1

Subject: Computer control of processes

Code: EI414

Scheme &Solutions (November-2016)

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Q.No:1

(a) to (l) carries 12 marks, one mark for each question

Q.No:2

(a) Diagram -3M, Explanation-3M

(b) Controllaabity-2M, Obsevabity-2M, Example- 2M

Q.No:3

(a) Hardware requirements-3M, Software requirements-3M

(b) Conversion Procedure-4M, Example-2M

Q.No:4

(a) First order system without delay-2M, First order system with delay-2M

(b) Necessary conditions-2M, Jury’s table formation-4M, Sufficient conditions-2M

Q.No:5

(a) Stability in Z-domain-4M

(b) Importance of digital controllers-3M, Dahlin’s specifications-2M, Controller function-3M

Q.No:6

(a) Diagram-3M, Explanation-5M

(b) PLC-ladder programming explanation-2M, Basic Ladder instructions-2M

Q.No:7

(a) Diagram-2M, Explanation-2M, Applications-2M

(b) Configuration-3M, Communication facilities-3M

Q.No:8

(a) Steps-3M, Steps with example-3M

(b) Diagram-3M, Explanation-3M

Q.No:9

(a) Expert system-1M, Explanation with digrams-5M

(b) AI definition-1M, Explanation with diagram-5M

3

1. Answer all questions

Each question carries one mark 1x12=12M

(a) Controllability:

A control system is said to be completely controllable if it is possible to transfer the state from any

arbitrary initial state to any desired state in a finite time period

(OR)

A control system is controllable if every state variable can be controlled in a finite time period by some

unconstrained control signal

(b) State transition matrix (STM):

It is defined as the matrix which satisfies the linear homogeneous state equation and is given by the

expression Atet )( for Continuous time system, kAk )( for discrete-time system

(c) Elements of computer aided process control:

Computer and control algorithm

DAC & Hold devices

Sampler & ADC

Final control element (Control valve)

Process and Measuring elements

(d) System Modelling:

Description of the system with a set of mathematical equations is referred as system modelling

(e) Use of Modified Z-transforms:

To obtain the values of the response between sampling instants

To analyze sampled data control systems containing transportation lag. i.e. dead-time

(f) Representation of pure delay element:

stdeAsM

sY .

)(

)( td = dead-time

(g) Programming languages of PLC:

Low level languages: Ladder programming, Boolean mnemonics

High level languages: Functional blocks, English like statements

4

(h) Advantages of DCS:

Incremental system growth

Inexpensive

Excellent computing power

Flexible & Reliable

Up gradation cost is less

(i) PLC SCAN:

The process of reading the inputs, executing the control application program and updating the output is

known as PLC SCAN

(j) System Identification:

The determination of a mathematical model of a system is known as system identification

(k) * Suitable control system to control the dynamics of aeroplane:

Optimal control (* Denotes out of syllabus)

(l) *Application of Fuzzy controller:

To control a large class of processes which may be time-varying, non-linear with variety of disturbances

(* Denotes out of syllabus)

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UNIT-I

2. (a) With neat sketch, explain the functional block diagram of a computer control

System 6M

The computer receives the measurement from the process at discrete times and

Gives control signal value also at discrete –times

We need a Hardware element DAC between digital computer and process

Sampler produces a sequence of sampled values at particular time instants

An ADC converts the value in to digital value

The control commands produced by the computer program should be converted

from discrete-time to continuous- time signals. This is achieved by ZOH

(b) Differentiate the controllability and observability of a system with an example 6M

A control system is said to be completely controllable if it is possible to transfer the

state from any arbitrary initial state to any desired state in a finite time period

Controllability: The necessary and sufficient condition for the system be completely

Controllable is that n x n controllability matrix (U) has rank equal to n

i.e. ].........,[;)( 1gFFggUnU n

Observability: The system is said to be observable if every initial state i.e. x (0)

can be determined from observations of y(k) over a finite number of sampling periods

The necessary and sufficient condition for the system to be completely observable is that

n x n observability matrix (V) has rank equal to n

Computer

& Control

Algorithm

DAC

Hold

Control

valve

Process

Measuring

Element

Sampler

&ADC

Set point

6

i.e.

1

2

:

;)(

nCF

CF

CF

C

VnV

Example:

)(11)(

)(1

0)(

10

11)1(

kxky

kukxkx

Here 2;11;1

0;

10

11

nCgF

01

11

10],[

UU

FggU

i.e. Rank of U is equal to 2, so the system is controllable

01

01

11

VV

V

i.e. Rank of V is equal to 2, so the system is observable

(OR)

3. (a) Enumerate the hardware and software requirements for computer based process

control 6M

Hardware requirements:

Multiplexers & De Multiplexers

ADC, DAC&Latches

Computer

Mass storage, operator console, Alarms

Line printers, CRTs...etc

Software requirements:

For the computer to perform controller functions the error must be determined and the

controller mode equations solved to determine the necessary output to the final control

element.

The above said operations are performed by the software

The computer must have program to solve the equations of P, I, D modes.

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Nature of the software:

Soft ware format: The particular format of the software used for the computer controller

depends on the computer and its programming language

Examples: flow chart, Machine language, Assembly language

Input data operations: The first phase for software involved in the computer based

controllers to input the data and put them in to a suitable for controller mode evaluation

Controller modes: The operation of controller action is entirely taken over by software

with in the computer i.e. by programs

(b) Describe the conversion of state variable representation to transfer function model,

with an example 6M

State Model is combination of State and output equations. For continuous-time system the state

model is given by: )2(.........)(.)()(

)1(......)()()(

tuDtCxty

tButAxtx

State Model is combination of State and output equations. For Discrete-time system the state

model is given by: )4(...........)(.)()(

)3(......)()()1(

kuDkCxky

kgukFxkx

For continuous-time system we use Laplace transforms and for discrete-time system we use

Z-transforms

For Continuous-time system state model:

Take Laplace transform on both sides for equations (1) & (2) we get the following expressions:

DBAsICfunctiontransfer

sU

sY

sDUsBUAsICsDUsCXsYtuDtCxLTtyLT

sBUAsIsXsBUsAXssXtButAxLTtxLT

1

1

1

)()(

)(

)()()()()()()}(.)({)}({

)()()()()()()}()({)}({

(or)

For Discrete-time system state model:

Take the Z-transform on both sides for equations (3) & (4) we get the following expressions:

DgFzICzU

zY

zDUzgUFzICzDUzCXkuDkCxZkyZ

zgUFzIzXzgUzFXzzXkgukFxZkxZ

1

1

1

)()(

)(

)()()()()(})(.)({)}({

)()()()()()(})()({)}1({

Note: Take your own example and obtain the transfer function

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UNIT-II

4. (a) Explain the mathematical modelling of I order system with and without pure

delay 4M

The transfer function for the first order without delay continuous-time system is given by

the expression

)1(1)(

)(

s

k

sM

sY

The equation (1) is rearranged as follows

)2()()()(1)(

)(skMssYsY

s

k

sM

sY

Take the inverse Laplace transform for the equation (2)

)3()()()}({)}()({ 11 tkmdt

dytyskMLssYsYL

Equation (3) is the differential equation that describes the continuous-time I order system

Conversion of Continuous-time system into Discrete-time system:

)()(

1()(1)(

)()(

nmtm

nynyTdt

tdy

nyty

With the above replacements equation (3) becomes:

TnmkT

nyny

nmT

kTny

Tnynkmnyny

Tny

);()1()(

)()1()()(1()(1

)(

The transfer function for the first order with delay continuous-time system is given by the

expression

)1(1)(

)(

s

ke

sM

sY std

The equation (1) is rearranged as follows

)2()()()(1)(

)(sMkessYsY

s

ke

sM

sY stst

d

d

Take the inverse Laplace transform for the equation (2)

)3()()()}({)}()({ 11 dst

ttkmdt

dytysMkeLssYsYL d

Equation (3) is the differential equation that describes the continuous-time I order system with

delay

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Conversion of Continuous-time system into Discrete-time system:

)()(

1()(1)(

)()(

od nnmttm

nynyTdt

tdy

nyty

With the above replacements equation (3) becomes:

TnnmkT

nyny

nnmT

kTny

Tnynnkmnyny

Tny

o

oo

);()1()(

)()1()()(1()(1

)(

(b) Using Jury stability criteria check the stability of the sampled data control

system represented by the characteristic equation:

Z4 -1.368Z3+0.4Z2 +0.08Z+0.002 = 0 8M

F (Z) = Z4 -1.368Z3+0.4Z2 +0.08Z+0.002 = 0

For the given problem n= 4

Number of rows in the Jury’s table: 2n-3= 2(4)-3 = 5

Number of sufficient conditions = n-1 = 4-1= 3

Necessary conditions:

F (1)>0 and (-1) n F (-1)>0

F (1) = 1-1.368+0.4+0.08+0.002 = 0.114

Therefore F (1)>0

(-1) n F (-1) = (-1)4[1+1.368+0.4-0.08+0.002] = 2.69

Therefore (-1) n F (-1) >0

Both the necessary conditions are satisfied

Sufficient conditions:

These are obtained from the Jury’s table. For the given problem the Jury’s table is given

Below:

Row z0 z1 z2 z3 z4

1 0.002 0.08 0.4 -1.368 1

2 1 -1.368 0.4 0.08 0.002

3 -0.9999 1.368 -0.3992 -0.0827

4 -0.0827 -0.3992 1.368 -0.9999

5 0.9929 -1.4008 0.5122

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The sufficient conditions for the given problem:

True

True

True

5122.09999.05122.09999.0

0827.09999.00827.09999.0

1002.0

Both the necessary and sufficient conditions are satisfied. So the given system is stable

(OR)

5. (a) Write short note on stability in Z-domain 4M

The following figure shows the Block diagram of the closed loop sampled data control

system

The transfer function of the closed loop sampled data control system is given by the

expression:)(1

)(

)}()({1

)}({

)(

)(

zGH

zG

sHsGZ

sGZ

zR

zC

The characteristic equation for the system is: 0)(1 zGH

If the roots of the characteristic equation (poles) are inside the unit circle the system is

stable

If the roots of the characteristic equation (poles) are outside the unit circle the system is

unstable

If the roots of the characteristic equation (poles) lies on the unit circle the system is

marginally stable

G(s)

H(s)

r (t) c (t)

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(b) Explain the importance of digital controllers and derive the equation for Dahlin’s

Algorithm 8M

Importance of digital controllers:

Integral part of the sampled data control system

Digital controllers are easily programmable and so they are more versatile

Low cost, High speed and High accuracy

Digital coded signals can be easily stored transmitted, processed as desired

The system performance can be modified as desired

Dahlin’s Algorithm:

Pulse transfer function: )()(1

)()(

)(

)(

zDzHGp

zHGpzD

zR

zY

D (z) is controller transfer function. From the above equation controller transfer function is

obtained and is given by:

)()(

1

)()(

)(

1)(

zRzY

zRzY

zHGpzD

For Dahlin’s Algorithm the closed loop response is first specified in continuous-time

domain and then the corresponding behaviour in discrete-time domain

The closed loop response desired is given by that of fist order with time delay process

)1(

)(1

)(1)(

)(

ss

esY

ssRand

s

e

sRsY

ss

Here µ is time constant and ϴ is the dead time of the response. For these

specifications the controller transfer function

/

)1(1

)1(

1

1)(

1)(

t

k

k

eb

ba

azbz

az

zHGpzD

UNIT-III

HGp(z) D(z) R (z) Y (z)

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6. (a) With a neat sketch, explain the basic elements of distributed control system (DCS) 8M

DCS is Layered structure. Here each layer consists of group of tasks to be performed on the lower

layer by

getting instructions from the higher level

Level-1:

It is known as Field level

It consists of specially designed dedicated controllers.

The functions of this level are: Emergency shutdown, Provides manual backup when other

layers are failed

Level-2:

It is area control level. It consists of a direct digital control link. Functions of this Level are:

Process data collection

Message for alarm signals

Open loop and Closed loop control

Hardware test

Level-3: It is known as plant control level. It consists of supervisory control link

Functions of this level are:

Optimal control

Adaptive control

Level-4:

It is known as plant management level. Functions of this level are:

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Resource scheduling

Production planning

Management functions related to process or production management.

(b) Write short note on PLC Ladder programming 4M

It is one of the popular programming language for PLC’s.

It consists of symbols for specific functions.

This approach consists several Ladder instructions for various operations

Basic Ladder Instructions:

1. Normally open contact:

2. Normally closed contact:

3. Turn on output:

4. Turn off output

(OR)

7. (a) With the help of neat schematic, explain the principle of PLC. State the applications of PLC.

Power Supply:

Provides the voltage needed to run the primary PLC components

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I/O Module:

Provides signal conversion and isolation between the internal logic- level signals inside the PLC

and the field’s high level signal.

Processor:

Provides intelligence to command and govern the activities of the entire PLC systems.

Programming device:

Used to enter the desired program that will determine the sequence of operation and control of

process equipment or driven machine.

PLC Applications:

PLC’s are used in many industries for one or more of the functional areas.

Sequence control, timing, counting, and data calculation.

Quick change of machine or process logic, to manufacture different items using the same

machine or process equipment.

Auto compilation of production/consumption/Down time/Maintenance data.

Batch or Continuous process control.

Open loop or feedback control, process data acquisition and display

Precise motion/Position control

(b) Write short note on Configuration and Communication facilities for DCS 6M

Configuration:

Distributed control systems are of 2 types: (i) Bus based DCS (ii) Non-Bus based DCS

For the Non-Bus based DCS, the configuration is done by the manufacturer. For the Bus-based

DCS the configuration is done via key board through an operating system

Configuration of DCS consists of two Steps:

Step-1: Configuration of operating system

Step-2: Configuration of controller functions

Step:1 The configuration of operating system is done to perform certain tasks such as:

Composition between group and overview displays

Period for trends

Message tables

High way station priorities

Step: 2 Configuration of controller function is carried out to obtain complete control strategy. The

configuration is obtained via software modules which are stored in the memory. These modules

are called as Algorithms. Linking of these algorithms is performed via addresses. This is known

as Software wiring.

Communication facilities:

The communication paths are essential for DCS

The path must be implemented from top to the bottom of the system, to transfer any necessary

control action down to the individual final elements or to collect the actual information from

them

Short and long distance buses will be used in a combination to cover a big area of distributed

plants like steel, chemical, mining industry

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For short distances, the star connection is applicable which usually combined with bus and

point-point connection will result in to a mesh network.

At level-1 usually point-point connection will be applied to connect individual sensors and

actuators to analog and digital inputs and outputs of the local field station.

Recent advance for this is PROFIBUS (process field bus).

For shorter distances an instrumentation bus IEEE-488 can be applied.

UNIT-IV

8. (a) Explain the basic steps in System identification 6M

System Identification:

The determination of a mathematical model of a system is known as system identification.Mathematical

models deals with Linear Time invariant Single input Single output discrete time System is described

by difference equation. The proposed method is for estimating he coefficients of difference equations

A set of “N” linear algebraic equations are formulated “N” is the number of measurements.

Derive the canonical equation whose solution yields the Parameter estimate of θ (N) ϴ is the vector

parameter

under identification

Estimate of initial condition of the dynamic equation is required.

(N+n measurements are taken and N+n equations are formed where n is the order of the difference

equation).

Example:

Consider the simple first order system

y(k) + a1 y(k-1) = b1u(k-1)

Definition of the problem:

For the given input sequence u(-1) , u(0) , u(1), u(2) , -----, u(n-1)

The output is y(0), y(1), y(2), …. , y(n)

Find out the parameter a1 and b1 and the initial condition y(-1)

Step: 1

y(0) + a1y(-1)= b1 u(-1)

y(1) + a1y(0) = b1 u(0)

Y(2) + a1y(1) = b1 u(1)

.

.

.

.

y(N) + a1y(N-1) = b1u(N-1)

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The equation 1 is used to determine the initial condition and remaining equations are used to estimate

the unknown a1 and b1.

Representing the above equations in matrix form

The equation represents the set of N linear algebraic equations for N measurements with unknowns a1 and

b1 .

Suppose an error occurs while taking the measurements, then input and output relation becomes

Y = Φθ+e error = e=y- Φθ

Cost function (J):

J = eTe

Where e is the N dimensional error vector

Step: 2

Determination of canonical form of the equation:

We know that e = y-Φθ

J = (y – Φθ)T (y-Φθ)

(y – Φθ)T = (y-Φθ) (y-Φθ) is a similar matrix

J = (y – Φθ)2

Step: 3

To determine the solution of canonical form of equation:

)(..)2()1( NeeeeT

)1()1(

)1()1(

)0()0(

,,,,&

)(

.

.

)2(

)1(

3

1

1

NuNy

uy

uy

b

a

Ny

y

y

y

y

)(

.

)3(

)2(

)1(

Ne

e

e

e

e

)](.......)3()2()1([ 2222 NeeeeJ

N

k

keJ

1

2 )(

)(0

)()(

)(2

))((2

formcanonocalyJ

AA

y

yJ

TT

TTT

T

17

Step: 4

Initial condition estimation:

From (1) we have got

y(0) + a1y(-1) = b1u(-1)

y(-1) = [b1u(-1) – y(0)] / a1

Where it is assumed that condition a1 ≠ 0 & u(-1) is a known

(b) Write about Self tuning regulator with an example 6M

In this approach the two functional blocks are used: (i) Parameter estimator (ii) Parameter adaptor

The parameter estimator continuously estimates the system parameters based on the system input

N

k

N

kT

N

k

N

k

N

k

N

kT

TT

kuky

kyky

y

kukuky

kukyky

y

1

1

1

0

21

0

1

0

1

0

2

1

)1()(

)1()(

)()()(

)()()(

)(

18

and output.

The parameter adaptor continuously adjusts the control parameters so that the performance of Indies

is maximized.

The following method are used for parameter estimation

Least square method

Method of instrumental variables

Maximum likelihood method

Stochastic approximation method

(OR)

9. (a) Describe the role of expert system in process control 6M

Expert or knowledge based systems are artificial intelligent systems that capture the expertise of

Human experts who are knowledgeable in a particular application domain. The information presented

to expert system may be

Functional

Experimental

Incomplete

Judgmental

Speculative

Uncertain

Fuzzy

Intuitive

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Expert controller:

Expert system should satisfy the following to use it in process control:

Ability to control a large class of processes which may be time varying, non-linear,

with variety of disturbances.

Requirement of minimum prior knowledge about the process.

Ability to improve its performance with time as it is required more and more

knowledge

Ability to monitor the performance of the system and detect problems with sensors

actuators,

and other components

(b) Explain about artificial intelligence and its applications 6M

Artificial intelligence:

May be defined as the branch of computer science which deals with the software and hardware

techniques

to solve symbolic problems as against the problems solved by EDP machines.

Difference in the working of conventional EDP system and AI system:

20

Artificial intelligence systems can understand screw driver and relate it to screw or conveyer

belt. These systems can understand the meaning of scheduling and relate to operation sequence.

AI programs contain information like data, collection of separate words or characters and

knowledge. Like inter connection of words, structure and relation map. The three outputs of AI

system are Opinions, analysis and recommendation.

AI systems are divided as follows:

Natural language systems

Perception system for vision, speech, and touch

Expert or knowledge based system

Natural language systems:

AI systems that should be able to understand the natural languages as humans being understand.

Such systems have easy means of communication with computer.

Perception systems for vision, speech, and touch:

They can interpret visual scene or make inference about quality or physical orientation of the object.

Vision system of robot falls into this category.

21

Expert or knowledge based systems:

These are artificial intelligent systems that capture the expertise of human experts who are

knowledgeable

in a particular application domain. The information presented to expert system may be

Functional

Experimental

Incomplete

Judgmental

Speculative

Uncertain

Fuzzy

Intuitive

Expert system should satisfy the following to use it in process control:

Ability to control a large class of processes which may be time varying, non-linear,

with variety of disturbances.

Requirement of minimum prior knowledge about the process.

Ability to improve its performance with time as it is required more and more

knowledge

Ability to monitor the performance of the system and detect problems with sensors

actuators, and other components

22

General structure of the expert system:

23

Interface engine: It tries to determine the knowledge which solves the problem in hand.

Explanator: Explain the reason for concluding the particular solution. The explanation given by the

explanator may be reviewed by the human expert. If the human expert does not accept the conclusion

then he provides some more knowledge in terms of rules, facts, and heuristics.

Learner:

Learner accepts the new knowledge and stores in its data base.

Solution reviewer:

It relooks into the problem in hand and tries to find another solution

Note: The process of reasoning, solution explanation, learning, and solution review may be repeated till

the human being accepts the Solution determined by interface engine.

24

Scheme &Solutions of EI 414

Prepared By:

P. Vinodh Babu

Associate Professor

Department of EIE

Bapatla Engineering College

Bapatla-522102