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5. Electrical Elements Modeling

6. Mechanical Elements Modeling

Muhammad Faizal bin Ismail

Dept. of Electrical Engineering

PPD, UTHM

ifaizal@uthm.edu.my

013-7143106

5. Electrical Elements Modeling

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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

++=

++=

3

cont.

Laplace transform the equation:

4

)()(

tidt

tdvC C =

Therefore,

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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

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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.

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Example 2 (cont.)

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Now, writing the mesh equations,

Nodal Analysis

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Potentiometer

• A potentiometer is used to measure a linear or rotational

displacement.

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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

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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

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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.

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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.

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 =++

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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 ).

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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 =−∴

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))()(()(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

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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 :,)(

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Modeling 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

0θ=

( )dt

dBttK i

0

0)()(

θθθ −−=

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cont.

• From Newton’s second law for rotational system,

• Therefore,

• Transforming equation above, we get:

• Transfer function of system:

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,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

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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)

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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:-

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+

-

B

Kt

R+LsΘi(s) +

-

Ksn

s

Θo(s)Kp

Va(s)1

J1eqs+B1eq

TL(s)

+-