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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
professional services.
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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 nonlinear algebraic equations, single and multistep methods for differential equations.
Transform Theory: Fourier transform,Laplace transform, Ztransform.
ELECTRICAL ENGINEERING
Electric Circuits and Fields: Network graph, KCL, KVL, node and mesh analysis, transient response of dc and ac networks; sinusoidal steadystate analysis, resonance, basic filter concepts; ideal current and voltage sources, Thevenin’s, Norton’s and Superposition and Maximum Power Transfer theorems, twoport networks, three phase circuits; Gauss Theorem, electric field and potential due to point, line, plane and spherical charge distributions; Ampere’s and BiotSavart’s laws; inductance; dielectrics; capacitance.
Signals and Systems: Representation of continuous and discretetime signals; shifting and scaling operations; linear, timeinvariant 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; autotransformer; 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; perunit quantities; bus impedance and admittance matrices; load flow; voltage control; power factor correction; economic operation; symmetrical components; fault analysis; principles of overcurrent, 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; steadystate errors; Routh and Niquist techniques; Bode plots; root loci; lag, lead and leadlag 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; Qmeters; 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; multivibrators; sample and hold circuits; A/D and D/A converters; 8bit 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 subdivided 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 Ztransform SS 178
SS 7 The ContinuousTime Fourier Transform SS 222
SS 8 The ContinuousTime 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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CS 1.3 The valid root locus diagram is
CS 1.4 A openloop polezero plot is shown below
The general shape of the root locus is
CS 1.5 A openloop polezero plot is shown below
CS 1 Root Locus Technique CS 11PE 11 Root Locus Technique PE 1EF 11 Root Locus Technique EF 1
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GATE EE vol1Electric circuit & Field, Electrical & electronic measurement
GATE EE vol2Analog electronics, Digital electronics, Power electronics
GATE EE vol3Control systems, Signals & systems
GATE EE vol4Electrical machines, Power systems
Engineering mathematics, General AptitudeThe general shape of the root locus is
CS 1.6 A openloop polezero plot is shown below
The general shape of the root locus is
CS 12 Root Locus Technique CS 1PE 1 Root Locus Technique PE 12EF 1 Root Locus Technique EF 12
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CS 1.7 A openloop polezero plot is shown below
The general shape of the root locus is
CS 1.8 The forwardpath openloop 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 forwardpath openloop transfer function of as ufb system is
( )G s ( )( )s
K s14
2
2
=++
For this system root locus is
CS 1 Root Locus Technique CS 13PE 13 Root Locus Technique PE 1EF 13 Root Locus Technique EF 1
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GATE EE vol1Electric circuit & Field, Electrical & electronic measurement
GATE EE vol2Analog electronics, Digital electronics, Power electronics
GATE EE vol3Control systems, Signals & systems
GATE EE vol4Electrical machines, Power systems
Engineering mathematics, General Aptitude
CS 1.10 The forwardpath openloop 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 openloop polezero plot is shown below
CS 14 Root Locus Technique CS 1PE 1 Root Locus Technique PE 14EF 1 Root Locus Technique EF 14
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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 forwardpath 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
CS 1 Root Locus Technique CS 15PE 15 Root Locus Technique PE 1EF 15 Root Locus Technique EF 1
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GATE EE vol1Electric circuit & Field, Electrical & electronic measurement
GATE EE vol2Analog electronics, Digital electronics, Power electronics
GATE EE vol3Control systems, Signals & systems
GATE EE vol4Electrical machines, Power systems
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 closedloop transfer function will have a pole on the real axis at 5 , will be ____
CS 1.19 The openloop 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 forwardpath transfer function is
( )G s ( )( )
( )s s
K s3 5
6= + ++
The breakaway point and breakin points are located respectively as(A) ,4.273
(B) 7.73,4.27
(C) 4.27,3
(D) 4.27,7.73
CS 16 Root Locus Technique CS 1PE 1 Root Locus Technique PE 16EF 1 Root Locus Technique EF 16
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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 forwardpath 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 ?
CS 1 Root Locus Technique CS 17PE 17 Root Locus Technique PE 1EF 17 Root Locus Technique EF 1
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GATE EE vol1Electric circuit & Field, Electrical & electronic measurement
GATE EE vol2Analog electronics, Digital electronics, Power electronics
GATE EE vol3Control systems, Signals & systems
GATE EE vol4Electrical machines, Power systems
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 openloop transfer function
( )G s ( )( )
,s sK s
0> >2 ba b a=
++
The valid rootloci 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 breakin point
CS 1.27 Consider the ufb system shown below
CS 18 Root Locus Technique CS 1PE 1 Root Locus Technique PE 18EF 1 Root Locus Technique EF 18
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The rootloci, as a is varied, will be
Common Data For Q. 27 and 28 :The forwardpath transfer function of a ufb system is
( )G s ( )
( )( )s s
K s s1
32a=
+ +
CS 1.28 The rootloci for 0K > with 5a = is
CS 1.29 The rootloci for 0>a with 10K = is
CS 1 Root Locus Technique CS 19PE 19 Root Locus Technique PE 1EF 19 Root Locus Technique EF 1
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GATE EE vol2Analog electronics, Digital electronics, Power electronics
GATE EE vol3Control systems, Signals & systems
GATE EE vol4Electrical machines, Power systems
Engineering mathematics, General Aptitude
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 jwaxis
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 closedloop system is ( 1)( 2) 0s s s K+ + + = . The centroid of the asymptotes in rootlocus 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 ListI with ListII 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
CS 20 Root Locus Technique CS 1PE 1 Root Locus Technique PE 20EF 1 Root Locus Technique EF 20
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CS 1.34 The rootlocus 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 openloop 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 ?
CS 1 Root Locus Technique CS 21PE 21 Root Locus Technique PE 1EF 21 Root Locus Technique EF 1
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GATE EE vol2Analog electronics, Digital electronics, Power electronics
GATE EE vol3Control systems, Signals & systems
GATE EE vol4Electrical machines, Power systems
Engineering mathematics, General Aptitude
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 closedloop 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
CS 22 Root Locus Technique CS 1PE 1 Root Locus Technique PE 22EF 1 Root Locus Technique EF 22
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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 ____
CS 1 Root Locus Technique CS 23PE 23 Root Locus Technique PE 1EF 23 Root Locus Technique EF 1
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GATE EE vol2Analog electronics, Digital electronics, Power electronics
GATE EE vol3Control systems, Signals & systems
GATE EE vol4Electrical machines, Power systems
Engineering mathematics, General AptitudeCS 1.46 A ufb system has an open loop transfer function
G s H s^ ^h h s sK s
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
G s H s^ ^h h s sK s
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 jwaxis at s j2 5!=2. Gain margin for K 18= is 20 dB.
3. Gain margin for K 1800= is 20 dB4. 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?
CS 24 Root Locus Technique CS 1PE 1 Root Locus Technique PE 24EF 1 Root Locus Technique EF 24
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CS 1.51 The open loop transfer function of a control system is
G s H s^ ^h h s s s s
K s1 4 16
12=
 + ++
^ ^
^
h h
h
Consider the following statements for the system1. Root locus of the system cross jwaxis for .K 35 7=2. Root locus of the system cross jwaxis 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
****************
CS 1 Root Locus Technique CS 25PE 25 Root Locus Technique PE 1EF 25 Root Locus Technique EF 1
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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 polezero 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 polezero 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 polezero 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 Polezero 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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Polezero 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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Engineering mathematics, General Aptitude
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 polezero 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!=Polezero 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
G s H s^ ^h h s s s s
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
Polezero 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
G s H s^ ^h h s s s s
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= , 5The 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
G s H s^ ^h h s s s s
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 225Now, 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
Characteristic equation,
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=
Centroid As Re ReSum of Sum of
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.
CS 44 Root Locus Technique CS 1PE 1 Root Locus Technique PE 44EF 1 Root Locus Technique EF 44
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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
Centroid As Re ReSum of Sum of
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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Engineering mathematics, General Aptitude
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
Characteristic equation,
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
Centroid As Re ReSum of Sum of
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 943Thus 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,
G s H s^ ^h h s sK s
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 jwaxis at s j2 5!= .The gain margin for K 18= is given by,
CS 50 Root Locus Technique CS 1PE 1 Root Locus Technique PE 50EF 1 Root Locus Technique EF 50
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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 4725Point .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 jwaxis is determined using routh’s array which is shown below
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GATE EE vol3Control systems, Signals & systems
GATE EE vol4Electrical machines, Power systems
Engineering mathematics, General Aptitude
s4 1 12 K
s3 3 K 16
s2 K3
52  K
s1
KK K
5259 8322
 + 
s0 K
The root locus cross the jwaxis, 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 jwaxis two times and the break points are given by
solution of dsdK 0= .
We can directly check option using polezero 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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