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