MECHATRONICS

Learning Resources
Unit wise
Compiled by
Dr. A. P. Sathiyagnanam
Assistant Professor
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Unit I & II
INTRODUCTION
What Is Mechatronics?
The term mechatronics used for this integration of microprocessor control systems, electrical
systems and mechanical systems. A mechatronics system is not just a marriage of electrical and mechanical
systems and is more than just a control system; it is complete integration of all them.
Open Loop and Closed Loop System
An open – loop system the speed of rotation of the shaft might be determined solely by the initial
setting of a knob which affects the voltage applied to the motor. Any changes in the supply voltage, the
characteristics of the motor as a result of temperature changes, or the shaft load will change the shaft speed
but not to be compensated for. There is no feedback loop.
With a closed-loop system, however, the initial setting of the control knob will be for a particular
shaft speed and this will be maintained by feedback, regardless of any changes in supply voltage, motor
characteristics or load. In an open-loop control system the output from the system has no effect on the input
signal. In a closed-loop control system the output does have an effect on the input signal, modifying it to
maintain an output signal at the required value.

Figure 1.1
Kirchhoff’s Law or Low of Governing
Law 1: The total current flowing towards a junction is equal to the total flowing from the junction, i.e. the
algebraic sum of the currents at the junction is zero.
Law 2: In a closed circuit or loop, the algebraic sum of the potential differences across each part of the
circuit is equal to the applied e.m.f.
A convenient way of using law 1 is called node analysis since the law is applied to each principal
node of a circuit, a node being a point of connection or junction between building blocks or circuit elements
and principal node being one where three or more branches of the circuit meet. A convenient way of using
law 2 is called mesh analysis since the law is applied to each mesh, a mesh being a closed path or loop which
contains no other loop.

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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Figure 1.2

SYSTEM MODELS

MECHANICAL TRANSLATIONAL SYSTEMS

The model of mechanical translational systems can be obtained by using three basic elements Mass,
Spring and Dash-Pot. These three elements represent three essential phenomena which occur in various
ways in mechanical systems.
List of Symbols Used In Mechanical Translational System
x = Displacement, m.
𝑑𝑑𝑑𝑑
v = Velocity, m/sec2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑥𝑥
a= = Acceleration, m/sec2
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2
f = Applied force, N (Newton’s)
fm = Opposing force offered by mass of the body, N
fk = Opposing force offered by the elasticity of the body (spring), N
fb = Opposing force offered by the friction of the body (dash – pot), N
M = Mass, kg
K = Stiffness of spring, N/m
B = Viscous friction co – efficient, N – sec/m
Note: Lower case letters are functions of time.
Mass
Consider an ideal mass element shown in figure1.3which has negligible friction and elasticity. Let a
force be applied on it. The mass will offer an opposing force which is proportional to acceleration of the
body.
Let f = Applied force
fm= Opposing force due to mass
Here fmα a
Figure 1.3

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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𝑑𝑑 2 𝑥𝑥 𝑑𝑑 2 𝑥𝑥
fm α orfm = M
𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑 2
𝑑𝑑 2 𝑥𝑥
By Newton’s second law, f =fm = M ………….. (1)
𝑑𝑑𝑑𝑑 2
Dash-Pot Figure 1.4
Consider an ideal frictional element dashpot shown in figure 1.4 which has negligible mass and
elasticity. Let a force be applied on it. The dash-pot will offer an opposing force which is proportional to
velocity of the body.
Let f = Applied force
fb= Opposing force due to friction
Here, fbα v
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
fbα or fb = B
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
By Newton’s second law, f =fb = B ……. (2)
𝑑𝑑𝑑𝑑
When the dashpot has displacement at both ends as shown in figure 1.5 the opposing force is
proportional to differential velocity.
𝑑𝑑(𝑥𝑥 1 −𝑥𝑥 2 ) 𝑑𝑑(𝑥𝑥 1 −𝑥𝑥 2 )
fb α 𝑑𝑑𝑑𝑑
; 𝑓𝑓𝑏𝑏 = B
𝑑𝑑𝑑𝑑
𝑑𝑑(𝑥𝑥 1 −𝑥𝑥 2 )
∴ 𝑓𝑓 = 𝑓𝑓𝑏𝑏 = 𝐵𝐵 ….… (3)
𝑑𝑑𝑑𝑑

Figure 1.5
Spring
Consider an ideal elastic element spring shown in figure 1.6 which has negligible mass and friction.
Let a force be applied on it. The spring will offer an opposing force which is proportional to displacement of
the body.
Let f = Applied force
𝑓𝑓𝑘𝑘 = opposing force due to elasticity
Here 𝑓𝑓𝑘𝑘 ∝ 𝑥𝑥𝑥𝑥𝑥𝑥𝑓𝑓𝑘𝑘 = 𝐾𝐾𝐾𝐾 Figure 1.6
By Newton’s second law, f =𝑓𝑓𝑘𝑘 = 𝐾𝐾𝐾𝐾 .……. (4)

When the spring has displacement at both ends as shown in figure 1.7 the opposing force is proportional to
differential displacement.
𝑓𝑓𝑘𝑘 ∝ (𝑥𝑥1 − 𝑥𝑥2 )
𝑓𝑓𝑘𝑘 = 𝐾𝐾 (𝑥𝑥1 − 𝑥𝑥2 )
∴ 𝑓𝑓 = 𝑓𝑓𝑘𝑘 = 𝐾𝐾 (𝑥𝑥1 − 𝑥𝑥2 ) ………(5)
Figure 1.7
1] Problems
Write the differential equations governing the mechanical system shown in figure 1 and determine the
transfer function.

Figure 1
SOLUTION

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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In the given system, applied force f(t) is the input and displacement x is the output.
Let Laplace transform of f (t) =𝐿𝐿[𝑓𝑓(𝑡𝑡)] = 𝐹𝐹(𝑠𝑠)
And Laplace transform of x = L [x] = X (s)
𝑋𝑋(𝑠𝑠)
Hence the required transfer function is
𝐹𝐹(𝑠𝑠)
The system has two nodes and they are mass M1and M2. The free body diagram of mass M1is shown in
figure 2. The opposing forces acting on mass M1 are marked as 𝑓𝑓𝑚𝑚1 , 𝑓𝑓𝑏𝑏1 , 𝑓𝑓𝑏𝑏 , 𝑓𝑓𝑘𝑘1 and 𝑓𝑓𝑘𝑘
𝑑𝑑 2 𝑥𝑥 1 𝑑𝑑𝑑𝑑 1
𝑓𝑓𝑚𝑚1 = 𝑀𝑀1 ; 𝑓𝑓𝑏𝑏1 = 𝐵𝐵1 ; 𝑓𝑓𝑘𝑘1 = 𝐾𝐾1 𝑥𝑥1
𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑
𝑑𝑑
𝐵𝐵 (𝑥𝑥 − 𝑥𝑥2 ); 𝑓𝑓𝑘𝑘 = 𝐾𝐾 (𝑥𝑥1 – 𝑥𝑥)
𝑑𝑑𝑑𝑑 1
By Newton’s second law
𝑓𝑓𝑚𝑚1 + 𝑓𝑓𝑏𝑏1 + 𝑓𝑓𝑏𝑏 + 𝑓𝑓𝑘𝑘1 + 𝑓𝑓𝑘𝑘 = 0

𝑑𝑑 2 𝑥𝑥 1 𝑑𝑑𝑑𝑑 1 𝑑𝑑
∴ 𝑀𝑀1 𝑑𝑑𝑑𝑑 2
+ 𝐵𝐵1 𝑑𝑑𝑑𝑑
+ 𝐵𝐵
𝑑𝑑𝑑𝑑
�𝑥𝑥1 – 𝑥𝑥� + 𝐾𝐾1 𝑥𝑥1 + 𝐾𝐾�𝑥𝑥1 – 𝑥𝑥� = 0……….. (1)

On taking Laplace transform with zero initial conditions

𝑀𝑀1 𝑠𝑠 2 𝑋𝑋1 (𝑠𝑠) + 𝐵𝐵1 𝑠𝑠𝑋𝑋1 (𝑠𝑠) + 𝐵𝐵𝐵𝐵[𝑋𝑋1 (𝑠𝑠)– 𝑋𝑋(𝑠𝑠)] + 𝐾𝐾1 𝑋𝑋1 (𝑠𝑠) + 𝐾𝐾[𝑋𝑋1 (𝑠𝑠)– 𝑋𝑋(𝑠𝑠)] = 0
𝑋𝑋1 (𝑠𝑠)[𝑀𝑀1 𝑠𝑠 2 + (𝐵𝐵1 + 𝐵𝐵)𝑠𝑠 + (𝐾𝐾1 + 𝐾𝐾)]– 𝑋𝑋(𝑠𝑠)[𝐵𝐵𝐵𝐵 + 𝐾𝐾] = 0

𝑋𝑋1 (𝑠𝑠)[𝑀𝑀1 𝑠𝑠 2 + (𝐵𝐵1 + 𝐵𝐵)𝑠𝑠 + (𝐾𝐾1 + 𝐾𝐾)] = 𝑋𝑋(𝑠𝑠)[𝐵𝐵𝐵𝐵 + 𝐾𝐾]

𝐵𝐵𝐵𝐵+𝐾𝐾
∴ X1 (s) = X (s) …………. (2)
𝑀𝑀1 𝑠𝑠 2 + (𝐵𝐵1 +𝐵𝐵)𝑠𝑠+(𝐾𝐾1 + 𝐾𝐾)

Note: Laplace transform of 𝑥𝑥1 = 𝐿𝐿[𝑥𝑥] = 𝑋𝑋1 (𝑠𝑠)

The free body diagram of mass M2is shown in figure 3. The opposing forces acting on M2 are marked
as fm2, fb2, fb and fk.

𝑑𝑑 2 𝑥𝑥 𝑑𝑑𝑑𝑑
Fm2 = M2 ; 𝑓𝑓𝑏𝑏2 = 𝐵𝐵2
𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑
𝑑𝑑
𝑓𝑓𝑏𝑏 = 𝐵𝐵 (𝑥𝑥 − 𝑥𝑥1 ) ; 𝑓𝑓𝑘𝑘 = 𝐾𝐾(𝑥𝑥 − 𝑥𝑥1 )
𝑑𝑑𝑑𝑑

By Newton’s second law

Figure 3

𝑓𝑓𝑚𝑚2 + 𝑓𝑓𝑏𝑏2 + 𝑓𝑓𝑏𝑏 + 𝑓𝑓𝑘𝑘 = 𝑓𝑓(𝑡𝑡)

𝑑𝑑 2 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑
𝑀𝑀2
𝑑𝑑𝑑𝑑 2
+ 𝐵𝐵2
𝑑𝑑𝑑𝑑
+ 𝐵𝐵 (𝑥𝑥 − 𝑥𝑥1 ) + 𝐾𝐾(𝑥𝑥 − 𝑥𝑥1 ) = 𝑓𝑓(𝑡𝑡)
𝑑𝑑𝑑𝑑
……….. (3)

On taking Laplace transform with zero initial conditions

𝑀𝑀2 𝑠𝑠 2 𝑋𝑋(𝑠𝑠) + 𝐵𝐵2 𝑠𝑠𝑠𝑠(𝑠𝑠) + 𝐵𝐵𝐵𝐵[𝑋𝑋(𝑠𝑠) − 𝑋𝑋1 (𝑠𝑠)] + 𝐾𝐾[𝑥𝑥(𝑠𝑠) − 𝑋𝑋1 (𝑠𝑠)] = 𝐹𝐹(𝑠𝑠)
𝑋𝑋(𝑠𝑠)[𝑀𝑀2 𝑠𝑠 2 + (𝐵𝐵2 + 𝐵𝐵)𝑠𝑠 + 𝐾𝐾] − 𝑋𝑋1 (𝑠𝑠)[𝐵𝐵𝐵𝐵 + 𝐾𝐾] = 𝐹𝐹(𝑠𝑠) ………… (4)
Substituting for X1 (s) from equation (2) in equation (4) we get,

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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(Bs +K)2
X(s) [𝑀𝑀2 𝑠𝑠 2 + (𝐵𝐵2 + 𝐵𝐵)𝑠𝑠 + 𝐾𝐾]– 𝑋𝑋(𝑠𝑠) M 2 = 𝐹𝐹(𝑠𝑠)
1 s + (B 1 + B)s+K 1 + K)

[𝑀𝑀1 𝑠𝑠 2 + 𝐵𝐵1 + 𝐵𝐵)𝑠𝑠+(𝐾𝐾1 + 𝐾𝐾)] [𝑀𝑀2 𝑠𝑠 2 + (𝐵𝐵2 + 𝐵𝐵)𝑠𝑠 + 𝐾𝐾] −(𝐵𝐵𝐵𝐵+𝐾𝐾)2

X(s)� �= F(s)
𝑀𝑀1 𝑠𝑠 2 + (𝐵𝐵1 +𝐵𝐵) 𝑠𝑠 +(𝐾𝐾1 + 𝐾𝐾)

𝑋𝑋(𝑠𝑠) 𝑀𝑀1 𝑠𝑠 2 + (𝐵𝐵1 + 𝐵𝐵)𝑠𝑠 + (𝐾𝐾1 + 𝐾𝐾)

∴ =
𝐹𝐹(𝑠𝑠) [𝑀𝑀1 𝑠𝑠 2 + (𝐵𝐵1 + 𝐵𝐵)𝑠𝑠 + (𝐾𝐾1 + 𝐾𝐾) ] [𝑀𝑀2 𝑠𝑠 2 + (𝐵𝐵2 + 𝐵𝐵)𝑠𝑠 + 𝐾𝐾 ] − (𝐵𝐵𝐵𝐵 + 𝐾𝐾)2 ]
RESULT
The differential equations governing the system are
𝑑𝑑 2 𝑥𝑥 1 𝑑𝑑𝑑𝑑 𝑑𝑑
1. 𝑀𝑀1 𝑑𝑑𝑑𝑑 2
+ 𝐵𝐵1 𝑑𝑑𝑑𝑑1 + 𝐵𝐵 (𝑥𝑥1 − 𝑥𝑥) + 𝐾𝐾1 𝑥𝑥1 + 𝐾𝐾(𝑥𝑥1 − 𝑥𝑥) =0
𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑
2. 𝑀𝑀2 2 + 𝐵𝐵2 + 𝐵𝐵 (𝑥𝑥 − 𝑥𝑥1 ) + 𝐾𝐾(𝑥𝑥 − 𝑥𝑥1 ) = 𝑓𝑓(𝑡𝑡)
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
The transfer function of the system is
𝑋𝑋(𝑠𝑠) 𝑀𝑀1 𝑠𝑠 2 +(𝐵𝐵1 +𝐵𝐵)𝑠𝑠 +(𝐾𝐾1 +𝐾𝐾)
=
𝐹𝐹(𝑠𝑠) [𝑀𝑀1 𝑠𝑠 2 + (𝐵𝐵1 +𝐵𝐵)𝑠𝑠 +(𝐾𝐾1 +𝐾𝐾)][𝑀𝑀2 𝑠𝑠 2 +(𝐵𝐵2 +𝐵𝐵)𝑠𝑠 +𝐾𝐾]− (𝐵𝐵𝐵𝐵 + 𝐾𝐾)2

MECHANICAL ROTATIONAL SYSTEMS

The model of rotational mechanical systems can be obtained by using three elements, moment
inertia [J] of mass, das-pot with rotational frictional coefficient [B] and torsional spring with stiffness [K].
List of Symbols Used In Mechanical Rotational System
θ = Angular displacement, rad
𝑑𝑑 θ
𝑑𝑑𝑑𝑑
= Angular velocity, rad/sec
𝑑𝑑 2 θ
= Angular acceleration, rad/sec2
𝑑𝑑𝑑𝑑 2
T = Applied torque, N-m
J = Moment of inertia, kg-m2/rad
B = Rotational frictional coefficient, N-m / (rad/sec)
K = Stiffness of the spring, N-m/rad
Mass
Consider an ideal mass element shown figure 1.8which has negligible friction and elasticity. The
opposing torque due to moment of inertia is proportional to the angular acceleration.
Let T = Applied torque
𝑇𝑇𝑗𝑗 = Opposing torque due to moment of inertia of
the body.
𝑑𝑑 2 θ 𝑑𝑑 2 θ
Here 𝑇𝑇𝑗𝑗 ∝ 𝑑𝑑 𝑡𝑡
or 𝑇𝑇𝑗𝑗 = 𝐽𝐽 𝑑𝑑𝑑𝑑 2 Figure 1.8
By Newton’s second law
𝑑𝑑 2 θ
T =𝑇𝑇𝑗𝑗 = 𝐽𝐽 2 ………. (1)
𝑑𝑑𝑑𝑑
Dash pot
Consider an ideal frictional element dash pot shown in
figure which has negligible moment of inertia and elasticity. Let
a torque be applied on it. The dash pot will offer on opposing
torque which is proportional to the angular velocity of the
body.
Let T = Applied torque

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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𝑇𝑇𝑏𝑏 = Opposing torque due to friction

𝑑𝑑θ 𝑑𝑑θ
𝑇𝑇𝑏𝑏 α 𝑜𝑜𝑜𝑜𝑇𝑇𝑏𝑏 = 𝐵𝐵
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
By Newton’s second law
𝑑𝑑 θ
T = 𝑇𝑇𝑏𝑏 = 𝐵𝐵 ……….. (2)
𝑑𝑑𝑑𝑑
When the dash point has angular displacement at both ends as shown in fig. the opposing torque is
proportional to the differential angular velocity.
𝑑𝑑 𝑑𝑑
𝑇𝑇𝑏𝑏 α (θ1 − θ2 ) 𝑇𝑇𝑏𝑏 = 𝐵𝐵 (θ1 − θ2 )
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑑𝑑
∴ 𝑇𝑇 = 𝑇𝑇𝑏𝑏 = 𝐵𝐵 (θ1 − θ2 ) ………… (3)
𝑑𝑑𝑑𝑑
Spring
Consider an ideal elastic element, torsional spring as shown in fig. which has negligible moment of
inertia and friction. Let a torque be applied on it. The torsional spring will offer an opposing torque which is
proportional to angular displacement of the body.
Let T = Applied torque.
𝑇𝑇𝑘𝑘 = Opposing torque due to elasticity
𝑇𝑇𝑘𝑘 αθ 𝑇𝑇𝑘𝑘 = 𝐾𝐾θ
By Newton’s second law
𝑇𝑇 = 𝑇𝑇𝑘𝑘 = 𝐾𝐾θ ………… (4)
When the spring has angular displacement at both ends shown in fig. the opposing torque is
proportional to differential angular displacement.
𝑇𝑇𝑘𝑘 α(θ1 − θ2 )
𝑇𝑇𝑘𝑘 = 𝐾𝐾 (θ1 − θ2 )
∴ 𝑇𝑇 = 𝑇𝑇𝑘𝑘 = 𝐾𝐾 (θ1 − θ2 ) ………. (5)
Figure 1.9
2] Problem
Write the differential equations governing the mechanical
rotational system shown in figure 1. Obtain the transfer function of the system.

(Applied torque) (Output)

Figure 1
SOLUTION
In the given system, applied torque T is the input and angular displacement θ is the output.
Let Laplace transform of T = L [T] =T(s)
And Laplace transform of θ = L [θ] = θ(s)
θ (𝑠𝑠)
Hence the required transfer function is
𝑇𝑇(𝑠𝑠)
The system has two nodes and they are masses with
moment of inertia 𝐽𝐽1 and𝐽𝐽2 . The differential equations governing
the system are given by torque balance equations at these
nodes.
Let the angular displacement of mass with moment of
inertia 𝐽𝐽1 beθ1 . The Laplace transform of θ1 = L [θ1 ] = θ1 (s). The Figure 2
free body diagram of 𝐽𝐽1 is shown fig. The opposing torques
acting on 𝐽𝐽1 are marked as

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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𝑇𝑇𝑗𝑗 1 and𝑇𝑇𝑘𝑘
𝑑𝑑2 θ1
𝑇𝑇𝑗𝑗 1 = 𝐽𝐽1
𝑑𝑑𝑑𝑑 2

∴ 𝑇𝑇𝑘𝑘 = 𝐾𝐾 (θ1 − θ)
By Newton’s second law,
𝑇𝑇𝑗𝑗 1 + 𝑇𝑇𝑘𝑘 = 𝑇𝑇
2
𝑑𝑑 θ1
𝐽𝐽1 + 𝐾𝐾(θ1 − θ) = 𝑇𝑇
𝑑𝑑𝑑𝑑 2

𝑑𝑑 2 θ1
𝐽𝐽1
𝑑𝑑𝑑𝑑 2
+ 𝐾𝐾 θ1 − 𝐾𝐾θ = 𝑇𝑇 ………….. (1)
On taking Laplace transform with zero initial conditions,
𝐽𝐽1 𝑠𝑠 2 θ1 (𝑠𝑠) + 𝐾𝐾 θ1 (𝑠𝑠) − 𝐾𝐾θ(𝑠𝑠) = 𝑇𝑇(𝑠𝑠)
(𝐽𝐽1 𝑠𝑠 2 + 𝐾𝐾)𝜃𝜃1 (𝑠𝑠) − 𝐾𝐾𝐾𝐾(𝑠𝑠) = 𝑇𝑇(𝑠𝑠) ………… (2)
The free body diagram of mass with moment of inertia 𝐽𝐽2 is shown in figure 3. The opposing torques
acting on 𝐽𝐽2 are marked as 𝑇𝑇𝑗𝑗 2 , 𝑇𝑇𝑏𝑏 and 𝑇𝑇𝑘𝑘 .
𝑑𝑑2 𝜃𝜃 𝑑𝑑𝑑𝑑
𝑇𝑇𝑗𝑗 2 = 𝐽𝐽2 2 ; 𝑇𝑇𝑏𝑏 = 𝐵𝐵
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

and 𝑇𝑇𝑘𝑘 = 𝐾𝐾(𝜃𝜃 − 𝜃𝜃1 )

By Newton’s second law,

𝑇𝑇𝑗𝑗 2 + 𝑇𝑇𝑏𝑏 + 𝑇𝑇𝑘𝑘 = 0
𝑑𝑑2 𝜃𝜃 𝑑𝑑𝑑𝑑 Figure 3
∴ 𝐽𝐽2 + 𝐵𝐵 + 𝐾𝐾(𝜃𝜃 − 𝜃𝜃1 ) = 0
𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝑑𝑑
𝐽𝐽2 + 𝐵𝐵 + 𝐾𝐾𝐾𝐾 − 𝐾𝐾𝜃𝜃1 = 0 ………… (3)
𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑

On taking Laplace transform with zero initial conditions,

𝐽𝐽2 𝑠𝑠 2 𝜃𝜃(𝑠𝑠) + 𝐵𝐵𝐵𝐵𝐵𝐵(𝑠𝑠) + 𝐾𝐾𝐾𝐾(𝑠𝑠) − 𝐾𝐾𝜃𝜃1 (𝑠𝑠) = 0
(𝐽𝐽2 𝑠𝑠 2 + 𝐵𝐵𝐵𝐵 + 𝐾𝐾)𝜃𝜃(𝑠𝑠) − 𝐾𝐾𝜃𝜃1 (𝑠𝑠) = 0
( 𝒋𝒋𝟐𝟐 𝒔𝒔𝟐𝟐 +𝑩𝑩𝑩𝑩+𝑲𝑲)
𝜃𝜃1 (𝑠𝑠) =
𝐾𝐾
θ (𝑠𝑠) ………… (4)
Substituting for 𝜃𝜃1 (s) from equation (4) in equation (2) we get,

( 𝑱𝑱𝟐𝟐 𝒔𝒔𝟐𝟐 + 𝑩𝑩𝑩𝑩 + 𝑲𝑲)

(𝐽𝐽1 𝑠𝑠 2 + 𝐾𝐾) 𝜃𝜃(𝑠𝑠) − 𝐾𝐾𝐾𝐾(𝑠𝑠) = 𝑇𝑇(𝑠𝑠)
𝐾𝐾
( 𝑱𝑱𝟏𝟏 𝒔𝒔 + 𝑲𝑲 ) �𝑱𝑱𝟐𝟐 𝒔𝒔 + 𝑩𝑩𝑩𝑩 + 𝑲𝑲� − 𝑲𝑲𝟐𝟐
𝟐𝟐 𝟐𝟐
� � 𝜃𝜃(𝑠𝑠) = 𝑇𝑇(𝑠𝑠)
𝐾𝐾

𝜃𝜃(𝑠𝑠) 𝐾𝐾
∴ 𝑇𝑇(𝑠𝑠) = ………….. (5)
( 𝐽𝐽 1 𝑠𝑠 2 + 𝐾𝐾 ) (𝐽𝐽 2 𝑠𝑠 2 + 𝐵𝐵𝐵𝐵 +𝐾𝐾)−𝐾𝐾 2
RESULT
The differential equations governing the system are
𝑑𝑑 2 𝜃𝜃1
1. 𝐽𝐽1 𝑑𝑑𝑑𝑑 2
+ 𝐾𝐾𝜃𝜃1 − 𝐾𝐾𝐾𝐾 = 0

𝑑𝑑 2 𝜃𝜃1 𝑑𝑑𝑑𝑑
2. 𝐽𝐽2 𝑑𝑑𝑑𝑑 2
+ 𝐵𝐵 + 𝐾𝐾𝐾𝐾 − 𝐾𝐾𝜃𝜃1 = 0
𝑑𝑑𝑑𝑑

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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𝜃𝜃(𝑠𝑠) 𝐾𝐾
The transfer function of the system is = (𝐽𝐽 1 𝑠𝑠 2 + 𝐾𝐾)(𝐽𝐽 2 𝑠𝑠 2 + 𝐵𝐵𝐵𝐵+𝐾𝐾)−𝐾𝐾 2
𝑇𝑇(𝑠𝑠)

ELECTRICAL SYSTEMS
The models of electrical systems can be obtained by using resistor, capacitor and inductor. The
current-voltage relation of resistor, inductor and capacitor are given in table 1.1. For modeling electrical
network or equivalent circuit is formed by using R, L and voltage or current source.
The differential equations governing the electrical systems can be formed by writing Kirchoff’s
current law equations by choosing various nodes in the network or Kirchoff’s voltage law equations by
choosing various closed path in the network. The transfer function can be obtained by taking Laplace
transform of the differential equations and rearranging them as a ratio of output to input.
Table 1.1: Current-voltage relation of R, L and C
Voltage across the Current through the
Element
element element

𝑣𝑣(𝑡𝑡)
𝑣𝑣(𝑡𝑡) = 𝑅𝑅𝑅𝑅(𝑡𝑡) 𝑖𝑖(𝑡𝑡) =
𝑅𝑅

𝑑𝑑 1
𝑣𝑣(𝑡𝑡) = 𝐿𝐿 𝑖𝑖(𝑡𝑡) 𝑖𝑖(𝑡𝑡) = ∫𝑣𝑣(𝑡𝑡)𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 𝐿𝐿

1 𝑑𝑑𝑑𝑑(𝑡𝑡)
𝑣𝑣𝑣𝑣 = ∫𝑖𝑖(𝑡𝑡)𝑑𝑑𝑑𝑑 𝑖𝑖(𝑡𝑡) = 𝐶𝐶
𝑐𝑐 𝑑𝑑𝑑𝑑
3] Problem
Obtain the transfer function of the electrical network shown in figure 1

SOLUTION
In the given network input is e(t) and output is V2 (t)
Let Laplace transform of e(t) = L[e(t)] = E(s)
Laplace transform of V2(t) = L[V2(t)] = V2(s)
𝑉𝑉2 (𝑠𝑠)
The transfer function of the network is Figure 1
𝐸𝐸(𝑠𝑠)

Transform the voltage source in series with resistance R1 into equivalent current source as shown in
figure. The network has two nodes. Let the node voltage be v1 and v2. The Laplace transform of node
voltages v1 and v2 are v1(s) and v2(s) respectively. The
differential equations governing the network are given by the
Kirchoff’s current law equations at these nodes.

At node 1, by Kirchoff’s current law

𝑉𝑉1 𝑑𝑑𝑑𝑑1 𝑣𝑣1 −𝑣𝑣2 𝑒𝑒
+ C1 ……..(1)
𝑅𝑅1 𝑑𝑑𝑑𝑑 + 𝑅𝑅2 = 𝑅𝑅1
On taking Laplace transform with zero initial conditions
𝑉𝑉1 𝑉𝑉1 (𝑆𝑆) 𝑉𝑉2 (𝑆𝑆) 𝐸𝐸(𝑆𝑆)
𝑅𝑅1
+ C1sV1(s) + - =
𝑅𝑅2 𝑅𝑅2 𝑅𝑅1
1 1 𝑉𝑉2 (𝑆𝑆) 𝐸𝐸(𝑆𝑆)
V1(S) � + 𝑠𝑠𝐶𝐶1 + � - 𝑅𝑅 = 𝑅𝑅 …...… (2)
𝑅𝑅1 𝑅𝑅2 2 1
At node 2, by Kirchoff’s law

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

Figure 2
 [9]

𝑉𝑉2 − 𝑉𝑉1 𝑑𝑑𝑑𝑑

𝑅𝑅2
+ C2 dt2 = 0 ………. (3)
On taking Laplace transform of equation (3) with zero initial condition
𝑉𝑉2 (𝑠𝑠) 𝑉𝑉1 (𝑠𝑠)
- + C2S V2(s)= 0
𝑅𝑅2 𝑅𝑅2
𝑉𝑉1 (𝑠𝑠) 𝑉𝑉2 (𝑠𝑠) 1
= + C2S V2(s)= � + 𝑠𝑠𝐶𝐶2 � V2 (s)
𝑅𝑅2 𝑅𝑅2 𝑅𝑅2
∴V1(S) = [1 + sC2R2]V2(s)
Substituting for V1(s) from equation (4) in equation (2)
1 1 𝑉𝑉2 (𝑠𝑠) 𝐸𝐸(𝑠𝑠)
(1 + sR2C2) V2(s) = �𝑅𝑅 + 𝑠𝑠𝐶𝐶1 + 𝑅𝑅 � - 𝑅𝑅2
=
𝑅𝑅1
1 2
(1+𝑠𝑠𝑅𝑅2 𝐶𝐶2 )(𝑅𝑅2 +𝑅𝑅1 +𝑠𝑠𝐶𝐶1 𝑅𝑅1 𝑅𝑅2 )− 𝑅𝑅1 𝐸𝐸(𝑠𝑠)
� 𝑅𝑅1 𝑅𝑅2
�V2 (s)= 𝑅𝑅
1
𝑉𝑉2 (𝑠𝑠) 𝑅𝑅2
∴ = �(1+𝑠𝑠𝑅𝑅 )(𝑅𝑅 �
𝐸𝐸(𝑠𝑠) 𝐶𝐶
2 2 2 +𝑅𝑅1 +𝑠𝑠𝐶𝐶1 𝑅𝑅1 𝑅𝑅2 )− 𝑅𝑅1 1

RESULT
The (node basis) differential equations governing the electrical network are

𝑉𝑉1 𝑑𝑑𝑑𝑑1 𝑣𝑣1 −𝑣𝑣2 𝑒𝑒

1. + C1
𝑅𝑅1 𝑑𝑑𝑑𝑑 + 𝑅𝑅2 = 𝑅𝑅1

𝑉𝑉2 − 𝑉𝑉1 𝑑𝑑𝑑𝑑2

2. + C2 = 0
𝑅𝑅2 dt

𝑉𝑉2 (𝑠𝑠) 𝑅𝑅
3. = �(1+𝑠𝑠𝑅𝑅 𝐶𝐶 )(𝑅𝑅 +𝑅𝑅 2+𝑠𝑠𝐶𝐶 𝑅𝑅 𝑅𝑅 )− �
𝐸𝐸(𝑠𝑠) 2 2 2 1 1 1 2 𝑅𝑅1 1

Thermal System

List of symbols used in thermal systems

q = Heat flow rate, Kcal/sec
θ1 = Absolute temperature of emitter, °K
θ2 = Absolute temperature of receiver, °K
∆θ = Temperature difference, °C
A = Area normal to heat flow, m3
K = Conduction or convection coefficient, Kcal/sec-°C
Kr = Radiation coefficient, Kcal/sec-°C
H = K/A = Convection coefficient, Kcal/m-sec-°C
K = Thermal conductivity, Kcal/m-sec °C
∆X = Thickness of conductor, m
R = Thermal resistance, °C-sec/Kcal
C = Thermal capacitance, Kcal/°C

Heat flow rate

Thermal systems are those that involve the transfer of heat from one substance to another. There
are three different ways of heat flow from one substance to another. They are conduction, convection and
radiation.

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For conduction,
𝐾𝐾𝐾𝐾
Heat flow rate, q = K∆θ = ………………(1)
∆𝑋𝑋
For convection,
Heat flow rate, q = K∆θ= HA∆θ …………….. (2)
For radiation,
Heat flow rate, q = Kf(𝜃𝜃14 − 𝜃𝜃24 )
If θ1>>θ2 then, q = Kr𝜃𝜃̅ 4 ………………. (3)
1
Where 𝜃𝜃̅ 4 = (𝜃𝜃14 − 𝜃𝜃24 ) 4

Note: 𝜃𝜃̅ 4 is called effective temperature difference of the emitter and receiver.
Basic elements of thermal system
The models of thermal system are obtained by using thermal resistance and capacitance which are
the basic elements of the thermal system.
The thermal resistance and capacitance are distributed in nature. But for simplicity in analysis
lumped parameter mode is used. In lumped parameter model it is assured that the substances that are
characterized by resistance to heat flow have negligible heat capacitance and the substances that are
characterized by heat capacitance have negligible resistance to heat flow.
The thermal resistance, R for heat transfer between two substances is defined as the ratio of change
in temperature and change in heat flow rate.
Change in Temperature , °C
Thermal resistance, R =
Cha nge in heat flow rate , Kcal /sec
For conduction or convection,
Heat flow rate, q = K∆θ
On differentiating we get,
dq = K d(∆θ)
𝑑𝑑(∆𝜃𝜃) 1
∴ =
𝑑𝑑𝑑𝑑 𝐾𝐾
𝑑𝑑(∆𝜃𝜃 )
But thermal resistance, R =
𝑑𝑑𝑑𝑑
1
∴ Thermal resistance, R = for conduction …………… (1)
𝐾𝐾
1 1
= = for convection ………… (2)
𝐾𝐾 𝐻𝐻𝐻𝐻
For radiation,
Heat flow rate, q = Kr𝜃𝜃̅ 4
On differentiating we get
dq = Kr 4 𝜃𝜃̅ 3 d 𝜃𝜃̅

𝑑𝑑(∆𝜃𝜃) 1
∴ =
𝑑𝑑𝑑𝑑 𝐾𝐾

�
𝑑𝑑𝜃𝜃
But thermal resistance, R =
𝑑𝑑𝑑𝑑
1
∴ Thermal resistance, R = �3
………… (3)
𝐾𝐾𝑟𝑟 4𝜃𝜃
(for radiation)
Thermal capacitance, C is defined as the ratio of change in heat stored and changes in temperature

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Change in Temperature , °C
Thermal capacitance, C =
Change in heat flow rate , Kcal /sec
Let M = Mass of substance considered, kg
Cp = Specific heat of substance, Kcal/kg-°C
Now, Thermal capacitance, C = Mcp ………… (4)

EXAMPLE OF THERMAL SYSTEM

Consider a simple thermal system shown in figure
1.10. Let us assume that the tank is insulated to eliminate
heat loss of the surrounding air, there is no heat storage in
the insulation and liquid in the tank is kept at uniform
temperature by perfect mixing with the help of a stirrer.
Thus, a single temperature is used to describe the
temperature of the liquid in the tank and of the out flowing
liquid. The transfer function of this system can be derived
as shown below.
Let
𝜃𝜃̅𝑖𝑖 = Steady state temperature of inflowing liquid, °C
𝜃𝜃0̅ = Steady state temperature of out flowing liquid, Figure 1.10 Thermal system
G = Steady state liquid flow rate, kg/sec
M = Mass of liquid in tank, kg.
c = Specific heat of liquid, Kcal/kg °C
R = Thermal resistance, °C – sec/Kcal.
C = Thermal capacitance, Kcal/°C
Q = Steady state heat input rate. Kcal/sec
Let us assume that the temperature of inflowing liquid is kept constant. Let the heat input rate to
the system supplied by the heater is suddenly changed from 𝑄𝑄� to 𝑄𝑄� +qi. Due to this, the heat output flow
rate will gradually change from 𝑄𝑄� to 𝑄𝑄� + q0. The temperature of the out flowing liquid will also be changed
from 𝜃𝜃0̅ to 𝜃𝜃̅0 + θ.
For this system the equation for q0, C and R are obtained as follows,
Change in output heat flow rate, q0
= Liquid flow rate, G × Specific heat of liquid,c× Changing temperature, θ
= Gcθ ………… (1)
Thermal capacitance, C = Mass, M × specific heat of liquid, c
= Mc ………… (2)
Change in Temperature , θ
Thermal resistance, R =
Change in heat flow rate , q 0
θ
= ………… (3)
q0
On substituting for q0 from equation (1) in equation (3) we get,

θ 1
R = = ………… (4)
q0 Gc
In this system, the rate of change of temperature is directly proportional to change in heat input
rate.
dθ
∴ αqi – q0
dt
The constant of proportionality in the capacitance C of the system.

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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dθ
∴C α qi – q0 ………… (5)
dt
Equation (5) is the differential equation governing the system. Since equation (5) is of first order equation,
the system is first order system.
θ θ
From equation (1.20), R = , ∴ q0 = ….….. (6)
q0 R
On substituting for q0 from equation (6) in equation (5) we get,
dθ θ
C = qi -
dt R
dθ Rq i −θ
C = qi -
dt R
dθ
RC = Rqi -θ
dt
dθ
RC +θ = Rqi ……….. (7)
dt
𝑑𝑑𝑑𝑑
Let, L[θ] = θ(s); L� �= sθ(s) and L[qi] = Qi(s)
𝑑𝑑𝑑𝑑
On taking Laplace transform of equation (7) we get,
RC s θ(s) + θ(s) = R Qi(s)
θ(s) [sRc + 1} = R Qi(s)
𝜃𝜃(𝑠𝑠)
is the required transfer function of the system
𝑄𝑄1 (𝑠𝑠)
1
𝜃𝜃(𝑠𝑠) 𝑅𝑅 𝑅𝑅
∴ (𝑠𝑠) = = 1 = 𝐶𝐶
1 ……… (8)
𝑄𝑄1 𝑠𝑠𝑠𝑠𝑠𝑠+1 𝑅𝑅𝑅𝑅�𝑠𝑠+ � 𝑠𝑠+
𝑅𝑅𝑅𝑅 𝑅𝑅𝑅𝑅

FLUID SYSTEM BUILDING BLOCKS

In fluid flow systems there are three basic building block which
can be considered to be the equivalent of electrical resistance,
capacitance and inductance. For such systems (figure 8.19) the input,
the equivalent of the electrical current, is the volumetric rate of flow q,
and the output, the equivalent of electrical potential difference, is
pressure difference (p1 – p2). Fluid systems can be considered to fall Fi 1 11 Fl id t bl k
into two categories: Hydraulic, where the fluid is a liquid and is dement to be incompressible; and
pneumatic, where it is a gas which can be compressed and consequently shows a density changes.

ELECTRICAL EQUIVALENT FOR MECHANICAL SYSTEMS

Transfer function of armature controlled dc motor

The speed of dc motor is directly proportional to armature voltage and inversely proportional to flux
is field winding. In armature controlled dc motor the desired speed is obtained by varying the armature
voltage. This speed control system is an electro-mechanical control system. The electrical system consists of
the armature and the field circuit but for analysis purpose, only the armature circuit is considered because
the field is excited by a constant voltage. The mechanical system consists of the rotating part of the motor
and load connected to the shaft of the motor. The armature controlled dc motor speed control system is
shown in figure.

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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Figure 1.12Armature controlled DC motor

Let
Ra = Armature resistance, Ω
La = Armature inductance, H
ia = Armature current, A
va = Armature voltage, V
eb = Back e.m.f, V
Kt = Torque constant, N-m/A
T = Torque developed by motor, N-m
θ = Angular displacement of shaft, rad.
J = Moment of inertia of motor and load, kg-m2/rad.
B = Frictional coefficient of motor and load, N-m(rad./sec)
Kb = Backe.m.f constant, V/(rad./sec)

The equivalent circuit of armature is shown in figure 1.13

By Kirchoff’s voltage law, we can write
𝑑𝑑𝑖𝑖𝑎𝑎
Ia Ra+ La + eb = va ………….(1)
𝑑𝑑𝑑𝑑
Torque of dc motor is proportional to the product of flux
and current. Since flux is constant in this system, the Figure 1.13 Equivalent circuit of armature
torque is proportional to ia alone.
T ∝ia
∴ Torque, T = Ktia ……… (2)
The mechanical system of the motor is shown in figure 1.14. The
differential equation governing the mechanical system of motor is given
by Figure 1.14
d2θ dθ
J +B =T ………… (3)
dt 2 dt
The back e.m.f for dc machine is proportional to speed (angular velocity) of shaft
dθ dθ
∴eb∝ ; Back e.m.f, eb = Kb ………..(4)
dt dt
The Laplace transform of various times domain signals involved in this system are shown below.
L[Va] = Va(s) L[eb] = Eb(s) L[T] = T(s)
L[ia] =Ia(s) L[θ] = θ(s)
The differential equations governing the armature controlled dc motor speed control system are
di a d2θ dθ
ia R a + L a + eb = v a J +B =T
dt dt 2 dt
dθ
T =Ktia eb = Kb
dt

On taking Laplace transform of the system differential equations with zero initial conditions we get

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

 [14]

Ia(s) Ra + LasIa(s) + Eb(s) = Va(s) ………….. (5)

T(s) =KtIa(s) ………….. (6)
Js2θ(s) + B sθ(s) = T(s) ………….. (7)
Eb(s) =Kbsθ(s) ………….. (8)
On equating equations (6) and (7) we get,
KtIa(s) = (Js2 + Bs) θ(s)
(Js 2 +Bs )
Ia(s) = θ(s) ………….. (9)
Kt
Equation (5) can be written as
(Ra +sLa) Ia(s) +Eb(s) = Va(s) ………….. (10)
Substituting for Eb(s) and Ia(s) from equation (8) and (9) respectively in equation (10),
(Js 2 +Bs )
(Ra +sLa) θ(s) + Kbsθ(s) =Va(s)
Kt
(R a +sL a )�Js 2 +Bs �K b K t s
� �θ(s)= Va(s)
Kt
The required transfer function is θ(s)/Va(s)
θ(s) Kt
∴ = ………….. (11)
V a (s) (R a +sL a ) (Js 2 +Bs )+K b K t s
Kt
=
R a Js 2 + R a Bs +L a Js 3 +L a Bs 2 +K b K t s
Kt
=
[JL a s 2 +(JR a +BL a )s+(BR a +K b K t )]
K t /JL a
= JR +B L a B R a +K b K t ………….. (12)
a�s 2 +� a �s+( )�
JL a JL a
The transfer function of armature controlled dc motor can be expressed in another standard from as
shown below
θ(s) Kt
=
V a (s) (R a +sL a ) (Js 2 +Bs )+K b K t s
Kt
= sL Js 2
R a � a +1� Bs � �+K b K t s
Ra Bs
K t /R a B
= K +K ………….. (13)
s�(1+sT a )(1+sT m )+� b t ��
Ra B
Where, La/Ra = Ra = Electrical time constant
And J/B = Tm = Mechanical time constant
Transfer function of field controlled dc motor
The speed of dc motor is directly proportional to armature voltage and inversely proportional to flux.
In field controlled dc motor the armature voltage is kept constant and the speed is varied by varying the flux
of the machine. Since flux is directly proportional to field current, the flux is varied by varying field current.
The speed control system is an electromechanical control system. The electrical system consists of armature
and field circuit but for analysis purpose, only field circuit is considered because the armature is excited by a
constant voltage. The mechanical system consists of the rotating part of the motor and the load connected
to the shaft of the motor. The field controlled dc motor speed control system is shown in figure 1.15.

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Figure 1.15 Field controlled DC motor

Let
Rf = Field resistance, Ω
Lf = Field inductance, H
if = Field current, A
vf = Field voltage, V
T = Torque developed by motor, N-m
Ktf = Torque constant, N-m/A
J = Moment of inertia of motor and load, kg=m3/rad.
B = Frictional coefficient of motor and load, N-m/(rad./sec)
The equivalent circuit of field is shown in figure 1.16
By Kirchoff’s voltage lad, we can write
𝑑𝑑𝑖𝑖𝑓𝑓
Rf if+ Lf = vf …….. (1)
𝑑𝑑𝑑𝑑
The torque of dc motor is proportional to product of flux and
armature current. Since armature current is constant in this system, the
torque is proportional to flux alone, but flux is preoperational to field
current.
T ∝ef, ∴ Torque, T = Kif if ….. (2)
The mechanical system of the motor is shown in figure 1.17. The Figure 1.16 Equivalent circuit
differential equation governing the mechanical system of the motor is
given by
d2θ dθ
J +B =T ……. (3)
dt 2 dt
The Laplace transform of various time domain signals involved Figure 1.17
in this system are shown below
L[if] = If(s) ; L[T] = T(s); L[vf] = Vf(s); L[θ] = θ(s)
The differential equations governing the field controlled dc motor are
di d2θ dθ
Rf if + Lf f = vf; T = Ktf if ; J +B =T
dt dt 2 dt
On taking Laplace transform of the system differential equation, we get
Rf If (s) + Lf s If(s) = Vf (s) ……. (4)
T(s) =Ktf If(s) ……. (5)
2
Js θ(s) +Bsθ(s) = T(s) ……. (6)
Equation (5) and (6) we get,
Ktf If(s) = Js2θ(s) + Bsθ(s)
(𝐽𝐽𝐽𝐽 +𝐵𝐵)
If(s) = s θ(s) ……. (7)
𝐾𝐾𝑡𝑡𝑡𝑡
The equation (4) can be written as

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(Rf +sLf) If(s) +Eb(s) = Vf(s) ……. (8)

On Substituting for Ib(s) from equation (7) in equation (8) we get,
(Js +B)
(Rf +sLf) s θ(s) = Vf(s)
K tf
θ(s) K tf
= s(R f +sL f ) (B+sJ )
V f (s)
K tf
= sL
sR f �1+ f �(B+sJ)
Rf
Km
= s(1+sT f )(1+sT m )
……. (9)
Where, Motor gain constant, Km =Ktf/RfB
Field time constant, Tf= Lf/Rf
Mechanical time constant, Tm = J/B

Electrical analogous of mechanical translational systems

Systems remain analogous as long as the differential equations governing the systems or transfer
functions are of identical from. The electric analogue of any other kind of system is of great importance
since it is easier to construct electrical models and analyse them.
The three basic elements mass, dash-pot and spring that are used in modeling mechanical
translational systems are analogous to resistance, inductance and capacitance of electrical systems. The
input force in mechanical system is analogous to either voltage source or current source in electrical
systems. The output velocity (first derivative of displacement) in mechanical system is analogous to either
current or voltage in an element in electrical system. Since the electrical systems has two types of inputs
either voltage or current source, there are two types of analogies: force-voltage analogy and force-current
analogy.
Force-voltage analogy
The force balance equations of mechanical elements and their analogous electrical elements in
force-voltage analogy are shown in table 1.2.
The following points serve as guide lines to obtain electrical analogous of mechanical systems based
on force-current analogy.
1. In electrical systems element in parallel will have same voltage, likewise in mechanical systems, the
elements have same force are said to be in parallel.
2. The elements have same velocity in mechanical system should have analogous same voltage in
electrical analogous system.
3. Each node (meeting point of elements) in the mechanical system corresponds to a node in electrical
system. A mass is considered as a node.
4. The number of nodes in electrical analogous is same as that of the number of nodes (masses) in
mechanical system. Hence the number of node voltages and system equations will be same as that
of the number of velocities of (nodes) masses in mechanical system.
5. The mechanical driving sources (forces) and passive elements connected to the node (mass) in
mechanical system should be represented by analogous elements connected to a node in electrical
system.

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Table 1.2
Mechanical system Electrical system
Input: Force : Force Input : Voltage source
Output : Velocity Output : Current through the element
|→ x
𝑑𝑑𝑑𝑑
|→ v =
𝑑𝑑𝑑𝑑

e = Rj
𝑑𝑑𝑑𝑑
f= B = Bv
𝑑𝑑𝑑𝑑

|→ x
𝑑𝑑 2 𝑥𝑥 𝑑𝑑𝑑𝑑
|→ a = =
𝑑𝑑𝑡𝑡 2 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑
e=L
𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝑥𝑥 𝑑𝑑𝑑𝑑
f = 𝑑𝑑𝑡𝑡 2 = M 𝑑𝑑𝑑𝑑

|→ x = ∫ 𝑣𝑣𝑣𝑣𝑣𝑣
|→ v

1
e = ∫ 𝑖𝑖𝑖𝑖𝑖𝑖
𝐶𝐶
f = Kx = K∫ 𝑣𝑣𝑣𝑣𝑣𝑣

Table 1.3 shows the list of analogous quantities in force-voltage analogy

Item Mechanical systems Electrical system
(mesh basis system)
Independent variable Force, f Voltage, e
(input)
Dependent variable Velocity, v Current, i
(output) Displacement, x Charge, q
Dissipative element Frictional coefficient of Resistance, R
dashpot, B
Storage element Mass, M Inductance, L
Stiffness of spring, K Inverse of capacitance, I/C
Physical law Newton’s second law Kirchoff’s voltage law
∑ 𝐹𝐹= 0 ∑ 𝑉𝑉= 0
Changing the level of Lever Transformer
independent variable 𝑓𝑓1 𝑙𝑙 1
=
𝑒𝑒1 𝑁𝑁1
=
𝑓𝑓2 𝑙𝑙 2 𝑒𝑒2 𝑁𝑁2

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6. The mechanical driving sources (force) and passive elements connected to the node (mass) in
mechanical system should be represented by analogous elements in a closed loop in analogous
electrical system.
7. The element connected between two (nodes) masses in mechanical system is represented as a
common element between two meshes in electrical analogous system.
FORCE- CURRENT ANALOGY
The force balance equations of mechanical elements and their analogous electrical elements in
force-current analogy are shown in table 1.4. The table 1.5 shows the list of analogous quantities in force-
current analogy
Table 1.4
Mechanical system Electrical system
Input: Force : Force Input : Current source
Output : Velocity Output : Voltage across the element
|→ x
𝑑𝑑𝑑𝑑
|→ v = 𝑑𝑑𝑑𝑑

1
I= v
𝑑𝑑𝑑𝑑 𝑅𝑅
f= B = Bv
𝑑𝑑𝑑𝑑

|→ x
𝑑𝑑 2 𝑥𝑥 𝑑𝑑𝑑𝑑
|→ a = =
𝑑𝑑𝑡𝑡 2 𝑑𝑑𝑑𝑑

1
i = ∫ 𝑣𝑣𝑣𝑣𝑣𝑣
𝐿𝐿

𝑑𝑑 2 𝑥𝑥 𝑑𝑑𝑑𝑑
f = 𝑑𝑑𝑡𝑡 2 = M 𝑑𝑑𝑑𝑑

|→ x = ∫ 𝑣𝑣𝑣𝑣𝑣𝑣
|→ v

𝑑𝑑𝑑𝑑
i=C
𝑑𝑑𝑑𝑑
f = Kx = K∫ 𝑣𝑣𝑣𝑣𝑣𝑣

The following points serve as guide lines to obtain electrical analogous of mechanical systems based on
force-current analogy.

1. In electrical systems element in parallel will have same voltage, likewise in mechanical systems, the
elements have same force are said to be in parallel.
2. The elements have same velocity in mechanical system should have analogous same voltage in
electrical analogous system.
3. Each node (meeting point of elements) in the mechanical system corresponds to a node in electrical
system. A mass is considered as a node.

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4. The number of nodes in electrical analogous is same as that of the number of nodes (masses) in
mechanical system. Hence the number of node voltages and system equations will be same as that
of the number of velocities of (nodes) masses in mechanical system.
5. The mechanical driving sources (forces) and passive elements connected to the node (mass) in
mechanical system should be represented by analogous elements connected to a node in electrical
system.
6. The element connected between two nodes (masses) in mechanical system is represented as a
common element between two nodes in electrical analogous elements connected to a node in
electrical system.
Table 1.5
Item Mechanical systems Electrical system
(node basis system)
Independent variable Force, f Current, i
(input)
Dependent variable Velocity, v Voltage, V
(output) Displacement, x Flux, φ
Dissipative element Frictional coefficient of Conductance G = 1/R
dashpot, B
Storage element Mass, M Capacitance, C
Stiffness of spring, K Inverse of capacitance, I/L
Physical law Newton’s second law Kirchoff’s voltage law
∑ 𝐹𝐹= 0 ∑ 𝑖𝑖= 0
Changing the level of Lever Transformer
independent variable 𝑓𝑓1 𝑙𝑙 1
=
𝑖𝑖1 𝑁𝑁2
=
𝑓𝑓2 𝑙𝑙 2 𝑖𝑖2 𝑁𝑁1

4] Problems
Write the differential equations governing the mechanical system shown in figure 1. Draw the force-
voltage and force-current electrical analogous circuits and verify by writing mesh and node equations.

SOLUTION
The given mechanical system has to
nodes (masses). The differential equations
governing the mechanical system are given by
force balance equations at these nodes. Let
the displacements of masses M1 and M2 be x1
and x2 respectively. The corresponding
velocities by v1 and v2.
The free body diagram of M1 is shown Figure 1
in figure 2. The opposing forces are marked as
fm1, fb1, fb12 and fk1
𝑑𝑑 2 𝑥𝑥 1 𝑑𝑑𝑥𝑥 𝑑𝑑
fml = M1= 𝑑𝑑𝑡𝑡 2
; fb1 = B1 𝑑𝑑𝑑𝑑1 ; fb12 = B12 (x1 – x2) and
𝑑𝑑𝑑𝑑
fk1 = K1 (x1 – x2)
By Newton’s second law, fm1 + fb1 + fb12 + fk1 = f(t)
𝑑𝑑 2 𝑥𝑥 1 𝑑𝑑𝑥𝑥 𝑑𝑑
M1 +B1 1 + B12 (x1 – x2) + K1 (x1 – x2) = f(t) ………. (1)
𝑑𝑑𝑡𝑡 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
Figure 2

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The free body diagram of M2 is shown in figure 3

The opposing forces are marked as fm2, fb2, f=, fk1 and fk2.
𝑑𝑑 2 𝑥𝑥 𝑑𝑑𝑥𝑥
fm2 = M2 𝑑𝑑𝑡𝑡 22 ; fb2 = B2 𝑑𝑑𝑑𝑑2 ;
𝑑𝑑
fb12 = B12 (x2 – x1); fk1 = K1 (x2 – x1) and fk2 = K2 x2
𝑑𝑑𝑑𝑑
By Newton’s second law, fm2 + fb2 + fk2 + fb12 + fk1 = 0
𝑑𝑑 2 𝑥𝑥 𝑑𝑑𝑥𝑥 𝑑𝑑
M2 𝑑𝑑𝑡𝑡 22 + B2 𝑑𝑑𝑑𝑑2 + K2x2 + B12 (x2 – x1) + K1 (x2 – x1) =0
𝑑𝑑𝑑𝑑
On replacing the displacement by velocity in the differential equations
(1) and (2) of the mechanical systems we get,
Figure 3
𝑑𝑑2 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
�𝑖𝑖. 𝑒𝑒, 2 = ; = 𝑣𝑣 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 = � 𝑣𝑣𝑣𝑣𝑣𝑣�
𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 1
M1 +B1v1 + B12 (v1 – v2) + K1 ∫(𝑣𝑣1 − 𝑣𝑣2 )dt = f(t)
𝑑𝑑𝑑𝑑
𝑑𝑑𝑣𝑣
M2 2 + B2v2 + K2∫ 𝑣𝑣2 dt+ B12 (v2 – v1) + K1∫(𝑣𝑣2 − 𝑣𝑣1 )dt=0
𝑑𝑑𝑑𝑑
Force-voltage analogous circuit
The given mechanical system has two nodes (masses). Hence the force-voltage analogous electrical
circuit will have two meshes.
The force applied to mass, M1 is represented by a voltage source in first mesh. The elements M1, B1,
K1 and B12 are connected to first node. Hence they are represented by analogous elements in mesh 1
forming a closed path. The elements K1, B12, M2, K2 and B2 are connected to second node. Hence they are
represented by analogous element in mesh 2 forming a closed path.
The elements K1 and B12 are common between node 1 and 2 and so they are represented by
analogous element as common elements between
two meshes. The force-voltage electrical analogous
circuit is show in figure 1.8.4.
The electrical analogous elements for the
elements of mechanical system are given below.

f(t) → e(t) M1→ L1 B1→ R1 K1→ I/C1

v1→ i1 M2→ L2 B2→ R2 K2→ 1/C2
Figure 4 Force-voltage electrical analogous circuit
v2→ i2 B12→ R12

Figure 6
Figure 5
The mesh basis equations using Kirchoff’s voltage law for the circuit shown in figure 4 are given
below.
𝑑𝑑𝑑𝑑 1 1
L1 + R1 i1 + R12(i1 – i2) + ∫ (i1 – i2) dt = e(t) ……… (5)
𝑑𝑑𝑑𝑑 𝐶𝐶1
𝑑𝑑𝑑𝑑 2 1 1
L2 + R2 i2 + ∫ i2dt+R12 (i1 – i2) + ∫ (i2 – i1) dt = 0 ……… (6)
𝑑𝑑𝑑𝑑 𝐶𝐶2 𝐶𝐶1

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It is observed that the mesh basis equations (5) and (6) are similar to the differential equations (3)
and (4) governing the mechanical system
Force-current analogous circuit
The given mechanical system has two nodes (masses). Hence the force-current analogous electrical
circuit will have two nodes.
The force applied to mass M1 is represented as a current source connected to node 1 in analogous
electrical circuit. The elements M1, B1, K1 and B12, are connected to first node. Hence they are represented
by analogous elements connected to node 1 in analogous electrical circuit. The elements K1, B12, M2, K2 and
B2 are connected to second node. Hence they are represented by analogous elements as elements
connected to node 2 in analogous electrical circuit.
The elements K1 and B12 are common between node 1 and 2 and so they are represented by
analogous elements as common element between two nodes in analogous circuit. The force-current
electrical analogous circuit is shown in figure 7.
The electrical analogous elements for the elements of mechanical system are given below
f(t) →I (t) M1→C1 B1→I/R1 K1→ I/L1
v1→v1 M2→C2 B2→I/R2 K2→ 1/L2
v2→v2 B12→I/R12

Figure 7 Force-current electrical analogous circuit

Figure 9
Figure.8

The node basis equations using Krichoff’s current law for the circuit shown in figure 7 are given
below.
𝑑𝑑𝑑𝑑 1 1 1 1
C1 + v + (v1 – v2) + ∫ (v1 – v2) dt = i(t) ……… (5)
𝑑𝑑𝑑𝑑 𝑅𝑅1 1 𝑅𝑅2 𝐿𝐿1
𝑑𝑑𝑑𝑑 2 1 1 1 1
C2 + + v2+ ∫ v2dt+ (v2 – v1) + ∫ (v2 – v1) dt = 0 …… (6)
𝑑𝑑𝑑𝑑 𝑅𝑅2 𝐿𝐿2 𝑅𝑅2 𝐿𝐿1
It is observed that the node basis equations (7) and (8) are similar to the differential equations (3)
and (4) governing the mechanical system.

PROPERTIES OF SIGNAL FLOW GRAPH

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The basic properties of signal flow graph are the following:

i) The algebraic equations which are used to construct signal flow graph must be in the form of cost
and effect relationship
ii) Signal flow graph is applicable to linear systems
iii) A node in the signal flow graph represents the variable or signal
iv) A node adds the signals of all incoming branches and transmits the sum to all outgoing branches
v) A mixed node which has both incoming and outgoing signals can be treated as an output node by
adding an outgoing branch of unity transmittance.
vi) The signals travel along branches only in the marked direction and when it travels it gets multiplied
by the gain or transmittance of the branch.
vii) The signal flow graph of system is not unique. By rearranging the system equations different types
of signal flow graphs can be drawn for a given system

SIGNAL FLOW GRAPH ALGEBRA

Signal flow graph for a system can be reduced to obtain the transfer function of the system using the
following rules. The guideline in developing the rules for signal flow graph algebra is that the signal at a
node is given by same of all incoming signals.
Rule 1: Incoming signal to a node through a branch is given by the product of a signal at previous node and
the gain of the branch.
Examples

Rule 2: Cascaded branches can be combined to give a signal branch whose transmittance is equal to the
product of individual branch transmittance.
Examples

Rule 3: Parallel branches may be represented by single branch whose transmittance is the sum of individual
branch transmittances.
Examples

Rule 4: A mixed node can be eliminated by multiplying the transmittance of outgoing branch (from the
mixed node) to the transmittance of all incoming branches to the mixed node.
Examples

Rule 5: A loop may be eliminated by writing equations at the input and output node and rearranging the
equations to find the ratio of output to input. This ratio gives the gain of resultant branch.

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Examples

Proof:
x2 = ax1 + cx3
x3 = bx2
Put x2 = ax1 + cx3 in the equation for x3
∴x3 = b (ax1 + cx3)
x3 = abx1 + cx3
x3 – bc x3 = abx1
x3(1-bc) = ab x1
x3 ab
=
x1 1−bc

SINGLE FLOW GRAPH REDUCTION

The single flow graph of a system can be reduced either by using the rules of the single flow graph
algebra (i.e.) by writing equations at every node and then rearranging these equations to get the ratio of
output and input (transfer function)
The signal flow graph reduction by above method will be time consuming and tedious. S. J. Mason
has developed a simple procedure to determine the transfer function of the system represented as a signal
flow graph. He has developed a formula called by his name Mason’s gain formula which can be directly used
to find the transfer function of the system.
Mason’s Gain Formula
The Mason’s gain formula is used to determine the transfer function of the system from the signal
flow graph of the system.
Let R(s) = Input to the system
and C(s) = Output of the system
𝐶𝐶(𝑠𝑠)
Transfer function of the system, T(s) =
𝑅𝑅(𝑠𝑠)
Manson’s gain formula states the overall gain of the system [transfer function] as follows,
1
Overall gain, T = Σ𝑘𝑘 𝑃𝑃𝑘𝑘 ∆𝑘𝑘
∆
Where, T = T(s) = Transfer function of the system.
𝑃𝑃𝑘𝑘 = Forward path gain of 𝐾𝐾 𝑡𝑡ℎ forward path
Sumof gain products of all possible
∆= 1– (Sum of individual loop gains) + � �
combinations of two non − touching loops
Sum of gain products ofa ll possible
-� �+ ………………………………….
combinations of three non − touching loops

∆𝑘𝑘 = ∆For that part of the graph which is not touching 𝐾𝐾 𝑡𝑡ℎ forward path

5] Problem
Construct a signal flow graph for armature controlled dc motor

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SOLUTION
The differential equations governing the armature controlled dc motor are
𝑑𝑑𝑑𝑑 𝑎𝑎
𝑉𝑉𝑎𝑎 = 𝑖𝑖𝑎𝑎 𝑅𝑅𝑎𝑎 + 𝐿𝐿𝑎𝑎 + 𝑒𝑒𝑏𝑏 ………(1)
𝑑𝑑𝑑𝑑
𝑇𝑇 = 𝐾𝐾𝑡𝑡 𝑖𝑖𝑎𝑎 ……… (2)
𝑑𝑑𝑑𝑑
𝑇𝑇 = 𝐽𝐽 + 𝐵𝐵𝐵𝐵 ……… (3)
𝑑𝑑𝑑𝑑
𝑒𝑒𝑏𝑏 = 𝐾𝐾𝑏𝑏 𝜔𝜔 ………. (4)
𝑑𝑑𝑑𝑑
𝜔𝜔 = ………. (5)
𝑑𝑑𝑑𝑑
On taking Laplace transform of equations (1) to (5) we get,

𝑉𝑉𝑎𝑎 (𝑠𝑠) = 𝐼𝐼𝑎𝑎 (𝑠𝑠)𝑅𝑅𝑎𝑎 + 𝐿𝐿𝑎𝑎 𝑠𝑠𝐼𝐼𝑎𝑎 (𝑠𝑠) + 𝐸𝐸𝑏𝑏 (𝑠𝑠) ........ (6)
T(s) = 𝐾𝐾𝑡𝑡 𝐼𝐼𝑎𝑎 (𝑠𝑠) ……... (7)
T(s) = 𝐽𝐽𝐽𝐽𝐽𝐽(𝑠𝑠) + 𝐵𝐵𝐵𝐵(𝑠𝑠) ……… (8)
𝐸𝐸𝑏𝑏 (𝑠𝑠) = 𝐾𝐾𝑏𝑏 𝜔𝜔(𝑠𝑠) ……… (9)
𝜔𝜔(𝑠𝑠) = 𝑠𝑠𝑠𝑠 (𝑠𝑠) .….…(10)
The input and output variables of armature controlled dc motor are armature voltage 𝑉𝑉𝑎𝑎 (s) and
angular displacement𝜃𝜃𝑠𝑠 respectively. The variables 𝐼𝐼𝑎𝑎 (𝑠𝑠), 𝑇𝑇(𝑠𝑠), 𝐸𝐸𝑏𝑏 (𝑠𝑠)𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(𝑠𝑠)are intermediate variables.

The equations (6) to (10) are rearranged and the individual signal flow graph are shown in figure 1 to
figure5

𝑉𝑉𝑎𝑎 (𝑠𝑠) − 𝐸𝐸𝑏𝑏 (𝑠𝑠) = 𝐼𝐼𝑎𝑎 (𝑠𝑠) [𝑅𝑅𝑎𝑎 + 𝑠𝑠𝐿𝐿𝑎𝑎 ]

1
∴ 𝐼𝐼𝑎𝑎 (𝑠𝑠) = [𝑉𝑉 (𝑠𝑠) − 𝐸𝐸𝑏𝑏 (𝑠𝑠)]
𝑅𝑅𝑎𝑎 + 𝑠𝑠𝐿𝐿𝑎𝑎 𝑎𝑎

Figure 1
𝑇𝑇(𝑠𝑠) = 𝐾𝐾𝑡𝑡 𝐼𝐼𝑎𝑎 (𝑠𝑠)

Figure 2
T(s) = 𝜔𝜔(𝑠𝑠)[ 𝐽𝐽𝑠𝑠 + 𝐵𝐵]
1
∴ 𝜔𝜔(𝑠𝑠) = 𝑇𝑇(𝑠𝑠)
𝐽𝐽𝑠𝑠 + 𝐵𝐵 Figure 3
𝐸𝐸𝑏𝑏 (𝑠𝑠) = 𝐾𝐾𝑏𝑏 𝜔𝜔(𝑠𝑠)

Figure 4
𝜔𝜔(𝑠𝑠) = 𝑠𝑠𝑠𝑠(𝑠𝑠)
1
∴ 𝜃𝜃(𝑠𝑠) = 𝜔𝜔(𝑠𝑠)
𝑠𝑠
Figure5

The overall single flow graph of armature controlled dc motor is obtained by interconnecting the
individual signal flow graphs shown in figure 1 to figure 5. The overall signal flow graph is shown in figure 6.

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Figure 6 Signal flow graph of armature controlled DC motor

6] Problems

Find the overall transfer function of the syxtem whose signal flow graph is shown in figure.

Figure 1
SOLUTION

I) Forward path gains

There are two forward paths, ∴K = 2
Let forward path gains be P1 and P2

Figure 2 Forward path – I

Figure3 Forward patch – 2

Gain of forward path – 1, P1 = G1 G2 G3 G4 G5

Gain of forward path – 2, = G4 G5 G6

II) Individual Loop Gain

There are three idividual loops. Let individual loop gains be P11, P21 and P13

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Figure 4 Loop –I Figure5 Loop – 2 Figure 6 Loop – 3

Loop gain of indicidual loop – 1, P11 = - G2 H1
Loop gain of individual loop – 2, P21 = - G2 G3 H2
Loop gain of individual loop – 3, P31 = - G5 H3
III) Gain Products Of Two Non Touching Loops
There are two combinations of two non-touching loops. Let the gain products of two non touching loops be
P12 and P22.

Figure 7 Firstcombinations of two non-touching loops

Figure 8 Second combination of two non touching loops

Gain product of first combination of two non touching loops
P12 = P11 P31
= (-G2H1) (-G5 H5)
= G2 G5 H1 H3
Gain product of second combination of two non touching loops
P22 = P21 P31
= (-G2 G3 H2) (-G5 H3)
= G2 G3 G5 H2 H3
IV) Calculation Of ∆ And ∆K
∆ = 1 – (P11 + P21 + P31) + (P12 + P22)
= 1 – (–G2 H1 – G2 G3 H2 – G5 H3) + (G2 G5 H1 H3 + G2 G3 G5 H2 H3)
= 1 + G2 H1 + G2 G3 H2 + G5 H3 + G2 G5 H1 H3 + G2 G3 G5 H2 H4
∆1= 1, Since there is no part of graph which is not touching with first
forward path.
The part of the graph which is non touching with second forward path is shown in Figure 9
figure 9
∆2 = 1 – P11 = 1 – (–G2 H1)
= 1 + G2 H1

V) Tranfer Function, T

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1
By Mason’s gain formaula the transfer function T = ∑𝑘𝑘 PK∆K
∆
1
= ∆(P1∆1 + P2∆2)
(Here K = 2, since we have only to forwar patch)

𝐺𝐺1 𝐺𝐺2 𝐺𝐺3 𝐺𝐺4 𝐺𝐺5 +𝐺𝐺4 𝐺𝐺5 𝐺𝐺6 (1+𝐺𝐺2 𝐻𝐻1 )
∴ T =
1+ 𝐺𝐺2 𝐻𝐻1 + 𝐺𝐺2 𝐺𝐺3 𝐻𝐻2 + 𝐺𝐺5 𝐻𝐻3 + 𝐺𝐺2 𝐺𝐺5 𝐻𝐻1 𝐻𝐻3 + 𝐺𝐺2 𝐺𝐺3 𝐺𝐺5 𝐻𝐻2 𝐻𝐻3

𝐺𝐺1 𝐺𝐺2 𝐺𝐺3 𝐺𝐺4 𝐺𝐺5 +𝐺𝐺4 𝐺𝐺5 𝐺𝐺6 + 𝐺𝐺2 𝐺𝐺4 𝐺𝐺5 𝐺𝐺6 𝐻𝐻1
=
1+ 𝐺𝐺2 𝐻𝐻1 + 𝐺𝐺2 𝐺𝐺3 𝐻𝐻2 + 𝐺𝐺5 𝐻𝐻3 + 𝐺𝐺2 𝐺𝐺5 𝐻𝐻1 𝐻𝐻3 + 𝐺𝐺2 𝐺𝐺3 𝐺𝐺5 𝐻𝐻2 𝐻𝐻3

𝐺𝐺2 𝐺𝐺4 𝐺𝐺5 [𝐺𝐺1 𝐺𝐺3 +𝐺𝐺6 / 𝐺𝐺2 + 𝐺𝐺6 𝐻𝐻1 ]

=
1+ 𝐺𝐺2 𝐻𝐻1 + 𝐺𝐺2 𝐺𝐺3 𝐻𝐻2 + 𝐺𝐺5 𝐻𝐻3 + 𝐺𝐺2 𝐺𝐺5 𝐻𝐻1 𝐻𝐻3 + 𝐺𝐺2 𝐺𝐺3 𝐺𝐺5 𝐻𝐻2 𝐻𝐻3

BLOCK DIAGRAM
A control system may consist of a number of components. It control engineering to show the
function performed by each component, we commonly use a diagram called the block diagram. A black
diagram of a system is a pictorial representation of the functions performed by each component and of the
flow signals. Such a diagram depicts the interrelationships that exist among the various components. The
elements of a block diagram are block, branch point and summing point.
Block
In a block diagram all system variable are linked to each other
through functional blocks. The functional block or simply block is a symbol
for the mathematical operation on the input signal to the block that
procures the output. The transfer functions of the components are usually
entered in the corresponding blocks, which are connected by arrows to
indicate the direction of the flow of signals. Figure 1shows the functional
Figure 1 Functionalblock
block of the block diagram.
The arrowhead pointing towards the block indicates the input, and arrowhead leading away from
the block represents the output. Such arrows are referred to as signals. The output signal from the block is
given by the product of input signal and transfer function in the block.
Summing point
Summing points are used to add two or more signals in the system.
Referring to figure 2, a circle with a cross is the symbol that indicates a
summing operation.
Figure 2 Summing point
The plus or minus sign at each arrowhead indicates whether the
signal isto be added or subtracted. It is important that the quantities being
added or subtracted have the same dimensions and the same units.
Branch point
A branch point is a point from which the signal from a block goes
concurrently to other blocks or summing points.
Constructing block diagram for control systems Figure 3 Branch point

A control system can be represented diagrammatically by block diagram. The differential equations
governing the system are used to construct the block diagram. By taking Laplace transform the differential
equations are converted to algebraic equations. The equations will have variables and constants. From the
working knowledge of the system the input and output variables are identified and the block diagram for
each equation can be draw. Each equation gives one section of block diagram. The output of one section

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will be input for another section. The various sections are interconnected to obtain the overall block
diagram of the system.
7] Problems
Construct the block diagram of armature controlled dc motor.
SOLUTION
The differential equations governing the armature controlled dc motor are
𝑑𝑑𝑑𝑑 𝑎𝑎
𝑉𝑉𝑎𝑎 = 𝑖𝑖𝑎𝑎 𝑅𝑅𝑎𝑎 + 𝐿𝐿𝑎𝑎
𝑑𝑑𝑑𝑑
+ 𝑒𝑒𝑚𝑚 ………. (1)
𝑇𝑇 = 𝐾𝐾𝑡𝑡 𝑖𝑖𝑎𝑎 ……... (2)
𝑑𝑑𝑑𝑑
𝑇𝑇 = 𝐽𝐽 + 𝐵𝐵𝐵𝐵 …….… (3)
𝑑𝑑𝑑𝑑
𝑒𝑒𝑏𝑏 = 𝐾𝐾𝑏𝑏 𝜔𝜔 ………. (4)
𝑑𝑑𝑑𝑑
𝜔𝜔 = ………. (5)
𝑑𝑑𝑑𝑑
On taking Laplace transform of equation (1) we get,
𝑉𝑉𝑎𝑎 (𝑠𝑠) = 𝐼𝐼𝑎𝑎 (𝑠𝑠)𝑅𝑅𝑎𝑎 + 𝐿𝐿𝑎𝑎 𝑠𝑠𝐼𝐼𝑎𝑎 (𝑠𝑠) + 𝐸𝐸𝑏𝑏 (𝑠𝑠) ……... (6)
In equation (6), 𝑉𝑉𝑎𝑎 (s) and𝐸𝐸𝑏𝑏 (𝑠𝑠)are inputs and 𝐼𝐼𝑎𝑎 (𝑠𝑠) is the out
put. Hence the equation (6) is rearranged and the block diagram for this
equation is shown in figure 1
𝑉𝑉𝑎𝑎 (𝑠𝑠) − 𝐸𝐸𝑏𝑏 (𝑠𝑠) = 𝐼𝐼𝑎𝑎 (𝑠𝑠)[𝑅𝑅𝑎𝑎 + 𝑠𝑠𝑠𝑠𝑎𝑎 ] Figure 1
1
∴ 𝐼𝐼𝑎𝑎 (𝑠𝑠) = [𝑉𝑉𝑎𝑎 (𝑠𝑠) − 𝐸𝐸𝑏𝑏 (𝑠𝑠)]
𝑅𝑅𝑎𝑎 + 𝑠𝑠𝐿𝐿𝑎𝑎

On taking Laplace transform of equation (2)we get,

𝑇𝑇(𝑠𝑠) = 𝐾𝐾𝑡𝑡 𝐼𝐼𝑎𝑎 (𝑠𝑠) ……... (7)
In equation (7), 𝐼𝐼𝑎𝑎 (𝑠𝑠) is theinput and T(s) is the output. The block diagram for
Figure 2
this equation is shown figure.
On taking Laplace transform of equation (3) we get,
𝑇𝑇(𝑠𝑠) = 𝐽𝐽𝐽𝐽𝐽𝐽(𝑠𝑠) + 𝐵𝐵𝐵𝐵(𝑠𝑠) …….. (8)
In equation (8), T(s) is the input and 𝜔𝜔(𝑠𝑠) is the output. Hence the
equation (8) is rearranged and the block diagram for this equation is shown in Figure 3
figure.
𝑇𝑇(𝑠𝑠) = (𝐽𝐽𝐽𝐽 + 𝐵𝐵)𝜔𝜔(𝑠𝑠)
1
∴ 𝜔𝜔(𝑠𝑠) = 𝑇𝑇(𝑠𝑠)
𝐽𝐽𝐽𝐽 + 𝐵𝐵
On taking Laplace transform of equation (4) we get,
𝐸𝐸𝑏𝑏 (𝑠𝑠) = 𝐾𝐾𝑏𝑏 𝜔𝜔(𝑠𝑠) …….. (9) Figure 4

In equation (9) 𝜔𝜔(𝑠𝑠) is the input and 𝐸𝐸𝑏𝑏 (s) is the output. The block diagram for
this equation is shown in figure.
On taking Laplace transform of equation (5) we get,
𝜔𝜔(𝑠𝑠) = 𝑠𝑠𝑠𝑠(𝑠𝑠) ..…. (10) Figure 5
In equation (10), 𝜔𝜔(𝑠𝑠) is the input and θ(s) is the output. Hence equation (10) is
rearranged and the block diagram for this equation is shown in figure.

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1
𝜃𝜃(𝑠𝑠) = 𝜔𝜔(𝑠𝑠)
𝑠𝑠
The overall block diagram of armature controlled dc motor is obtained by connecting the various
sections shown in figure 1 to figure 5. The overall block diagram is shown in figure 6.

Figure 6 Block diagram of armature controlled dc motor

Block Diagram Reduction

The block diagram can be reduced to find the overall transfer function of the system. The following
rules can be used for block diagram reduction. The rules are framed such that any modification made on the
diagram does not alter the input output relation.
RULES OF BLOCK DIAGRAM ALGEBRA
1. Combining the blocks in cascade

2. Combining parallel blocks (or combining feed forward paths)

3. Moving the branch point ahead of the block

4. Moving the branch point before the block

5. Moving the summing point a head of the block

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6. Moving the summing point before the block

7. Interchanging summing point

8. Splitting summing points

9. Combining summing points

10. Elimination of feed back loop

Proof: C = (R – CH) G

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C = RG – CHG
C + CHG = RG
C(1 + HG) = RG
C G
=
R 1+GH
Also

8] Problem
Reduce the block diagram shown in figure 1 and find C/R

Figure1
SOLUTION
Step 1: Move the branch point after the block

Step 2: Eliminate the feedback path and combining blocks in cascade

Step 3: Combining parallel blocks

Step 4: combining blocks in cascade

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RESULT
C G 1 G 2 +G 3
The overall transfer function of the system, R
= 1+G 1 H

ROUTH HURWITZ CRITERION

The Routh stability criterion is based on ordering the coefficients of the characteristic equation,
𝑎𝑎0 𝑠𝑠 𝑛𝑛 + 𝑎𝑎2 𝑠𝑠 𝑛𝑛−1 + 𝑎𝑎2 𝑠𝑠 𝑛𝑛−2 + … … . . + 𝑎𝑎𝑛𝑛−1 𝑠𝑠 + 𝑎𝑎𝑛𝑛 = 0,where 𝑎𝑎0 > 0 into a schedule, called the Routh
array as shown below.
𝑠𝑠 𝑛𝑛 ∶ 𝑎𝑎0 𝑎𝑎2 𝑎𝑎4 𝑎𝑎6 𝑎𝑎8 ….
𝑠𝑠 𝑛𝑛−1 ∶ 𝑎𝑎1 𝑎𝑎3 𝑎𝑎5 𝑎𝑎7 𝑎𝑎9 ….
𝑠𝑠 𝑛𝑛−2 ∶ 𝑏𝑏0 𝑏𝑏1 𝑏𝑏2 𝑏𝑏3 𝑏𝑏4 ….
𝑠𝑠 𝑛𝑛−3 ∶ 𝑐𝑐0 𝑐𝑐1 𝑐𝑐2 𝑐𝑐3 𝑐𝑐4 ….
⋮
𝑠𝑠1 ∶ 𝑔𝑔𝑜𝑜
𝑠𝑠0 ∶ ℎ0
The Routh stability criterion can be stated as follows.
“The necessary and sufficient condition for stability is that all of the elements in the first column of
the Routh array be positive. If this condition is not met, the system is unstable and the number of single
changes in the elements of the first column of the Routh array corresponds to the number of roots of the
characteristic equation in the right half of the s-plane”
Note: If the order of sign of first column elements is +, +, - , + and +. Then + to – is considered as one sign
change and – to + as another sign change.
CONSTRUCTION OF ROUTH ARRY
Let the characteristic polynomial be
𝑎𝑎0 𝑠𝑠 𝑛𝑛 + 𝑎𝑎1 𝑠𝑠 𝑛𝑛−1 + 𝑎𝑎2 𝑠𝑠 𝑛𝑛−2 + 𝑎𝑎3 𝑠𝑠 𝑛𝑛−3 +. . . 𝑎𝑎𝑛𝑛−1 𝑠𝑠1 + 𝑎𝑎𝑛𝑛 𝑠𝑠 0
The coefficients of the polynomial are arranged in two rows as shown below.
𝑠𝑠 𝑛𝑛 ∶ 𝑎𝑎0 𝑎𝑎2 𝑎𝑎4 𝑎𝑎6 ….
𝑠𝑠 𝑛𝑛−1 ∶ 𝑎𝑎1 𝑎𝑎3 𝑎𝑎5 𝑎𝑎7 ….
If n is even then 𝑠𝑠 row is formed by coefficient of even order terms and 𝑠𝑠 𝑛𝑛−1 row is formed by
𝑛𝑛

coefficients of odd order terms (i.e., coefficients of odd powers of s).

If n is odd, then 𝑠𝑠 𝑛𝑛 row is formed by coefficients of odd order term and 𝑠𝑠 𝑛𝑛−1 row is formed by
coefficients of even order terms (i.e., coefficients of even powers of s)
The other rows of routh array upto𝑠𝑠 0 row can be formed by the following procedure. Each row of
Routh array is constructed by using the elements of previous two rows.
Consider two consecutive rows of Routh array as shown below.

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𝑠𝑠 𝑛𝑛−1 ∶ 𝑥𝑥0 𝑥𝑥1 𝑥𝑥2 𝑥𝑥3 𝑥𝑥4 ……

𝑠𝑠 𝑛𝑛−𝑥𝑥−1 : 𝑦𝑦0 𝑦𝑦1 𝑦𝑦2 𝑦𝑦3 𝑦𝑦4 ……
Let the next row be,
𝑠𝑠 𝑛𝑛−𝑥𝑥−2 ∶ 𝑧𝑧𝑜𝑜 𝑧𝑧1 𝑧𝑧2 𝑧𝑧3 ……..
The elements of 𝑠𝑠 𝑛𝑛−𝑥𝑥−2 row are given by,
𝑥𝑥 𝑥𝑥1
(−1) �𝑦𝑦0 𝑦𝑦1 � 𝑦𝑦0 𝑥𝑥1 − 𝑦𝑦1 𝑥𝑥0
0
𝑧𝑧0 = =
𝑦𝑦0 𝑦𝑦0
𝑥𝑥 𝑥𝑥2
(−1) �𝑦𝑦0 𝑦𝑦2 � 𝑦𝑦0 𝑥𝑥2 − 𝑦𝑦2 𝑥𝑥0
0
𝑧𝑧1 = =
𝑦𝑦0 𝑦𝑦0
𝑥𝑥 𝑥𝑥3
(−1) �𝑦𝑦0 𝑦𝑦3 � 𝑦𝑦0 𝑥𝑥3 − 𝑦𝑦3 𝑥𝑥0
0
𝑧𝑧2 = =
𝑦𝑦0 𝑦𝑦0
𝑥𝑥 𝑥𝑥4
(−1) �𝑦𝑦0 𝑦𝑦4 � 𝑦𝑦0 𝑥𝑥4 − 𝑦𝑦4 𝑥𝑥0
0
𝑧𝑧3 = =
𝑦𝑦0 𝑦𝑦0
The elements 𝑧𝑧0, 𝑧𝑧1, 𝑧𝑧2 , 𝑧𝑧3, …… are computed until an element equals to zero or for all possible
computations as shown above.
I the process of constructing Routh array the missing terms are considered as zeros. Also, all the
elements of any row can be multiplied or divided by a positive constant to simplify the computational work.
In the construction of Routh array one may come across the following three cases.
Case 1: Normal Routh array (Non-zero elements in the first column ofrouth array)
Case 2: A row of all zeros
Case 3: A first element of a row is zero but some or other elements are not zero.
Case 1: Normal Routh array (Non-zero elements in the first column of
routh array)
9] Problem
Using Routh criterion, determine the stability of the system represented by the characteristic
equation, S4 + 8s3 + 18s2 + 16s + 5 = 0. Comment on the location of the roots of characteristic equation.
SOLUTION
The characteristic equation of the system is S4 + 8s3 + 8s2 + 16s + 5 = 0
The given characteristic equation is 4th order equation and so it has 4 roots. Since the highest power
of s is even number, form the first row of routh array using the coefficients of even powers of s and from the
second row using the coefficient of odd powers of s.
s4 : 1 18 5 …. Row – 1
3
s : 8 16 …. Row – 2
3
The elements of s row can be divided by 8 to simplify the computations.

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s4 : 11 18 5 …. Row – 1 1×18−2×1 1×5 − 0×1

s2 : 1 1
s3 : 11 2 …. Row – 2 s2 : 16 5

s2 : 16
16 5 …. Row – 3 s1 :
16×2−5×1
16
s1 : 1.6875 ≈ 1.7
s1 : 1.7
1.7 …. Row – 4

s0 : 55 …. Row – 5 1.7×5−0×16
s0 : 1.7
s0 : 5
s3 : 88 16 …. Row – 2

Column – 1

On examining the elements of first column of routh array it is observed that all the elements are
positive and there is no sign change. Hence all the roots are lying on the left half or s-plane and the system
is stable.
RESULT
1. Stable system
2. All the four roots are lying on the left half of s-plane.

Case 2: A row of all zeros

10] Problem
Construct Routh array and determine the stability of the system whose characteristic equation is s6 +
2s + 8s + 12s3 + 20s2 + 16s + 16 = 0. Also determine the number of roots lying on right half of s-plane, left
5 4

half of s-plane and on imaginary axis.

SOLUTION
The characteristic equation of the system is s6 + 2s5 + 8s4 + 12s3 + 20s2 + 16s + 16 = 0.
The given characteristic polynomial is 6th order equation and so it has 6 roots. Since the highest power of s is
even number from the first row of routh array using the coefficients of even powers of s and form the
second row using the coefficients of odd powers of s.
s6 : 1 8 20 16 …. Row – 1

s5 : 2 12 16 …. Row – 2

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The elements of s5 row can be divided by 2 to simplify the computations.

s6 : 1 8 20 16 …. Row – 1

s5 : 1 6 8 …. Row – 2

s4 : 1 6 8 …. Row – 3

s3 : 0 0 …. Row – 4

s3 : 1 3 …. Row – 4

s2 : 3 8 …. Row – 5

s1 : 0.33 …. Row – 6

s0 : 8 8 …. Row – 7

Column – 1

1×8−6×1 1×20 − 8×1 1×16 − 0×1

s4 : 1 1 1
s4 : 2 12 16
divide by 2
s4 : 1 6 8

1×6−6×1 1×8 − 8×1

s1 : 1 1

s1 : 0 0

The auxiliary equation is A = s4 + 6s2 + 8. On

differentiating A with respect to s we get
𝑑𝑑𝑑𝑑
: = 4s3 + 12s
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
The coefficients of are used to form s3 row
𝑑𝑑𝑑𝑑
s3 : 4 12
divide by 4
s3 : 1 3

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1×6−3×1 1×8 − 0×1

s2 : 1 1
s2 : 3 8

3×3−8×1
s1 : 3
s1 : 0.33

0.33×8−0×3
S0 : 0.33

S0 : 8

On examining the elements of 1st column of routh array it is observed that there is no sign change.
The rows with all zeros indicate the possibility of roots on imaginary axis. Hence the system is limitedly or
marginally stable.
The auxiliary polynomial is
S4 + 6s2 + 8 = 0
Let s2 = x
∴x2 + 6x + 8 = 0
−6±√62 −4×8
The roots of quadratic are, x =
2
= -3 ± 1 = 2 or -4
The roots of auxiliary polynomial is, s = ±√𝑥𝑥 = ±√−2 and ±√−4
= +j√𝑥𝑥, -j√−2 +j2 and -j2
Theroots of auxiliary polynomial are also roots of characteristic equation. Hence 4 roots are lying on
imaginary axis and the remaining two roots are lying on the left half of S - plane.
RESULT
1. The system is limitedly or marginally stable
2. Four roots are lying on imaginary axis and the remaining two roots are lying on the left half of s-
plane.

Case 3: A first element of a row is zero but some or other elements

are not zero
11] Problem
Construct Routh array and determine the stability of the system represented by the characteristic
equation s5 + s4 + 2s3 + 2s2 + 3s + 5 = 0. Comment of the location of the roots of characteristic equation.
SOLUTION
The characteristic equation of the system is s5 + s4 + 2s3 + 2s2 + 3s + 5 = 0

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The given characteristic polynomial is 5th order equation and so it has 5 roots. Since the highest
power of s is odd number, form the first row of routh array using the coefficients of odd powers of s and
from the second row using the coefficients of even powers of s.

s5 : 1 2 3 …. Row – 1
4
s : 1 2 5 …. Row – 2
3
s : ∈ -2 …. Row – 3
2∈+2
s2 : 5 …. Row – 4
∈
−�5∈2 +4∈+4�
s1 : 2∈+2
…. Row – 5
s0 : 5 …. Row – 6

On letting ∈→ 0, we get
s5 : 1 2 3 …. Row – 1
4
s : 1 2 5 …. Row – 2
s3 : 0 -2 …. Row – 3
2
s : ∞ 5 …. Row – 4
1
s : -2 …. Row – 5
0
s : 5 …. Row – 6

Column - 1

1×2−2×1 1×3 − 5×1

s2 :
1 1
s3 : 0 -2
Replace: 0 by ∈
s3 : ∈ -2

∈×2−(−2×1) ∈×5 − 0×1

s2 : ∈ ∈
2∈ +2
s2 : ∈
5

2∈ +2
1 ×(−2)−(−5×∈)
s : ∈
2∈ +2
∈

−(5∈2 +4∈+4)
s1 : 2∈+2

S0 : 5

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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On observing the elements of first column of routh array, it is found that three are two sign changes.
Hence two roots are lying on the right half of s-plane and the system is unstable. The remaining three roots
are lying on the lift half of s-plane.
RESULT
1. The system is unstable.
2. Two roots are lying on the right half of s-plane and three roots are lying on the left of s-plane.

12] Problem
By routh stability criterion determine the stability of the system represented by the characteristic
equation 9s5 – 20s4 + 10s3 – s2 – 9s – 10 = 0. Comment on the location of roots of characteristic equation.
SOLUTION
The characteristic polynomial of the system is 9s5 – 20s4 + 10s3 – s2 – 9s – 10 = 0
On examining the coefficients of the characteristic polynomial, it is found that some of the
coefficients are negative and so some roots will lie on the right of s-plane. Hence the system is unstable.
The routh array can be constructed to find the number of roots lying on right half of s-plane.
9s5 – 20s4 + 10s3 – s2 – 9s – 10 = 0
The given characteristic polynomial is 5th order equation and so it has 5 roots. Since the highest
power of s is odd number, form the first row of routh array using the coefficients of odd powers of s and
form the second row using coefficients of even powers of s.

s5 : 9 10 -9 …. Row – 1
4
s : -20 -1 -10 …. Row – 2
3
s : 9.55 -13.5 …. Row – 3
2
s : -29.3 -10 …. Row – 4
1
s : -16.8 …. Row – 5
s0 : -10 …. Row – 6

Column - 1

−20×10−(−1)×9 −20×(−9)–(−10)×9
s3 : −20 1
s3 : 9.55 -13.5

9.55×(−1)−(−13.5)×(−20) 9.55×(− 10)

s2 : 9.55 9.55
s2 : -29.3 -10

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

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−29.3×(−13.5)−(−10)×9.55
s1 : −29.3
s1 : -16.8

−16.8×(−10)
s0 : −16.8
s1 : 10

By examining the elements of 1st column of routh array it is observed that there are three sign
changes and so three roots are lying on the right half of s-plane and the remaining two rrots are lying on the
left half of s-plane.
RESULT
1. The system is unstable
2. Three roots are lying on the right half and two roots are lying on the left half of s-plane.
FREQUENCY RESPONSE PLOTS
Frequency response analysis of control systems can be carried either analytically or graphically. The
various graphical techniques available for frequency response analysis are
1. Bode plot
2. Polar plot (or Nyquist plot)
3. Nichols plot
The Bode plot, Polar plot and Nichols plot are usually drawn for open loop systems. From the open
loop response plot the performance and stability of closed loop system are estimated. The M and N circles
and Nichols chart are used to graphically determine the frequency response of unity feedback closed loop
system from the knowledge of open loop response.
The frequency response plots are used to determine the frequency domain specifications, to study
the stability of the systems and to adjust the gain of the system to satisfy the desired specifications.
POLAR PLOT
The polar plot of a sinusoidal transfer function 𝐺𝐺(𝐽𝐽𝐽𝐽) is a plot of the magnitude of 𝐺𝐺(𝑗𝑗𝑗𝑗) versus the
phase angle of𝐺𝐺(𝑗𝑗𝑗𝑗)on polar coordinates as 𝜔𝜔 is varied from zero to infinity. Thus the polar plat is the locus
of vectors |𝐺𝐺(𝐽𝐽𝐽𝐽)| < 𝐺𝐺(𝑗𝑗𝑗𝑗)𝑎𝑎𝑎𝑎𝑎𝑎 is varied from zero to infinity. The polar plat is also called Nyquist plot.
The polar plot is usually plotted on a polar graph sheet. The polar graph sheet has concentric circles
and radial lines. The circles represent the magnitude and the radial lines represent the phase angles. Each
point on the p9olar graph has a magnitude of a point is given by the value of the circle passing through that
point and the phase angle is given by the radial line passing through that point. In polar graph sheet a
positive phase angle is measured in anticlockwise from the reference axis (0°) and a negative angle is
measured clockwise from the reference axis (0°).
Alternatively, if 𝐺𝐺(𝑗𝑗𝑗𝑗) can be expressed in rectangular coordinates as,
𝐺𝐺(𝑗𝑗𝑗𝑗) = 𝐺𝐺𝑅𝑅 (𝑗𝑗𝑗𝑗) + 𝑗𝑗𝐺𝐺𝐼𝐼 (𝑗𝑗𝑗𝑗)
Where 𝐺𝐺𝑅𝑅 (𝑗𝑗𝑗𝑗) = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅(𝑗𝑗𝑗𝑗)
and𝐺𝐺𝐼𝐼 (𝑗𝑗𝑗𝑗) = 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼(𝑗𝑗𝑗𝑗),
then the polar plot can be plotted in ordinary graph sheet between
𝐺𝐺𝑅𝑅 (𝑗𝑗𝑗𝑗)𝑎𝑎𝑎𝑎𝑎𝑎𝐺𝐺𝐼𝐼 (𝑗𝑗𝑗𝑗)as𝜔𝜔 is varied from 0 to∞.

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To plot the polar plot, first compute the magnitude and

phase of G (jω) for various values of ω and tabulate them. Usually
the choices of frequencies are corner frequencies and frequencies
around corner frequencies. Choose proper scale for the
magnitude circles. Fix all the points on polar graph sheet and join
the points by smooth curve. Write the frequency corresponding
to each point of the plot.
To plot the polar plot on ordinary graph sheet, compute
the magnitude and phase for various values ofω. Then convert
the polar coordinates to rectangular coordinates using P → R
conversion (polar to rectangular conversion) in the calculator.
Sketch the polar plot using rectangular coordinates.
Figure 2.1
For minimum phase transfer function with only poles, the
type number of the system determines at what quadrant the polar plot starts and the order of the system
determines at what quadrant the polar plot ends.
(note: The minimum phase systems are systems with all poles and zeros on the left half of s-plane)

Figure 2.2 Start of polar plot Figure 2.3 End of polar plot
Typical sketches of polar plot
Type: 0, Order : 1
1
G(s) =
1+𝑠𝑠𝑠𝑠
1
G(jω) = 1+𝑗𝑗𝑗𝑗𝑗𝑗

Type: 1 Order:2
1
G(s) = 𝑠𝑠(1+𝑠𝑠𝑠𝑠)
1
G(jω) = 𝑗𝑗𝑗𝑗 (1+𝑗𝑗𝑗𝑗𝑗𝑗 )

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Type: 0 Order:2
1
G(s) = (1+𝑠𝑠𝑇𝑇 )(1+𝑠𝑠𝑇𝑇
1 2)
1
G(jω) = (1+𝑗𝑗𝑗𝑗 𝑇𝑇1 )(1+𝑗𝑗𝑗𝑗 𝑇𝑇2 )

Type: 0,Order:3
1
G(s) = (1+𝑠𝑠𝑇𝑇 )(1+𝑠𝑠𝑇𝑇 )(1+𝑠𝑠𝑇𝑇 )
1 2 3
1
G(jω) = (1+𝑗𝑗𝑗𝑗 𝑇𝑇1 )(1+𝑗𝑗𝑗𝑗 𝑇𝑇2 )(1+𝑗𝑗𝑗𝑗 𝑇𝑇3 )

Type: 1, Order:3
1
G(s) = 𝑠𝑠(1+𝑠𝑠𝑇𝑇 )(1+𝑠𝑠𝑇𝑇 )
1 2
1
G(jω) =
𝑗𝑗𝑗𝑗 (1+𝑗𝑗𝑗𝑗 𝑇𝑇1 )(1+𝑗𝑗𝑗𝑗 𝑇𝑇2 )

Type: 2, Order:4
1
G(s) = 𝑠𝑠 2 (1+𝑠𝑠𝑇𝑇 )(1+𝑠𝑠𝑇𝑇 )
1 2
1
G(jω) = (𝑗𝑗𝑗𝑗 )2 (1+𝑗𝑗𝑗𝑗 𝑇𝑇1 )(1+𝑗𝑗𝑗𝑗 𝑇𝑇2 )

Type: 0,Order:3
1
G(s) = 𝑠𝑠 2 (1+𝑠𝑠𝑇𝑇 )(1+𝑠𝑠𝑇𝑇 )(1+𝑠𝑠𝑇𝑇 )
1 2 3
1
G(jω) =
(𝑗𝑗𝑗𝑗 )2 (1+𝑗𝑗𝑗𝑗 𝑇𝑇1 )(1+𝑗𝑗𝑗𝑗 𝑇𝑇2 )(1+𝑗𝑗𝑗𝑗 𝑇𝑇3 )

Figure 2.4 Figure 2.5

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DETERMINATION OF GAIN MARGIN AND PHASE MARGINE FROM

POLAR PLOT
The gain margin is defined as the inverse of the magnitude of 𝐺𝐺(𝑗𝑗𝑗𝑗) at phase crossover frequency.
The phase crossover frequency is the frequency at which the phase of 𝐺𝐺(𝑗𝑗𝑗𝑗) is 180°
Let the polar plot cut the 180°axis at point B and the magnitude circle passing through the point B
1
be𝐺𝐺𝐵𝐵 . Now the Gain margin, 𝐾𝐾𝑔𝑔 = 𝐺𝐺 .If the point B lies within unity circle then the Gain margin is positive
𝐵𝐵
otherwise negative. (If the polar plat is drawn in ordinary graph sheet using rectangular coordinates then the
point B is the cutting point of 𝐺𝐺(𝑗𝑗𝑗𝑗) locus with negative real axis and 𝐾𝐾𝑔𝑔 = 1/|𝐺𝐺𝐵𝐵 | where 𝐺𝐺𝐵𝐵 is the
magnitude corresponding to point B).

Figure 2.6 Polar plot showing positive Figure 2.7 Polar plot showing negative
gain margin and phase margin. gain margin and phase margin.

1 1
Gain margin, 𝐾𝐾𝑔𝑔 = Gain margin, 𝐾𝐾𝑔𝑔 =
𝐺𝐺𝐵𝐵 𝐺𝐺𝐵𝐵

Phase margin, 𝛾𝛾 = 180° + φ𝑔𝑔𝑔𝑔 Phase margin, 𝛾𝛾 = 180° + φ𝑔𝑔𝑔𝑔

The phase margin is defined as, phase margin, 𝛾𝛾 = 180 + φ𝑔𝑔𝑔𝑔 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒φ𝑔𝑔𝑔𝑔 is the phase angle of 𝐺𝐺(𝑗𝑗𝑗𝑗)
at gain crossover frequency. The gain crossover frequency is the frequency at which the magnitude of 𝐺𝐺(𝑗𝑗𝑗𝑗)
is unity.
Let the polar plot cut the unity circle at point A as shown in fig. Now the phase marginγ is given by
∠AOP, ie. If ∠AOP is below –180°axis then the phase margin is positive and if it is above- 180°axis then the
phase margin is negative.
13] Problem
The open loop transfer function of a unity feedback system is given by G(s) = 1/s (1+s) (1+2s). Sketch
the polar plot and determine the gain margin and phase margin.
SOLUTION
Give that, G(s) = 1/s (1+s) (1 + 2s)
Put s - jω
1
∴ G (jω) =
𝑗𝑗𝑗𝑗 (1+𝑗𝑗𝑗𝑗 )(1+𝑗𝑗 2𝜔𝜔)

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The corner frequencies are ωcl = ½ = 0.5 rad/sec and ωc2 = 1 rad/sec. The magnitude and phase
angle of G(jω) are calculated for the corner frequencies and for frequencies around corner frequencies and
tabulated in table 1. Using polar to rectangular conversion, the polar coordinates and tabulated in table 2.
The polar plot using polar coordinates is sketched on a polar graph sheet as shown in figure 1.
1
G (jω) =
𝑗𝑗𝑗𝑗 (1+𝑗𝑗𝑗𝑗 )(1+𝑗𝑗 2𝜔𝜔 )
1
=
𝜔𝜔∠90°�1+𝑗𝑗 𝜔𝜔 2 𝑡𝑡𝑡𝑡𝑡𝑡 −1 𝜔𝜔 √1+4𝜔𝜔 2 ∠𝑡𝑡𝑡𝑡𝑡𝑡 −1 2𝜔𝜔
1
= ∠ − 90° − 𝑇𝑇𝑇𝑇𝑇𝑇−1 𝜔𝜔 − 𝑡𝑡𝑡𝑡𝑡𝑡−1 2𝜔𝜔
𝜔𝜔 �(1+𝜔𝜔 2 )(1+4𝜔𝜔 2 )
1 1
∴|𝐺𝐺(𝑗𝑗𝑗𝑗)|= =
𝜔𝜔 �(1+𝜔𝜔 2 )(1+4𝜔𝜔 2 ) 𝜔𝜔 √1+4𝜔𝜔 2 +𝜔𝜔 4
1
=
𝜔𝜔√1+5𝜔𝜔 2 +4𝜔𝜔 4
∠G(jω) = - 90° - tan-1ω - tan-1 2ω
Table 1: Magnitude and phase of G(jω) at various frequencies
ω
0.35 0.4 0.45 0.5 0.6 0.7 1.0
Rad/sec
|G(jω)| 2.2 1.8 1.5 1.2 0.9 0.7 0.3
∠G(jω) -179.5
-144 -150 -156 -162 -171 -198
deg ≈ -180

Table 2: Real and imaginary part of g(jω) at various frequencies

ω
0.35 0.4 0.45 0.5 0.6 0.7 1.0
Rad/sec
GR(jω) -1.78 -1.56 -1.37 -1.14 -0.89 -0.7 -0.29
GI(jω) -1.29 -0.9 -0.61 -0.37 -0.14 0 0.09

RESULT

Gain margin, Kg = 1.4286

Phase margin, γ= +12°

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,

 [44]

Figure1 Polar plot of G(jω) = I/jω(1 + jω) (1 + j2ω) [using polar coordinates)

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 [45]

14] Problem
The open loop transfer function of a unity feedback system is given by G(s) = 1/s2(1+s) (1 + 2s).
Sketch the polar plot and determine the gain margin and phase margin.
Solution
Given that, G(s)= 1/s2(1 + s) (1 + 2s)
Put s = jω
1
G(jω) =
(𝑗𝑗𝑗𝑗 )2 (1+𝑗𝑗𝑗𝑗 )(1+𝑗𝑗 2𝜔𝜔)

The corner frequencies are ωc1 = 1 rad/sec and ωc2 = 2 rad/sec. The magnitude and phase angle of
G(jω) are calculated for the corner frequencies and frequencies around corner frequencies and tabulated in
table 1. Using the polar to rectangular conversion, the polar coordinates listed in table 1 are converted to
rectangular coordinates and tabulated in table 2. The polar plot using polar coordinates is sketched on a
polar graph sheet as shown in figure 1. The polar plot using rectangular coordinates is sketched on an
ordinary graph sheet as shown in figure 2.
1
G(jω) =
(𝑗𝑗𝑗𝑗 )2 (1+𝑗𝑗𝑗𝑗 )(1+𝑗𝑗 2𝜔𝜔 )
1
=
𝜔𝜔 2 ∠180 °√1+𝜔𝜔 2 ∠𝑡𝑡𝑡𝑡𝑡𝑡 −1 𝜔𝜔 √1+4𝜔𝜔 2 ∠𝑡𝑡𝑡𝑡𝑡𝑡 −1 2𝜔𝜔
1 -1 -1
G(jω) = ∠ (-180 – tan ω - tan 2ω)
𝜔𝜔 2 √1+𝜔𝜔 2 √1+4𝜔𝜔 2
1 1
|G(jω)| = =
𝜔𝜔 2 √1+𝜔𝜔 2 √1+4𝜔𝜔 2 𝜔𝜔 2 �(1+𝜔𝜔 2 )(1+4𝜔𝜔 2 )
1
=
𝜔𝜔 2 √1+5𝜔𝜔 2 + 4𝜔𝜔 4
∠G(Jω) = -180° - tan-1ω - tan-1 2ω

Table 1: Magnitude and phase plot of G(jω) at various frequencies

ω 0.45 0.5 0.55 0.6 0.65 0.7 0.75 1.0
Rad/sec
|G(jω)| 3.3 2.5 1.9 1.5 1.2 0.97≈1 0.8 0.3
∠G(jω) -246 -251 -256 -261 -265 -269 -273 -288
deg

Table 2: Real and imaginary part of g(jω) at various frequencies

ω 0.45 0.5 0.55 0.6 0.65 0.7 0.75 1.0
Rad/sec
GR(jω) -1.34 -0.81 -0.46 -0.23 -0.1 -0.02 0.04 0.09
GI(jω) 3.01 2.36 1.84 1.48 1.2 1.0 0.8 0.29

RESULT
Gain margin, Kg = 0
Phase margin, γ= -90°

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Figure 1 Polar plot of G(jω) = 1/(jω)2 (1 + jω) (1 + j2ω) (1 + j2ω), Using polar coordinates)

BODE PLOT
The step by step procedure for plotting the magnitude plot is given below.
Step 1: Convert the transfer function into Bode form or time constant form.
The Bode form of the transfer function is
𝐾𝐾(1+𝑠𝑠𝑇𝑇1 )
G(s)= 𝑠𝑠2 𝑠𝑠
𝑠𝑠(1+𝑠𝑠𝑇𝑇2 )� 1+ 2 + 2ζ �
𝜔𝜔 𝑛𝑛 𝜔𝜔 𝑛𝑛

𝐾𝐾(1+𝑗𝑗𝑗𝑗 𝑇𝑇1 )
G(Jω) 𝜔𝜔 2 𝜔𝜔
𝑗𝑗𝑗𝑗 (1+𝑗𝑗𝑗𝑗 𝑇𝑇2 )� 1+ 2 + 𝑗𝑗 2ζ �
𝜔𝜔 𝑛𝑛 𝜔𝜔 𝑛𝑛

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Step 2: List the corner frequencies in the increasing order and prepare a table as shown below.

Corner frequency Slope Change in slope

Term
rad/sec db/dec db/dee

In the above table enter K or 𝐾𝐾/(𝑗𝑗𝑗𝑗)𝑛𝑛 or 𝐾𝐾(𝑗𝑗𝑗𝑗)𝑛𝑛 as the first term and the other terms in the
increasing order of corner frequencies. Then enter the corner frequency, slope contributed by each term and
change in slope at every corner frequency.
Step 3 : Choose an arbitrary frequency 𝜔𝜔 which is lesser than the lowest corner frequency.
Calculate the db magnitude of K or 𝐾𝐾/(𝑗𝑗𝑗𝑗)𝑛𝑛 or 𝐾𝐾(𝑗𝑗𝑗𝑗)𝑛𝑛 at 𝜔𝜔𝑙𝑙 and at the lowest corner
frequency.
Step 4 : Then calculate the gain (db magnitude) at every corner frequency one by one by using
the formula,

Gain at 𝜔𝜔𝑦𝑦 = change in gain from 𝜔𝜔𝑥𝑥 𝑡𝑡𝑡𝑡𝜔𝜔𝑦𝑦 + 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝜔𝜔𝑥𝑥

𝜔𝜔
= �𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝜔𝜔𝑥𝑥 𝑡𝑡𝑡𝑡𝜔𝜔𝑦𝑦 × log 𝜔𝜔 𝑦𝑦 � + 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝜔𝜔𝑥𝑥
𝑥𝑥

Step 5 : Choose an arbitrary frequency 𝜔𝜔ℎ which is greater than the highest corner
frequency. Calculate the gain at 𝜔𝜔ℎ by using the formula in step 4.
Step 6 : In a semi log graph sheet mark the required range of frequency on x-axis
(ordinary scale) after choosing proper scale.
Step 7 : Mark all the points obtained in steps 3, 4, and 5 on the graph and join the points
by straight lines. Mark the slope at every part of the graph.
(Note : The magnitude plot obtained above is an approximate plot. If an exact plot is
needed then appropriate correction should be made at every corner
frequencies)
Procedure for Phase Plot of Bode Plot
The phase plot is and no approximations are made while drawing the phase plot. Hence the exact
phase angles of G(jω) are computed for various values of 𝜔𝜔 and tabulated. The choices of frequencies are
preferably the frequencies chosen for magnitude plot. Usually the magnitude plot and phase plot are drawn
in a single semilog-sheet on a common frequency scale.
Take another y-axis in the graph where the magnitude plot is drawn and in this y-axis mark the
desired range of phase angles after choosing proper scale. From the tabulated values of 𝜔𝜔and phase angles,
mark all the points on the graph. Join the points by a smooth curve.

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Determination of Gain Margin and Phase Margin from Bode Plot

The gain margin in db is given by the negative of db magnitude of 𝐺𝐺(𝑗𝑗𝑗𝑗)at the phase cross-over
frequency, 𝜔𝜔𝑝𝑝𝑝𝑝 . the𝜔𝜔𝑝𝑝𝑝𝑝 is the frequency at which phase of 𝐺𝐺(𝑗𝑗𝑗𝑗)is −180°. If the db magnitude of
𝐺𝐺(𝑗𝑗𝑗𝑗)𝑎𝑎𝑎𝑎 𝜔𝜔𝑝𝑝𝑝𝑝 is negative then gain margin is positive and vice versa.
Let φ𝑔𝑔𝑔𝑔 be the phase angle of 𝐺𝐺(𝑗𝑗𝑗𝑗) at gain cross over frequency 𝜔𝜔𝑔𝑔. The𝜔𝜔𝑔𝑔𝑔𝑔 is the frequency at
which the db magnitude of 𝐺𝐺(𝑗𝑗𝑗𝑗) is zero. Now the phase margin 𝛾𝛾 is given by, 𝛾𝛾 = 180° + φ𝑔𝑔𝑔𝑔 . Ifφ𝑔𝑔𝑔𝑔 is less
negative than −180° than phase margin is positive and vice versa.
The positive and negative gain margins are illustrated in figure.

Figure2.8 Bode plot showing phase margin (γ) and gain margin (Kg)
15] Problem
Sketch Bode plot for the following transfer function and determine the system gain K for the gain cross over
frequency to be 5 rad/sec
𝐾𝐾𝐾𝐾 2
G (s) = (1+0.2𝑠𝑠)(1+0.02𝑠𝑠)

SOLUTION
𝐾𝐾(𝑗𝑗𝑗𝑗 )2
The sinusoidal transfer function G(jω) = (1+0.2𝑗𝑗𝑗𝑗 )(1+0.02𝑗𝑗𝑗𝑗 )
(𝑗𝑗𝑗𝑗 )2
Let K =1, ∴G(jω) = (1+0.2𝑗𝑗𝑗𝑗 )(1+0.02𝑗𝑗𝑗𝑗 )

MAGNITUDE PLOT
The corner frequencies are
ωc1 = 1/0.2 = 5 rad/sec and ωc2 = 1/0.02 = 50 rad/sec
The various terms of G(jω) are listed in table 1 in the increasing order of their corner frequency. Also
the table shows the slope contributed by each term and the change in slope at the corner frequency.
Table 1
Term Corner frequency Slope Change in slope

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rad/sec db/sec db/dec

(jω)2 - +40 ------
1
1 + 𝑗𝑗 0.2𝜔𝜔 ωc1=
1
= 5 - 20 40 – 20 = 20
0.2
1
1 - 20 20 – 20 = 0
1 + 𝑗𝑗 0.02𝜔𝜔 ωc2= 0.02 = 50

Choose a low frequency ωl such that ωl<ωc1 and choose a high frequency ωh such that ωh>ωc2.
Let ωl = 0.5 rad/sec and ωh = 100 rad/sec
Let A = |𝐺𝐺(𝑗𝑗𝑗𝑗)| in db
Let us calculate A at ωl, ωc1, ωc2 and ωh.
At ω = ωl, A = 20 log|(𝑗𝑗𝑗𝑗)2 | = 20 log (ω)2 = 20 log (0.5)2 = - 12 db
At ω = ωc1, A = 20 log|(𝑗𝑗𝑗𝑗)2 | = 20 log (ω)2 = 20 log (5)2 = 28 db
𝜔𝜔
At ω = ωc2, A = �𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝜔𝜔𝑐𝑐1 𝑡𝑡𝑡𝑡 𝜔𝜔𝑐𝑐2 × 𝑙𝑙𝑙𝑙𝑙𝑙 𝜔𝜔 𝑐𝑐2 � + 𝐴𝐴(𝑎𝑎𝑎𝑎 ω= ωc 1 )
𝑐𝑐1
50
= 20 x log = + 28 = 48 db
5
𝜔𝜔 ℎ
At ω = ωh, A = �𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝜔𝜔𝑐𝑐2 𝑡𝑡𝑡𝑡 𝜔𝜔ℎ × 𝑙𝑙𝑙𝑙𝑙𝑙
𝜔𝜔 𝑐𝑐2
� + 𝐴𝐴(𝑎𝑎𝑎𝑎 ω= ωc 2 )
100
= 20 x log = 50
+ 48 = 48 db
Let the points a, b, c and d be the points corresponding to frequencies ωl, ωc1, ωc2 and ωh
respectively on the magnitude plot. In a semilog graph sheet choose a scale of 1 unit = 10 db on y – axis. Fix
the point a, b, c and d on the graph. Joint the points by straight lines and mark the slope on the respective
region.
PHASE PLOT
The phase angle of G(jω) as a function of ω is given by
φ = ∠g(Jω) = 180° - tan-1 0.2ω - tan-1 0.02ω
The phase angle of G(jω) are calculated for various values of ω and listed in table 2.
Table 2.
ω tan-1 0.2ω tan-1 0.02ω φ = ∠G(jω)
rad/sec Deg deg deg
0.5 5.7 0.6 173.7 ≈ 174
1 11.3 1.1 167.6 ≈ 168
5 45 5.7 129.3 ≈ 130
10 63.4 11.3 105.3 ≈ 106
50 84.3 45 50.7 ≈ 50
100 87.1 63.4 29.5 ≈ 30
One the same semilog sheet choose a scale of 1 unit = 20°, on the y- axis on the right side of semilog
sheet. Mark the calculated phase angel on the graph sheet. Joint the points by a smooth curve.

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(1+0.2𝑗𝑗𝑗𝑗 )(1+0.02𝑗𝑗𝑗𝑗 )
𝐾𝐾(𝑗𝑗𝑗𝑗 )2
Figure 1 Bode plot of transfer function G(jω) =

Calculation of K
Given that the gain crossover frequency is 5 rad/sec. At ω = 5 rad/se the gain is 28 db. If gain
crossover frequency is 5 rad/sec then at that frequency the db gain should be zero. Hence to every point of
magnitude plot a db gain of – 28 db should be added. The addition of – 28db shifts the plot down wards.
The corrected magnitude plot is obtained by shifting the plot with K = 1 by 28db downwards. The magnitude
correction is independent of frequency. Hence the magnitude of -28db is contributed by the term K. The
value of K is calculated by equation 20 logK to -28 db.

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∴ 20 log K = - 28 db
28
Log K = -28/20 ∴ K = 10− 20 = 0.0398
The magnitude plot with K = 1 and 0.0398 and the phase plot are shown in figure
Note
The frequency ω = 5 rad/sec is a corner frequency. Hence in the expect plot the db gain at ω = 5
r5ad/sec will be 3 db less than the approximate plot. Therefore for exact plot the 20 log K will contribute a
gain of – 25dp
∴ 20 log K = - 25 db
25
Log K = -25/20 ∴ K = 10− 20 = 0.562
16] Problem
Sketch the bode plot for the following transfer function and determine phase margin and gain
margin.

75(1+0.2𝑠𝑠)
G(s) =
𝑠𝑠(𝑠𝑠 2 +16𝑠𝑠+100)

SOLUTION
The sinusoidal transfer function G(jω) is obtained by replacing s by jω in the given s-domain
transfer function after converting it to bode from or time constant from.
75(1+0.2𝑠𝑠)
Given that G(s) = 𝑠𝑠(𝑠𝑠 2 +16𝑠𝑠+100)

On comparing the quadratic factor of G(s) with standard from of quadratic factor we can estimate ζ and ωn
∴s2 + 16s + 100 = s2 + 2ζωns + 𝜔𝜔𝑛𝑛2
On comparing we get,
+ 𝜔𝜔𝑛𝑛2 = 100 ;∴ωn = 10
2ζωn = 16
16
∴ζ =
2𝜔𝜔 𝑛𝑛
16
= = 0.8
2×10
75(1+0.2𝑠𝑠)
G(s) = 𝑠𝑠2 16𝑠𝑠
𝑠𝑠×100 � + +1�
100 100
0.75(1+0.2𝑠𝑠)
=
𝑠𝑠 (1+0.01𝑠𝑠 2 +0.16𝑠𝑠)

0.75(1+0.2𝑗𝑗𝑗𝑗 ) 0.75(1+𝑗𝑗 0.2𝜔𝜔 )

∴G(jω) = =
𝑗𝑗𝑗𝑗 (1+0.01(𝑗𝑗𝑗𝑗 )2 +0.16𝑗𝑗𝑗𝑗 ) 𝑗𝑗𝑗𝑗 (1−0.01𝜔𝜔 2 +𝑗𝑗 0.16𝜔𝜔 )

Magnitude plot
1
The corner frequencies are ωc1 = = 5 rad/sec and ωc2 =ωn = 10 rad/sec
0.2
Note: For the quadratic factor the corner frequency is ωn

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The various terms of G(Jω) are listed in table 1 in increasing order of their corner frequencies. Also
the table shows the slop contributed by each term and the change in slop at the corner frequency.
Table 1
Term Corner frequency Slope Change in slop
rad/sec dp/dec db/dec
0.75
- -20
𝑗𝑗𝑗𝑗
1
1+ j0.2 ω ωc1 = 0.2 = 5 20
-20 + 20 = 0
1
ωc2 = ωn =10 -40 0 – 40 = -40
1 − 0.01𝜔𝜔 2 + 0.16𝜔𝜔

Choose a low frequency ωl such that ωl<ωcl and choose a high frequency ωh such that ωh>ωc2
Let ωl = 0.5 rad/sec and ωh = 20 rad/sec
Let A = |G(jω)| in db; Let us calculate A at ωl, ωcl, ωc2 and ωh.
0.75 0.75
At, ω = ωl, A = 20 log � �= 20 log = 3.5 db
𝑗𝑗𝑗𝑗 0.5
𝜔𝜔
At, ω= ωcl, A = �𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝜔𝜔𝑐𝑐𝑐𝑐 𝑡𝑡𝑡𝑡 𝜔𝜔𝑐𝑐2 × 𝑙𝑙𝑙𝑙𝑙𝑙 𝜔𝜔 𝑐𝑐2 � + A (atω = ωcl)
𝑐𝑐1
10
=0 x log 5 + (-16.5) = - 16.5 db
𝜔𝜔
At, ω= ωh, A = �𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝜔𝜔𝑐𝑐2 𝑡𝑡𝑡𝑡 𝜔𝜔ℎ × 𝑙𝑙𝑙𝑙𝑙𝑙 𝜔𝜔 ℎ � + A (atω = ωc2)
𝑐𝑐2
20
= -40 x log 10 + (-16.5) = - 28.5 db
Let the point a, b, c and d be the points corresponding to frequencies ωl, ωc1, ωc2 and ωh respectively
on the magnitude plot. In a semi log graph sheet choose a scale of 1 unit = 5 dp on y-axis. The frequencies
are marked in decades from 0.1 to 100 rad/sec on logarithmic scales in x-axis. Fix the points a, b, c and d on
the graph. Join the points by straight lines and mark the slop on the respective region.
PHASE PLOT
The phase angle of G(jω) as a function of ω is given by
0.16𝜔𝜔
φ = ∠G(jω) = tab-1 0.2ω - 90° -tan-1 For ω ≤ ωn
1−0.01𝜔𝜔 2
0.16𝜔𝜔
φ= ∠G(jω) = tab-1 0.2ω - 90° -tan-1 + 180° For ω ≤ ωn
1−0.01𝜔𝜔 2
The phase angle of G(jω) are calculated for various values of ω and listed in table 2
Note: in quadratic factors the angle varies from 0° to 180°. But the calculator calculates tan-1 only between
0° to 90°. Hence a correction factor of 180° should be added to the phase angle after corner frequency.
On the same semilog sheet choose a scale of 1 unit = 20° on the y-axis on the right side of semi log
sheet. Mark the calculated phase angle on the graph sheet. Joint the points by a smooth curve.

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𝑗𝑗𝑗𝑗 (1−0.01𝜔𝜔 2 +𝑗𝑗 0.16𝜔𝜔 )

0.75(1+𝑗𝑗 0.2𝜔𝜔 )
Figure 1: Bode plot of transfer function, G(jω) =

Table 2
ω Tan-1 0.2 ω 𝟎𝟎.𝟏𝟏𝟏𝟏𝝎𝝎
-tan-1𝟏𝟏−𝟎𝟎.𝟎𝟎𝟎𝟎𝝎𝝎𝟐𝟐 φ = ∠G(jω)
rad/sec deg.
0.5 5.7 4.6 -88.9 ≈-88
1 11.3 9.2 -87.9 ≈ -88
5 45 46.8 -91.8 ≈ -92
10 63.4 90 -116.6 ≈-116
20 75.9 -46.8 + 180 = 133.2 -147.3 ≈ -148
50 84.3 -18.4 + 180 = 161.6 -167.3≈ -168
100 87.1 -9.2 + 180 = 170.8 -173.7 ≈ -174

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The magnitude plot and the phase plot are shown in figure 1 from the figure 1 we find that the
phase angle at gain crossover frequency (ωgc) = φgc = 88°
∴ Phase Margin, γ = 180° + φgc = 180° - 88° = 92°
Here, Gain margin = + ∞.
The phase plot crosses - 180° only at infinity. The |G(jω)| at infinity is - ∞ db. Hence gain margin is + ∞.
NICHOLS PLOT
The Nichols plot is a frequency response plot of the open loop transfer function of a system. The
Nichols plot is a graph between magnitude of G(j𝜔𝜔) in db and the phase of G(j𝜔𝜔) in degree, plotted on a
ordinary graph sheet.
To plot the Nichols plot, first compute the magnitude of G(j𝜔𝜔)in db and phase of G(j𝜔𝜔) in deg for
various values of 𝜔𝜔 and tabulate them. Usually the choices of frequencies are corner frequencies. Choose
appropriate scales for magnitude on y-axis and phase an x-axis. Fix all the points on ordinary graph sheet and
join the points by smooth curve. Write the frequency corresponding to each point of the plot.
In another method, first the Bode plot of G(j𝜔𝜔) is sketched. From the Bode plot the magnitude and
phase for various values of frequency, 𝜔𝜔 are noted and tabulated. Using these values the Nichols plot is
sketched as explained earlier.
Determination of Gain Margin and Phase Margin From Nichols Plot
The gain margin in db is given by the negative of db magnitude of G(j𝜔𝜔) at the phase crossover
frequency, 𝜔𝜔𝑝𝑝𝑝𝑝 .
The 𝜔𝜔𝑝𝑝𝑝𝑝 is the frequency at which phase of G(j𝜔𝜔) is –180°. If the db magnitude of G(j𝜔𝜔) at 𝜔𝜔𝑝𝑝𝑝𝑝 is
negative then gain margin is positive and vice versa.
Let φ𝑔𝑔𝑔𝑔 be the phase angle of G(j𝜔𝜔) at gain cross over frequency 𝜔𝜔𝑔𝑔𝑔𝑔 . The 𝜔𝜔𝑔𝑔𝑔𝑔 is the frequency at
which the db magnitude of G(j𝜔𝜔) is zero. Now the phase margin,𝛾𝛾 is given by 𝛾𝛾 = 180° + φ𝑔𝑔𝑔𝑔 . If φ𝑔𝑔𝑔𝑔 is less
negative than –180° then phase margin is positive and vice versa.
The positive and negative gain margins are illustrated in figure1

Figure 1 Nichols plot showing phase margin and gain margin

Gain Adjustment In Nichols Plot

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In the open loop transfer function, G(j𝜔𝜔) the constant K contributes only magnitude. Hence by
changing the value of K the system gain can be adjusted to meet the desired specifications. The desired
specifications are gain margin and phase margin.
In a system transfer function if the value of K required to be estimated to satisfy a desired
specification then draw the Nichols plot of the system with K = 1. The constant K can add 20log to every
point of the plot. Due to this addition the Nichols plot will shift vertically up or down. Hence shift the plot
vertically up or down tomeet the desired specification. Equate the vertical distance by which the Nichols plot
is shifted to 20 log K and solve for K.
Let x = change in db (x is positive if the plot is shifted up and vise versa)
Now, 20 log K = x
𝑋𝑋
log K =
20
𝑥𝑥
∴ 𝐾𝐾 = 1020
17] Problem
Consider a unit feedback system have an open loop transfer function
𝐾𝐾(1+10𝑠𝑠)
G(s) = . Sketch the Nichols plot and determine the value of K so that (i) Gain margin is 10db,
𝑠𝑠 2 (1+𝑠𝑠)(1+2𝑠𝑠)
(ii) Phase margin is 10°
SOLUTION
𝐾𝐾(1+10𝑠𝑠)
Given that G(s) =
𝑠𝑠 2 (1+𝑠𝑠)(1+2𝑠𝑠)
The sinusoidal transfer function G(jω) is obtained by letting to s = jω. Also put K = 1
(1+𝑗𝑗10𝜔𝜔)
∴G(jω) =
(𝑗𝑗𝑗𝑗 )2 (1+𝑗𝑗𝑗𝑗 )(1+𝐽𝐽2𝜔𝜔 )

�1+(10𝜔𝜔 )2 ∠𝑡𝑡𝑡𝑡𝑡𝑡 −1 10𝜔𝜔

=
𝜔𝜔 2 ∠180 ° √1+𝜔𝜔 2 ∠𝑡𝑡𝑡𝑡𝑡𝑡 −1 𝜔𝜔 �1+(2𝜔𝜔 2 ∠𝑡𝑡𝑡𝑡𝑡𝑡 −1 2𝜔𝜔

√1+100𝜔𝜔 2
G(jω) =
𝜔𝜔 2 �1+𝜔𝜔 2 𝐽𝐽 √1+4𝜔𝜔 2
√1+100𝜔𝜔 2
∴|𝐺𝐺(𝑗𝑗𝑗𝑗)|= 20 log
𝜔𝜔 2 √1+𝜔𝜔 2 √1+4𝜔𝜔 2
∠G(jω) tan-1 10ω - 180°- tan-1ω - tan-1 2ω
The magnitude of G(jω in db and phase of G(jω) in deg are calculated for various values of ω and
listed in the following table. The Nichols plot of G(jω) with K = 1 is sketched as shown in figure 4.13.1
ω 0.2 0.4 0.6 0.8 1.0 1.5 2.0 3.0 4.0
rad/sec
|𝐺𝐺(𝑗𝑗𝑗𝑗)| 34.1 25.4 19.3 14.3 10 1.4 -5.3 -15.2 -22.5
db
∠G(jω) -150 -164 -181 -194 -204 -222 -232 -244 -250

From the Nichols plot the gain margin and phase margin of the system when K = 1 are
Gain margin = - 19.5 db
Phase margin = - 45°
Gain Adjustment for Required Gain Margin

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For a gain margin of 10 db, the magnitude of G(jω) should be – 10 db, when the phase is - 180°. Hence if we
add -29.5 db to every point of G(jω), then the plot shifts down wards and it will cross - 180° axis at a
magnitude of -10db. The magnitude correction is independent of frequency and so this gain can be
contributed by the term K. Let this value of K be K1. The value of K1 is calculated by equating 20 log K1 to -
29.5 db.
∴ 20 log K1 = -29.5 db
Log K1 = -29.5/20
−29.5
K1 = 10 20 = 0.0335
Gain adjustment for required phase margin
Let φgc2 = phase of G(jω) at gain crossover frequency for a phase margin of 10°
∴ Phase margin, γ2 = 180 + φgc2
∴φgc2 =γ2 - 180° = 10 – 180 = -170°
When K = 1, the magnitude of G(jω) is +23 db corresponding to phase of - 170°. But for a phase margin of
10°, this gain should be made zero. Hence if we add -23 db to every point of G(jω) locus then the plot shifts
downwards and it will cross - 170° axis at magnitude of 0 db. The magnitude correction is independent of
frequency and so this gain can be contributed by the term K. Let this value of K be K2. The value of K2 is
calculated by equation 20 log K2 to – 23 db.
∴ 20 log K2 = –23
Log K2 = –23/20
−23
K2 =10 20 = 0.07

RESULT

When K = 1,
Gain margin = –19.5db
Phase margin = –45°
For gain margin of 10db, K = K1 = 0.0335
For a phase margin of 10°, K = K2 = 0.07

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(𝑗𝑗𝑗𝑗 )2 (1+𝑗𝑗𝑗𝑗 )(1+𝐽𝐽2𝜔𝜔 )

(1+𝑗𝑗 10𝜔𝜔 )
Figure1 Nichols plot of G(jω) =

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Unit III
Sensors and Transducers
Sensors are devices which produce a proportional output signal (mechanical, electrical, magnetic, etc.)
when exposed to a physical phenomenon (pressure, temperature, displacement, force, etc.). Many devices
require sensors for accurate measurement of pressure, position, speed, acceleration or volume. Transducers
are devices which converts an input of one form of energy into an output of another form of energy. The term
transducer is often used synonymously with sensors. However, ideally, the word 'transducer' is used for the
sensing element itself whereas the term 'sensor' is used for the sensing element plus any associated signal
conditioning circuitry. Typically, a transducer may include a diaphragm which moves or vibrates in response
to some form of energy, such as sound.
Some common examples of transducers with diaphragms are microphones, loudspeakers, thermometers,
position and pressure sensors. Sensors are transducers when they sense one form of energy input and output
in a different form of energy. For example, a thermocouple responds to a temperature change (thermal energy)
and outputs a proportional change in electromotive force (electrical energy). Therefore, a thermocouple can
be called a sensor and or transducer.
Figure illustrates a sensor with sensing process in terms of energy conversion. The form of the output
signal will often be a voltage analogous to the input signal, though sometimes it may be a wave form whose
frequency is proportional to the input or a pulse train containing the information in some other
form.
Sensor

Input signal/energy Transducer

output Signal conditioning Output signal/energy
Transducer
(from variables Circuitry
Or parameters)

Selection of Sensors
In selecting a sensor for a particular application there are a number of factors that need to be
considered:
1. The nature of the measurement required, e.g., the variable to be measured, its nominal value, the
range of values, the accuracy required, the required speed of measurement, the reliability required,
the environmental conditions under which the measurement is to be made.
2. The nature of the output required from the sensor, this determining the signal conditioning
requirements in order to give suitable output signals from the measurement.
3. The possible sensors can be identified taking into account such factors as their range, accuracy,
linearity, speed of response, reliability, maintainability, life, power supply requirements, ruggedness,
availability and cost.
The selection of sensors cannot be taken in isolation from a consideration of the form of output that
is required from the system after signal conditioning and thus there has to be a suitable marriage between
sensor and signal conditioner.
Classification and characteristics of Sensors
Sensors are generally classified into two types based on its power requirement: passive and active. In
active sensors, the power required to produce the output is provided by the sensed physical phenomenon
itself (Examples: thermocouples, photovoltaic cells, piezoelectric transducers, thermometer etc.) whereas the
passive sensors require external power source (Examples: resistance thermometers, potentiometric devices,

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differential transformers, strain gage etc.). The active sensors are also called as self-generating transducers.
Passive sensors work based on one of the following principles: resistance, inductance and capacitance.
Sensors can also be classified as analog or digital based on the type of output signal. Analog sensors
produce continuous signals that are proportional to the sensed parameter. These sensors generally require
analog-to-digital conversion before sending output signal to the digital controller (Examples:
potentiometers, LVDTs (linear variable differential transformers), load cells, and thermistors, bourdon tube
pressure sensor, spring type force sensors, bellows pressure gauge etc.). Digital sensors on the other hand
produce digital outputs that can be directly interfaced with the digital controller (Examples: incremental,
encoder, photovoltaic cells, piezoelectric transducers, phototransistors, photodiodes etc.). Often, the digital
outputs are produced by adding an analog-to-digital converter to the sensing unit. If many sensors are
required, it is more economical to choose simple analog sensors and interface them to the digital controller
equipped with a multi-channel analog-to-digital converter.
Another way of classifying sensor refers to as primary or secondary sensors. Primary sensors produce
the output which is the direct measure of the input phenomenon. Secondary sensors on the other hand
produce output which is not the direct representation of the physical phenomenon. Mostly active sensors are
referred as primary sensors where as the passive sensors are referred as secondary sensors.
1. Static Characteristics
Static character sties of an instrument are the parameters which are more or less constant or varying
very slowly with time. The following characteristics are static characteristics.
Range
Every sensor is designed to work over a specified range i.e. certain maximum and minimum values.
The design ranges are usually fixed, and if exceeded, result in permanent, damage to or destruction of it
sensor. For example, a thermocouple may have a range of-100 to 1260 °C.
Span
It represents the highest possible input value which can be applied to the sensor without causing
unacceptably large inaccuracy. Therefore, it is the difference between maximum and minimum values of the
quantity to be measured.
Span = Maximum value of the input - Minimum value of the input
Errors
Error is the difference between a measured value and the true input value.
Error = Measured value - True input value
Accuracy
A very important characteristic of a sensor is accuracy which really means inaccuracy. Inaccuracy is
measured as a ratio of the highest deviation of a value represented by the sensor to the ideal value. The
accuracy of a sensor is inversely proportional to error, i.e., a highly accurate sensor produces low errors.
Sensitivity
Sensor sensitivity is defined as the change in output per change in input. The factor may be constant
over the range of the sensor (linear), or it may vary (nonlinear).
Change in output ∆𝜃𝜃𝑜𝑜
Sensitivity = =
Change in output ∆𝜃𝜃 𝑖𝑖
When an instrument consists of different elements connected in series and have static sensitivities of S1, S2,
S3, ... etc, then the overall sensitivity is expressed as follows.

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𝜃𝜃1 𝜃𝜃2 𝜃𝜃3

S1 = , S2 = , S3 = , ….
𝜃𝜃 𝑖𝑖 𝜃𝜃 𝑖𝑖 𝜃𝜃 𝑖𝑖
𝜃𝜃0 𝜃𝜃1 𝜃𝜃2 𝜃𝜃3
Overall sensitivity, S = = × × × …. = S1 × S2 × S3 ……
𝜃𝜃 𝑖𝑖 𝜃𝜃 𝑖𝑖 𝜃𝜃 𝑖𝑖 𝜃𝜃 𝑖𝑖
Hysteresis
Hysteresis is defined as the maximum differences in output
for a given input when this value is approached from the opposite
direction. It is a phenomenon which shows different outputs when
loading and unloading. Simply, hysteresis means that both the
loading and unloading curves do not coincide. Figure 1 shows that
the deviation of unloading from loading condition due to hysteresis
effect.
Linearity
Linearity of a sensor refers to the output that is directly Figure 1 Hysteresis effects
proportional to input over its entire range, so that the slope of a
graph of output versus input describes a straight line. If the response
of the system to input A is output A, and the response to input B is
output B, then the response to input C (= input A + input B) will be
output C ( = output A + output B).
Non-linearity
Non-linearity of a sensor refers to the output that is not
proportional to input over its entire range, so that the slope of a
graph of output versus input describes a curve. Non-linearity error is
the deviation of output curve from a specified straight line as shown Figure 2 Non-linearity errors
in Figure 2.
Repeatability and reproducibility
Repeatability may be defined as the ability of the sensor to give same output reading when the same
input value is applied repeatedly under the same operating conditions.
Reproducibility may be defined as the degree of closeness among the repeated measurements of the
output for the same value of input under the same operating conditions at different times.
Stability
Stability means the ability of the sensor to indicate the same output over a period of time for a constant
input.
Dead band/time
Dead band of a sensor is the range of input values for which the instrument does not respond. The dead band
is typically a region of input close to zero at which the output remains zero.
Dead time is the time taken by the sensor from the application of input to begin
its response and change.
Resolution
Resolution is defined as the smallest change that can be detected by a sensor. It can also be defined
as the minimum value of the input required to cause an appreciable change or an increment in the output.

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Zero Drift
Drift is the variation of change in output for a given input over a period of time. When making a measurement
it is necessary to start at a known datum, and it is often convenient to adjust the output of the instrument
to zero at the datum. The signal level may vary from its set zero value when the sensor works. This
introduces an error into the measurement equal to the amount of variation or drift. Zero drift may result from
changes of temperature, electronics stabilizing, or aging of the transducer or electronic components.
Output impedance
Impedance is the ratio of voltage and current flow for a sensor. Two types of impedance are important
in sensor applications: input impedance and output impedance. Input impedance is a measure of how much
current must be drawn to power a sensor. Output impedance is a measure of a sensor's ability to provide
current for the next stage of the system.
2. Dynamic characteristics
Sensors and actuators respond to inputs that change with time. Any system that changes with time is
considered a dynamic system. Dynamic characteristics of aninstrument are the parameters which are
varying with time. The following characteristics are dynamic characteristics.
Response time
The time taken by a sensor to approach its true output when subjected to a step input is sometimes
referred to as its response time. It is more usual, however, to quote a sensor as having a flat response
between specified limits of frequency. This is known as the frequency response, and it indicates that if the
sensor is subjected to sinusoidally oscillating input of constant amplitude, the output will faithfully reproduce
a signal proportional to the input.
Time constant
It is the time taken by the system to reach 63.2% of its final output signal amplitude i.e. 62.3% of
response time. A system having smaller time constant reaches its final output faster than the one with larger
time constant. Therefore possesses higher speed of response.
Rise time
It is the time taken by the system to reach 63.2% of its final output signal.
Setting time
It is the time taken by a sensor to be within a close range of its steady state value.
TYPES OF SENSORS
1. Inductive Displacement Sensors
The most widely used variable-inductance displacement transducer in industry is LVDT (Linear Variable
Differential Transformer). It is a passive type sensor. It is an electro-mechanical device designed to produce
an AC voltage output proportional to the relative displacement of the transformer and the ferromagnetic
core.
The physical construction of a typical LVDT consists of a movable core of magnetic material and three
coils comprising the static transformer as shown in Figure 1.39. One of the three coils is the primary coil or
excitation coil and the other two are secondary coils or pick-up coils. An AC current (typically 1 kHz) is passed
through the primary coil, and an AC voltage is induced in the secondary coils. The magnetic core inside the
coil winding assembly provides the magnetic flux path liking the Primary and secondary Coils.

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When the magnetic core is at the centre position or null

position the output voltages are equal and opposite in polarity
and, therefore, the output voltage is zero. The Null Position of an
LVDT is extremely stable and repeatable. When the magnetic core
is displaced from the Null Position, a certain number of coil
windings are affected by the proximity of the sliding core and thus
an electromagnetic imbalance occurs. This imbalance generates a
differential AC output voltage across the secondary coil which is
linearly proportional to the direction and magnitude of the
displacement. The output voltage to displacement plot is a
straight line within a specified range. Beyond the nominal range,
the output deviates from a straight line in a gentle curve as shown
in Figure 1.40. Figure 3 LVDE (Linear variable
The Rotational Variable Differential Transformer (RVDT) is differential transformer)
used to measure rotational angles and
operates under the same principles as
the LVDT sensor. Whereas the LVDT
uses a cylindrical iron core, the RVDT
uses a rotary ferromagnetic core. A
schematic of RVDT is shown in Figure
1.41.
Calculation of output voltage
Motion of a magnetic core
changes the mutual inductance of two
Figure 4 LVDT output
secondary coils relative to a primary coil
Primary coil voltage: Vin = sin(ωt)
Secondary coils induced emf:
V1 = k1 sin(ωt) and
V2 = k2 sin(ωt)
The value of k1 and k2 depend on the amount
of coupling between the primary and the
secondary coils, which is proportional to the
position of the coil.
When the coil is in the central position, k1=k2
Vout = V1 – V2 = 0
When the coil is displaced x units, k1 ≠ k2
Figure 5 RVDT (Rotary variable
Vout = (k1— k2) sin(ωt)
differential transformer)
Positive or negative displacements are
determined from the phase of Vout.
Applications
LVDT can be used to measure the displacement, deflection, position and profile of a workpiece.
Advantages
 Relative low cost due to its popularity.
 Solid and robust, capable of working in a wide variety of environments.

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 No friction resistance, since the iron core does not contact the transformer coils, resulting in an
infinite (very long) service life.
 High signal to noise ratio and low output impedance.
 Negligible hysteresis.
 Short response time, only limited by the inertia of the iron core and the rise time of the amplifiers.
 No permanent damage to the LVDT if measurements exceed the designed range.
 It can operate over a temperature range of - 265 °C to 600°C.
 High sensitivity up to 40 V/mm.
 Less power consumption (less than 1W)
Disadvantages:
 The performance of these sensors is likely affected by vibration etc.
 Relatively large displacements are required for appreciable output.
 Not suitable for fast dynamic measurements because of mass of the core.
 Inherently low in power output.
 Sensitive to stray magnetic fields but shielding is not possible.

2. Inductive proximity sensor

Inductive proximity sensors are today the most commonly employed industrial sensors for detection of
ferrous metal objects over short distances. Inductive proximity sensors operate under the electrical principle
of inductance. Inductance is the phenomenon where a fluctuating current, which by definition has a
magnetic component, induces an electromotive force (emf) in a target object.

Figure 6 Inductive proximity sensor

An inductive proximity sensor has four components; the induction coil, oscillator, detection circuit and
output circuit as shown in Figure 6. The oscillator generates a fluctuating magnetic field the shape of a
doughnut around the winding of the coil that locates in the device's sensing face. When a metal object moves
into the magnetic field of detection, eddy circuits build up in the metallic object. These eddy currents
produce a secondary magnetic field that interacts with field of the probe, thereby loading the probe
oscillator. The effective impedance of the probe coil changes, resulting in an oscillator frequency shift (or
amplitude change). The sensor's detection circuit monitors the oscillator's strength and triggers an output
signal from the output circuitry proportional to the sensed gap between probe and target.
3. Pyro-electric Sensors
The pyroelectric sensor is made of a crystalline material that generates a surface electric charge when
exposed to heat source. Example of crystalline material is lithium tantalite. When this type of material is
heated below a temperature known as Curie point, a large spontaneous electrical polarization is exhibited
from the material in response to a temperature change. The change in polarization is observed as an
electrical voltage signal if electrodes are placed on opposite faces of a thin slice of the material. Figure 7 (a)
shows that the charges in the pyroelectric material are balanced if there is no infrared radiation from the
heat source falls on the materials surface. When the material is exposed to the infrared radiation from the

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heat source the charges in the pyroelectric material are not balance and hence there is some excess charge
in the material as shown in Figure 7 (b).

Figure 7 Pyroelectric effect

The design can be thought of as a typical form of a capacitor circuit. Figure 8 shows the equivalent
circuit of a pyroelectric sensor. It essentially consists of a capacitor charged by the excess charge with a
resistance R to represent the internal leakage combined with the input resistance of an external circuit.
For detection of a human motion or intrusion in the country borders, the
pyroelectric are sensors used. In such, applications, the sensing element
has to differentiate between general background heat radiation and a
moving heat source. Therefore, a single pyroelectric sensor is not capable to
use and dual pyroetectric sensors are used as shown in Figure 9. In this dual
pyroelectric sensor, the sensing element has the one front electrode and
two back electrodes. When two sensors are connected, both sensors Figure 8 Equivalent
circuit of a pyroelectric
receive the same heat signal and their outputs are cancelled. When a heat
sensor
source moves from its position the heat radiation
moves from one of the sensing elements to the
other. Then the current alternates in one direction
first and then reversed to the other direction. When
the amount of infrared radiation from heat source
striking the crystal, the electric charge also changes
and can then be measured with a sensitive FET
device built into the sensor.

4. Force Sensors
Force sensors are used in many mechanical equipments and aggregates for an accurate
determination of forces applied in the system. The
force sensor outputs an electrical signal Figure 9 Dual pyroelectric sensor
corresponding to the force applied. Force sensors are
commonly used in many applications such as automotive brakes, suspension, transmission, speed control,
lifts, aircrafts, digital weighing systems etc. Most of the force sensor uses displacement as the measure of
the force. The simplest form of force sensor is the spring balance in which a force is applied to the one end
of the spring causes displacement of the spring. This displacement is the measure of the force applied. A
common force sensor is a strain gauge load cell which is explained under.
5. Strain gauge load cell
A load cell is an electromechanical transducer that converts load acting on it into an analog electrical
signal. Load cells provide accurate measurement of compressive and tensile loads. Load cells commonly
function by utilizing an internal strain gauge that measure deflection. Because the modulus of elasticity of a
load cell is constant the amount of strain can be calibrated to determine the force upon the load cell.
Typically the force creates the train in the load cell which is measured by strain gauge transducer.

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Strain gauge is attached to the object or the

strained element where the force is being applied. As the
object is stressed due to the applied force, the resulting
strain deforms the strain gauge attached with it. This
causes an increase in resistivity of the gauge which
produces electrical signal proportional to the
deformation. The measurement of resistivity is the
measure of strain which in turn gives the measurement
of force or load applied on the object. The change of
resistance is generally very small and is usually measured
using a Wheatstone bridge circuit where the strain
gauges are connected into the circuit. The strain gauges
are serving as resistors in the circuit. The Wheatstone
bridge circuit produces analog electrical output signal. In
a typical strain gauge load cell for measuring force, four
strain gauges are attached to the surface of the
counterforce and are electrically connected in a full
Wheatstone bridge circuit as shown in Figure 10. Load
cells have different shapes (cylindrical tubes, rectangular Figure 10 Strain gauge load cell
or square beams, and shaft) for different applications
and load requirements to ensure that the desired component of force is measures, thus strain gauges having
different shapes are positioned in various orientations upon the load cell body. The different configurations
of strain gauges are already discussed under strain gauges displacement sensors.
6. Fluid Pressure Sensors
Pressure is an expression of the force required to stop a fluid from expanding, and is usually expressed in
terms of force per unit area. A pressure sensor measures pressure of gases or liquids. These sensors
generate a signal as a function of the pressure applied by the fluid. Pressure sensors are used in many
applications such as automotive vehicles, hydraulic systems, engine testing etc. Pressure sensors may
required to measure different types of pressures: 1. Absolute pressure where the pressure is measured
relative to the perfect vacuum or zero-pressure, 2. Gauge pressure where the pressure is measured relative
to the atmospheric pressure, and 3. Differential pressure where a pressure difference is measured. The
devices which are used to measure fluid pressure in industrial processes are:
1. Diaphragm pressure sensor
2. Capsule pressure sensor
3. Bellows pressure sensor
4. Bourdon tube pressure sensor
5. Piezoelectric sensor
6. Tactile Sensor
The construction and working principle of these sensors are explained here.
7. Diaphragm pressure sensor
The diaphragm pressure sensor uses the elastic deformation of a diaphragm (i.e. membrane) to
measure the difference between an unknown pressure and a reference pressure. Diaphragm is a thin circular
elastic membrane made of generally silicon as show in Figure 11. As pressure changed, the diaphragm
moves, and this motion is the measure of differential pressure. Diaphragms are popular because they
require less space and the motion they produce is sufficient for operating electronic transducers. They also
are available in a wide range of materials for corrosive service applications.

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Figure 11. Diaphragm

A typical diaphragm pressure gauge contains a chamber divided by a diaphragm, as shown in the Figure 12.
One side of the diaphragm is open to the external targeted pressure. PExt, and the other side is connected to
a known pressure, PRefr. The pressure difference, PExt, PRef, mechanically deflects the diaphragm.

Figure 12 Typical diaphragm pressure gauge

The diaphragm deflection can be measured in any number of ways. For example, i can be detected
via a mechanically-coupled indicating needle, an attached strain gauge [refer Figure 13 (a)], a linear variable
differential transformer (LVDT) [refer Figure 13 (b)], or with many other displacement/velocity sensors. Once
known, the deflection can be converted to a pressure loading using plate theory.

Figure 13

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Strain gauge arrangement consists of four strain gauges with, two measuring the strain in a
circumferential direction while the remaining two measure strains in a radial direction. The four strain
gauges are connected to form the arms of a Wheatstone bridge. The sensitivity of pressure gauges using
LVDTs is good and, therefore, stiff primary sensors with very little movement can be used to reduce
environmental effects. Frequency response is also good.
Advantages:
 Much faster frequency response than U tubes.
 Accuracy up to ±0.5% of full scale.
 Good linearity when the deflection is no larger than the order of the diaphragm thickness.
Disadvantages:
 More expensive than other pressure sensors.

8. Capsule pressure sensor

In order to improve the sensitivity, two corrugated diaphragms are combined by arranging these in
back-to-back and sealed together at the periphery to obtain shell like shape as shown in Figure 1.65. These
are called as capsules. One of the diaphragms is provided with a central reinforced port to allow the pressure
to be measured, and the other is linked to a mechanical element. The difference in pressure between inner
and outer surface of the capsule produces displacement. These capsules can also be attached with the LVDT
as described in the diaphragm pressure gauge.

Figure 14 Capsule pressure sensor

9. Bellows pressure sensor
The bellows is a one-piece, collapsible, seamless metallic unit that has deep folds formed from very
thin-walled tubing. It looks like a stake of capsules. It is more sensitive than the diaphragm and capsule
pressure sensors. The diameter of the bellows ranges from 1.2 to 30 cm and may have as many as 24 folds.
System pressure is applied to the internal volume of the bellows. As the inlet pressure varies, the bellows
will expand or contract. The moving end of the bellows is connected to a mechanical linkage assembly. The
deflection can be measured in any number of ways. For example, it can be detected via a mechanically-
coupled indicating needle [refer Figure 15 (a)], a linear variable differential transformer (LVDT) as described
in the diaphragm pressure gauge [refer Figure 13 (b)], a potentiometer [refer Figure 14 (b)], or with many
other displacement sensors. As the bellows and linkage assembly moves, either an electrical signal is
generated or a direct pressure indication is provided. Figure 14 shows a bellows pressure sensing element
along with the potentiometer.
The potentiometric bellows pressure sensor provides a simple method for obtaining an electronic
output from a mechanical pressure gauge. The device consists of a precision potentiometer, whose wiper
arm is mechanically linked bellows or Bourdon-element. The movement of the wiper arm across the
potentiometer converts the mechanically detected sensor deflection into a resistance measurement, using a
Wheatstone bridge circuit.

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Figure 15 Bellower pressure sensor

The flexibility of a metallic bellows is similar in character to that of a helical, coiled compression
spring. Up to the elastic limit of the bellows, the relation between increments of load and deflection is linear.
In practice, the bellows must always be opposed by a spring, and the deflection characteristics will be the
resulting force of the spring and bellows.
10. Bourdon tube pressure sensor
The bourdon tube pressure instrument is one of the oldest pressure sensing instruments in use today. It
is widely used in applications where inexpensive static pressure measurements are needed, the bourdon
tube consists of a thin-walled C-shaped tube that is flattened diametrically on opposite sides to produce a
cross-sectional area elliptical in shape, having two long flat sides and two short round sides. The tube is bent
lengthwise into an arc of a circle of 270 to 300 degrees. Bourdon tube is open to external pressure input on
one end and is coupled mechanically to an indicating needle on the other end as shown schematically in
Figure 16. Pressure applied to the inside of the tube causes distention of the flat sections and tends to
restore its original round cross-section. This change in cross-section causes the tube to straighten slightly.
Since the tube is permanently fastened at one end, the tip of the tube traces a curve that is the result of the
change in angular position with respect to the center. Within limits, the movement of the tip of the tube can
then be used to position a pointer or to develop an equivalent electrical signal to indicate the value of the
applied internal pressure.
The deflection of the Bourdon tube can be measured in any number of ways. For example, it can be
detected via a mechanically-coupled indicating needle [refer Figure 16], a linear variable differential
transformer (LVDT) as described in the diaphragm pressure gauge [refer Figure 12 (b)], a potentiometer
[refer Figure15(b)], or with many other displacement sensors.

Figure 16 C-shaped Bourdon tube pressure sensor

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To increase their sensitivity, Bourdon tube elements can be extended into spirals or helical coils [Figures
17 (a) and (b)]. This increases their effective angular length and therefore increases the movement at their
tip, which in turn increases the resolution of the transducer.

Figure 17 Spiral and helical coil Bourdon tubes

Advantages:
 Portable
 Convenient to use
 No leveling required
Disadvantages:
 Limited to static or quasi-static measurements.
 Accuracy may be insufficient for many applications. A mercury barometer can be used to calibrate
and check Bourdon Tubes.
11. Piezoelectric sensors
A piezoelectric sensor is a device that uses the piezoelectric effect to measure pressure, acceleration,
strain or force. When pressure, force or acceleration is applied to piezoelectric materials such as quartz
crystal, PZT ceramic, tourmaline, gallium phosphate, and lithium sulfate, an electrical charge is developed
across the crystal that is proportional to the force applied (Figure 18 (a)). When pressure is applied to a
crystal, it is elastically deformed. This deformation results in a flow of electric charge (which lasts for a
period of a few seconds). The" resulting electric signal can be measured as an indication of the pressure
which was applied to the crystal.

Figure 18 Piezoelectric sensors

The net electrical charge (q) produced in the crystal is proportional to the deformation of the crystal
(x) due to the applied pressure and the stiffness of the material (k). Since the deformation is proportional to
the applied pressure or force (P), the net electric charge is given by the equation:
q = k × x =S× P
where S is the charge sensitivity.
The piezoelectric sensors are attached with the diaphragm pressure sensing element to measure the
pressure as shown in Figure 18 (b).

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The output electrical signal of the piezoelectric sensor is related to the t mechanical force or
pressure as if it had passed through the equivalent circuit as shown in Figure 19. The model of the equivalent
circuit includes the following components:
C represents the capacitance of the sensor surface itself;
R is the insulation leakage resistance of the transducer; and
q is the charge generator
If the sensor is connected to a load resistance, this also acts in parallel
with the insulation resistance.
The fundamental difference between these piezoelectric sensors and
static-force devices such as strain gauges is that the electric signal generated Figure 19 Equivalent
by the piezoelectric sensors decays rapidly. This characteristic makes these circuit of Piezoelectric
sensors unsuitable for the measurement of static forces or pressures but sensor
useful for dynamic measurements.
Piezoelectric pressure sensors do not require an external excitation source and are very rugged.
These sensors, however, do require charge amplification circuitry and very susceptible to shock and
vibration.
The desirable features of piezoelectric sensors include their rugged construction, small size, high
speed, and self-generated signal. On the other hand, they are sensitive to temperature variations and
require special cabling and amplification.
12. Tactile sensors
Tactile pressure sensors are used to detect
the pressure distribution between a sensor and a
target. They are often used on the robot grippers or
flat tactile arrays to identify whether the finger is in
touch with the target object or not. These sensors are
also used in touch screen display of laptops, ATM
machines, mobiles etc. Most tactile pressure sensors
use resistive-based technologies where the sensor
acts as a variable resistor in an electrical circuit. A
small deflection of the diaphragm causes implanted
resistors to exhibit a change in resistance value. The
sensor converts this change in resistance into a
voltage that is interpreted as a continuous and linear
Figure 20 Tactile pressure sensor
pressure reading. When tactile pressure sensors are
unloaded, their resistance is very high. When force is applied, their resistance decreases. Pressure sensitive
film is used to create a direct, visual image of the pressure distribution. Active pressure sensor arrays consist
of multiple sensing elements packaged in a single sensor.
There are many different forms of tactile
sensors. One form of tactile pressure sensor includes
upper and lower conductive layers separated by an
intermediate insulating layer which is formed as a
separating mesh (Refer to Figure 20). The upper
conductive layer is of negligible resistance. The lower
conductive layer is formed of a plurality of conductive
strips (A-F) separated by insulating strips. Each Figure 21 PVDF Tactile sensor
conductive strip (A-F) has a known resistance. An
electrical signal is applied to the conductive strips (A-F) in turn and the electrical path between the upper
and lower conductive layers then determined. The electrical resistance of the conductive path establishes

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the location of the pressure point at which bridging occurs and from this it is possible to establish the
location and size of the pressure area.
Figure 22 shows another form of tactile sensor. It uses piezoelectric material of polyvinylidene fluoride
(PVDF) film. Two layers of PVDF films are used and they are separated by a soft film which transmits
vibrations. When the alternating voltage is supplied in the lower PVDF film it results in mechanical
oscillations of the film. The intermediate film transmits these vibrations to the upper PVDF film. Due to the
piezoelectric effect the vibrations formed cause an-alternating voltage to be produced across the upper film.
So, pressure is applied to the upper PVDF film and its vibrations affect the output voltage.
SWITCHES
Proximity Switches
There are a number of forms of switch which can be activated by the presence of an object in order
to give a proximity sensor with an output which is cither ON or OFF.
The micro switch is a small electrical switch which requires physical contact and a small operating
force to close the contact and a small operating force to close the contacts. For example, in the case of
determining the presence of an item on a conveyor belt, this might be actuated by the weight of the item on
the belt depressing the belt and hence a spring-loaded platform then closing the switch. Figure 3.1 shows
examples of ways such switches can be actuated.

Figure 22. Various types of switches

Figure 23 shows the basic form of a reed switch. It consists
of two magnetic switch contacts scaled in a gas tube. When a
magnet is brought close to the switch, the magnetic reeds are
attracted to each other and close the switch contacts. It is a non-
contact proximity switch. Such a switch is very widely used for
checking the closure of doors. It is also used with such devices as
tachometers which involve the rotation of a toothed wheel past the
reed switch. If one of the teeth has magnet attached to it, then
every time it passes the switch it will momentarily close the Figure 23 Reed switch
contacts and hence produce a current/voltage pulse in the
associated electrical circuit.

Figure 24. Using photoelectric sensor to detect objects

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Photosensitive devices can be used detect the presence of an

opaque object by it breaking a beam of light, or infrared radiation, falling on
such a device or by detecting the light reflected back by the object (figure
24)
Inputting Data by Switches
Mechanical switches consist of one or more pairs of contacts which
can be mechanically closed or opened and in doing so make or break
electrical circuits. Thus 0 or 1 signals can be transmitted by the act of
opening or closing a switch.
Mechanical switches are specified in terms of their number of poles Figure 25. Switches
and throws. Poles are the number f separate circuits that can be completed
by the same switching action and throws are the number of individual
contacts for each pole. Figure 25(a) shows a single pole-single throw (SPST)
switch. Figure 25(b) a single pole-double throw (DPDT) switch and figure
25(c) double pole-double throw switch.
De-bouncing
A problem that occurs with mechanical switches is switch bounce.
When a mechanical switch is switched to close the contacts, we have one Figure 26 Switch bounce on
contact being moved towards the other. It hits the other and, because the closing a switch
contacting elements are elastic, bounces. It may bounce a number of times
(figure 26) before finally settling to its closed state after, typically, some 20
m. Each of the contacts during this bouncing time can register as a separate
contact. Thus, to a microprocessor, it might appear that perhaps two or
more separate switch actions have occurred. Similarly, when a mechanical
switch is opened, bouncing can occur. To overcome this problem either
hardware or software can be used.
With software, the microprocessor is programmed to detect if the
switch is closed and then wait, say 20 m. After checking that bouncing has
ceased and the switch is in the same closed position, the next part of the Figure 27 Debouncing a SPDT
program can take place. switch
The hardware solution to the bounce problem is based on the use
of a flip-flop. Figure 27 shows a circuit for debouncing a SPDT switch which
is based on the use of a SR flip-flop. As shown we have S at 0 and R at 1
with an output of 0. When the switch is moved to its lower position,
initially S becomes 1 and R becomes 0. This gives an output of 1. Bouncing
in changing S from 1 to 0 to 1 to 0, etc. gives no change in the output. Such
a flip-flop can be derived from two NOR or two NAND gates. A SPDT switch Figure 28. Debouncing a SPDT
can be de-bounced by the use of a D flip-flop figure 28 shows the circuit. switch
The output from such a flip-flap only changes when the clock signal
changes. Thus by choosing a clock period which is greater than the time for which the bounces last, say
20m, then the bounce signals will be ignored.
Keypads
A keypad is an array of switches, perhaps the keyboard of a computer or the touch input membrane
pad for some device such as a microwave oven. A contact type key of the form generally used with a
keyboard is shown in figure 29(a) is built up from two wafer-thin plastic films on which conductive layers
have been printed. These layers are separated by a spacer layer. When the switch area of the membrane is
pressed, the top contact layer closes with the bottom one to make the connect in and then opens when the
pressure released.

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While each switch in such arrays could be connected to

individually give signals when closed, a more economical method is to
connect them in an array in that an individual output is not needed for
each key but each key gives a unique row-column combination. Figure
29 shows the connections for a 16 way keypad.

SIGNAL CONDITIONING
The output signal from the sensor of a measurement system
has generally to be processed in some way to make it suitable for the
next stage of the operation. The signal may b, for example, too small
and have to be amplified, contain interference which has to be Figure 29 (a) contact key,
removed, be non-linear and require linearization, be analogue and have (b) membrane key
to be made digital, be digital and have to be made analogue, be a
resistance change and have t be made into a current change, be a voltage change and have to be made into
a suitable size current change, etc. All these change can be referred to as signal conditioning. For example,
the output from a thermocouple is a small voltage, a few mill volts. A signal conditioning module might then
be used to convert this into a suitable size current signal, provide noise rejection, linearization and cold
junction compensation (i.e., compensating for the cold junction not being at 0°C)
Signal-conditioning processes
The following are some of the processes that can occur in conditioning a signal:
1. Protection to prevent damage to the next element e.g., a microprocessor, as a result of high current
or voltage. Thus there can be series current-limiting resistors, fuses to break if the current is too
high, polarity protection and voltage limitation circuits.
2. Getting the signal into the right type of signal. This can mean making the signal into a dc voltage or
current. Thus, for example, the resistance change of a strain gage has to be converted into a
voltage change. This can be done by the use of a Wheatstone bridge and using the out-of-balance
voltage. It can mean making the signal digital or analogue.
3. Getting the level of the signal right. The signal from athermocouple might be just a few millivolts.
If the signal is to be fed into an analogue-to-digital converter for inputting to a microprocessor then
it needs to be made much larger, volts rather than millivolts. Operational amplifiers are widely used
for amplification.
4. Eliminating or reducing noise. For example, filters might be used to eliminate mains noise from a
signal.
5. Signal manipulation, e.g., making it a linear function of some variable. The signals from some
sensors, e.g., a flow meter, are non-linear and thus a signal conditioner might be used so that the
signal fed on to the next element is linear.
The Operational Amplifier
The basis of many signal conditioning modules is the operational
amplifier. The operational amplifier is a high gain dc amplifier, the gain
typically being of the order of 100,000 or more, than is supplied as an
integrated circuit on a silicon chip. It has two inputs, known as the
inverting input (-) and the non-inverting input (+). The output depends on
the connections made to these inputs. There are other inputs to the
operational amplifier, namely a negative voltage supply, a positive voltage Figure 30 Pin connections for a
supply and two inputs termed offset null, these being to enable 741 operational amplifier

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corrections to be made for the non-idle behavior of the amplifier. Figure 30 shows the pin connections for a
741 type operational amplifier
The following indicates the types of circuits that might be used with operational amplifiers when
used as signal conditioners. For more details the reader is referred to more specialist texts.
Inverting amplifier
Figure 31 shows the connections made to the amplifier when used as an inverting amplifier. The
input is taken to the inverting input through a resistor R1with the non-inverting input being connected to
ground. A feedback path is provided from the output via the resistor R1 to the inverting input. The
operational amplifier has a voltage gain of about 1,00,000 and the change in output voltage is typically
limited to about ± 10 V. The input voltage must then be between +0.0001 V and – 0.0001 V. This is virtually
zero and so point x is at virtually earth potential. For this reason it is called a virtual earth. The potential
difference across R1 is (Vin – Vx). Hence, for an ideal operational amplifier with an infinite gain, and hence Vx
= 0, the input potential Vin can be considered to be across R1. Thus
Vin = I1 R1
The operational amplifier has a very high impedance between its
input terminals; for a 741 about 2 MΩ. Thus virtually no current flows
through X into it. For an ideal operational amplifier the input impedance is
taken to be infinite and so there is no current flow through X. Hence the
current I1 through R1 must be the current through R2. The potential
difference across R2 is (Vx – Vout) and thus, since Vx is zerofor the ideal
amplifier, the potential difference across R2 is – Vout. Thus Figure 31 Inverting amplifier
–Vout = I1 R2
Dividing these two equations:
𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑅𝑅2
Voltage gain of circuit = = −
𝑉𝑉𝑖𝑖𝑖𝑖 𝑅𝑅1
Thus the voltage gain of the circuit is determined solely by the relative values of R2 and R1. The
negative sign indicates that the output is inverted, i.e., 180° out of phase, with respect to the input. To
illustrate the above, consider an inverting operational amplifier circuit which has a resistance of 1 MΩ. What
is voltage gain of the circuit?
𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑅𝑅2 10
Voltage gain of circuit = = − = − = – 10
𝑉𝑉𝑖𝑖𝑖𝑖 𝑅𝑅1 1

Non-inverting amplifier
Figure 32 shows the operational amplifier connected as a non
inverting amplifier. The output can be considered to be taken from across a
potential divider circuit consisting of R1 in series with R2. The voltage Vxis
then the fraction R1l(R1+ R2) of the output voltage. Figure 32 Non inverting
𝑅𝑅1
Vx= Vout
𝑅𝑅1 + 𝑅𝑅2
Since there is virtually no current through the operational amplifier
between the two inputs three can be virtually no potential different between
them. Thus, with the ideal operational amplifier, we must have Vx = Vin.
Hence
𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑅𝑅1 + 𝑅𝑅2 𝑅𝑅2
Voltage gain of circuit = = =1+
𝑉𝑉𝑖𝑖𝑖𝑖 𝑅𝑅1 𝑅𝑅1
Figure 33 Voltage follower
A particular form of this amplifier is when the feedback loop is a short circuit,

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i.e., R2 = 0. Then the voltage gain is 1. The input to the circuit is into a large resistance, the input resistance
typically being 2 MΩ. The output resistance, the resistance between the output terminal and the ground
line is, however, much smaller, e.g., 75Ω. Thus the resistance in the circuit that follows is a amplifier is
referred to as a voltage follower, figure 33 showing the basic circuit.
Summing amplifier
Figure 34 shows the circuit of a summing amplifier. As with the inverting amplifier, X is a virtual
earth. Thus the sum of the currents entering X must equal that leaving it. Hence
I = IA + IB + IC
But IA = VA/RA, IB = VB/RB, IB = VB/RB. Also we must have the same current I passing through the feedback
resistor. The potential difference across R2 is (Vx – Vout). Hence, since Vx can be assumed to be zero, it is –
Vout and so I – Vout/R2. Thus
𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 𝑉𝑉𝐴𝐴 𝑉𝑉𝐵𝐵 𝑉𝑉𝐶𝐶
= + +
𝑅𝑅2 𝑅𝑅𝐴𝐴 𝑅𝑅𝐵𝐵 𝑅𝑅𝐶𝐶
The output is thus the scaled sum of the inputs, i.e.,
𝑅𝑅2 𝑅𝑅2 𝑅𝑅
Vout= –� 𝑉𝑉 + 𝑉𝑉𝐵𝐵 + 𝑅𝑅 2 𝑉𝑉𝐶𝐶 �
𝑅𝑅𝐴𝐴 𝐴𝐴 𝑅𝑅𝐵𝐵 𝐶𝐶

If RA = RB = RC = R1 then
𝑅𝑅1
Vout = – (VA + VB + VC)
𝑅𝑅2 Figure 34 Summing amplifier
To illustrate the above, consider the design of a circuit that can be used t produce an output voltage
which is the average of the input voltages from three sensors. Assuming that an inverted output is
acceptable, a circuit of the form shown in figure 34 can be used. Each of the three inputs must be scaled to
1/3 to give an output of the average. Thus a voltage gain of the circuit of 1/3 for each of the input signals is
required. Hence if the feedback resistance is 4 kΩ the resistors in each input arm will be 12 kΩ.
Integrating amplifier
Consider an inverting operational amplifier circuit with the feedback being via a capacitor, as
illustrated in figure 35. Current is the rate of movement of change q and since for a capacitor the charge
q = Cv, where v is the voltage across it, then the current through the capacitor I = dq/dt = C dv/dt. The
potential difference across C is (Vx – Vout) and since Vx is effectively zero, being the virtual earth, it is – Vout.
Thus the current through the capacitor is – CdVout/dt. But this is also the current through the input
resistance R. Hence
𝑉𝑉𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑𝑜𝑜𝑜𝑜𝑜𝑜
=–C
𝑅𝑅 𝑑𝑑𝑑𝑑
Rearranging this gives
1
dVout= –� � Vindt
𝑅𝑅𝑅𝑅
Integrating both sides give
1 𝑡𝑡 2
Vout (t2) = Vout(t1) = – ∫ 𝑉𝑉 𝑑𝑑𝑑𝑑 Figure 35 Integrating amplifier
𝑅𝑅𝑅𝑅 𝑡𝑡 1 𝑖𝑖𝑖𝑖
Vout (t2) is the output voltage at time t2 and Vout(t1) is the output voltage at time t1. The output is
proportional to the integral of the input voltage, i.e., the area under a graph of input voltage with time.
A differentiation circuit can be produced if the capacitor and resistor are interchanged in the circuit
for the integrating amplifier.

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Differential amplifier
A differential amplifier is one that amplifies the difference between two input voltages. Figure 36
shows the circuit. Since there is virtually no current through the high resistance in the operational amplifier
between the two input terminals, there is no potential drop and thus both the inputs X will be at the same
potential. The voltage V2 is across resistors R1 and R2 in series. Thus the potential Vx at X is
𝑉𝑉𝑋𝑋 𝑅𝑅2
=
𝑉𝑉2 𝑅𝑅1 + 𝑅𝑅2
The current through the feedback resistance must be equal to that from V1 through R1. Hence
𝑉𝑉1 −𝑉𝑉𝐴𝐴 𝑉𝑉𝑋𝑋 −𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜
=
𝑅𝑅1 𝑅𝑅2
This can be rearranged to give
𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 1 1 𝑉𝑉1
=Vx� + �-
𝑅𝑅2 𝑅𝑅2 𝑅𝑅1 𝑅𝑅2
Hence substituting for Vx using the earlier equation
𝑅𝑅2
Vout = (V2 – V1)
𝑅𝑅1
The output is thus a measure of the difference between the two Figure 36 Differential amplifier
input voltages.
As an illustration of the use of such a circuit with a sensor, the difference in voltage between the
e.m.f.s of the two junctions of the thermocouple is being amplified. The values of R1 and R2 can, for
example, be chosen to give a circuit with an output of 10 mV for a temperature difference between the
thermocouple junctions of 10°C if such a temperature difference produces an e.m.f. difference between the
junctions of 530 µV. For circuit we have
𝑅𝑅2
Vout = (V2 – V1)
𝑅𝑅1

Logarithmic amplifier
Some sensors have outputs which are non-linear. For example the output from a thermocouple is
not a perfectly linear function of the temperature difference between its junctions. A signal conditioner
might then be used to linearise the output from such a sensor. This can be done using an operational
amplifier circuit which is designed to have non-linear relationship between its input and output so that when
its input is non-linear the output is linear. This is achieved by a suitable choice of component for the
feedback loop.
The logarithmic amplifier shown in figure 3.37 is an example of such a signal conditioner. The
feedback loop contains a diode (or a transistor with a grounded base). The diode has non-linear
characteristic. It might be represented by V = C in I, where C is a constant. Then, since the current through
the feedback loop is the same as the current through the input resistance and the potential difference across
the diode is – Vout, we have
Vout= – C in (Vin/R) = K In Vin
Were K is some constant. However, if the input Vin is provided by a
sensor with an input t, where Vin = eat, with A and a being constants, then
Vout = K In Vin = K In (A eat) = K In A + Kat
The result is linear relationship between Voutand t.
Digital Signals Figure 37 Logarithmic
The output from most sensors tends to be in analogue from. Where amplifier
a microprocessor is used as part of the measurement or control system the

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analogue output from the sensor has to be converted into a digital from before it can be used as an input to
the microprocessor. Likewise, most actuators operate with analogue inputs and so the digital output from a
microprocessor has to be converted into an analogue form before it can be used as an input by the actuator.
The binary system is based on just the two symbols or states 0 and 1. These are termed binary digits
or bits. When number is represented by this system, the digit position in the number indicates the weight
attached to each digit, the weight increasing by a factor of 2 as we proceed from right to left:
---- 23 22 21 20
bit 3 bit 2 bit 1 bit 0
for example, the decimal number 15 is 20 + 21 + 22 + 23 = 1111 in the binary system. In binary number the
bit 0 is termed the least significant bit (LSB) and the highest bit the most significant bit (MSB). The
combination of bits to represent a number is termed a word. Thus 1111 is a four – bit word. The term byte is
used for a group of 8 bits.
Analogue-to-digital conversion
Analogue-to-digital conversion involves converting analogue signals into binary words. The basic
elements of analogue-to-digital conversion.

Input, Sample and Analogue-to- Output

Analogue hold digital converter
Digital
signal signal

The procedure used is that a clock supplies regular time signal pulses
to the analogue-to-digital converter (ADC) and every time it receives a pulse
it sample the analogue signal. Figure 38 illustrates this analogue-to-digital
conversion by showing the types of signals involved at the various stages.
Figure 38a shows the analogue signal and figure 38b the clock signal which
supplies the time signals at which the sampling occurs. The result of the
sampling is a series of narrow pulses (figure 3.38d). A simple and hold unit is
necessary becomes the analogue-to-digital converter requires a finite
amount of time, termed the conversation time, to convert the analogue
signal into a digital one.
The relationship between the sampled and held input and the output
for an analogue-to-digital converter is illustrated by the graph shown in figure
39 for a digital output which is restricted to three bits. With three bits there
are 23 = 8 possible output levels. Thus, since the output of the ADC to
represent the analogue input can be only one of these eight possible levels is
termed the quantization interval. Thus for the ADC given in figure 39, the
quantization interval is 1 V. Because of the step-like nature of the
relationship, the digital output is not always proportional to the analogue
input and thus there will be error, this being termed the quantization error.
When the input is center over the interval the quantization error is zero, the Figure 38 signals
maximum error being equal to one-half of the interval or 4 ½ bit.
The word length possible determines the resolution of the element, i.e., the smallest change in input
which will result in a change in the digital output. The smallest change in digital output is one bit in the least
significant bit position in the word, i.e., the far right bit. Thus with a word length of n bits the full-scale
analogue input VFS is divided into 28 pieces and so the minimum change in input that can be detected, i.e.,
the resolution, is VFS/28.

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Digital-to-analogue conversion
The input to a digital-to-analogue converter (DAC) is a binary word; the output is an analogue signal
that represents the weighted sum of the non-zero bits represented by the word. Thus, for example, an input
of 0010 must give an analogue output which is twice that given by an input of 0001.
Consider the situation where a microprocessor gives an output of an 8-bit word. This is fed through
an 8 bit digital-to-analogue converter to a control valve. The control valve requires 6.0 V to be fully open. If
the fully open state is indicated by 11111111 what will be the output to the valve for a change of 1 bit?
The full-scale output voltage of 6.0 V will be divided into 28 intervals. A change of 1 bit is thus a
change in the output voltage of 6.0/28 = 0.023 V

Figure 39 Input-output for a DAC Figure 40 Input-output for an ADC

Figure 41 illustrates this for an input to a DAC with a resolution of 1 V for unsigned binary words.
Each additional bit increase the output voltage by 1 V.
Digital-to analogue converters
A simple form of a digital-to-analogue converter uses a summing amplifier to from the weighted sum
of the all the non-zero bits in the input word (figure 41). The reference voltage is connected to the resistors
by means of electronic switches which respond to binary 1. The values of the input resistances depend on
which bit in the word a switch is responding to, the valve of the resistor for successive bit from the LSB being
halved. Hence the sum of the voltages is a weighted sum of the digits in the word. Such a system is referred
to as a weighted-resistor network.

A problem with the weighted-resistor network

is that accurate resistances have to be used for each of
the resistors and it is difficult to obtain such resistors
over the wide range needed. As a result this form of
DAC tends be limited to 4 bit conversions.
Another, more commonly used, version uses a
R-2R ladder network (figure 42). This overcomes the
problem of obtaining accurate resistances over a wide
range of valves, only two valves being required. The
output voltage is generated by switching sections of the Figure 41 Weighted-resistor DAC
ladder to either the reference voltage or 0 V according
to whether there is a 1 or 0 in the digital input.

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Figure 42 R-2R Ladder DAC

Figure 43 shows details of the GEC Plessey ZN55D 8 bit latched input digital-to-analogue converter
using a R-2R ladder network. After the conversion is complete, the 8 bit result is placed in an internal latch
until the next conversion is complete. Date is held is the latch when ENABLE is high, the latch being side to
be transparent when ENABLE is low. A latch is just a device to retain the output until a new one replaces it.
When a DAC without a latch would be connected via a peripheral interface adapter (PIA), shows how the
ZN558D might be used with a microprocessor when the output is required to be a voltage which varies
between zero and the reference voltage, this being termed unipolar operation. With Brefin = 2.5 V, the
output range is +5 V when R1 = 8 Ω and R2 = 8 kΩ and the range is +10V when R1 = 16 kΩ and R2 = 5.33 kΩ.

Figure 43 ZN558D ADC

Analogue-To-Digital Converters
The input to an analogue-to-digital converter is an analogue signal and the output is binary word
that represents the level of the input signal. There are a number of analogue-to-digital converter, the most
common being successive approximations, ramp, dual ramp and flash.
Successive approximations ADC
Successive approximations are probably the most commonly used method figure 44 illustrates the
subsystems involved. A voltage is generated by a clock emitting a regular sequence of pluses which are
counted, in a binary manner, and the resulting binary word converted into an analogue voltage by a digital-
to-analogue converter. This voltage rises in steps and is compared with the analogue input voltage from the
sensor. When the clock generated voltage passes the input analogue voltage the pulses from the clock are
stopped from being counted by a gate being closed.

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Figure 44 Successive approximations ADC

The output from the counter at that time is then a digital representation of the voltage. While the
comparison could be accomplished by starting the count at 1, the less significant bit that is less than the
analogue value, these adding successive lesser bits for which the total does not exceed the analogue value.
For example, we might start the comparison with 1000. If this is too large we try 0100. If this is too small we
then try 0110. If this is too large we try 0101. With an n-bit word it only takes n steps to make the
comparison. Thus if the clock has a frequency f, the time between pulses is 1/f. Hence the time take to
generate the word, i.e., the conversion time, is n/f.
Ramp ADC
The ramp form of analogue-to-digital
converter involves as analogue voltage which is
increased at a constant rate, a so-called ramp
voltage and applied to a comparator where it is
compared with the analogue voltage from the
sensor. The time taken for the ramp voltage to
increase to the value of the sensor voltage will
depend on the size of the sampled analogue
voltage. When the ramp voltage starts, a gate is Figure 45 Ramp ADC
opened which starts a binary counter counting
the regular pulses from a clock. When the two voltages are equal, the gate closes and the word indicated by
the counter is the digital representation of the sampled analogue voltage. Figure 45 indicates the
subsystems involved in the ramp from of analogue-to-digital converter.
Dual ramp ADC

Figure 46 Dual ramp ADC

The dual ramp converter is more common than the single ramp. Figure 46 shows the basic circuit.
The analogue voltage is applied to an integrator which drives a comparator. The output from the
comparator goes high as soon as the integrator output is more than a few millivolts. When the comparator
output is high, an AND gate passes pulses to a binary counter. The counter counts pulses until it overflows.
The counter then resets to zero, sends a signal to a switch which disconnects the unknown voltage and
connects a reference voltage, and starts counting again. The polarity of the reference voltage is opposite to

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that of the input voltage. The integrator voltage then

decrease at a rate proportional to the reference voltage.
When the integrator output low and so switching the
clock off. The count is then a measure of the analogue
input voltage. Duel ramp analogue-to-digital action
averages out random negative and positive
contributions over the sampling period. They are
however, very slow.
Flash ADC
The flash analogue-to-digital converter is very Figure 47 Flash ADC
fast. For an n-bit converter, 28 – 1 separate voltage
comparators are used in parallel, with each having the
analogue input voltages as one input. A reference
voltage is applied to a ladder of resistors so that the
voltage applied as the other input to each comparator is
one bit larger in size then the voltage applied to the
previous comparator in the ladder. Thus when the
analogue voltage is applied to the ADC, all those
comparators for which the analogue voltage is greater
than the reference voltage of a comparator will give a Figure 48 Multiplexer
high output and those for which it is less will be low. The resulting outputs are fed in parallel to a logic gate
system which translates them into a digital word.
Multiplexers
A multiplexer is a circuit that is able to have inputs of data from a number of sources and then, by
selecting an input channel, give an output from just one of them. In applications where there is a need for
measurements to be made at a number of different locations, rather than use a separate ADC and
microprocessor for each measurement, a multiplexer can be used to select each input in turn and switch it
through a signal ADC and microprocessor. The multiplexer is essentially an electronic switching device which
enables each of the inputs to be sampled in turn.

Figure 49 Two channel multiplexer

As an illustration of the types of analogue multiplexers available, the DG508ACI has eight input
channels with each channel having a 3-bit binary address for selection purposes. The transition time
between taking samples is 0.6 µs.
Figure 44 shows the basic principle of a multiplexer which can be used to select digital data inputs;
for simplicity only a two input channel system is shown. The logic level applied to the select-input
determines which AND gate is enabled so that its data input passes through the OR gate to the output. A
number of forms of multiplexers are available in integrated packages. The 151 types enable on line from
eight to be selected, the 153 type one line from four inputs which are supplied as data on two lines each, the
157 types one line from two inputs which are supplied as data on four lines.

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Unit IV
ELECTRICAL ACTUATION SYSTEM

Basic Principles
Figure 1 shows the basic principle of the dc motor, a loop of
wire which is free to rotate in the field of a permanent magnet. When
a current is passed through the coil, the resulting forces acting on its
sides at right angle to the field cause forces to act on those sides to
give rotation. However, for the rotation to continue, when the coil
passes through the vertical position the current direction through the
Figure 1 DC motor basics
coil has to be reversed.
In the conventional dc motor coils of wire are mounted in steps
on a cylinder of magnate material called to armature. The armature is
mounted on bearings and is free to rotate. It is mounted in the
magnetic field produced by field poles. These may be for small motors,
Permanent magnets or electromagnets with their magnetism produced
by a current through the field cols. Figure 2 show the basic principle of
a four-pole dc motor with the magnetic field produced by current
carrying coils. The ends of each armature coil are connected to
adjacent segments of a segmented ring called the commutator with
Figure 4.2 DC motor
electrical contacts made to the segments through carbon contacts
called brushes. As the armature rotates, the commutator reverses the current in each coil as it move
between the field poles. This is necessary if the force acting on the coil is to remain acting in the same
direction and so the rotations continue. The direction of rotation of the DC motor can be reversed by
reversing either the armature current or the field current.
Permanent Magnet Dc Motor
Consider a permanent magnet dc motor the permanent magnet giving a
constant value of flux density. For an armature conductor of length L and
carrying a current i the force resulting from a magnetic flux density B at right
angles to the conductor is BiL. (Figure 3) With N such conductors the force is
NBiL. The forces result in torque T and the coil axis of Fb, with b being the
breadth of the coil. Thus:
Torque T = NbbLi = ktj Figure 3 Armature
Where,kt is the torque constant. Since an armature coil is rotating in a
magnetic field, electromagnetic induction will occur and a back e.m.f will be
induced. The back e.m.f. vb is proportional to the rate at which the flux linked
by the coil changes and hence, for a constant magnetic field, is proportional to
the angular velocity ω of the rotation. Thus:
Back e.m.f. vb = kvω
Where kv is the back e.m.f. constant Figure 4 Equivalent circuit
We can consider a dc motor to have the equivalent circuit shown in figure 4. i.e. the armature coil
being represented by a resistor R is series with an inductance L in series with a source of back e.m.g. If we
neglect the inductance of the armature coil then the voltage providing the current i through the resistance
is the applied voltage V minus the back e.m.f. i.e., V – vb. Hence:

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𝑉𝑉−𝑣𝑣𝑏𝑏 𝑉𝑉−𝑘𝑘 𝑣𝑣 (𝑡𝑡)

i = =
𝑅𝑅 𝑅𝑅
The torque T is thus:
𝑘𝑘 𝑡𝑡
T = kgi = [V – kvω]
𝑅𝑅
Graphs of the torque against the rotational speed ω are a series of
straight lines for different voltage values (figure 5). The starting torque, i.e. Figure 5 Torque-speed
the torque when ω = 0, is thus proportional to the applied voltage, the non- characteristic
load speed is proportional to the applied voltage and the torque decreases with increasing speed.
As an example, a small permanent magnet motor S6M41 by PMI motors has kt = 3.01 N cm/A.
kv = 3.15 V/krpm, a terminal resistance of 1.207Ω and an armature resistance of 0.940 Ω.
DC motors with field coils
DC motors with field coils are classified as series, shunt compound and separately excited according to
how the filed windings and armature windings are connected (figure 7.26)
1. Series wound motor
With the series wound motor the armature and fields
coils are in series. Such a motor exerts the highest
starting torque and has the greater no-load speed. With
light loads there is a danger that a series would motor
might run at too high a speed. Reversing the polanty of
the supply to the coils has no effect on the direction of
rotation of the motor, it will continue rotating in the same
direction since both the field and armature currents have
been reversed.
2. Shunt wound motor
With the shunt would motor the armature and field
coils are in parallel. It provides the lowest starting torque
a much lower no-load speed and has good speed
regulation. Because of this almost constant speed
regardless of load, shunt would motors are very widely Figure 6 DC motors wound
used. To reverse the direction of rotation, either the
armature or field supplied must be reversed. For this reason, the separately excited windings are preferable
for such a situation.
3. Compound motor
The compound motor has two field windings, one in series with the armature
and one in parallel. Compound wound motors aim to get the best features of the
series and shunt would motors, namely a high starting torque and good speed
regulation.
4. Separately excited motor
The separately excited motor has separate control of the armature and field
currents and can be considered to be a special case of the shunt wound motor.
Figure 7 indicates the torque-speed characteristics of the above motors. The
speed of such dc motor can be changed by either changing the armature current Figure 7 Torque-speed
or the field current. Generally it is the armature current that is varied. The characteristics
choice of motor will depend on its application. For example, with a robot

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manipulator, the robot wrist might use a series would motor because the
speed decreases as the load increases. A shunt would motor would be used
where a constant speed was required, regardless of the load.
Brushless permanent magnet dc motors
A problem with dc motors is the they require a commutator and
brushes in order to periodically reverse the current through each armature
coil. The brushes make sliding contacts with the commutator and as a
consequence sparks jump between to two and they suffer wear. Brushes
thus have to be periodically changed and the commutator resurfaced. To
avoid such problems brushless motors have been designed. Figure 8 Commutator

Essentially they consist of a sequence of stator coils and permanent

magnet rotor. A current carrying conductor in a magnetic field
experiences a force likewise, as a consequence of Newton’s third law of
motion, the magnet will also experience an opposite and equal force. With
the conventional dc motor the magnet is fixed and the current-carrying
conductors mad to move. With the brushless permanent magnet dc motor
the reverse is the case, the current carrying conductors are fixed and the
magnet moves. The rotor is a ferrite or ceramic permanent magnet figure
9 shows the basic form of such a motor. The current to the stator oils is
electronically switched by transistors in sequence round the coils, the
Figure 9 Brushless permanent
switching being controlled by the position of the rotor so that there are
magnet dc motor
always forces acting on the magnet causing it to rotate in the same
direction. Hall sensor are generally used to sense to position of the rotor
and initiate the switching by the transistors, the sensors being positioned
around the stator.
Figure 8 shows the transistor switching circuits that might be used
with the motor shown in figure 7. To switch the coils in sequence we need
to supply signals to switch the transistors on in the right sequence. This is
provided by the outputs from the three sensors operating though a
decoder circuit to give the appropriate base currents. Thus when the rotor
is in the vertical position, i.e., 0°, there is an output from sensor c but non
from ‘a’ and ‘b’. This is used to switch on transistors A+ and B- for the
rotor in the 60° position there are signals from the sensors b and c and
transistors A+ and C- are switched on. The entire circuit for controlling
such a motor is available on the single integrated circuit.
AC motors
Alternating current motors can be classified in the groups single
phase and polyphase with each group being further subdivided into
induction and synchronous motors. Single-phase motor lend to be used
for low power requirements while polyphase motors are used for higher Figure 10 Single-phase
powers. Induction motors tend to be cheaper than synchronous motors induction motor
and are thus very widely used.
The single-phase squirrel-cage condition motor consists of a squirrel-cage motor, this being copper
or aluminium bars that fit into slots in end rings to form complete electrical circuits (figure 10). There are no
external electrical connections to the rotor. The basic motor consists of this rotor with a stator having a set
of windings. When an alternating current passes through the stator windings and alternating magnetic field
is proceed. As a result of electromagnetic inductione.m.f.s are induced in the conductors of the rotor and
currents flow in the rotor. Initially when the rotor is rotor is stationary, the forces on the current carrying

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conduction of the rotor in the magnetic field of the stator are such as to result
is in not torque. The motor is not self-starting. A number of methods are
used to make the motor self-starting and give this initial impetus to start it,
one is to use an auxiliary staring winding to give the rotor an initial push. The
rotor rotates at a speed determined by the frequency of the alternating
current applied to the stator. For a constant frequency supply to a two-pole
single-phase motor the magnetic field will alternate at this frequency. This
speed of rotation of the magnetic field is termed the synchronous sped. The
rotor will never quit match this frequency of rotation, typically differing from Figure 11 Three-phase
it by about 1 to 3%. This difference is termed slip. Thus for a 50 Hz supply the induction motor
speed of rotation of the rotor will be almost 50 revolutions per second.
The three-phase induction motor (figure 12) is similar to the single-phase
induction motor but has a stator with three windings located 120° apart, each
winding being connected to one of the three lines of the supply. Because the
three phases reach their maximum currents at different times, the magnetic
field can be considered to rotate round the stator poles, completing one
rotation in one fuel cycle of the current. The rotation of the field is much
smoother than with the signal-phase motor. The three-phase motor has a
great advantage over the single-phase motor of being self-starting. The
direction of rotation is reversed by interchanging any two of the line Figure 12 Three-phase
connections, this changing the direction of rotation of the magnetic field. synchronous motor

Synchronous motors have stators similar to those described

above for induction motors to a rotor which is a permanent magnet
(figure 12). The magnetic field produced by the stator rotates and so
the magnet rotors with it. With one pair of poles per phase of the
supply, the magnetic field rotates through 360° in one cycle of the
supply and so the frequency of rotation with this arrangement is the
same as the frequency of the supply. Synchronous motors are used
when a precise speed is required. They are not self-starting and some
system has to be employed to start them.
AC motors have the great advantage over dc motors of being
cheaper, more rugged, reliable and maintenance free. However sped
control is generally more compels than with dc motors and as a
consequence a speed-controlled dc drive generally works out cheaper
than a speed controlled ad drive through the price difference is
steadily dropping as a result of technological developments and the
reduction in price of solid-state devices. Speed control of ac motors is
based around the precision of a variable frequency supply, since the
speed of such motors is determined by the frequency is constant.
Thus to maintain a constant torque at the different speeds when the
frequency is varied the voltage applied to the stator less also to be
varied, with one method, the ac is first rectified to dc by a converter
and then inverted back to ac again but at a frequency that can be
selected (figure 13) another method that is often used for operating
Figure 13 Variable reluctance
slow-speed motors is the cycloconverter. The converts ac at one stepper motor
frequency directly to ac at another frequency with the intermediate
dc conversion.

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Stepper motors
The stepper motor is a device that produces rotation through equal angles, the so-called steps for
each digital pulse supplied to its input. Thus, for example, if with such a motor 1 pulse produces a rotation
of 6° then 60 pulses will produce a rotation through 360°. There are a number of forms of stepper motor.
Variable reluctance stepper
Figure 13 shows the basic form of the variable reluctance stepper motor. With this form the rotor is
made of soft steel and is cylindrical with four poles, i.e. fewer poles than on the stator. When an opposite
pair of windings has current switched to them a magnetic field is produced with line of force which pass
from the stator poles though the nearest set of poles on the rotor. Since lines of force can be considered to
be rather like elastic thread and always trying to shorten themselves, the rotor will move until the rotor and
stator poles line up. This is termed the position of minimum reluctance. This form of stepper generally gives
step angles of 7.5° or 15°.
Permanent magnet stepper
Figure 14 shows the basic form of the permanent
magnet motor. The motor shown has a stator with four poles.
Each pole is wound with a field winding, the coils on opposite
pairs of poles being in series. Current is supplied from a dc
source to the windings through switches. The rotor is a
permanent magnet and thus when a pair of stator poles has a
current switched to it, the rotor will move to line up with it.
Thus for the currents giving the siltation shown in the figure
the rotor moves to the 45° position. If the current is then
switched so that the polarities are reversed, the rotor will
move a further 45° in order to line up again. Thus by switching
the currents through the coils the rotor rotates in 45° steps. Figure 14 Permanent magnate
With this of motor, step angles are commonly 1.8°, 7.5°, 15°, stepper motor
30°, 34° or 90°.
Hybrid stepper
Hybrid stepper motors combine the features of both the variable
reluctance and permanent magnet motors, having a permanent magnet
encased in iron caps which are cut to have teeth (figure 15). The rotor sets
itself in the minimum reluctance position in response to a pair of stator coils
being energized. Typical step angles are 0.9° and 1.8°. Such stepper motors Figure 15 Hybrid motor
are extensively used in high accuracy positioning applications, e.g in computer rotor
hard disc drive.
STEPPER MOTOR SPECIFICATIONS
The following are some of the terms commonly used in specifying stepper motors
1. Phase
This term refers to the number of independent windings on the stator, e.g. a four-phase motor. The
current required per phase and its resistance and inductance will be specified so that the controller
switching output is specified Two-phase motors, e.g., figure 16 tend to be used in light-duty applications,
three-phase motors tend to be variable reluctance steppers e.g., figure 7.40, and four-phase motors tend to
be used for higher power applications
2. Step angle
This is the angle through which the rotor rotates for one switching change for the stator coils.

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3. Holding torque
This is the maximum torque that can be
applied to powered motor without moving it from
its rest position and causing spindle rotation.
4. Pull-in-torque
This the maximum torque against which a motor
will start for a given pulse rate and reach
synchronism without losing a step.
5. Pull-out torque
This the maximum torque that can be applied to
a motor, running at a given stepping rate, without
Figure 16 Stepper motor characteristics
losing synchronism
6. Pull-in-rate
This is the maximum switching rate at which a loaded motor can start without losing a step.
7. Pull-out rate
This is the switching rate at when a loaded motor will remain in synchronism as the switching rate is
reduced.
8. Slew range
This is the range of switching rates between pull-in and pull-out within which the motor runs in
synchronism but cannot start up or reverse.
Pneumatic and Hydraulic Systems
Most of mechatronics systems work based oil motion or action by means of sort. This motion or
actuation is caused either by a torque or force from which displacement and acceleration can be obtained.
To obtain this force or acceleration, abators are mainly used. Actuator is a device which provides enough
force needed start the mechatronics systems. At the same, power should be supplied to the actuator to
activate it. The power supplied to actuators might be anyone of the flowing forms such as compressed air,
pressurized fluid, electric power and mechanical power. If compressed air is supplied to the system, it is
called as pneumatic system. But if pressurized fluid is supplied to flow the system, it is called is hydraulic
system. Electrical actuator system is with electrical power and mechanical system is with mechanical power.
Among these, hydraulic and pneumatic systems are quiet effective and efficient way of getting motion and
action which can be used in mechatronics systems.
1. Introduction to Pneumatic Systems
Fluid power technology over the years has continuous development involved the applications of
pneumatic and hydraulic systems in several areas, like
(i) Manufacturing,
(ii) Process industries,
(iii) Transportation systems, and
(iv) Utilities.
The fluid power systems are also used:
(i) to carry out mechanical works such as linear, swivel and rotary motion for plant equipment and
machinery, clamping, shifting and positioning, packaging, feeding sorting, stamping, drilling,
turning, milling and sawing etc.
(ii) to obtain control application such as controlling of plant, process and equipment to take necessary

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corrective action,
(iii) to measure process parameter to act on necessary output.
Advantages of fluid systems:
(i) Air is available every where in enormous quantities.
(ii) Transporting air and hydraulic fluid will be easy through pipe line over large distances,
(iii) Storing of compressed air will be easy in a reservoir and removed as required. Hydraulic oil can be
stored in accumulators,
(iv) Compressed air is too sensitive with temperature fluctuations but hydraulic fluids are insensitive,
(v) Compressed air offers minimal risk of explosion or fire,
(vi) The construction of components in fluid system is simple in construction and cheap.
Disadvantages:
(i) Good preparation of compressed air and hydraulic fluid required to remove the dirt and
condensate,
(ii) Speed fluctuation will always be with pneumatic systems,
(iii) The working pressure of compressed air is limited to 6-7 bar.
(iv) The exhaust air will release with very high noise thereby leading noise pollution,
(v) Producing compressed air and hydraulic fluid are expensive.
2. Pneumatic systems
In pneumatic systems, force is produced by gas. It is mainly by air pressure acting on the surface of a
piston or valve.
Compressed air is produced in a compressor and stored in a receiver. From compressor, it is send to
valves which control the direction of fluid flow. Also, flow control valves control the amount of power
produced by the cylinders. The force acting on the piston is given by the equation:
Force = Pressure x Area = p x A
3. Basic Elements of Pneumatic System
The basic components of a fluid power system are essentially the same, regardless of whether the system
uses a hydraulic or a pneumatic medium. A pneumatic system essentially has the following components as
shown in Figure 17
(i) Compressor and Motor
(ii) Pressure relief valve and Check valve
(iii) Cooler, filter and water trap
(iv) Air receiver
(v) Directional control valves
(vi) Actuator or pneumatic cylinder

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Figure 17 Basic elements of a pneumatic system

The fresh atmosphere air is not sent directly to the compressor to use in pneumatic systems. First, it
is filtered by filters. Then, filtered atmospheric air is supplied to the compressor through silencer to reduce
noise level. Then it is compressed.
Pressure relief valve is used to avoid the damage of compressor due to excess pressure raise in the
system. Check valve is a one-way valve that allows pressurized air to enter the pneumatic system, but
prevents backflow and loss of pressure into the compressor when it is stopped.
The cooler is used to cool the compressed air which is usually very hot. The filter is used to remove
contamination in the compressed air and water trap is used to remove water particles.
The pressurized air is stored in a device called an air receiver, preventing surges in pressure and
relieving the duty cycle of the compressor.
Directional control valves are used to control flow of pressurized air from the source to the selected
port. These valves can be actuated either manually or electrically.
Actuator or pneumatic cylinder converts energy stored in the compressed air into mechanical
motion.
4. Hydraulic Systems
A hydraulic system uses force which is applied at one point and transmitted to-another point using
an incompressible fluid, plant equipment and machinery. In this type of machine, high-pressure liquid called
hydraulic fluid is transmitted throughout the machine to various hydraulic motors and hydraulic cylinders.
The fluid is controlled directly or automatically by control valves and distributed through hoses and tubes.
The popularity of hydraulic machinery is due to large amount of power that can be transferred through small
tubes and flexible hoses, and the high power density and wide array of actuators that can make use of this
power.
5. Basic Elements of Hydraulic System
The necessary components of any hydraulic systems are
(i) Hydraulic pump unit
(ii) Control valves
(iii) Reciprocating or rotary unit
 Hydraulic pump unit
A pump is a device in which mechanical energy is converted into fluid energy. The pump is
connected with the reservoir called fluid tank.
 Control valves
The flow of pressurized fluid by a pump is controlled by the following valves such as:

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(i) Pressure relief valves control the fluid pressure.

(ii) Non return valve controls the back flow of fluid.
(iii) Directional control valves control the direction of fluid.
 Hydraulic actuator or cylinder
The actuator is the element which converts hydraulic power into mechanical power. The pressurized
fluid by the pump is supplied to either rotary pump or hydraulic cylinder based on the type of motion
needed. Rotary pumps are used to get rotary motion and hydraulic cylinder is used to obtain linear motion.

Figure 18 Basic elements of a hydraulic system

Working of hydraulic system (Figure 18)
The fluid stored in the tank is send to the filter to remove dust and foreign particles. After the fluid is
filtered, it is sucked by the pump which is driven by a motor. During pumping, the p. assure of the fluid will
increase and it is released with high pressure to the accumulator through non return valve. One pressure
relief valve is connected at the exit to control the delivery pressure of fluid. The fluid with high pressure is
supplied to the hydraulic cylinder through directional control valve.
6. Hydraulic Accumulator
A hydraulic accumulator is an energy storage device. It is a pressure storage reservoir in which a fluid
is held under pressure by compressed gas or a spring or a raised weight. The main reasons that an
accumulator is used in a hydraulic system are:
(i) the pump does not need to be so large to cope with extremes of demand,
(ii) the supply circuit can respond more quickly to any temporary demand and to smooth pulsations.
Compressed gas accumulators are the most common type
which is shown in Figure 19. A compressed gas accumulator
consists of a cylinder with two chambers that are separated by a
totally enclosed bladder. One chamber contains hydraulic fluid
and is connected to the hydraulic line. The other chamber
contains an inert gas (mostly nitrogen) under pressure that
provides the compressive force on the hydraulic fluid. As the
volume of the compressed gas changes the pressure of the gas,
and the pressure on the fluid, changes inversely.
7. Hydraulic Pumps
In general, a pump is a device which converts the mechanical
energy supplied into hydraulic energy by lifting water to higher
levels. Here, hydraulic energy refers ID potential and kinetic Figure 19 Hydraulic accumulator
energy of a liquid. Hydraulic pumps are the energy-absorbing

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machines. Since, it requires mechanical power to drive. Lifting of water

to higher levels is carried out by the various actions of pumps such as
centrifugal action, reciprocating action etc., The symbol of a pump is
shown in Figure 20.
Power required by a pump
The motor power required to derive a pump is given by Figure 20 Symbol of a pump
𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 × 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷
Power, P = =
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 × 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 × 𝑙𝑙
Power = [ ∴force = Pressure × area]
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑙𝑙
Q = A × V = Area ×
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
Power, P = p × Q
8. Advantages and Disadvantages of Hydraulic Systems
Advantages of hydraulic systems
1. It is easy to produce and transmit hydraulic power.
2. Hydraulic systems are uniform and smooth.
3. Balancing hydraulic forces is easier.
4. Weight-to-power ratio is less.
5. It is easy to maintain.
6. Systems are cheaper.
7. Hydraulic systems are safe and compact.
8. Frictional resistance is less.
Disadvantages of hydraulic systems
1. Manufacturing cost of the system is quiet high.
2. Hydraulic elements should be kept free from dirt, corrosion, rust etc.,
3. Petroleum based hydraulic systems more prone to fire hazards.
4. Hydraulic power is not readily available as pneumatic
Microcomputer Structure
Computers have three sections: a control processing unit (CPU) to recognize and carry out program
instructions, input and output circuitry interfaces to handle communications between the computer and the
outside world, and memory to hold the program instructions and data. Digital signal move from one section
to another along paths called buses. A bus, in the physical sense, is just a number of conductors along which
electrical signals can be carried. It might be tracks on a printed circuit board or wires in a ribbon cable. The
data associated with the processing function of the CPU is carried by the data bus, the information for the
address of a specify memory location for the accessing of stored data is carried by the address bus and the
signal relating to control actions are carried by the control bus. Figure 21 illustrates the general
arrangement. In some cause a microprocessor chip constitutes just the CPU while in other cause it might
have all the components necessary for the complete computer on one chip. Microprocessor which have
memory and various input/output arrangements all on the same chip are called microcontrollers. They are
effectively a microcomputer on a single chip.

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Figure 21 General form of a computer

Buses
The data bus is used to transport a word to or from the CPU and the memory or the input/output
interfaces. Word length used may be 4, 8, 16, 32 or 64. Each wire in the bus carries a binary signal, i.e., a 0
or a 1. Thus with a for-wire bus we might have the word 1010 being carried, each bit being carried by a
separate wire in the bus as
Word Bus wire
0 (least significant bit) Firstdata bus wire
1 Second data bus wire
0 Third data bus wire
1(most significant bit) Fourth data bus wire
The more wires the data bus has the longer the word length that can be used. The range of values
which a single item of data can have is restricted to that which can be represented by the word length. Thus
with a word of length 4 bits the number of values is 24 = 16. Thus if the data is to represent, say, a
temperature, then the range of possible temperatures must be divided into 16 segments if we are to
represent that range by a 4-bit word. The earliest microprocessors are still widely used in such devices as
toys, washing machines and domestic central heating controllers. They were followed by 8-bit
microprocessor, e.g, the Motors 6800, the Intel 8085A and the Zilog Z80. Now, 16-bit, 32-bit and 64-bit
microprocessor are available, however, 8-bit microprocessor are still widely used for controllers.
The address bus carriers signals which indicate where data is to be found and so the selectionof
certain memory location or input or output ports. When a particular address is selectedby its address being
placed on the address bus, only that location is open to the communicationfor the CPU. The CPU is thus able
to communicate with just one location at a time. A computer with an 8-bit data bus has typically a 1-bit
wide address bus, i.e 16 wirers. This size of address bus enable 216 location to be addressed 216 is 65 536
location sand is usually written as 64K, where K is equal to 1024. The more memory that can be addressed
the greater the volume of data that can be stored and the larger and more sophisticated the programs that
can be used.
The control bus is the means by which signals are sent to synchronies the separate elements. The
system clock signals are carried by the control bus. These signals generate time intervals during which
system operations can take place. The CUP sends some control signals to other elements to indicate the type
of operation being performed e.g. whether it needs to READ (receive) a signal or WRITE (send) a signal.
CPU
The CPU is the section of the processor which processes the data, fetching instruments from
memory, decoding them and executing them. It can be considered to consist of a control unit, arithmetic
and logic unit (ALU) and registers (figure 15.2). it is the bit which is the microprocessor.
The control unit determines the timing and sequence of operations. It generates the time signals
used to fetch a program instrument form memory and execute it. The Motorola 6800 used a clock with a
maximum frequency of 1 MHz, i.e., a clock period of 1 µs, and instructions require between two and twelve

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clock cycles. Operations involving the microprocessor are reckoned in terms of the number of cycles they
take. The arithmetic and logic unit is responsible for performing the actual data manipulation. Internal data
that the CPU is currently using in temporarily held in a group of registers while instructions are being
execute.

Status register
Registers Timing and
control
Program counter Accumulator

Memory address Status register Instruction register

Figure 22 Factures of a CPU
There are number of types of register, the number, size and type of registers varying from one
microprocessor to another. Te following are common types of registers.
1. Accumulator
The accumulator register (AO is where data for an input to the arithmetic and logic unit is
temporarily stored. In order for the CPU to be able to access, i.e. read, instructions or data in the memory it
has to supply the address of the required memory word using the address bus. When this has been done,
the required instructions or data can be read into the CPU using the data bus. Since only one memory
location can be addressed at once, temporary storage has to be used when, for example, numbers are
combined. For example, in the addition of two numbers, one of the numbers is fetched from one address
and placed in the accumulator register while the CPU fetches the other number form the other memory
address. Then the two numbers can be processed by the arithmetic and logic section of the CPU. The result
is then transferred back into the accumulator register. The accumulator register is thus a temporary holding
register for data to be operated on by the arithmetic and logic unit and also, after the operation, the register
for holding the results. It is thus involved in all data transfers associated with the execution of arithmetic
and logic operations.
2. Status register, or condition code register or flag register
The contains information concerning the result of the latest process carried out in the arithmetic and
logic unit. It contains individual bits with each bit having special significance. The bits are called flags. The
status of the latest operation is indicated by each flag with each flag being set or reset to indicate a specific
status. For example, they can be used to indicate whether the last operation resulted in a negative result, a
zero result, a carry output occurs an overflow occurs or the program is to be allowed to be interrupted to
allow as external event to occur.
3. Program counter register (PC) or instruction pointer (IP)
This is the register used to allow the CPU to keep track of its position in a program. This register
contains the address of the memory location that contains the next program instruction. As each instruction
is executed the program counter register is updated so that it contains the address of the memory location
where the next instruction to be executed is stored. The program counter is incremented each time so that
the CPU executes instructions sequentially unless an instruction, such as JUMP or BRANCH, changes the
program counter out of the sequence.
4. Memory address register (MAR)
The contains the address of data. Thus, for example, in the summing of two number the memory
address register is loaded with the address of the first number. The data at the address is then moved to the
accumulator. The memory address of the second number is then loaded into the memory address register.

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The data at this address is then added to the data in the accumulator. The result is then stored in a memory
location addressed by the memory address register.
5. Instruction register (IR)
This stores an instruction after fetching an instruction from the memory, the CPU stores it in the
instruction register. It can then be decoded and used to execute an operation.
6. General-purpose registers
These may serve as temporary storage for data or addresses and be used in operations involving
transfers between various other registers.
7. Stack pointer register (SP)
The contents of this register depends on the microprocessor concerned. For example, the Motorola
6800 microprocessor (figure 21) has two accumulator registers, a status register, and index register a stack
pointer register and a program counter register. The status register has flag bits to show negative, zero,
carry, overflow, half-carry and interrupt. The Motorola 6802 is similar but includes a small amount of RAM
and a built-in clock generator.
The Intel 8085A microprocessor (figure 24) has sic general-purpose registers, a stack pointer, a
program counter and two temporary registers.

Figure 23 Motorola 6800 architecture

MEMORY
The memory unit stores binary data and takes the form of one or more integrated circuits. The data
may be program instruction codes or number being operated on. The size of the memory is determined by
the number of wires in the address bus. The memory elements is a unit consist essentially of large number
of storage cells with each cell capable of storing either a 0 or a 1 bit. The storage cells are grouped in
locations with each location capable of storing one word. In order to access the stored word, each location
is identified by a unique address. Thus with a 4-bit address bus we can have 16 different address with each
perhaps, capable of storing one byte, i.e., a group of eight bits.

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The size of a memory unit is specified in terms of the number of storage locations available; 1 k is 210
= 1024 locations and thus a 4 K memory has 4096 locations. There are a number of forms of memory unit:
1. ROM
2. PROM
3. EPROM
4. EEPROM
5. RAM
ROM
For data that is stored permanently a memory device called a read-
only memory (ROM) is used. ROMs are programmed with the required
contents during the manufacture of the integrated circuit. Not data can that
be written into this memory while the memory chip is in the computer. The
data can only be read and is used for fixed programs such as computer
operating systems and programs for dedicated microprocessor applications.
Figure 25 ROM chip
They do not lose their memory when power is removed. Figure 15.5 shows
the pin connections of a typical ROM chip which is capable of storing 1 K x 8
bits.
PROM
The term programmable ROM (PROM) is used for ROM clips that can
be programmed by the user. Initially every memory cell has a fusible link
which keeps its memory at 0. The 0 is permanently changed to 1 by sending a
current through the fuse to permanently open it. One the fusible like has
been opened the data is permanently stored in the memory and cannot be
further changed.
EPROM
The term erasable and programmable ROM (EPROM) is used for
Figure 26 RAM chip
ROMs that can be programmed and their contents altered. A typical EPROM
chip contains a series of small electronic circuits, cell, which can store change. The program is stored by
applying voltage to the integrated circuit connection pins and producing a pattern of charged and uncharged
cells. The pattern remains permanently in the chip until erased by shining ultraviolet light through a quartz
window on the top of the device. This cases all the cells to become discharged. The chip can then by
reprogrammed. The Intel 2716 EPROM has 11 address pins and a single chip enable pin which is active when
taken low.

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EEPROM
Electrically enable PROM (EEPROM) is similar to EPROM. Erasure is by applying a relatively high
voltage rather than using ultraviolet light.
RAM
Temporary data, i.e. data currently being operated on, is stored in a read/write memory referred to
as a random-access memory (RAM). Such a memory can be read or written to figure 15.6 shows the typical
pin connections for a 1 k x 8-bit RAM chip. The Motorola 6810 RAM chip has seven address pins and six chip
select pins of the which four are active when low and two active when high and all must be made
simultaneously activate enable the RAM.
Microcontroller
For a microprocessor to give a working microcomputer system which can be used for control,
additional chips are necessary, e.g., memory device for program and data storage and input/output ports to
a flow it to communicate with the external world and receive signals from it. The microcontroller is the
integration of a microprocessor with memory and input/output interfaces, and other peripherals such as
timers, on a signal chip. Figure 28 shows the general block diagram of a microcontroller.
The general microcontroller has pins for external connections of inputs and outputs, power clock
and control signals. The pins for the inputs and outputs are grouped into units called input/output ports.
Usually such ports have eight line in order to be able to transfer an 8-bit word of data. Two ports may be
used for a 16-bit word, one to transmit the lower 8 bits and the other the upper & bits. The ports can be
input only, output only or programmable to be either input or output.
The Motorola 68HC11 and the Intel 8051 are example of 8-bit microcontrollers in that the data path
is 8 bits wide. The Motorola 68HC16 is an example of a 16-bit microcontroller and the Motorola 68300 a 32-
bit microcontroller.

Figure 28 Block diagram of a microcontroller

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Selecting a microcontroller
In selecting a microcontroller the following factors need to be considered:
1. Name of input/output pins
How many input/output pins are going to be needed for the tank concerned?
2. Interfaces required
What interfaces are going to be required? For example, is pulse width modulation required? Many
microcontrollers have PWM outputs, e.g., the PICI7C42 has two.
3. Memory requirements
What size memory is required for the task?
4. The number of interrupts required
How many events will need interrupt?
5. Processing speed required
The microprocessor takes time to execute instructions this time being determined by the processor
clock.

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Unit V

Temperature Measurement System

As a brief indication of how a microcontroller might be used, figure 1 shows the main elements of a
temperature measuring system using a MC568HC11. The temperature sensor gives a voltage proportional to
the temperature. The output from the temperature sensor is connected to an ADC input line of the
microcontroller. The microcontroller is programmed to convert the temperature into a BDC output which
can be used to switch on the elements of a two-digital seven-element display. However, because the
temperature may be fluctuating it is necessary to use a storage register which can hold data long enough for
the display to be read. The storage register, 74HCT273, is an octal D-type flip-flop which is reset on the next
positive-going edge of the clock input from the microcontroller.

Figure 1. Temperature measurement system

Domestic Washing Machine
Figure 2 shows how a microcontroller might be used as the controller for a domestic washing
machine. The microcontroller often used is the Motorola M68HCIO5B6; this is simpler and cheaper than the
Motorola M68HC11microcontroller discussed earlier in this chapter and is widely used for low cost
applications. The inputs from the sensors for water temperature and motor speed are via the analogue-
digital input port. Port A provides the output for the various actuators used to control the machine and also
the input for the water level switch. Port B gives outputs the display. Port C gives outputs to the display and
also receives inputs from the keyboard used to input to the machine the various program selections. The
PWM section of the timer provides a pulse width modulated signal to control the motor speed. The entire
machine program is interrupted and stopped if the door of the washing machine is opened.

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Figure 2. Washing machine

Programming
A commonly used method for the development of program follows the steps:
1. Define the problem, stating quite clearly what function the program is
to perform, the inputs and outputs required, any constraints
regarding speed of operation, accuracy., memory size, etc.
2. Define the algorithm to be used. An algorithm is the sequence of
steps which define a method of solving the problem.
3. For systems with fewer then thousand of instructions a useful aid is to
represent the algorithm by means of a flow chart. Figure 3 shows the
standard symbols used in the preparation of flow charts. Each step of
an algorithm is represented by one or more of these symbols and
linked together by lines to represent the program flow. Another
useful design tool is pseudo code. Pseudo code is way of describing
the steps in an algorithm in an informal way which can later the
translated in to a program. Figure 3. Flow chart
Programmable Logic Controllers symbols
A programmable logic controller (PLG) can be defined as a digital electronics device that uses a
programmable memory to store instructions and to implement functions such as logic, sequencing, timing,
counting and arithmetic in order to control machines and process. The term logic is used because the
programming is primarily concerned wi5th implementing logic and switching operations. Inputs devices, e.g.
switches, and output devices, e.g. motors, being controlled are connected to the PLC and these the
controller monitors the inputs and outputs according to this program stored in the PLC by the operator and
so controls the machine or process. Originally they were designed as a replacement for hard-wired relay and
timer logic control systems. PLCs have the great advantage that it is possible to modify a control system
without having to rewire the connections to the input and output devices, the only requirement being that

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an operator has to key in a different set of instructions. The result is a flexible system which can be used to
control system which very quite widely in their nature and complexity.
PLCs are simpler to computers but have certain features w3hich are specific to their use as
controllers. These are:
1. They are rugged and designed to withstand vibrations, temperature, humidity and noise.
2. They are easily programmed and have an easily understood programming language. Programming is
primarily concerned with logic and switching operations.
Basic Structure
Figure 4 shows the basic internal structure of a PLC. It consists essentially of a central processing unit
(CPU), memory, and input/output circuitry. The CPU controls and processes all the operations within the
PLC. It is supplied with a clock with a frequency of typically between 1 and 8 MHz. This frequency determines
the operating speed of the PLC and provides the timing and synchronization for all elements in the system. A
bus system carries information and data to and from the CPU, memory and input/output units. There are
several memory elements a system ROM to give permanent storage for the operating system and fixed
data, RAM for the user’s program, and temporary buffer stores for the input/output channels.

Architecture of a PLC
The programs in RAM can be changed by the user. However, to prevent the loss of the programs
when the power supply is switched off, a battery is likely to be used in the PLC to maintain the RAM
contents for a period of time. After a program has been developed in RAM it may be loaded in to an EPROM
memory chip and so made permanent. Specification for small PLCs often specify the program memory size in
the terms of the number of program steps that can be stored. A program step is an instruction for some
event to occur. A program task might consist of a number of steps and could be, for example: examine the
state of switch A, examine the switch B, if A and B are closed then energies solenoid P which then might
result in the operation of some actuator. When this happens another task might then be started. Typically
the number steps that can be handled by a small PLC is of the order of 300 to 1000, which is generally more
than adequate for most control situations.

Figure 4. Architecture of a PLC

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The input/output unit provides the interface between the

system and the outside world. Programs are centered into the
input/output unit from the panel which can vary from small keyboards
with liquid crystal displays to those using a visual display unit (VDU)
with keyboard and screen display. Alternatively the programs can be
entered in to the system by means of a link to a personal computer
(PC) which is loaded with an appropriate software package.
The input/output channels provide signal conditioning and
isolation functions so that sensors and actuators can be generally Figure 5. Input channel
directly connected to them without the need for other circuitry.
Figure 19.3 shows the basic form of an input channel. Common input
voltages are 5 V and 24 V.
Programming
PLC programming based on the use of ladder diagrams involves writing a program in a similar
manner to drawing a switching circuit. The ladder diagram consists of two vertical lines representing the
power rails. Circuits are connected as horizontal lines, i.e. the rungs of the ladder, between these two
verticals Figure 19.6 shows the basic standard symbols that are used and an example of rungs in ladder
diagram.
In drawing the circuit line for a rung, inputs most always precede outputs and there must be at least
one output on each line. Each rung must start with an input or a series of inputs and end with an output.
The inputs and outputs are numbered, the notation used depending on the PLC manufacturer, e.g.
the Mitsubishi F series of PLCs pressed input elements by an X and output elements by a Y and uses the
following numbers:
Inputs X400 – 407, 410 – 413
X500 – 507, 510 – 513
(24 possible inputs)

Outputs Y430 – 437

Y350 – 537
(16 possible outputs)

Figure 6. Ladder diagram

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To illustrate the drawing of a ladder diagram, consider a situation

where the output from the PLC is to energies a solenoid when a normally
open start switch connected to the input is activated by being closed (Figure
7a). The program required is shown in Figure 7b. Starting with the input, we
have the normally open symbol ll. This might have an input address X400.
The line terminates with the output, the solenoid, with the symbol O. This
might have the output address Y430. To indicate the end of the program the
end rung is marked. When the switch is closed the solenoid is activated. This
might, for example, be a solenoid valve which opens to allow water to enter
a vessel.
Figure 7. Switch controlling
Another example might be an on-off temperature control in which a solenoid
the input goes from low to high when the temperature sensor reaches the
set temperature. The output is then to go from on to off. The temperature sensor shown in the figure is a
thermistor connected in a bridge arrangement with out put to an amplifier connected as a comparator. The
program shows the input as a normally closed pair of contacts, so giving the on signal and hence an output.
When the contacts are opened to give the off signal then the output is switched off.
Such ladder programs can be entered from special keypads or selected from a monitor screen by
using a mouse. They can also be specified by using a mnemonic language. However they are centered, the
programs are then translated by the PLC into machine language for the benefit of the microprocessor and its
associated elements.
Designing
This chapter is a brief review of the design process and brings together many of the topics discussed in
this book in the consideration of both traditional and mechatronics solutions to design problems and case
studies. The design processes can be considered as a number of stages:
1. The need
The design process begins with a need from, perhaps, a customer or client. This may be identified by
market research being used to establish the needs of potential costumers.
2. Analysis of the problem
The first stage in developing a design is to find out the true nature of the problem, i.e. analyzing it. This is
an important stage in that not defining the problem accurately can lead to wasted time on designs that will
not fulfill the need.
3. Preparation of a specification
Following the analysis a specification of the requirements can be prepared. This will state the problem,
any constraints placed on the solution, and the criteria which may be used to judge the quality of the design.
In stating the problem, all the functions required of the design, together with any desirable features, should
be specified. Thus there might be a statement of mass, dimensions, types and range of motion required,
accuracy, input and output requirement of elements, interfaces, power requirements, operating
environment, relevant standards and codes of practice, etc.
4. Generation of possible solutions
This is often termed the conceptual stage. Outline solutions are prepared which are worked out in
sufficient detail to indicate the means of obtaining each of the required functions, e.g. approximate sizes,
shapes, materials and costs. It also means finding out what has been done before for similar problems, there
is no sense in reinventing the wheel.
5. Selections of a suitable solution
The various solutions are evaluated and the most suitable one selected.

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6. Production of a detailed design

The detail of the selected design has now to be worked out. This might require the production of
prototypes or mock-ups in order to determine the optimum details of a design.
7. Production of working drawings
The selected design is then translated into working drawings, circuit diagrams, etc. so that the item can
be made. In should not be considered that each stage of the design process just flows on stage by stage.
There will often be the need to return to an earlier stage and give it further consideration. Thus when at the
stage of generating possible solutions there might be a need to go back and reconsider the analysis of the
problem.
Designing Mechatronics Systems
The design process consists of the following stages (refer Figure 8):
Stage 1: Need for design
The design process begins with a need. Needs are usually arise from dissatisfaction with an existing
situation. Needs may come from inputs of operating or service personal or from a customer through sales or
marketing representatives. They may be to reduce cost, increase reliability or performance or just change
because of public has become bored with the product.
Stage 2: Analysis of problem
Probably the most critical step in a design process is the analysis of the problem i.e., to find out the
true nature of the problem. The true problem is not always what it seems to be at the first glance. Its
importance is often overlooked because this stage requires such a small part of the total time to create
the final design. It is advantageous to define the problem as broadly as possible. If the problem is not
accurately defined, it will lead to a waste of time on designs and will not fulfill the need.
Stage 3: Preparation of specification
The design must meet the required performance specifications. Therefore, specification of the
requirements needs to be prepared first. This will state the problem definition of special technical terms,
any constraints placed on the solution and the criteria that will be used to evaluate the design. Problem
statement includes all the functions required of the design, together with any desirable features. The
following are some of the statements about the problem:
 Mass and dimensions of design.
 Type and range of motion required.
 Accuracy of the element.
 Input and output requirements of elements.
 Interfaces.
 Power requirements.
 Operating environment.
 Relevant standards and code of practice, etc.
Stage 4: Generation of possible solution
This stage is often known as conceptualisation stage. The conceptulisation step is to determine the
elements, mechanisms, materials, process of configuration that in some combination or other result in a
design that satisfies the need. This is the key step for employing inventiveness and creativity.
A vital aspect of this step is synthesis. Synthesis is the process of taking elements of the concept and
arranging them in the proper order, sized and dimensioned in the proper way. Outline solutions are
prepared for various possible models which are worked out in sufficient details to indicate the means of
obtaining each of the required functions.

Stage 5: Selection of suitable solution or Evaluation

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This stage involves a thorough analysis of the design. The evaluation stage involves detailed
calculation, often computer calculation of the performance of the design by using an analytical model. The
various solutions obtained in stage 4 are analysed and the most suitable one is selected.

Figure 8. Stages in designing mechatronics systems

Stage 6: Production of detailed design
The detail of selected design has to be worked out. It might have required at extensive simulated
service testing of an experimental model or a full size prototype in order to determine the optimum details
of design.
Stage 7: Production of working drawing
The finalised drawing must be properly communicated to the person who is going to manufacture. The
communication may be oral presentation or a design report. Detailed engineering drawings of each
components and the assembly of the machine with complete specification for the manufacturing process are
written in the design report.
Stage 8: Implementation of design
The components as per the drawings are manufactured and assembled as a whole system.
Case Studies
Mpcnatronics systems are widely used now a day in many industries. Some of the examples are
explained here.
1. Case Study 1: Pick and Place Robot
The basic form of a pick and place robot is shown in Figure 9. The robot has three axes about which
motion can occur. The following movements are required for this robot.
1. Clockwise and anticlockwise rotation of the robot unit on its base.
2. Linear movement of the arm horizontally i.e., extension or contraction of arm.

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3. Up and down movement of the arm and

4. Open and close movement of the gripper.
The foresaid movements can be obtained by
pneumatic cylinders which are operated by solenoid
valves with limit switches. Limit switches are used to
indicate when a motion is completed.
The clockwise, rotation of the robot unit on
its base can be obtained from a piston and cylinder
arrangement during pistons forward movement.
Similarly counter clockwise rotation can be obtained
during backward movement of the piston in cylinder.
Linear movement of the arm can result during
forward and backward movement of the piston in a
cylinder.
The upward movement of the arm can result
from forward movement of the piston in a cylinder
whereas downward movement from its retardation. Figure 9. Basic from of a pick and place robot
The griper can also be operated in a similar way as
explained above i.e., gripper is opened during
forward movement of the piston and closed during
backward movement of the piston in the cylinder.
Figure 10 shows a mechanism used for this purpose.
A microcontroller used to control the
solenoid valves of various cylinders is shown in Figure
11. The micro controller used of this purpose is
M68HC11 type. A software program is used to control
the robot.
TRIAC optoisolator consists of LED and TRIAC. If
the input of the LED is 1, it glows and activates the
TRIAC to conduct the current to the solenoid valve.
Otherwise TRIAC will not conduct die current to the
solenoid valve. Figure 10. Gripper mechanism of a robot
2. Case Study 2: Automatic Car Park System
Consider an automatic car park system with barriers operated by coin inserts. The system uses a PLC for
its operation. There are two barriers used namely in barrier and out barrier. In barrier is used to open when
the correct money is inserted while out barrier opens when a car is detected in front of it. Figure 5.18 shows
a schematic arrangement of an automatic car park barrier. It consists of a barrier which is pivoted at one
end, two solenoid valves A and B and a piston cylinder arrangement.
A connecting rod connects piston and barrier as shown in Figure 12. Solenoid valves are used to control
the movement of the piston. Solenoid A is used to move the piston upward in turn barrier whereas solenoid
B is used to move the piston downward. Limit switches are used to detect me foremost position of the
barrier. When current flows through solenoid A, the piston in the cylinder moves upward and causes die
barrier to rotate about its pivot and raises to let a car through.

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Figure 11 Microcontroller circuit for pick and place robot

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Figure 12 Microcontroller circuit for pick and place robot

When the barrier hits the limit switch, it will turns on the timer to give a required time delay. After
that time delay, the solenoid B is activated which brings the barrier downward by an operating piston in the
cylinder. This principle is used for both the barriers.
3. Case study 3: Engine Management System
An electronic engine management system is made up of sensors, actuators, and related wiring that is
tied into a central processor called microprocessor or microcomputer (a smaller version of a computer).
Electronic management systems monitor and gather data from a number of sensors in the engine
and continuously adjust the fuel supply and injection timing. This minimizes emissions and maximizes fuel
efficiency and engine output at any given workload. The electronic engine management generally consists of
the following basic components: An electronic control unit (ECU), a fuel delivery system (typically fuel
injection), an ignition system and a number of sensors. Figure 13 shows the various components in the
typical engine management system.
Electronics control unit (ECU):
The sensors provide feedback to the ECU to indicate how the engine is running so that die ECU can
make the necessary adjustments to the operation of the fuel delivery and/or ignition system.
Fuel delivery system:
This system consist high pressure fuel pump which is mounted in or near the tank. The fuel line from
the pump passes through a filter before it runs forward to the engine bay. The fuel line connects to a fuel rail
that feeds each of the injectors. At the end of the rail is a fuel pressure regulator, with surplus fuel heading
back to the tank in the return line.
Ignition system:
Ignition system consists of ignition coil, distributor and spark plug. These components are connected
with the ECU to receive the signal for proper timed operation.
Various sensors:
Engine sensors fall into five broad categories: Throttle-Position Sensors, Exhaust Gas Oxygen
Sensors, Manifold Absolute Pressure Sensors, Temperature Sensors and Speed/Timing Sensors. All these
sensor functions are centrally controlled by microcontroller as shown in Figure 14.

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Figure 13 components of engine management system

a. Throttle-Position sensors:
A throttle-position sensor sends the signal to ECU about the throttle opening and the force applied
by the driver. Then the ECU controls the fuel delivery and spark timing based on the throttle position. Two
common throttle-position sensors are potentiometric and Hall-effect sensors.

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Figure 14 Interfacing of sensor with controller in engine management system

b. Exhaust Gas Oxygen (EGO) Sensors:
Exhaust gas oxygen (EGO) sensors are placed within the engine's exhaust system. The amount of
oxygen in the exhaust gas indicates whether or not the ECU has directed the fuel delivery system to provide
the proper' air-to-fuel ratio. If the relative amount of air is too high or too low, engine power, smoothness,
fuel efficiency and emissions will all suffer.
c. Manifold Absolute Pressure (MAP) Sensors:
Manifold Absolute Pressure (MAP) Sensors measure the degree of vacuum in the engine's intake
manifold. The amount of vacuum depends on engine rpm and throttle opening. The most common MAP
sensors are piezoresistive .and variable capacitor sensors.
d. Temperature Sensors:
Temperature sensors are used to report engine temperature to the driver/operator via dash panel-
mounted temperature gauge, report engine temperatures to the ECU to activate/de-activate cooling fans in
water-cooled engines, to richen fuel mixtures for easier starting in cold weather and to lean-out mixtures
for maximum fuel economy. Two common temperature sensors are thermistors or thermodiodes.
e. Engine Speed/Timing Sensors:
Speed/timing sensors provide information to the ECU regarding engine speed and the crank
position. This information is used by the ECU to control fuel and ignition, as well as to make sure that engine
speed does not exceed safe operating limits. It is also used to control the fuel injectors and spark plugs.
Most common speed/timing sensors are variable reluctance, optical crankshaft position and Hall-effect
sensors.

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f. Exhaust gas regulation (EGR) Valve Position Sensor:

The signal from EGR valve position sensor is used to adjust the air fuel mixture. The exhaust gases
introduced by the EGR valve into the intake manifold reduce the available oxygen and thus less fuel is
needed in order to maintain low hydro carbon level in the exhaust.
g. Mass Air flow (MAF) sensor:
MAF sensor is used to measure engine load to squirt in the right amount of petrol, and fire the spark
at just the right moment. The amount of power being" developed depends on how much air the engine is
breathing. Most common airflow sensors are Hot Wire Airflow sensor and Vane Airflow Meter.
h. Knock Sensor:
The knock sensor is used to identify the sounds of knocking and sends signal to ECU to avoid knocking. It
is screwed into the engine block and is designed to separate out the special noise which means that knocking
is occurring. Many Electronic Fuel Injection (EFI) engines run ignition timing very close to knocking.

B.E Mechanical, VIII(Sem), Mechatronics MEEC802/PMEEC603 Compiled By Dr. A. P. Sathiyagnanam,