3-1 Control System Presentation

8/6/2019 3-1 Control System Presentation

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Control System Topics SLIDE NO-introduction to control systems 3

- feedback in control systems (click on any line to read the contents)19

Laplace transforms 41-Transfer Functions and its properties 80

-Mason's Gain Formula 89

-Time Domain Analysis and steady state errors 113

- Concept of BIBO and Routh Harwitz criterion for stability 127

-State variable Analysis 156

An introduction to controllers 169

- Root locus technique 193

-Frequency Domain Analysis(polar and Bode plot) 206

-Digital Control Systems 238

Z - Transforms 261

1


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D.C. MOTOR SPEED CONTROL 272Compensation 327

Control system in Biotechnology 350

2


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INTRODUCTIONTO

CONTROLSYSTEMSBY

V DINESH BHARGAV ( 07359 )MECHANICAL ENGINEERING


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The following presentation consists of:What is control system?

History of the control systems,Introductory concepts and terms,Types of control systems,Comparison between open loop and

closed loop control systems,Transfer function and its properties,Types of signals and

Examples.


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WHAT IS CONTROL SYSTEM?

Control system is an amalgamation orcombination of different physical components relatedor connected together in such a manner to regulate,

direct , control itself to obtain desired output.In the following slides the terms used in

control system are explained.


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HISTORY OF CONTROL SYSTEMS:-Historically, the wide practical applications of control

systems first took place in the process areas likepetroleum processing, etc. Most were developed by thesuccessful realization by the intuitive and experimentallyoriented engineering methods.

The first known control system was the centrifugal

governor used for speed control of steam engine inventedbyJames Watt.

Later, people like Minorsky, Hazen and Nyquistcontributed to the development of the control systems.

In the year 1922, Minorsky invented automatic

controller for the ships.In the year 1932, Nyquist developed the criterion forstability of the closed loop systems.


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In the year 1934, Hazen developed the theory of servomechanism, which is a comprehensive mathematical theoryaimed at applications where in there is need to control

mechanical motions viz., displacement, velocity, acceleration.Around the second world war, the technical needs of militarylike automatic airplane pilots , radar antennae etc, led to thescientific approach in this field.

In the year 1940, Bode developed bode plot which made

linear closed loop control systems that satisfied performancecriteria possible.

In the year 1950, Evans developed root locus. Between 1960to 1980, the control of deterministic and stochastic controlsystems were achieved.

By the year 1980, research was being done modern controlsystems and digital control systems. From the year 2000,developments were made in the field of micro and nanocontrol systems.


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WATTS CENTRIFUGAL GOVERNOR


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INTRODUCTORY CONCEPTSAND TERMS:-System:

It is an arrangement or combination of different physicalcomponents that are connected or related together toform an entire unit to achieve a certain objective.

Control:

It is the process of regulation, direction and tracking.

Plant:

It is the portion of the system which needs to becontrolled, also called as controlled system.

Process:

It is a natural or artificial, voluntary progressing,continuous operation or development marked by gradualchanges that succeed one another in relatively fixed wayto obtain a desired result.


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Input:

It is a signal or excitation that is applied to a control system to get aspecified result. It is of two types:

Manipulated input: Input which is subject to our control. Disturbance input: These are undesirable effects beyond our control

generated from surrounding environment.

Output:

It is the actual response from the control system due to theapplication of specified input .

Disturbance:

It is any signal that has an adverse impact on the system.

Controller:

It is the element of the system itself or an external agency whichcontrol the plant or process.

The words controller, regulator, tracking system are used indifferent situations.


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o Controller:

The primary role of the controller is the clevermanagement of manipulated variables so as to

counteract the effects of disturbance for the specifiedoutput.

o Regulator:

When the desired value of the controlled outputs is

more or less fixed and the main problem is to rejectdisturbance effects, the control system is called aregulator.

o Tracking system:

In the follow up or tracking system , the controlledoutput is required to follow or track a time varyingcommand input.


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Command input

Controller

Process

Manipulated variable

Controlled output

Disturbance

Block diagram representing various parts of control systems


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TYPES OF CONTROL SYSTEMS:-Based on various parameters, control systems areclassified in different ways as follows:

1.Open loop and closed loop control systems:

The systems in which output doesnt have any impact onthe system are called open loop control systems.

The systems which maintains a prescribed relationshipbetween the output and the reference input by comparingand using the difference as a means of control usingfeedback is closed loop control system.

Feedback:

The arrangement of a system by which output is compared

with a input to reduce error to make system more stable iscalled feedback.


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Controlled

outputCommand

inputController Process

Manipulatedvariable

DisturbanceOpen loop control systems:

Commandinput

Controller Process

Manipulatedvariable

Controlledoutput

Disturbance

MeasurementFeedback signal

Closed loop control systems:

2 Stochastic and deterministic control systems:


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2. Stochastic and deterministic control systems:

If the output is known from input or externaldisturbances it is known as deterministic controlsystems.

If the output is unpredictable it is known as stochastic

control system.3. Linear and non linear control system:

Linear control system is the one which obeys theprinciple of superposition and homogeneity.

i.e., f( x+y ) = f( x ) + f( y ) ; f( kx ) = k f( x )

Non linear control system is the one which does notobey the above principle. Linearization over desiredranges of operation is possible.

4. Time varying and time invariant control systems:

When the parameters of the model vary with time it is

called time varying control system. Eg: mass changesshould be considered in guidance of rocket systems.

If the parameters of the model doesnt vary with time ,they are called time invariant control systems.


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5. Distributed and lumped parameter models:

In a distributed parameter models, the variables in a systemare distributed in space and vary with both the spatialcoordinates and time. These models consists of partialdifferential equations w.r.t time and space.

In lumped parameter models, the matter is assumed to belumped at some discrete points of space. These models usedifferential equations w.r.t time.

6. SISO and MIMO systems:

If the system has only single input and single output , it iscalled single input single output system.

If the system has multiple inputs and outputs, it is calledmultiple input and multiple output system.

7. According to application:

the control systems may be electrical, mechanical, thermal,pneumatic, business, managerial control systems.

COMPARISON BETWEEN OPEN


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COMPARISON BETWEEN OPENLOOP AND CLOSED LOOP CONTROLSYSTEM:-

Depending on the application of control system, one mayeither use open loop control system or closed loop control

system. The selection of the control system is also based upon

sensitivity criteria and economic considerations.In the closed loop control systems , the effects offeedback depend upon the kind of feedback used . There aretwo types namely , positive feedback and negative feedback

The main differences between open loop and closed control


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Open loop controlsystem

Closed loop controlsystem

It is sensitive to external disturbances andinternal variations. It is insensitive to external disturbancesand internal variations.

To get accurate results ,we need to usehighly expensive components.

To get accurate results ,we can useinexpensive components also.

It is used when inputs are known andthere are minimal disturbances.

It is used where high sensitivity is required.

Calibration is easy . Recalibration isneeded from time to time.

Calibration is difficult. Generallyrecalibration is not needed.

Generally very easy to build. Design of the system is much morecomplicated.

Generally ,the problem of oscillations isnot witnessed.

To achieve accurate results , it may tend toovercorrect errors leading to oscillations of

the system.

High in manufacturing , maintainance costand power consumption.

Low in manufacturing , maintainance costand power consumption.

Stability until unless disturbance is given. Wider range of stability.

The main differences between open loop and closed controlsystems are:-


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FEEDBACK IN CONTROL SYSTEMSBy

V.V.Kaushik Sridhar,

3/4 B.Tech, MME,

Roll no. 7532


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Based on whether feedback is given to a system ornot, control systems can be classified into:

1. Open control system (in which no feedback is

given).2. Closed control system (in which feedback is given)

Both the above mentioned control systems havetheir own advantages and disadvantages.


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WHAT IS FEEDBACK?

The arrangement of a system by which output iscompared with the input to make the systemmore stable or error-free is known as feedback.

Feedback can also be defined as an arrangementin which a part of the output is given back or fed-

back into the input.


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Based on the definition of feedback given inthe previous slide, one can define open loopand closed loop control systems as given

below:An open loop system may be defined as one

that is activated by a single signal i.e. theinput and the output has no effect on the

system.A closed loop system can be defined as onethat is driven by two signals- one the inputsignal and the other, a signal called thefeedback derived from the output of thesystem.


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BLOCK DIAGRAM OF OPEN ANDCLOSED LOOP CONTROL SYSTEMS

Closed Loop Control System


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TYPES OF FEEDBACK

There are two types of feedback namely positivefeedback and negative feedback.

Negative feedback occurs when the output of asystem acts to oppose changes to the input of thesystem; with the result that the changes areattenuated. If the overall feedback of the system isnegative, then the system will tend to be stable.

Positive feedback, sometimes referred to as"cumulative causation", is a feedback loop system

in which the system responds to perturbation inthe same direction as the perturbation.

AND NEGATIVE FEEDBACK


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AND NEGATIVE FEEDBACKSYSTEMS

Positive feedback1.Very rarely used.

2.Transfer function is givenby:

T.F.= G(s)1-G(s)H(s)

3. In this feedback, errorcannot be reduced but

overall gain of thesystem increases.

Negative feedback1.Very widely used.

2.Transfer function is givenby:

T.F.= G(s)1+G(s)H(s)

3. In this feedback, errorcan be reduced but

overall gain of thesystem decreases.


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EFFECTS OF FEEDBACK

Gain is reduced in ve feedback and increased in+ve feedback. Parameter variation is reduced. Sensitivity is increased.

The above effects of feedback help us to listdown the various advantages and disadvantagesof open loop and closed loop control systems.

ADVANTAGES OF OPEN LOOP AND


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ADVANTAGES OF OPEN LOOP ANDCLOSED LOOP CS

Open loop CS1. Easy to design.

2. Cost factor- moreeconomical.

3. More stable (nooscillations tilldisturbances).

4. No error signal.

5. Easy to maintain.

Closed loop CS1. More accurate.

2. Can accomodate anydisturbance.

3. Wider stability.

DISADVANTAGES OF OPEN LOOP


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DISADVANTAGES OF OPEN LOOPAND CLOSED LOOP CS

Open loop CS1. Cannot accommodate

disturbances.

2. Poor performance

under disturbances.3. Re- calibration

needed from time totime.

Examples: Driving a

scooter, immersion

heater

Closed loop CS1. Complex design.

2. Cost factor-Costly.

3. Difficult to calibrate.

4. Oscillations.

Examples: Iron box, Airconditioner.


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EXAMPLES AND APPLICATIONS OFFEEDBACK CONTROL SYSTEMS

Open loop control systems(non-feedback):Most automatic toasters are open loop control

systems because they are controlled by a timer.This time required to make the toast should beestimated by the user, who is not a part of the

system. Control over the quality of the toast (theoutput) is removed once the time, which is botthe input and the control action , has been set.

The time is typically set by calibrated means or aswitch.



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An autopilot mechanism and the airplane itcontrols is an example of a closed loop controlsystem. Its purpose is to maintain a specified airplaneheading, despite atmospheric changes. It performsthis task by continuously measuring the airplaneheading, and automatically adjusting the airplanecontrol surfaces ( rudder, ailerons) so as to bring the

actual airplane into correspondence with the specifiedheading. The human pilot or operator who presets theautopilot is not a part of the control system.


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BLOCK

D

IAGRAM

OFA

N AUTOPILO

T

SYSTEM

TH

ATUSES

FEED

BACK

CONTR

OL


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Positive feedback control systems:

Audio feedback or acoustic feedback is

a common example of positive feedback. Itis the familiar squeal that results whensound from loudspeakers enters a closely-placed microphone and gets amplified, and

as a result the sound gets louder and louder.To avoid this condition, the microphone

must be prevented from "hearing" itsown loudspeaker.

Example of a biological positive feedback loop is the


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Example of a biological positive feedback loop is theprocess of blood clotting. The loop is initiated wheninjured tissue releases signal chemicals that activateplatelets in the blood. An activated platelet releases

chemicals to activate more platelets, causing a rapidcascade and the formation of a blood clot.


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Negative feedback control systems: 1. A negative feedback amplifier, or more commonly simply

a feedback amplifier, is an amplifier which uses negative feedback toimprove performance (gain stability, linearity, frequencyresponse, step response) and reduce sensitivity to parametervariations due to manufacturing or environmental uncertainties.

2. Its operation is most easily seen in its use by James Watt to controlthe speed of his steam engine. Two heavy balls on an upright framerotate at the same speed as the engine. As their speed increases they

move outwards due to the centrifugal force. This causes them to lift amechanism which closes the steam inlet valve and the engine slows.When the speed of the engine falls too far, the balls will move in theopposite direction and open the steam valve.

3 Many biological process (e g in the human


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3. Many biological process (e.g., in the humananatomy) use negative feedback. Examples ofthis are numerous, from the regulating of body

temperature, to the regulating ofblood glucose levels. The disruption offeedback loops can lead to undesirable results:in the case of blood glucose levels, if negative

feedback fails, the glucose levels in the bloodmay begin to rise dramatically, thus resultingin diabetes.


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FEEDB

AC

KLOOP

IN

AB

IOLOGICAL

C

ONTROL

SYST

E M

EXAMPLES:-


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There are many examples for both open loop and closed loop control systems both in natureand which are man made .

A MAN DRIVING A CAR:-Consider the possibility in which a man drives a car blind folded . Then he is safe as

long as there is no traffic on the road . But without his eyes while driving a car in traffic , hecant sense the mechanical motions like distance , speed , acceleration of his own vehicle andthat of others also . But when he is able to see , he can sense these motions . This information

is sent to brain which commands the limbs to act accordingly .In the example discussed above , brain acts like a controller.

The motions which are sensed are the feedback signals . Therefore the case in which the mandrives with his eyes closed is an example for open loop control system . And the case in which

the man drives normally is an example for closed loop control system .


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RESIDENTIAL HEATING SYSTEM:-Consider the case of the residential heating system , which can be

controlled by using both open loop and closed loop control system .

Consider the heating which makes use of open loop control system, whose basic structure is shown below;

Temperature

set point Valve Roomradiator

Steam

Outdoortemperature

Open loop model:

Controlled ro

temperature


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In this system , the indoor temperature is variable which is to be controlledand the outdoor temperature is main disturbance input.

The desired is set on a calibrated dial , which positions the valvethat admits the steam for circulation through the radiator . The valve dialis calibrated at a certain temperature . When the value changes

significantly the controlled temperature will deviate from the desiredvalue by a large error and hence precise control isnt realized . The openloop model is used for residential areas where control isnt very stringent


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The open loop model can be converted to a closed loop model by adding the function ofmeasurement of the controlled temperature. Comparison of measured and desired values of

temperature makes the control logic element to make changes in the heating rate so as to obtain

the desired value .To control the above system , a thermostat with a bimetallic temperature sensor and asnap action switch to control a simple solenoid actuated steam valve is often used .This

controls the power supply to the solenoid . Generally , two cases arise when the measuredtemperature isnt equal to the desired value .

If T , represent desired and measured temperatures respectively , thenWhen T i.e., T is negative then switch opens , solenoid de-energized , temperature

falls down due to heat transfer to surroundings .


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LAPLACE TRANSFORMS

V.Vasudeven07360Mechanical department


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WHAT ARE LAPLACE TRANSFORMS?


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A Laplace transform is a type ofintegral transform.

Plug one function in0

s te dt

( ) f t

Get another function out

( )F s=

The new function is in a different domain.


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( )F s is the Laplace transform of ( ). f t

Write

{ }( ) ( ), f t F s=L

0s te dt

( )f t ( )F s=When

{ }

{ }

( ) ( ),

( ) ( ), etc.

y t Y s

x t X s

=

=

L

L


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A Laplace transform is an example of an improper integral : one ofits limits is infinite.

0 0

( ) lim ( )

h

s t s t

he f t dt e f t dt

=

Define


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A CALCULATION

Let0 if

( )1 if

t cu t c

t c


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c

1

t

0 if( )

1 if

t cu t c

t c


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{ }0

1 1

( ) ( ) lim

lim lim ( )

h s t s t

hc

h s c s t s h s c

s sch h

u t c e u t c dt e dt

ee e e s

= = =

= =

L

Calculating the Laplace transform of theHeaviside function is almost trivial.

Remember that ( )u t c is zero untilthen its one.

,t c=

We can use Laplace transforms to turn ani i i l l bl


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initial value problem

" 3 ' 4 ( 1)

(0) 1, '(0) 2

y y y t u t y y

+ = = =

into an algebraic problem

2

2 1( )*( 3 4) ( 1) ss

s e

Y s s s s +

+ + + =

Solve for y(t)

Solve for Y(s)

Laplace transforms are particularly effectiveon differential equations with forcing functions


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1

1

Asawtooth function

t

that are piecewise, like the Heaviside function,and other functions that turn on and off.

I V P


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I.V.P.

Laplace transform

Algebraic Eqn


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If we solve the algebraic equation

2

2 2

( 1) ( 1)( )

( 3 4)

s s s s e eY s

s s s

+ =

+ and find the inverse Laplace transform of the solution, Y(s), we have the

solution to the I.V.P.

Algebraic Expression


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Algebraic Expression

Soln. to IVP

Inverse

Laplacetransform

The inverse Laplace transform of


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is

4 43 32 15 80 4 16

4325 5

( ) ( 1)( + ( ) )

( )( ( ) )

t tee

t t

y t u t e e t

u t e e

=

2

2 2( 1) ( 1)( )

( 3 4)

s s s s e eY s s s s

+ = +

Th


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4 43 32 15 80 4 16

4325 5

( ) ( 1)( + ( ) )

( )( ( ) )

t tee

t t

y t u t e e t

u t e e

=

is the solution to the I.V.P.

" 3 ' 4 ( 1)

(0) 1, '(0) 2

y y y t u t

y y

+ =

= =

Thus


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HOW DO WE TRANSFORM ANDIFFERENTIAL EQUATION?

we need several nice properties of Laplace


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we need several nice properties of Laplacetransforms that may not be readily apparent.

First, Laplace transforms, and inversetransforms, are linear:

{ } { } { }{ } { } { }1 1 -1

( ) ( ) ( ) ( ) ,( ) ( ) ( ) ( )

cf t g t c f t g t cF s G s c F s G s

++

L = L +L

L = L +L

for functionsf(t),g(t), constant c, andtransformsF(s), G(s).

there is a very simple relationshipb h L l f f iSecond,


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between the Laplace transform of a givenfunction and the Laplace transform of that

functions derivative.

{ } { }

{ } { }2'( ) ( ) (0),

''( ) ( ) (0) '(0)

f t s f t f

f t s f t s f f

L = L

L = L

and

These show when we apply differentiationby parts to the integral defining the transform.

DIFFERENTIAL EQUATIONS


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DIFFERENTIAL EQUATIONS

10

)0(')0()('" KyKytrbyayy ===++

)()]0([)]0(')0([ 2 sRbYysYaysyYs =++

)()()()]0(')0()[()(

1)(

2

sQsRsQyyassY

basssQ

+++=++

=

)()( 1 YLty =

1st step

2nd step

3rd

step


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Now we know there are rules that letus determine the Laplace transform

of an initial value problem, but...


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HOW DO WE FIND INVERSELAPLACE TRANSFORMS?

First you must know that Laplace transforms


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First you must know that Laplace transformsare one-to-one on continuous functions.

In symbols

{ } { }( ) ( ) ( ) ( ) f t g t f t g t =L =L

whenfandgare continuous.

That means that Laplace transforms areinvertible.

INVERSE LAPLACE TRANSFORMS


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INVERSE LAPLACE TRANSFORMS

If { }( ) ( ), f t F s=L

{ }1 12( ) ( )c i

s t

i c i F s e F s ds

+

= L

then { }-1 ( ) ( ), F s f t =L where

An inverse Laplace transform is an improper


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An inverse Laplace transform is an impropercontour integral, a creature from the world

of complex variables.

Thats why you dont see them naked very often. You usually just see what theyyield, the output.

In practice, Laplace transforms and inverseLaplace transforms are obtained using tables

and computer algebra systems.

IMPULSE?


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IMPULSE?

An impulse is the effect of a force that acts over a very short time interval.

Engineers and physicists use theDirac delta function to model impulses.

A lightning strike creates an electricalimpulse.

The force of a major leaguers batstriking a baseball creates a mechanical

impulse.

THE DIRAC DELTA FUNCTION


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THE DIRAC DELTA FUNCTION

This so-called quasi-function was createdby P.A.M. Dirac, the inventor of quantum

mechanics.

0( ) 0 ( ) 1t a t a t a dt = =when and

People use this thing all the time. Youneed to be familiar with it.


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THE LAPLACE TRANSFORM OF THE

DIRAC DELTA FUNCTION

{ ( )} 1/ a s L t a e =

DIRAC DELTA (UNIT IMPULSE) FUNCTION


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DIRAC DELTA (UNIT IMPULSE) FUNCTION

Impulse of f(t), for

Generalized function

Laplace transform

= =

=0

1)(andotherwise0

if)( at

atat

aseatL = )}({

10 ttt

t t


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LAPLACE TRANSFORMS


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Def:

Inverse:

Linearity:

Shifting Theorom:

0),(for)()()(0

>==

ttfdttfefLsFst

)()( 1 FLtf =

)}({)}({)}()({ tgbLtfaLtbgtafL +=+

)}({)(

)()}({

1asFLtfe

asFtfeL

at

at

=

=

REVIEW: PARTIAL FRACTIONS

C I t d l f t


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n

nk

kk

n

k

m

m

m

ps

A

ps

A

ps

A

ps

A

ps

A

pspsps

asas

+

+

+

+

=

++

)()()()(

)()()(

2

2

1

1

1

1

12

1

11

21

0

1

1

Case I: unrepeated real factor

Case II: repeated real factor

Case III: unrepeated complex factor

Case IV: repeated complex factor

n

nkk

kk

nk

m

m

m

ps

A

ps

A

cbss

BsA

cbss

BsA

cbss

BsA

pspscbss

asas

+

+

+++

+++

++

+++

=

++

++

)()()()(

)()()(

2

2

2

11

12

1212

2

1111

22

0

1

1

EXISTENCE FOR LAPLACE TRANSFORM


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EXISTENCE FOR LAPLACE TRANSFORM

piecewise continuous on every finite inteval in therange of

satisfy

for some constant kandM.Conclusion:

The Laplace transform of exists for alls > k

0allfor|)(| tMetf kt

)(tf

)(tf

0t

)(tf

Condition:


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TRANSFORM OF INTEGRALS


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s

sF

s

fLdfL

t)()(

})({0

==

})(

{)( 1

0s

sFLdf

t

=

CONVOLUTIONfi i i


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Definition

Property

Application

==t

dtgftgfth0

)()())(()(

)}()({))(*( 1 sGsFLtgf=

)()()(

1)(

2

sQsRsY

basssQ

=

++=

=t

dttqty

0

)()()(

0)0('0)0()('" ===++ yytrbyayy1. Solve

2. Calculate integral

List of Laplace Transformsf(t) L(f) f(t) L(f)


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f(t) L(f) f(t) L(f)

1

11/s 7

cos t2 t 1/s2 8 sin t

3 t2 2!/s3 9 cosh at

4 tn(n=0, 1,)

10 sinh at

5 ta

(a positive)

11 eat cos t

6 eat 12 eat sin t

1

!+n

s

n

1

)1(+

+a

s

a

as

1

22 +s

s

22

+s

22

as

s

22as

a

22

)( +

as

as

22)(

+as

LINEAR SYSTEM APPLICATIONS


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Definition: A linear mapping of a set of input functions

into a corresponding set of output functions

Transfer function: ratio of the Laplace transform of theoutput of the system to that of the input of the system

)}({)( txfty =

TRANSFER FUNCTION PROPERTIES


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TRANSFER FUNCTION PROPERTIES

Completely characterizes the systems (initiallyrelaxed)

Independent of inputs and outputs

Equals the Laplace transform of the impulseresponse of system

Can be used to find the response to another systeminput (using convolution or partial fraction)


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THANK YOU

TRANSFER FUNCTION AND ITSPROPERTIES:-

Transfer function of a system is a mathematical


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ymodel and an operational method of expressing

differential equations that relates the output variableto the input variable. It is the property of the systemitself which is independent of the magnitude of theinput or the driving function.

Transfer function is the ratio of Laplace transform

of output to the Laplace transform of input with initialconditions set to zero.

Transfer function= C(s)/R(s), initial conditions zero. Poles and zeros of transfer function:

C(s)/R(s) = ( bosn + b1sn-1 + _ _ _ + bn ) / ( aosm + a1sm-1 + _ _ _ + am )= N(s) / P(s).

The poles of the transfer function are the roots ofthe equation P(s)=0. The equation P(s)=0 is calledth h t i ti ti f th t


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the characteristic equation of the system.

The zeros of the transfer function are the roots ofthe equation N(s)=0.Order and type of the system:

C(s)/R(s) = ( bosn + b1s

n-1 + _ _ _ + bn ) / ( aosm +

a1sm-1

+ _ _ _ + am )*sr

= N(s) / P(s)

Order of the system is the degree of the equationP(s)=0.

Type of the system is the number of poles at theorigin.

TYPES OF INPUT SIGNALS


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TYPES OF INPUT SIGNALS

submitted by Tushar Ranjan Moharana

Date 9th September 2009

Assignment-I

BASIC TYPES


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Basically inputs are classified into four groups :-

1. Step input

2. Ramp input3. Parabolic input

4. Impulse input


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STEP INPUTMagnitude of input is constant(let 0) for certain time and suddenly increases tosome other value (let 1) and maintained at that value.

Use : The functioning of automobiles on smooth road is tested by using

this input through

Simulation

Used to test signal forElectronic equipment.

By giving step input signal

We can determine the

response of system when

it is switched on or off.

X axis (time)

RAMP INPUT


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Impulse varies linearly with time.It is used in the horizontal deflection plates of a cathode rayoscilloscope.

It moves the electron beam horizontally with a constant speedas the vertical deflection

plate trace the wave

form

X axis (time)


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PARABOLIC

INPUT

Magnitude of input signal is proportional to

square of the timeUsed in accelerating speed of vehicle etc.

X axis (time)

INPUT


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Impulse have high magnitude but acts for lesstime X axis (time)

It is used to test the

response of equipments tosudden fluctuation in the

inputBy giving impulse input

to any system we can

determine its behavior to anyother signal

Parabolic signal:

f(t)=t2/2, when t > 0

=0, otherwise.


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Ramp signal and parabolic signal are used for the analysis

of faster signals. Sinusoidal signal:

f(t)=sin(t)*(t)

Parabolic SignalRamp Signal

Step Input Signal

Impulse Signal Sinusoidal Signal

MASONS GAINFORMULA


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FORMULA

NAME-SINGO SABATINIHEMBRAM

BRANCH-METALLURGICALAND MATERIALSENGINEERING

ROLL NUMBER- 7528

MASONS GAINFORMULA


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FORMULATo reduce a block diagram, one may use the Masons gain formula(also called Masons rule).

Mason first derived the idea using what he called a signal-flowgraph, which is a different

graphical representation of a block diagram. A signal-flow graph isdrawn with paths (lines) and

nodes. The transfer functions in a block diagram become the pathsand the variables in between

the blocks become the nodes. The input and output variables of ablock diagram are designated the

source and sink nodes

Signal flow graph is a diagramwhich represents a set of simultaneouslinear algebraic equation When applying


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linear algebraic equation When applyingthe signal flow graph method to controlsystem we must first transform lineardifferential equation into algebraicequation in s.

Signal flow graph consist of a network, inwhich nodes are connected by directedbranches.

Each node represents a system variable and

each branch connected between two nodesacts as a signal multiplier.

Note that signal flows only in one direction.

The signal flow graph depicts the flow of signals

from one point of a system to another and gives the relationshipamong the signals.

Signal flow graph contains essentially the same information as a


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block diagram. The advantage of using

signal flow graph is to represent a control system is that a gainformula or Masons gain formula is available

which gives the relationships among the system variableswithout requiring a reduction of the graph.


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TERMINOLOGY


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The gain of the branch is a transmittance

Branch :It is a directed line segmentjoining two nodes.

Transmittance :It is a gain between the two

nodes.

Feed back loop: If a loop originates andterminates at the same node ,is known as

feed back loop.

Self loop: A loop that consists of only one

node where it originates and terminates.

Chain node: In this node no incoming or outgoing .Here x2 and x3are chain nodes.

Flow of signals: Source to sink.

Feed back: Sink to sour.


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Dummy nodes: If incoming and outgoing branches exists at the

first and last nodes representing input and output variables can notbe considered input and output nodes.

In determining forward and backward paths self loops are notconsidered.

In such cases separate input and output nodes are created byadjoining branches with unity gain.

Path : It is a traversal ofconnected branches in thedirection of the branch arrows.


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Path gain: The product of allgains going to the forwardpath.Loop :It is a closed path .Loop gain :It is the product of thebranch transmittance of a loop.Non touching loops : Loopsare non-touching if they do notposses any common nodes.

Non touching loopx1

x x3 x4


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Touching loop

2

x2x1 x3 x4x5

Forward path : Af d th i th


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forward path is a pathfrom an input node toan output node which

does not cross anynodes more then once.Forward path gain : It is the

product of the branchtransmittances of a forward


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TRANSMISSION RULE: x2 x5=dx1

a x4=cx1


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x5 d b x2=ax1

c x3=bx1

x4

MULTIPLICATION RULE:

a b c abc

x3x1

x2x1

x1 x2

x3 x4

SFG VS BLOCK DIAGRAMS


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Source orinput node

Sink or outputnode

Branch

Node

Input signal Output signal

BlockSignal

Signal Flow Graph Block Diagram

PROPERTIES OF SIGNALFLOW GRAPH


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1. A branch indicates the functional dependence of one signal

upon another. A signal passes through only

in the direction specified by the arrow of the branch .

2. A node adds the signals of all incoming branches and

transmits this sum to all outgoing branches.

3. A mixed node which has both incoming and outgoing

branches may be treated as an output node by adding anoutgoing branch of unity transmittance.

4. For a given system, signal flow graph is not unique. Manydifferent signal flow graphs can be drawn for a given system bywriting the system equations differently.

Utility of Signal FlowGraphs Alternative to block diagram approach


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Alternative to block diagram approach

- may be better for complex systems.- good for highly interwoven systems.

- system variables represented as nodes, branches(lines) betweennodes show relationships between system variables.

The flow graph gain formula (Mason)allows the system transferfunction to be directly computed without manipulation or reduction

of the diagram.

We now need to define somemore terms and show how eachof these quantities can be


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of these quantities can be

calculated:System determinant:

= 1 (sum of all individual loop gains)

+ (sum of the products of the gains of all possible two loops thatdo not touch each other)

(sum of the products of the gains of all possible three loops

that do not touch each other)

+ and so forth with sums of higher number of non-touching loopgains.

EXAMPLEG6


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Step 1: forward paths=2

G1 G

2

G4G3 G

5-H1 -H2

-

H3

G1G2G3G4G

5

G6

1 1

P1=G1G2G3G4G5

P2=G6

Step 2: No. of loops=4


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Sum of product gain of two non touching loops=G4G6H2H3 +G2G6H1H3 + G2G4H1H2

Step 3: three non touching loops=1

=L1L2L4= -G2G4G6H1H2H3-H1

G2L1

G4

-H2L2

G1G2G3G4G5

L3 -H3

G6

L4

-H4

PRODUCT GAIN:

=1 ( - G2H1-G4H2-G1G2G3G4G5H3-G6H3)


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G1G2G3G4G5H3-G6H3)

+(G2G6H1H3+G4G6H2H3+G2G4H1H2)

- (-G2G4G6H1H2H3)

1=Product gain(excluding P1)=12=Product gain(excludinP2)=1-(-G2H1-G4H2)+

(G2G4G6H1H2H3)=1+G2H1+G4H2+G2G4H1H2

TRANSFER FUNCTION:The ratio of laplace transform ofoutput to laplace transform of


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output to laplace transform of

input at zero initials.

T.F=C(S)/R(S)

at zero initials

C(S)/R(S)=((P1*1)+(P2*2))/

HOW TO DEVLOP SFG FOR THE SYSTEM REPRESENTED BY THESEEQUATIONS .

5-X1-2X2-3X3=0

-2X1+X2+1.5X4=0


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-2X2+X3-3X1=0.5X3X4=2X3

Arrange the above equations :

X1=5-2X2-3X3

X2=2X1-1.5X4

X3=0.5X3+3X1+2X2X4=2X3

HERE 5 IS INPUT.

-1.5


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X1 XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX4

2

5

0.5

-2

-3

X2 X3

2

3

DISADVANTAGESMissing a critical loop or


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g p

probability of missing acritical loop ,reduces the

accuracy.Does not give a idea ofcause and effect of theoverall system.

FLOW GRAPH SUMMARY Block diagrams and signal flow graphs allow visualrepresentation of complex dynamic systems.

Through Masons gain formula, we can compute the input to


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Through Mason s gain formula, we can compute the input to

output transfer function of the overall system.Note how easily other transfer functions can be obtained e.g.noise response they all share the same .


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TIME DOMAIN ANALYSIS

The variation of output/response of a system wrttime is called time response. It has two parts-

Steady state response and transient response.


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STEADY STATE RESPONSE:THERESPONSE WHICH REMAINS EVEN


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RESPONSE WHICH REMAINS EVEN

AFTER THE TRANSIENTS HAVE DIEDOUT.

From this response we can observe-

1.Time taken by the system to reach steady state value.

2.Whether there error is existing in the output or not

STEADY STATE ERROR IN TIMEDOMAIN ANALYSIS


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One of the objectives of most control systems is that thesystem output response follows a specific reference signalaccurately in the steady state.

Steady-state error is the difference between the outputand the reference in the steady state.

Steady-state errors in control systems are almostunavoidable and generally derive from the imperfections,frictions, and the natural composition of the system.

In the design problem, one of the objectives is to keep thesteady-state error below a certain tolerable value.

The error in control system is defined ase(t) = r(t) - b(t)

or

E(s) = R(s) - B(s) =R(s) - H(s)C(s)


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The steady-state error is then given by:

The steady-state error can be evaluated using the final-value theorem as:


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This shows that the steady-state error depends on thereference input R(s) and the loop transfer function

G(S)H(S)

The openloop transfer function, G(S)H(S), may bewritten as

Steady-state error due to a step inputSteady

-state error due to a step input

For R(s)


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G(s)H(s) as the positional error constant, then

We can summarize the steady-state error due to a step input as follows:

Type-0 system

(or higher)Type-1 system


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Steady-state errors: (a) step

input, (b) ramp input

Steady-state error due to a ramp inputSteady

-state error due to a ramp input

For R(S)= , we have


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=K lim limv

K ;n=0,1,2,..n-1s 0 s 0 s

sG(s)H(s)=

If we defineIf we define = velocity error constant= velocity error constant

Therefore,

Hence, we have

Laplace transform of r(t) is given by:

The input is described by:

Steady-state error due to a parabolic inputSteady-

state error due to a parabolic input


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Hence, the following conclusions can be made:

The steady-state error could be written as:

Defining the acceleration error constant as:

The steady-state error of the system is:

Steady-state errors for various inputs and systemtypesaretabulated below.


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The error constants for non-unity feedback systems may be obtained byreplacing G(s) by G(s)H(s).

Systems of type higher than two are not employed for two reasons:


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Systems of type higher than two are not employed for two reasons:1. The system is difficult to stabilize.

2. The dynamic errors for such systems tend to be larger than thosetypes-0, -1 and -2.

GENERALISED COEFFICIENT METHODBy using convolution theorem and taylors seriesexpansion we can write steady state error as


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Es=k0R(t)+k1dR(t)/dt+..+kn d^nR(t)/dts

Where kn= lt s 0 d^n(1/1+G(s)H(s))

ds^n

IMPORTANT INFORMATIONABOUT VARIATION OF ERRORWITH TIME.


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Demerits:

# It gives error of only

standard signals.# the value of error at zeroor infinity is not given by

this method.# Applicable only for stables stems.


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CONCEPT OF BIBO

By : Alok Ranjan

WHY INVESTIGATE STABILITY?

The issue of ensuring the stability of a closed-


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loop system is the most important to controlsystem design. An unstable feedback system isof no practical value. A stable system shouldexhibit a bounded output if the correspondinginput is bounded. This is known as BIBO stability.

CONCEPT OF BIBO

A system is defined to be BIBO stable if every


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bounded input to the system results in abounded output over the time interval zero toinfinity.

BIBO stability is based on transfer functionrepresentation.

For Linear-Time-invariant (LTI) system, if theimpulse response is absolutely integrable,thesystem is BIBO stable.

Impulse response approach to zero when time

lead to infinite.

CRITERION FOR BIBO STABILITY FOR LTICSYSTEMS


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x(t) y(t)

y(t) = h(t)*x(t)

= x(t- )d

= d

h(t)

If x(t) is bounded, i.e. < K< ,then

< K < , and K d


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Hence for BIBO stability.d 0 If this condition is


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characteristic equation be positive if a0 > 0. If this condition is

not met, the system is unstable and number of sign changes of

the terms 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.


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Special cases:

Occasionally, in applying the Routh stability criterion,

certain difficulties arise causing the breakdown of the Rouths

test. The difficulties encountered are generally of the following

types:

Difficulty 1: When the first term in any row is zero while rest of


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Difficulty 1: When the first term in any row is zero while rest of

the row has at least one non zero term.

Because of this zero term, the terms in the next row

become infinity and Rouths test breaks down. The following

method is used to overcome this difficulty:

Substitute a small positive number for the zero and proceed toevaluate the rest of the array. Then examine the signs of the first columnby limiting 0.

Ex:-3) Consider the following with characteristic equations5 +s4 +2s3 + 2s2 + 3s + 5 = 0

The Routh array for this system is as below:1s5 : 2 3


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s4

s3

s0

s1

s2

2

-2

5

2* +2

5

-4-4 -522+2

-2

5

:

:

:

:

:

1

0

It is seen that since the first element in the third row is zero it isreplaced by and proceeded. There are two sign changes in thefirst column. Therefore the given system is unstable having two

poles in the right half ofs-plane.

Difficulty 2 :When all the elements in any row of the Routh array

are zero.

This condition indicates that there are symmetrically

located roots in the s-plane .The polynomial whose coefficients

are the elements of the row just above the row of zeros in the


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are the elements of the row just above the row of zeros in theRouth array is called an auxiliary polynomial. Because of a zero

row, the Rouths test breaks down. This situation is overcome by

replacing the row of zeros in the Routh array by a row of

coefficients of the polynomial generated by taking the first

derivative of the auxiliary polynomial.

Ex:-4) Consider the following with characteristic equations6 +2s5 + 8s4 +12s3 + 20s2 + 16s + 16 = 0

The Routh array for this system is as below:

s5

1

12 16:

s6

2

: 8 20 16


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s

s4

s3

12

0

2

0

1216

::

:

216

Since the terms in the third row are zero the Rouths test breaks.

Now the auxiliary polynomial is formed from the coefficients ofthes4-row, which is given by,

A(s) = 2s4 + 12s2 + 16

The derivative of the polynomial with respect tos is

dds

A(s) = 8s3 + 24s


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APPLICATIONS OF ROUTH-HURWITZ CRITERION TOLINEAR FEED BACK SYSTEMS:

The Routh stability criterion is frequently used for the

determination of the condition of stability of linear feedback


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determination of the condition of stability of linear feedbacksystems as shown in the in the following example:

Ex -5):-

Consider the closed loop transfer function of the system as

C(s)/R(s)=K/{s*(s2 + s + 1)*(s + 4)+K}

Then , the characteristic equation becomes,

s*(s2 + s + 1)*(s + 4)+K=0

s4 + 5s3 + 5s2 + 4s + K=0

s4 1 5

4

K:

Applying Routh-Hurwitz stability criterion for the abovesystem:


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s3

s0

s1

s2

45

K

21/5

84/5 5K21/5

:

:

:

:

K

Since for a stable system, the signs of elements ofthe first column of the Routh array should be all positive ,the condition of system stability requires that,K > 0 and

(84/5-5K) > 0

Therefore for stability , K should lie in the range84/25 > K > 0


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Therefore for stability , K should lie in the range84/25 > K > 0When K=84/25 , there will be a zero at the first entry in

the fourth row of the Routh array . This corresponds to thepresence of a pair of symmetrical roots . Therefore

K=84/25, will cause sustained self oscillations in theclosedloop system .

LIMITATIONS OF ROUTH-HURWITZCRITERION:

Routh Hurwitz criterion has the following disadvantages :

It gives only the number of roots in the right half of the s-

plane . It i i f ti d th l f th t


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p It gives no information as regards the values of the roots

and does not distinguish between real and complex roots .

It cant be used for characteristic equations containingexponential , trigonometric coefficients .

It is very complicated . It is time consuming for higher ordersystems .

It becomes tough to predict marginal stability for biggersystems .

STATE VARIABLE METHODS


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PRESENTED BY

K.RAJASEKHARA REDDY(7837)

BIOTECHNOLOGY

DEFINITIONS:


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STATE VARIBLES IN A SYSTEM

systemm inputs p outputs


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system

Statevaribles(x1,x2,x3)

STANDARD STATE FORM EQUATION


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STANDARD STATE FORM EQUATION


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EXAMPLE WITH GENERAL CIRCUIT

CIRCUIT DIAGRAM


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EQUATIONS:

i1+i2+i3=0

i1+i2+c dv/dt =0 L1 di1/dt +i1R1+(e-V)=0


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L1 di1/dt +i1R1+(e V) 0 L2 di2/dt +i2R2-V=0dV/dt=x1=(-1/c)(i1+i2)

di1/dt=x2=(-1/L1)(i1R1+e-V)di2/dt =x3=(-1/L2)(i2R2-V)

STATE EQUATION FROM PREVIOUSEQUATIONS

x1 0 -1/c -1/c x1 o

x2 = 1/L1 R1/L1 0 x2 + -1/L1 ux3 1/L2 0 R2/L2 x3 0


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x3 1/L2 0 R2/L2 x3 0

Consider the voltage and current in R2 as outputvariables y1,y2

y=Cx+Du

INPUT STATE OUTPUT EQUATION

y1 0 0 R2 x1

= x2y2 0 0 1 x3


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y2 0 0 1 x3

TRANSFER FUNCTION FROM STATEVARIABLE MODELS

x=Ax+Bu

y=Cx+DuTaking laplace transform on both sides


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g pS*x(s)-x(0)=A*x(s)+B*u(s) (sI-A)*x(s)=B*u(s)

x(s)=(1/(sI-A))*B*u(s)y(s)=c/(sI-A)*B*u(s)+D*u(s)Transfer function=y(s)/x(s)=(c/(sI-A))*B+D

WHY WE GO FOR STATE VARIABLETECHNIQUE


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Due to many demerits of transfer function we gofor state variable method.

DEMERITS OF TRANSFER FUNCTION

Applicable only for linear systemsAnalysis of multiple i/p system become tediousApplicable only for time invarient system


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Initial conditions need to be zeroDoes not provide any information about the

internal structure of the system Issues with stability there are many cases where

the analysis using transfer function gives overallstable system but the system elements show thetendency to exceed the specified ratings.

More of hit and trial error method.


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AN INTRODUCTION TO P ANDPI CONTROLLERS.By:-SHASHANK KUMAR

B.TECH, MECHANICAL ENGINEERING

ROLL NO. 07352

WHAT IS A CONTROLLER ?

A controller compares the actual value of a system with thereference input i.e. the desired value , measures the deviation

and then produces a control signal in accordance with theerror signal so as to reduce the deviation to zero or a


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error signal so as to reduce the deviation to zero or aminimum value.

This action is called the control action.

DIFFERENT TYPES OF CONTROLLER

The different types of controller used today in industrialapplications can be categorized as follows :-

1. Two-position or on-off controllers.


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2. Proportional controllers.

3. Integral controllers.

4. Proportional-plus-integral controllers.

5. Proportional-plus-derivative-controllers.

6. Proportional-plus-integral-plus-derivative controllers.

PROPORTIONAL CONTROLLER

A proportional controller is a type of linear feedback control system which followsthe following relation between input and output:-

U(t)= ;here e(t) is the error signal which is the difference between the reference value and


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the process variable.

in laplace form, this is

The proportional mode adjusts the output signal in direct proportion to thecontroller input. The adjustable parameter to be specified is the proportionalcontroller gain kp.

The proportional controller is basically an amplifier with an adjustable gain. An example is a driver driving a car , if he is over speeding he reduces the

accelerator, if he is at a lower speed he accelerates the car.,

A SIMPLE PROPORTIONAL

CONTROLLER

e(s)c(s)


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ADVANTAGES OF THE PROPORTIONALCONTROLLER

1. It does not require a very precise analytical

modeling of the system being controlled.2. It is very simple to implement


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y p p

DRAWBACKS OF THE PROPORTIONALCONTROLLER

1. A proportional controller reduces error but does noteliminate it (unless the process has naturally integrating

properties).2 It causes non zero steady state error


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2. It causes non-zero steady state error.

3. It does not provide any protection against oscillations.

PROPOTIONAL AND INTEGRALCONTROLLER

CONTROLLER PART


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THE SPECIFICATION OF VARIOUS

CONSTANTS ARE AS FOLLOWS


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Kp =PROPOTIONAL GAINKi =INTEGRAL GAIN

E(t) =ERROR FUNCTIONTi= INTEGRAL TIME

THE EQUATION RELATED WITH IT ARE AS FOLLOWS :

M(t)=Kp e(t) + Kp/Ti e(t) dt

0

t


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TRANSFER FUNCTION

M(s)/E(s)=Kp(1+1/Ti s)

WHERE Kp : PROPOTIONAL GAIN OF THE CONTROLLER

Ti : INTEGRAL TIME

THE OUTPUT GRAPH OF PICONTROLLER


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ADVANTAGES OF PI CONTROLLER

The integral term of the PI controller causes the steadystate error to be zero for a step input

LIMITATION OF PI CONTROLLER


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The problem with using a PI controller isthat it introduces a phase-lag. This means

that the phase margin (a measure ofstability) decreases. So careful designconsiderations with respect to the gain mustbe considered.

APPLICATION OF PI CONTROLLERS

THEY ARE USED IN FINDING AIR FLOWSMOTOR POSITIONING IN ROBOTS ETC


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PROPORTIONAL PLUSDERIVATIVE CONTROLLERBy

Akhilesh Kumar

Proportional-Derivative control is useful for fast response controllers that donot need a steady-state error of zero. Proportional controllers are fast.

Derivative controllers are fast. The two together is very fast.

INTRODUCTION:


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1. Proportional action

Provides an instantaneous response to the control error.

This is useful for improving the response of a stable system but cannotcontrol an unstable system by itself.

The gain is the same for all frequencies leaving the system with a nonzero

steady-state error.

2. Derivative action:

Derivative action acts on the derivative or rate of change of the control error.

This provides a fast response, as opposed to the integral action, but cannot

accommodate constant errors (i.e. the derivative of a constant, nonzero error is 0). Derivatives have a phase of +90 degrees leading to a predictive response.However,


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p g g p pderivative control will produce large control signals in response to high frequency

control errors.Combination of proportional and derivative:

Proportional-Derivative or PD control combines proportional control andderivative control in parallel.

where

is the proportional control gain andis the derivative gain or derivative time.

DESIGN:

KpR(s) C(s)E(s)+ _ + +Wns(s+2*Wn*e)

The block diagram of PD controller is shown below:


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Kd s

Output, C(s)= (Kp + sKd)E(s)

OP-AMP REALIZATION OF PD CONTROLLER:

R1

R2

R

R

EE


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Transfer function of the circuit : E/E= R2/R1 + R3*C*s

R

R3

C

AN PRACTICAL EXAMPLE:Consider the figure shown below:

controller J

Ia= constant

TwTm,Bm

Rf

LfV


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B

Br

The figure shows an electromechanical servo where

Tm = Motor TorqueJ = Inertia

Tw = Torque disturbance

V = Viscous friction coefficient


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The damper torque on J, due to V, behaves exactly like a derivative control mode in that italways opposes velocity dB/dt with a strength proportional to dB/dt, making motion less

oscillatory.

Solving the system and considering second order characteristic equation, we get

Wn = (K/J) and e = (Td/2) * (K/J)

If Td > 0, motor torque is felt by the load similar to mechanical viscousdamping torque. The derivative control is hence interpreted as pseudo-friction

effect.

The block diagram of the given system is as shown below:

Tw


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KpKc(1+Td*

s)

1/R

f

Kt1/Js*

s

Kp

BBr+

_

++

If we change the system by using a proportional controller in the forward path andadding derivative mode at the controlled variable B.

And comparing derivative control applied to system error with that to the controlledvariable. We find the similar effect on characteristic equation. However , derivative

control has more violent response to sudden changes due to derivative term Brspresence.

If we study the effect of derivative control on steady state errors (say for a unit step


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If we study the effect of derivative control on steady state errors (say for a unit stepdisturbance with Br= zero, we conclude that :

The performance is unaffected by the derivative mode.

However, there is an indirect method for gaining improvements in steady state accuracyi.e. by proper choice of derivative mode allows the use of higher values of proportional

gain, thereby improving the steady state accuracy.


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APPLICATIONS:

1. Fast steering mirror.

2. Helicopter Hover.


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3. Inverted pendulum.

4. Mixing tank.5. Vibration Isolation.

6. Simple offline least squares system identification.

ROOT LOCUS(THEORY)


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ROOT LOCUS(THEORY)

Consider the following standard control system,

C(s)G(s)

R(s) error

1+ KG(s)H(s)

Where K is varied


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G(s)

H(s)

Transience and stability of a system depends upon its CLOSED LOOPPOLES.

To build a better system, movement of poles can be adjusted by modifyingthe system parameters.

The ROOT LOCUS method thus is of help.


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What is a ROOT LOCUS?

Plot of the LOCI of the closed loop poles as a function of the open loopgain (K), where K is varied from infinity to infinity

When,K is varied from infinity to zero, it is a direct root locus.

K is varied from zero to infinity, it is a inverse root locus.

There are 8 steps to constructing a ROOT LOCUS and must be followedin order to reach a conclusion

RULE 1:

Root locus is always symmetrical to the x-axis or the real axis.

CONSTRUCTION OF A ROOT LOCUS


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Roots are either real or complex conjugate pairs or both.

Hence the roots or poles and zeroes of the equation are found first.Say,

G(s)=_________(s+1)___(s^2 + 4s +5)(s+3)

So the poles of the equation ares+3=0, s=-3

s^2 +4s+5=0 s=-2+I and -2-IAnd the poles are

RULE 2: Number of LOCI

If the number of poles be n and the number of zeroes be m.

If n>m , then

Total number of loci= n

Each locus starts from a pole and ends in a zero.


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So, number of loci that start from a pole and end in infinity is n-m.

E.g. in the previous caseTotal number of loci=3

And number of loci starting from a pole and end in infinity is 3-2=1

Rule 3: Real axis loci


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Some of the loci will lie on the real axis.

A point on the real axis will lie on the root locus if and only if,

Number of poles and zeroes to the right is odd

RULE 4: Angle of asymptotes

Mostly, number of poles > number of zeroes. i.e. n>mHence, n-m branches move to infinity, along the asymptote.

Asymptote: It is defined as a line on the root locus whichconverges at infinity.


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The angle of asymptote is given by,Theta= (2*q + 1)*180 / n-m for K>0

Theta= (2*q)*180 / n-m for K


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e.g. In the above exampleG(s)=_________(s+1)___

(s^2 + 4s +5)(s+3)

sum(real part of poles) = -2+(-2)+(-3)= -7sum(real parts of zeroes) = -1

n-m =2Centroid= (-7+1)/2= -3

RULE 6: break away and break-in point

Break-in point is defined as the point where the root locus enters real axis.

Break away point is defined as the point where the root locus comes out of the realaxis.

Break away & break in points are points on the real axis at which multiple roots ofthe characteristic equation occur.


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Observation: If there are 2 adjacently placed poles on the real axis and the real axisis a part of the root locus, hence minimum 1 break away point exists between the 2

poles.

1+ G(s)H(s)=0; we get a relation in terms of K.

-dk/ds=0, gives the break away point.e.g. G(s)H(s)=K / s^2 + 4s+5

Hence,s^2+4s+5+K=0, and K=(-s^2+4s+5)

-dk/ds=0, 2s+4=0

RULE 7: Intersection of root locus with jw axis.

To calculate the intersection with jw axis, this is followed

1.Construct 1+G(s)H(s)=02.Develop the routh array in terms of K

3.Find K value for which routh array contains a term of zeroes.4.Frame auxiliary equation with the help of coefficients from the row above the row ofzeroes.

5 Substitute the K value found in step 2 equate the equation to zero find the value of


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5.Substitute the K value found in step 2, equate the equation to zero, find the value ofs.

The roots of the equation give the intersection points.

e.g. say G(s)H(s)= K/s*s+1*s+3Characteristic equation will be s^3+ 4s^2 +3s+K=0,

Routh array s^3: 1 3s : 4 K

s^0: (12-K)/4..hence K-12=0 K=12Substituting K=12 in the auxiliary equation, 4s^2+K=0,

hence we get K=(+or-) root(3)i , which are the intersection points.

RULE 8: Angle of departure and angle of arrival.

The root locus always leaves a complex pole, at an angle known as the departureangle given by,

Theta=180+ arg( G(s)H(s) )

The root locus always arrives at a complex zero, at an angle known as the arrivalangle.Theta=180- arg( G(s)H(s) ),


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Where G(s)H(s) is the angle excluding the zero where the angle has the becalculated

Steps of a ROOT LOCUS method:

1.Determine the branch number ending at infinity of the loci using rule 1.

2.Plot the poles and zeroes.

3.Find the real axis loci using dark lines.

4.Find the asymptotes and their angles using rule 3.


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5.Using rule 4 determine the asymptotes centre and draw step 4 and step 5.

6.Calculate the break in or breakaway points.

7.Calculate angle of departure and angle of arrival using the rule.

8.Determine the jw crossover if root locus has complex poles and zeroes.

ByK S Venkatakrishnan

7328Mechanical Engineering


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FREQUENCYFREQUENCY

DOMAIN ANALYSISDOMAIN ANALYSISBY

T V D .GOWTHAM

ROLL NO -7356

B.TECH

MECHANICAL ENGINEERING

FREQUENCY RESPONSE

The term frequency response means the steadystate response of a system to a sinusoidal input

In frequency response analysis we vary the


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In frequency response analysis,we vary thefrequency of input signal over a certain range

and study the resulting response

Frequency response requirements differdepending on the application.

Frequency response curves are often used toindicate the accuracy of electronic components


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y por systems

WHY FREQUENCY RESPONSE ANALYSISCOMES INTO PICTURE

The frequency response of a control systempresents a qualitative picture of the transient

response the frequency response allows us to understand

a circuit's response to more complex inputs


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Time domain analysis is limited to only smallersystems,the stability of larger systems is difficultto analyse

Generation of sinusoidal signals is very easy

Frequency response measurements can be useddirectly to quantify system performance and

design control systems.


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FREQUENCY RESPONSE FOR A GIVENSINUSOIDAL INPUT


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The sinusoidal transfer function G(jw) ,the ratio ofY(jw) to X(jw),is a complex quantity and can berepresented by magnitude and phase angle withfrequency as a parameter

A negative phase angle is called phase lag andpositive phase angle is called phase lead


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p p g pThe sinusoidal transfer function of linear system

can be obtained by substituting jw in S of atransfer function

REPRESENTATION OF FREQUENCY RESPONSECHARACTERISTICS IN GRAPHICAL FORMS

There are three commonly used representations ofsinusoidal transfer functions:-

Bode plots


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Bode plots Polar plots Nyquist plots

1. INTRODUCTION2. ILLUSTRATION3. GUIDELINES FOR A POLAR PLOT4 SIMPLE PLOTS

POLAR PLOT


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4. SIMPLE PLOTS

5. STABILITY OF POLAR PLOT6. ADVANTAGES OF FREQUENCY

RESPONSE OVER TIME RESPONSEANALYSIS

A POLAR PLOT IS DEFINED AS THE LOCUS OF GH(JW) PHASOR

IN POLAR REPRESENTATION WHEN W CHANGES FROM INFINITYTO ZERO AND THE PLOT IS IN GH(JW) PLANE

Here the plot is between IG(jw)I & IG(jw) as a function ofw

This is used in determining the stability of closed loopsystem from its open loop frequency response .For closedloop system stability, open loop transfer function is G(s)H(s) instead of G(s) unless H(s) is 1.Unless stated abovethe open loop tranfer function is G(s)


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the open loop tranfer function is G(s).

ILLUSTRATION OF POLAR PLOTS

G(s)H(s) = K (s+a)/s(s+b) (s+c)

Replacing s by jw ,we haveG(jw)H(jw)= K(a+jw)/jw(b+jw)(c+jw)

= lGH(jw)l F(w) is called polar representation

lGH(jw)l is called magnitude and F(w) is called phase angle

lGH(jw)l = K(a^2+w^2)^.5 /jw(b^2+w^2)^.5 (c^2+w^2)^.5;


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F(w) = arg(a+jw)-arg(b+jw)-arg(c+jw)-90

Let us suppose GH(s) =K/(s+4)(s+6)(s+9)putting s =jw

GH(jw) = K/(jw+4)(jw+6)(jw+9)= K/{(16+w^2)^.5 arg(4+jw) (36+w^2)^.5

arg(6+jw) (81+w^2)^.5 arg(9+jw)}Next step is to variation of w from infinity to 0

GUDELINES TO DRAW THE POLAR PLOTS1.IDENTIFY G(s)H(s)

2.APPLY W TENDS TO INFINITY BY CONSIDERINGONLY HIGHER ORDER TERMS

3.APPLY W TENDS TO ZERO BY CONSIDERING ONLY

LOWER ORDER TERMS4.CONSIDER W=1 AS A TRAIL POINT AND GIVE THELOCATION OF ITS MAGNITUDE AND PHASE ANGLE


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5.NOW GIVE THE APPROXIMATE SHAPE OF POLAR

PLOT IN GH(w) PLANE6.APPLY POLAR PLOT ANALYSIS TO IDENTIFY THEOBSERVED STABILITY AND DEGREE OF STABILITY

OF CONTROL SYSTEM

FOR G(s) = K/[(s+4)(s+6)(s+9)] POLAR PLOT IS


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Similarly for the following we get the corresponding graphs in polar planeand X-Y PLANE

. G(s)H(s)= (1+sT)^-1


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SIMPLE PLOTS OF THE FOLLOWING FUNCTIONS

Polar plot of G(s)=1/(S+2)

First substitutes =j in G(s)

2tan)(

4

1)(

1

2

=+

= jGjG


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polar plot for the system with

)2)(1(

10)(

++

=

sss

sG

)2)(1(

10)(

++=

jjjjG

)2()1(

10)(

22++

=

jG

11 tan2

tan90)( = jG

The plot is shown in Fig


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STABILITY ANALYSIS OF POLAR PLOTS

HERE WE ARE CONSIDERING CHARACTERISTIC EQUATION1+GH(jw) = 0 WHERE (-1,J0) IS CONSIDERED AS ORIGIN AND AS

A CRITICAL POINT W.R.T CLOSED LOOP STABILITYIF POLAR PLOT PASSES THROUGH (-1,J0) POINT SYSTEM IS

CRITICALLY STABLEIF (-1,JO)IS NOT ENCLOSED BY PLOT SYSTEM IS ABSOLUTELYSTABLE

IF (-1,J0) IS ENCLOSED BY THE PLOT THEN THE SYSTEMIS


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IF ( 1,J0) IS ENCLOSED BY THE PLOT THEN THE SYSTEMISUNSTABLE

CONCEPT OF ENCLOSURE : A POINT OR REGION IS SAID TO BEENCLOSED BY POLAR PLOT IF THE POINT OR REGION LIES ON

THE LHS OF DIRECTION OF JOURNEY OF PLOT STARTING FROMW =INFINITY TO W=0

FOR FINDING OUT STABILITY

CONSIDER THE EXPRESSION AND RATIONALISE IT EQUATE THE IMAGINARY VALUE T0 ZERO TO FIND OUT PHASECROSSOVER FRQUENCY FIND OUT GH(jw) AND EQUATE IT TO a+JOSUPPOSING CRITICAL K=SOME VALUE WE CALCULATE a

IT IS SAIDlal = 1 SYSTEM IS CRITICALLY STABLE

lal < 1 SYSTEM IS STABLElal >1 SYSTEMIS UNSTABLE

WHERE a=(CRITICAL K)/(ACTUAL K)GAIN MARGIN ALSO DETERMINES THE STABILITY : GM=-20 loglal


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GM= 6dB IS THE BEST RANGE GM=0 Db CRITICALLY STABLE

GM= -6dB UNSTABLE

HENCE POLAR PLOT AND CONCEPT OF ENCIRCLEMENT ARE ABLE TOPREDICT THE STABILITY OF A SYSTEM.

FURTHER THIS PORTION OF POLAR PLOT IS USED IN NYQUIST CRITERION

PRESENTATION ON BODE PLOTS

By

Jaivee T.Joseph

O7324

Mechanical dept


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Mechanical dept.

B. Tech

BODE PLOTS

A Bode plot is a graph of the logarithm of the transferfunction of a linear, time-invariant system versusfrequency, plotted with a log-frequency axis, to show thesystem's frequency response It is usually a combination


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system s frequency response. It is usually a combinationof a Bode magnitude plot (usually expressed as dB of

gain) and a Bode phase plot (the phase is the imaginarypart of the complex logarithm of the complex transfer

function).

EXPRESSING IN DB

Given the tranfer function:

)1/)((

)1/()(

+

+=

pjwjw

zjwKjwG B


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)1/)(( +pjwjw

20log|G(jw) = 20logK + 20log|(jw/z+1)| - 20log|jw| -20log|jw/p + 1|

The gain term, 20logKB, is just so manydB and this is a straight line on Bode paper,independent of omega (radian frequency).

The term, - 20log|jw| = - 20logw, when plottedon semi-log paper is a straight line sloping at-20dB/decade. It has a magnitude of 0 at w = 1.

Poles, zeros and bode plots


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0

20

-20

=1

-20db/dec

wlg

The term, - 20log|(jw/p + 1), is drawn with the

following approximation: If w < p we use theapproximation that 20log|(jw/p + 1 )| = 0 dB,a flat line on the Bode. If w > p we use the

approximation of 20log(w/p), which slopes at-20dB/dec starting at w = p. Illustrated below.It is easy to show that the plot has an error of

-3dB at w = p and 1 dB at w = p/2 and w = 2p.One can easily make these corrections if it is

appropriate.


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0

20

-20

-40

= p

-20db/dec

wlg

20

+20db/dec

When we have a term of 20log|(jw/z + 1)| weapproximate it be a straight line of slop 0 dB/decwhen w < z. We approximate it as 20log(w/z)

when w > z, which is a straight line on Bode paperwith a slope of + 20dB/dec. Illustrated below.


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0

-20

-40

= z

+20db/dec

wlg

STRAIGHT-LINE AMPLITUDE PLOT

At every value of s where =xn

(a zero), increasethe slope of the line by per decade.

At every value of s where =yn

(a pole), decrease

the slope of the line by per decade.The initial value of the graph depends on the

boundaries. The initial point is found by putting theinitial angular frequency into the function andfinding |H(j)|.
http://en.wikipedia.org/wiki/Decade_%28log_scale%29http://en.wikipedia.org/wiki/Decade_%28log_scale%29


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finding |H(j)|.The initial slope of the function at the initial value

depends on the number and order of zeros and polesthat are at values below the initial value, and arefound using the first two rules.

STRAIGHT-LINE PHASE PLOT To draw the phase plot, for each pole and zero: if A is positive, start line (with zero slope) at 0 degrees if A is negative, start line (with zero slope) at 180 degrees at every =x

n(for stable zeros Re(z) < 0), increase the slope by

degrees per decade, beginning one decade before =xn (E.g.: ) at every =y

n(for stable poles Re(p) < 0), decrease the slope by

degrees per decade, beginning one decade before =yn

(E.g.: )

"unstable" (right half plane) poles and zeros (Re(s) > 0) haveopposite behavior


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flatten the slope again when the phase has changed by degrees(for a zero) or degrees (for a pole),

After plotting one line for each pole or zero, add the lines together toobtain the final phase plot; that is, the final phase plot is thesuperposition of each earlier phase plot.

ADVANTAGES

In the absence of a computer, a bode plotcan be sketched by approximating the

magnitude and phase with right linesegments.

Gain crossover, phase crossover, gainmargin and phase margin are more easily


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margin and phase margin are more easily

determined on the bode plot than the otherplots

For design purposes, the affect of addingcontrollers and their parameters are more

easily visualized here.

DISADVANTAGES

Absolute and relative stability of only minimumphase systems can be determined from bodeplots

They are useful only for stability studies ofsystems with minimum phase loop transferfunctions


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ADVANTAGES OF FREQUENCY RESPONSE OVER TIME

RESPONSE ANALYSIS

Here a system may be designed so that effects of undesirable noises arenegligible and such analysis and design canbe extended to certainnonlinear systems

In design problems, there are no unified methods of arriving at adesignated system that meets time-domain performance specifications. Onthe other hand, in frequency domain, a wealth of graphical and othertechniques are available that are useful for system analysis and design,irrespective of the order of the system


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irrespective of the order of the system

We can use the data obtained from measurements on the physical systemwithout deriving its mathematical model Frequency response tests are simple and canmake use readily available

sinusoidal signal generators

DIGITAL CONTROLSYSTEM


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Ashique Poyilil

7308

Mechanical engineering

238

Digital Control System - (DCS) A digital computer usedfor real-time control of a dynamic system, usually in anindustrial environment, possibly as part of aSupervisory Control and Data Acquisition (SCADA)system.

A DCS samples feedback from the system undercontrol and modifies the control signals in an attemptto achieve some desired behaviour.


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239

Analysis of such digital-analogue feedbacksystems can i

3-1 Control System Presentation

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