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Control Systems DAE 32103
Citation preview
18/03/2014
1
5. Electrical Elements Modeling
6. Mechanical Elements Modeling
Muhammad Faizal bin Ismail
Dept. of Electrical Engineering
PPD, UTHM
013-7143106
5. Electrical Elements Modeling
2
18/03/2014
2
Example – RLC Network
Determine the transfer function of the circuit.
Solution:
All initial conditions are zero. Assume the output is vc(t).
The network equations are
)()(
)()(
)()(
tvdt
tdiLRtitv
tvvvtv
C
CLR
++=
++=
3
cont.
Laplace transform the equation:
4
)()(
tidt
tdvC C =
Therefore,
18/03/2014
3
RLC network
Control Systems Engineering, Fourth
Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
RLC network
Control Systems Engineering, Fourth
Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
Block diagram of series RLC electrical
network
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Laplace-transformed network
Control Systems Engineering, Fourth
Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
RLC network
a. Two-loop electrical
network;
b. transformed
two-loop
electrical
network;
c. block diagram
RLC network
18/03/2014
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Block diagram of the network
Control Systems Engineering, Fourth
Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.
RLC network
RLC network – Example 1
Answer = 81516
5
)(
)()(
2++
==sssV
sVsG
i
o
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Example 2
Problem: Find the transfer function, G(s) = VL(s)/V(s). Solve the problem two ways –
mesh analysis and nodal analysis. Show that the two methods yield the
same result.
11
Example 2 (cont.)
12
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Now, writing the mesh equations,
Nodal Analysis
13
Potentiometer
• A potentiometer is used to measure a linear or rotational
displacement.
14
Linear Rotational
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Rotational Potentiometer
• The output voltage,
• Where Kp is the constant in V/rad.
• Where θmax is the maximum value for θ(t).
• The Laplace transform of the equation is
15
Tachometer
• The tachometer produces a direct current voltage which is
proportional to the speed of the rotating axis
16
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Operational Amplifier (Op-Amp)
17
DC Motor
• Applications e.g. tape drive, disk drive, printer, CNC machines, and robots.
• The equivalent circuit for a dc motor is
18
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DC Motor (cont.)
Reduced block diagram
19
The transfer function
(consider TL(t) equals to zero)
6. Mechanical Elements Modeling
The motion of mechanical elements can be
described in various dimensions, which are:
1. Translational.
2. Rotational.
3. Combination of both.
20
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Translation
• The motion of translation is defined as a
motion that takes place along or curved path.
• The variables that are used to describe
translational motion are acceleration, velocity,
and displacement.
21
Translational Mechanical System
22
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Example 1
Find the transfer function for the spring-mass-damper system shown
below.
Solution:
1. Draw the free-body diagram of a system and assume the mass is traveling
toward the right.
23Figure 2.4 a. Free-body diagram of mass, spring, and damper system;
b. transformed free-body diagram
cont.
2. From free-body diagram, write differential equation of motion using Newton’s
Law. Thus we get;
3. Laplace transform the equation:
4. Find the transfer function:
)()()()(
2
2
tftKxdt
tdxf
dt
txdM v =++
)()()()(2
sFsKXssXfsXMs v =++
24
KsfMssF
sXsG
v ++==
2
1
)(
)()(
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Example 2
Find the transfer function, xo(s)/xi(s) for the spring-mass system.
Solution:
• The ‘object’ of the above system is to force the mass (position xo(t)) to follow a command position xi(t).
• When the spring is compressed an amount ‘x’m, it produces a force ‘kx’ N ( Hooke’s Law ).
25
cont.
• When one end of the spring is forced to move an amount xi(t), the other end will move and the net compression in the spring will be
x(t) = xi(t) – xo(t)
• So the force F acting on the mass are,
• From Newton’s second law of motion, F = ma
• Therefore,
• Transforming the equation:
NtXotXiktF ))()(()( −=
2
2
))()((dt
XodmtXotXik =−∴
26
))()(()(2 sXosXiksXoms −=
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Example 3
Find the transfer function for the spring-mass with viscous frictional damping.
Solution:
The friction force produced by the dash pot is proportional with velocity,
which is; ƒ = viscous frictional constant N/ms-1
,0
dt
dXfFd =
27
cont.
The net force F tending to accelerate the mass is F= Fs – FD,
F = k ( Xi(t) – Xo(t) ) – ƒ
Free Body Diagram,
From N II,
F = ma
Laplace transform,
Ms2Xo(s) = k[Xi(s) – Xo(s)] – BsXo(s)
dt
dXo
28
F=ma
K(Xi-Xo)m ƒ
dt
dXo
2
0
2
0)()((
dt
xdm
dt
dXoBtXtXk i =−−
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Rotational Mechanical System
• 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,
J = Inertia
T = Torque
θ = Angular Displacement
ω = Angular Velocity
where Newton’s second law for rotational system are,
onacceleratiangularwhereJTTorque∑ == αα :,)(
29
Modeling of Rotational
Mechanical System
30
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16
Example 1
Rotary Mechanical System
31
cont.
• The shaft has a stiffness k, which means, if the shaft is twisted through an angle θ, it will produce a torque kθ, where K – (Nm/rad).
• For system above the torque produce by flexible shaft are,
Ts = K (θi (t)-θo(t)) Nm
• The viscous frictional torque due to paddle
• Therefore the torque required to accelerating torque acting on the mass is
Tr = Ts - TD
dt
dBTD
0θ=
( )dt
dBttK i
0
0)()(
θθθ −−=
32
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cont.
• From Newton’s second law for rotational system,
• Therefore,
• Transforming equation above, we get:
• Transfer function of system:
33
,αJT =2
2
,dt
dJTrwhere oθ
=
( )dt
dBttk
dt
dJ i
0
02
0
2
)()(θ
θθθ
−−=
( ) ( ) )()()(2 sBsssksJs ooio θθθθ −−=
KBsJs
K
s
s
i ++=
2
0
)(
)(
θ
θ
Example 2
Closed Loop Position Control System
34
Ks
Load
va(t)
MotorAmplifier Gears
Load
HandwheelPotentiometer
Kp
Error Detector
θi
θo
e(t)
R L
θm(t)
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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:-
1. A pair of potentiometers acts as an error-measuring device.
2. For input potentiometer, vi(t) = kpθi(t)
3. For the output potentiometer, vo(t) = kpθo(t)
4. The error signal, Ve(t) = Vi(t) – Vo(t) = kpθi(t) - kpθo(t) (1)
5. This error signal are amplified by the amplifier with gain constant, Ks. Va(t) = K s Ve(t) (2)
35
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:-
36
+
-
B
Kt
R+LsΘi(s) +
-
Ksn
s
Θo(s)Kp
Va(s)1
J1eqs+B1eq
TL(s)
+-