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GATEELECTRICAL ENGINEERING

Vol 3 of 4

Second Edition

GATEELECTRICAL ENGINEERING

Vol 3 of 4

RK Kanodia Ashish Murolia

NODIA & COMPANY

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 any information herein, and NODIA & COMPANY nor its author shall be responsible for any error, omissions, or damages 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

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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 and data 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 and improper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series. Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’s theorems.

Differential equations: First order equation (linear and nonlinear), Higher order linear differential equations with constant coefficients, Method of variation of parameters, Cauchy’s and Euler’s equations, Initial and boundary value problems, Partial Differential Equations and variable separable method.

Complex variables: Analytic functions, Cauchy’s integral theorem and integral formula, Taylor’s 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-step methods 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, transient response of dc and ac networks; sinusoidal steady-state analysis, resonance, basic filter concepts; ideal current and voltage sources, Thevenin’s, Norton’s and Superposition and Maximum Power Transfer theorems, two-port networks, three phase circuits; Gauss Theorem, electric field and potential due to point, line, plane and spherical charge distributions; Ampere’s and Biot-Savart’s laws; inductance; dielectrics; capacitance.

Signals and Systems: Representation of continuous and discrete-time signals; shifting and scaling operations; linear, time-invariant and causal systems; Fourier series representation of 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, generator characteristics, armature reaction and commutation, starting and speed control of motors; three phase induction motors – principles, types, performance characteristics, starting and speed control; single phase induction motors; synchronous machines – performance, regulation and parallel operation of generators, motor starting, characteristics and applications; servo and 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; power factor correction; economic operation; symmetrical components; fault analysis; principles of over-current, differential and distance protection; solid state relays and digital protection; circuit breakers; system stability concepts, swing curves and equal area criterion; HVDC transmission 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-lag compensation; 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, BJT, FET; amplifiers – biasing, equivalent circuit and frequency response; oscillators and feedback amplifiers; operational amplifiers – 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; triggering circuits; phase control rectifiers; bridge converters – fully controlled and half controlled; principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.

***********

PREFACEThis book doesn’t make promise but provides complete satisfaction to the readers. The market scenario is confusing and readers don’t find the optimum quality books. This book provides complete set of problems appeared in competition exams as well as fresh set of problems.

The book is categorized into units which are then sub-divided into chapters and the concepts of the problems are addressed in the relevant chapters. The aim of the book is to 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 view of preparation duration and time taken for solving each problem in the examination. So the problems solving methods is the books are those which take the least distance to the solution.

But however to make a comment that this book is absolute for GATE preparation will be an inappropriate one. The theory for the preparation of the examination should be followed from the standard books. But for a wide collection of problems, for a variety of problems 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 seven number of chapters on an average each of which contains 40 problems which are selected so as to avoid unnecessary redundancy and highly needed completeness.

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

R. K. KanodiaAshish Murolia

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

***********

CS 1ROOT 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)

CS 10 Root Locus Technique CS 1PE 1 Root Locus Technique PE 10EF 1 Root Locus Technique EF 10

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GATE Electrical Engineering-2015in 4 Volumes

by R. K. Kanodia & Ashish Murolia

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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



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GATE EE vol-1Electric circuit & Field, Electrical & electronic measurement

GATE EE vol-2Analog electronics, Digital electronics, Power electronics

GATE EE vol-3Control systems, Signals & systems

GATE EE vol-4Electrical machines, Power systems

Engineering mathematics, General AptitudeThe general shape of the root locus is




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

( )G s ( )( )s s

K s s8 252 6

2=+ ++ +

The root locus for this system is

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

( )G s ( )( )s

K s14

2

2

=++

For this system root locus is


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Engineering mathematics, General Aptitude

CS 1.10 The forward-path open-loop transfer function of a ufb system is

( )G s ( )

sK s 1

2

2

= +

The root locus of this system is

Common Data For Q. 11 and 12:A open-loop pole-zero plot is shown below


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CS 1.11 The transfer function of this system is

(A) ( )( )

( )s s

K s s3 2

2 22

+ ++ +

(B) ( )( )( )s s

K s s2 23 2

3 + ++ +

(C) ( )( )

( )s s

K s s3 2

2 22

+ +- +

(D) ( )( )( )s s

K s s2 23 2

2 - ++ +

CS 1.12 The break point is(A) breakaway at .1 29s =-

(B) breakin at .2 43s =-(C) breakaway at .2 43s =-

(D) breakin at .1 29s =-

Common Data For Q. 13 and 14:A root locus of ufb system 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 ufb system is

( )G s ( )( )

( )s s s

K s3 2 2

22=

+ + ++

The angle of departure from the complex poles is ____

CS 1.16 Consider the feedback system shown below


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Engineering mathematics, General AptitudeFor this system root locus is

Common Data For Q. 17 and 18: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 on the 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 s4 12 20

8= + + ++

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

CS 1.20 For a ufb system forward-path transfer function is

( )G s ( )( )

( )s s

K s3 5

6= + ++

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


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

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

K1 2

= + +The root locus plot of above system is

CS 1.22 A ufb system has forward-path transfer function

( )G s sK

2=

The root locus plot is

CS 1.23 For the ufb system, shown below consider two point

s1 2 3j=- + and 2s j2

12 =- +

Which of the above point lie on root locus ?


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Engineering mathematics, General Aptitude(A) Both s1 and s2 (B) s1 but not s2

(C) s2 but not s1 (D) neither s1 nor s2

CS 1.24 A ufb system has open-loop transfer function

( )G s ( )( )

,s sK s

0> >2 ba b 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 Ks K2 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

K4 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 ufb system shown below


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The root-loci, as a is varied, will be

Common Data For Q. 27 and 28 :The forward-path transfer function of a ufb system is

( )G s ( )

( )( )s s

K s s1

32a=-

+ +

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

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


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

( )s s

K s2 4

6= + ++

, consider the following characteristic of the root locus :1. It has one asymptotes

2. It has intersection with jw-axis

3. 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 ufb system is

( )G s ( )( )( )

( )s s s s

K s1 2 4

3= + + ++

The angles of asymptotes are

(A) 0, ,2p p (B) 0, ,3

234p p

(C) , ,3 35p p p (D) None of the above

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

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

( )( )s s s s s

K s s s20 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


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CS 1.34 The root-locus of a ufb system is shown below

The open loop transfer function is

(A) ( )( )s s s

K1 3+ + (B)

( )( )

s sK s

13

++

(C) ( )( )

s sK s

31

++

(D) ( )( )s s

Ks1 3+ +

CS 1.35 The characteristic equation of a linear control system is 5 9 0s Ks2 + + = . The root loci of the system is

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

( )G s ( )s s s

K7 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 sK s

31= +-

. Which is the correct root locus diagram ?


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

( )G s ( )s s

K s2232

=++^ h

From 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 the s -plane

(B) one real root is found on the right half of the s -plane

(C) the root loci cross the jw axis 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< <-(C) Two of its roots tend to infinity along the asymptotes [ ]Re s 1=-(D) It may have complex roots in the right half plane.

CS 1.40 Figure shows the root locus plot (location of poles not given) of a third order system whose open loop transfer function is

(A) sK

3 (B) ( )s s

K12 +

(C) ( )s s

K12 +

(D) ( )s s

K12 -

Common Data For Q. 41 to 43 :The open loop transfer function of a unity feedback system is given by

( )G s ( )( )

( )s s s

s2 10

2 a= + ++

; 0>a


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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) .1 056- , .3 471- (B) 2.112, 6.943- -(C) 1.056, 6.943- - (D) 1.056, 6.943-

CS 1.44 For the characteristic equation s s Ks K23 2+ + + 0= , the root locus of the system as K varies from zero to infinity is

CS 1.45 The open loop transfer function of a ufb system is

G s H s^ ^h h s sK s

91

2=++

^

^

h

h

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


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Engineering mathematics, General AptitudeCS 1.46 A ufb system has an open loop transfer function


11

=-+

^

^

h

h

Root 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


23

=++

^

^

h

h

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 h s s s s

K2 2 6 102 2=

+ + + +^ ^h h

The 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 ufb system is given below.

G s H s^ ^h h s s s

K4 5

=+ +^ ^h h

Consider 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 20 dB-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 h s s

K1 5

=+ +^ ^h h

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


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


K s1 4 16

12=

- + ++

^ ^

^

h h

h

Consider the following statements for the system1. Root locus of the system cross jw-axis for .K 35 7=2. Root locus of the system cross jw-axis for .K 23 3=3. Break away point is .s 0 45=4. Break in point is .s 2 26=-Which of the following statement is correct ?(A) 1, 3 and 4 (B) 2, 3 and 4

(C) 3 and 4 (D) all

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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.

CS 1.2 Correct option is (D).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 f is 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 @


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Df 0! c=So, departure angle for pole P0 is 0c, So root locus branch will depart at 0c. Only option (A) satisfy this condition.

CS 1.4 Correct option is (A).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.

CS 1.6 Correct option is (A).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 c c= -

+= =^ h

; q 1=

Hence, root locus plot will be as shown below.


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Engineering mathematics, General AptitudeCS 1.7 Correct option is (A).

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 Correct option is (A).

CS 1.8 Correct option is (C).Forward Path open loop transfer function of given ufb system is

G s^ h s s

K s s8 252 6

2=+ ++ +^ ^h h

Characteristic equation is

s s8 252 + + 0=

s j28 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 ufb system is,

G s^ h s

K s14

2

2

=++^ h

Here, poles s 12 + 0= s j1!=and zeros s 42 + 0= s j2!=


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Pole-zero plot is

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

CS 1.10 Correct option is (A).Forward Path one loop transfer function of given ufb system is

G s^ h s

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 =- and 3s =-

Transfer function of the system is

G s^ h s s

K s j s j2 3

1 1 1 1=

- - - -- + - -

^ ^

^ ^

h h

h h

" "

" "

, ,

, ,


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s s

K s j s j2 3

1 1 1 1=

+ +- - - +^ ^

^ ^

h h

h h" ", ,

s s

K s j2 31 12 2

=+ +- -

^ ^

^ ^

h h

h h" ,

or G s^ h s s

K s s2 3

2 22

=+ +- +

^ ^

^

h h

h

CS 1.12 Correct option is (C).

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-^ h and 3-^ h on real 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-^ h and 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 IIThe point, at which multiple roots are present, are known as break point. These

are obtained from the dsdK 0=

Here, characteristic equation is

G s H s1 + ^ ^h h 0=

s s

K s s1

2 32 22

++ +- +

^ ^

^

h h

h 0=

K s ss s

2 22 3

2=- +

- + +^

^ ^

h

h h

s ss s

2 25 6

2

2

=- +

- + +^ h ...(1)

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

dsdK

s ss s s s s s

2 22 2 2 5 5 6 2 2

02 2

2 2

=- +

- - + + + + + -=

^

^ ^ ^ ^

h

h 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 .1 45- .Given root locus of ufb system is shown below


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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^ h s s

K s s1 23 5

=+ +- -

^ ^

^ ^

h h

h h

Characteristic equation

G s H s1 + ^ ^h h 0=

s s

K s s1

1 23 5

++ +- -

^ ^

^ ^

h h

h h 0=

K s s

s s8 153 2

2

2

=- +

- + +^

^

h

h ...(1)

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

dsdK

s ss s s s s s

8 158 15 2 3 3 2 2 8

02 2

2 2

=- +

- - + + + + + -=

^

^ ^ ^ ^

h

h h h h

s s11 26 612 - - 0=or s .3 9=+ and .s 1 45=-Thus .s 1 45=- is break away point and .s 3 9=+ is break in point.

CS 1.14 Correct option is (C).Break away and Break in points always satisfy characteristic equation.Substituting .s 1 45=- in eq (1), we get

K . .

. .

1 45 8 1 45 15

1 45 3 1 45 22

2

=- - - +

- - + - +

^ ^

^ ^

h h

h h

6

8

@

B

..

28 70250 2475

=--^ h

.8 62 10 3#= -

CS 1.15 Correct answer is 108.4 .Forward path transfer function of given ufb system is

G s^ h s s s

K s3 2 2

22=

+ + ++

^ ^

^

h h

h

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 P1 is given by :

Df 180! c f= +6 @


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Engineering mathematics, General Aptitudewhere f is net angle contribution at pole P1 due 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 21

P31f = -

f 9tan tan1 0 211 1c= - +- -

: D

Departure angle :

Df 180! c f= +6 @

1tan tan180 1 90 21 1! c c= + - -- -

: D

.180 45 90 26 56!= + - -6 @

Df .108 4! c=Hence, departure angle for pole P1 is 108.4c+ and departure angle for pole P2 is

108.4c- because P1 and P2 are complex conjugate.

CS 1.16 Correct option is (A).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 s

K s1 3

2=

+ +- +

^ ^

^

h h

h

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

CS 1.17 Correct option is (A).Given root locus is shown below


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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

K4 2 22=

+ + +^ ^h h

s s s

K6 10 83 2=

+ + +Closed loop transfer function,

T s^ h 1

s s sK

s s sK

6 10 8

6 10 8

3 2

3 2=

++ + +

+ + +

s s s K

K6 10 83 2=

+ + + +Characteristics equation :

6 10 8s s s K3 2+ + + + 0=Routh array is shown below

s3 1 10

s2 6 K8 +

s1 K6

52 -

s0 K8 +

When root locus cut will cut imaginary axis if element in s1 is zero.

Thus K6

52 - 0= & K 52=

and then from auxiliary equation we have

s6 8 522 + +^ h 0=

s2 660 10=- =-

s .j3 162!=

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

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

Ks s0=

intPr intProduct of phasors drawn from at that po

oduct of phasors drawn from at that poOLZOLP=

Here no any zero is present.

Hence, K = Product of phasors drawn from OLP at that point


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Engineering mathematics, General Aptitude K

s 5=- PP PP PP3 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 s4 12 20

8=

+ + ++

^ ^ ^

^

h h h

h

Here, Zeros : Z 8=- Poles : P 01 = ; P 42 =- ; P 123 =- ; P 204 =-Value of K at 10s =- is :

Ks 10=-

1010

Phasors drawn from atPhasors drawn from at

OLZ sOLP s

pp

==-=-

^

^

h

h

Pole-zero plot is shown below.

Ks 10=-

PZ

PP PP PP PP1 2 3 4# # #=^

^ ^ ^ ^

h

h h h h

210 6 2 10 600# # #= =

K 600=

CS 1.20 Correct option is (D).For given system, forward path transfer function,

G s^ h s s

K s3 5

6=

+ ++

^ ^

^

h h

h

Characteristic equation for closed loop transfer function,

G s H s1 + ^ ^h h 0=

or s s

K s1

3 56

++ +

+^ ^

^

h h

h 0=

or K s

s s6

8 152

=+

- + +^

^

h

h

differentiating w.r.t to s and equating to zero, we have

dsdK

ss s s s

66 2 8 8 15

02

2

=+

- + + + + +=

^

^ ^ ^

h

h h h

or s s12 332 + + 0=or s .7 73=- and .s 4 27=-The root locus is shown below. Here we can easily say that s .4 27=- is break

away point and s .7 73=- point.


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CS 1.21 Correct option is (B).Open loop transfer function :

G s H s^ ^h h s s s

K1 2

=+ +^ ^h h

In 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 root locus diagram.Now characteristic equation ,

G s H s1 + ^ ^h h 0=

s s s

K11 2

++ +^ ^h h

0=

s s s K3 23 2+ + + 0=Routh array is shown below.

s3 1 2s2 3 K

s1 K3

6-

s0K

At K 6= , s1 row is zero, thus using auxiliary equation we get,

s3 62 + 0=

s j 2!=Root locus cut on jw axis at s j 2!= for K 6= . Thus option (B) is correct because (D) does not cut jw axis.

CS 1.22 Correct option is (D).For given ufb system, forward transfer function,

G s^ h sK

2=

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-^ h

Here, 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=


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Engineering mathematics, General AptitudeHence, root locus plot will be as shown below.

CS 1.23 Correct option is (C).For given ufb system, open loop transfer function is,

G s^ h s s

K s s1 23 4

=+ ++ +

^ ^

^ ^

h h

h 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 h

or 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 ss s

11

f==

^ ^h h , we have

1f 1 2 3 4q q q q= + + +

tan tan tan23

13 90 1

31 1 1c= + - - -- - -

b l

tan tan tan23 3 90 180 31 1 1c c= + - - -- - -

^ h

. . .56 30 71 56 270 71 56c= + - +or 1f .70 56 180! c!=-Hence, point s1 not lie on root locus.

Point : s j22

12 =- + :-


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Using phase condition G s H ss s2=

^ ^h h 2f= we have,

2f 31 2 4q q q q= + - -

tan tan tan2 2

12

1 90 1802

11 1 1c c= + - - -- - -c m

tan tan tan2 2

12

1 90 1802

11 1 1c c= + - - +- - -

2f 180c=- Here phase condition is satisfied.Hence, point s2 lies on root locus.

CS 1.24 Correct option is (A).For given ufb system, open loop transfer function is

G s^ h s sK s

2 ba

=++

^

^

h

h; 0> >b a

Here, zero : Z a=- Poles : s 0= , 0; s b=-Departure angle at double poles on origin is :

f rq2 1 180c

=+^ h

; r 2= , q 0= , 1

f 90c= and 270cTo get intersection with imaginary axis we use Ruth array as shown below.characteristic equation is

s s Ks K3 2b a+ + + 0=Routh Array is shown below

s3 1 K

s2b Ka

s1

bb a-

s0Ka

Here, for any value of K , sl row of Routh array will not be zero. Thus system is stable for all positive value of K and hence root locus does not cross jw axis. Therefore root locus completely lies in left half of s -plane. Based on these above result we say that Correct option is (A).

CS 1.25 Correct option is (C).Characteristic equation of given feedback control system is

s s s s Ks K4 4 11 30 42 2 2+ + + + + +^ ^h h 0=

or s s s s

K s1

4 4 11 304

2 2

2

++ + + +

+^ ^

^

h h

h 0= ...(1)

Characteristic equation is,

G s H s1 + ^ ^h h 0= ...(2)Comparing, eq (1) and (2) we get open loop transfer function as


K s4 4 11 30

42 2

2

=+ + + +

+^ ^

^

h h

h

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

K s2 2 5 6

42

= + + + ++^ h


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Engineering mathematics, General AptitudeOpen loop poles are

s ,2 2=- - and s 6=- , 5-The point at which asymptotes meet (centroid) is given by:

As Re ReSum of Sum of

P ZP Z

=--

^ h

6 6@ @

Here, open loop zeros : s 4 02 + = s j2!=

As 4 22 2 5 6 0

= -- - - - -^ h

.215 7 5= - =-

It intersects on real axis. So point is: . ,7 5 0-^ h

CS 1.26 Correct option is (D).Open loop transfer function for given system is


K4 4 52=

+ + +^ ^h h

Characteristic equation,

s s s s

K14 4 52+

+ + +^ ^h h 0=

K s s s s4 4 52=- + + +^ ^h h ...(1)Differentiating wrt s and equating to zero we have

dsdK s s s s s s2 4 4 5 4 2 4 02 2=- + + + + + + =^ ^ ^ ^h h h h8 B

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

or s 2=- and .s 0 775=- , .3 225-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,

ds

d K2

2

s s s s2 8 5 2 4 82=- + + + + +^ ^ ^h h h6 @

s s6 24 212=- + +^ h

For .s 0 775=- and .s 3 225=- ,

ds

d K2

2

. 06 0 <=- ; .s 0 775=- , .3 225- are maxima points.

Hence, s .0 775=- and .s 3 225=- are break away points.

For s 2=- :

ds

d K2

2

3 0>=+ ; s 2=- is minima points.

Thus s 2=- is break in point.Hence, there is two break away points . , .s 0775 3 225=- -^ h and one break in point.

CS 1.27 Correct option is (B).For a ufb system, forward transfer function is :


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G s^ h s s

1a

=+^ h

Method I :Characteristic equation :

G s H s1 + ^ ^h h 0=

s s

1 1a

++^ h

0=

s s 12 a+ + 0=

s

s112

a++^ h

0=

Open loop transfer function, as a is varied is

G s H s^ ^h h s

s12

a=+

Here, zero : s 0=and poles : s 1 02 + = s j& !=Root locus will be :

Method II :Closed loop transfer function :

T s^ h G s H sG s

s s1 11

2 a=

+=

+ +^ ^

^

h h

h

G s H sG s

1 + ^ ^

^

h h

h

ss

s1

1

11

2

2

a=+

+

+

G s H s^ ^h h s

s12

a=+

CS 1.28 Correct option is (A).Forward path transfer function of given ufb system is

G s^ h s s

K s s1

32

a=

-+ +^

^ ^

h

h h; 5a = and K 0>

or G s^ h s s

K s s1

5 32=-

+ +^

^ ^

h

h 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.


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Engineering mathematics, General AptitudeAngle of asymptotes,

af P Z

q2 1 180c=

-+

^

^

h

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

10 1 180

180c

c=+

=^ h

Only option (A) satisfies these conditions.

CS 1.29 Correct option is (C).Forward path transfer function :

G s^ h s s

K s s1

32

a=

-+ +^

^ ^

h

h h; K 10= and 0>a


G s H s1 + ^ ^h h 0=

s ss s

11

10 32

a+

-+ +^

^ ^

h

h h 0=

s s s s10 3 33 2 a a- + + + +^ h6 @ 0= 10s s s s10 29 32 a+ + + +^ ^h h 0= ...(1)

s s s

s1

10 2910 3

2

a+

+ ++

^

^

h

h 0= ...(2)

Open loop gain as a is varied,

G s H s^ ^h h s s s

s10 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 , s1 row can not be zero. Hence, root locus does not intersect jw axis for 0>a .

CS 1.30 Correct option is (B).Open loop transfer function given

G s H s^ ^h h 2s s

K s4

6=

+ ++

^ ^

^

h h

h

Poles : 2P = ; s 2=- and s 4=- Zeros : Z 1= ; s 6=- No. of asymptotes P Z= -


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2 1 1= - = Thus (1) is correct.Now characteristic,

s s K s2 4 6+ + + +^ ^ ^h h h 0=or 8 6s K s K62 + + + +^ 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 jw axis because s1 row will not be zero. 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 K1 2+ + +^ ^h h 0=

or 1 2s s sK1 +

+ +^ ^h h 0=

G s H s1 + ^ ^h h 0=Comparing we get open loop transfer function,

G s H s^ ^h h s s s

K1 2

=+ +^ ^h h

Poles : s 0= , s 1=- , s 2=-Zeros : No zero

Centroid As Re ReSum of Sum of

P ZP Z

= --6 6@ @

33 0

0 1 2 03= -

- - -=-^ ^h h

1=-

CS 1.32 Correct option is (C).Forward path transfer function of given ufb system is,

G s^ h s s s s

K s1 2 4

3=

+ + ++

^ ^ ^

^

h h h

h


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Engineering mathematics, General AptitudeHere 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=

-+

^

^

h

h; 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

q

q

60 0

180 13

35

c

c

c

p== == =

= =

p

p

Z

[

\

]]

]]

CS 1.33 Correct option is (B).Open loop transfer function,

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

K s s s20 50 4 5

10 20 5002

2

=+ + + +

+ + +^ ^ ^

^ ^

h h h

h 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=-Angle of asymptotes,

af P Zq2 1 180c

= -+^ h

; P Z 2- = , q 0= , 1

20 1 180 90c

c=+

=^ h; q 0=

22 1 180

270c

c=+

=^ h; q 1=


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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 also in this region. Only one break away point will there.

CS 1.34 Correct option is (B).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 sK s

13

++

.

CS 1.35 Correct option is (D).

Characteristic equation, s Ks5 92 + + 0=

sK s1

95

2++

0=

G s H s1 + ^ ^h h 0=Open loop transfer function is,

G s H s^ ^h h s

Ks9

52=+

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).

CS 1.36 Correct answer is 10.Open loop transfer function,

G s_ i s s s

K7 122=

+ +^ h; H s 1=^ h

If point : s j1 1=- + lies on root locus, then it satisfies the characteristic equation

G s H s1 + ^ ^h h 0=

s s s

K17 122+

+ +^ h 0=

s s s K7 123 2+ + + 0= ...(1)

Substituting s j1 1=- + we have

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


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Engineering mathematics, General Aptitude K10- + 0= K 10=+

CS 1.37 Correct option is (C).Open loop transfer function of given ufb system is,

G s^ h s sK s

31

=+-

^

^

h

h

s sK s

31

=+

- -^

^

h

h

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 count of 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.

CS 1.38 Correct option is (A).Open loop transfer function given is

G s^ h ( )s s

K s2232

=++^ h

Root 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 ZP 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= , 270cFro above analysis Root locus plot will be as given below.


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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.

CS 1.39 Correct option is (C).Characteristic equation of a closed loop system is :

s s s K s1 3 2+ + + +^ ^ ^h h h 0= ; K 0>

3s s s

K s1

12

++ +

+^ ^

^

h h

h 0=

Open loop transfer function is,

G s H s^ ^h h s s s

K s1 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 ZP Z

= --6 6@ @

3 10 1 3 2

22= -

- - - -=-^ ^h h

1=-

Break away point lie in the range Re s1 0< <- 6 @

Two of its roots tends to infinite along the asymptotes Re s 1=-6 @ . Root locus lies in only left half of s -plane.

CS 1.40 Correct option is (A).Given Root locus plot,


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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^ h sK

3=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 hh tan x

y3 1=- -a k

tan3 131=- -

c m

3 60 180# c c=- =-

CS 1.41 Correct option is (B).Open loop transfer function of given ufb system is

G s^ h s s s

s2 10

2 a=

+ ++

^ ^

^

h h

h


s s s

s1

2 102 a

++ +

+^ ^

^

h h

h 0=

2s s s s10 2 2a+ + + +^ ^h h 0= s s s12 22 23 2 a+ + + 0=

s s s

12 222

3 2a+

+ + 0=

Open loop transfer function as a varied,

G s H s^ ^h h s s s12 22

23 2

a=+ +

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

af P Zq2 1 180c

= -+^ h

; P Z 3- = ; q 0= , 1, 2

30 1 180

60c

c=+

=^ h for q 0=

32 1 180

180c

c=+

=^ h for q 1=


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4 1 180300

cc=

+=^ h

for q 2=

af 60c= , 180c, 300c

CS 1.42 Correct option is (C).Intercept point of asymptotes


P ZP Z

= --6 6@ @

Poles : s 0= , s 2=- , s 10=-Zeros : No any zero.

As 3 00 2 10 0

4= -- - -

=-^ h

CS 1.43 Correct option is (C).

We have K s s s

212 223 2

=- + +^ h

dsdK 0=

s s2

3 24 220

2

&- + +

=^ h

or s s3 24 222- - - 0=or s .1 056=- , .6 943-Thus break point are s .1 056=- , and .6 943- .

CS 1.44 Correct option is (A).Given characteristics equation is

s s Ks K23 2+ + + 0= s s K s2 13 2+ + +^ h 0=

s sK s

121

2+++

^

^

h

h 0=

Open loop transfer function,


21

2=++

^

^

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=^ h at 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

2180

90c

c=+

=^ h; q 0=

2 1

2180

270c

c=+

=^ h; q 1=

Centroid (intercept point of asymptotes on real axis)

As Re ReSum of Sum of

P ZP Z

= --6 6@ @


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3 10 0 2 1

21= -

+ - - -=-^ ^h h

.0 5=-

Root locus is as shown below.

CS 1.45 Correct answer is 27.The root locus plot give the location of the closed loop poles for different values of parameter gain K .The characteristic equation is

s sK s

191

2+++

^

^

h

h 0=

or s s Ks K93 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 sK s

111

+-+

^

^

h

h 0=

or K ss s

11

=- +-^ h

The break points are given by solution of dsdK 0=

Hence, dsdK ds

dss s

11

0= +- -

=^ h; E

or s s s s1 2 1 1+ - - -^ ^ ^h h h 0=or s s s s s2 2 12 2- + - - + 0=or s s2 12 + - 0=

or s .2

2 2 828!= - .1 1 414!=- ...(1)


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The root locus is given below

From eq (1),

Centre of circle ,1 0= -^ h

and radius of circle 1.414 2= =

CS 1.47 Correct option is (C).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 tan31 1

sw

sw

+ -- -a ak k tan180 2

1c sw= + +

-a k ...(1)

or tan tan tan31 1

sw

sw

+ -- -a ak k: D tan tan180 2

1c sw= + +

-a k: D

or 1 3

3

sw

sw

sw

sw

+ +

+ -

a ak k

1 0 2

0 2

sw

sw

=- +

+ +^ ah k

33

2s s ww

+ +-

^ h 2s

w= +

or 3 2s- +^ h 3 2s s w= + +^ h

or 6 92 2s s w+ + +^ h 6 9=- +

or 3 2 2s w+ +^ h 32= ^ h

This is the equation of circle.

CS 1.48 Correct option is (D).Given open loop transfer function is

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

K1 1 3 3

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

The root locus start from

s1 j1=- + s2 j1=- - s3 j3=- - s4 j3=- +Since, there is no zero, all root loci end at infinity.


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Engineering mathematics, General AptitudeNo. 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 ZP ZS S

= --6 6@ @

24 01 1 3 3 0

= -- + - + - + - -

=-^ ^ ^ ^h h h h

The root locus is shown below.

There is no breakaway point.

CS 1.49 Correct option is (D).The characteristic equation of the system is

G s H s1 + ^ ^h h 0=

or s s s

K14 5

++ +^ ^h h

0=

or s s s K9 203 2+ + + 0= ...(1)Intersection of root loci with jw axis is determine using Routh’s array.Routh’s array is shown below

s3 1 20

s2 9 K

s1 K9

180 -

s0 K

The critical gain before the closed loop system goes to instability is K 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,


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GM (in dB) log KK20 c

10= b l 20 20log dB18180

10= =

The gain margin for K 1800= is given by

GM (in dB) 20 20log dB1800180

10= =-b l

The break away point is given by solution of dsdK 0= .

From eq. (1),

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

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

or s .6

18 9 165!= -

.4 5275=- and .1 4725-Point .s 1 4725=- lies on root locus. So, break away point is .s 1 4725=- .The value of K at break away point is

K s s s4 5= + +^ ^h h at .s 1 4725=- .13 128=

CS 1.50 Correct answer is 29.First we check if point lies on root locus. For this we use angle criterion

G s H ss s0=

^ ^h h 180!=

Since G s H ss j3 5=- +

^ ^h h j j

K3 5 1 3 5 5

=- + + - + +^ ^h h

j jK

2 5 2 5=

- + +^ ^h h

so, G s H ss j3 5=- +

^ ^h h tan tan25

251 1=- - -- -

b bl l

tan tan180 25

251 1c=- + -- -

180c=-

Angle criterion satisfy.Now, using magnitude condition we have

G s H ss j3 5=- +

^ ^h h 1=

or j jK

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

or K4 25 4 25+ +

1=

or K 29=

CS 1.51 Correct option is (D).The characteristic equation of the system is

s s s s

K s1

1 4 161

2+- + +

+^ ^

^

h h

h 0= ...(1)

or s s s K s K3 12 164 3 2+ + + - +^ h 0=Intersection of root loci with jw-axis is determined using routh’s array which is shown below


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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 s1 row is complete zero.

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

K .2

59 12 37!= .35 7= , 23.3

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

solution of dsdK 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 .0 45= is break away point

and s .2 26=- is break in point

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

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