Chapter 2 – Analog Control System (cont.)
Electrical Elements ModellingMechanical Elements Modelling
4. Electrical Elements Modelling
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:
)()(
tidt
tdvC C
Therefore,
4
Potentiometer
A potentiometer is used to measure a linear or rotational displacement.
Linear Rotational
5
Rotational Potentiometer
The output voltage,
Where Kp is the constant in V/rad.
Where max is the maximum value for (t). The Laplace transform of the equation is
6
Tachometer
The tachometer produces a direct current voltage which is proportional to the speed of the rotating axis
7
Operational Amplifier (Op-Amp)
8
DC Motor Applications e.g. tape drive, disk drive, printer, CNC machines, and robots. The equivalent circuit for a dc motor is
9
DC Motor (cont.)Reduced block diagram
The transfer function
(consider TL(t) equals to zero)
10
Example 1
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 1 (cont.)
12
Now, writing the mesh equations,
Nodal Analysis
13
5. Mechanical Elements Modelling
The motion of mechanical elements can be described in various dimensions, which are:
1. Translational.
2. Rotational.
3. Combination of both.
14
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.
15
Translational Mechanical System
16
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.
Figure 2.4 a. Free-body diagram of mass, spring, and damper system;b. transformed free-body diagram 17
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
KsfMssF
sXsG
v
2
1
)(
)()(
18
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 ).19
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
))()(()(2 sXosXiksXoms
20
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 ,0dt
dXfFd
21
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
F=maK(Xi-Xo)
m ƒ dt
dXo
20
2
0 )()((dt
xdm
dt
dXoBtXtXk i
22
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 = InertiaT = Torqueθ = Angular Displacementω = Angular Velocity
where Newton’s second law for rotational system are,
onacceleratiangularwhereJTTorque :,)(
23
Modelling of Rotational Mechanical System
24
Example 1
Rotary Mechanical System
25
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
0
dt
dBttK i
00 )()(
26
cont.
From Newton’s second law for rotational system,
Therefore,
Transforming equation above, we get:
Transfer function of system:
,JT 2
2
,dt
dJTrwhere o
dt
dBttk
dt
dJ i
002
02
)()(
)()()(2 sBsssksJs ooio
KBsJs
K
s
s
i
20
)(
)(
27
Example 2
Closed Loop Position Control System
Ks
Load
va(t)
MotorAmplifier Gears
Load
HandwheelPotentiometer
Kp
Error Detector
i
o
e(t)
R L
m(t)
28
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)
29
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:-
+
-
B
Kt
R+Lsi(s) +
-
Ksns
o(s)
Kp
Va(s)
1J1eqs+B1eq
TL(s)
+
-
30
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