meeting w4 chapter 2 part 2
TRANSCRIPT
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Chapter 2 – Analog Control System (cont.)
Electrical Elements ModellingMechanical Elements Modelling
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4. Electrical Elements Modelling
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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
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cont.
Laplace transform the equation:
)()(
tidt
tdvC C
Therefore,
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Potentiometer
A potentiometer is used to measure a linear or rotational displacement.
Linear Rotational
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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
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Tachometer
The tachometer produces a direct current voltage which is proportional to the speed of the rotating axis
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Operational Amplifier (Op-Amp)
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DC Motor Applications e.g. tape drive, disk drive, printer, CNC machines, and robots. The equivalent circuit for a dc motor is
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DC Motor (cont.)Reduced block diagram
The transfer function
(consider TL(t) equals to zero)
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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.
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Example 1 (cont.)
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Now, writing the mesh equations,
Nodal Analysis
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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.
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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.
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Translational Mechanical System
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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.
Figure 2.4 a. Free-body diagram of mass, spring, and damper system;b. transformed free-body diagram 17
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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
)(
)()(
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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 ).19
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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
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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 ,0dt
dXfFd
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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
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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 = InertiaT = Torqueθ = Angular Displacementω = Angular Velocity
where Newton’s second law for rotational system are,
onacceleratiangularwhereJTTorque :,)(
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Modelling of Rotational Mechanical System
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Example 1
Rotary Mechanical System
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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 )()(
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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
)(
)(
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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)
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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)
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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)
+
-
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