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GATE
ELECTRICAL ENGINEERINGVol 3 of 4
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Second Edition
GATEELECTRICAL ENGINEERING
Vol 3 of 4
RK Kanodia
Ashish Murolia
NODIA & COMPANY
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GATE Electrical Engineering Vol 3, 2eRK Kanodia & Ashish Murolia
Copyright By NODIA & COMPANY
Information contained in this book has been obtained by author, from sources believes to be reliable.However, neither NODIA & COMPANY nor its author guarantee the accuracy or completeness of anyinformation herein, and NODIA & COMPANY nor its author shall be responsible for any error, omissions, ordamages arising out of use of this information. This book is published with the understanding that NODIA &COMPANY and its author are supplying information but are not attempting to render engineering or other
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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 anddata interpretation.
ENGINEERING MATHEMATICS
Linear Algebra:Matrix Algebra, Systems of linear equations, Eigen values and eigen vectors.
Calculus:Mean value theorems, Theorems of integral calculus, Evaluation of definite andimproper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourierseries. Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes,Gauss and Greens theorems.
Differential equations: First order equation (linear and nonlinear), Higher order linear
differential equations with constant coefficients, Method of variation of parameters, Cauchysand Eulers equations, Initial and boundary value problems, Partial Differential Equationsand variable separable method.
Complex variables: Analytic functions, Cauchys integral theorem and integral formula,Taylors and Laurent series, Residue theorem, solution integrals.
Probability and Statistics: Sampling theorems, Conditional probability, Mean, median,mode and standard deviation, Random variables, Discrete and continuous distributions,Poisson,Normal and Binomial distribution, Correlation and regression analysis.
Numerical Methods: Solutions of non-linear algebraic equations, single and multi-stepmethods for differential equations.
Transform Theory:Fourier transform,Laplace transform, Z-transform.
ELECTRICAL ENGINEERING
Electric Circuits and Fields:Network graph, KCL, KVL, node and mesh analysis, transientresponse of dc and ac networks; sinusoidal steady-state analysis, resonance, basic filterconcepts; ideal current and voltage sources, Thevenins, Nortons and Superposition andMaximum Power Transfer theorems, two-port networks, three phase circuits; Gauss Theorem,electric field and potential due to point, line, plane and spherical charge distributions;Amperes and Biot-Savarts laws; inductance; dielectrics; capacitance.
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Signals and Systems:Representation of continuous and discrete-time signals; shifting andscaling operations; linear, time-invariant and causal systems; Fourier series representationof continuous periodic signals; sampling theorem; Fourier, Laplace and Z transforms.
Electrical Machines:Single phase transformer equivalent circuit, phasor diagram, tests,
regulation and efficiency; three phase transformers connections, parallel operation; auto-transformer; energy conversion principles; DC machines types, windings, generatorcharacteristics, armature reaction and commutation, starting and speed control of motors;three phase induction motors principles, types, performance characteristics, starting andspeed control; single phase induction motors; synchronous machines performance, regulationand parallel operation of generators, motor starting, characteristics and applications; servoand stepper motors.
Power Systems:Basic power generation concepts; transmission line models and performance;cable performance, insulation; corona and radio interference; distribution systems; per-
unit quantities; bus impedance and admittance matrices; load flow; voltage control; powerfactor correction; economic operation; symmetrical components; fault analysis; principles ofover-current, differential and distance protection; solid state relays and digital protection;circuit breakers; system stability concepts, swing curves and equal area criterion; HVDCtransmission and FACTS concepts.
Control Systems: Principles of feedback; transfer function; block diagrams; steady-state errors; Routh and Niquist techniques; Bode plots; root loci; lag, lead and lead-lagcompensation; state space model; state transition matrix, controllability and observability.
Electrical and Electronic Measurements:Bridges and potentiometers; PMMC, moving iron,dynamometer and induction type instruments; measurement of voltage, current, power,energy and power factor; instrument transformers; digital voltmeters and multimeters;phase, time and frequency measurement; Q-meters; oscilloscopes; potentiometric recorders;error analysis.
Analog and Digital Electronics:Characteristics of diodes, BJ T, FET; amplifiers biasing,equivalent circuit and frequency response; oscillators and feedback amplifiers; operationalamplifiers characteristics and applications; simple active filters; VCOs and timers;combinational and sequential logic circuits; multiplexer; Schmitt trigger; multi-vibrators;sample and hold circuits; A/ D and D/A converters; 8-bit microprocessor basics, architecture,programming and interfacing.
Power Electronics and Drives:Semiconductor power diodes, transistors, thyristors, triacs,GTOs, MOSFETs and IGBTs static characteristics and principles of operation; triggeringcircuits; phase control rectifiers; bridge converters fully controlled and half controlled;principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.
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PREFACE
This book doesnt make promise but provides complete satisfaction to the readers. Themarket scenario is confusing and readers dont find the optimum quality books. This book
provides complete set of problems appeared in competition exams as well as fresh set ofproblems.
The book is categorized into units which are then sub-divided into chapters and theconcepts of the problems are addressed in the relevant chapters. The aim of the book isto avoid the unnecessary elaboration and highlights only those concepts and techniques
which are absolutely necessary. Again time is a critical factor both from the point of viewof preparation duration and time taken for solving each problem in the examination. Sothe problems solving methods is the books are those which take the least distance to thesolution.
But however to make a comment that this book is absolute for GAT E preparation willbe an inappropriate one. T he theory for the preparation of the examination should befollowed from the standard books. But for a wide collection of problems, for a variety ofproblems and the efficient way of solving them, what one needs to go needs to go through
is there in there in the book. Each unit (e.g. Networks) is subdivided into average sevennumber of chapters on an average each of which contains 40 problems which are selectedso as to avoid unnecessary redundancy and highly needed completeness.
I shall appreciate and greatly acknowledge the comments and suggestion from the users ofthis book.
R. K . Kanodia
Ashish Murolia
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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
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CS 1
ROOT LOCUS TECHNIQUE
CS 1.1 Form the given sketch the root locus can be
CS 1.2 Consider the sketch shown below
The root locus can be
(A) (1) and (3)
(B) (2) and (3)
(C) (2) and (4)
(D) (1) and (4)
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CS 10 Root Locus Technique CS 1
PE 1 Root Locus Technique PE 10
EF 1 Root Locus Technique EF 10
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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
CS 1.5 A open-loop pole-zero plot is shown below
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CS 1 Root Locus Technique CS 11
PE 11 Root Locus Technique PE 1
EF 11 Root Locus Technique EF 1
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The general shape of the root locus is
CS 1.6 A open-loop pole-zero plot is shown below
The general shape of the root locus is
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CS 12 Root Locus Technique CS 1
PE 1 Root Locus Technique PE 12
EF 1 Root Locus Technique EF 12
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CS 1.7 A open-loop pole-zero plot is shown below
The general shape of the root locus is
CS 1.8 The forward-path open-loop transfer function of a ufbsystem is
( )G s ( )( )
s s
K s s
8 25
2 62= + +
+ +
The root locus for this system is
CS 1.9 The forward-path open-loop transfer function of as ufbsystem is
( )G s ( )
( )
s
K s
1
42
2
=+
+
For this system root locus is
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CS 1 Root Locus Technique CS 13
PE 13 Root Locus Technique PE 1
EF 13 Root Locus Technique EF 1
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CS 1.10 The forward-path open-loop transfer function of a uf bsystem is
( )G s ( )
s
K s 12
2
= +
The root locus of this system is
Common Data For Q. 11 and 12:
A open-loop pole-zero plot is shown below
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CS 14 Root Locus Technique CS 1
PE 1 Root Locus Technique PE 14
EF 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 s
3 22 22
+ +
+ + (B)
( )
( )( )
s s
K s s
2 2
3 23+ +
+ +
(C)( )( )
( )s s
K s s
3 22 22
+ +- + (D)
( )( )( )s s
K s s
2 23 2
2- ++ +
CS 1.12 The break point is(A) breakaway at .129s =-
(B) breakin at .243s =-
(C) breakaway at .243s =-
(D) breakin at .129s =-
Common Data For Q. 13 and 14:
A root locus of ufbsystem is shown below
CS 1.13 The breakaway point is _ _ _ _
CS 1.14 At breakaway point the value of gain K is
(A) 24
(B) 7.4 10 3# -
(C) 8.6 10 3# -
(D) 29
CS 1.15 The forward-path transfer function of a ufbsystem is
( )G s ( )( )
( )
s s s
K s
3 2 2
22= + + +
+
The angle of departure from the complex poles is _ _ _ _
CS 1.16 Consider the feedback system shown below
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CS 1 Root Locus Technique CS 15
PE 15 Root Locus Technique PE 1
EF 15 Root Locus Technique EF 1
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For 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 onthe real axis at 5- , will be _ _ _ _
CS 1.19 The open-loop transfer function a system is
( ) ( )G s H s ( )( )( )
( )s s s s
K s
4 12 208
=+ + +
+
A closed loop pole will be located at 10s=- , when the value of K is _ _ _
CS 1.20 For a uf bsystem forward-path transfer function is
( )G s ( )( )
( )s s
K s
3 56
=+ +
+
The breakaway point and break-in points are located respectively as
(A) ,4.273
(B) 7.73,4.27
(C) 4.27,3
(D) 4.27,7.73
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CS 16 Root Locus Technique CS 1
PE 1 Root Locus Technique PE 16
EF 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
K
1 2=
+ +
The root locus plot of above system is
CS 1.22 A uf bsystem has forward-path transfer function
( )G s s
K2=
The root locus plot is
CS 1.23 For the ufbsystem, shown below consider two point
s1 2 3j=- + and 2s j2
12 =- +
Which of the above point lie on root locus ?
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CS 1 Root Locus Technique CS 17
PE 17 Root Locus Technique PE 1
EF 17 Root Locus Technique EF 1
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(A) Both s1and s2 (B) s1but not s2
(C) s2but not s1 (D) neither s1nor s2
CS 1.24 A uf bsystem has open-loop transfer function
( )G s ( )
( ),
s s
K s0> >2 b
ab a=
+
+
The valid root-loci for this system is
CS 1.25 The characteristic equation of a feedback control system is given by
( 4 4)( 11 30) 4s s s s K s K 2 2 2+ + + + + + 0=where 0K > . In the root locus of this system, the asymptotes meet in s-plane at(A) ( 9.5,0)-
(B) ( 5.5,0)-
(C) ( 7.5,0)-
(D) None of the above
CS 1.26 The root locus of the system having the loop transfer function
( ) ( )G s H s ( )( )s s s s
K
4 4 52=
+ + +has
(A) 3 breakaway point
(B) 3 breakin point
(C) 2 breakin and 1 breakaway point
(D) 2 breakaway and 1 break-in point
CS 1.27 Consider the ufbsystem shown below
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CS 18 Root Locus Technique CS 1
PE 1 Root Locus Technique PE 18
EF 1 Root Locus Technique EF 18
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The root-loci, as ais varied, will be
Common Data For Q. 27 and 28 :
The forward-path transfer function of a ufbsystem is
( )G s ( )
( )( )
s s
K s s
1
32
a=
-
+ +
CS 1.28 The root-loci for 0K > with 5a = is
CS 1.29 The root-loci for 0>a with 10K = is
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CS 1 Root Locus Technique CS 19
PE 19 Root Locus Technique PE 1
EF 19 Root Locus Technique EF 1
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CS 1.30 For the system ( ) ( )G s H s ( )( )
( )s s
K s
2 46
=+ +
+, consider the following characteristic
of the root locus :1. It has one asymptotes
2. It has intersection with jw-axis3. It has two real axis intersections.
4. It has two zeros at infinity.
The root locus have characteristics
(A) 1 and 2 (B) 1 and 3
(C) 3 and 4 (D) 2 and 4
CS 1.31 The characteristic equation of a closed-loop system is ( 1)( 2) 0s s s K + + + = .The centroid of the asymptotes in root-locus will be _ _ _ _
CS 1.32 The forward path transfer function of a uf bsystem is
( )G s ( )( )( )
( )s s s s
K s
1 2 43
=+ + +
+
The angles of asymptotes are
(A) 0, ,2p
p (B) 0, ,32
34p p
(C) , ,3 3
5pp
p (D) None of the above
CS 1.33 Match List-I with List-I I in respect of the open loop transfer function
( ) ( )G s H s ( )( )( )
( )( )
s s s s s
K s s s
20 50 4 5
10 20 5002
2
=+ + + +
+ + +
List I (Types of Loci) List II (Numbers)
P. Separate Loci 1. One
Q. Loci on the real axis 2. Two
R. Asymptotes 3. Three
S. Break away points 4. Five
P Q R S(A) 4 3 1 1
(B) 4 3 2 1(C) 3 4 1 1(D) 3 4 1 2
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CS 20 Root Locus Technique CS 1
PE 1 Root Locus Technique PE 20
EF 1 Root Locus Technique EF 20
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CS 1.34 The root-locus of a uf bsystem is shown below
The open loop transfer function is
(A)( )( )s s s
K
1 3+ + (B)
( )( )
s s
K s
13
+
+
(C) ( )
( )
s s
K s
3
1
+
+
(D) ( )( )s s
K s
1 3+ +
CS 1.35 The characteristic equation of a linear control system is 5 9 0s K s2+ + = . Theroot loci of the system is
CS 1.36 A unity feedback control system has an open-loop transfer function
( )G s ( )s s s
K
7 122=
+ +
The gain K for which 1 1s j=- + will lie on the root locus of this system is
_ _ _
CS 1.37 An unity feedback system is given as ( )G s ( )( )
s s
K s
31
=+
-. Which is the correct root
locus diagram ?
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CS 1 Root Locus Technique CS 21
PE 21 Root Locus Technique PE 1
EF 21 Root Locus Technique EF 1
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CS 1.38 The open loop transfer function ( )G s of a uf bsystem is given as
( )G s ( )s s
K s
2232
=+
+^ hFrom the root locus, at can be inferred that when K tends to positive infinity,
(A) three roots with nearly equal real parts exist on the left half of thes-plane
(B) one real root is found on the right half of the s-plane(C) the root loci cross the jwaxis for a finite value of ; 0K K!
(D) three real roots are found on the right half of the s-plane
CS 1.39 The characteristic equation of a closed-loop system is
( 1)( 3) ( 2) 0, 0s s s K s k >+ + + + = .Which of the following statements is true ?
(A) Its root are always real
(B) It cannot have a breakaway point in the range [ ]Re s1 0< a
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CS 22 Root Locus Technique CS 1
PE 1 Root Locus Technique PE 22
EF 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) .1056- , .3471- (B) 2.112, 6.943- -
(C) 1.056, 6.943- - (D) 1.056, 6.943-
CS 1.44 For the characteristic equation s s K s K 23 2+ + + 0= , the root locus of the systemas K varies from zero to infinity is
CS 1.45 The open loop transfer function of a ufbsystem is
G s H s ^ ^h hs s
K s
9
12= +
+
^^
hh
In the root locus of the system as parameter K is varied from 0 to 3, the gainK when all three roots are real and equal is _ _ _ _
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CS 1 Root Locus Technique CS 23
PE 23 Root Locus Technique PE 1
EF 23 Root Locus Technique EF 1
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CS 1.46 A uf bsystem has an open loop transfer function
G s H s ^ ^h hs s
K s
1
1=
-
+
^
^
h
hRoot locus for the system is a circle. Centre and radius of the circle are respectively(A) ,0 0^ h, 2(B) ,0 0^ h, 2(C) ,1 0-^ h, 2(D) ,1 0-^ h, 2
CS 1.47 The open loop transfer function of a system is
G s H s ^ ^h hs s
K s
2
3=
+
+
^^
hh
The root locus of the system is a circle. The equation of circle is(A) 4 42 2s w+ + =^ h (B) 3 32 2s w- + =^ h(C) 3 32 2
2s w+ + =^ ^h h
(D) 4 22 2 2s w- + =^ ^h h
CS 1.48 Consider the open loop transfer function of a system shown below
G s H s ^ ^h hs s s s
K
2 2 6 102 2=
+ + + +^ ^h hThe break away point in root locus plot for the system is/ are(A) 3 real
(B) only real
(C) 1 real, 2 complex
(D) None
CS 1.49 The open loop transfer function of a ufbsystem is given below.
G s H s ^ ^h hs s s
K
4 5=
+ +^ ^h hConsider the following statements for the system.
1. Root locus plot cross jw-axis at s j2 5!=
2. Gain margin for K 18= is 20 dB.
3. Gain margin for K 1800= is 20dB-
4. Gain K at breakaway point is 13.128
Which of the following is correct ?
(A) 1 and 2
(B) 1, 2 and 3
(C) 2, 3 and 4
(D) All
CS 1.50 The open loop transfer function of a system is
G s H s ^ ^h hs s
K
1 5=
+ +^ ^h hWhat is the value of K , so that the point s j3 5=- + lies on the root locus?
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CS 24 Root Locus Technique CS 1
PE 1 Root Locus Technique PE 24
EF 1 Root Locus Technique EF 24
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Sample
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CS 1.51 The open loop transfer function of a control system is
G s H s ^ ^h hs s s s
K s
1 4 16
12= - + +
+
^ ^
^
h h
hConsider the following statements for the system1. Root locus of the system cross jw-axis for .K 357=
2. Root locus of the system cross jw-axis for .K 233=
3. Break away point is .s 045=
4. Break in point is .s 226=-
Which of the following statement is correct ?(A) 1, 3 and 4 (B) 2, 3 and 4
(C) 3 and 4 (D) all
****************
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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 fis net angle contribution at this pole due to all other poles and zeros.
f Z Pf f= -
f 90 90 90 90c c c c= - + - +6 6@ @ f 180c=-
Departure angle Df 180 180! c c= -6 @
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Df 0! c=
So, departure angle for pole P0is 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 cc=
-
+= =
^ h; q 1=
Hence, root locus plot will be as shown below.
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CS 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 Correctoption is (A).
CS 1.8 Correct option is (C).
Forward Path open loop transfer function of given uf bsystem is
G s^ hs s
K s s
8 25
2 62= + +
+ +^ ^h h
Characteristic equation is
s s8 252+ + 0=
s j2
8 64 100 4 3! !=- - =-
s j4 3=- + ; s j4 3=- -
Thus pole are s j4 3=- + ; s j4 3=- - and zeros are 2s=- ; 6s=-
Plot of Pole-zero is shown below
Root locus will start from poles and ends with zeros.
Only option (C) satisfy above condition.
CS 1.9 Correct option is (B).
Forward path open loop transfer function of given uf bsystem is,
G s
^ h
s
K s
1
42
2
=
+
+^ h
Here, poles s 12+ 0=
s j1!=
and zeros s 42+ 0=
s j2!=
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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 uf bsystem is
G s^ hs
K s 12
2
= +^ h
Zeros are : s 12+ 0=
s j1!=
Poles are : s 0= ; s 0=
Location of poles and zeros is shown below
Hence option (B) and (D) may not be correct option. A point on the real axis
lies on the root locus if the total no. of poles and zeros to the right of this point
is odd. This is not satisfied by (C) because at origin there are double pole. Thus
remaining Correct option is (A).
CS 1.11
Correct option is (C).Given open loop pole zero plot is :
From above plot zeros are 1 1s j= + and 1 1s j= - and poles are 2s=- and3s=-
Transfer function of the system is
G s^ hs s
K s j s j
2 3
1 1 1 1=
- - - -
- + - -
^ ^^ ^
h hh h
" "" "
, ,, ,
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s s
K s j s j
2 3
1 1 1 1=
+ +
- - - +
^ ^^ ^
h hh h" ", ,
s s
K s j
2 3
1 12 2
= + +
- -
^ ^^ ^h hh h" ,or G s^ h
s s
K s s
2 3
2 22=
+ +
- +
^ ^^
h hh
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-^ hand 3-^ honreal axis. Thus s 1.29=- can not be break point because it does not lie on root
locus. Thus possible break point is 2.43s =- which lies between 2- and 3- .
On root locus it may be seen easily that 2.43s =- lie on root locus and locus
start from pole 2-^ hand 3-^ h. Thus at 2.43s =- it must break apart.Thus this point is break away point.
Gain K will be maximum at break away pointand minimum at break in point.
We can also check maxima and minima for gain K
Method II
The point, at which multiple roots are present, are known as break point. These
are obtained from theds
dK 0=
Here, characteristic equation is
G s H s 1+ ^ ^h h 0=
s s
K s s1
2 3
2 22+
+ +
- +
^ ^^
h hh 0=
K s s
s s
2 2
2 32= - +
- + +
^^ ^
hh h
s s
s s
2 2
5 62
2
=- +
- + +^ h ...(1)
Now, differentiating eq (1) w.r.t sand equating zero we have
ds
dK s s
s s s s s s
2 22 2 2 5 5 6 2 2 02 2
2 2=
- +
- - + + + + + -=^^ ^ ^ ^hh h h h
s s7 8 222+ - 0=
which gives 1.29s=+ and 2.43s=- out of which 2.43s=- is break point.
CS 1.13 Correct answer is .145- .
Given root locus of uf bsystem is shown below
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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^ hs s
K s s
1 2
3 5=
+ +
- -
^ ^^ ^
h hh h
Characteristic equation
G s H s 1+ ^ ^h h 0=
s s
K s s1
1 2
3 5+
+ +
- -
^ ^^ ^
h hh h
0=
K s s
s s
8 15
3 22
2
=- +
- + +
^
hh ...(1)
Differentiating eq. (1) wrt to sand equating to zero we have
dsdK
s s
s s s s s s
8 15
8 15 2 3 3 2 2 802 2
2 2
=- +
- - + + + + + -=^
^ ^ ^ ^h
h h h h s s11 26 612- - 0=
or s .39=+ and .s 145=-
Thus .s 145=- is break away point and .s 39=+ is break in point.
CS 1.14 Correct option is (C).
Break away and Break in points always satisfy characteristic equation.
Substituting .s 145=- in eq (1), we get
K . .
. .
145 8 145 15
145 3 145 22
2
=- - - +
- - + - +
^ ^^ ^
h hh h
68
@B
..
28702502475
=- -^ h
.862 10 3#= -
CS 1.15 Correct answer is 108.4 .
Forward path transfer function of given ufbsystem is
G s^ hs s s
K s
3 2 2
22= + + +
+
^ ^^
h hh
Here zero is s 2=- and poles are s 3=- and s j1 1!
=-Pole-zero Plot is shown below
Angle of departure at pole P1is given by :
Df 180! c f= +6 @
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where fis net angle contribution at pole P1due to all other poles and zeros.
f Z Pf f= -
Z P P1 2 3f f f= - +
6 @where, Z1f tan 11= - ; 90P2 cf = ; tan 21P3 1f = - f 9tan tan1 0
211 1
c= - +- -: DDeparture angle :
Df 180! c f= +6 @ 1tan tan180 1 90
21 1
! c c= + - -- -: D .180 45 90 2656!= + - -6 @ Df .1084! c=
Hence, departure angle for pole P1is 108.4c
+ and departure angle for pole P2is108.4c- because P1and P2are complex conjugate.
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 sK s
1 3
2= + +
- +
^ ^h hhRoot locus is plotted for K 0= to K 3= . But here, K is negative. Thus we will
plot for K 3=- to K 0= . This is called complementary root locus.
Hence, the root locus on the real axis is found to the left of an even countof real
poles and real zeros of G s^ h. Plot will start from pole and ends on zero. ThusCorrect option is (A).
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
K
4 2 22=
+ + +^ ^h h
s s s
K
6 10 83 2=
+ + +
Closed loop transfer function,
T s^ h1
s s s
Ks s s
K
6 10 8
6 10 8
3 2
3 2
=+
+ + +
+ + +
s s s K
K
6 10 83 2=
+ + + +
Characteristics equation :
6 10 8s s s K 3 2+ + + + 0=
Routh array is shown below
s3 1 10
s2 6 K8+
s1 K
652-
s0 K8+
When root locus cut will cut imaginary axis if element in s1is zero.
Thus K6
52- 0= &K 52=
and then from auxiliary equation we have
s6 8 522+ +^ h 0= s2
660 10=- =-
s .j3162!=
CS 1.18 Correct answer is 17.Using Pole - zero plot :
Gain K at any s s0=^ hpoint on root locus is given by : K
s s0=
intPr intProduct of phasorsdrawnfrom at that po
oduct of phasorsdrawnfrom at that poOL Z
OL P=
Here no any zero is present.
Hence, K =Product of phasors drawn fromOL Pat that point
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Ks 5=-
PP P P PP 3 1 2# #= ^ ^ ^h h h 1 4 1 4 12 2# #= + +
4 1 172= + =
K 17=
CS 1.19 Correct answer is 600.
Open loop transfer function of given system is
G s H s ^ ^h hs s s s
K s
4 12 20
8=
+ + +
+
^ ^ ^^
h h hh
Here, Zeros : Z 8=-
Poles : P 01 = ; P 42 =- ; P 123 =- ; P 204 =-
Value of K at 10s=- is :
Ks 10=-
10
10
Phasorsdrawnfrom at
Phasorsdrawnfrom at
OL Z s
OL P s
p
p=
=-
=-
^ hhPole-zero plot is shown below.
Ks 10=-
PZ
PP PP PP PP 1 2 3 4# # #= ^
^ ^ ^ ^h
h h h h
2
10 6 2 10 600# # #= =
K 600=
CS 1.20 Correct option is (D).
For given system, forward path transfer function,
G s^ h s sK s
3 5
6= + +
+
^ ^^ h hhCharacteristic equation for closed loop transfer function,
G s H s 1+ ^ ^h h 0=or
s s
K s1
3 5
6+
+ +
+
^ ^^h h
h 0=
or K s
s s
6
8 152=
+
- + +
^^
h h
differentiating w.r.t to sand equating to zero, we have
ds
dK s
s s s s
6
6 2 8 8 1502
2
=
+
- + + + + +=
^^ ^ ^
hh h h
or s s12 332+ + 0=
or s .773=- and .s 427=-
The root locus is shown below. Here we can easily say that s .427=- is break
away point and s .773=- point.
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CS 1.21 Correct option is (B).
Open loop transfer function :
G s H s ^ ^h hs s s
K
1 2=
+ +^ ^h hIn option (A) and pption (C), the root locus on the real axis is found to the left
of an even count of real poles and real zeros of GH . Hence, these can not rootlocus diagram.
Now characteristic equation ,
G s H s 1+ ^ ^h h 0=
s s s
K11 2
++ +^ ^h h 0=
s s s K 3 23 2+ + + 0=
Routh array is shown below.
s3
1 2
s2 3 K
s1 K
36-
s0
K
At K 6= , s1row is zero, thus using auxiliary equation we get,
s3 62+ 0=
s j 2!=
Root locus cut on jwaxis at s j 2!= for K 6= . Thus option (B) is correct
because (D) does not cut jwaxis.
CS 1.22 Correct option is (D).
For given uf bsystem, forward transfer function,
G s^ hs
K2=
Angle of departure or angle of asymptote for multiple poles is,
af rq2 1 180c
= +^ h
;
where r =no. of multiple poles
q 0= , 1, 2, .... r 1-^ hHere, r 2= ; (2 multiple poles at origin) q 0= , 1
af 20 1 180
90c
c= +
=^ h
for q 0=
and af 22 1 180
270c
c= +
=^ h
for q 1=
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Hence, root locus plot will be as shown below.
CS 1.23 Correct option is (C).
For given ufbsystem, open loop transfer function is,
G s
^ h
s s
K s s
1 2
3 4=
+ +
+ +
^ ^^ ^
h hh h
Given two point, s1 j2 3=- + ; s j22
12 =- +
If any point lie on root locus, it satisfies the characteristic equation. Thus we have
q s^ h 1 G s H s 0= + =^ ^h hor G s H s ^ ^h h 1= (Magnitude)and G s H s ^ ^h h 180! c= (Phase)Point: 2 3s j1 =- +
Now we use the phse condition G s H s s s
11
f==
^ ^h h , we have 1f 1 2 3 4q q q q= + + +
tan tan tan23
13 90
131 1 1
c= + - --
- - - b l tan tan tan
23 3 90 180 31 1 1c c= + - - -- - -^ h
. . .5630 7156 270 7156c= + - +
or 1f .7056 180! c!=-
Hence, point s1not lie on root locus.
Point : s j22
12 =- + :-
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Using phase condition G s H s s s2=
^ ^h h 2f= we have, 2f 31 2 4q q q q= + - -
tan tan tan2 2
1
2
1
90 180 2
11 1 1c c
= + - - -
- - -
c m tan tan tan
2 21
21 90 180
211 1 1
c c= + - - +- - -
2f 180c=- Here phase condition is satisfied.
Hence, point s2lies on root locus.
CS 1.24 Correct option is (A).
For given uf bsystem, open loop transfer function is
G s^ hs s
K s2 b
a=
+
+
^
^
h
h; 0> >b a
Here, zero : Z a=-
Poles : s 0= , 0; s b=-
Departure angle at double poles on origin is :
fr
q2 1 180c=
+^ h; r 2= , q 0= , 1
f 90c= and 270c
To get intersection with imaginary axis we use Ruth array as shown below.
characteristic equation is
s s K s K 3 2b a+ + + 0=
Routh Array is shown below
s3
1 K
s2
b Ka
s1
b
b a-
s0
Ka
Here, for any value of K , slrow of Routh array will not be zero. Thus system
is stable for all positive value of K and hence root locus does not cross jwaxis.
Therefore root locus completely lies in left half of s-plane.Based on these above result we say that Correct option is (A).
CS 1.25 Correct option is (C).
Characteristic equation of given feedback control system is
s s s s K s K 4 4 11 30 42 2 2+ + + + + +^ ^h h 0=or
s s s s
K s1
4 4 11 30
42 2
2
++ + + +
+
^ ^^h h
h 0= ...(1)
Characteristic equation is,
G s H s 1+ ^ ^h h 0= ...(2)Comparing, eq (1) and (2) we get open loop transfer function as G s H s ^ ^h h
s s s s
K s
4 4 11 30
42 2
2
=+ + + +
+
^ ^^h h
h
( )( )( )( )s s s s
K s
2 2 5 642
=+ + + +
+^ h
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Open loop poles are
s ,2 2=- - and s 6=- , 5-
The point at which asymptotes meet (centroid) is given by:
As Re ReSumof Sumof
P Z
P Z=
-
-
^ h6 6@ @Here, open loop zeros : s 4 02+ =
s j2!=
As 4 22 2 5 6 0
=-
- - - - -^ h
.215 75=- =-
It intersects on real axis. So point is: . ,75 0-^ hCS 1.26 Correct option is (D).
Open loop transfer function for given system is
G s H s ^ ^h hs s s s
K
4 4 52=
+ + +^ ^h hCharacteristic equation,
s s s s
K14 4 52
++ + +^ ^h h 0=
K s s s s 4 4 52=- + + +^ ^h h ...(1)Differentiating wrt sand equating to zero we have
dsdK s s s s s s 2 4 4 5 4 2 4 02 2=- + + + + + + =^ ^ ^ ^h h h h8 B
or s s s s s 2 4 4 5 42 2- + + + + +^ ^h h 0=or s s s2 2 8 52- + + +^ ^h h 0= ...(2)or s 2=- and .s 0775=- , .3225-
Now, we check for maxima and minima value of gain K at above point. If gain
is maximum, then that point will be break away point. If gain is minimum, then
that point will be break in point.
Again differentiating equation (2) with respect to s, we get,
dsd K
2
2
s s s s 2 8 5 2 4 82
=- + + + + +^ ^ ^h h h6 @ s s6 24 212=- + +^ hFor .s 0775=- and .s 3225=- ,
ds
d K2
2
. 060 =+ ; s 2=- is minima points.
Thus s 2=- is break in point.
Hence, there is two break away points . , .s 0775 3225=- -^ hand one break inpoint.
CS 1.27 Correct option is (B).
For a uf bsystem, forward transfer function is :
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Sample
Chapterof
GATEElectr
icalEnginee
ring,
Volum
e-3 G s^ h s s
1a
=+^ h
Method I :
Characteristic equation : G s H s 1+ ^ ^h h 0=
s s1 1
a+
+^ h 0= s s 12 a+ + 0=
s
s112
a+
+^ h 0=Open loop transfer function, as ais varied is
G s H s ^ ^h hs
s
12a
=+
Here, zero : s 0=and poles : s 1 02+ = s j& !=
Root locus will be :
Method II :
Closed loop transfer function :
T s^ hG s H s
G s
s s1 11
2 a=
+ =
+ +^ ^^h hh
G s H s
G s
1+ ^ ^^h hh
s
ss
1
1
11
2
2
a=
+
+
+
G s H s ^ ^h hs
s
12a
=+
CS 1.28 Correct option is (A).
Forward path transfer function of given ufbsystem is
G s^ hs s
K s s
1
32
a=
-
+ +
^^ ^
hh h
; 5a = and K 0>
or G s^ hs s
K s s
1
5 32= -
+ +
^^ ^
hh h
Zeros : s 5=- , s 3=- Poles : s 0= , s 1= , s 1=-
Locus branches start from pole and ends on zeros or infinite along asymptote.
Here, no. of asymptotes P Z 3 2 1= - = - =
Only option (A) has one asymptotes.
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SampleChapt
erofGATEElectricalEn
gineering,V
olume-3
GATE EE vol-1
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GATE EE vol-3
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Angle of asymptotes,
af P Z
q2 1 180c=
-
+
^
^
h
h; q 0= , 1, 2, .... P Z 1- -^ h
1
0 1 180180
cc=
+=^ h
Only option (A) satisfies these conditions.
CS 1.29 Correct option is (C).
Forward path transfer function :
G s^ hs s
K s s
1
32
a=
-
+ +
^^ ^
hh h
; K 10= and 0>a
Characteristic equation,
G s H s 1+ ^ ^h h 0=
s s
s s1
1
10 32
a+
-
+ +
^^ ^
hh h
0=
s s s s 10 3 33 2 a a- + + + +^ h6 @ 0= 10s s s s 10 29 32 a+ + + +^ ^h h 0= ...(1)
s s s
s1
10 29
10 32
a+
+ +
+
^ ^
hh
0= ...(2)
Open loop gain as ais varied,
G s H s ^ ^h hs s s
s
10 29
10 32
a=
+ +
+
^
^
h
h ...(3)
Here, no. of asymptotes P Z= -
3 1 2= - =
So, two root branches will go to infinite along asymptotes as "3a .
Now using equation (1) we have
s s s10 29 10 303 2 a a+ + + +^ h 0=Routh array is shown below
s3 1 ( )29 10a+
s2 10 30a
s1 ( )29 7a+
s0 30a
For 0>a , s1row can not be zero. Hence, root locus does not intersect jwaxis
for 0>a .
CS 1.30 Correct option is (B).
Open loop transfer function given
G s H s ^ ^h h 2s sK s
46=
+ ++
^ ^^ h hh Poles : 2P= ; s 2=- and s 4=-
Zeros : Z 1= ; s 6=-
No. of asymptotes P Z= -
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Sample
Chapterof
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icalEnginee
ring,
Volum
e-3
2 1 1= - = Thus (1) is correct.
Now characteristic,
s s K s 2 4 6+ + + +
^ ^ ^h h h 0=
or 8 6s K s K 62+ + + +^ h 0=Making Routh array we have
s2 1 K8 6+
s1 K6+
s0 K8 6+
Root locus is plotted for K 0= to 3. i.e. K 0> .
Here, for K 0> root locus does not intersect jwaxis because s1row will not bezero. Thus (2) is incorrect.
Here, poles are two and zero is one. Hence one imaginary zero lies on infinite.
Thus (4) is incorrect.
Therefore (B) must be correct option. But we check further as follows.
Root locus will be as given below
It has two real axis intersections. Thus (3) is correct.
CS 1.31 Correct answer is 1- .
Characteristic equation of given closed loop system is
s s s K 1 2+ + +^ ^h h 0=or
1 2s s sK1+
+ +
^ ^h h 0=
G s H s 1+ ^ ^h h 0=Comparing we get open loop transfer function,
G s H s ^ ^h hs s s
K
1 2=
+ +^ ^h hPoles : s 0= , s 1=- , s 2=-
Zeros : No zero
Centroid As Re ReSumof Sumof
P Z
P Z=
-
-6 6@ @
3
3 0
0 1 2 0
3=
-
- - -=-
^ ^h h 1=-
CS 1.32 Correct option is (C).
Forward path transfer function of given ufbsystem is,
G s^ hs s s s
K s
1 2 4
3=
+ + +
+
^ ^ ^^
h h hh
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SampleChapt
erofGATEElectricalEn
gineering,V
olume-3
GATE EE vol-1
Electric circuit & Field, Electrical & electronic measurementGATE EE vol-2
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GATE EE vol-3
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GATE EE vol-4
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Here zeros : Z 1= ; s 3=-
and poles : P 4= ; s 0= , s 1=- , s 2=- , s 4=-
Angle of asymptotes ,
af P Z
q2 1 180c=
-
+
^^ hh ; q 0= , 1, 2,..... P Z 1- -^ h
4 1
0 1 1803
180 60c
c=-
+= =^
^h
h; q 0=
4 1
2 1 180180
cc=
-
+=^
^h
h; q 1=
4 1
4 1 180300
cc=
-
+=^
^h
h; q 2=
Thus af
;
;300 ; 2
q
60 0
180 1
3
35
c
c
cp=
= =
= == =
p
p
Z
[\
]]
]]
CS 1.33 Correct option is (B).
Open loop transfer function,
G s H s ^ ^h hs s s s s
K s s s
20 50 4 5
10 20 5002
2
=+ + + +
+ + +
^ ^ ^^ ^
h h hh h
Separate loci =No. of open loop poles 5=
Asymptotes =No. of OLP No. of OLZ 5 3 2= - =
At this point we say that Correct option is (B). But we check further as follows. Loci on real axis =no. of poles that lie on real axis
3= ; (s 0= , s 20=- , s 50=- )
Open loop zeros : s 10=- , 10s j20!=-
Open loop zeros : 0s= , s 20=- , s 50=- , s j2 1!=-
Centroid As 5 30 20 50 2 2 10 10 10
=-
- - - - - - - -^ ^h h 22=-
Angle of asymptotes,
af P Zq2 1 180
c=
-+^ h ; P Z 2- = , q 0= , 1
2
0 1 18090
cc=
+=
^ h; q 0=
2
2 1 180270
cc=
+=
^ h; q 1=
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Sample
Chapterof
GATEElectr
icalEnginee
ring,
Volum
e-3
The root locus is shown below.
Here, Root locus lie only region on real axis that is in left of an odd count of real
poles and real zeros.
Hence, Root locus lies between 20- and 50- and break away point will be alsoin this region. Only one break away point will there.
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 s
K s
13
+
+.
CS 1.35 Correct option is (D).Characteristic equation, s K s5 92+ + 0=
s
K s19
52+ +
0=
G s H s 1+ ^ ^h h 0=Open loop transfer function is,
G s H s ^ ^h hs
K s
952= +
OLP :s j3!= and OLZ : s 0=
Option (A) and option (B) are incorrect because root locus are starting from zeros.On real axis loci exist to the left of odd number of real poles and real zeros.
Hence only Correct option is (D).
CS 1.36 Correct answer is 10.
Open loop transfer function,
G s_ is s s
K
7 122=
+ +^ h ; H s 1=^ hIf point : s j1 1=- + lies on root locus, then it satisfies the characteristic equation
G s H s 1+ ^ ^h h 0=
s s sK17 122
++ +^ h 0=
s s s K 7 123 2+ + + 0= ...(1)
Substituting s j1 1=- + we have
j j j K1 7 1 12 13 2- + + - + + - + +^ ^ ^h h h 0=
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erofGATEElectricalEn
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olume-3
GATE EE vol-1
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GATE EE vol-3
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GATE EE vol-4
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K10- + 0=
K 10=+
CS 1.37 Correct option is (C).
Open loop transfer function of given uf bsystem is,
G s^ hs s
K s
3
1=
+
-
^^
hh
s s
K s
3
1=
+
- -
^
hh
OLP : s 0= , s 3=-
OLZ : s 1=
Here, Gain K is negative, so root locus will be complementary root locus and is
found to the left of an even countof real poles and real zeros of GH .
Hence Option (A) and Option (D) are incorrect. Option (B) is also incorect
because it does not satisy this condition. Hence (C) is correct.
CS 1.38 Correct option is (A).
Open loop transfer function given is
G s^ h( )s s
K s
2232
=+
+^ hRoot locus starts at OLP : s 0= , s 0= , s 2=-
No. of asymptotes ,
# #OLP OLZ- =No. of OLP No. of OLZ
3 1 2= - =Angle of asymptotes :
af P Zq2 1 180c
=-
+^ h; P Z 2- = , q 0= , 1
1. af 20 1 180
90c
c= +
=^ h
for q 0=
2. af 22 1 180
270c
c= +
=^ h
for q 1=
Centroid As Re ReSumof Sumof
P Z
P Z=
-
-6 6@ @
0 2
3 132
32
=-
- - -
=-^ bh l
Angle of departure at double pole (at origin)
Df rq2 1 180c
= +^ h
; r 2= , q 0= , 1
90c= , 270c
Fro above analysis Root locus plot will be as given below.
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icalEnginee
ring,
Volum
e-3
From root locus it can be observe easily that for all values of gain K (K 0= to
3) root locus lie only in left half of s-plane.
CS 1.39 Correct option is (C).
Characteristic equation of a closed loop system is :
s s s K s 1 3 2+ + + +^ ^ ^h h h 0= ; K 0>
3s s s
K s1
1
2+
+ +
+
^ ^^h h
h 0=
Open loop transfer function is,
G s H s ^ ^h hs s s
K s
1 3
2=
+ +
+
^ ^^h h
h
OLP : s 0= , s 1=- , s 3=- and OLZ : s 2=-
Pole zero plot is shown below.
No. of asymptotes P Z= - 3 1 2= - =
Angle of asymptotes : af P Zq2 1 180c
=-
+^ h; P Z 2- = , q 0= , 1
af 90c= and 270c
Centroid As Re ReSumof Sumof
P Z
P Z=
-
-6 6@ @
3 1
0 1 3 222
=-
- - - -=-
^ ^h h1=-
Break away point lie in the range Re s1 0<
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SampleChapt
erofGATEElectricalEn
gineering,V
olume-3
GATE EE vol-1
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GATE EE vol-3
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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^ hs
K3=
Now we verify the above result as follows.
Using phase condition we have G s H s s s0=
^ ^h h 180! c=From given plot, for a given point on root locus we have
G s H s ,s 1 3=
^ ^ ^h h h tan xy3 1=- - a k tan3
131=- - c m
3 60 180# c c=- =-
CS 1.41 Correct option is (B).
Open loop transfer function of given uf bsystem is
G s^ hs s s
s
2 10
2 a=
+ +
+
^ ^^h h
hCharacteristic equation,
s s s
s1
2 10
2 a+
+ +
+
^ ^^h h
h 0=
2s s s s 10 2 2a+ + + +^ ^h h 0= s s s12 22 23 2 a+ + + 0=
s s s1
2 222
3 2a
++ +
0=
Open loop transfer function as avaried,
G s H s ^ ^h hs s s12 22
23 2
a=
+ +
No. of poles : P 3= and no. of zeros : Z 0=
Angle of asymptotes,
af
P Z
q2 1 180c=
-
+^ h;
P Z3
- =;
q0
=, 1, 2
3
0 1 18060
cc=
+=
^ hfor q 0=
3
2 1 180180
cc=
+=
^ hfor q 1=
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Sample
Chapterof
GATEElectr
icalEnginee
ring,
Volum
e-3 3
4 1 180300
cc=
+=
^ hfor q 2=
af 60c= , 180c, 300c
CS 1.42 Correct option is (C).
Intercept point of asymptotes
Centroid As Re ReSumof Sumof
P Z
P Z=
-
-6 6@ @Poles : s 0= , s 2=- , s 10=-
Zeros : No any zero.
As 3 00 2 10 0
4=-
- - -=-
^ h
CS 1.43 Correct option is (C).
We have K s s s
212 223 2
=- + +^ h
ds
dK 0= s s
23 24 22
02
&- + +
=^ h
or s s3 24 222- - - 0=
or s .1056=- , .6943-
Thus break point are s .1056=- , and .6943- .
CS 1.44 Correct option is (A).
Given characteristics equation is
s s K s K 23 2+ + + 0=
s s K s 2 13 2+ + +^ h 0=
s s
K s1
2
12+ +
+
^^
hh 0=
Open loop transfer function,
G s H s ^ ^h hs s
K s
2
12= +
+
^
^
h
h
OLZ : s 1=- ; No. of zeros : 1Z=
OLP : s 0= , s 0= , s 2=- ; No. of Poles : P 3=
Root loci starts K 0=^ hat s 0= , s 0= and s 2=-One of root loci terminates at s 1=- and other two terminates at infinity.
No. of asymptotes : P Z 2- =
Angle of asymptotes :
af P Zq2 1 180c
=-
+^ h; P Z 2- = , q 0= , 1
0 1
2
18090
cc=
+=
^ h; q 0=
2 1
2180
270c
c= +
=^ h
; q 1=
Centroid (intercept point of asymptotes on real axis)
As Re ReSumof Sumof
P Z
P Z=
-
-6 6@ @
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GATE EE vol-1
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3 1
0 0 2 121
=-
+ - - -=-
^ ^h h .05=-
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 s
K s1
9
12+ +
+
^^
hh 0=
or s s K s K 93 2+ + + 0= ...(1)
For all the roots to be equal and real, we require s P 3+^ h 3 3s Ps P s P 03 2 2 3= + + + = ...(2)On comparing equation (1) and (2), we get
P3 9=
or P 3=
and K P3=
3 3= ^ h 27=
CS 1.46 Correct option is (D).Open loop poles are : s 0= and s 1= and open loop zero is s 1=-
The characteristic equation is
s s
K s1
1
1+
-
+
^^
hh 0=
or K s
s s
11
=-+
-^ h
The break points are given by solution ofds
dK 0=
Hence,ds
dK dsd
s
s s
11
0=+
- -=
^ h; Eor s s s s 1 2 1 1+ - - -^ ^ ^h h h 0=or s s s s s 2 2 12 2- + - - + 0=
or s s2 12+ - 0=
or s .2
2 2828!=
- .1 1414!=- ...(1)
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Sample
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icalEnginee
ring,
Volum
e-3
The root locus is given below
From eq (1),
Centre of circle ,1 0= -^ hand radius of circle 1.414 2= =
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 tan3
1 1
sw
sw
+ -- -a ak k tan180 2
1c
sw
= ++
- a k ...(1)
or tan tan tan3
1 1
sw
sw
+ -- -a ak k: D tan tan180 21c s w= + +- a k: D
or1
3
3
sw
sw
sw
sw
++
+ -
a ak k
1 02
02
sw
sw
=-
+
++
^ ah k
33
2s s ww
+ +
-
^ h 2sw
=+
or 3 2s- +^ h 32
s s w= + +^ hor 6 92 2s s w+ + +^ h 6 9=- +or 3 2 2s w+ +^ h 3 2= ^ hThis is the equation of circle.
CS 1.48 Correct option is (D).
Given open loop transfer function is
G s H s ^ ^h hs j s j s j s j
K
1 1 3 3=
+ + + - + + + -^ ^ ^ ^h h h hThe root locus start from
s1 j1=- +
s2 j1=- -
s3 j3=- -
s4 j3=- +
Since, there is no zero, all root loci end at infinity.
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olume-3
GATE EE vol-1
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GATE EE vol-3
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GATE EE vol-4
Electrical machines, Power systemsEngineering mathematics, General Aptitude
No. of open loop poles P 4= and No. of open loop zeros Z 0=
So, no. of asymptotes is 4 with angles of
Af 45c= , 135c, 225c, 315c
45! c= , 135! c
Centroid : Point of intersection of asymptotes with real axis is
As Re Re
P Z
P ZS S=
-
-6 6@ @
24 0
1 1 3 3 0=
-
- + - + - + - -=-
^ ^ ^ ^h h h hThe root locus is shown below.
There is no breakaway point.
CS 1.49 Correct option is (D).
The characteristic equation of the system is
G s H s 1+ ^ ^h h 0=or
s s s
K14 5
++ +^ ^h h 0=
or s s s K 9 203 2+ + + 0= ...(1)
Intersection of root loci with jw axis is determine using Rouths array.
Rouths array is shown below
s3 1 20
s2 9 K
s1 K
9180-
s0 K
The critical gain before the closed loop system goes to instability isK 180c=
and the auxiliary equation is
s9 1802+ 0=
or s2 20=-
or s j2 5!=
Hence, root loci intersect with jw-axis at s j2 5!= .
The gain margin for K 18= is given by,
8/9/2019 GATE EE vol-3
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CS 50 Root Locus Technique CS 1
PE 1 Root Locus Technique PE 50
EF 1 Root Locus Technique EF 50
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Volum
e-3 GM (in dB) log K
K20 c10= b l 20 20log dB18180
10= =
The gain margin for K 1800= is given by
GM (in dB) 20 20log dB180018010= =-b l
The break away point is given by solution ofds
dK 0= .
From eq. (1),
K s s s9 203 2=- + +^ hor
dsdK s s3 18 20 02=- + + =^ h
or s .6
18 9 165!=
-
.45275=- and .14725-
Point .s 14725=- lies on root locus. So, break away point is .s 14725=- .The value of K at break away point is
K s s s4 5= + +^ ^h h at .s 14725=- .13128=
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 s s s0=
^ ^h h 180!=Since G s H s
s j3 5=- +^ ^h h
j j
K
3 5 1 3 5 5=
- + + - + +
^ ^h hj j
K
2 5 2 5=
- + +^ ^h hso, G s H s
s j3 5=- +^ ^h h tan tan25 251 1=- - -- -b bl l
tan tan18025
251 1
c=- + -- -
180c=-
Angle criterion satisfy.
Now, using magnitude condition we have
G s H s s j3 5=- +^ ^h h
1=
orj j
K
2 5 2 5- + +^ ^h h 1=or K
4 25 4 25+ + 1=
or K 29=
CS 1.51 Correct option is (D).
The characteristic equation of the system is
s s s s
K s
1 1 4 16
12+ - + +
+
^ ^h hh 0= ...(1)or s s s K s K 3 12 164 3 2+ + + - +^ h 0=Intersection of root loci with jw-axis is determined using rouths array which is
shown below
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CS 1 Root Locus Technique CS 51
PE 51 Root Locus Technique PE 1
EF 51 Root Locus Technique EF 1
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Electrical machines, Power systemsEngineering 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 jw-axis, if s1row is complete zero.
i.e. K K59 8322- + - 0=
or K K59 8322- + 0=
K .259 12 37!
= .357= , 23.3
Hence, root locus cross jw-axis two times and the break points are given by
solution ofds
dK 0= .
We can directly check option using pole-zero plot :
Hence, break away point will lie on real axis from s 0= to 1 and break in point
will lie on real axis for s 1< - .
Hence, s .045= is break away point
and s .226=- is break in point
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