
=

Linearized state
Linearized state-
-space model
space model –
– Example 1
Example 1
The parameter of the tank:
u(t)
qin
The parameter of the tank:
3
2
2
8
0
/
150
100
,
1
C
V
k
cm
A
cm
a =
=
y(t)
u(t)
qout 2
3
sec
/
981
8
.
0
,
.
sec
/
150
cm
g
C
V
cm
k D
=
=
=
‘ Nonlinear state equation:
⎩
⎨
⎧
=
=
))
(
)
(
(
)
(
))
(
),
(
(
)
(
t
u
t
h
t
y
t
u
t
t
x
x
f
x
&
⎩ = ))
(
),
(
(
)
( t
u
t
h
t
y x
)
(
9465
0
)
(
3544
0
)
(
)
(
2
)
( 1 k
t
gx
aCD
f
where
)
(
9465
.
0
)
(
3544
.
0
)
(
)
(
)
,
( 1
1
t
u
t
x
t
u
A
k
A
g
u D
+
−
=
+
−
=
x
f
)
(
))
(
),
(
( 1 t
x
t
u
t
h =
x
)
(
))
(
),
(
( 1

Linearized state
Linearized state-
-space model
space model –
Example 1 (cont’)
Linearize the system around y = 20cm:
Linearize the system around y = 20cm:
‘ The equilibrium point:
20
20
1 =
x
0
5
.
1
3544
.
0
)
,
( 1 =
+
−
= u
x
u
x
f 9465
.
0
=
u
⇒

Linearized state
Linearized state-
-space model
space model –
Example 1 (cont’)
‘ The matri of the lineari ed state space model:
0396
.
0
2
2
1
−
=
−
=
∂
∂
= D
x
A
g
aC
x
f
A 5
.
1
1
=
=
∂
∂
=
A
k
u
f
B
‘ The matrix of the linearized state-space model:
2 )
(
1
)
(
1
∂ u
u
x
A
x ,
x
,
x )
(
)
(
∂ u
u A
u ,
x
,
x
1
=
∂
=
h
C 0
=
∂
∂
=
h
D
)
(
1
∂ u
x ,
x )
(
∂ u
u ,
x
‘ The lineari ed state eq ation describing the s stem aro nd the
‘ The linearized state equation describing the system around the
equilibrium point y=20cm is:
⎧ +
= )
(
~
5
1
)
(
~
0396
0
)
(
~ t
u
t
t x
x
&
)
(
)
(
2
)
,
( 1
t
u
A
k
A
t
gx
aC
u D
+
−
=
x
f
⎩
⎨
⎧
=
+
−
=
)
(
~
)
(
~
)
(
5
.
1
)
(
0396
.
0
)
(
t
t
y
t
u
t
t
x
x
x
)
(
))
(
),
(
( 1 t
x
t
u
t
h =
x

Linearized state
Linearized state-
-space model
space model –
– Example 2
Example 2
The parameters of the robot:
The parameters of the robot:
2
C
02
0
5
0
1
.
0
,
2
.
0
,
5
.
0
m
kg
J
kg
M
kg
m
m
l
m
l =
=
=
m
l
2
sec
/
81
.
9
,
005
.
0
.
02
.
0
,
5
.
0
m
g
B
m
kg
J
kg
M
=
=
=
=
u θ
‘ Nonlinear state equation :
⎩
⎨
⎧
=
=
))
(
),
(
(
)
(
))
(
),
(
(
)
(
t
u
t
h
t
y
t
u
t
t
x
x
f
x
&
⎩ ))
(
),
(
(
)
( t
u
t
h
t
y x
⎥
⎤
⎢
⎡ )
(
2 t
x
where:
⎥
⎥
⎦
⎢
⎢
⎣ +
+
+
−
+
+
−
=
)
(
)
(
1
)
(
)
(
)
(
cos
)
(
)
(
)
,
(
2
2
2
1
2
t
u
ml
J
t
x
ml
J
B
t
x
ml
J
g
Ml
ml
u C
x
f
)
(
))
(
),
(
( 1 t
x
t
u
t
h =
x

Linearized state
Linearized state-
-space model
space model –
Example 2 (cont’)
Linearize the system around the equilibrium point y = π/6 (rad):
Linearize the system around the equilibrium point y = π/6 (rad):
‘ Finding the equilibrium point:
6
/
1 π
=
x
0
1
)
(
)
(
2
⎥
⎤
⎢
⎡
+ B
g
Ml
ml
x
u
x
f ⇒ ⎨
⎧ = 0
2
x
0
)
(
1
)
(
cos
)
(
)
(
)
,
(
2
2
2
1
2
=
⎥
⎥
⎦
⎢
⎢
⎣ +
+
+
−
+
+
−
=
u
ml
J
x
ml
J
B
x
ml
J
g
Ml
ml
u C
x
f ⇒
⎩
⎨
= 2744
.
1
u
Then the equilibrium point is:
⎥
⎤
⎢
⎡
=
⎥
⎤
⎢
⎡
=
6
/
1 π
x
x ⎥
⎦
⎢
⎣
⎥
⎦
⎢
⎣ 0
2
x
x
2744
.
1
=
u

Linearized state
Linearized state-
-space model
space model –
Example 2 (cont’)
‘ The s stem matri aro nd the eq ilibri m point:
‘ The system matrix around the equilibrium point:
⎥
⎦
⎤
⎢
⎣
⎡
= 12
11
a
a
a
a
A
0
)
(
1
1
11 =
∂
∂
=
u
x
f
a
,
x
1
)
(
2
1
12 =
∂
∂
=
x
f
a
⎦
⎣ 22
21 a
a
)
( u
,
x
)
(
1
2
)
(
1
2
21 )
(
sin
)
(
)
( C
t
x
ml
J
Ml
ml
x
f
a
+
+
=
∂
∂
=
)
(
2 u
,
x
)
(
)
(
1 )
( u
u
ml
J ,
x
,
x
2
2
22
)
( ml
J
B
x
f
a
+
−
=
∂
∂
=
⎥
⎥
⎤
⎢
⎢
⎡
+
=
)
(
1
)
(
)
(
)
(
)
(
)
(
2
B
g
Ml
ml
t
x
u C
x
f
)
(
)
(
2 )
( u
u
ml
J
x ,
x
,
x
+
∂
⎥
⎦
⎢
⎣ +
+
+
−
+
+
− )
(
)
(
1
)
(
)
(
)
(
cos
)
(
)
(
)
,
(
2
2
2
1
2
t
u
ml
J
t
x
ml
J
B
t
x
ml
J
g
Ml
ml
u C
x
f

Linearized state
Linearized state-
-space model
space model –
Example 2 (cont’)
‘ The inp t matri aro nd the eq ilibri m point:
‘ The input matrix around the equilibrium point:
⎥
⎦
⎤
⎢
⎣
⎡
= 1
b
b
B
0
1
∂f
b
⎦
⎣ 2
b
0
)
(
1
1 =
∂
=
u
u
b
,
x
1
f
∂
2
)
(
2
2
1
ml
J
u
f
b
u +
=
∂
∂
=
,
x
⎥
⎥
⎤
⎢
⎢
⎡
+
=
)
(
1
)
(
)
(
)
(
)
(
)
(
2
B
g
Ml
ml
t
x
u C
x
f
⎥
⎦
⎢
⎣ +
+
+
−
+
+
− )
(
)
(
1
)
(
)
(
)
(
cos
)
(
)
(
)
,
(
2
2
2
1
2
t
u
ml
J
t
x
ml
J
B
t
x
ml
J
g
Ml
ml
u C
x
f

Linearized state
Linearized state-
-space model
space model –
Example 2 (cont’)
‘ The output matrix around the equilibrium point:
‘ The output matrix around the equilibrium point:
1
1
1 =
∂
∂
=
x
h
c
[ ]
2
1 c
c
=
C 0
)
(
2
2 =
∂
∂
=
x
h
c
)
(
1
∂ u
x ,
x )
(
2 u
,
x
1
d
=
D 0
)
(
1 =
∂
∂
=
u
h
d
)
(
∂ u
u ,
x
‘ Then the linearized state equation is:
⎩
⎨
⎧ +
=
)
(
~
)
(
~
)
(
~
)
(
~
)
(
~
)
(
~
t
t
t
t
u
t
t
D
C
B
x
A
x
&
⎩ +
= )
(
)
(
)
( t
u
t
t
y D
x
C
⎥
⎤
⎢
⎡
=
1
0
A ⎥
⎤
⎢
⎡
=
0
B [ ]
0
1
=
C 0
=
D
⎥
⎦
⎢
⎣
=
22
21 a
a
A ⎥
⎦
⎢
⎣
=
2
b
B [ ]
0
1
C 0
=
D
)
(
)
,
( 1 t
x
u
h =
x

Regulating nonlinear system around equilibrium point
Regulating nonlinear system around equilibrium point
‘ Drive the nonlinear system to the neighbor of the 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
‘ Around the equilibrium point, use a linear controller to maintain the
system around the equilibrium point.
Linear
r(t) Nonlinear
system
+
−
y(t)
Linear
control u(t)
e(t)
system
ON-OFF
Mode
select

IntroCtrlSys_Chapter2.pdf

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