GATE ELECTRICAL ENGINEERING Vol 3 of 4
Second Edition
GATE ELECTRICAL ENGINEERING Vol 3 of 4
RK Kanodia Ashish Murolia
NODIA & COMPANY
GATE Electrical Engineering Vol 3, 2e RK 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 non-linear algebraic equations, single and multi-step methods for differential equations. Transform Theory: Fourier transform,Laplace transform, Z-transform.
ELECTRICAL ENGINEERING Electric Circuits and Fields: Network graph, KCL, KVL, node and mesh analysis, transient response of dc and ac networks; sinusoidal steady-state analysis, resonance, basic filter concepts; ideal current and voltage sources, Thevenin’s, Norton’s and Superposition and Maximum Power Transfer theorems, two-port networks, three phase circuits; Gauss Theorem, electric field and potential due to point, line, plane and spherical charge distributions; Ampere’s and Biot-Savart’s laws; inductance; dielectrics; capacitance.
Signals and Systems: Representation of continuous and discrete-time signals; shifting and scaling operations; linear, time-invariant and causal systems; Fourier series representation of continuous periodic signals; sampling theorem; Fourier, Laplace and Z transforms. Electrical Machines: Single phase transformer – equivalent circuit, phasor diagram, tests, regulation and efficiency; three phase transformers – connections, parallel operation; 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 over-current, differential and distance protection; solid state relays and digital protection; circuit breakers; system stability concepts, swing curves and equal area criterion; HVDC transmission and FACTS concepts. Control Systems: Principles of feedback; transfer function; block diagrams; steadystate errors; Routh and Niquist techniques; Bode plots; root loci; lag, lead and lead-lag compensation; state space model; state transition matrix, controllability and observability. Electrical and Electronic Measurements: Bridges and potentiometers; PMMC, moving iron, dynamometer and induction type instruments; measurement of voltage, current, power, energy and power factor; instrument transformers; digital voltmeters and multimeters; phase, time and frequency measurement; Q-meters; oscilloscopes; potentiometric recorders; error analysis. Analog and Digital Electronics: Characteristics of diodes, BJT, FET; amplifiers – biasing, equivalent circuit and frequency response; oscillators and feedback amplifiers; operational amplifiers – characteristics and applications; simple active filters; VCOs and timers; combinational and sequential logic circuits; multiplexer; Schmitt trigger; multi-vibrators; sample and hold circuits; A/D and D/A converters; 8-bit microprocessor basics, architecture, programming and interfacing. Power Electronics and Drives: Semiconductor power diodes, transistors, thyristors, triacs, GTOs, MOSFETs and IGBTs – static characteristics and principles of operation; triggering circuits; phase control rectifiers; bridge converters – fully controlled and half controlled; principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.
***********
PREFACE This book doesn’t make promise but provides complete satisfaction to the readers. The market scenario is confusing and readers don’t find the optimum quality books. This book provides complete set of problems appeared in competition exams as well as fresh set of problems. The book is categorized into units which are then sub-divided into chapters and the concepts of the problems are addressed in the relevant chapters. The aim of the book is to avoid the unnecessary elaboration and highlights only those concepts and techniques which are absolutely necessary. Again time is a critical factor both from the point of view of preparation duration and time taken for solving each problem in the examination. So the problems solving methods is the books are those which take the least distance to the solution. But however to make a comment that this book is absolute for GATE preparation will be an inappropriate one. The theory for the preparation of the examination should be followed from the standard books. But for a wide collection of problems, for a variety of problems and the efficient way of solving them, what one needs to go needs to go through is there in there in the book. Each unit (e.g. Networks) is subdivided into average seven number of chapters on an average each of which contains 40 problems which are selected so as to avoid unnecessary redundancy and highly needed completeness. I shall appreciate and greatly acknowledge the comments and suggestion from the users of this book.
R. K. Kanodia Ashish Murolia
CONTENTS CS
CONTROL SYSTEM
CS 1
Transfer Function
CS 3
CS 2
Stability
CS 52
CS 3
Time Response
CS 92
CS 4
Root Locus Technique
CS 137
CS 5
Frequency Domain Analysis
CS 180
CS 6
Design of Control Systems
CS 226
CS 7
State Variable Analysis
CS 242
CS 8
Gate Solved Questions
CS 298
SS
SIGNALS & SYSTEMS
SS 1
Continuous Time Signals
SS 3
SS 2
Continuous Time Systems
SS 37
SS 3
Discrete Time Signals
SS 78
SS 4
Discrete Time Systems
SS 107
SS 5
The Laplace Transform
SS 147
SS 6
The Z-transform
SS 178
SS 7
The Continuous-Time Fourier Transform
SS 222
SS 8
The Continuous-Time Fourier Series
SS 263
SS 9
Sampling and Signal Reconstruction
SS 302
SS 10
Gate Solved Questions
SS 323 ***********
CS 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)
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
10 PECS EF 11 CS 1.3
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 10 1 EF 10
The valid root locus diagram is
i. n o c . a i d o n . w w w CS 1.4
A open-loop pole-zero plot is shown below
The general shape of the root locus is
CS 1.5
in . o c . a i d o n . w w w
A open-loop pole-zero plot is shown below
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GATE EE vol-3 Control systems, Signals & systems
GATE EE vol-4 CS PE 11 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 11 EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
in . o
c . ia
d o
n . w w
w CS 1.6
A open-loop pole-zero plot is shown below
The general shape of the root locus is
i. n
o .c
a i d
o n
. w w
w
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Sample Chapter of GATE Electrical Engineering, Volume-3
The general shape of the root locus is
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
12 PECS EF 11 CS 1.7
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 12 1 EF 12
A open-loop pole-zero plot is shown below
i. n o c . a i d o n . w w w
The general shape of the root locus is
CS 1.8
The forward-path open-loop transfer function of a ufb system is K (s + 2) (s + 6) G (s) = s2 + 8s + 25 The root locus for this system is
in . o c . a i d o n . w w w CS 1.9
The forward-path open-loop transfer function of as ufb system is K (s2 + 4) G (s) = (s2 + 1) For this system root locus is
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GATE EE vol-3 Control systems, Signals & systems
GATE EE vol-4 CS PE 13 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 13 EF 13
Root Locus Technique Root Locus Technique Root Locus Technique
c . ia
d o
n . w w
CS 1.10
w
The forward-path open-loop transfer function of a ufb system is K (s2 + 1) G (s) = s2 The root locus of this system is
i. n
o .c
a i d
o n
. w w
Common Data For Q. 11 and 12:
A open-loop pole-zero plot is shown below
w
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Sample Chapter of GATE Electrical Engineering, Volume-3
in . o
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
14 PECS EF 11 CS 1.11
CS 1.12
Root Locus Technique Root Locus Technique Root Locus Technique
The transfer function of this system is K (s2 + 2s + 2) (A) (s + 3) (s + 2) K (s2 - 2s + 2) (C) (s + 3) (s + 2)
PE CS 14 1 EF 14
K (s + 3) (s + 2) (s3 + 2s + 2) K (s + 3) (s + 2) (D) (s2 - 2s + 2) (B)
i. n o c . a i d o n . w w w
The break point is (A) breakaway at s =- 1.29 (B) breakin at s =- 2.43 (C) breakaway at s =- 2.43 (D) breakin at s =- 1.29
Common Data For Q. 13 and 14:
A root locus of ufb system is shown below
CS 1.13
The breakaway point is ____
CS 1.14
At breakaway point the value of gain K is (A) 24 (B) 7.4 # 10-3 (C) 8.6 # 10-3 (D) 29
CS 1.15
The forward-path transfer function of a ufb system is K (s + 2) G (s) = (s + 3) (s2 + 2s + 2) The angle of departure from the complex poles is ____
CS 1.16
Consider the feedback system shown below
in . o c . a i d o n . w w w Buy Online: www.nodia.co.in
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GATE EE vol-4 CS PE 15 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 15 EF 15
Root Locus Technique Root Locus Technique Root Locus Technique
in . o
c . ia
d o
n . w w
Common Data For Q. 17 and 18: A root locus for a ufb system is shown below
w CS 1.17
The root locus crosses the imaginary axis at (A) ! j 3.162 (B) ! j 2.486 (C) ! j 4.564 (D) None of the above
CS 1.18
The value of gain for which the closed-loop transfer function will have a pole on the real axis at - 5 , will be ____
CS 1.19
The open-loop transfer function a system is K (s + 8) G (s) H (s) = s (s + 4) (s + 12) (s + 20)
i. n
o .c
a i d
o n
A closed loop pole will be located at s =- 10 , when the value of K is ___ CS 1.20
. w w
For a ufb system forward-path transfer function is K (s + 6) G (s) = (s + 3) (s + 5)
w
The breakaway point and break-in points are located respectively as (A) 3, 4.27 (B) 7.73, 4.27 (C) 4.27, 3 (D) 4.27, 7.73
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Sample Chapter of GATE Electrical Engineering, Volume-3
For this system root locus is
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
16 PECS EF 11 CS 1.21
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 16 1 EF 16
The open loop transfer function of a system is given by K G (s) H (s) = s (s + 1) (s + 2) The root locus plot of above system is
i. n o c . a i d o n . w w w
CS 1.22
CS 1.23
A ufb system has forward-path transfer function G (s) = K2 s The root locus plot is
in . o c . a i d o n . w w w
For the ufb system, shown below consider two point s1 =- 2 + j 3 and s2 =- 2 + j 1 2
Which of the above point lie on root locus ?
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GATE EE vol-4 CS PE 17 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 17 EF 17
Root Locus Technique Root Locus Technique Root Locus Technique
CS 1.24
(B) s1 but not s2 (D) neither s1 nor s2
A ufb system has open-loop transfer function K (s + a) G (s) = 2 ,b > a > 0 s (s + b) The valid root-loci for this system is
in . o
c . ia
d o
n . w w
w CS 1.25
The characteristic equation of a feedback control system is given by (s2 + 4s + 4) (s2 + 11s + 30) + Ks2 + 4K = 0 where K > 0 . 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
i. n
CS 1.26
o .c
a i d
The root locus of the system having the loop transfer function K has G (s) H (s) = s (s + 4) (s2 + 4s + 5) (A) 3 breakaway point (B) 3 breakin point (C) 2 breakin and 1 breakaway point (D) 2 breakaway and 1 break-in point
o n
CS 1.27
. w w
w
Consider the ufb system shown below
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Sample Chapter of GATE Electrical Engineering, Volume-3
(A) Both s1 and s2 (C) s2 but not s1
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
18 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 18 1 EF 18
The root-loci, as a is varied, will be
i. n o c . a i d o n . w w w
Common Data For Q. 27 and 28 :
The forward-path transfer function of a ufb system is K (s + a) (s + 3) G (s) = s (s2 - 1) CS 1.28
CS 1.29
The root-loci for K > 0 with a = 5 is
in . o c . a i d o n . w w w
The root-loci for a > 0 with K = 10 is
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CSPE 1 19 EF 19
Root Locus Technique Root Locus Technique Root Locus Technique
CS 1.30
c . ia
K (s + 6) For the system G (s) H (s) = , consider the following characteristic (s + 2) (s + 4) of the root locus : 1. It has one asymptotes 2. It has intersection with jw-axis 3. It has two real axis intersections. 4. It has two zeros at infinity. The root locus have characteristics (A) 1 and 2 (B) 1 and 3 (C) 3 and 4 (D) 2 and 4
d o
n . w w
w
CS 1.31
The characteristic equation of a closed-loop system is s (s + 1) (s + 2) + K = 0 . The centroid of the asymptotes in root-locus will be ____
CS 1.32
The forward path transfer function of a ufb system is K (s + 3) G (s) = s (s + 1) (s + 2) (s + 4) The angles of asymptotes are (B) 0, 2p , 4p (A) 0, p , p 2 3 3 (C) p , p, 5p (D) None of the above 3 3
i. n
CS 1.33
o .c
a i d
Match List-I with List-II in respect of the open loop transfer function K (s + 10) (s2 + 20s + 500) G (s) H (s) = s (s + 20) (s + 50) (s2 + 4s + 5) P. Q. R. S. (A) (B) (C) (D)
o n
List I (Types of Loci) Separate Loci Loci on the real axis Asymptotes Break away points P Q R 4 3 1 4 3 2 3 4 1 3 4 1
w
. w w
List II (Numbers) 1. One 2. Two 3. Three 4. Five
S 1 1 1 2
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Sample Chapter of GATE Electrical Engineering, Volume-3
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GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
20 PECS EF 11 CS 1.34
Root Locus Technique Root Locus Technique Root Locus Technique
The root-locus of a ufb system is shown below
i. n o c . a i d o n . w w w
The open loop transfer function is K (A) s (s + 1) (s + 3) K (s + 1) (C) s (s + 3) CS 1.35
CS 1.36
CS 1.37
PE CS 20 1 EF 20
(B)
K (s + 3) s (s + 1)
(D)
Ks (s + 1) (s + 3)
The characteristic equation of a linear control system is s2 + 5Ks + 9 = 0 . The root loci of the system is
in . o c . a i d o n . w w w
A unity feedback control system has an open-loop transfer function K G (s) = 2 s (s + 7s + 12) The gain K for which s =- 1 + j1 will lie on the root locus of this system is ___ K (1 - s) An unity feedback system is given as G (s) = . Which is the correct root s (s + 3) locus diagram ?
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GATE EE vol-3 Control systems, Signals & systems
GATE EE vol-4 CS PE 21 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 21 EF 21
in . o
c . ia
The open loop transfer function G (s) of a ufb system is given as K ^s + 23 h G (s) = 2 s (s + 2) 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 K; K ! 0 (D) three real roots are found on the right half of the s -plane
d o
n . w w
w
CS 1.39
The characteristic equation of a closed-loop system is s (s + 1) (s + 3) + K (s + 2) = 0, k > 0 .Which of the following statements is true ? (A) Its root are always real (B) It cannot have a breakaway point in the range - 1 < Re [s] < 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
i. n
o .c
o n
. w w
(A) K3 s
K (C) 2 s (s + 1)
a i d
K s2 (s + 1) K (D) 2 s (s - 1) (B)
w
Common Data For Q. 41 to 43 : The open loop transfer function of a unity feedback system is given by 2 (s + a) ; a>0 G (s) = s (s + 2) (s + 10)
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Sample Chapter of GATE Electrical Engineering, Volume-3
CS 1.38
Root Locus Technique Root Locus Technique Root Locus Technique
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
22 PECS EF 11 CS 1.41
CS 1.42
CS 1.43
CS 1.44
Root Locus Technique Root Locus Technique Root Locus Technique
Angles of asymptotes are (A) 60c, 120c, 300c (C) 90c, 270c, 360c
PE CS 22 1 EF 22
(B) 60c, 180c, 300c (D) 90c, 180c, 270c
i. n o c . a i d o n . w w w
Intercepts of asymptotes at the real axis is (A) - 6
(B) - 10 3
(C) - 4
(D) - 8
Break away points are (A) - 1.056 , - 3.471 (C) - 1.056, - 6.943
(B) - 2.112, - 6.943 (D) 1.056, - 6.943
For the characteristic equation s3 + 2s2 + Ks + K = 0 , the root locus of the system as K varies from zero to infinity is
in . o c . a i d o n . w w w
CS 1.45
The open loop transfer function of a ufb system is K ^s + 1h G ^s h H ^s h = 2 s ^s + 9h In the root locus of the system as parameter K is varied from 0 to 3, the gain K when all three roots are real and equal is ____
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CSPE 1 23 EF 23
A ufb system has an open loop transfer function K ^s + 1h G ^s h H ^s h = s ^s - 1h Root locus for the system is a circle. Centre and radius of the circle are respectively (A) ^0, 0h, 2
in . o
(B) ^0, 0h, 2 (C) ^- 1, 0h, 2 (D) ^- 1, 0h,
CS 1.47
c . ia
2
d o
The open loop transfer function of a system is K ^s + 3h G ^s h H ^s h = s ^s + 2h The root locus of the system is a circle. The equation of circle is (A) ^s + 4h2 + w2 = 4 (B) ^s - 3h2 + w2 = 3
n . w w
w
(C) ^s + 3h2 + w2 = ^ 3 h (D) ^s - 4h2 + w2 = ^2 h2
2
CS 1.48
Consider the open loop transfer function of a system shown below K G ^s h H ^s h = 2 ^s + 2s + 2h^s2 + 6s + 10h 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. K G ^s h H ^s h = s ^s + 4h^s + 5h Consider the following statements for the system. 1. Root locus plot cross jw-axis at s = ! j2 5 2. Gain margin for K = 18 is 20 dB. 3. Gain margin for K = 1800 is - 20 dB 4. Gain K at breakaway point is 13.128 Which of the following is correct ? (A) 1 and 2 (B) 1, 2 and 3 (C) 2, 3 and 4 (D) All
i. n
o .c
a i d
o n
. w w
w
CS 1.50
The open loop transfer function of a system is K G ^s h H ^s h = ^s + 1h^s + 5h What is the value of K , so that the point s =- 3 + j5 lies on the root locus? Buy Online: www.nodia.co.in
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Sample Chapter of GATE Electrical Engineering, Volume-3
CS 1.46
Root Locus Technique Root Locus Technique Root Locus Technique
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
24 PECS EF 11 CS 1.51
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 24 1 EF 24
The open loop transfer function of a control system is K ^s + 1h G ^s h H ^s h = s ^s - 1h^s2 + 4s + 16h Consider the following statements for the system 1. Root locus of the system cross jw-axis for K = 35.7 2. Root locus of the system cross jw-axis for K = 23.3 3. Break away point is s = 0.45 4. Break in point is s =- 2.26 Which of the following statement is correct ? (A) 1, 3 and 4 (B) 2, 3 and 4 (C) 3 and 4 (D) all
i. n o c . a i d o n . w w w ****************
in . o c . a i d o n . w w w Buy Online: www.nodia.co.in
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GATE EE vol-4 CS PE 25 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 25 EF 25
Root Locus Technique Root Locus Technique Root Locus Technique
CS 1.1
in . o
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.
c . ia
d o
n . w w
w
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
i. n
o .c
The angle of departure of the root locus branch from a complex pole is given by
a i d
fD = !6180c + f@ where f is net angle contribution at this pole due to all other poles and zeros.
o n
. w w
w
f f f Departure angle fD
= fZ - fP = 6- 90c + 90c@ - 690c + 90c@ =- 180c = !6180c - 180c@
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Sample Chapter of GATE Electrical Engineering, Volume-3
SOLUTION
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
26 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 26 1 EF 26
fD = ! 0c 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
i. n o c . a i d o n . w w w
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 : ^2q + 1h 180c ; q = 0, 1 fa = P-Z ^0 + 1h 180c fa = = 90c; q = 0 2
in . o c . a i d o n . w w w
^2 + 1h 180c 3 # 180c = = 270c; q = 1 2 P-Z Hence, root locus plot will be as shown below. and
fa =
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GATE EE vol-4 CS PE 27 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 27 EF 27
Correct option is (A). An open loop pole-zero plot is given as :
in . o
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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
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Correct option is (C). Forward Path open loop transfer function of given ufb system is K ^s + 2h^s + 6h G ^s h = 2 s + 8s + 25 Characteristic equation is s2 + 8s + 25 = 0 s = - 8 ! 64 - 100 =- 4 ! j3 2 s =- 4 + j3 ; s =- 4 - j3 Thus pole are s =- 4 + j3 ; s =- 4 - j3 and zeros are s =- 2 ; s =- 6 Plot of Pole-zero is shown below
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Root locus will start from poles and ends with zeros. Only option (C) satisfy above condition. CS 1.9
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Correct option is (B). Forward path open loop transfer function of given ufb system is, K ^s2 + 4h G ^s h = s2 + 1 Here, poles s2 + 1 = 0 s = ! j1 2 and zeros s +4 = 0 s = ! j2
w
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Sample Chapter of GATE Electrical Engineering, Volume-3
CS 1.7
Root Locus Technique Root Locus Technique Root Locus Technique
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
28 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 28 1 EF 28
Pole-zero plot is
i. n o c . a i d o n . w w w
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 K ^s2 + 1h G ^s h = s2 2 Zeros are : s +1 = 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
in . o c . a i d o n . w w w
Correct option is (C). Given open loop pole zero plot is :
From above plot zeros are s = 1 + j1 and s = 1 - j1 and poles are s =- 2 and s =- 3 Transfer function of the system is K "s - ^1 + j1h, "s - ^1 - j1h, G ^s h = "s - ^- 2h, "s - ^- 3h,
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GATE EE vol-4 CS PE 29 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 29 EF 29
Root Locus Technique Root Locus Technique Root Locus Technique
or CS 1.12
in . o
c . ia
Correct option is (C). Method i :
d o
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 ^- 2h and ^- 3h 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 s =- 2.43 which lies between - 2 and - 3 . On root locus it may be seen easily that s =- 2.43 lie on root locus and locus start from pole ^- 2h and ^- 3h. Thus at s =- 2.43 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
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Method II
The point, at which multiple roots are present, are known as break point. These are obtained from the dK = 0 ds Here, characteristic equation is 1 + G ^s h H ^s h = 0 K ^s2 - 2s + 2h =0 1+ ^s + 2h^s + 3h -^s + 2h^s + 3h K = ^s2 - 2s + 2h -^s2 + 5s + 6h ...(1) = 2 s - 2s + 2 Now, differentiating eq (1) w.r.t s and equating zero we have 2 2 dK = -^s - 2s + 2h^2s + 5h + ^s + 5s + 6h^2s - 2h = 0 2 ds ^s2 - 2s + 2h 7s2 + 8s - 22 = 0 which gives s =+ 1.29 and s =- 2.43 out of which s =- 2.43 is break point.
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CS 1.13
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Correct answer is - 1.45 . Given root locus of ufb system is shown below
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Sample Chapter of GATE Electrical Engineering, Volume-3
K "^s - 1h - j1, "^s - 1h + j1, ^s + 2h^s + 3h K "^s - 1h2 - ^ j1h2, = ^s + 2h^s + 3h K ^s2 - 2s + 2h G ^s h = ^s + 2h^s + 3h =
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
30 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 30 1 EF 30
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 K ^s - 3h^s - 5h G ^s h = ^s + 1h^s + 2h Characteristic equation
i. n o c . a i d o n . w w w
1 + G ^s h H ^s h = 0 K ^s - 3h^s - 5h 1+ =0 ^s + 1h^s + 2h -^s2 + 3s + 2h K = 2 ^s - 8s + 15h Differentiating eq. (1) wrt to s and equating to zero we have 2 2 dK = -^s - 8s + 15h^2s + 3h + ^s + 3s + 2h^2s - 8h = 0 2 ds ^s2 - 8s + 15h 11s2 - 26s - 61 = 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 -8^- 1.45h2 + 3 ^- 1.45h + 2B K = 6^- 1.45h2 - 8 ^- 1.45h + 15@ ^- 0.2475h == 8.62 # 10-3 28.7025
CS 1.15
Correct answer is 108.4 . Forward path transfer function of given ufb system is K ^s + 2h G ^s h = ^s + 3h^s2 + 2s + 2h Here zero is s =- 2 and poles are s =- 3 and s =- 1 ! j1 Pole-zero Plot is shown below
...(1)
in . o c . a i d o n . w w w
Angle of departure at pole P1 is given by : fD = !6180c + f@ Buy Online: www.nodia.co.in
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CSPE 1 31 EF 31
Root Locus Technique Root Locus Technique Root Locus Technique
in . o
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fD = !6180c + f@
= !:180c + tan 1 - 90c - tan-1 1 D 2 -1
d o
= !6180 + 45 - 90 - 26.56@
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fD = ! 108.4c 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
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We redraw the block diagram after moving take off point as shown below
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Forward path transfer function is - K ^s + 2h G ^s h = ^s + 1h^s + 3h 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).
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CS 1.17
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Correct option is (A). Given root locus is shown below
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Sample Chapter of GATE Electrical Engineering, Volume-3
where f is net angle contribution at pole P1 due to all other poles and zeros. f = fZ - fP = fZ1 - 6fP2 + fP3@ where, fZ1 = tan-1 1; fP2 = 90c; fP3 = tan-1 1 2 -1 -1 1 f = tan 1 - :90c + tan D 2 Departure angle :
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
32 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 32 1 EF 32
It does not have any zero and have poles at s =- 4 ands =- 1 ! j1. Thus open loop transfer function, K G ^s h = + s 4 s ^ h^ 2 + 2s + 2h K = 3 2 s + 6s + 10s + 8 Closed loop transfer function, K 3 2 s 6 s + + 10s + 8 T ^s h = K 1+ 3 s + 6s2 + 10s + 8 K = 3 s + 6s2 + 10s + 8 + K Characteristics equation :
i. n o c . a i d o n . w w w
s3 + 6s2 + 10s + 8 + K = 0 Routh array is shown below s3 s
2
1
10
6
8+K
s1
52 - K 6
s0
8+K
When root locus cut will cut imaginary axis if element in s1 is zero. 52 - K = 0 & K = 52 Thus 6 and then from auxiliary equation we have 6s2 + ^8 + 52h = 0
in . o c . a i d o n . w w w s2 =- 60 =- 10 6 s = ! j3.162
CS 1.18
Correct answer is 17. Using Pole - zero plot :
Gain K at any ^s = s 0h point on root locus is given by : Product of phasors drawn from OLP at that point K s=s = Product of phasors drawn from OLZ at that point 0
Here no any zero is present. Hence, K = Product of phasors drawn from OLP at that point
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CSPE 1 33 EF 33
s =- 5
= ^PP3h # ^PP1h # ^PP2h
= 1 # 42 + 1 # = 42 + 1 = 17 K = 17
CS 1.19
42 + 1
in . o
Correct answer is 600. Open loop transfer function of given system is K ^s + 8h G ^s h H ^s h = s ^s + 4h^s + 12h^s + 20h Here, Zeros : Z =- 8 Poles : P1 = 0 ; P2 =- 4 ; P3 =- 12 ; P4 =- 20 Value of K at s =- 10 is : p ^Phasors drawn from OLP at s =- 10h K s =- 10 = p ^Phasors drawn from OLZ at s =- 10h Pole-zero plot is shown below.
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K
CS 1.20
^PP1h # ^PP2h # ^PP3h # ^PP4h s =- 10 ^PZ h = 10 # 6 # 2 # 10 = 600 2 K = 600 =
i. n
Correct option is (D). For given system, forward path transfer function, K ^s + 6h G ^s h = + s ^ 3h^s + 5h Characteristic equation for closed loop transfer function,
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1 + G ^s h H ^s h = 0 K ^s + 6h or 1+ =0 ^s + 3h^s + 5h -^s2 + 8s + 15h or K = ^s + 6h differentiating w.r.t to s and equating to zero, we have 2 dK = -^s + 6h^2s + 8h + ^s + 8s + 15h = 0 ds ^s + 6h2 or s2 + 12s + 33 = 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.
o n
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Sample Chapter of GATE Electrical Engineering, Volume-3
K
Root Locus Technique Root Locus Technique Root Locus Technique
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
34 PECS EF 11
CS 1.21
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 34 1 EF 34
i. n o c . a i d o n . w w w
Correct option is (B). Open loop transfer function : G ^s h H ^s h =
K s ^s + 1h^s + 2h 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 , 1 + G ^s h H ^s h = 0 K =0 1+ s ^s + 1h^s + 2h s3 + 3s2 + 2s + K = 0 Routh array is shown below. s3
1
2
2
3
K
1
s
6-K 3
s0
K
s
At K = 6 , s1 row is zero, thus using auxiliary equation we get, 3s2 + 6 = 0
in . o c . a i d o n . w w w
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 = K2 s Angle of departure or angle of asymptote for multiple poles is, ^2q + 1h 180c ; fa = r where r = no. of multiple poles q = 0 , 1, 2, .... ^r - 1h Here, r = 2 ; (2 multiple poles at origin) q = 0, 1 ^0 + 1h 180c fa = = 90c for q = 0 2 and
fa =
^2 + 1h 180c = 270c for q = 1 2
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CSPE 1 35 EF 35
Root Locus Technique Root Locus Technique Root Locus Technique
in . o
CS 1.23
c . ia
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Correct option is (C). For given ufb system, open loop transfer function is, K ^s + 3h^s + 4h G ^s h = ^s + 1h^s + 2h Given two point, s1 =- 2 + j3 ; s2 =- 2 + j 1 2 If any point lie on root locus, it satisfies the characteristic equation. Thus we have
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or and Point: s1 =- 2 + j3
q ^s h = 1 + G ^s h H ^s h = 0 (Magnitude) G ^s h H ^s h = 1 (Phase) G ^s h H ^s h = ! 180c
Now we use the phse condition G ^s h H ^s h s = s = f1 , we have f1 = q1 + q2 + q3 + q4 = tan-1 3 + tan-1 3 - 90c - tan-1 b 3 l 2 1 -1 = tan-1 3 + tan-1 3 - 90c - ^180c - tan-1 3h 2
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= 56.30 + 71.56 - 270c + 71.56 or f1 =- 70.56 ! ! 180c Hence, point s1 not lie on root locus. Point : s2 =- 2 + j 1 :2
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Sample Chapter of GATE Electrical Engineering, Volume-3
Hence, root locus plot will be as shown below.
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
36 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 36 1 EF 36
G ^s h H ^s h s = s = f2 we have, f2 = q1 + q2 - q3 - q4 = tan-1 1 + tan-1 1 - 90c - c180c - tan-1 1 m 2 2 2 2 = tan-1 1 + tan-1 1 - 90c - 180c + tan-1 1 2 2 2 2 f2 =- 180c Here phase condition is satisfied. Hence, point s2 lies on root locus.
Using phase condition
2
CS 1.24
i. n o c . a i d o n . w w w
Correct option is (A). For given ufb system, open loop transfer function is K ^s + ah ; b>a>0 G ^s h = 2 s ^s + b h Here, zero : Z =- a Poles : s = 0 , 0; s =- b Departure angle at double poles on origin is : ^2q + 1h 180c ; r = 2, q = 0, 1 f = r
f = 90c and 270c To get intersection with imaginary axis we use Ruth array as shown below. characteristic equation is s3 + bs2 + Ks + Ka = 0 Routh Array is shown below s3
1
K
2
b
Ka
1
s
b-a b
s0
Ka
s
in . o c . a i d o n . w w w
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
^s2 + 4s + 4h^s2 + 11s + 30h + Ks2 + 4K = 0 K ^s2 + 4h or =0 1+ 2 ^s + 4s + 4h^s2 + 11s + 30h Characteristic equation is, 1 + G ^s h H ^s h = 0 Comparing, eq (1) and (2) we get open loop transfer function as K ^s2 + 4h G ^s h H ^s h = 2 ^s + 4s + 4h^s2 + 11s + 30h K ^s2 + 4h = (s + 2) (s + 2) (s + 5) (s + 6)
...(1) ...(2)
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CSPE 1 37 EF 37
Root Locus Technique Root Locus Technique Root Locus Technique
s =- 2, - 2 and s =- 6 , - 5 The point at which asymptotes meet (centroid) is given by: Sum of Re 6P @ - Sum of Re 6Z @ sA = ^P - Z h 2 Here, open loop zeros : s + 4 = 0 s = ! j2 ^- 2 - 2 - 5 - 6h - 0 sA = 4-2 = - 15 =- 7.5 2 It intersects on real axis. So point is: ^- 7.5, 0h
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CS 1.26
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Correct option is (D). Open loop transfer function for given system is K G ^s h H ^s h = s ^s + 4h^s2 + 4s + 5h Characteristic equation, K =0 1+ s ^s + 4h^s2 + 4s + 5h K =- s ^s + 4h^s2 + 4s + 5h Differentiating wrt s and equating to zero we have dK =- 2s + 4 s2 + 4s + 5 + s2 + 4s 2s + 4 = 0 h^ h ^ h^ hB 8^ ds
w
...(1)
or -^2s + 4h^s2 + 4s + 5 + s2 + 4s h = 0 or ...(2) -^s + 2h^2s2 + 8s + 5h = 0 or s =- 2 and s =- 0.775 , - 3.225 Now, we check for maxima and minima value of gain K at above point. If gain is maximum, then that point will be break away point. If gain is minimum, then that point will be break in point. Again differentiating equation (2) with respect to s , we get, d 2 K =- 2s2 + 8s + 5 + s + 2 4s + 8 h ^ h^ h@ 6^ ds2 2 =-^6s + 24s + 21h For s =- 0.775 and s =- 3.225 , d 2 K =- 6.0 < 0 ; s =- 0.775 , - 3.225 are maxima points. ds2 Hence, s =- 0.775 and s =- 3.225 are break away points. For s =- 2 : d 2 K =+ 3 > 0 ; s =- 2 is minima points. ds2 Thus s =- 2 is break in point. Hence, there is two break away points ^s =- .0775, - 3.225h and one break in point.
i. n
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CS 1.27
Correct option is (B). For a ufb system, forward transfer function is :
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Sample Chapter of GATE Electrical Engineering, Volume-3
Open loop poles are
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
38 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
G ^s h = Method I :
PE CS 38 1 EF 38
1 s ^s + ah
i. n o c . a i d o n . w w w
Characteristic equation :
1 + G ^s h H ^s h = 0 1 =0 1+ s ^s + ah s 2 + as + 1 = 0 =0 1 + 2as ^s + 1h Open loop transfer function, as a is varied is G ^s h H ^s h = 2as s +1 Here, zero : s = 0 and poles : s2 + 1 = 0 & s = ! j Root locus will be :
Method II :
Closed loop transfer function : G ^s h T ^s h = = 2 1 1 + G ^ s h H ^ s h s + as + 1 1 2 G ^s h + s 1 = a 1 + G ^s h H ^s h 1+ 2 s s +1 G ^s h H ^s h = 2as s +1 CS 1.28
in . o c . a i d o n . w w w
Correct option is (A). Forward path transfer function of given ufb system is K ^s + ah^s + 3h ; a = 5 and K > 0 G ^s h = s ^s2 - 1h K ^s + 5h^s + 3h or G ^s h = s ^s2 - 1h 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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GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement
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GATE EE vol-4 CS PE 39 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 39 EF 39
Root Locus Technique Root Locus Technique Root Locus Technique
^2q + 1h 180c ; q = 0 , 1, 2, .... ^P - Z - 1h ^P - Z h ^0 + 1h 180c = = 180c 1 Only option (A) satisfies these conditions. fa =
CS 1.29
in . o
c . ia
Correct option is (C). Forward path transfer function : K ^s + ah^s + 3h ; K = 10 and a > 0 G ^s h = s ^s2 - 1h Characteristic equation,
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1 + G ^s h H ^s h = 0 10 ^s + ah^s + 3h 1+ =0 s ^s2 - 1h s3 - s + 10 6s2 + ^a + 3h s + 3a@ = 0 s ^s2 + 10s + 29h + a10 ^s + 3h = 0 a10 ^s + 3h 1+ 2 =0 s ^s + 10s + 29h Open loop gain as a is varied, a10 ^s + 3h G ^s h H ^s h = 2 s ^s + 10s + 29h Here, no. of asymptotes = P - Z = 3-1 = 2 So, two root branches will go to infinite along asymptotes as a " 3. Now using equation (1) we have
w
s3 + 10s2 + ^29 + 10ah s + 30a = 0 Routh array is shown below s3
1
(29 + 10a)
s2
10
30a
s1
(29 + 7a)
s0
30a
...(1) ...(2)
...(3)
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For a > 0 , s1 row can not be zero. Hence, root locus does not intersect jw axis for a > 0 . CS 1.30
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Correct option is (B). Open loop transfer function given K ^s + 6h G ^s h H ^s h = ^s + 2h^s + 4h Poles : P = 2 ; s =- 2 and s =- 4 Zeros : Z = 1; s =- 6 No. of asymptotes = P - Z
w
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Sample Chapter of GATE Electrical Engineering, Volume-3
Angle of asymptotes,
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
40 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
= 2-1 = 1
PE CS 40 1 EF 40
Thus (1) is correct.
Now characteristic,
^s + 2h^s + 4h + K ^s + 6h = 0 or s2 + ^6 + K h s + 8 + 6K = 0 Making Routh array we have s2 s1 s0
i. n o c . a i d o n . w w w 1
8 + 6K
6+K
8 + 6K
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
in . o c . a i d o n . w w w
s ^s + 1h^s + 2h + K = 0 K or =0 1+ s ^s + 1h^s + 2h 1 + G ^s h H ^s h = 0 Comparing we get open loop transfer function, K G ^s h H ^s h = s ^s + 1h^s + 2h Poles : s = 0 , s =- 1, s =- 2 Zeros : No zero Sum of Re 6P @ - Sum of Re 6Z @ Centroid sA = P-Z ^0 - 1 - 2h - ^0 h = =- 3 =- 1 3-0 3
CS 1.32
Correct option is (C). Forward path transfer function of given ufb system is, K ^s + 3h G ^s h = s ^s + 1h^s + 2h^s + 4h Buy Online: www.nodia.co.in
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GATE EE vol-4 CS PE 41 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 41 EF 41
Root Locus Technique Root Locus Technique Root Locus Technique
c . ia
^2 + 1h 180c = 180c; q = 1 ^4 - 1h ^4 + 1h 180c = 300c; q = 2 = ^4 - 1h Z60c = p ; q=0 3 ] fa = [180c= p; q = 1 ]300c= 5p ; q = 2 3 \ =
d o
n . w w
Thus
CS 1.33
in . o
w
Correct option is (B). Open loop transfer function, K ^s + 10h^s2 + 20s + 500h G ^s h H ^s h = s ^s + 20h^s + 50h^s2 + 4s + 5h 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 , s =- 10 ! j20 Open loop zeros : s = 0 , s =- 20 , s =- 50 , s =- 2 ! j1
i. n
o .c
a i d
o n
. w w
^0 - 20 - 50 - 2 - 2h - ^- 10 - 10 - 10h sA = 5-3 =- 22 Angle of asymptotes, ^2q + 1h 180c ; P - Z = 2, q = 0, 1 fa = P-Z ^0 + 1h 180c = = 90c; q = 0 2 Centroid
w
=
^2 + 1h 180c = 270c; q = 1 2
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Sample Chapter of GATE Electrical Engineering, Volume-3
Here zeros : Z = 1; s =- 3 and poles : P = 4 ; s = 0 , s =- 1, s =- 2 , s =- 4 Angle of asymptotes , ^2q + 1h 180c ; q = 0 , 1, 2,..... ^P - Z - 1h fa = ^P - Z h ^0 + 1h 180c 180 = = 60c; q = 0 = 3 ^4 - 1h
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
42 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 42 1 EF 42
The root locus is shown below.
i. n o c . a i d o n . w w w
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 + s =- 1 and 0, and ends at - 3 and 3. Hence poles are - 1, 0 , and zeros are - 3, 3. So transfer function is
CS 1.35
K (s + 3) . s (s + 1)
Correct option is (D). Characteristic equation,
s2 + 5Ks + 9 = 0 5s = 0 1+ K 2 s +9 1 + G ^s h H ^s h = 0 Open loop transfer function is, G ^s h H ^s h = 52 Ks s +9
in . o c . a i d o n . w w w
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,
K ; H ^s h = 1 s ^s + 7s + 12h If point : s =- 1 + j1 lies on root locus, then it satisfies the characteristic equation G_s i =
2
1 + G ^s h H ^s h = 0 =0 1+ 2 K s ^s + 7s + 12h s3 + 7s2 + 12s + K = 0 Substituting s =- 1 + j1 we have
...(1)
^- 1 + j h3 + 7 ^- 1 + j h2 + 12 ^- 1 + j h + K = 0
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GATE EE vol-4 CS PE 43 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 43 EF 43
Root Locus Technique Root Locus Technique Root Locus Technique
CS 1.37
in . o
Correct option is (C). Open loop transfer function of given ufb system is, - K ^s - 1h K ^1 - s h = G ^s h = s ^s + 3h s ^s + 3h 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.
c . ia
d o
CS 1.38
n . w w
Correct option is (A). Open loop transfer function given is K ^s + 23 h G ^s h = 2 s (s + 2) Root locus starts at OLP : s = 0 , s = 0 , s =- 2 No. of asymptotes , #OLP - #OLZ = No. of OLP – No. of OLZ
w
= 3-1 = 2 Angle of asymptotes : ^2q + 1h 180c ; P - Z = 2, q = 0, 1 fa = P-Z ^0 + 1h 180c 1. = 90c for q = 0 fa = 2
^2 + 1h 180c = 270c for q = 1 2 Sum of Re 6P @ - Sum of Re 6Z @ Centroid sA = P-Z ^0 - 2h - b- 23 l = =- 2 3-1 3 Angle of departure at double pole (at origin) ^2q + 1h 180c ; r = 2, q = 0, 1 fD = r
i. n
fa =
2.
o .c
a i d
o n
. w w
= 90c, 270c Fro above analysis Root locus plot will be as given below.
w
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Sample Chapter of GATE Electrical Engineering, Volume-3
- 10 + K = 0 K =+ 10
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
44 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 44 1 EF 44
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
i. n o c . a i d o n . w w w
Correct option is (C). Characteristic equation of a closed loop system is : s ^s + 1h^s + 3h + K ^s + 2h = 0 ; K > 0 K ^s + 2h 1+ =0 s ^s + 1h^s + 3h Open loop transfer function is, K ^s + 2h G ^s h H ^s h = s ^s + 1h^s + 3h 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 ^2q + 1h 180c Angle of asymptotes : ; P - Z = 2, q = 0, 1 fa = P-Z fa = 90c and 270c Sum of Re 6P @ - Sum of Re 6Z @ Centroid sA = P-Z ^0 - 1 - 3h - ^- 2h = =- 2 =- 1 3-1 2 Break away point lie in the range - 1 < Re 6s @ < 0
in . o c . a i d o n . w w w
Two of its roots tends to infinite along the asymptotes Re 6s @ =- 1. 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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GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement
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GATE EE vol-4 CS PE 45 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 45 EF 45
Root Locus Technique Root Locus Technique Root Locus Technique
c . ia
d o
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 = K3 s Now we verify the above result as follows. Using phase condition we have G ^s h H ^s h s = s = ! 180c
n . w w
w
0
From given plot, for a given point on root locus we have y G ^s h H ^s h s = 1, 3 =- 3 tan-1 a k x ^ h =- 3 tan-1 c
3 1 m
=- 3 # 60c =- 180c CS 1.41
Correct option is (B). Open loop transfer function of given ufb system is 2 ^s + ah G ^s h = s ^s + 2h^s + 10h Characteristic equation, 2 ^s + ah 1+ =0 s ^s + 2h^s + 10h s ^s + 2h^s + 10h + 2s + 2a = 0 s3 + 12s2 + 22s + 2a = 0 2a =0 1+ 3 s + 2s2 + 22s Open loop transfer function as a varied, 2a G ^s h H ^s h = 3 s + 12s2 + 22s No. of poles : P = 3 and no. of zeros : Z = 0 Angle of asymptotes, ^2q + 1h 180c ; P - Z = 3 ; q = 0 , 1, 2 fa = P-Z ^0 + 1h 180c = = 60c for q = 0 3
i. n
o .c
a i d
o n
. w w
w
=
^2 + 1h 180c = 180c for q = 1 3
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Sample Chapter of GATE Electrical Engineering, Volume-3
in . o
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
46 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
=
PE CS 46 1 EF 46
^4 + 1h 180c = 300c for q = 2 3
fa = 60c, 180c, 300c
CS 1.42
i. n o c . a i d o n . w w w
Correct option is (C). Intercept point of asymptotes Sum of Re 6P @ - Sum of Re 6Z @ Centroid sA = P-Z Poles : s = 0 , s =- 2 , s =- 10 Zeros : No any zero. ^0 - 2 - 10h - 0 sA = =- 4 3-0
CS 1.43
Correct option is (C). We have
K =
-^s3 + 12s2 + 22s h 2
2 dK = 0 & -^3s + 24s + 22h = 0 2 ds
or - 3s2 - 24s - 22 = 0 or s =- 1.056 , - 6.943 Thus break point are s =- 1.056 , and - 6.943 . CS 1.44
Correct option is (A). Given characteristics equation is s3 + 2s2 + Ks + K = 0 s3 + 2s2 + K ^s + 1h = 0 K ^s + 1h 1+ 2 =0 s ^s + 2h Open loop transfer function, K ^s + 1h G ^s h H ^s h = 2 s ^s + 2h OLZ : s =- 1; No. of zeros : Z = 1 OLP : s = 0 , s = 0 , s =- 2 ; No. of Poles : P = 3 Root loci starts ^K = 0h 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 : ^2q + 1h 180c ; P - Z = 2, q = 0, 1 fa = P-Z ^0 + 1h 180c = = 90c; q = 0 2
in . o c . a i d o n . w w w
^2 + 1h 180c = 270c; q = 1 2 Centroid (intercept point of asymptotes on real axis) Sum of Re 6P @ - Sum of Re 6Z @ sA = P-Z =
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GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement
GATE EE vol-2 Analog electronics, Digital electronics, Power electronics
GATE EE vol-3 Control systems, Signals & systems
GATE EE vol-4 CS PE 47 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 47 EF 47
Root Locus Technique Root Locus Technique Root Locus Technique
in . o
c . ia
d o
CS 1.45
n . w w
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 K ^s + 1h 1+ 2 =0 s ^s + 9h or ...(1) s3 + 9s2 + Ks + K = 0 For all the roots to be equal and real, we require ...(2) ^s + P h3 = s3 + 3Ps2 + 3P2 s + P3 = 0 On comparing equation (1) and (2), we get
w
3P = 9 P =3 K = P3 = ^3h3 = 27
or and
CS 1.46
i. n
o .c
Correct option is (D). Open loop poles are : s = 0 and s = 1 and open loop zero is s =- 1 The characteristic equation is K ^s + 1h 1+ =0 s ^s - 1h s ^s - 1h or K =s+1 The break points are given by solution of dK = 0 ds - s ^s - 1h d dK Hence, = ; =0 s+1 E ds ds
a i d
o n
or or or
w
. w w
or
^s + 1h^2s - 1h - s ^s - 1h = 0 2s2 - s + 2s - 1 - s2 + s = 0 s2 + 2s - 1 = 0 s = - 2 ! 2.828 =- 1 ! 1.414 2
...(1)
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Sample Chapter of GATE Electrical Engineering, Volume-3
^0 + 0 - 2h - ^- 1h =- 1 =- 0.5 3-1 2 Root locus is as shown below. =
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
48 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 48 1 EF 48
The root locus is given below
i. n o c . a i d o n . w w w
From eq (1), Centre of circle
= ^- 1, 0h
and radius of circle = 1.414 = CS 1.47
2
Correct option is (C). The angle criterion of root locus is
G ^s h H ^s h = 180c or, ^s + 3h - s - ^s + 2h = 180c Substitute s = s + jw , we get ^s + jw + 3h - ^s + jwh - ^s + jw + 2h = 180c tan-1 a w k - tan-1 a w k = 180c + tan-1 a w k s+3 s s+2 or tan :tan-1 a w k - tan-1 a w kD = tan :180c + tan-1 a w kD s+3 s s+2 w -w 0+ w s + 3 s s+2 or = w w 1 +a 1 - ^0 ha w k s + 3 ka s k s+2
or or
...(1)
in . o c . a i d o n . w w w w - 3w 2 = s+2 s ^s + 3h + w - 3 ^s + 2h = s ^s + 3h + w2 ^s2 + 6s + 9h + w2 =- 6 + 9
or ^s + 3h2 + w2 = ^ 3 h This is the equation of circle.
2
CS 1.48
Correct option is (D). Given open loop transfer function is G ^s h H ^s h =
K ^s + 1 + j h^s + 1 - j h^s + 3 + j h^s + 3 - j h The root locus start from s1 =- 1 + j s2 =- 1 - j s 3 =- 3 - j s 4 =- 3 + j Since, there is no zero, all root loci end at infinity. Buy Online: www.nodia.co.in
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GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement
GATE EE vol-2 Analog electronics, Digital electronics, Power electronics
GATE EE vol-3 Control systems, Signals & systems
GATE EE vol-4 CS PE 49 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 49 EF 49
Root Locus Technique Root Locus Technique Root Locus Technique
in . o
c . ia
d o
n . w w
w
There is no breakaway point. CS 1.49
Correct option is (D). The characteristic equation of the system is
1 + G ^s h H ^s h = 0 K or =0 1+ s ^s + 4h^s + 5h or s3 + 9s2 + 20s + K = 0 Intersection of root loci with jw axis is determine using Routh’s array. Routh’s array is shown below
i. n
s3
1
20
s2
9
K
o .c
180 - K 9
1
s
K
a i d
o n
. w w
s0
...(1)
The critical gain before the closed loop system goes to instability is Kc = 180 and the auxiliary equation is or
w
9s2 + 180 = 0 s2 =- 20
or s = ! j2 5 Hence, root loci intersect with jw-axis at s = ! j2 5 . The gain margin for K = 18 is given by,
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Sample Chapter of GATE Electrical Engineering, Volume-3
No. of open loop poles P = 4 and No. of open loop zeros Z = 0 So, no. of asymptotes is 4 with angles of fA = 45c, 135c, 225c, 315c = ! 45c, ! 135c Centroid : Point of intersection of asymptotes with real axis is S Re 6P @ - S Re 6Z @ sA = P-Z ^- 1h + ^- 1h + ^- 3h + ^- 3h - 0 =- 2 = 4-0 The root locus is shown below.
GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia
Sample Chapter of GATE Electrical Engineering, Volume-3
50 PECS EF 11
Root Locus Technique Root Locus Technique Root Locus Technique
PE CS 50 1 EF 50
GM (in dB) = 20 log 10 b Kc l = 20 log 10 180 = 20 dB 18 K The gain margin for K = 1800 is given by GM (in dB) = 20 log 10 b 180 l =- 20 dB 1800 The break away point is given by solution of dK = 0 . ds From eq. (1),
i. n o c . a i d o n . w w w K =-^s3 + 9s2 + 20s h dK =- 3s2 + 18s + 20 = 0 ^ h ds s = - 18 ! 9.165 6
or or
=- 4.5275 and - 1.4725 Point s =- 1.4725 lies on root locus. So, break away point is s =- 1.4725 . The value of K at break away point is K = s ^s + 4h^s + 5h 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 H ^s h
s = s0
= ! 180
Since
G ^s h H ^s h s =- 3 + j5 =
so,
G ^s h H ^s h
s =- 3 + j5
K ^- 3 + j5 + 1h^- 3 + j5 + 5h K = ^- 2 + j5h^2 + j5h =- tan-1 b 5 l - tan-1 b 5 l -2 2 =- 180c + tan-1 5 - tan-1 5 2 2
in . o c . a i d o n . w w w =- 180c
Angle criterion satisfy. Now, using magnitude condition we have G ^s h H ^s h s =- 3 + j5 = 1
or or or CS 1.51
K =1 ^- 2 + j5h^2 + j5h K =1 4 + 25 4 + 25 K = 29
Correct option is (D). The characteristic equation of the system is K ^s + 1h ...(1) =0 1+ s ^s - 1h^s2 + 4s + 16h or s 4 + 3s3 + 12s2 + ^K - 16h s + K = 0 Intersection of root loci with jw-axis is determined using routh’s array which is shown below
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GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement
GATE EE vol-2 Analog electronics, Digital electronics, Power electronics
GATE EE vol-3 Control systems, Signals & systems
GATE EE vol-4 CS PE 51 EF 11 Electrical machines, Power systems Engineering mathematics, General Aptitude
CSPE 1 51 EF 51
Root Locus Technique Root Locus Technique Root Locus Technique
1
12
s3
3
K - 16
s2
52 - K 3
K
s1
- K2 + 59K - 832 52 - K
s0
K
K
in . o
c . ia
d o
The root locus cross the jw-axis, if s1 row is complete zero. i.e. - K2 + 59K - 832 = 0 or K2 - 59K + 832 = 0 K = 59 ! 12.37 = 35.7 , 23.3 2 Hence, root locus cross jw-axis two times and the break points are given by solution of dK = 0 . ds We can directly check option using pole-zero plot :
n . w w
w
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
i. n
o .c
*************
a i d
o n
. w w
w
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Sample Chapter of GATE Electrical Engineering, Volume-3
s4
