control system - chapter 2
DESCRIPTION
Electrical Control SystemTRANSCRIPT
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CHAPTER
MATHEMATICAL MODELLING OF
DYNAMIC SYSTEM
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INTRODUCTION
• A mathematical model of a dynamic system isdefined as a set of equation ,i.e differential equation,that represent the dynamic of the system accuratelyor at least fairly well.
• Note that the a mathematical model is not unique toa given system and may be represent in manydifferent ways, therefore, may have manymathematical model depending on one’s
perspective.
• In this chapter we obtain mathematical models thatrelate the outputs of the system to its input.
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INTRODUCTION (cont.)
• To develop a mathematical model we
need to use the fundamental physical law
of science and engineering.
– To model electrical networks we need apply
Ohm’s law & kirchhoff’s law.
– To model mechanical system we need applyNewton’s Law.
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INTRODUCTION (cont.)
• The relation between the input and output of the
system can be described by differential
equation.
Output, c(t)Input, r(t)
System
Figure 2.1
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INTRODUCTION (cont.)
• In practice, the complexity of the system
requires some assumptions in the
determination of the mathematical model.
• The equations of the mathematical model
may be solved using mathematical tools
such as the Laplace Transform.
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2.1 Review of Laplace Transform
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• The Laplace transform method substitutes relatively
easily solved algebraic equations for the more difficult
differential equations.
• The Laplace transformation for a function of time, is;
LAPLACE TRANSFORM (cont.)
where s= σ+ j ω, a complex variable. While j = −1 , σ =Re {s} (real part)and ω = Im {s} (imaginary part).
• The Eq. 2.1 allows us to convert f(t) F(s).
0
)()( dt et f s F st =£{f(t)} Eq 2.1
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LAPLACE TRANSFORM (cont.)
EXAMPLE Eq 2.1:
Find the Laplace transform of
Sol:
)()( t u Aet f at
t a s st at st )(
a s
A
e
a s
A t t a s
0)(
0 0 0
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• Table 2.1 shows the conversion of f(t) to F(s) for specific cases derived
using Eq.2.1
Table 2.1: Laplace Transform Table
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LAPLACE TRANSFORM (cont.)
• To transform F(s) to f(t), the inverse Laplace transform is
used.
• The inverse Laplace transform is written as:
1 j st
• However Eq. 2.2 requires complex integration to find f(t)
given F(s)
• The easy way to find f(t) given F(s) is by using Laplace
Transform Table, Table 2.1.
2 j j
10
q .
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LAPLACE TRANSFORM (cont.)
Example 2.2 (Find f(t) given F(s) using table ILT):
Find the inverse Laplace transform of 21
3
1)(
s s F
Solution:
From laplace table
)()()3(
1
)()(
1
3
2
2
t f t ute s
There fo re
t utea s
t
at
11
at m
m e
m
t
a s
!)(
1
I L TT a b l eF ro m: N o te
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2.3 Laplace transform theorems
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LAPLACE TRANSFORM (cont.)
Laplace Transform theorems, Table 2.2 is addition to the Laplace
transform table, Table 2.1 to assist in transforming between f(t) and
F(s).
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LAPLACE TRANSFORM (cont.)
Table 2.1: Laplace Transform Theorems
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LAPLACE TRANSFORM (cont.)
2.3.1 Partial- Fraction Expansion method• Used to find the inverse Laplace transform of a complicated
system.
• Have 4 case:
Case1: Roots of Denominator of F(s) are Real and Distinct
)(..........
)()()(
)..().........()(
)(
)(
)(
)(
21
21
n
n
s s
N
s s
B
s s
A s F
s s s s s s
sC
s R
sC
s F
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LAPLACE TRANSFORM (cont.)
Case2: Roots of Denominator of F(s) are Real and Repeated
r
n s s s s s s
sC
s R
sC s F
)..().........()(
)(
)(
)()(
21
r
n s s
N
s s
B
s s
A s F
)(..........
)()()(
21
Case3: Roots of Denominator of F(s) are Complex or imaginary
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LAPLACE TRANSFORM (cont.)
Case 4:
Partial Fraction can be used when the order numerator
N(s) is less than the order denominator D(s). If the order
numerator is greater than or equal to the order
denominator, then the numerator must be divided by
denominator successively until the result has a
remainder whose numerator is of order less than its
denominator.
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LAPLACE TRANSFORM (cont.)
Note:
• Example all of the cases partial-fraction
expansion are shown on extra note .
• Example of the application of Laplace
transform, inverse Laplace transform
and partial-fraction expansion for solving differential equation are shown
on extra note given. ( extra note 2 )18
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MODELLING OF PHYSICAL
SYSTEM
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2.4 Modeling of Electrical System
Elements
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Modeling of Electrical System Elements
• A mathematical model of an electrical circuit can be
obtained by Kirchhoff’s laws.
Figure Electronic Components
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Modeling of Electrical System Elements. (cont)
Example 1 :
Obtain the transfer function of the RC network shown below.
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Modeling of Electrical System Elements. (cont)
Solution:
Network equation
V1(t) = i(t)R + V2(t)
)()( 2 t dV C t I
By using laplace transform, transfer function obtained is,
/1
/1
1
1
1
1
)(
)()(
1
2
s s RCs sV
sV sG
where = RC, the time constant of the network
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Modeling of Electrical System Elements. (cont)
Example 2 :
Construct the block diagram of the system and obtain the transfer
function of the RLC network shown below.
Fig : Series RLC network
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Modeling of Electrical System Elements. (cont)
Solution
• The network equations are
» eq1
)()( t it dvC C
»
» eq 2
)()(
)()( t vdt
t di L Rt it v C
)()()()( t v Rt it vdt t di L C
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Modeling of Electrical System Elements. (cont)
Laplace transformed the eq. 1 and eq. 2, we get
eq
eq 4
)()( s I sCsVc
LsI(s) = V(s) – I(s)R – V C (s)
Draw block diagram of the RLC network using eq 3 and eq 4
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Modeling of Electrical System Elements. (cont)
we get, Block diagram of the series RLC network
+
-
1
Ls
1
Cs
-
V(s)V C (s)+ LsI(s) I(s)
I(s)R V C (s)
The transfer function of the system is:
1
1
)(
)(2
RCs LCs sV
sV c
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Modeling of Electrical System Elements. (cont)
Exercise
Construct a block diagram of this system in the Figure below and
obtain the transfer function .
)()(
)(2
0
R Ls RLCs
R
sV
sV
i
solution
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Modeling of Electrical System Elements. (cont)
Example: Refer to extra note 3
.
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Modeling of Electrical System Elements.
Potentiometer
potentiometer is used to measure a linear or rotational
displacement. . There are two types of potentiometers which is;
1. rotational potentiometers
2. linear potentiometers.
Fig: Rotational potentiometer
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Modeling of Electrical System Elements.
Potentiometer . (cont.)
For a rotational potentiometer
e(t) α θ(t)
Therefore, e(t) = Kpθ(t); where Kp = Potentiometer constant V/rads
Laplace transform of the equation is
E(s) = K pθ(s)
Block diagram of the system
K pE(s)θ(s)
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Modeling of Electrical System Elements.
Tachometer .
Tachometer produce a voltage proportional to the angular velocity of the
generator shaft.
et(t) m(t)
et(t) = K k m(t)
Inverse laplace transform, we get,
E t (s) = K k
m(s)
Where, Kk = tachometer constant
K k
E t (s)m(s)
Fig: Block diagram of a tachometer 32
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Modeling of Electrical System Elements.
Operational Amplifier (Op-Amps)
The output of the amplifier
Vo = KsVi(t)
Where, Ks = amplifier gain, and Vi = amplifier input
Laplace transform of the equation is
Vo(s) = KsVi(s)
Block diagram of the system
K sV 0 (s)V i (s)
Fig: Block diagram of an Op Amp
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2.5 MODELLING OF MECHANICAL
SYSTEM
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MODELLING OF MECHANICAL SYSTEM
The motion of mechanical elements can
be described in various dimensions, which
are:
. .
2. Rotational.
3. Gear System.
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Translational Mechanical ystem
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Modeling of Translational Mechanical System.
The motion of translation is defined as a
motion that takes place along or curved .
describe translational motion are
acceleration, velocity, and displacement.
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Modeling of Translational Mechanical System.
(cont.)
Figure below shows the translational mechanical system
Figure: Translational Mechanical Components38
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Modeling of Translational Mechanical System.
(cont.)
Example:
Find the transfer function of X(s)/F(s) , for system shown below
Figure : Spring-Mass-Damper System
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Modeling of Translational Mechanical System.
(cont.)
Solution
Draw the free-body diagram of a system and assume the mass is
traveling toward the right.
Figure 2.4 a. Free-body diagram of mass, spring, and damper system;
b. transformed free-body diagram
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Modeling of Translational Mechanical System.
(cont.)
From free-body diagram, write differential equation of motion using
Newton’s Law. Thus we get;
)()()()(
2
2
t f t Kxdt
t dx f
dt
t xd M
v
By taking the Laplace transform
)()()()(2 s F s KX s sX f s X Ms v
Therefore the transfer function of system is;
K s f Ms s F
s X sG
v
2
1
)(
)()(
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Modeling of Translational Mechanical System.
(cont.)
Spring-mass system
Example : Obtain the transfer function between Xo(s)/Xi(s)
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How to solve: Refer to extra note 4
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Modeling of Translational Mechanical System.
(cont.)
Spring-mass with viscous frictional damping
Example : Obtain the transfer function between Xo(s)/Xi(s)
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How to solve: Refer to extra note 4
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Modeling of Translational Mechanical System. (cont.)
Obtain the transfer function between X22(s)/F(s)
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Modeling of Translational Mechanical System. (cont.)
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Modeling of Translational Mechanical System. (cont.)
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How to solve: Refer to extra note 4
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Rotational Mechanical ystem
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Rotational Mechanical System. (Cont)
The rotational motion can be defined as motion about a fixed axis.
The extension of Newton’s Law of motion for rotational motion states
that the algebraic sum of moments or torque about a fixed axis is
equal to the product of the inertia and the angular acceleration about
the axis where,
T = Torque
θ = Angular Displacement
ω = Angular Velocity
where Newton’s second law for rotational system are,
o nacceleratiangular where J T Torque :,)(
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Modeling of Rotational Mechanical System. (Cont)
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Modeling of Rotational Mechanical System. (Cont)
Obtain the transfer function between θ2(s)/T(s)
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Modeling of Rotational Mechanical System. (Cont)
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Modeling of Rotational Mechanical System. (Cont)
52
How to solve: Refer to extra note 5
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Obtain the transfer function between θ2(s)/T(s)
Modeling of Rotational Mechanical System. (Cont)
53
How to solve: Refer to extra note 5
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Gear system
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Gear system
Gear system is a mechanical device that
transmit energy from one part of the
system to another in such away that force,
, ,
altered. These devices can also be
regarded as matching devices used to
attain maximum power transfer.
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Gear system. (Cont.)
Figure below show two gears are coupled together
Where,
T = torque
θ = angular displacement
N = teeth number
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Gear system. (Cont.)
from front view
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Gear system. (Cont.)
For ideal case, inertia and friction are neglected. The
relationship between T1 and T2, θ1 and θ2 and N1 and N2
are derived from the following facts;
1. The number of teeth on the surface of the gears is proportional
to the radius r 1 and r 2 of the ears that is
r 1N2 = r 2N1
2. The distance traveled along the surface of each gear is the same,
thus,
θ1r 1 = θ2r 2
3. The work done by one gears is same to that other since there areassumed to be no losses. thus,
T1θ1 = T2θ2
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Gear system. (Cont.)
If the angular velocities of two gears ω1 and ω2 are brought into the
picture, Therefore
1
2
2
1
2
1
NT
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Gear system. (Cont.)
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Find transfer function between θ 1(s)/T 1(s)
Gear system. (Cont.)
61How to solve: Refer to extra note 6
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Find transfer function between θ 2 (s)/T 1(s)
Gear system. (Cont.)
62
How to solve: Refer to extra note 6
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Gear system. (Cont.)
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Find transfer function between θ 1(s)/T 1(s)
Gear system. (Cont.)
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Find transfer function between θ 2 (s)/T 1(s)
Gear system. (Cont.)
65
How to solve: Refer to extra note 6
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2.6 Modelin of Electromechanical S stem Elements.
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Modeling of Electromechanical System Elements.
(cont.)
Direct Current (DC) Motor
• Applications of DC motor e.g. tape drive, disk drive,
printer, CNC machines, and robot.
• The DC motor is represented by a circuits shown on
the next slide.
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Modeling of Electromechanical System Elements.
(cont.)
Magnetic flux
R Li a(t)
eb(t)v a(t)
Armature circuit
T L(t)
J m,
T m(t)
m
m(t)
rigid
Load
J L
68
BL
Bm
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Modeling of Electromechanical System Elements.
(cont.)
Let:
Ra = armature resistance
La = armature inductance
ia(t) = armature current m
Va(t) = armature input voltage
eb(t) = back emf
ωm(t) = motor angular velocity
θm(t) = motor angular displacementJm= moment of inertia of motor + loadBm=viscous frictional constant of motor + load
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Modeling of Electromechanical System Elements.
(cont.)
Two method to control DC motor, which are:
1. Armature control
2. Field control
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Modeling of Electromechanical System Elements.
(cont.) (Armature control)
Field current is fix and variable voltage
is applied to armature circuit.
For the given DC servo unit , find θθmm(s)/(s)/VVaa(s),(s), assume thatassume that KKtt is torqueis torque
constant andconstant and KKbb isis emf emf constantconstant..
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Modeling of Electromechanical System Elements.
(cont.) (Armature control)
There are 3 conditions of armature, which
are:
. n- oa con on
2. No-load condition
3. Combining with gear system
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Modeling of Electromechanical System Elements.
(cont.) (Armature control)(In load Condition)(In load Condition)
a) Find Armature circuit equation
s
b
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b) Find force equation applied to the motor
Modeling of Electromechanical System Elements.
(cont.) (Armature control)(In load Condition)(In load Condition)
L
2
L
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)3()s(IK )s(T atm
3) Find relationship between torque(TTmm(s)(s)) and armature current(IIaa(s)(s))
Modeling of Electromechanical System Elements.
(cont.) (Armature control)(In load Condition)(In load Condition)
4) Find relationship between motor emf(EEbb(s)(s)) and motor shaft velocity (ωωmm(s(s))))
)4()s(sK )s(E mbb
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s
b
So now, we have four equations which are:
L
2
L
Modeling of Electromechanical System Elements.
(cont.) (Armature control)(In load Condition)(In load Condition)
)3()s(IK )s(T atm
)4()s(sK )s(E mbb
From (3) and (2)
K
s
t
L
2
L
a
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Assume :
Insert (4) and (5) to equation (1)
s
K
s
m
t
L
2
L
a
Modeling of Electromechanical System Elements.
(cont.) (Armature control)(In load Condition)(In load Condition)
s
L
2
L
So that :
s
K
Z
b
t
a
m
sK K Z
K
)s(V
)s(
bta
t
a
m
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t
a
m
K s
K
ANSWER :ANSWER :
Modeling of Electromechanical System Elements.
(cont.) (Armature control)(In load Condition)(In load Condition)
t
a
m
K s
K
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For No-load condition and Combining with
Modeling of Electromechanical System Elements.
(cont.) (Armature control)(no load Condition)(no load Condition)
gear sys em p ease see e ex ra no e .
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ExerciseExercise Armature control
Find θL(s)/Va(s)
L(t)
J m , Bm
T m(t)
m(t)
m(t)
Load
J L, BL
N1
N2
b
2
2
1
L
2
2
1
L
t
2
1
a
L
K
N
N
B
N
N
J s
K
N
N
ANSWER :ANSWER :
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Exercise
Find transfer function between θ L(s)/E a(s)
81
2
2
1
2
2
1
2
1
)()(
)(
N
N B B Deq
N
N J J Jeq
KtKb Deq Jeqs Ra s
Kt N
N
s Ea
s
La
La
L
Ans:
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Modeling of Electromechanical System Elements.
(cont.) (Field control)
Armature current is fix and variable
voltage is applied to field circuit.
For the given DC servo unit , find
mm f f ,, f f
constantconstant..
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Modeling of Electromechanical System Elements.
(cont.) (Field control)
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Modeling of Electromechanical System Elements.
(cont.) (Field control)
1) Find Field circuit equation
f
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Modeling of Electromechanical System Elements.
(cont.) (Field control)
2) Find force equation applied to the motor
L
2
L
m
2
m
or or
wherewhere
Lam JJJ Lam BBB andand85
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Modeling of Electromechanical System Elements.
(cont.) (Field control)
)3()s(IK )s(T f f m
3) Find relationship between torque(TTmm(s)(s)) and field current(IIf f (s)(s))
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Modeling of Electromechanical System Elements.
(cont.) (Field control)
m
2
m
So now, we have three equations which are:
)3()s(IK )s(T f f m
f
From (3) and (2)
K
s
f
m
2
m
f
From (4) and (1)
K
s
f
f
m
2
m
f
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2
m
f
f
m
B
K
ANSWER :ANSWER :
Modeling of Electromechanical System Elements.
(cont.) (Field control)
2
m
f
f
m
B
K
Where :Where :
Lam
JJJ
Lam BBB
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2.7 Modeling of speed and Position Control System
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Modeling closed loop Position Control System
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Modeling closed loop Position Control System
(cont.)
Figure below shows the closed loop position control system.
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Modeling closed loop Position Control System
(cont.)
The objective of this system is to control the position of the
mechanical load in according with the reference position
The operation of this system is as follows:-
A pair of potentiometers acts as an error-measuring device. For input potentiometer, vi(t) = kpθi(t)
For the output potentiometer, vo(t) = kpθo(t)
The error signal,
Ve(t) = Vi(t) – Vo(t)
= kpθi(t) - kpθo(t) --------------------( 1
This error signal are amplified by the amplifier with gain constant, Ks.
Va(t) = K s Ve(t) -----------------------------( 2 )
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Modeling closed loop Position Control System
(cont.)
Transforming equations ( 1 ) and ( 2 ):-
Ve(s) = Kpθi(s) - Kpθo(s) -----------------( 3 )
By using the mathematical models developed previously for motor and
gear the block diagram of the position control system is shown below:-
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Modeling closed loop Position Control System
(cont.)
Transfer function
Let
bt eqeq
t
a
m
K K B s J R Ls
K
sV
s
))(()(
)(
11
J J ieq and B Bieq
Therefore
bt s
t
a
m
K K B Js R L
K
sV
s
))(()(
)(
For small motors, the armature inductance, L is small and can be neglected
Where, L = 0
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Modeling closed loop Position Control System
(cont.)
Therefore,
)()(
)(
bt
t
a
m
K K BR sRJ
K
sV
s
K t
1
s K K BR
RJ
bt
bt
1
sm
m
where,
bt
t m
K K BR K K
bt
m K K BR
RJ
= motor again
= motor time constant
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Modeling closed loop Position Control System
(cont.)
The simplified block diagram.
where,
sT
K
sV
s
m
m
a
m
1)(
)(
From the block diagram we get,
)1()(
sT s
n K K K sG
m
m s p
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Modeling closed loop Position Control System
(cont.)
Let, n K K K K m s p
)1()(
sT s
K sG
m
so,
,
)1(1
)1(
)(
)(
sT s
K
sT s
K
s
s
m
m
i
o
mm
m
m
m
T
K s
T s
T
K
K s sT
K
K sT s
K
1
)1(
22
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Modeling closed loop Position Control System
(cont.)
From the equation above, if θi(s) given, therefore,
θo(t)=
K s sT
s K
m
i
2
1 )(
£
From that, the value K are not fixed and can be change,
therefore the system response are depends on the value K.
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Modeling closed loop Position Control sys em w ve oc y ee ac
99
Modeling closed loop Position Control system with
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Modeling closed loop Position Control system with
velocity feedback
(cont.)
A tachometer is coupled to the motor shaft so that a voltage signal
which is proportional to motor speed can be feedback as shown
below
100
Modeling closed loop Position Control system
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Modeling closed loop Position Control system
with velocity feedback
(cont.)
From block diagram above, the transfer function for
g m sm
m s
g m s
m
m s
e
m
K K K sT
K K
K K K
sT
K K
sV
s
11
)1(
)(
)(
m sT 1
Therefore the forward path transfer function of the system is,
)1()1(
)(
g m sm g m sm
m s p
K K K sT s
K
K K K sT s
n K K K sG
101
Modeling closed loop Position Control system
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Modeling closed loop Position Control system
with velocity feedback
(cont.)
Therefore the transfer function of system is,
)1(1
)1(
)(
)(
g m sm
g m sm
i
o
K K K sT s
K
K
K K K sT s
K
s
s
g m sm s s
K s K K K s
K
g m sm
)1(2
102
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Modeling closed loop Velocity Control System
103
M d li l d l V l it C t l S t
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Modeling closed loop Velocity Control System
(cont.)
Figure below shows Schematic Diagram for speed control system.
K g
104
K s
R L
m(t)
Tachometer
+E
-E
K p
v e(t)v a(t)
v i (t)
M d li l d l V l it C t l S t
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Modeling closed loop Velocity Control System
(cont.)
• Input Voltage (Input Velocity)
• Precision control of the angular velocity of an inertia load driven directly by an
armature controlled dc motor can be achieved by comparing an input voltage,
Vi representing a demanded speed and derived from a source of constant
voltage Vs via a potentiometer with a feedback voltage Vg derived from a
iiiV
tachometer coupled to the motor shaft.
)()()( t V t V t V g ie
Therefore we get,
and,
)()( t V K t V e sa
105
M d li l d l V l it C t l S t
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Modeling closed loop Velocity Control System
(cont.)
The block diagram of the system are shown below.
106
M d li l d l V l it C t l S t
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Modeling closed loop Velocity Control System
(cont.)
Transfer function for
bt mm
t
a
m
K K B s J Ls R
K
sV
s
))(()(
)(
For small motors, the armature inductance L can be neglected where L=0
,
bt mm
t
a
m
K K B s J R
K
sV
s
)()(
)(
)( bt mm
t
K K RB s RJ
K
1
s K K RB
RJ
K K RB
K
bt m
m
bt m
t
107
M d li l d l V l it C t l S t
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Modeling closed loop Velocity Control System
(cont.)
Let
m
bt m
t K K K RB
K
= motor gain
m
bt m
m
K K RB
= motor time constant
Therefore,
m
m
a
m
s K
sV s
1)()(
108
Modeling closed loop Velocity Control System
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Modeling closed loop Velocity Control System
(cont.)
The overall transfer function
g m s
m
m s
p
i
m
K K K
s
K K
K
s
s
1
)(
)(
m s 1
)1()(
)(
g m sm
m s p
i
m
K K K s
K K K
s
s
109
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Did u know what is Speed Regulation?
(refer to extra note 8)
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