GATE EE vol-3

8/9/2019 GATE EE vol-3

1/50

GATE

ELECTRICAL ENGINEERINGVol 3 of 4


2/50

Second Edition

GATEELECTRICAL ENGINEERING

Vol 3 of 4

RK Kanodia

Ashish Murolia

NODIA & COMPANY


3/50

GATE Electrical Engineering Vol 3, 2eRK Kanodia & Ashish Murolia

Copyright By NODIA & COMPANY

Information contained in this book has been obtained by author, from sources believes to be reliable.However, neither NODIA & COMPANY nor its author guarantee the accuracy or completeness of anyinformation herein, and NODIA & COMPANY nor its author shall be responsible for any error, omissions, ordamages arising out of use of this information. This book is published with the understanding that NODIA &COMPANY and its author are supplying information but are not attempting to render engineering or other

professional services.

MRP 530.00

NODIA & COMPANYB 8, Dhanshree Ist, Central Spine, Vidyadhar Nagar, J aipur 302039Ph : +91 141 2101150,www.nodia.co.inemail : [email protected]

Printed by Nodia and Company, J aipur


4/50

SYLLABUS

GENERAL ABILITY

Verbal Ability : English grammar, sentence completion, verbal analogies, word groups,instructions, critical reasoning and verbal deduction.

Numerical Ability :Numerical computation, numerical estimation, numerical reasoning anddata interpretation.

ENGINEERING MATHEMATICS

Linear Algebra:Matrix Algebra, Systems of linear equations, Eigen values and eigen vectors.

Calculus:Mean value theorems, Theorems of integral calculus, Evaluation of definite andimproper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourierseries. Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes,Gauss and Greens theorems.

Differential equations: First order equation (linear and nonlinear), Higher order linear

differential equations with constant coefficients, Method of variation of parameters, Cauchysand Eulers equations, Initial and boundary value problems, Partial Differential Equationsand variable separable method.

Complex variables: Analytic functions, Cauchys integral theorem and integral formula,Taylors and Laurent series, Residue theorem, solution integrals.

Probability and Statistics: Sampling theorems, Conditional probability, Mean, median,mode and standard deviation, Random variables, Discrete and continuous distributions,Poisson,Normal and Binomial distribution, Correlation and regression analysis.

Numerical Methods: Solutions of non-linear algebraic equations, single and multi-stepmethods for differential equations.

Transform Theory:Fourier transform,Laplace transform, Z-transform.

ELECTRICAL ENGINEERING

Electric Circuits and Fields:Network graph, KCL, KVL, node and mesh analysis, transientresponse of dc and ac networks; sinusoidal steady-state analysis, resonance, basic filterconcepts; ideal current and voltage sources, Thevenins, Nortons and Superposition andMaximum Power Transfer theorems, two-port networks, three phase circuits; Gauss Theorem,electric field and potential due to point, line, plane and spherical charge distributions;Amperes and Biot-Savarts laws; inductance; dielectrics; capacitance.


5/50

Signals and Systems:Representation of continuous and discrete-time signals; shifting andscaling operations; linear, time-invariant and causal systems; Fourier series representationof continuous periodic signals; sampling theorem; Fourier, Laplace and Z transforms.

Electrical Machines:Single phase transformer equivalent circuit, phasor diagram, tests,

regulation and efficiency; three phase transformers connections, parallel operation; auto-transformer; energy conversion principles; DC machines types, windings, generatorcharacteristics, armature reaction and commutation, starting and speed control of motors;three phase induction motors principles, types, performance characteristics, starting andspeed control; single phase induction motors; synchronous machines performance, regulationand parallel operation of generators, motor starting, characteristics and applications; servoand stepper motors.

Power Systems:Basic power generation concepts; transmission line models and performance;cable performance, insulation; corona and radio interference; distribution systems; per-

unit quantities; bus impedance and admittance matrices; load flow; voltage control; powerfactor correction; economic operation; symmetrical components; fault analysis; principles ofover-current, differential and distance protection; solid state relays and digital protection;circuit breakers; system stability concepts, swing curves and equal area criterion; HVDCtransmission and FACTS concepts.

Control Systems: Principles of feedback; transfer function; block diagrams; steady-state errors; Routh and Niquist techniques; Bode plots; root loci; lag, lead and lead-lagcompensation; state space model; state transition matrix, controllability and observability.

Electrical and Electronic Measurements:Bridges and potentiometers; PMMC, moving iron,dynamometer and induction type instruments; measurement of voltage, current, power,energy and power factor; instrument transformers; digital voltmeters and multimeters;phase, time and frequency measurement; Q-meters; oscilloscopes; potentiometric recorders;error analysis.

Analog and Digital Electronics:Characteristics of diodes, BJ T, FET; amplifiers biasing,equivalent circuit and frequency response; oscillators and feedback amplifiers; operationalamplifiers characteristics and applications; simple active filters; VCOs and timers;combinational and sequential logic circuits; multiplexer; Schmitt trigger; multi-vibrators;sample and hold circuits; A/ D and D/A converters; 8-bit microprocessor basics, architecture,programming and interfacing.

Power Electronics and Drives:Semiconductor power diodes, transistors, thyristors, triacs,GTOs, MOSFETs and IGBTs static characteristics and principles of operation; triggeringcircuits; phase control rectifiers; bridge converters fully controlled and half controlled;principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.

***********


6/50

PREFACE

This book doesnt make promise but provides complete satisfaction to the readers. Themarket scenario is confusing and readers dont find the optimum quality books. This book

provides complete set of problems appeared in competition exams as well as fresh set ofproblems.

The book is categorized into units which are then sub-divided into chapters and theconcepts of the problems are addressed in the relevant chapters. The aim of the book isto avoid the unnecessary elaboration and highlights only those concepts and techniques

which are absolutely necessary. Again time is a critical factor both from the point of viewof preparation duration and time taken for solving each problem in the examination. Sothe problems solving methods is the books are those which take the least distance to thesolution.

But however to make a comment that this book is absolute for GAT E preparation willbe an inappropriate one. T he theory for the preparation of the examination should befollowed from the standard books. But for a wide collection of problems, for a variety ofproblems and the efficient way of solving them, what one needs to go needs to go through

is there in there in the book. Each unit (e.g. Networks) is subdivided into average sevennumber of chapters on an average each of which contains 40 problems which are selectedso as to avoid unnecessary redundancy and highly needed completeness.

I shall appreciate and greatly acknowledge the comments and suggestion from the users ofthis book.

R. K . Kanodia

Ashish Murolia


7/50

CONTENTS

CS CONTROL SYSTEM

CS 1 Transfer Function CS 3

CS 2 Stability CS 52

CS 3 Time Response CS 92

CS 4 Root Locus Technique CS 137

CS 5 Frequency Domain Analysis CS 180

CS 6 Design of Control Systems CS 226

CS 7 State Variable Analysis CS 242

CS 8 Gate Solved Questions CS 298

SS SIGNALS & SYSTEMS

SS 1 Continuous Time Signals SS 3

SS 2 Continuous Time Systems SS 37

SS 3 Discrete Time Signals SS 78

SS 4 Discrete Time Systems SS 107

SS 5 The Laplace Transform SS 147

SS 6 The Z-transform SS 178

SS 7 The Continuous-Time Fourier Transform SS 222

SS 8 The Continuous-Time Fourier Series SS 263

SS 9 Sampling and Signal Reconstruction SS 302

SS 10 Gate Solved Questions SS 323

***********


8/50

CS 1

ROOT LOCUS TECHNIQUE

CS 1.1 Form the given sketch the root locus can be

CS 1.2 Consider the sketch shown below

The root locus can be

(A) (1) and (3)

(B) (2) and (3)

(C) (2) and (4)

(D) (1) and (4)


9/50


PE 1 Root Locus Technique PE 10

EF 1 Root Locus Technique EF 10

www.nod

ia.co.in

Buy Online: www.nodia.co.in

*Shipping Free* *Maximum Discount*

www.

nodia.

co.in

GATE Electrical Engineering-2015in 4 Volumes

by R. K. Kanodia & Ashish Murolia

Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

CS 1.3 The valid root locus diagram is

CS 1.4 A open-loop pole-zero plot is shown below

The general shape of the root locus is



10/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt

erofGATEElectricalEn

gineering,V

olume-3

GATE EE vol-1

Electric circuit & Field, Electrical & electronic measurementGATE EE vol-2

Analog electronics, Digital electronics, Power electronics

GATE EE vol-3

Control systems, Signals & systems

GATE EE vol-4

Electrical machines, Power systemsEngineering mathematics, General Aptitude





11/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3



CS 1.8 The forward-path open-loop transfer function of a ufbsystem is

( )G s ( )( )

s s

K s s

8 25

2 62= + +

+ +

The root locus for this system is

CS 1.9 The forward-path open-loop transfer function of as ufbsystem is

( )G s ( )

( )

s

K s

1

42

2

=+

+

For this system root locus is


12/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


CS 1.10 The forward-path open-loop transfer function of a uf bsystem is

( )G s ( )

s

K s 12

2

= +

The root locus of this system is

Common Data For Q. 11 and 12:

A open-loop pole-zero plot is shown below


13/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

CS 1.11 The transfer function of this system is

(A)( )( )

( )s s

K s s

3 22 22

+ +

+ + (B)

( )

( )( )

s s

K s s

2 2

3 23+ +

+ +

(C)( )( )

( )s s

K s s

3 22 22

+ +- + (D)

( )( )( )s s

K s s

2 23 2

2- ++ +

CS 1.12 The break point is(A) breakaway at .129s =-

(B) breakin at .243s =-

(C) breakaway at .243s =-

(D) breakin at .129s =-


A root locus of ufbsystem is shown below

CS 1.13 The breakaway point is _ _ _ _

CS 1.14 At breakaway point the value of gain K is

(A) 24

(B) 7.4 10 3# -

(C) 8.6 10 3# -

(D) 29

CS 1.15 The forward-path transfer function of a ufbsystem is

( )G s ( )( )

( )

s s s

K s

3 2 2

22= + + +

+

The angle of departure from the complex poles is _ _ _ _

CS 1.16 Consider the feedback system shown below


14/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


For this system root locus is


A root locus for a ufb system is shown below

CS 1.17 The root locus crosses the imaginary axis at(A) 3.162j!

(B) 2.486j!

(C) 4.564j!

(D) None of the above

CS 1.18 The value of gain for which the closed-loop transfer function will have a pole onthe real axis at 5- , will be _ _ _ _

CS 1.19 The open-loop transfer function a system is

( ) ( )G s H s ( )( )( )

( )s s s s

K s

4 12 208

=+ + +

+

A closed loop pole will be located at 10s=- , when the value of K is _ _ _

CS 1.20 For a uf bsystem forward-path transfer function is

( )G s ( )( )

( )s s

K s

3 56

=+ +

+

The breakaway point and break-in points are located respectively as

(A) ,4.273

(B) 7.73,4.27

(C) 4.27,3

(D) 4.27,7.73


15/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

CS 1.21 The open loop transfer function of a system is given by

( ) ( )G s H s ( )( )s s s

K

1 2=

+ +

The root locus plot of above system is

CS 1.22 A uf bsystem has forward-path transfer function

( )G s s

K2=

The root locus plot is

CS 1.23 For the ufbsystem, shown below consider two point

s1 2 3j=- + and 2s j2

12 =- +

Which of the above point lie on root locus ?


16/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


(A) Both s1and s2 (B) s1but not s2

(C) s2but not s1 (D) neither s1nor s2

CS 1.24 A uf bsystem has open-loop transfer function

( )G s ( )

( ),

s s

K s0> >2 b

ab a=

+

+

The valid root-loci for this system is

CS 1.25 The characteristic equation of a feedback control system is given by

( 4 4)( 11 30) 4s s s s K s K 2 2 2+ + + + + + 0=where 0K > . In the root locus of this system, the asymptotes meet in s-plane at(A) ( 9.5,0)-

(B) ( 5.5,0)-

(C) ( 7.5,0)-

(D) None of the above

CS 1.26 The root locus of the system having the loop transfer function

( ) ( )G s H s ( )( )s s s s

K

4 4 52=

+ + +has

(A) 3 breakaway point

(B) 3 breakin point

(C) 2 breakin and 1 breakaway point

(D) 2 breakaway and 1 break-in point

CS 1.27 Consider the ufbsystem shown below


17/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

The root-loci, as ais varied, will be

Common Data For Q. 27 and 28 :

The forward-path transfer function of a ufbsystem is

( )G s ( )

( )( )

s s

K s s

1

32

a=

-

+ +

CS 1.28 The root-loci for 0K > with 5a = is

CS 1.29 The root-loci for 0>a with 10K = is


18/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


CS 1.30 For the system ( ) ( )G s H s ( )( )

( )s s

K s

2 46

=+ +

+, consider the following characteristic

of the root locus :1. It has one asymptotes

2. It has intersection with jw-axis3. It has two real axis intersections.

4. It has two zeros at infinity.

The root locus have characteristics

(A) 1 and 2 (B) 1 and 3

(C) 3 and 4 (D) 2 and 4

CS 1.31 The characteristic equation of a closed-loop system is ( 1)( 2) 0s s s K + + + = .The centroid of the asymptotes in root-locus will be _ _ _ _

CS 1.32 The forward path transfer function of a uf bsystem is

( )G s ( )( )( )

( )s s s s

K s

1 2 43

=+ + +

+

The angles of asymptotes are

(A) 0, ,2p

p (B) 0, ,32

34p p

(C) , ,3 3

5pp

p (D) None of the above

CS 1.33 Match List-I with List-I I in respect of the open loop transfer function

( ) ( )G s H s ( )( )( )

( )( )

s s s s s

K s s s

20 50 4 5

10 20 5002

2

=+ + + +

+ + +

List I (Types of Loci) List II (Numbers)

P. Separate Loci 1. One

Q. Loci on the real axis 2. Two

R. Asymptotes 3. Three

S. Break away points 4. Five

P Q R S(A) 4 3 1 1

(B) 4 3 2 1(C) 3 4 1 1(D) 3 4 1 2


19/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

CS 1.34 The root-locus of a uf bsystem is shown below

The open loop transfer function is

(A)( )( )s s s

K

1 3+ + (B)

( )( )

s s

K s

13

+

+

(C) ( )

( )

s s

K s

3

1

+

+

(D) ( )( )s s

K s

1 3+ +

CS 1.35 The characteristic equation of a linear control system is 5 9 0s K s2+ + = . Theroot loci of the system is

CS 1.36 A unity feedback control system has an open-loop transfer function

( )G s ( )s s s

K

7 122=

+ +

The gain K for which 1 1s j=- + will lie on the root locus of this system is

_ _ _

CS 1.37 An unity feedback system is given as ( )G s ( )( )

s s

K s

31

=+

-. Which is the correct root

locus diagram ?


20/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


CS 1.38 The open loop transfer function ( )G s of a uf bsystem is given as

( )G s ( )s s

K s

2232

=+

+^ hFrom the root locus, at can be inferred that when K tends to positive infinity,

(A) three roots with nearly equal real parts exist on the left half of thes-plane

(B) one real root is found on the right half of the s-plane(C) the root loci cross the jwaxis for a finite value of ; 0K K!

(D) three real roots are found on the right half of the s-plane

CS 1.39 The characteristic equation of a closed-loop system is

( 1)( 3) ( 2) 0, 0s s s K s k >+ + + + = .Which of the following statements is true ?

(A) Its root are always real

(B) It cannot have a breakaway point in the range [ ]Re s1 0< a


21/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

CS 1.41 Angles of asymptotes are(A) 60 ,120 ,300c c c (B) , ,60 180 300c c c

(C) , ,90 270 360c c c (D) , ,90 180 270c c c

CS 1.42 Intercepts of asymptotes at the real axis is

(A) 6- (B)310

-

(C) 4- (D) 8-

CS 1.43 Break away points are(A) .1056- , .3471- (B) 2.112, 6.943- -

(C) 1.056, 6.943- - (D) 1.056, 6.943-

CS 1.44 For the characteristic equation s s K s K 23 2+ + + 0= , the root locus of the systemas K varies from zero to infinity is

CS 1.45 The open loop transfer function of a ufbsystem is

G s H s ^ ^h hs s

K s

9

12= +

+

^^

hh

In the root locus of the system as parameter K is varied from 0 to 3, the gainK when all three roots are real and equal is _ _ _ _


22/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


CS 1.46 A uf bsystem has an open loop transfer function

G s H s ^ ^h hs s

K s

1

1=

-

+

^

^

h

hRoot locus for the system is a circle. Centre and radius of the circle are respectively(A) ,0 0^ h, 2(B) ,0 0^ h, 2(C) ,1 0-^ h, 2(D) ,1 0-^ h, 2

CS 1.47 The open loop transfer function of a system is

G s H s ^ ^h hs s

K s

2

3=

+

+

^^

hh

The root locus of the system is a circle. The equation of circle is(A) 4 42 2s w+ + =^ h (B) 3 32 2s w- + =^ h(C) 3 32 2

2s w+ + =^ ^h h

(D) 4 22 2 2s w- + =^ ^h h

CS 1.48 Consider the open loop transfer function of a system shown below

G s H s ^ ^h hs s s s

K

2 2 6 102 2=

+ + + +^ ^h hThe break away point in root locus plot for the system is/ are(A) 3 real

(B) only real

(C) 1 real, 2 complex

(D) None

CS 1.49 The open loop transfer function of a ufbsystem is given below.

G s H s ^ ^h hs s s

K

4 5=

+ +^ ^h hConsider the following statements for the system.

1. Root locus plot cross jw-axis at s j2 5!=

2. Gain margin for K 18= is 20 dB.

3. Gain margin for K 1800= is 20dB-

4. Gain K at breakaway point is 13.128

Which of the following is correct ?

(A) 1 and 2

(B) 1, 2 and 3

(C) 2, 3 and 4

(D) All

CS 1.50 The open loop transfer function of a system is

G s H s ^ ^h hs s

K

1 5=

+ +^ ^h hWhat is the value of K , so that the point s j3 5=- + lies on the root locus?


23/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

CS 1.51 The open loop transfer function of a control system is


K s

1 4 16

12= - + +

+

^ ^

^

h h

hConsider the following statements for the system1. Root locus of the system cross jw-axis for .K 357=

2. Root locus of the system cross jw-axis for .K 233=

3. Break away point is .s 045=

4. Break in point is .s 226=-

Which of the following statement is correct ?(A) 1, 3 and 4 (B) 2, 3 and 4

(C) 3 and 4 (D) all

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


24/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


SOLUTION

CS 1.1 Correct option is (D).

Option (A) :

Root locus is always symmetric about real axis. This condition is not satisfied for

option (A). A point on the real axis lies on the root locus if the total no. of poles

and zeros to the right of this point is odd. This is also not satisfied by (A). Thus

option (A) is not a root locus diagram.

Option (B) & Option (C) :

These does not satisfy the condition that, a point on the real axis lies on the root

locus if the total no. of poles and zeros to the right of this point is odd. Thus,option (B) and (C) are also not root locus diagram.

Option (D) :

This is symmetric about real axis and every point of locus satisfy the condition

that no. of poles and zeros in right of any point on locus be odd.


Here, option (2) and option (3) both are not symmetric about real axis. So, both

can not be root locus.

CS 1.3 Correct option is (A).

Here pole-zero location is given as

The angle of departure of the root locus branch from a complex pole is given by Df 180! c f= +6 @where fis net angle contribution at this pole due to all other poles and zeros.

f Z Pf f= -

f 90 90 90 90c c c c= - + - +6 6@ @ f 180c=-

Departure angle Df 180 180! c c= -6 @


25/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

Df 0! c=

So, departure angle for pole P0is 0c, So root locus branch will depart at 0c. Only

option (A) satisfy this condition.


Given open loop pole-zero plot is :

No. of poles P 2=

No. of zeros Z 1=

No. of branches of root locus is equal to =no. of poles 2= . Thus (B) and (D)

are not correct.

The branch of root locus always starts from open loop pole and ends either at an

open loop zero (or) infinite. Thus (C) is incorrect and remaining (A) is correct.

CS 1.5 Correct option is (C).

Root locus plot starts from poles and ends at zeros (or) infinite. Only option (C)satisfy this condition. No need to check further.


Root locus always starts from open loop pole, and ends at open loop zero (or)

infinite. Only option (A) satisfy this condition.

We can find the root locus of given plot as follows :

Here, no. of poles 2=

and no. of zeros 0=

So, no. of asymptotes P Z 2= - =and angle of asymptotes is :

af P Zq2 1 180c

=-

+^ h; q 0= , 1

af 20 1 180

90c

c= +

=^ h

; q 0=

and af P Z2 1 180

23 180 270#

c cc=

-

+= =

^ h; q 1=

Hence, root locus plot will be as shown below.


26/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4



An open loop pole-zero plot is given as :

Root locus always starts from open loop poles and end at open loop zeros or

infinite along with asymptotes. Thus option (C) and (D) are wrong.

A point on the real axis lies on the root locus if the total no. of poles and zeros to

the right of this point is odd. This is not satisfied by (B). Thus remaining Correctoption is (A).


Forward Path open loop transfer function of given uf bsystem is

G s^ hs s

K s s

8 25

2 62= + +

+ +^ ^h h

Characteristic equation is

s s8 252+ + 0=

s j2

8 64 100 4 3! !=- - =-

s j4 3=- + ; s j4 3=- -

Thus pole are s j4 3=- + ; s j4 3=- - and zeros are 2s=- ; 6s=-

Plot of Pole-zero is shown below

Root locus will start from poles and ends with zeros.

Only option (C) satisfy above condition.

CS 1.9 Correct option is (B).

Forward path open loop transfer function of given uf bsystem is,

G s

^ h

s

K s

1

42

2

=

+

+^ h

Here, poles s 12+ 0=

s j1!=

and zeros s 42+ 0=

s j2!=


27/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

Pole-zero plot is

Root locus will start from poles and ends with zeros. Thus (B) is correct.


Forward Path one loop transfer function of given uf bsystem is

G s^ hs

K s 12

2

= +^ h

Zeros are : s 12+ 0=

s j1!=

Poles are : s 0= ; s 0=

Location of poles and zeros is shown below

Hence option (B) and (D) may not be correct option. A point on the real axis

lies on the root locus if the total no. of poles and zeros to the right of this point

is odd. This is not satisfied by (C) because at origin there are double pole. Thus

remaining Correct option is (A).

CS 1.11

Correct option is (C).Given open loop pole zero plot is :

From above plot zeros are 1 1s j= + and 1 1s j= - and poles are 2s=- and3s=-

Transfer function of the system is

G s^ hs s

K s j s j

2 3

1 1 1 1=

- - - -

- + - -

^ ^^ ^

h hh h

" "" "

, ,, ,


28/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


s s

K s j s j

2 3

1 1 1 1=

+ +

- - - +

^ ^^ ^

h hh h" ", ,

s s

K s j

2 3

1 12 2

= + +

- -

^ ^^ ^h hh h" ,or G s^ h

s s

K s s

2 3

2 22=

+ +

- +

^ ^^

h hh


Method i :

Root locus lie on real axis where no. of poles and zeros are odd in number from

that right side.

Hence, from given pole-zero plot root locus lie between poles 2-^ hand 3-^ honreal axis. Thus s 1.29=- can not be break point because it does not lie on root

locus. Thus possible break point is 2.43s =- which lies between 2- and 3- .

On root locus it may be seen easily that 2.43s =- lie on root locus and locus

start from pole 2-^ hand 3-^ h. Thus at 2.43s =- it must break apart.Thus this point is break away point.

Gain K will be maximum at break away pointand minimum at break in point.

We can also check maxima and minima for gain K

Method II

The point, at which multiple roots are present, are known as break point. These

are obtained from theds

dK 0=

Here, characteristic equation is

G s H s 1+ ^ ^h h 0=

s s

K s s1

2 3

2 22+

+ +

- +

^ ^^

h hh 0=

K s s

s s

2 2

2 32= - +

- + +

^^ ^

hh h

s s

s s

2 2

5 62

2

=- +

- + +^ h ...(1)

Now, differentiating eq (1) w.r.t sand equating zero we have

ds

dK s s

s s s s s s

2 22 2 2 5 5 6 2 2 02 2

2 2=

- +

- - + + + + + -=^^ ^ ^ ^hh h h h

s s7 8 222+ - 0=

which gives 1.29s=+ and 2.43s=- out of which 2.43s=- is break point.

CS 1.13 Correct answer is .145- .

Given root locus of uf bsystem is shown below


29/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

Here, root locus branches meet between 1- and 2- and go apart. Hence, break

away point will lie between 1- and 2- ; .

For this system zeros are s 3= and s 5= ; poles are s 1=- and s 2=- .

Transfer function of given system will be

G s^ hs s

K s s

1 2

3 5=

+ +

- -

^ ^^ ^

h hh h

Characteristic equation

G s H s 1+ ^ ^h h 0=

s s

K s s1

1 2

3 5+

+ +

- -

^ ^^ ^

h hh h

0=

K s s

s s

8 15

3 22

2

=- +

- + +

^

hh ...(1)

Differentiating eq. (1) wrt to sand equating to zero we have

dsdK

s s

s s s s s s

8 15

8 15 2 3 3 2 2 802 2

2 2

=- +

- - + + + + + -=^

^ ^ ^ ^h

h h h h s s11 26 612- - 0=

or s .39=+ and .s 145=-

Thus .s 145=- is break away point and .s 39=+ is break in point.


Break away and Break in points always satisfy characteristic equation.

Substituting .s 145=- in eq (1), we get

K . .

. .

145 8 145 15

145 3 145 22

2

=- - - +

- - + - +

^ ^^ ^

h hh h

68

@B

..

28702502475

=- -^ h

.862 10 3#= -

CS 1.15 Correct answer is 108.4 .

Forward path transfer function of given ufbsystem is

G s^ hs s s

K s

3 2 2

22= + + +

+

^ ^^

h hh

Here zero is s 2=- and poles are s 3=- and s j1 1!

=-Pole-zero Plot is shown below

Angle of departure at pole P1is given by :

Df 180! c f= +6 @


30/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


where fis net angle contribution at pole P1due to all other poles and zeros.

f Z Pf f= -

Z P P1 2 3f f f= - +

6 @where, Z1f tan 11= - ; 90P2 cf = ; tan 21P3 1f = - f 9tan tan1 0

211 1

c= - +- -: DDeparture angle :

Df 180! c f= +6 @ 1tan tan180 1 90

21 1

! c c= + - -- -: D .180 45 90 2656!= + - -6 @ Df .1084! c=

Hence, departure angle for pole P1is 108.4c

+ and departure angle for pole P2is108.4c- because P1and P2are complex conjugate.


The given system is shown below

We redraw the block diagram after moving take off point as shown below

Forward path transfer function is

G s^ h s sK s

1 3

2= + +

- +

^ ^h hhRoot locus is plotted for K 0= to K 3= . But here, K is negative. Thus we will

plot for K 3=- to K 0= . This is called complementary root locus.

Hence, the root locus on the real axis is found to the left of an even countof real

poles and real zeros of G s^ h. Plot will start from pole and ends on zero. ThusCorrect option is (A).


Given root locus is shown below


31/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

It does not have any zero and have poles at s 4=- ands j1 1!=- .

Thus open loop transfer function,

G s

^ h

s s s

K

4 2 22=

+ + +^ ^h h

s s s

K

6 10 83 2=

+ + +

Closed loop transfer function,

T s^ h1

s s s

Ks s s

K

6 10 8

6 10 8

3 2

3 2

=+

+ + +

+ + +

s s s K

K

6 10 83 2=

+ + + +

Characteristics equation :

6 10 8s s s K 3 2+ + + + 0=

Routh array is shown below

s3 1 10

s2 6 K8+

s1 K

652-

s0 K8+

When root locus cut will cut imaginary axis if element in s1is zero.

Thus K6

52- 0= &K 52=

and then from auxiliary equation we have

s6 8 522+ +^ h 0= s2

660 10=- =-

s .j3162!=

CS 1.18 Correct answer is 17.Using Pole - zero plot :

Gain K at any s s0=^ hpoint on root locus is given by : K

s s0=

intPr intProduct of phasorsdrawnfrom at that po

oduct of phasorsdrawnfrom at that poOL Z

OL P=

Here no any zero is present.

Hence, K =Product of phasors drawn fromOL Pat that point


32/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


Ks 5=-

PP P P PP 3 1 2# #= ^ ^ ^h h h 1 4 1 4 12 2# #= + +

4 1 172= + =

K 17=

CS 1.19 Correct answer is 600.

Open loop transfer function of given system is


K s

4 12 20

8=

+ + +

+

^ ^ ^^

h h hh

Here, Zeros : Z 8=-

Poles : P 01 = ; P 42 =- ; P 123 =- ; P 204 =-

Value of K at 10s=- is :

Ks 10=-

10

10

Phasorsdrawnfrom at

Phasorsdrawnfrom at

OL Z s

OL P s

p

p=

=-

=-

^ hhPole-zero plot is shown below.

Ks 10=-

PZ

PP PP PP PP 1 2 3 4# # #= ^

^ ^ ^ ^h

h h h h

2

10 6 2 10 600# # #= =

K 600=


For given system, forward path transfer function,

G s^ h s sK s

3 5

6= + +

+

^ ^^ h hhCharacteristic equation for closed loop transfer function,

G s H s 1+ ^ ^h h 0=or

s s

K s1

3 5

6+

+ +

+

^ ^^h h

h 0=

or K s

s s

6

8 152=

+

- + +

^^

h h

differentiating w.r.t to sand equating to zero, we have

ds

dK s

s s s s

6

6 2 8 8 1502

2

=

+

- + + + + +=

^^ ^ ^

hh h h

or s s12 332+ + 0=

or s .773=- and .s 427=-

The root locus is shown below. Here we can easily say that s .427=- is break

away point and s .773=- point.


33/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3


Open loop transfer function :

G s H s ^ ^h hs s s

K

1 2=

+ +^ ^h hIn option (A) and pption (C), the root locus on the real axis is found to the left

of an even count of real poles and real zeros of GH . Hence, these can not rootlocus diagram.

Now characteristic equation ,

G s H s 1+ ^ ^h h 0=

s s s

K11 2

++ +^ ^h h 0=

s s s K 3 23 2+ + + 0=

Routh array is shown below.

s3

1 2

s2 3 K

s1 K

36-

s0

K

At K 6= , s1row is zero, thus using auxiliary equation we get,

s3 62+ 0=

s j 2!=

Root locus cut on jwaxis at s j 2!= for K 6= . Thus option (B) is correct

because (D) does not cut jwaxis.


For given uf bsystem, forward transfer function,

G s^ hs

K2=

Angle of departure or angle of asymptote for multiple poles is,

af rq2 1 180c

= +^ h

;

where r =no. of multiple poles

q 0= , 1, 2, .... r 1-^ hHere, r 2= ; (2 multiple poles at origin) q 0= , 1

af 20 1 180

90c

c= +

=^ h

for q 0=

and af 22 1 180

270c

c= +

=^ h

for q 1=


34/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


Hence, root locus plot will be as shown below.


For given ufbsystem, open loop transfer function is,

G s

^ h

s s

K s s

1 2

3 4=

+ +

+ +

^ ^^ ^

h hh h

Given two point, s1 j2 3=- + ; s j22

12 =- +

If any point lie on root locus, it satisfies the characteristic equation. Thus we have

q s^ h 1 G s H s 0= + =^ ^h hor G s H s ^ ^h h 1= (Magnitude)and G s H s ^ ^h h 180! c= (Phase)Point: 2 3s j1 =- +

Now we use the phse condition G s H s s s

11

f==

^ ^h h , we have 1f 1 2 3 4q q q q= + + +

tan tan tan23

13 90

131 1 1

c= + - --

- - - b l tan tan tan

23 3 90 180 31 1 1c c= + - - -- - -^ h

. . .5630 7156 270 7156c= + - +

or 1f .7056 180! c!=-

Hence, point s1not lie on root locus.

Point : s j22

12 =- + :-


35/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

Using phase condition G s H s s s2=

^ ^h h 2f= we have, 2f 31 2 4q q q q= + - -

tan tan tan2 2

1

2

1

90 180 2

11 1 1c c

= + - - -

- - -

c m tan tan tan

2 21

21 90 180

211 1 1

c c= + - - +- - -

2f 180c=- Here phase condition is satisfied.

Hence, point s2lies on root locus.


For given uf bsystem, open loop transfer function is

G s^ hs s

K s2 b

a=

+

+

^

^

h

h; 0> >b a

Here, zero : Z a=-

Poles : s 0= , 0; s b=-

Departure angle at double poles on origin is :

fr

q2 1 180c=

+^ h; r 2= , q 0= , 1

f 90c= and 270c

To get intersection with imaginary axis we use Ruth array as shown below.

characteristic equation is

s s K s K 3 2b a+ + + 0=

Routh Array is shown below

s3

1 K

s2

b Ka

s1

b

b a-

s0

Ka

Here, for any value of K , slrow of Routh array will not be zero. Thus system

is stable for all positive value of K and hence root locus does not cross jwaxis.

Therefore root locus completely lies in left half of s-plane.Based on these above result we say that Correct option is (A).


Characteristic equation of given feedback control system is

s s s s K s K 4 4 11 30 42 2 2+ + + + + +^ ^h h 0=or

s s s s

K s1

4 4 11 30

42 2

2

++ + + +

+

^ ^^h h

h 0= ...(1)

Characteristic equation is,

G s H s 1+ ^ ^h h 0= ...(2)Comparing, eq (1) and (2) we get open loop transfer function as G s H s ^ ^h h

s s s s

K s

4 4 11 30

42 2

2

=+ + + +

+

^ ^^h h

h

( )( )( )( )s s s s

K s

2 2 5 642

=+ + + +

+^ h


36/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


Open loop poles are

s ,2 2=- - and s 6=- , 5-

The point at which asymptotes meet (centroid) is given by:

As Re ReSumof Sumof

P Z

P Z=

-

-

^ h6 6@ @Here, open loop zeros : s 4 02+ =

s j2!=

As 4 22 2 5 6 0

=-

- - - - -^ h

.215 75=- =-

It intersects on real axis. So point is: . ,75 0-^ hCS 1.26 Correct option is (D).

Open loop transfer function for given system is


K

4 4 52=

+ + +^ ^h hCharacteristic equation,

s s s s

K14 4 52

++ + +^ ^h h 0=

K s s s s 4 4 52=- + + +^ ^h h ...(1)Differentiating wrt sand equating to zero we have

dsdK s s s s s s 2 4 4 5 4 2 4 02 2=- + + + + + + =^ ^ ^ ^h h h h8 B

or s s s s s 2 4 4 5 42 2- + + + + +^ ^h h 0=or s s s2 2 8 52- + + +^ ^h h 0= ...(2)or s 2=- and .s 0775=- , .3225-

Now, we check for maxima and minima value of gain K at above point. If gain

is maximum, then that point will be break away point. If gain is minimum, then

that point will be break in point.

Again differentiating equation (2) with respect to s, we get,

dsd K

2

2

s s s s 2 8 5 2 4 82

=- + + + + +^ ^ ^h h h6 @ s s6 24 212=- + +^ hFor .s 0775=- and .s 3225=- ,

ds

d K2

2

. 060 =+ ; s 2=- is minima points.

Thus s 2=- is break in point.

Hence, there is two break away points . , .s 0775 3225=- -^ hand one break inpoint.


For a uf bsystem, forward transfer function is :


37/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3 G s^ h s s

1a

=+^ h

Method I :

Characteristic equation : G s H s 1+ ^ ^h h 0=

s s1 1

a+

+^ h 0= s s 12 a+ + 0=

s

s112

a+

+^ h 0=Open loop transfer function, as ais varied is

G s H s ^ ^h hs

s

12a

=+

Here, zero : s 0=and poles : s 1 02+ = s j& !=

Root locus will be :

Method II :

Closed loop transfer function :

T s^ hG s H s

G s

s s1 11

2 a=

+ =

+ +^ ^^h hh

G s H s

G s

1+ ^ ^^h hh

s

ss

1

1

11

2

2

a=

+

+

+

G s H s ^ ^h hs

s

12a

=+


Forward path transfer function of given ufbsystem is

G s^ hs s

K s s

1

32

a=

-

+ +

^^ ^

hh h

; 5a = and K 0>

or G s^ hs s

K s s

1

5 32= -

+ +

^^ ^

hh h

Zeros : s 5=- , s 3=- Poles : s 0= , s 1= , s 1=-

Locus branches start from pole and ends on zeros or infinite along asymptote.

Here, no. of asymptotes P Z 3 2 1= - = - =

Only option (A) has one asymptotes.


38/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


Angle of asymptotes,

af P Z

q2 1 180c=

-

+

^

^

h

h; q 0= , 1, 2, .... P Z 1- -^ h

1

0 1 180180

cc=

+=^ h

Only option (A) satisfies these conditions.


Forward path transfer function :

G s^ hs s

K s s

1

32

a=

-

+ +

^^ ^

hh h

; K 10= and 0>a

Characteristic equation,

G s H s 1+ ^ ^h h 0=

s s

s s1

1

10 32

a+

-

+ +

^^ ^

hh h

0=

s s s s 10 3 33 2 a a- + + + +^ h6 @ 0= 10s s s s 10 29 32 a+ + + +^ ^h h 0= ...(1)

s s s

s1

10 29

10 32

a+

+ +

+

^ ^

hh

0= ...(2)

Open loop gain as ais varied,

G s H s ^ ^h hs s s

s

10 29

10 32

a=

+ +

+

^

^

h

h ...(3)

Here, no. of asymptotes P Z= -

3 1 2= - =

So, two root branches will go to infinite along asymptotes as "3a .

Now using equation (1) we have

s s s10 29 10 303 2 a a+ + + +^ h 0=Routh array is shown below

s3 1 ( )29 10a+

s2 10 30a

s1 ( )29 7a+

s0 30a

For 0>a , s1row can not be zero. Hence, root locus does not intersect jwaxis

for 0>a .


Open loop transfer function given

G s H s ^ ^h h 2s sK s

46=

+ ++

^ ^^ h hh Poles : 2P= ; s 2=- and s 4=-

Zeros : Z 1= ; s 6=-

No. of asymptotes P Z= -


39/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

2 1 1= - = Thus (1) is correct.

Now characteristic,

s s K s 2 4 6+ + + +

^ ^ ^h h h 0=

or 8 6s K s K 62+ + + +^ h 0=Making Routh array we have

s2 1 K8 6+

s1 K6+

s0 K8 6+

Root locus is plotted for K 0= to 3. i.e. K 0> .

Here, for K 0> root locus does not intersect jwaxis because s1row will not bezero. Thus (2) is incorrect.

Here, poles are two and zero is one. Hence one imaginary zero lies on infinite.

Thus (4) is incorrect.

Therefore (B) must be correct option. But we check further as follows.

Root locus will be as given below

It has two real axis intersections. Thus (3) is correct.

CS 1.31 Correct answer is 1- .

Characteristic equation of given closed loop system is

s s s K 1 2+ + +^ ^h h 0=or

1 2s s sK1+

+ +

^ ^h h 0=

G s H s 1+ ^ ^h h 0=Comparing we get open loop transfer function,

G s H s ^ ^h hs s s

K

1 2=

+ +^ ^h hPoles : s 0= , s 1=- , s 2=-

Zeros : No zero

Centroid As Re ReSumof Sumof

P Z

P Z=

-

-6 6@ @

3

3 0

0 1 2 0

3=

-

- - -=-

^ ^h h 1=-


Forward path transfer function of given ufbsystem is,

G s^ hs s s s

K s

1 2 4

3=

+ + +

+

^ ^ ^^

h h hh


40/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


Here zeros : Z 1= ; s 3=-

and poles : P 4= ; s 0= , s 1=- , s 2=- , s 4=-

Angle of asymptotes ,

af P Z

q2 1 180c=

-

+

^^ hh ; q 0= , 1, 2,..... P Z 1- -^ h

4 1

0 1 1803

180 60c

c=-

+= =^

^h

h; q 0=

4 1

2 1 180180

cc=

-

+=^

^h

h; q 1=

4 1

4 1 180300

cc=

-

+=^

^h

h; q 2=

Thus af

;

;300 ; 2

q

qq

60 0

180 1

3

35

c

c

cp=

= =

= == =

p

p

Z

[\

]]

]]


Open loop transfer function,

G s H s ^ ^h hs s s s s

K s s s

20 50 4 5

10 20 5002

2

=+ + + +

+ + +

^ ^ ^^ ^

h h hh h

Separate loci =No. of open loop poles 5=

Asymptotes =No. of OLP No. of OLZ 5 3 2= - =

At this point we say that Correct option is (B). But we check further as follows. Loci on real axis =no. of poles that lie on real axis

3= ; (s 0= , s 20=- , s 50=- )

Open loop zeros : s 10=- , 10s j20!=-

Open loop zeros : 0s= , s 20=- , s 50=- , s j2 1!=-

Centroid As 5 30 20 50 2 2 10 10 10

=-

- - - - - - - -^ ^h h 22=-


af P Zq2 1 180

c=

-+^ h ; P Z 2- = , q 0= , 1

2

0 1 18090

cc=

+=

^ h; q 0=

2

2 1 180270

cc=

+=

^ h; q 1=


41/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

The root locus is shown below.

Here, Root locus lie only region on real axis that is in left of an odd count of real

poles and real zeros.

Hence, Root locus lies between 20- and 50- and break away point will be alsoin this region. Only one break away point will there.


The loci starts from 1s+ =- and 0, and ends at 3- and 3. Hence poles are

1,0- , and zeros are 3, 3- .

So transfer function is( )( )

s s

K s

13

+

+.

CS 1.35 Correct option is (D).Characteristic equation, s K s5 92+ + 0=

s

K s19

52+ +

0=

G s H s 1+ ^ ^h h 0=Open loop transfer function is,

G s H s ^ ^h hs

K s

952= +

OLP :s j3!= and OLZ : s 0=

Option (A) and option (B) are incorrect because root locus are starting from zeros.On real axis loci exist to the left of odd number of real poles and real zeros.

Hence only Correct option is (D).



G s_ is s s

K

7 122=

+ +^ h ; H s 1=^ hIf point : s j1 1=- + lies on root locus, then it satisfies the characteristic equation

G s H s 1+ ^ ^h h 0=

s s sK17 122

++ +^ h 0=

s s s K 7 123 2+ + + 0= ...(1)

Substituting s j1 1=- + we have

j j j K1 7 1 12 13 2- + + - + + - + +^ ^ ^h h h 0=


42/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


K10- + 0=

K 10=+


Open loop transfer function of given uf bsystem is,

G s^ hs s

K s

3

1=

+

-

^^

hh

s s

K s

3

1=

+

- -

^

hh

OLP : s 0= , s 3=-

OLZ : s 1=

Here, Gain K is negative, so root locus will be complementary root locus and is

found to the left of an even countof real poles and real zeros of GH .

Hence Option (A) and Option (D) are incorrect. Option (B) is also incorect

because it does not satisy this condition. Hence (C) is correct.


Open loop transfer function given is

G s^ h( )s s

K s

2232

=+

+^ hRoot locus starts at OLP : s 0= , s 0= , s 2=-

No. of asymptotes ,

# #OLP OLZ- =No. of OLP No. of OLZ

3 1 2= - =Angle of asymptotes :

af P Zq2 1 180c

=-

+^ h; P Z 2- = , q 0= , 1

1. af 20 1 180

90c

c= +

=^ h

for q 0=

2. af 22 1 180

270c

c= +

=^ h

for q 1=


P Z

P Z=

-

-6 6@ @

0 2

3 132

32

=-

- - -

=-^ bh l

Angle of departure at double pole (at origin)

Df rq2 1 180c

= +^ h

; r 2= , q 0= , 1

90c= , 270c

Fro above analysis Root locus plot will be as given below.


43/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

From root locus it can be observe easily that for all values of gain K (K 0= to

3) root locus lie only in left half of s-plane.


Characteristic equation of a closed loop system is :

s s s K s 1 3 2+ + + +^ ^ ^h h h 0= ; K 0>

3s s s

K s1

1

2+

+ +

+

^ ^^h h

h 0=

Open loop transfer function is,

G s H s ^ ^h hs s s

K s

1 3

2=

+ +

+

^ ^^h h

h

OLP : s 0= , s 1=- , s 3=- and OLZ : s 2=-

Pole zero plot is shown below.

No. of asymptotes P Z= - 3 1 2= - =

Angle of asymptotes : af P Zq2 1 180c

=-

+^ h; P Z 2- = , q 0= , 1

af 90c= and 270c


P Z

P Z=

-

-6 6@ @

3 1

0 1 3 222

=-

- - - -=-

^ ^h h1=-

Break away point lie in the range Re s1 0<


44/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


From given plot, we can see that centroid (point where asymptotes intersect on

real axis) is origin and all three root locus branches also start from origin and

goes to infinite along with asymptotes. Therefore there is no any zero and three

poles are at origin.

So option (A) must be correct.

G s^ hs

K3=

Now we verify the above result as follows.

Using phase condition we have G s H s s s0=

^ ^h h 180! c=From given plot, for a given point on root locus we have

G s H s ,s 1 3=

^ ^ ^h h h tan xy3 1=- - a k tan3

131=- - c m

3 60 180# c c=- =-


Open loop transfer function of given uf bsystem is

G s^ hs s s

s

2 10

2 a=

+ +

+

^ ^^h h

hCharacteristic equation,

s s s

s1

2 10

2 a+

+ +

+

^ ^^h h

h 0=

2s s s s 10 2 2a+ + + +^ ^h h 0= s s s12 22 23 2 a+ + + 0=

s s s1

2 222

3 2a

++ +

0=

Open loop transfer function as avaried,

G s H s ^ ^h hs s s12 22

23 2

a=

+ +

No. of poles : P 3= and no. of zeros : Z 0=


af

P Z

q2 1 180c=

-

+^ h;

P Z3

- =;

q0

=, 1, 2

3

0 1 18060

cc=

+=

^ hfor q 0=

3

2 1 180180

cc=

+=

^ hfor q 1=


45/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3 3

4 1 180300

cc=

+=

^ hfor q 2=

af 60c= , 180c, 300c


Intercept point of asymptotes


P Z

P Z=

-

-6 6@ @Poles : s 0= , s 2=- , s 10=-

Zeros : No any zero.

As 3 00 2 10 0

4=-

- - -=-

^ h


We have K s s s

212 223 2

=- + +^ h

ds

dK 0= s s

23 24 22

02

&- + +

=^ h

or s s3 24 222- - - 0=

or s .1056=- , .6943-

Thus break point are s .1056=- , and .6943- .


Given characteristics equation is

s s K s K 23 2+ + + 0=

s s K s 2 13 2+ + +^ h 0=

s s

K s1

2

12+ +

+

^^

hh 0=


G s H s ^ ^h hs s

K s

2

12= +

+

^

^

h

h

OLZ : s 1=- ; No. of zeros : 1Z=

OLP : s 0= , s 0= , s 2=- ; No. of Poles : P 3=

Root loci starts K 0=^ hat s 0= , s 0= and s 2=-One of root loci terminates at s 1=- and other two terminates at infinity.

No. of asymptotes : P Z 2- =

Angle of asymptotes :

af P Zq2 1 180c

=-

+^ h; P Z 2- = , q 0= , 1

0 1

2

18090

cc=

+=

^ h; q 0=

2 1

2180

270c

c= +

=^ h

; q 1=

Centroid (intercept point of asymptotes on real axis)

As Re ReSumof Sumof

P Z

P Z=

-

-6 6@ @


46/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


3 1

0 0 2 121

=-

+ - - -=-

^ ^h h .05=-

Root locus is as shown below.


The root locus plot give the location of the closed loop poles for different values

of parameter gain K .

The characteristic equation is

s s

K s1

9

12+ +

+

^^

hh 0=

or s s K s K 93 2+ + + 0= ...(1)

For all the roots to be equal and real, we require s P 3+^ h 3 3s Ps P s P 03 2 2 3= + + + = ...(2)On comparing equation (1) and (2), we get

P3 9=

or P 3=

and K P3=

3 3= ^ h 27=

CS 1.46 Correct option is (D).Open loop poles are : s 0= and s 1= and open loop zero is s 1=-

The characteristic equation is

s s

K s1

1

1+

-

+

^^

hh 0=

or K s

s s

11

=-+

-^ h

The break points are given by solution ofds

dK 0=

Hence,ds

dK dsd

s

s s

11

0=+

- -=

^ h; Eor s s s s 1 2 1 1+ - - -^ ^ ^h h h 0=or s s s s s 2 2 12 2- + - - + 0=

or s s2 12+ - 0=

or s .2

2 2828!=

- .1 1414!=- ...(1)


47/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3

The root locus is given below

From eq (1),

Centre of circle ,1 0= -^ hand radius of circle 1.414 2= =


The angle criterion of root locus is

G s H s ^ ^h h 180c=or, s s s3 2+ - - +^ ^h h 180c=Substitute s js w= + , we get

j j j3 2s w s w s w+ + - + - + +

^ ^ ^h h h 180c=

tan tan3

1 1

sw

sw

+ -- -a ak k tan180 2

1c

sw

= ++

- a k ...(1)

or tan tan tan3

1 1

sw

sw

+ -- -a ak k: D tan tan180 21c s w= + +- a k: D

or1

3

3

sw

sw

sw

sw

++

+ -

a ak k

1 02

02

sw

sw

=-

+

++

^ ah k

33

2s s ww

+ +

-

^ h 2sw

=+

or 3 2s- +^ h 32

s s w= + +^ hor 6 92 2s s w+ + +^ h 6 9=- +or 3 2 2s w+ +^ h 3 2= ^ hThis is the equation of circle.


Given open loop transfer function is

G s H s ^ ^h hs j s j s j s j

K

1 1 3 3=

+ + + - + + + -^ ^ ^ ^h h h hThe root locus start from

s1 j1=- +

s2 j1=- -

s3 j3=- -

s4 j3=- +

Since, there is no zero, all root loci end at infinity.


48/50




www.n

odia

.co.in



w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olume-3

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


No. of open loop poles P 4= and No. of open loop zeros Z 0=

So, no. of asymptotes is 4 with angles of

Af 45c= , 135c, 225c, 315c

45! c= , 135! c

Centroid : Point of intersection of asymptotes with real axis is

As Re Re

P Z

P ZS S=

-

-6 6@ @

24 0

1 1 3 3 0=

-

- + - + - + - -=-

^ ^ ^ ^h h h hThe root locus is shown below.

There is no breakaway point.


The characteristic equation of the system is

G s H s 1+ ^ ^h h 0=or

s s s

K14 5

++ +^ ^h h 0=

or s s s K 9 203 2+ + + 0= ...(1)

Intersection of root loci with jw axis is determine using Rouths array.

Rouths array is shown below

s3 1 20

s2 9 K

s1 K

9180-

s0 K

The critical gain before the closed loop system goes to instability isK 180c=

and the auxiliary equation is

s9 1802+ 0=

or s2 20=-

or s j2 5!=

Hence, root loci intersect with jw-axis at s j2 5!= .

The gain margin for K 18= is given by,


49/50




www.nod

ia.co.in



www.

nodia.

co.in



Sample

Chapterof

GATEElectr

icalEnginee

ring,

Volum

e-3 GM (in dB) log K

K20 c10= b l 20 20log dB18180

10= =

The gain margin for K 1800= is given by

GM (in dB) 20 20log dB180018010= =-b l

The break away point is given by solution ofds

dK 0= .

From eq. (1),

K s s s9 203 2=- + +^ hor

dsdK s s3 18 20 02=- + + =^ h

or s .6

18 9 165!=

-

.45275=- and .14725-

Point .s 14725=- lies on root locus. So, break away point is .s 14725=- .The value of K at break away point is

K s s s4 5= + +^ ^h h at .s 14725=- .13128=


First we check if point lies on root locus. For this we use angle criterion

G s H s s s0=

^ ^h h 180!=Since G s H s

s j3 5=- +^ ^h h

j j

K

3 5 1 3 5 5=

- + + - + +

^ ^h hj j

K

2 5 2 5=

- + +^ ^h hso, G s H s

s j3 5=- +^ ^h h tan tan25 251 1=- - -- -b bl l

tan tan18025

251 1

c=- + -- -

180c=-

Angle criterion satisfy.

Now, using magnitude condition we have

G s H s s j3 5=- +^ ^h h

1=

orj j

K

2 5 2 5- + +^ ^h h 1=or K

4 25 4 25+ + 1=

or K 29=


The characteristic equation of the system is

s s s s

K s

1 1 4 16

12+ - + +

+

^ ^h hh 0= ...(1)or s s s K s K 3 12 164 3 2+ + + - +^ h 0=Intersection of root loci with jw-axis is determined using rouths array which is

shown below


50/50




www.n

odia

.co.in

w

ww.

nodia.c

o.in

SampleChapt


gineering,V

olu

GATE EE vol-1



GATE EE vol-3


GATE EE vol-4


s4 1 12 K

s3 3 K 16-

s2 K3

52- K

s1

KK K

5259 8322

-- + -

s0 K

The root locus cross the jw-axis, if s1row is complete zero.

i.e. K K59 8322- + - 0=

or K K59 8322- + 0=

K .259 12 37!

= .357= , 23.3

Hence, root locus cross jw-axis two times and the break points are given by

solution ofds

dK 0= .

We can directly check option using pole-zero plot :

Hence, break away point will lie on real axis from s 0= to 1 and break in point

will lie on real axis for s 1< - .

Hence, s .045= is break away point

and s .226=- is break in point

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

GATE EE vol-3

Documents

Transcript of GATE EE vol-3