ARYA INSTITUTE OF ENGINEERING & TECHNOLOGY, JAIPUR

DEPARTMENT OF ELECTRICAL ENGINEERINGLAB-MANUALVI SEM EE

6EE07

CONTROL SYSTEM LABRAJASTHAN TECHNICAL UNIVERSITY SYLLABUS6EE07 CONTROL SYSTEM LAB-III1. Defining Systems in TF, ZPK form2. (a) Plot step response of a given TF and system in state-space. Take different values of damping ratio and wn natural undamped frequency. (b) Plot ramp response.3. For a given 2nd order system plot step response and obtain time response specification.4. To design 1st order R-C circuits and observe its response with the following inputs and trace the curve. (a) Step (b) Ramp (c) Impulse5. To design 2nd order electrical network and study its trarient response for step input and following cases. (a) Under damped system (b) Over damped System. (c) Critically damped system6. To Study the frequency response of following compensating Networks, plot the graph and final out corner frequencies. (a) Lag Network (b) Lead Network (c) Log-lead Network.7. To draw characteristics of a.c servomotor8. To perform experiment on Potentiometer error detector.9. Check for the stability of a given closed loop system.10. Plot bode plot for a 2nd order system and find GM and PM.ARYA INSTITUTE OF ENGG.& TECH.,JAIPUR

EXPERIMENTS LIST

6EE07 CONTROL SYSTEM LAB-IIIROTOR # 1

1. To study the comparison of following power electronics devices regarding ratings, performance characteristics and applications: Power Diode, Power Transistor, Thyristor, DIAC, TRIAC, GTO, MOSFET, MCT and SIT.

2. To determine V-I characteristics of SCR and measure forward breakdown voltage, latching and holding currents.

3. Find V-I characteristics of TRIAC and DIAC.

4. Find output & Transfer characteristics of MOSFET and IGBT.

5. Find UJT static emitter characteristics and study the variation in peak point and valley point

6. Study and test firing circuits for SCR-R, RC and UJT firing circuits.

ROTOR # 2

7. Study and test 3-phase diode bridge rectifier with R and RL loads. Study the effect of filters.

8. Study and obtain waveforms of single-phase half wave controlled rectifier with and without filters. Study the variation of output voltage with respect to firing angle.

9. Study and obtain waveforms of single-phase half controlled bridge rectifier(Semi Converter) with R and RL loads. Study and show the effect of freewheeling diode.

10. Study and obtain waveforms of single-phase full controlled bridge converter with R and RL loads. Study and show rectification and inversion operations with and without freewheeling diode.

11. Control the speed of a dc motor using single-phase half controlled bridge rectifier and full controlled bridge rectifier. Plot armature voltage versus speed characteristics.LAB ETHICS

DOs

1. Enter the lab on time and leave at proper time.

2. Keep the bags outside in the racks.

3. Utilize lab hours in the corresponding experiment.

4. Shut down the computers before leaving the lab.

5. Maintain the decorum of the lab.

Donts

1. Dont bring any external material in the lab.

2. Dont make noise in the lab.

3. Dont bring the mobile in the lab. If extremely necessary then keep ringers off.

4. Dont enter in server room without permission of lab incharge.

5. Dont litter in the lab.

6. Dont delete or make any modification in system files.

7. Dont carry any lab equipments outside the lab

We need your full support and cooperation for smooth functioning of the lab.

INSTRUCTIONS WHEN WORKING AT COMPUTER LABBEFORE ENTERING IN THE LAB

1. All the students are supposed to prepare the theory regarding the next program.

2. Students are supposed to bring the practical file and the lab copy.

3. Previous program should be written in the practical file.

4. Algorithm & Program of the current program should be written in the lab copy.

5. Any student not following these instructions will be denied entry in the lab and Sessional Marks will be affected.

WHILE WORKING IN THE LAB1. Adhere to experimental schedule as instructed by the faculty.

2. Get the previously executed program signed by the faculty.

3. Get the output of current program checked by the faculty in the lab copy.

4. Each student should work on his assigned computer at each turn of the lab.

5. Take responsibility of valuable accessories.

6. Concentrate on the assigned practical and dont play games.

7. If anyone is caught red-handed carrying any equipment of the lab, then he will have to face serious consequences.

INSTRUCTIONS WHEN WORKING AT CONTROL SYSTEM LABBEFORE ENTERING IN THE LAB

6. All the students are supposed to prepare the theory regarding the present Experiment.

7. Students are supposed to bring the practical file and the lab copy.

8. Previous experiment should be written in the practical file.

9. Object, Apparatus Table & Brief Theory of the current practical should be written in the lab copy.

10. Any student not following these instructions will be denied entry in the lab and Sessional Marks will be affected.

WHILE WORKING IN THE LAB

8. Adhere to experimental schedule as instructed by the faculty.

9. Record the observations in lab copy & checked by the faculty

10. Each student should work on his assigned table of the lab.

11. Take responsibility of valuable accessories.

12. Concentrate on the assigned practical and be careful.

13. If anyone is caught red-handed carrying any equipment of the lab, then he will have to face serious consequences.

EXPERIMENT # 1

OBJECT: INTRODUCTION TO MATLAB COMPUTING SOFTWARE.1.1 What is MATLAB?

MATLAB is a software package for high-performance numerical computation and visualization. It provides an interactive environment with hundreds of built-in functions for technical computation, graphics, and animation. Best of all, it also provides easy extensibility with its own high-level programming language. The name MATLAB stands for MATrix LABoratory. MATLAB is the product of the Math Works Inc.

1.2 Functions of MATLAB

MAT Labs built-in functions provide excellent tools for linear algebra computations, data analysis, signal processing, optimization, numerical solutions of ordinary differential equations (ODEs), quadrature, and many other types of scientific computations. Most of these functions use state-of-the art algorithms. There are numerous functions for 2-D and 3-D graphics as well as for animation. Also, for those who cannot do without their FORTRAN or C codes, MATLAB even provides an external interface to run those programs from within MATLAB. The user, however, is not limited to the built-in functions; he can write his own functions in the MATLAB language. Once written, these functions behave just like the built-in functions. MAT Labs language is very easy to learn and to use.

There are also several optional Toolboxes available from the developers of MATLAB. These Toolboxes are collections of functions written for special applications such as Symbolic Computation, Image Processing, Statistics, Control System Design, Neural Networks, etc.

1.3 General Ideas of MATLAB

Here we discuss some basic feature and commands. To begin, let us look at the general structure of the MATLAB environment.

1.3.1 MATLAB Windows

On almost all systems, MATLAB works through three basic windows, which are discussed below

1. Command window: This is main window. It is characterized by the MATLAB command prompt >>. When you launch the application program, MATLAB puts you in this window. All commands, including those for running user-written programs, are typed in this window at the MATLAB prompt. In MATLAB 6, this window is a part of the MATLAB window that contains four other smaller windows. Launch Pad: This sub window lists all MATLAB related applications and toolboxes that are installed on your machine. You can launch any of the listed applications by double clicking on them.

Workspace: This sub window lists all variables that you have generated so far and shows their type and size. You can do various things with these variables, such as plotting, by clicking on a variable and these variables, such as plotting, by clicking on a variable and then using the right button on the mouse to select your option.

Command History: All commands typed on the MATLAB prompt in the command window get recorded, even across multiple sessions (you worked on Monday, then on Thursday, and then on next Wednesday, and so on), in this window. You can select a command from this window with the mouse and execute it in the command window by double clicking on it. You can also select a set of commands from this window and create an M-file with the right click of the mouse (and selecting the appropriate option from the menu).

Current Directory: This is where all files from the current directory are listed. You can do file navigation here. You also have several options of what you can do with a file once you select it (with a mouse click). To see the options, click the right button of the mouse after selecting a file. You can run M-files, rename them, delete them, etc.

2. Graphics window: The output of all graphics commands typed in the command window are flushed to the graphics or Figure window, a separate gray window with (default) white background color. The user can create as many figure windows as the system memory will allow.

3. Edit window: This is where you write, edit, create, and save your own programs in files called M-files. You can use any text editor to carry out these tasks. On most systems, MATLAB provides its own built-in editor. However, you can use your own editor by typing standard file-editing command that you normally use on your system. From within MATLAB, the command is typed at the MATLAB prompt following the special character !. The exclamation character prompts MATLAB to return the control temporarily to the local operating system, which executes the command the following the ! character. After the editing is completed, the control is returned to MATLAB. For example, on UNIX systems, typing! VI myprogram.m at the MATLAB prompt (and hitting the return key at the end) invokes the VI editor on the file myprogram.m Typing! Emacs myprogram.m invokes the emacs editor.

1.3.2 Input-Output: - MATLAB supports interactive computation, taking the input from the screen, and flushing the output to the screen. In addition, it can read input files and write output files. The following features hold for all forms of input-output:

Data type: The fundamental data type in MATLAB is the array. It encompasses several distinct data objects-integers, doubles (real numbers), matrices character strings, structures and cells. In most cases, however, you never have to worry about the data type or the data object declarations. For example, there is no need to declare variables as real or complex. When a real number is entered as the value of a variable, MATLAB automatically sets the variable to be real (double). Dimensioning: Dimensioning is automatic in MATLAB. No dimension statements are required for vectors or arrays. You can find the dimension of an existing matrix or a vector with the size and length commands. Case sensitivity: MATLAB is case-sensitive; that is, it differentiates betn the lowercase and uppercase letters. Thus a and A are different variables. Most MATLAB commands and built-in functions calls are typed in lowercase letters. You can turn case sensitivity on and off with the casesen command. However we do not recommend it. Output display: The output of every\ command is displayed on the screen unless MATLAB is directed otherwise. A semicolon at the end of a command suppresses the screen output, except for graphic and on line help commands. Command history: MATLAB saves previously typed commands in a buffer. These commands can be recalled with the up-arrow key (). This helps in editing previous commands. You can recall previous command by typing the first few characters and then pressing the key. Alternatively, you can copy and paste commands from the Command History sub window where all your commands from even previous sessions of MATLAB are recorded and listed. On most Unix system, MATLABs command-line editor also understands standard emacs key bindings1.3.3 File types: - MATLAB has three types of files for storing information: M-files are standard ASCII text files, with a .m extension to the filename. There are two types of these files: script files and function files. Most programs you write in MATLAB are saves as M-files. All built-in functions in MATLAB are M-files, most of which reside on your computer in precompiled format. Some built-in functions are provided with source code in readable M-files so that can be copied and modified.

Mat-files are binary data files, with a .mat extension to the filename. Mat-files are created by MATLAB when you save data with the save command. The data is written in a special format that only MATLAB can read. Mat-files can be loaded into MATLAB with the load command.

Mex-files are MATLAB-callable FORTRAN and C programs, with a .mex extension to the filename. Use of these files requires some experience with MATLAB and a lot of patience.

1.4 Platform dependence

One of the best features of MATLAB is its platform-independence. Once you are in MATLAB, for most part, it does not matter which computer you are on. Almost all commands work same way. MATLAB supports almost every computational platform. In addition to windows, MATLAB available for AIX, Digital Unix, HP UX, IRIX, IRIX 64, Linux and Solaris operating system. Older version of MATLAB is available for additional platform such as Mac OS and open VMS.

1.5 General commands of MATLAB

1.5.1 Workspace information

who

lists variables currently in the workspace

whos

lists variables currently in the workspace with their size

what

lists m-, mat-, and mex-files on the disk

clear

clears the workspace, all variables are removed

clear x y zclears only variables x, y and z

clear allclears all variables and functions from workspace

clc

clears command window, command history is lost

home

same as clc

clf

clears figure window

1.5.2 Directory information

pwd

shows the current working directory

cd

changes the current working directory

dir

lists contents of the current directory

ls

lists contents of the current directory, same as dir

path

gets or sets MATLAB search path

editpath modifies MATLAB search path

copyfile copies a file

mkdir

creates a directory

1.5.3 General information

computer tells you the computer type you are using

clock

gives you wall clock time and date as a vector

date

tells you the date as a string

more

controls the paged output according to the screen size

ver

gives the license and the version information about MATLAB installed on

your computer

bench

benchmarks your computer on running MATLAB compared to other

computer

1.5.4 Termination

control-clocal abort, kills the current command execution

quit

quits MATLAB

exit

same as quit

RESULT:- We have successfully studied the basics of the MATLAB programing Software.

VIVA-VOCA QUESTIONS/ANSWERS:Q. 1 What is MATLAB?

Ans. MATLAB is a numerical computing environment and programming language. MATLAB is a software package for high-performance numerical computation and visualization. MATLAB means MATrix+LABoratory.Q.2 Who invented the MATLAB software?

Ans. MATLAB was invented in the late 1970s by Cleve Moler, then the chairman of the computer science department at the University of New Mexico. He designed it to give it to his students access to LINPACK and EISPACK without having to learn Fortran. It soon spreads to other universities and found a strong audience within the applied mathematics community. The MATLAB was rewritten in C and founded the MathWorks Incorporated in 1984 to continue its development.

Q.3 Describe the functions of MATLAB.

Ans. MATLAB allows easy matrix manipulation, plotting of functions and data, implementation of algorithms, creation of user interfaces, and interfacing with programs in other languages. Although it is numeric only, an optional toolbox uses the MuPAD symbolic engine, allowing access to computer algebra capabilities. An additional package, Simulink, adds graphical multidomain simulation and Model-Based Design for dynamic and embedded systems.

Q.4 Which is the latest version of the MATLAB software?

Ans. The latest version of the MATLAB is in year 2009 R2009a, 7.8. The first version of MATLAB was in year 1984 R-1.0.Q.5 Write down application of MATLAB software.

Ans. MATLAB was first adopted by control design engineers, Little's specialty, but quickly spread to many other domains. It is now also used in education, in particular the teaching of linear algebra and numerical analysis, and is popular amongst scientists involved with image processing.EXPERIMENT # 2

OBJECT: - To design 1st order R-C circuits and observes its response with the following inputs and traces the curve.

(a) Step input

(b) Ramp input

(c) Impulse inputApparatus required: -

PC with MATLAB 7.0.4 software.theory:-

First order system:For a linear variant system, when the power of s in denominator is one then it is called First order system.

Block diagram representation is shown in fig 2.1:

Fig 3.1 Block Diagram Ist order R-C circuit.For a R-C series circuit, shown in figure.2.2

Fig 2.2 Series RC network circuit.

By applying KVL in inner loop.

Taking Laplace transform .

(1)Apply KVL in outer loop.

(2)Divide equation 2 by equation 1.

Or transfer function =

Put RC=T

We conclude that this is a Ist order transfer function. So R-C circuit is Ist order system. With transfer function.

Case (1):

For unit step input response

Taking inverse Laplace

for t (0

Graph of C(t) V/S t is shown in fig 3.3.

Fig 3.3 unit step response curveCase (2):

For unit ramp input response of 1st order.

R(s) = 1/S2

By taking Laplace transform

for t(0

Error signal

Graph of C(t) V/S t for Ramp input is shown in Fig 3.4.

Fig 3.4 unit ramp response curveCase (3):

For unit impulse input response of 1st order system. R(s)=1

Taking Laplace transform.

,for t(0

The response curve of unit impulse input shown in fig 3.5.

Fig 3.5 unit impulse response curveSOURCE CODE:

Case (1):For unit step input response

1) t=0:.01:6

2) T=2

3) c=(1-exp(-t/T))

4) plot(t,c)

5) grid on

Fig 3.6 shows MATLAB output for unit step response

Fig.3.6 MATLAB output for unit step response

Case (2):

For unit ramp input response

1) t=0:.01:6

2) T=2

3) c=(t-T+T*exp(-t/T))

4) plot(t,c)

5) grid on

fig 3.7 shows MATLAB output for unit ramp response

Fig 3.7 MATLAB output for unit ramp response

Case (3):

For unit impulse input response

1) t=0:.01:6

2) T=2

3) c=(exp(-t/T))/T

4) plot(t,c)

5) grid on

fig3.8 shows MATLAB output for unit impulse response

Fig.3.8 MATLAB output for unit impulse responseResult:-We have successfully design 1st order R-C circuits and observe its response with the following inputs and trace the curve.

(a) Step input

(b) Ramp input

(c) Impulse inputDiscussion of the result:-

1. a series R-C circuit is a first order system.2. An important property of a linear-time variant system:a) for unit step input

, t0

ess= e() = 0.

B) for unit ramp input

,

t0

es s= T.

C) For unit impulse input

,

t(0So we conclude that response of the system varies as we vary the input.

VIVA-VOCA QUESTIONS/ANSWERS:Q.1)Define first order system?ANS.1) For a linear variant system, when the power of s in denominator is one then it is called First order system

Q.2)What is the value of r(t) for unit ramp input?

ANS.2) r(t) =t.

EXPERIMENT # 3OBJECT: - DESIGN THE SECOND ORDER ELECTRICAL NETWORK AND STUDY IT TRANSIENT RESPONSE FOR STEP INPUT UNDER FOLLOWING CASES. CASE 1: Undamped (=0)

CASE 2: Underdamped (=0.5)

CASE 3: Critical damped (=1)

CASE 4: Over damped (=1.5)

APPARATUS REQUIRED: - PC with MATLAB 7.0.4 software.THEORY:-

6.1 TIME RESPONSE OF SECOND ORDER SYSTEM: - The block diagram of second order control system is in fig.6.1.

Fig.3.1

. (3.1)

From equation (3.1)

For unit step input

. (3.2)

. (3.3)

Break the equation (6.3) by partial fraction, put . (3.4)

Multiply equation (3.4) by s and put s=0

Multiply equation (3.4) both side by [(s+n) 2-(n) 2] and put

Equation 3.3 can be written as

. (3.5)

Laplace inverse pf equation (6.5)

. (3.6)

Put

Put the values of wn and

. (3.7)

The error signal for the system

. (3.8)

The steady state value of c (t)

Therefore at steady-state there is no error between input and output.

n = natural frequency of oscillation or undamped natural fraquency

n = damped frequency of oscillation

n = damping factor or actual damping or damping coefficient

3.1.1 (A) Underdamped case (01, equation (6.3) can be written as

. (3.17)

. (3.18)

Break the equation (3.18) by partial fraction

Equation (3.18) can be written as

.(3.19)

Multiply both the sides bys and put s = 0

A =1

Multiply both the sides of equation (6.19) by s+ n+d and put s- n-d

Put and simplify for B

Put the values of A, B and C in equation (3.19)

Put the value of d

. (3.20)

Inverse Laplace of equation (3.20)

. (3.21)

From equation (3.21) we get two times constant

From equation (3.21) we observe that when is greater than one there are two exponential term, the first term has a time constant T1 which is smaller than the time constant of other exponential term (having time constant T2 ), in other word we can say that the first exponential term decaying much faster than other exponential term. So, for time response we can neglect it, then

.(3.22)

and time constant

.(3.23)

For different values of the curves of C (t) is shown in Fig.3.5

From the curves it is clear that the overdamped systems are sluggish.

Figure 3.5:- Overdamped

3.2 SOURCE CODE:-

CASE 1:- Undamped zeta=0 n=6 num= [wn*wn] den= [1 2*zeta*wn wn*wn] g=tf (num, den)

step (g)

CASE 2:- Underdamped

zeta=0.5 n=6 num= [wn*wn] den= [1 2*zeta*wn wn*wn] g=tf (num, den) step (g)

CASE 3:- Critical damped

zeta=1 n=6 num= [wn*wn] den= [1 2*zeta*wn wn*wn] g=tf (num, den) step (g)

CASE 4:- Overdamped zeta=1.5 n=6 num= [wn*wn] den= [1 2*zeta*wn wn*wn] g=tf (num, den) step (g)

Case 1:- Output for the undamped case.

zeta = 0

n =6

num =36

den =1 0 36

36

Transfer function= --------

s2 + 36

Fig.3.6:- UndampedCase 2:- Output for the under damped case.

zeta=0.5000

n = 6

num =36

den =1 6 36

36

Transfer function = --------------

s2 + 6 s + 36

Fig.3.7:-Underdamped

Case 3:-Output for the critically damped case.

Zeta =1

n =6

num =36

den =1 12 36

36

Transfer function= -----------------

s2 + 12 s + 36

Fig.3.7 Critically damped

Case 4:- Output for the over damped case.

zeta = 1.5000

n = 6

num = 36

den = 1 18 36

36

Transfer function= -----------------

S2+ 18 s + 36

Fig. 3.8 Over damped

RESULT: - The response of the second order control system for unit step input in different cases is shown in fig- 3.5, 3.6, 3.7 & 3.8 respectively.RESULT ANALYSIS: -To obtain the output of the second order control system by calculation is difficult& it also takes so much time. By theoretical calculate we cannot get the accurate result while the use of the MAT LAB we can easily calculate the output response of the second order transfer function. It is very simple & accurate.VIVA-VOCA QUESTIONS/ANSWERS:Q.1 What is time response analysis?

Ans: If the input is given to a system, it result in response which generally varies with time or in other word is function of time. This response is called time response and the study of this response is known as time response analysis. The time response of any system exhibit two parts.

1) transient response

2) steady state response

Q.2 What is second order system?

Ans: -If the simplest transfer function has two poles i.e. highest power of s is two in denominator (same or different) then it is called the second order system

Q.3 What is under damped response?

Ans:- In this case output oscillates about its final value. These oscillate consisting of overshoot & undershoot & finally settle down as tends to infinity.

Q.4 What is over damped response?

Ans: In this case , output dose not overshoot, the value dictated by the input but it take relatively long time to reach its final value.

Q.5 What is critically damped response?

Ans: In this case, output reaches its final value in the minimum possible time without overshoot.

Q.6 What is undamped response?

Ans: In this case, the output dose not reach its final value but it exhibit the sustained oscillation with constant amplitude.

Q.7 Consider the following statement for an under damped.

(i) Peak overshoot in step input response reduces as damping is increased from 0.2 to 0.6.

(ii) Response peak in frequency response reduces as damping is increased from 0.2 to 0.6.

(a) Non of the above statement is true

(b) Statement (i) is true but statement (ii) is false.

(c) Statement (ii) is true but statement (i) is false.

(d) Both the statement are true.

Ans: (d)

Q.8 Undamped natural frequency n and response frequency r of a unity feedback system with open-loop transfer function.

; < 1/2

are related as

(a) n =r (b) n >r (c)n 0(1)

Ec(t)= |Ec(t)| Sin(180+wt) , for Ec(t) The moment of inertia and the coefficient of viscous friction at the motor shaft are respectively Jm &Fm.

The angular shift resulting in the motor shaft being 0m and the corresponding angular velocity being Wm.

Form the torque speed characteristics the dynamic equation relating the motor torque and the speed is formed below;

Tm = m.Wm + k.Ec .(6)

(i) When the speed is zero, the torque is (To) and this stalling torque is proportional to the control voltage Vo.

To=K Vc (4) or, K=To/Vc (5)

(ii) The slope of the torque speed characteristic is:( m= -To/WO (7) . Then, Wm=d0m/dt (8)

The Torque (T) generated is a function of the motor shaft angular speed 0 & the control volt Ec. The equation can be given as:( T = M.d0m/dt + K.Vc (9)

Kc= positive constant

The Torque balance equation for 2-phase Servomotor is:( T = J. d20/dt2 + B.d0/dt(10)

Where, J = moment of inertia of motor & load referred to motor shaft.

B = viscous friction coefficient of the motor & load referred to motor shaft

THE DIAGRAM OF A.C. SERVO-MOTOR:( Fig:-4.1

THE ROTORS OF THE SERVO-MOTOR: - THE TORQUE-SPEED CHARACTERISTIC OF SERVOMOTOR: -

Fig 4.3

Where, X/R=Small (linear) [V4>V3>V2>V1] X/R=Large (Non-linear)

RESULT: - We have successfully studied to the A.C. servomotor and in that control .voltage

Result in the development of the motor torque (Tm). The servomotor provides a large Torque at zero speed. This torque is necessary for rapid acceleration.

THE ANALYSIS OF RESULT:(IN THE 2-phase A.C. Servo-motor is also an Induction Motor having drag-up type rotor construction, The control voltage Ec(t) is applied to the to the control winding & a fixed voltage having a phase difference of 90 w.r.t. control winding voltage is applied to the reference winding. The control voltage result in the development of the motor Torque (Tm).

The Servo-motor provides a large at zero speed. This torque is necessary for rapid Acceleration.

SERVO-MOTORS are widely used in Radars, Electro-mechanical, Actuators, Computers, Machine tools, tracking and Quittance system, process controllers

and ROBOTS.VIVA-VOCA QUESTIONS/ANSWERS:Q:-1 WHATS KIND OF APPLICATION IN USED SERVO-MOTOR?

ANS:- Servo-motors are used in feedback control systems. Servo-motor

Have low rotor inertia & high speed of response. Its also called

Control motors.

Q:-2 WHAT IS THE RELATIONSHIP IN FEEDBACK?

ANS:- Servo-motors are used in feedback control system should have

Linear relationship between electrical control signal and rotor

Speed, torque speed characteristics should be linear.

Q:-3 WHAT IS THE RESPONSE OF A.C. SERVO-MOTOR?

ANS:- The response of Servo-motor should be fast and inertia should be

Low.

Q:-4 WHAT IS THE TYPES OF SERVO-MOTOR?

ANS:- The Servo-motor are classified as :-

A.C. Servomotor

D.C. Servomotor

Special Servo-motor

Q:-5 WHAT IS THE PARTS USED IN A.C. SERVO-MOTOR?

ANS:- The Servo-motor are two parts namely STATOR & ROTOR

A.C. Servo-motor is two phase induction motor.

Q:-6 What is the displaced angle of distributed winding?Ans:- The stator has two distributed winding. These winding are

Displaced from each other by 90 electrical .One is called main

Winding & other is called reference winding.

Q:-7 Whats values of depend on the torque-speed characteristics?Ans:-The torque-speed characteristics of two phase induction motor

Depends upon the ratio of reactance to resistance.

Q:-8 What is the types of torque speed characteristics?

ANS:- For high resistance and low reactance, the characteristics is linear

and for large ratio of X to R, it becomes of Non-linear.

Q:-9 Whats the various control voltage for the torque speed characteristics?Ans:- The torque speed characteristics for various control voltages are Almost linear.

[V4>V3>V2>V1]

Q:-10 Whats the types of rotor?Ans: - Rotor has two types.

Squirrel cage Rotor.

Drag cup type Rotor.

The squirrel cage rotor having large length & Small diameter.

So, its resistance is very high.

EXPERIMENT # 5

Object: - To perform experiment on Potentiometer error Detector.Apparatus Required: -

1) Experimental Board of Potentiometer.

2) Power supply (230 V).

3) Connecting leads.

Theory: -

IntroductionA potentiometer is an electromechanical transducer which converts the mechanical energy (Displacement) into electrical energy (Voltage). It is also called error detecting device.

Potentiometric transducers are relatively inexpensive and easy to apply.Precision potentiometers are simple rotary devices for obtaining shaft position information. The most straight forward application is the conversion of mechanical position to a voltage. Basically a precision potentiometer consists of a resistive element with a movable arm or slider in contact with the element. As the arm (slider) rotates, resistance varies between the end of the resistive element and the slider, indicating shaft position. The resistive element can be made of wire, conductive film or cermet element.

Potentiometers used for servomechanisms are generally above 7/8 to 3 and half inches in diameter. The early models were mostly of wire wound type. Current technology provides other choices such as stability, longer life and lower sensitivity to environment. Potentiometer can be excited with alternating and direct current. Single turn potentiometers have a rotation i.e. usually limited to 350 degrees.

Potentiometer CHARACTERISTICS.

A linear potentiometer produces a resistance change i.e. linearly related to the shaft position. A position of rotation will produces 50% of maximum resistance and a position of rotations will produce 75% of maximum resistance. Linearity is specified as the deviation (in percentage of total resistance) of the actual resistance. This is called normal or independent linearity.

Fig. 5.1 Potentiometer

Resolution in a potentiometer is minimum change of resistance output expressed as percentage of its total resistance. It is dependent of the number of turns of the wire per inch on the winding and the arch diameter of the slider. Noise in the potentiometer appears as spurious unwanted voltage. For wire wound potentiometer will cause a ripple voltage to appear at the slider as the shaft is rotated.

From figure No. 9.1 Under balance condition the ratio between output voltage (Eout) and input voltage (Ein) is given by

Where yi displacement from zero position

yt total length of transnational potentiometer.

Fig. 5.2 Plot Between Angular Position & Output Voltage

Similarly for rotational motion, the ratio between output voltage (Eout) and voltage (Ein) is given by

Where

i input angular displacement

t total travel of wiper.

Fig. 5.3 Double Ended arrangement of potentiometer

Potentiometer can be used as a error detector to compare the position of two remotely located shafts in figure 2. This type of arrangement is called the double ended arrangement.

In this arrangement two potentiometers are connected in parallel with their common point earthed. The output voltage (Eout) is given by

Eo=Kp (1 2)Where

The applied voltage is normally D.C. and polarity of output voltage describes the relative position of the shafts. In case of A.C., the phase difference will find the relative position of the two shafts.

Resolution for Potentiometer: -It is defined as the ratio of change in the output voltage in one step to the supply voltage.

Operating instructions1. Connect the mains cord to the supply line.

2. You can study the potentiometer as transducer by measuring voltage across the output points of the potentiometer and other the readings in the table.

3. Plot the graph for above readings.

4. Now you can study the system as null detector by keeping left hand potentiometer at zero degrees and treating it as reference, rotate the right hand potentiometer to read the error directly on the DPM. Note that these are ten turn potentiometers.5. You may also plot input position verses output position when meter is indicating null.

Some more important instruction.1. D.P.M. can be connected across the O/P and GND point of either potentiometer or it can be connected across the variable terminals of both pots for studying error detector function.2. If SW is OFF, you can measure angular displacement V/S resistance characteristics of potentiometer.3. By connecting 50K, 100K, 220K & 500K etc. you can also observe the loading error on the potentiometer sensor. You may connect there load resistances across the O/P terminals and GND terminal for each potentiometer or across the D.P.M. when error detector operation is being studied.4. Operate the potentiometer knobs very carefully.5. SW should be in on position for study of potentiometer.6. Cal. pot is adjust for 3.60 v D.C. across terminals 1&3 with SW On.

Fig. 5.4 Study Of Potentiometer

Fig 9.5 Study of PotentiometerOBESRVATION:-

SR. NO.Angular positionOutput Voltage

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Result: - After performing this experiment we have obtain the relation between angular

Position and output voltage which can be shown in graphical manner.

Analysis of Result: - we are obtained the graph between angular position and output voltage and this graph shows that the output voltage varies linearly with the angular position of the shaft.

APPLICATION OF POTENTIOMETER:-

A. Self-balancing potentiometer

The merits of dc potentiometer for accurate measurement of dc voltage have prompted the development of self-balancing (automatic) potentiometer for the measurement of dc voltages. They are widely used for indication and record of voltage developed by various transducers. The automatic balancing is provided be an ac servomotor. The servomotor drives the tapping point of the potentiometer, the writing mechanism, and the pointer for indication. The motor comes to rest when the unbalance voltage becomes zero. The block diagram is shown in figure No. 9.6. The output displacement and voltage feedback are related by the calibration constant of the potentiometer wire. The amplifier is tuned to 50 Hz and has high gain. The unbalance dc voltages are converted into proportional ac voltage of 50 Hz be means of a synchronous chopper. The forward path consist of the servomotor and the inertial load of the writing pen. By virtue of the negative feedback, the bandwidth of the system is increased, and the system is used to record faithfully voltages of frequency range dc to 5 Hz. The system is considered to function as voltage to displacement converter. The output shaft can be coupled to a shaft angle encoder it, in addition to analog indication, digital display is desired.

Fig. 5.5 Self Balancing Servo Operated Potentiometer

Fig. 5.6 Block Diagram of Servo Operated Potentiometer

Precaution: - when we perform the experiment on the potentiometer follow the following precautions.

1. Connection should be tight.

2. Connection should be proper.

3. First connect the complete circuit and then switch on power supply.

4. Switch off the power supply after taking readings.VIVA-VOCA QUESTIONS/ANSWERS:Q.1What is the transducer?

Ans.Transducer is a device which converts the non electrical input to electrical output.

Q.2What is the POT?

Ans.A resistive potentiometer used for the purpose of voltage division is called a POT.

Q.3In a resistive potentiometer high value of resistive of POT leads to:

Ans.High value of sensitivity.

Q.4Define the potentiometer.

Ans.A potentiometer is an instrument designed to measure an unknown voltage by comparing it with a known voltage.

Q.5A potentiometer is basically a :

Ans.Null type instrument.

Q.6When a potentiometer is used for measurement of voltage of an unknown source, the power consumed in the circuit of the unknown source under null condition.

Ans.Ideally zero.

TYPICAL TEST RESULT FOR POTENTIOMETER AS ERROR DETECTOR (Standard results as comes with manual):-

1) ANGULAR POSITION Vs OUTPUT VOLTAGESR.NO.ANGULAR POSITION

IN DEGREEOUTPUT VOLTAGE

IN MV.

1.1010

2.3029

3.6059

4.9090

5.120120

6.150150

7.180181

8.210211

9.240240

10.270269

11.300299

12.330329

13.360359

2) ANGULAR POSITION Vs RESISTANCE :-SR.NO.ANGULAR POSITION

IN DEGREERESISTANCE IN

K-Ohm

1.00.001

2.30.080

3.60.162

4.90.247

5.120.331

6.150.413

7.180.496

8.210.577

9.240.659

10.270.742

11.300.825

12.330.907

13.360.990

3) ERROR BETWEEN I/P & O/P ANGULAR POSITION:-

SR.NO.I/P ANGULAR

POSITION IN

DEGREEO/P ANGULAR

POSITION IN

DEGREEERROR

IN DEGREE

1.203010

2.608019

3.9012028

4.14017030

5.18022039

6.24026527

7.24032040

8.32035030

9.34036020

10.3503609

EXPERIMENT # 6OBJECT: - CHECK FOR THE STABILITY OF A GIVEN CLOSED LOOP SYSTEM.APPARATUS REQUIRED: - P. C. with MAT LAB 7.0.4 software.

THEORY:-

CONCEPT OF STABILITY

The concept of stability is very important to analyze and design the system. A system is said to be stable if its response cannot be made to increase indefinitely by the application of a bounded input excitation. If the output approaches towards infinite value for sufficiently large time, the system is said to be unstable.

A linear time invariant (LTI) system is stable if

(1) The system is excited by a bounded input, the output is bounded. (BIBO stability criteria)

(2) In the absence of the input, the output tends to zero(the equilibrium state of the system).

This is known as asymptotic stable.

NECESSARY BUT NOT SUFFICIENT CONDITIONS FOR STABILITY:-

Consider a system with characteristic equation

a0 sm+a1sm-1+.+am=0

(a)All the coefficients of the equation should have same sign,

(b)there should be no missing term.

If above two conditions are not satisfied the system will be unstable. But if all the coefficients have same sign and there is no missing term we have no guarantee that the system will be stable.

There are two methods for check the stability of a given closed loop.

ROUTH-HURWITZ CRITERION:-

Consider the following characteristics polynomial

a0s n+a1sn-1+.+an=0. ................... (1)

Where the coefficients a0, a1.an are all of the same sign and none is zero.

STEPArrange all the coefficients of above equ. (1) in two rows.

ROW 1 a0 a2 a4

ROW 2 a1 a3 a5

STEP From these two rows form a third row

ROW1 a0 a2 a4

ROW2 a1 a3 a5

ROW3 b1 b3 b5

Where, b1=-1a1 aoa2a1a3

b3=-1a1 a0a4a1a5

STEPFrom second and third row, form a fourth row

ROW1 a0 a2 a4

ROW2 a1 a3 a5

ROW3 b1 b3 b5

ROW4 c1 c3 c5

c1=-1b1 a1a3b1b3

c3=-1b1 a1a5b1b5

STEP Continue this procedure of forming a new rows

STATEMENT OF ROUTH-HURWITZ CRITERION

Routh-Hurwitz criterion states that the system is stable if and only if all the element in first column has the same algebraic sign. If all elements are not of the same sign then the number

Of sign changes of the elements in first column equals the number of roots of the characteristic equation in right half of the s-plane (or equals to the number of roots with positive real

Parts)

INPUT:- A closed loop control systems has the characteristic equation given by

S3+4.5S2+3.5S+1.5=0

Investigate the stability using Routh-Hurwitz criterion.

OUTPUT:-

S3 1 3.5

S2 4.5 1.5

S1 3.5

S0 1.5

No. of sign changes in first column=0

No. of roots in right half of S- plane=0

Hence, System is stable.

ROOT LOCUS METHOD:-

Root locus method is a graphical method in which roots of the characteristic equation are plotted in S-plane for the different values of parameter. The locus of the roots of the characteristic Equation when gain is varied from zero to infinity is called Root-locus.

RULES FOR CONSTRUCTION OF ROOT LOCII:-

Following are the rules to sketch the root locus plot.

Rule1: The root locus is symmetrical about the real axis.

Rule2: The root locii starts from an open loop pole with K=0 e.g. For the system having

G(s) H(s)=K(s+3)/(s+2)..(2)

Find the starting point of root locii.

Solution:-According to the rule the root locii starts from s=-2

Rule3: The root locii will terminate either on an open loop zeros or on infinity with K= e.g., Find the ending point of the root locii given in equation (2). According to the rule the root locii will terminate at s=-3

Rule4:-If N=No. of separate locii

P=No. of finite poles

Z=No. Of finite zeros then

Number of root locii will be equal to the no. Of poles if number of poles are more than number of zeros i.e.P>Z

N=P if p>z

If z>p, then number of root locii will be equal to the number of zeros.

If p=z, then No. of root locii=poles =zeros.

E.g.find the number of separate root locii for the system given by the equation (2).

Solution: p=1

Z=1

N=1

Rule5: ROOT LOCII ON THE REAL AXIS.

Any point on the real axis is a part of the root locus if and only if the number of poles and zeros to its right is odd.

Rule6:ASYMPTOTES

The branches of the root locus tend to infinity along a set of straight line called a asymptotes. These asymptotes making an angle with real axis and is given by

Q= (2K+1)1800/p-z where K=0,1,2,

The total number of asymptotes=p-z

e.g. If G(s) H(s)=K/s(s2+6s+10)(3)

p=3

z=0

No. of asymptotes=p-z=3-0=3

K=0 Q1= (2*0+1)*1800/3=600

K=1 Q2= (2*1+1)*1800/3=1800

K=2 Q3= (2*2+1)*1800/3=3000

Rule7: CENTROID OF ASYMPTOTES

The point of intersection of asymptotes with real axis is called centroid of asymptotes (SA) and is given by

SA= (sum of poles-sum of zeros)/p-z

E.g. Find the centroid of asymptotes of the system given by equation(3)

Solution: There are three poles at s1=0,s2=-3+j1,s3=-3-j1

No. of zeros=0

So, centroid SA=0-3+J1-3-J1-0/3=-2

Rule8: ANGLE OF DEPARTURE & ANGLE OF ARRIVAL OF THE ROOT LOCII

The angle of departure of the root locus from a complex pole is given by

SD=1800-(sum of the angles of vectors drawn to this pole from other poles) + (sum of angles of vectors drawn to this pole from the zeros).

The angle of arrival at a complex zero is given by

SA==1800-(sum of angles of vectors drawn to this zero from other zeros) + (sum of angles of vectors drawn to this zero from poles).

E.g. For

G(s) H(s) =k/s(s+6)(s2+4s+13)

Determine the angle of departure from complex poles.

SD=1800-(P1+P2+P2)

=1800-(1230+370+900) =-700

So, angle of departure at (-2+j3) =-700

So, angle of departure at (-2-j3)=700

Rule9: BREAKAWAY POINT ON REAL AXIS

If the root locus lies between two adjacent open loop poles on the real axis then there will be at least one breakaway point , because the roots move towards each other as K is increased and meet at a point. At this point K is maximum. If we increase the value of K between two poles the root locus breaks in two parts.

Similarly if root locus lies between two adjacent zeros on real axis, then there will be at least one break in point. If the root locus lies between an open loop pole and zero, then there will be no breakaway or break in point or may be both occur.

The breakaway or break in points can be determined from the roots of

Dk/ds=0

e.g. if G(s) H(s)=K/s(s2+6s+10)determine the breakaway point.

1+G(s) H(s)=1+k/s(s2+6s+10)

S (s2+6s+10)+k=0

Or, k=-s3-6s2-10s

dk/ds=-32-12s-10=0

s1=-1.1835 & s2=-2.815 are the breakaway points.

Rule10: The intersection of root locus branches with jw-axis can be determined through Routh Hurwitz criterion.

e.g. If G(s)H(s)=k/s(s2+6s+10).Find the intersection of root

locii with the imaginary axix.

Solution: The characteristics equation s3+6s+10s+k=0

S3 1 10

S2 6 k

S1 60-k/6

S0 k

Hence, we get a zero row if k=60

The auxiliary equation A(s)=6s2+k

6s2+k=0

6s2+60=0

S=+-j3.16

The root locus branches cross the imaginary axis at s=+-j3.16 for k=60.

INPUT: Plot the root locii for the closed loop control system and check the stalibity with

K

G(s)= ---------------------- , H(s)=1

s(s+1)(s2+4s+5)

Solution:Step1:Plot the poles and zeros s2+4s+5

S=-4s (16-20)/2=-2j1

Poles are at s1=0,s2=-1,s3=-2+j1& s4=-2-j1

Step2:The segment between s=0 & s=-1 is the part of root locus on real axis.

Step3:Number of root locii

Number of root loci N=P=4

Step4:Centroid of the asymptotes

SA=(sum of poles-sum of zeros )/p-z

= 0-1-2+j1-2-j1-0/4 = -5/4 = -1.25

Step5:Angle of asymptotes

Q=(2k+1)*180/p-z

K=0 Q1=450

K=1 Q2=1350

K=2 Q3=2250

K=3 Q4=3150

Step6: Breakaway point

The characteristic equation

1+G(s) H(s)=0

1+k/s(s+1)(s2+4s+5)=0

S4+5s3+9s2+5s+k=0

K=-(s4+5s3+9s2+5s)

dk/ds=-[4s3+15s2+18s+5]

4s3+15s2+18s+5=0

So, Breakaway point is s=-0.4

Step7: Angle of departure at the upper complex pole

SD=1800-(1540+1360+900)=-2000

Step8:Point of intersection on jw axis

S4+5s3+9s2+5s+k=0

S4 1 9 k

S3 5 5

S2 8 k

S1 ( 40-5K)/8

SO K

At 40-5k = 0 k = 8(for sustained oscillation)

The auxiliary equation A(s)=8s2+k

8s2+8=0

S=j1

Step9: From Routh table for stability

k>0

( 40-5k)/8>0

(40-5k)=0 or k=0

poles

k=k(n)

break

end

OUTPUT:

poles =

-10.0156 0.0078 + 9.5998i

0.0078 - 9.5998i

k = 71

Fig 6.1 A PLOT OF ROOT LOCUS

RESULT:- We have studied carefully and successfully and apply stability criterion , prove the stability of given closed loop function.

ANALYSIS OF RESULT:

1.I have studied the concept of stability is very important to analyse and design the system .

2.Any system is said to be stable if its response cannot be made to increase indefinitely by the application of a bounded input excited .

3.A linear time invariant (LTI) sytem is stable if

(a) The system is excited by a bounded input , the output is bounded(BIBO).

(b)In the absence of the input , output tends towards zero.

Some standard criteria must be study to check the stability.i.e.

(a)all the coefficient of the equestion should have same sign.

(b) There should be no missing term.

Otherwise system will be unstable.VIVA-VOCA QUESTIONS/ANSWERS: Q1 Why concept of stability is nessary?

Ans: Concept of stability is nessary to analyse and design the system.

Q.2 what is the meaning of stability?

Ans: A system is said to be stable if its response cannot be made to increase indefinite by the application of a bounded input excitation.

Q.3 What is linear time invariant(LTI)system.

Ans: A system which output is not vary with

time called LTI.

Q.4 What is the rule for stability of linear invareiant system?

Ans (a) The system is excitedby a bounded input, the output is bounded.

(b) In the absence of the input, the output tends towards zero.

Q.5 What is an asymptotic stability?

Ans: In the absence of input, the output tends towards zero(the equilibrium state of a system), is known as asymptotic stable.

Q.6 What are the necessary conditions for stability?

Ans: There are some necessary conditions for stability-

(a) All the coefficients of the should have same sign.

(b) There should be no missing term.

Q.7 What is the statement of Routh-Hurwitz criterion?

Ans: Routh-Hurwitz criterion states that the system is stable if and only if all the elements in the first column have the same algebraic sign. If all elements are not of the same sign then the number of sign changes of the elements in first column equals the characteristics equation in the right half of the s-plan.

Q.8 What is root locus?

Ans: The locus of the roots of the characteristics equation when gain varied from zero to infinity is called root locus.

Q.9 What is the meaning of equilibrium state ?

Ans: When input is absence then output tends to zero is called equilibrium state

EXPERIMENT # 7Object: -plot bode plot of given transfer function in second order control system.

Gs=25s2+4s+25

Apparatus required : - P C with MAT lab 7.0.4 softwareTheory: -Simplified and desirable form but actual solution of the problem must still be accomplished . solution of the problem may be either in the form of analysis or design for which following alternatives are possible.

Transformed differential equation keeping all initial condition zero may be obtained after manipulating the transfer function and the transient solution can also be obtained by inserting the proper boundary conditions and then using laplace transform inversion.

The transfer function can be graphically represented on the s-plane to evaluate the roots of the characteristic equation, open loop frequency response or closed loop frequency response through graphical manipulations.

The transfer function G(j) just by replacing s by j and which can then be used to obtain open loop and closed loop frequency response curve directly.

Advantages :

One of the advantage of frequency response method of control system analysis and design is that in this method sinusoidal signal is used as standard test signal which is easily available.

Frequency response test signal is simple and reliable.

Secondly, with the use of experimental data unknown transfer function of complicated components and system can be determine

Undesirable, noise is minimize

In frequency response method many powerful graphical tools are available such as Bode plot, magnitude versus phase plot and nyquist plot etc.

FREQUENCY TRANSFER FUNCTION

Frequency response test can be carried out through sinusoidal transfer function. Such a transfer function is defined to be the complex ratio of the steady state output sine wave to the input sine wave. So for linear components the frequency function is obtained from the given transfer function just by replacing s by j hence sinusodial transfer function and frequency transfer function is nothing but transfer function in which s is replaced by j.

For example: s transfer function is given as

Ts=KS+1ss+5s2+2s+1

Replacing s by j, we obtained sinusoidal or frequency transfer function as

Tj=kj+1jj+5j2+2j+1

BODE PLOT

Bode plot developed by H.W BODE can be simplified because they can be approximated as a sequence of straight line. Straight line approximation simplifies the evaluation of magnitude and phase frequency response. Bode plot consist of two plots. One is logarithm of magnitude versus frequency and other is phase angle versus frequency. Both the plots are drawn on single semi log graph paper. This plot can be sketched from their asymptotic properties as no detailed plotting is required. Bode plot are also known as asymptotic plots and bode plot can be sketched by approximating the magnitude and phase width asymptotic straight line which requires very less time. Some of the important parameters of frequency response like gain margin, phase margin, gain crossover frequency and phase crossover frequency can easily be determined from bode plot.

LIMITATION

Bode plot can be used to determined stability of only minimum phase system (minimum phase system is the system which has all its poles and zeros on the left half of s-plane.

Bode plot in comparison to polar plot can be more easily and quickly constructed.

For example

Gs=25s2+4s+25

Change the transfer function in s+1 form

=1s225+4s25+1

Putting an equation s=j

=11-225+0.15

We determine the values of magnitude in db and phase angles in degrees at different values of frequency

S.NOFACTORCORNER FREQUENCYMAGNITUDE IN DB

1.11-225+0.15=5Straight line of constant slope -20db/decade passing through 0 db line at =5

Calculate the phase angle

=-tan-10.1525-2

Calculate the value of for different values of .

S.NO

1.0.1-0.039

2.0.2-0.06

3.0.3-0.17

4.1-0.355

5.2-0.80

6.3-1.603

7.4-0.000872

8.50

9.64.57

10.72.50

11.817.06

SHAPE

Fig 7.1

SOURCE CODE: -num=[0 0 25];

den=[1 4 25];

g=tf(num,den)

bode(g)

grid on

[mag,phase,w]=bode(g)

OUTPUT: - The bode plot of the given transfer function is shown in figure. SHAPE

Fig 7.2

ANALYSIS OF RESULT:

Successfully done the bode plot of given transfer function and successfully plot the graph.

VIVA-VOCA QUESTIONS/ANSWERS:Q.1 which software is used in MAT LAB?

Ans: MAT LAB 7.0.4 software are used.

Q.2 why bode plot are used?

Ans: Bode plot can be used to determined stability of only minimum phase system (minimum phase system is the system which has all its poles and zeros on the left half of s-plane.

Q.3 which parameters are used in bode plot?

Ans: Some of the important parameters of frequency response like gain margin, phase margin, gain crossover frequency and phase crossover frequency can easily be determined from bode plot.

Q.4 what is the unit of frequency?

Ans: Unit of frequency is rad/sec.

Q.5 what is bode plot?

Ans: Bode plot can be simplified because they can be approximated as a sequence of straight lineis known as bode plot.

Q.6 How much plot consist of bode plote?Ans: Bode plot consist of two plots. One is logarithm of magnitude versus frequency and other is phase angle versus frequency. Both the plots are drawn on single semi log graph paper.

Q.7 In transfer function s is replaced in which form?

Ans: In transfer function s is replaced by j.

PAGE 0

AIET/Deptt. of EE/CS Lab/0

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