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INSTRUMENTATION & PROCESS

CONTROLClass Notes

Sanjay DalviAssistant ProfessorGharda Institute of TechnologyLavel, Ratnagiri, Maharashtra


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CONTENTS

PART I PROCESS CONTROL

1 Concepts of Control System 3

1.1 System 3

1.2 Steady state and Unsteady state 3

1.3 Process Control 31.4 Feedback and Feedforward 4

1.5 Transfer function for First Order system 4

1.5.1 Step input to first order system 5

1.5.2 Impulse input to first order system 6

1.5.3 Sinusoidal input to first order system 7

1.6 Transfer function for liquid level in tank 10

1.7 Transfer function for pure capacitive process 12

1.8 First order systems in series 13

1.8.1 Non-interacting 13

1.8.2 Interacting 16

1.9 Second order system 181.9.1 Types of second or higher order systems 18

1.9.2 Dynamic behaviour of second order system 18

2 Closed Loop Systems 25

iii
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iv CONTENTS

2.1 What is control system? 25

2.2 Feedback 25

2.3 Servo and Regulator Problem 26

2.4 Development of Block Diagram 26

2.4.1 Process 27

2.4.2 Measuring Element 29

2.4.3 Controller & final control element 29

3 Controllers 31

3.1 Proportional Control 31

3.2 ON-OFF Control 33

3.3 Proportional-Integral Control 33

3.4 Proportional-Derivative Control 33

3.5 Proportional-Integral-Derivative Control 34

3.6 Offset 35

3.6.1 Proportional control to first order system 363.6.2 PI control to first order system 37

4 Stability 39

4.1 Analysing response of a system 39

4.2 Routh-Hurwitz Criterion 41

4.3 Root-Locus analysis 43

4.4 Frequency response analysis 47

4.4.1 Frequency response of pure capacitive process 48

4.4.2 Frequency response of pure dead time process 49

4.4.3 Frequency response of second order system 49

4.5 Bode Diagram 50

4.5.1 Bode stability criterion 52

4.6 Nyquist plot 56

4.6.1 First order system 56

4.6.2 Second order system 57

4.6.3 Pure dead time 58

4.6.4 Nyquist stability criterion 58

5 Controller Tuning 63

5.1 Time integral performance of controllers 63

5.2 Selection of feedback controllers 645.2.1 Rules for selecting controller 65

5.2.2 Controller for different processes 66

5.3 Open-loop controller tuning 67

5.4 Ziegler-Nichols tuning 69


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CONTENTS v

6 Introduction to SciCos / XCos 71

6.1 Developing Model 72

6.2 Drawing Blocks 79

6.3 Connecting Blocks 79

6.4 Assigning parameters 806.5 Simulating Modelled diagram 81

6.6 Configuring CSCOPE 82

6.7 Configuring PID Control 83

6.8 Super-Blocks 83

6.9 Set Context 84

PART II INSTRUMENTATION

7 Flow measurement 87

7.1 Volumetric flowmeters 87

7.1.1 Rotameter 87

7.1.2 Turbine meter 87

7.1.3 Venturi meter 88

7.1.4 Orifice meter 89

7.1.5 Vortex meter 89

7.1.6 Electromagnetic flowmeter 90

7.1.7 Ultrasonic Flow Equipment 91

7.2 Mass flowmeters 92

7.2.1 Coriolis mass flowmeter 92

7.2.2 Thermal mass flowmeter 92

8 Temperature measurement 95

8.1 Thermometers 95

8.2 Thermocouples 97

8.2.1 Cold junction compensation 98

8.2.2 Types 98

8.3 Resistance Temperature Detector (RTD) 100

8.4 Thermisters 101

8.5 Pyrometers 101

9 Level measurement 105

9.1 Gauge glass 1059.2 Chain and tape float gauges 105

9.3 Lever and shaft float gauges 105

9.4 Displacer level measuring device 106

9.5 Head-pressure level gauges 106
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vi CONTENTS

9.6 Electrical type level gauges and switches 107

9.7 Capacitance probes 108

9.8 Other methods 108

PART III CONTROL STRATEGIES AND APPLICATIONS

10 Control Valve 111

10.1 Actuator 112

10.1.1 Pneumatic Actuators 113

10.1.2 Electric actuators 115

10.1.3 Hydraulic actuators 116

10.2 Positioner 117

10.3 Cage 118

10.4 Flow Characteristics and Valve Selection 118

10.5 Control valve sizing 119

10.5.1 Gas services 119

10.5.2 Liquid Service 121

11 Advanced control strategies 123

11.1 Feedforward Control 123

11.1.1 Comparison of Feedback and Feedforward Control 124

11.2 Ratio Control 126

11.2.1 Applications of Ratio control 126

11.3 Cascade control 127

11.4 Application of Feedforward, Feedback and Cascade control 129

11.5 Control strategy design at basic control level 129

11.6 Determine the different Variables 13011.6.1 Controlled variable (CVs) 131

11.6.2 Manipulated variables (MVs) 131

11.6.3 Measured variables 131

11.6.4 Disturbance variables (DVs) 132

11.7 Plantwide control design procedure 133

11.8 Some guidelines and recommendation for control system design 137

12 Microprocessor-based controller 139

12.1 Hardware 140

12.2 Distributed Control System (DCS) 140

12.2.1 Historical Review 14212.2.2 Modes of Computer control 143

12.2.3 Computer Control Networks 143

12.2.4 Small Computer Network 144

12.2.5 Commercial Distributed Control Systems 144
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CONTENTS vii

12.2.6 Description of the DCS elements 145

12.2.7 The advantages of DCS systems 146

12.3 Programmable Logic Controllers 147

12.4 Digital control software 147

References 149


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PART I

PROCESS CONTROL


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CHAPTER 1

CONCEPTS OF CONTROL SYSTEM

1.1 System

System in context to Process control is a set of equipments and devices interacting with

each other. This definition is considerably confusing and do not give exact idea. Therefore

lets consider a heating water bath, where water is heated to desired temperature usingheating element. Now the elements of the complete set-up consist of water tank with or

without agitator, heating element, temperature measuring device and temperature recorder

and/or controller. Combination of all these units is called as system.

1.2 Steady state and Unsteady state

If the process variables do not change with time then system is said to be at steady state.

Steady state models are always easier than unsteady state model to develop. But, unfortu-

nately all process control problems deal with unsteady state operations.

1.3 Process Control

As we discussed in previous section we have to deal with unsteady state operation which

are inherently unstable in nature and requires control. If we talk in simple language process

control subject deals with unsteady processes to make them steady.

Instrumentation & Process Control.

By Sanjay Dalvi Copyright 2012 Sanjay Dalvi

3

20/10/2012 Rev 0.1 (c) 2012 Sanjay Dalvi


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4 CONCEPTS OF CONTROL SYSTEM

1.4 Feedback and Feedforward

When process is required to be controlled, it requires some basic data regarding status

of the process. If data about the status of the system is provided to controller then this

arrangement is called as Feedback control.

Opposite to feedback control, status of input is supplied to controller so that correctiveaction can be taken before input disturbs the system. Feed forward arrangement is not

common and obvious.

1.5 Transfer function for First Order system

Before defining what is transfer function lets discuss the model for mercury thermometer.

Assumptions:

i) The resistance offered by the glass and mercury is negligible.

ii) At any instance the mercury assumes a uniform temperature throughout.

iii) The glass wall containing mercury does not expand or contract during the transientresponse.

Unsteady state energy balance,

{Input rate} {Output rate} = {rate of accumulation}

hA (To Tth) 0 =mCp dTthdt

(1.1)

For steady state conditions

hA (Tos Tths) = 0 (1.2)Subtract equation (1.2)from equation (1.1)

hA [(To Tos) (Tth Tths)] = mCp d (Tth Tths)dt (1.3)

Let deviation variables,

o = To Tosth= Tth Tths

therefore equation (1.3) becomes,

hA (o th) = mCp dthdt

(o

th) =

dth

dt

(1.4)

where,= mCphA Taking Laplace Transform of equation ( 1.4)

o(s) th(s) = sth(s)o(s) =th(s) ( s+ 1)



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TRANSFER FUNCTION FOR FIRST ORDER SYSTEM 5

th(s)

o(s) =

1

s+ 1 (1.5)

The parameter is called the time constant of the system and RHS of equation(1.5)is calledthetransfer functionof the system.

Transfer function=

LT of the deviation in thermometer reading

LT of the deviation in surrounding temperature

Response or output variableForcing function or input variable

G (s) = Y(s)

X(s)

Therefore response of the system,

Y (s) = G (s) X(s)

Above equation can be represented using block diagram,

G(s)X(s) Y(s)

Equation (1.5)represents first-order system as it is derived from first order linear differen-

tial equation.

1.5.1 Step input to first order system

Step input of sizeA can be represented in mathematical form as given below,

X= 0 fort


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Y(s) =A

s A

s + 1Inverting above equation,

Y (t) =A

1 e t

(1.8)

whereas fort


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Multiply equation (1.11) with (s + j) and sets= j

C2 = A

2j+ 2

=A/2

+j=

A/2 +j

j j

=A/2 ( j)

22 + 1

Multiply equation (1.11) with (s j ) and sets= j

C3 =A/2 j

=A/2 j

+j

+j

=A/2 ( +j)

22 + 1

Multiply equation (1.11) with (s+ 1) and sets= 1/

C1 = A 1 + j 1 j

= A 12 +

2 = A2

22 + 1

Y (s) = A2

2+ 1

1

s+ 1

A

2 (22 + 1)

js+ j

A

2 (22 + 1) +j

s j (1.12)

Taking inverse Laplace,

L1

A2

22 + 1

1

s+ 1

=

A

22 + 1e

t

L1

A

2 (22 + 1)

js + j

=

A ( j)2 (22 + 1)

ejt

L1

A

2 (22 + 1)

+j

s j

= A ( +j)

2 (22 + 1)ejt

therefore response equation becomes,

Y (t) = A

22 + 1et/

A2 (22 + 1)

( j) ejt + ( +j) ejt (1.13)



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TRANSFER FUNCTION FOR FIRST ORDER SYSTEM 9

Using the identity,

e(C1+jC2) t= ec1t (cos C2t +j sin C2t)

here in this case,C1 = 0andC2 = for third term,

ejt

= cos (t) +j sin(t)C2 = for second term,

ejt = cos (t) +j sin(t) = cos(t) j sin(t)therefore the square bracket of equation ( 1.13) is

cos (t) j sin(t) j cos(t) sin(t)+ cos(t) + j sin(t) +j cos(t) sin(t)

= 2 cos(t) 2sin(t)replacing square bracket of equation ( 1.13) we get,

Y (t) = A22 + 1

et/ A(22 + 1)

cos (t)

+ A

(22 + 1)sin (t) (1.14)

Using trigonometric identity,

p cos A + qsin A= r sin(A + )

where,r =

p2 + q2 andtan = p/qhere we have,

p= A22 + 1

q=

A

22 + 1

r=

(A)

2+ A2

(22 + 1)2

r = A22 + 1

= tan1 A

22 + 1

22 + 1

A

= tan1 () =

putting all these in equation (1.14) we get,

Y(t) = A

22 + 1et/+

A22 + 1

sin(t + ) (1.15)

Ast the first term of equation ( 1.15) vanishes

Y (t)|s= A22 + 1

sin(t + ) (1.16)



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Period Lag

Input

Output

Figure 1.1 Response of first order system to sinusoidal input

EXAMPLE 1.2

A mercury thermometer having a time constant of 0.1 min is placed in temperature

bath at 100C and allowed to come to equilibrium with the bath. At timet = 0, thetemperature of the bath begins to vary sinusoidally about its average temperature of

100C with amplitude of 2C. If the frequency of oscillation is10/cycles/min, whatis the phase lag and lag and amplitude of response?

Solution:

= 0.1min and amplitudeA= 2Csteady state temperature = 100C

f= 10 cycles/min= 2f= 2

10 = 20rad/min

Amplitude of the response,A

22 + 1=

2(0.1)2 400 + 1

= 0.896C

= tan1 () = tan1 (20 0.1) = 63.5 Phase lag= 63.5

As 1 cycle 360 10 minlag=

63.5

360

10= 0.0555 min(3.33 sec)

Lag= ||360f

where,in degrees.

1.6 Transfer function for liquid level in tank

Material balance across a tank,

Finin Foutout= d (V )dt



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TRANSFER FUNCTION FOR LIQUID LEVEL IN TANK 11

Fin

h V

FoutR

Figure 1.2 Liquid level in tank

for incompressible (liquid) fluid,

in= out=

dV

dt =Fin Fout

asV =Ahand lets considerFout = h/R

Adh

dt =Fin h

R (1.17)

At initial steady state i.e. att= 0

0 = Fins hsR

(1.18)

equation (1.17) (1.18)

Ad (h hs)

dt = (Fin Fins) (h hs)

R (1.19)

Using deviation variables as below,

F =Fin Fins (1.20)H=h hs (1.21)

now equation (1.19) can be written as,

AdH

dt = F H

R

Taking Laplace Transform (LT) of above equation,

AsH(s) = F(s) H(s) /R

H(s)

F(s) =

R

s+ 1 (1.22)



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where,=AROutflow can also be represented in terms of deviation variable as,

Fo = Fout Fouts= HR

taking LT of above equation,

Fo(s) =

H(s)R (1.23)

from equation (1.22)and (1.23) we get,

Fo(s)

F(s) =

1

s+ 1 (1.24)

1.7 Transfer function for pure capacitive process

Let consider a storage tank with one inflow stream and one outflow. Liquid withdrawn

from tank is through pump as shown in Fig. 1.3.Mass balance across the tank can be given

Fin

h V

Fout

Figure 1.3 Pure capacitive tank

by,

Fin Fo = A dhdt

(1.25)

Considering system was at steady state before disturbance (i.e. t 0). Therefore massbalance equation at steady state is,

Fin Fo = 0 (1.26)Subtracting equation (1.26) from equation (1.25),

F =AH

dt

Taking Laplace Transform of above equation,

F(s) = AsH(s)

H(s)

F(s) =

1

As (1.27)



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FIRST ORDER SYSTEMS IN SERIES 13

If input is change by unit step (F(t) = 1), transfer function of input is,

F(s) =1

s

H(s) = 1

As

H(t) = t

A

h (t) = hs+ t

A (1.28)

Equation (1.28) is output for the step change shows continuous growth in level and system

is non-regulating. System that have a limited change in output for a sustained change in

input are said to have regulation.

1.8 First order systems in series

When more than one tanks are connected in series then connection between the tanks either

can be interactingor non-interacting. This interaction is nothing but effect of capacitance

of one tank on other.

1.8.1 Non-interacting

q

h1 Tank 1

R1 q1

h2 Tank 2

q2R2

Figure 1.4 Non-interacting tanks

Mass balance across the tank - 1 and tank - 2 for constant density.

q q1 = A1 dh1dt

(1.29)

q1 q2 = A2 dh2dt

(1.30)



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Flow and head relationship is given by,

q1 = h1R1

(1.31)

q2 = h2R2

(1.32)

Put equation (1.31)in (1.29),

q h1R1

=A1dh1dt

(1.33)

at (t= 0),

qs h1sR1

= 0 (1.34)

Equation (1.33) (1.34)and using deviation variables,

Q= q qsH1 = h1 h1s

Q H1R1

=A1H1dt

Taking Laplace Transform,

Q (s) 1R1

H1(s) = A1sH1(s)

Rearranging above equation,H1(s)

Q (s) =

R11s+ 1

(1.35)

where,1 = A1R1 From equation(1.31) we can write,

Q1(s) =H1(s)

R1

Therefore equation(1.35) becomes,

Q1(s)Q (s) = 11s + 1 (1.36)

Similarly, put equation(1.32) in equation(1.30) so we get,

H2(s)

Q1(s) =

R22s+ 1

(1.37)

where,2 = A2R2 ReplaceQ1(s)of equation(1.37)with equation(1.36)

H2(s)

Q (s) =

R2

2s + 1

1

1s + 1

(1.38)

orQ2(s)

Q (s) =

1

2s + 1

1

1s+ 1

(1.39)

The overall transfer function is the product of two transfer functions of first order.

G1(s)Xo

G2(s)X1 X2



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EXAMPLE 1.3

Two non-interacting tanks are connected in series, as shown in Fig. 1.4. The time

constants are1 = 1and 2 = 0.5;R2 = 1. Sketch the response of the level in Tank 2 if a unit-step change is made in the inlet flow rate of Tank 1.

Solution:

For step change in inlet flow of Tank 1, Q (s) = 1/sTherefore equation( 1.38)is,

H2(s) =1

s

R2

2s+ 1

1

1s + 1

Taking partial fractions of RHS of above equation,

1

s (1s+ 1)(2s+ 1)=

C1s

+ C2

2s+ 1+

C31s+ 1

(1.40)

Multiply equation(1.40)withsand sets= 0

C1 = 1

Multiply equation(1.40)with(1s+ 1) and sets= 11

C2 = 21

1 2

Multiply equation(1.40)with(2s+ 1) and sets= 12

C3=

22

1 2

Put values ofC1,C2 and C3 in equation(1.40)

H2(s) = R2

1

s 1

(1 2)

1

s+ 11

+

2(1 2)

1

s+ 12

Inverting above equation,

H2(t) = R2 1 121

2

e t1

2 e

t2

1 Now, 2 = 0.5,1 = 1andR2 = 1

H2(t) = 1

2et e2t



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0 1 2 3 4 5

1.0

two tanks

one tank

H2(t)

t

1.8.2 Interacting

q

h1Tank 1 q1

R1

h2Tank 2

R2

q2

Figure 1.5 Interacting tanks

Mass balance across the tank - 1 and tank - 2 for constant density.

q q1 = A1 dh1dt

(1.41)

q1 q2 = A2 dh2dt

(1.42)

Flow and head relationship is given by,

q1 =h1 h2

R1(1.43)

q2 = h2R2

(1.44)



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Using deviation variables equation(1.41), (1.42), (1.43) and (1.44)will be,

Q Q1 = A1 dH1dt

(1.45)

Q1 Q2 = A2 dH2dt

(1.46)

Q1 =H1 H2

R1(1.47)

Q2 =H2R2

(1.48)

Transforming above equations,

Q (s) Q1(s) = A1sH1(s) (1.49)Q1(s) Q2(s) = A2sH2(s) (1.50)

R1Q1(s) = H1(s) H2(s) (1.51)R2Q2(s) = H2(s) (1.52)

ReplaceQ2(s)of equation(1.50)using equation(1.52)

R2Q1(s) H2(s) = A2R2sH2(s)

H2(s)

Q1(s) =

R22s+ 1

(1.53)

where,2 = A2R2Using equation(1.51)replaceH1(s)of equation(1.49)

Q (s) Q1(s) =A1s [R1Q1(s) + H2(s)]=1sQ1(s) +A1sH2(s)

Q (s) = (1s+ 1) Q1(s) + A1sH2(s)

Using equation(1.53)replaceQ1Q (s)of above equation,

Q (s) =

(1s + 1) (2s + 1) + A1R2s

R2

H2(s)

H2(s)

Q (s) =

R2(1s + 1) (2s + 1) + A1R2s

H2(s)

Q (s) =

R212s2 + (12+ A1R2) s+ 1

(1.54)

based on outflow,Q2(s)

Q (s)

=H2(s)

Q (s)

Q2(s)

H2(s)Using equation(1.54)and (1.52)

Q2(s)

Q (s) =

1

12s2 + (12+ A1R2) s + 1 (1.55)



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1.9 Second order system

A second order system is one whose output, is described by the solution of a second order

differential equation e.g.

a2d2y

dt2 + a1dy

dt + a0y = bf(t) (1.56)

Ifa0= 02

d2y

dt2 + 2

dy

dt + y = Kpx (t) (1.57)

where, 2 =a2/a0, 2=a1/a0 andKp = b/a0

=natural period of oscillation of the system

= damping factor

Kp = steady state or static or simply gain of the system

Equation(1.57) in terms of deviation variables,

2 d2Y

dt2 + 2

dY

dt + Y =KpX(t)

taking Laplace transform of above equation,

2s2Y (s) + 2sY (s) + Y (s) =KpX(s)

Y(s)

X(s)=

Kp(2s2 + 2 s+ 1)

(1.58)

1.9.1 Types of second or higher order systems

1. Multi-capacity processes: Process that consist of two or more capacities in series

through which material or energy must flow.

2. Inherently second order system: Material possessing inertia is subjected to accelera-

tion.

3. A processing system with its controller.

1.9.2 Dynamic behaviour of second order system

For unit step change in input,

X(s) =1

s

therefore equation(1.58)becomes,

Y (s) =

Kps (2s2 + 2s+ 1)

Letp1and p2 are the roots of the denominator of the above equation.

p1=

+

2 1

and p2 =

2 1



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SECOND ORDER SYSTEM 19

wherep1 and p2are called poles.

The roots of the polynomial in denominator are called poles.

The roots of the polynomial in numerator are called zeros.

Y(s) = Kp/

2

s (s p1) (s p2) (1.59)

Case A : When >1, two distinct and real poles

Case B : When= 1, two equal poles (multiple poles)

Case C : When 1

1

s (s p1) (s p2) =C1

s +

C2s p1 +

C3s p3 (1.60)

multiply equation(1.60)withs and sets= 0

C1 = 1

p1p2

multiply equation(1.60)with(s p1)and sets= p1

c2 = 1p1(p1 p2)

multiply equation(1.60)with(s p2)and sets= p2

C3 = 1p1(p1 p2)

putC1,C2, andC3in equation(1.60) and then put equation( 1.60)into equation(1.59)

Y(s) = Kp

2 1

p1p2s+

1

p1(p1

p2) (s

p1)

1p2(p1

p2) (s

p2)

(1.61)

inverting above equation

Y (t) = Kp

2

1

p1p2+

ep1t

p1(p1 p2) ep2t

p2(p1 p2)

(1.62)



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Now, using the values ofp1 and p2

p1p2=

+

2 1

2 1

=

1

2

p1 p2=

+

2 1

2 1

=2

2 1

p1(p1 p2) =2

2 1 + 2 2 1

2

p2(p1 p2) =2

2 1 2 2 1

2

Y(t) = Kp1

et e

21 t

2

2 1 2 (2 1)+

ete

21 t

2

2 1 2 (2 1)Now, leta= 2

2 1,b= 2 2 1, andc= 2 1 t

Y(t) = Kp

1 e t

(a + b) ec (a b) ec

a2 b2

=Kp

1 e t

a (ec ec) + b (ec + ec)

a2 b2

But,

sinh =e e

2 cosh =

e + e

2

therefore,

Y(t) =Kp

1 e t

2a sinh c + 2b cosh ca2 b2

(1.63)

Now replacinga,b, andc,

a2 b2 = 42 2 1 4 2 12 = 4 2 12a sinh c

a2 b2 =4

2 1

4 (2 1) sinh

2 1 t

=

2 1 sinh

2 1 t

2a cosh c

a2 b2 =4

2 14 (2 1)cosh

2 1

t

= cosh

2 1 t

Y (t) = Kp1 e

t

2 1sinh2 1

t

+ cosh2 1

t

(1.64)

The gain is,

Kp = (steady state output)

(steady state input)



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Y(t)

Kp =1

1

2 2 > 1 > 1

t

Figure 1.6 Response curve for overdamped and critically damped system

Case B: Critically damped response, when = 1 Poles will be,

p1 = 1

p2 = 1

therefore transfer function is,

Y (s) = Kp/

2

s (s + 1/)2 =

Kp

s (s+ 1)2 (1.65)

taking partial fraction,

1

s ( s+ 1)2 =

C1s

+ C2 s+ 1

+ C3

( s+ 1)2 (1.66)

multiply equation(1.66)withs and sets= 0,

C1 = 1

multiply equation(1.66)with( s+ 1)2

and sets= 1,C3 =

putC1and C3 in equation(1.66),

1 = (s+ 1)2 C1+ s ( s+ 1) C2+ sC3

=

2C1+ C2

s2 + (2 C1+ C2+ C3) s + C1

equating coefficients,

2C1+ C2 = 0

C2 = therefore equation(1.65)becomes,

Y (s) = Kp

1

s

s+ 1

( s+ 1)2

inverting above equation,

Y (t) = Kp

1 et/ t

et/

=Kp

1

1 +

t

et/

(1.67)



30/157


Case C: Underdamped response, when < 1 Partial fractions will be same as in

overdamped case i.e. equation(1.61), therefore we can use equation( 1.62).

Y (t) = Kp

2

1

p1p2+

ep1t

p1(p1 p2) ep2t

p2(p1 p2)

(1.62)

Let,p1 = a + bj and p2 = a bj therefore,p1p2 = a

2 + b2

p1 p2 = 2bjp1(p1 p2) = 2b (b aj)p2(p1 p2) = 2b (b + aj)

replacep1 andp2in equation(1.62),

Y (t) =Kp

2

1

a2 + b2 e

(a+bj)t

2b (b aj) e(abj)t

2b (b + aj)

=

Kp

2 (a2 + b2)

1 1

2b

(b + aj) e

(a+bj)t

+ (b aj) e(abj)t

Using identity,

e(a+bj)t =eat (cos bt +j sin bt)

e(abj)t =eat (cos(bt) +j sin(bt))=eat (cos bt j sin bt)

(b+ aj) e(a+bj)t + (b aj) e(abj)t = 2b cos bt 2a sin bt

Y (t) = Kp

2 (a2 + b2)

1 e

at

b (b cos bt a sin bt)

using trigonometric identity,

p cos A + qsin A= r sin(A + )

where,r =

p2 + q2 andtan = p/qHere we have,p= b,q= aandA= bt, therefore

r=

a2 +b2 and = tan1 (b/a) =

Y (t) = Kp

2 (a2 + b2)

1 e

at

b

a2 + b2 sin(bt + )

(1.68)

comparing roots discussed in overdamped section we can write,

a=

b=

1 2

therefore,

a2 + b2 = 1

2 and

ba

=

1 2



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Y(t) = Kp

1 +

11 2 e

t sin(t + )

(1.69)

where, =

1 2

=radian frequency

= tan1

1 2

=phase lag

Y (t)

Kp

t

A

BT

C

trise

5%

Figure 1.7 Response curve for underdamped system

Characteristics of underdamped response (Refer Fig.1.7)

1. Overshoot: Its the ratioA/B, whereB is the ultimate value of the response and Ais the maximum amount by which the response exceeds its ultimate value.

Overshoot= A

B= exp

1 2

2. Decay ratio: It is the ratioC/A(i.e. the ratio of the amount above the ultimate value

of two successive peaks).

Decay ratio= C

A= exp

2

1 2

= (Overshoot)

2

3. Period of oscillation (T) : It is time required for one complete oscillation.

The radian frequency= =

1 2

As = 2f

T = 2

1 2

4. Natural period of oscillation: Response will oscillate continuously if= 0.

G (s) = Kp/

s j 1

s +j 1

n= 1/

Tn= 2



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5. Response time: Ultimate value will reach only at t . Therefore the time neededto reach ultimate value within 5% is known as the response time.

6. Rise time: The time required for the response to reach its ultimate value for the first

time.

*******************



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CHAPTER 2

CLOSED LOOP SYSTEMS

Generally all controlled systems are closed loop systems. Here we will discuss some of

the commonly encountered chemical processes.

2.1 What is control system?

Lets consider a simple example of hot water bath. If we have to get hot water at specific

temperature, what we need is heater, temperature measuring device and controller as an

accessory to the water bath, as shown in Fig. 2.1. Temperature of water coming out from

tank will be affected by heat input from heater, temperature and flow rate of inflow water.

Now if we consider that water flow is steady, then outflow temperature is function of inflow

temperature and heat input from heater.

Now system described in Fig. 2.1can be simplified using block diagram as shown in

Fig.2.2.

2.2 Feedback

Feedback in control system can be of two types - negative and positive. Negative feedback

is difference between set point (TR) and measured variable (Tm). Negative feedback en-sures that the difference between TR andTm is used to adjust the control element so thatthe tendency is to reduce the error. The decrease in error would cause the controller and

final control element to decrease the load.



25



34/157

26 CLOSED LOOP SYSTEMS

Temperature

IndicatorController

Power

Heater

Temperaturesensor

Supply

Figure 2.1 Water heating bath

Contrroller Final Control Element Process

Measuring Element

TR

Set point

+

Comparator

error

Tm

Measured variable

Ti, Load

++ T

Controlled

Variable

Figure 2.2 Block diagram for hot water bath system

If the signal to the comparator were obtained by addingTR andTm, we would have apositive feedback system, which is inherently unstable. Positive feedback would never be

used intentionally in the system as it cause run away.

2.3 Servo and Regulator Problem

Sometimes, instead of load change, set point is required to be changed, this situation is

referred asServo Problem. Servo Problem is need of the process. For example in heating

water, bath temperature of feed-water may be constant but hot water is required at lower

temperature then controller should be set at lower temperature. Servo problem is rare in

process industry.. Examples are missile and aircraft tracking system.

Regulator problemis common in chemical industry, where load changes and set point

remain constant. Most of the process industry problems falls under this category.

2.4 Development of Block Diagram

Lets consider same example of heating water in bath. From Fig. 2.2we can prepare block

diagram as shown in Fig. 2.3.

For simplicity we will deal with each block separately. Objective of this system is

controlling the process, therefore first we will discuss process.



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DEVELOPMENT OF BLOCK DIAGRAM 27

Gc(s) GFC(s) Gp(s)

Gm(s)

TR

Tm

T

Ti

+

++

Figure 2.3 Transfer function bock diagram for heating bath

2.4.1 Process

Heating of water bath at steady flow conditions leaves energy balance for unsteady opera-

tion. General energy balance across the process block shows that two energy input streams

and one energy output stream. Therefore heat accumulation in process can be given as,

q+ mCp(Ti To) mCp(T To) = CpVdTdt

(2.1)

At steady state, (t=0),

qs+ mCp(Tis To) mCp(Ts To) = 0 (2.2)

Subtract equation (2.2)from equation (2.1),

q qs+ mCp[(Ti Tis) (T Ts)] =CpVd (T Ts)

dt (2.3)

Using deviation variable concept

Ti= Ti TisT=T TsQ= q qs

therefore equation (2.3) becomes,

Q + mCp

Ti T

=CpVdT

dt (2.4)

Taking Laplace Transform of above equation,

Q (s) + mCp Ti(s) T(s) =CpV sT(s)

T(s) =1/mCp

s+ 1Q (s) +

1

s+ 1Ti(s) (2.5)

where,=V /m



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Servo Problem: If change inQ (t)only, Ti(t) = 0

T(s)

Q (s)=

1mCp

s+ 1 (2.6)

Regulatory Problem : If change in Ti(t)only,Q (t) = 0

T(s)

Ti(s)=

1

s+ 1 (2.7)

Equation (2.5)can be represented on block diagram as shown in Fig. 2.4.

1

s + 1

1mCp

s + 1

Ti(s)

Q(s)+

+ T(s)

Figure 2.4 Block diagram for Process

If we rearrange equation (2.5) as,

T(s) =

Q (s) + mCpTi(s) 1mCp

s+ 1 (2.8)

mCpTi(s)

Q(s)+

+ P(s)

(b)

mCp

1mCp

s + 1

T(s)

Ti(s)

Q(s) ++

1mCp

s + 1

T(s)P(s)

(a)

(c)

Figure 2.5 Alternate Block Diagram for process

Let, Q (s) + mCpTi(s) =P(s)

T(s)

P(s)=

1mCp

s+ 1

this can be represented in terms of block diagram as shown in Fig. 2.5(a).



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DEVELOPMENT OF BLOCK DIAGRAM 29

But,P(s)can be represented as shown in Fig2.5(b). Therefore, equation (2.8) can berepresented using block diagram as shown in Fig. 2.5(c), and it is same as block diagram

shown in Fig. 2.4.

2.4.2 Measuring Element

As we have discussed in earlier chapter, thermometer is first order system, here we will

consider measuring element exhibit first order dynamic lag.

(b)

Measuring TmT

Element

(a)

1

ms + 1

Tm(s)T(s)

Figure 2.6 (a) Block diagram, (b) Transfer function block diagram for measuring element

T(s)

Tm(s)=

1

ms + 1 (2.9)

where, Tm= Tm Tms

m= time constant of measuring element.

2.4.3 Controller & final control element

For convenience, the block representing the controller and the final control element are

combined into one block.

Let assume proportional controller where control action is proportional to error.

q= Kc+ qo (2.10)

where, = TR Tm

TR = set point

Kc = proportional sensitivity or gain

qo = heat input when = 0

At steady state (t=0),

qs= qo (2.11)

Whereass = 0andqois constant.Subtract equation (2.11)from (2.10)

q qs= KcUsing deviation variables,

Q= Kc (2.12)

In above equation,

= = s= (TR Tm) (TRs Tms)= (TR TRs) (Tm Tms)=

TR Tm



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Taking Laplace Transform of above equations,

Q (s) = Kc (s) (2.13)

(s) =

TR(s) Tm(s)

(2.14)

Equation (2.13) and (2.14)can be represented as shown in Fig.2.7(a) and Fig.2.7(b). There-fore overall block diagram for controller is as shown in Fig. 2.7(c)

TR(s)

+

(b)

Tm(s)

(s)

KcQ(s)(s)

(a)

(c)

TR(s)

+

Tm(s)

(s)Kc

Q(s)

Figure 2.7 (a) Representation of equation (2.13), (b) Representation of equation (2.14)on block

diagram, (c) Overall block diagram for Controller

Finally, combining Fig.2.5, Fig.2.6(b) and Fig.2.7(c) we get block diagram for complete

system (see Fig.2.8).

mCp

1mCp

s + 1

T(s)

Ti(s)

+

+

TR(s)

+

(s)

KcQ(s)

1

ms+ 1

Tm(s)

Figure 2.8 Transfer function block diagram for heating water bath



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CHAPTER 3

CONTROLLERS

In this chapter, we shall present the transfer functions for the controllers frequently used

in industrial processes. Because the transducer and the converter will be lumped together

with the controller for simplicity, the result is that the input will be the measured variable

x(e.g. temperature, level, etc.) and the output will be a pneumatic signal p.

3.1 Proportional Control

The Proportional controller produces an output signal (pressure in the case of a pneumatic

controller, current or voltage for an electronic controller) that is proportional to the error .This action may be expressed as

p= Kc +ps (3.1)

where, p = output signal from controller, psig or ma

Kc = gain, or sensitivity

= error = set point measured variable

ps = constant

The units of set point and measured variable must be the same, since the error is the

difference between these quantities.

In a controller having adjustable gain, the value of the gain Kccan be varied by movinga knob in the controller. The value ofp s is the value of the output signal when E is zero,and in most controllers ps can be adjusted to obtain the required output signal when the

control system is at steady state and = 0.



31



40/157

32 CONTROLLERS

Let consider the deviation variable P =p ps, therefore equation (3.1) becomes,P(t) = Kc (t) (3.2)

Laplace Transform of equation ( 3.2) gives transfer function of Proportional Control

P(s) (s) =Kc (3.3)

The termproportional bandis commonly used among process control engineers in place

of the term gain. Proportional band (pb) is defined as the error (expressed as a percentage

of the range of measured variable) required to move the valve from fully closed to fully

open. A frequently used synonym is bandwidth.

EXAMPLE 3.1

A pneumatic proportional controller is used to control temperature within the range

of 60 to 100F. The controller is adjusted so that the output pressure goes from 3 psi

(valve fully open) to 15 psi (valve fully closed) as the measured temperature goes for

71 to 75F with the set point held constant. (a) Find the gain and the proportional

band. (b) assuming the proportional band of the controller is changed to 75%. Find

the gain and the temperature change necessary to cause a valve to go from fully open

to fully closed.

Solution:

(a) Proportional band and Gain:

Range of measured variable (fully open to fully closed) = 60 100 = 40FChange in measured variable (fully open to fully closed) = 71 75 = 4F

proportional band= 440 100 = 10%

Gain=P

=

15 375

71

= 3psi/F

(b) For 75% proportional band(pb):Change in measured temperatureT,

T =pb (range)= 0.75 40= 30F

Gain=15 3

30 = 0.4psi/F

From this example, we see that proportional gain corresponds inversely with proportional

band; thus

proportional gain 1

proportional band

Furthermore, the relation between proportional band (pb) in percentage andKc will be

Kc = 100

pb%



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ON-OFF CONTROL 33

3.2 ON-OFF Control

A special case of proportional control is on-off control. If the gainKc is made very high,the valve will move from one extreme position to the other if the measured variable deviates

only slightly from the set point. This very sensitive action is called on-off action because

the valve is either fully open (ON) or fully closed (OFF); i.e., the valve acts like a switch.This is a very simple controller and is exemplified by the thermostat used in a home-heating

system. The bandwidth of an on-off controller is approximately zero.

3.3 Proportional-Integral Control

This mode of control is described by the relationship,

p= Kc+Kc

I

t0

dt +ps (3.4)

where, Kc = gain, or sensitivity

I =integral time, minps = constant

The values ofKcand Imay be varied by two knobs in the controller. Integralcontrollersum-up all the error values over the period equal to integral time, and corrective action is

proportional to the sum-up value. Therefore smaller the Ifaster is the action taken byintegral controller.

Lets consider unit step change in error, i.e. = 1. The response of the system can begiven as,

p (t) = Kc+Kc

It +ps (3.5)

Equation (3.5)is linear equation with intercept (Kc +ps), as shown in Fig.3.1. Usingdeviation variable, P=p ps equation (3.4) will be

P(t) = Kc (t) + Kc

I

t

0

(t) dt

Transforming above equation,

P(s) =Kc (s) + KcIs

(s)

P(s)

(s) =Kc

1 +

1

Is

(3.6)

The reciprocal ofI,integral time is also refer asreset rate.

3.4 Proportional-Derivative Control


p= Kc + KcDd

dt+ps (3.7)



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34 CONTROLLERS

1

0

0

error

t

p

0t

ps

Kc

KcI

Response

Figure 3.1 Response of PI controller for step change in error

where, Kc = gain

I = derivative time, min

ps = constant

The values ofKc and D may be varied by two knobs in the controller. Derivativecontrol operates on the rate of change of error. Opposite to integral action smallerDmeans slow derivative action. Rate of change of error is calculated during the time period

D.Using deviation variable, P =p ps equation (3.7) will be

P(t) =Kc (t) + KcDd (t)

dt


P(s) = Kc (s) + KcDs (s)

P(s)

(s) =Kc(1 + Ds) (3.8)

3.5 Proportional-Integral-Derivative Control


p= Kc +Kc

I t

0

dt+ KcDd

dt+ps (3.9)

PID controller is three parameter control where separate knobs are available to vary Kc,I, andD. Now, using deviation variable, P=p ps equation (3.9)will be

P(t) = Kc (t) +Kc

I

t0

(t) dt+ KcDd (t)

dt



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OFFSET 35


P(s) = Kc (s) +KcIs

(s) + KcDs (s)

P(s)

(s) =Kc

1 +

1

Is+ Ds

(3.10)

3.6 Offset

The curves of Fig.3.2show the behaviour of a typical, feedback control system using dif-

ferent kinds of control when it is subjected to a permanent disturbance. The value of the

controlled variable is seen to rise at time zero owing to the disturbance. With no control,

this variable continues to rise to a new steady-state value. With control, after some time

the control system begins to take action to try to maintain the controlled variable close to

the value that existed before the disturbance occurred.

With proportional action only, the control system is able to arrest the rise of the con-trolled variable and ultimately bring it to rest at a new steady-state value. The difference

between this new steady-state value and the original value is calledoffset. For the particu-

lar system shown, the offset is seen to be only 22 percent of the ultimate change that would

have been realized for this disturbance in the absence of control.

As shown by the PI curve, the addition of integral action eliminates the offset; the

controlled variable ultimately returns to the original value. This advantage of integral

action is balanced by the disadvantage of a more oscillatory behaviour. The addition of

derivative action to the PI action gives a definite improvement in the response. The rise of

the controlled variable is arrested more quickly, and it is returned rapidly to the original

value with little- or no oscillation.

No Control

PropotionalPIPID

ControlledVariable

time

Figure 3.2 Response of a typical control system showing the effects of various modes of control.



44/157

36 CONTROLLERS

1/A

A

s + 1

Y(s)

P(s)

++

X(s)

+ Kc

Figure 3.3 Proportional controller to first order system

3.6.1 Proportional control to first order system

Referring to Figure(3.3) for set point change (i.e. P(s) = 0) transfer function for thesystem is,

Y (s)

X(s)=

AKc/ ( s+ 1)

1 +

AKc( s+ 1)

= AKc

s+ 1 + AKc=

A11s + 1

where,1 =

1 + AKc

A1 = AKc1 + AKc

for unit step change in set point,

X(s) = 1/s

Y(s) = A1

s (1s+ 1)

The ultimate value of response (Y ()) can be found out from above equation using finalvalue theorem.

Y() = limt

Y(t) = lims0

sY ()

= lims0

s A1

s (1s + 1)

=A1

Offset= Set point Ultimate value of the response= 1 AKc

1 + AKc

= 1

1 + AKc



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OFFSET 37

From above equation it is clear that increase in gain of proportional controller decrease the

offset. Referring to Figure(3.3) for load change (i.e. X(s) = 0) transfer function for thesystem is,

Y (s)

P(s) =

1/ (s+ 1)

1 + AKc( s+ 1)

= 1

s+ 1 + AKc=

A21s + 1

where, 1=

1 + AKcand A2=

1

1 + AKcunit step change in load,

P(s) = 1/s

Y(s) = A2

s (1s+ 1)

The ultimate value of response (Y(

)) can be found out from above equation using

final value theorem.

Y () = limt

Y (t) = lims0

sY ()

= lims0

s A2

s (1s + 1)

=A2

Offset= Set point Ultimate value of the response= 0 1

1 + AKc

= 11 +AKc

From above equation it is clear that increase in Kc decrease the offset. The negative signindicates magnitude of response is higher than set point. If we observed servo as well as

regulatory treatment discussed here, offset is always present for proportional control.

3.6.2 PI control to first order system

1/A

A

s + 1

Y(s)

P(s)

++

X(s)

+

Gc

Figure 3.4 PI controller to first order system



46/157


47/157

CHAPTER 4

STABILITY

When response of the system grow after addition of disturbance rather than coming back

to original position then system is said at unstable condition. Mathematically response of

the system can be analysed through roots of the open-loop transfer function.

4.1 Analysing response of a system

Consider a transfer function of a system is given by equation( 4.1).

G (s) =Q (s)

P(s)=

Q (s)

(s p1) (s p2) (s p3)m (s p4) (s p4) (s p5) (4.1)

Partial fractions of above equation may be,

G (s) = C1s p1 +

C2s p2 +

C31s p3 +

C32

(s p3)2+ + C3m

(s p3)m

+ C4s p4 +

C4s p4 +

C5s p5 (4.2)

Location of polesp1,p2,p3,p4, andp5are as shown in Fig.(4.1).

Analysis from Figure(4.1)



39



48/157

40 STABILITY

Imaginary axis

Real axis

p1 p5 (p3) p2

p4

p4

Figure 4.1 Location of poles on real and imaginary space

1. Real and distinct poles: p1and p2Real poles give rise to the term C1ep1t andC2ep2t. Now p1 is negative thereforeit shows decay and make the system stable. p2 is positive root and it increase theresponse and make the system unstable.

2. Multiple real poles: p3If single pole is repeated multiple times it give rise to term such as,

C31+

C321!

t +C33

2! t2 + + C3m

(m 1)! tm1

ep3t

for above term if,

(a) p3 0as t response increases and system becomes unstable.

(b) p3 < 0 as t response decreases and system becomes stable.

3. Complex conjugate poles: p 4and p4Complex conjugates always appear in pairs, i.e. p 4 = a +jb and p4 =a jb. Thisgive rise to the term eat sin(bt + ). Where,sin(bt + ) is periodic and the terma governs the stability of system.

(a) Whena >0, as t response grow and system becomes unstable.

(b) Whena


49/157

ROUTH-HURWITZ CRITERION 41

time

output

time

output

time

output

(a) (b)

(c)

4. Poles at origin:p5

Poles at origin (p5 = 0) give rise to the term C5 i.e. constant. Therefore, system givesconstant output at any t, and reaches to new steady state level.

EXAMPLE 4.1

Analyse the system whose open-loop transfer function is,

G (s) =10Kcs 1

Solution:

The characteristic equation of the open-loop transfer function will be,

1 + G (s) = 0

1 +10Kcs 1 = 0

s 1 + 10Kc= 0root of the above equation is p = 1 10Kc The gainKc will always be greater thanzero. For stable systemp


50/157

42 STABILITY

into following polynomial form:

1 + GpGfGcGm= a0sn +a1s

n1 + + an1s + an= 0

Leta0be positive. If it is negative, multiply both sides of the equation above by 1.

Test I: If any of the coefficientsa 1, a2, , an1, an is negative, there is at leastone root of the characteristic equation which has positive real part and the corresponding

system is unstable.

Test II: If all the coefficientsa0,a1,a2, ,an1,an are positive, then from the firsttest we cannot conclude anything about the location of the roots. Form the following array

(known as the Routh array)

Row 1 a0 a2 a4 a6 2 a1 a3 a5 a7 3 A1 A2 A3

4 B1 B2 B3 5 C1 C2 C3 ...

......

...

n + 1 W1 W2 where,

A1 =a1a2 a0a3

a1A2 =

a1a4 a0a5a1

A3 =a1a6 a0a7

a1

B1 =A1a3 a1A2

A1B2 =

A1a5 a1A3A1

C1 =B1A2 A1B2

B1C2 =

B1A3 A1B3B1

Examine the elements of the first column of array (a 0,a1,A1,B1,C1, ,W1).(a) If any of these element is negative, we have at least one root to the right side of the

imaginary axis and the system is unstable

(b) The number of sign changes in the elements of the first column is equal to the number

of roots to the right of imaginary axis.

(c) If nth row is zero then at least one root is on imaginary axis.

Therefore the system is stable if all the elements in the first order column of the Rouths

array are positive number.

EXAMPLE 4.2

Check stability of the feedback control system with following characteristic equation.

s3 + 2s2 + (2 + Kc) s+ Kc/I= 0



51/157

ROOT-LOCUS ANALYSIS 43

Solution:

Prepare Rouths array

Row1 1 2 + Kc

2 2 Kc/I

3 2 (2 + Kc) Kc/I

2 0

4 Kc/I 0

The elements of first column are,

1,2,2 (2 + Kc) Kc/I

2 ,Kc/I

all are always positive except third. The system is stable if,

2 (2 + Kc)>Kc

I

4.3 Root-Locus analysis

The Routh array is open-loop analysis and do not comment on closed loop response. Routh

Hurwitz criterion requires more mathematical calculation and require to find root of the

characteristic equation for every iteration. Root locus analysis determine the stability char-

acteristics of a closed loop system as the change in controller parameters (like Kc, I,andD).

Procedure for plotting root-locus diagram

1. Count number ofpoles(n) andzeroes(m) present in transfer function.

2. Calculate number ofbranchesof root loci emerge out (n m).3. Asymptotes: Loci originate from poles and may end asymptotically. If n poles and

m zeroes are present in function then (n m) loci approach asymptotically (n m)straight lines. The center of gravity of these ( n m) loci is given by,

=

nj=1

pj mi=1

zi

n m (4.3)

These asymptotic lines make angle with real axis,

k+1 = (2k+ 1)

n m (4.4)

4. Break away point: It is a point at which two root loci, emerging from adjacent poleson real axis intersects and then leave (or enter) the real axis at an angle /2. Atintersection,

mi=1

1

s zi =nj=1

1

s pj (4.5)



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44 STABILITY

5. Point of intersection on Imaginary axis: Arrange open loop transfer function (G)in standard form

G= KN

D

where,K= constant

N= (s z1) (s z2) (s zm)D= (s p1) (s p2) (s pn)

Write characteristic equation,

1 + G= 0

1 + KN

D = 0

D+ KN= 0 (4.6)

KN

D = 1

K|N||D|

= 1 (4.7)

Using equation(4.6) prepare Rouths array. Use the Rouths criterion on n th row.

i.e. if nth row is zero then at least one pair of root lie on imaginary axis. Calculate

intersection at imaginary axis by substituting s= 0 jb in characteristic equation.6. Draw a locus connecting break away point and intersection on imaginary axis and

following branches which are approaching asymptotes.

7. Loci originating from poles end up in adjacent zero if any.

8. For (n m) 2, the sum of roots of the characteristic equation is constant, real, andindependent ofK.

EXAMPLE 4.3

Plot the root locus diagram for the open-loop transfer function,

G= K

(s+ 1) (s + 2) (s + 3)

Solution:

By examining the transfer function we can summarise,

No.of poles,n = 3 (1,2,3)

No.of zeroes,m = 0

No. of branches =n= 3

No.of asymptots =n m= 3

Center of gravity using equation ( 4.3),

=

nj=1

pj mi=1

zi

n m =1 2 3 0

3 0 = 2



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ROOT-LOCUS ANALYSIS 45

Angles between asymptotes and real axis (eq.(4.4)) is

k+1 = (2k+ 1)

n mherek= 0, 1, 2

therefore =

3 , = 3

3 , and = 5

3

1

2

3

4

1

2

34

1234 1

Imaginary

Real

Break away point: One locus will emerge from 3andfollow real axis to asymptote.Two loci emerge from 1and 2towards each other will break away at an angle 2at point(4.5),

0 = 1

s p1 + 1

s p2 + 1

s p30 =

1

s + 1+

1

s+ 2+

1

s + 3

0 = 3s2 + 12s + 11

roots of above quadratic equation are,

s= 2.5773, 1.422as break away point will be between 1and 2break away point is 1.422.

1

2

3

4

1

2

3

4

1234 1

Imaginary

Real

1.422



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46 STABILITY

Point of intersection of two loci breaking away from real axis can be found out using

Rouths array,

D+ KN= 0

(s + 1) (s + 2) (s+ 3) + K+ 1 = 0

s3 + 6s2 + 11s + K+ 6 = 0

Routh s array,

Row 1 1 11

2 6 K+ 6

3 6 11 (K+ 6)

6 0

4 K+ 6

If(66 (K+ 6)) /6 = 0 the one root will be on imaginary axis, therefore whenK = 60loci will intersect imaginary axis. Point of intersection of imaginary axis iss= a +jb ands= a jb wherea= 0.Ifs= jbandK= 60, the characteristic equation becomes,

jb3 6b2 + 11jb + 66 = 066 6b2 +j 11b b3 = 0

but we know that real part is zero(i.e. a = 0),

66 6b2 = 0b = 3.32

similarly fors =jb, b =3.32 now as we know break away point and point ofintersectionwith imaginary axis we can draw approximate loci as shown in figure(4.2).

1

2

3

4

1

2

3

4

1234 1

Imaginary

Real

1.422

Figure 4.2 Root locus diagram for example(4.3)



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FREQUENCY RESPONSE ANALYSIS 47

4.4 Frequency response analysis

Consider a complex number, C= a +jb where,ais real part ofCand bis imaginary partofC. The modulus or absolute value or magnitude ofCis represented by |C| and definedby

|C| = a2 + b2 (4.8)The phase angle or argument ofCis represented by Corarg (C)and defined by

C= = tan1

b

a

(4.9)

Real

Imaginary

C= a+jb

a

b

|C|

Figure 4.3 Magnitude and direction of imaginary number

From fig.(4.3),

cos = a

|C| sin = b

|C|a = |c| cos b = |c| sin

therefore the complex number can be written as,

C= |C| cos +j|c| sin

We know that,

cos x=ejx + ejx

2 sin x=

ejx ejx2j

Then

C= |C|ej + ej

2 +j|C|e

j ej2j

= |C|ej + ej + ej ej

2

= |C|ej (4.10)LetD = a jb a conjugate ofCthen,

|C| = |D| and D= CConsider first order system with transfer function,

G (s) = Kpps + 1



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FREQUENCY RESPONSE ANALYSIS 49

AR= |G (j) | =

a2 + b2 =Kp

(4.15)

and phase shift,

= G (j) = tan1 (b/a)

= tan1

Kp/0

= tan1 ()

= 90 or /2 (4.16)

4.4.2 Frequency response of pure dead time process

The performance equation of dead time process is,

Y (t) =X(t d)

Taking Laplace transform of above equation,

L [Y (t)] = L [X(t d)]=esdLX(t)

Y (s) =esdX(s)

G (s) =esd

puts= j,

G (j) = ejd

comparing above equation with equation( 4.10) i.e.G (j) = |G (j) |ej

AR= 1 (4.17)

= d (4.18)

4.4.3 Frequency response of second order system

Transfer function of second order system as discussed in section( 1.9) can be given as,

G (s) = Kp

2s2 + 2 s+ 1

puts= j,

G (j) = Kp

1

22 +j2

1 22j2 (1

22)

j2

=Kp

1 22j2

(1 22)2 + (2 )2

= Kp

1 22

(1 22)2 + (2 )2j 2Kp

(1 22)2 + (2 )2



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BODE DIAGRAM 51

The frequency equal to1/pis calledcorner frequencywhere, two asymptotes inter-sect.

Deviation of AR at corner frequency is maximum.

Phase shift can be plotted for wide range from 0to . Phase lag changesfrom0

to 90

.

1

p

p

1

AR

1

p

0

45

90

Figure 4.4 Bode digram for first order system

Pure capacitive system: Amplitude ratio and phase shift equations for pure capacitive

system are,

AR=Kp

= 90

= 1

1AR

1

0

90

180

Figure 4.5 Bode digram for pure capacitive system



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52 STABILITY

4.5.1 Bode stability criterion

A control system is unstable if its open-loop frequency response exhibits an AR greater

than unity at the frequency for which the phase shift is 180. This frequency is termedthecross-over frequency(co).

Gain Margin(GM): It is a measure of difference betweenAR = 1and AR at cross-overfrequency.

GM= 1

ARco(4.21)

If Gain Margin is less than unity then system is unstable.

Phase Margin (PM): Degree of stability is expressed using phase margin.

P M= 180

phase lag in degrees for which the AR is unity

A negative phase margin indicates an unstable system.

Gain Margin greater than 1.7 and phase margin greater than30 is considered to be safefor control design.

1

AR

0

180

360

Gain Margin

Phase Margin

Figure 4.6 Measuring Gain Margin and Phase Margin on Bode Diagram

EXAMPLE 4.4

A control system using PI control is represented by the block diagram shown in figure

below. The transfer function describing the various blocks are as shown with Kc = 10,I= 1min,K1 = 0.8, andKv = 0.5. Determine the gain and phase margins.



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BODE DIAGRAM 53

Kc

1 +

1

Is

Kv

1

(s2 +s+ 2)(5s+ 2)

exp(0.8s)K1

R +

U

++ C

B

Solution:

Open-loop transfer function is

G (s) = Kc

1 +

1

Is

Kv

1

(s2 + s+ 2)(5s + 2)e0.8sK1

=

1 +

1

s

e0.8s

(0.5s2 + 0.5s + 1)(2.5s + 1)

= 1

2.5s + 1 1

0.5s2 + 0.5s + 1 e0.8s 1 +

1

s=G1G2G3G4

AR= AR1AR2AR3AR4

log AR= log AR1+ log AR2+ log AR3+ log AR4

and= 1+ 2+ 3+ 4

For composite graph slope of graph is summation of slope of individual graphs.

1) Consider first block,G1

G1 = 1

2.5s + 1Its a first order system therefore, amplitude and phase shift will be,

AR1 = 1

6.252 + 1 1 = tan1

(2.5)and corner frequency,

c1 = 1

=

1

2.5= 0.4rad/min

First order system has two asymptotes low frequency asymptote (LFA) and high fre-

quency asymptote (HFA),

when 0LFA will beAR1 1when HFA will beAR1 1log(2.5)

2) Consider second block,G2

G2 = 1

0.5s2 + 0.5s + 1

Its a second order system with= 1/2and= 1/ 22 therefore, amplitude andphase shift will be,

AR2 = 1

(1 0.52)2 + (0.5)22 = tan

1

0.51 0.52



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54 STABILITY

and corner frequency,

c2 = 1

=

2 = 1.414rad/min

Second order system has two asymptotes low frequency asymptote (LFA) and high

frequency asymptote (HFA),

when 0LFA will beAR2 1when HFA will beAR2 2log(0.707)3) Consider third block,G3

G3 = e0.8s

Its a pure dead time system therefore, amplitude and phase shift will be,

AR3 = 1 3 = 0.8

and no corner frequency as pure dead time system has only one asymptote

for = 0to AR will beAR3 = 14) Consider fourth block, G4

G4 =

1 +

1

s

Its a PI controller. Lets first find the equation for AR and phase shift for PI controller.

Puts= j in G4,

G (j ) = 1 + 1

j =

1 +j

j

=1 +j

j j

j

= 1 j1

AR4 =|G (j)

|= 1 + 1

2

= tan1

1

PI controller has two asymptotes low frequency asymptote (LFA) and high frequency

asymptote (HFA),

when 0LFA will beAR4 1log when HFA will beAR4 0

Summarising above calculations we can break complete Bode diagram into four seg-

ments base on corner frequencies as tabulated below.

in rad/min Slope of block Slope of Composite curve

1 2 3 40 to 0.4 0 0 0 -1 -1

0.4 to 1.0 -1 0 0 -1 -2

1.0 to 1.414 -1 0 0 0 -1

1.414 to -1 -1 0 0 -2



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BODE DIAGRAM 55

From above table we can plot Bode diagram for AR vs. frequency. From overall slope

it is clear that AR value reaches asymptotic values for low as well as high frequency

therefore select definite value for low frequency e.g. 0.01 and for high frequency e.g.

1000 and plot the graph.

0.01 0.1 1 10 100

0.1

1

10

100

AR

frequency for

phase margin

Similarly we can calculate phase shift for individual block and then for overall.

in rad/min 1 2 3 4

0.01 -1.43 -0.29 -0.01 -89.43 -91.16

0.40 -45.00 -12.30 -0.32 -68.20 -125.82

1.00 -68.20 -45.00 -0.80 -45.00 -159.00

1.41 -74.20 -90.00 -1.13 -35.30 -200.40

10.00 -87.70 -174.20 -8.00 -5.70 -275.60

100.00 -89.80 -179.40 -80.00 -0.57 -349.87

0.01 0.1 1 10 100

270

180

90

0

360

co



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56 STABILITY

4.6 Nyquist plot

A Nyquist plot is an alternate way to represent the frequency response characteristics of a

dynamic system. This is a plot of an imaginary part ofG (j)versus real part ofG (j).

Real

Imaginary

X

a1

b1

= 0

AR

1 1

= 1

Figure 4.7 Nyquist plot

Referring to Fig.(4.7), if point X is at = 1 then the distance between origin andpoint X is AR at = 1.

G (j) = a1+jb1

AR1 =

a21+ b21 = distance between X and origin

1 = tan1

b1a1

=angle between real axis and line joining X origin

The angle1 with the real axis is phase shift at = 1. The Nyquist plot contains the

same information as the pair of Bode plots for the same system.

4.6.1 First order system

As discussed in section(4.4)equations for AR and phase shift is same with Kp = 1,

AR= 11 + 22

= tan1 ( )

When = 0Amplitude Ratio,AR= 1and phase shift, = 0. Therefore the Nyquistplot originate on the real axis at a distance from origin equal to 1. When thenAR0 and 90. Therefore the end of the Nyquist plot is at the origin where thedistance from it is zero.

Since for every intermediate frequency0< AR < 1 and 90 <


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NYQUIST PLOT 57

1

1

11 Real

Imaginary

= 0

1

1

11 Real

Imaginary

= 0

1

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