www.stydyforfe.com
Study Guide for Fundamentals of Engineering (FE)
Electrical & Computer CBT Exam
Second Edition
Practice over 500 solved problems with detailed
solutions including Alternative-ltem Types
Wasim Asghar
PE, P. Eng, M. Eng
DISCLAIMER
This book is developed to assist reader in FE Electrical and Computer exam preparation. It has gone through multiple
review cycles to produce a high-quality text. However, there are no representations or warranties, express or
implied, about the completeness, accuracy, reliability, suitability or availability with respect to the information,
products or related graphics contained in this book for any purpose. The author does not accept any legal
responsibility for the content within. By using this book, the reader agrees to indemnify and hold harmless the
author and publisher from any damages claimed because of the content of this book.
NCEES® is a registered trademark of National Council of Examiners for Engineering and Surveying. NCEES® did not
partake in the development of this publication. NCEES® does not endorse or otherwise sponsor this publication and
makes no warranty, guarantee or representation, express or implied, as to its accuracy or content.
Copyright © 2018 by Wasim Asghar. All rights reserved. No part of this publication may be reproduced, stored in a
retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or
otherwise, without prior written permission of the author.
Printed by CreateSpace, An Amazon.com Company
ISBN-13: 978-1985699717
ISBN-10:1985699710
Table of Contents
Preface 1
About the author 3
Chapter # 1 - Mathematics 4
Problem Set # 1.1 - Algebra and Trigonometry 5
Problem Set # 1.2 - Complex Numbers 7
Problem Set # 1.3 - Discrete Mathematics and Progressions 8
Problem Set # 1.4 - Analytic Geometry 10
Problem Set # 1.5 - Calculus 12
Problem Set # 1.6 - Differential Equations 15
Problem Set # 1.7 - Matrix and Vector analysis 16
Chapter #2 - Probability and Statistics 18
Problem Set # 2.1 - Measures of central tendencies 19
Problem Set # 2.2 - Probability distributions 20
Problem Set # 2.3 - Expected values & Estimation for a single mean 21
Chapter # 3 - Ethics and Professional Practice 23
Problem Set # 3.1 - Codes of Ethics & NCEES® Model Law and Rules 24
Problem Set # 3.2 - Intellectual Property 26
Chapter # 4 - Engineering Economics 27
Problem Set # 4.1 - Time value of money 28
Problem Set # 4.2 - Cost estimation 29
Problem Set # 4.3 - Risk identification and analysis 30
Chapter # 5 - Properties of Electrical Materials 31
Problem Set # 5.1 - Chemical Properties 32
Problem Set # 5.2 - Electrical Properties 33
Problem Set # 5.3 - Mechanical Properties 34
Problem Set # 5.4 - Thermal Properties 35
Chapter # 6 - Engineering Sciences 36
Problem Set # 6.1 - Work, Energy, Power 37
Problem Set # 6.2 - Electrostatics 39
Problem Set # 6.3 - Capacitance 40
Problem Set # 6.4 - Inductance 41
Chapter # 7 - Circuit Analysis 42
Problem Set # 7.1 - Kirchoffs Laws - KCL, KVL43
Problem Set # 7.2 - Series / Parallel Equivalent Circuits 45
Problem Set # 7.3 - Thevenin and Norton Theorems 47
Problem Set # 7.4 - Waveform Analysis 49
Problem Set # 7.5 - Phasors 50
Problem Set # 7.6 - Impedance 52
Chapter # 8 - Linear Systems 54
Problem Set # 8.1 - Frequency / transient response 55
Problem Set # 8.2 - Resonance 57
Problem Set # 8.3 - Laplace Transform 59
Problem Set # 8.4 - Transfer functions 61
Problem Set # 8.5 - Two-Port Theory 63
Chapter# 9 - Signal Processing65
Problem Set # 9.1 - Continuous Time Convolution 66
Problem Set # 9.2 - Discrete Time Convolution 69
Problem Set # 9.3 - Z Transforms 71
Problem Set # 9.4 - Sampling 73
Problem Set # 9.5 - Filters 74
Chapter # 1 0 - Electronics 76
Problem Set # 10.1 - Solid-State Fundamentals 77
Problem Set # 10.2 - Diodes 78
Problem Set # 10.3 - BJTs 80
Problem Set # 10.4 - MOSFETs 83
Problem Set # 10.5 - Operational Amplifiers 86
Problem Set # 10.6 - Instrumentation 88
Problem Set # 10.7 - Power Electronics 90
Chapter # 1 1 - Power 91
Problem Set # 11.1 - Single Phase Power 92
Problem Set # 11.2 - Three Phase Power/Transmission & Distribution 94
Problem Set # 11.3 - Voltage Regulation 96
Problem Set # 11.4 - Transformers 97
Problem Set # 11.5 - Motors & Generators 98
Problem Set # 11.6 - Power Factor 99
Chapter # 12 - Electromagnetics 100
Problem Set # 12.1 - Maxwell Equations 101
Problem Set # 12.2 - Electrostatics / Magnetostatics 103
Problem Set # 12.3 - Transmission Lines and Wave Propagation 104
Problem Set # 12.4 - Electromagnetic compatibility 105
Chapter # 13 - Control Systems 106
Problem Set # 13.1 - Block Diagrams 107
Problem Set # 13.2 - Bode Plots 110
Problem Set # 13.3 - Steady Sate Errors 113
Problem Set # 13.4 - Routh-Hurwitz Criteria & System Stability 115
Problem Set # 13.5 - Root Locus 116
Problem Set # 13.6 - State Variables 121
Chapter # 1 4 - Communications 123
Problem Set # 14.1 - Amplitude Modulation 124
Problem Set # 14.2 - Angle Modulation 126
Problem Set # 14.3 - Pulse Code Modulation (PCM) & Pulse Amplitude Modulation (PAM) 127
Problem Set # 14.4 - Fourier Transforms 128
Problem Set # 14.5 - Multiplexing 129
Chapter # 1 5 - Computer Networks 130
Problem Set # 15.1 - Routing and Switching 131
Problem Set # 15.2 - Network topologies / Frameworks / Models 132
Problem Set # 15.3 - Local Area Networks 133
Chapter # 16 - Digital Systems 134
Problem Set # 16.1 - Number Systems 135
Problem Set # 16.2 - Boolean Logic 136
Problem Set # 16.3 - Logic Gates 137
Problem Set # 16.4 - Karnaugh Maps 139
Problem Set # 16.5 - Flip-flops and counters 141
Problem Set # 16.6 - State Machine Design 144
Chapter # 1 7 - Computer Systems 146
Problem Set # 17.1 - Architecture & Interfacing 147
Problem Set # 17.2- Microprocessor 149
Problem Set # 17.3 - Memory Technology and Systems 150
Chapter # 1 8 - Software Development 151
Problem Set # 18.1 - Algorithms 152
Problem Set # 18.2 - Data Structures 153
Problem Set # 18.3 - Software design methods/implementation/testing 154
Solutions 156
Chapter #1 - Mathematics 161
Chapter #2 - Probability and Statistics 184
Chapter #3 - Ethics and Professional Practice 192
Chapter # 4 - Engineering Economics 195
Chapter #5 - Properties of Electrical Materials 201
Chapter #6 - Engineering Sciences 205
Chapter #7 - Circuit Analysis 210
Chapter # 8 - Linear Systems 219
Chapter # 9 - Signal Processing 228
Chapter # 10 - Electronics 237
Chapter # 11 - Power 253
Chapter # 12 - Electromagnetics 264
Chapter # 13 - Control Systems 273
Chapter # 14 - Communications 284
Chapter # 15 - Computer Networks 290
Chapter # 16 - Digital Systems 293
Chapter #17 - Computer Systems 303
Chapter #18 - Software Development 308
Preface
'Practice makes perfect' is as applicable to passing NCEES® FE Exam as it is to anything else.
The biggest challenge involved in FE exam preparation is the breadth of knowledge required. However,
the silver lining is that exam questions may not be very complex. Therefore, it is im portant to gain
fundam ental understanding of all topics (more on exam taking strategy later). The intended audience
o f this book includes final year students, new graduates and seasoned professionals who have been
out o f school for a while.
What's new in the Second Edition?
Study Guide for Fundamentals of Engineering (FE) Electrical and Com puter CBT Exam has been well
reviewed by students. The author has made efforts to add challenging practice problems, introduce
new concepts, offer detailed solutions and update content in general based on latest exam form at and
student feedback.
Key changes in the second edition include:
•
100+ brand new practice problems.
•
Addition o f Mathem atics, Probability and Statistics, Ethics and Professional Practice &
Engineering Economics resulting in com plete coverage of all 18 exam sections.
•
Alternative-ltem Types (AITs).
•
Step-by-step solution of problems requiring detailed explanation.
•
Addition of figures and circuit diagrams in solutions to further understanding of students.
Best way to use this book
This book now covers all 18 sections of NCEES® FE Electrical and Com puter CBT exam. It is centered on
the idea of 'problem-based learning'. It is im portant to note that NCEES® FE Reference Handbook will
be the only reference material available to examinees during examination. As such, this book is
designed to develop reader's fam iliarity with the reference manual.
Students are suggested to conduct m ultiple reviews o f applicable NCEES® FE Reference Handbook
sections and understand theory behind relevant concepts and formulas using all available resources. It
is recommended to attempt problems from each chapter right after studying relevant concepts. You
may not be able to solve all problems in the first attempt. Therefore, it is suggested to make note of
concepts requiring further review. Once the underlying theory is understood, you should revisit the
problems and attem pt them again. After solving the problems, you are encouraged to review solutions
at the end of this book to reconfirm answers and methodology. In certain cases, there can be more
than one ways of solving same question, but an effort has been made to present efficient solving
technique. In cases involving unfamiliar concepts and theories, it is recommended to research such
content to gain necessary understanding. In fact, as part of exam preparation effort, students should
think about ways in which questions can be asked.
1
Copyrighted Material © 2018
Organization of this book
Every chapter starts with a reference to applicable sections and page numbers o f NCEES® FE Reference
Handbook, com m ents regarding difficulty level o f the section and topic specific tips and form ulas for
effective exam preparation. Solutions are grouped at the end for ease o f review. As noted earlier, this
book is especially designed to develop reader's fam iliarity with reference manual. Hopefully, after
solving all problem s and reviewing relevant solutions, students will be able to com fortably navigate
NCEES® FE Reference Handbook and recall applicable form ulas quickly during exam. Doing so w ill allow
them to save precious tim e on exam and help im prove their performance.
FE Electrical & Computer Exam Taking Strategies
It is strongly suggested to purchase latest NCEES® FE Electrical and Com puter sample exam from
www.ncees.org for additional practice. Author suggests adopting one o f the follow ing strategies for
taking FE Electrical and Com puter CBT exam:
Strategy # 1 - Three round knock-out
First Round - After reading the question classify it into one of the follow ing four categories
'Easy', 'M edium ', 'Difficult but solvable' or 'No clue'.
If it is 'Easy7or 'M e d iu m ' solve it right away otherw ise flag it and move on.
After com pleting first round you'll be left with 'Difficult but solvable' and 'No clue' questions.
Second Round - Go through the list of flagged questions and try to solve 'Difficult but solvable'
questions. Carefully remove the flags from solved questions.
After com pleting the second round, you'll have only 'N o clue' questions left.
Third Round - Depending on the amount o f tim e left in exam either try to solve the remaining
questions or apply elim ination method.
Under no circumstances should you leave any question unanswered.
Strategy # 2 - Relax, see and conquer
This strategy works best for bold exam takers. The idea is to go through all questions from get go to
gain big picture view o f the exam. It is advisable to flag all the tough questions during sequential
reading process but try not to solve them right away. After skimming through the entire exam section
(there is a morning and afternoon portion to exam), you will hopefully feel relaxed and confident
because the fear o f unknown will subside. Now you should start solving non-flagged questions
followed by flagged questions. Any leftover tim e should be spent rechecking your answers.
The author adopted strategy # 1 while taking FE Electrical and Computer CBT exam which he passed in
the first attem pt and strategy # 2 while taking PE - Power exam which was also passed in first attempt.
2
Copyrighted Material © 2018
FE Electrical & Computer On-demand Exam Preparation Course and Practice Exams
Visit www.studvforfe.com to learn about on-demand step-by-step exam preparation course covering
all exam sections. 20% discount is available on 'standard life-time course subscription' using discount
code 'SFE20'. Students can also benefit from 3 full-length practice exams for additional preparation.
Online Errata and Error Reporting
This book has undergone m ultiple review cycles and significant effort has been made to produce a
high-quality text. However, it is conceivable that certain errors might have gone unnoticed. Feel free to
report any errors, im provem ents or questions at comments(5)studvforfe.com. Please visit the website
www.studyforfe.com to view confirm ed errata online.
About the author
W asim Asghar is a licensed Professional Engineer in Texas (PE), Florida (PE) and Ontario (P. Eng) with
consulting experience in pow er system design, commissioning and plant engineering for leading clients
in various industries including energy, mining and manufacturing.
He holds Bachelor of Engineering - Electrical with distinction from M cM aster University, Hamilton,
Canada (2010) and M aster of Engineering - Power Systems from University of Toronto (2013) which
was com pleted with full-tim e work.
In 2014, he undertook a tw o-year international work assignment for a major project in Florida and
decided to pursue PE licensure in United States. The road to licensure was challenging prim arily
because o f lack of useful study material for FE and PE exams.
W asim passed both exams in first attempts. The lessons learned during exam preparation process
inspired him to w rite this book which is designed to help aspiring professional engineers better prepare
for FE Electrical and Com puter exam.
Acknowledgements
I am truly thankful to the support offered by these w onderful people:
•
M o ther - Farhat, for always believing in my abilities
•
Fath er-A sg h ar, for providing me the foundation to realize my goals
•
W ife - Amna, for being part of my dreams
•
Brother - Fahim, for providing continuous assistance during entire lifecycle o f this project
•
Uncle - Jawad, for being a great friend and m entor over the years
Dedication
This book is dedicated to my late aunt Nighat Parveen and late grandmother Nazira Begum who gave me
priceless love and affection during my childhood.
3
Copyrighted Material © 2018
Chapter # 1 - Mathematics
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
M athem atics
2 2 -3 6
Algebra and trigonom etry
Complex numbers
Discrete mathematics
Analytic geometry
Calculus
Differential equations
Linear algebra
Vector analysis
Facts about this section
•
1 1 -1 7 questions can be expected on the exam (according to NCEES® FE Specification).
•
Mathem atics is the most heavily weighted section on exam.
•
Difficulty level o f this section is rated 'M edium ' by author.
Tips for preparing this section
•
Understand concepts found on above mentioned pages of NCEES® FE Reference Handbook.
•
Some o f the most im portant concepts and equations of this section include:
■
Algebra - Equations of straight line (general, slope-intercept, point-slope, angle between
lines, distance between points, perpendicular line), Quadratic equation, Quadratic
surface, Logarithmic identities.
■ Complex Numbers - Rectangular form, Polar form, Conversions between Rectangular
and Polar forms, addition, subtraction, m ultiplication and division, Euler's identity.
■ Trigonom etry -T rig o n o m e tric functions, Law of Sines, Law of Cosines, Trigonom etric
identities and sim plification of a trigonom etric expression.
■ Discrete Math - Set theory, Functions and mapping methods, Directed graphs.
■ Analytic geom etry - M ensuration o f areas and volumes, Conic sections.
■ Calculus - Differential Calculus, Function max/min and derivatives, L'Hospital's rule,
Integral Calculus, Definite and Indefinite Integrals.
■
Differential equations - 1st order differential equation solutions - separation of variables
and integrating factor methods, 2nd order differential equation solutions.
■ M atrix and vector analysis - M atrix operations - addition, subtraction, m ultiplication,
determinant, transpose, inverse, Vector operations - dot product, cross product.
•
Solve problem sets on next pages and review solutions at the end of this book.
*Exam specification details can be found on pages 265-267 of NCEES® FE Reference Handbook.
4
Copyrighted Material © 2018
Problem Set # 1.1 - Algebra and Trigonometry
Consult NCEES® FE Reference Handbook - Pages 23-24 while solving these questions
Problem 1.1 a) Solve for x :
lo g 3(3 x - 12) - lo g 3(x) = 2
(A) 2
(B) -2
(C) 3
(D) 0
Problem 1.1 b) Solve for x:
ln (x ) + ln ( x + 8) = ln ( x — 12)
(A) 0
(B) -4
(C) -3
(D) 12
Problem 1.1 c) Solve for x:
lo g 4( x 2 + 8 x + 17) = 0
(A) 4
(B) 8
(C) -4
(D) 2
Problem 1.1 d) lo g 5 7 can be expressed a s ___________in lo g 10.
Problem 1.1 e) Simplify the follow ing trigonom etric expression:
s in 2 x (co t2 x + 1)
(A) csc2 x
(C) 0
(B) 1
(D) ta n 2 x
Problem 1.1 f) Simplify the follow ing trigonom etric expression:
c o t2 X
c o t2 X + 1
(A )s in 2 x
(B )co s2 x
(C) tan 2 x
(D) 1
5
Copyrighted Material © 2018
Problem 1.1 g) Sim plify the follow ing trigonom etric expression:
( s in x + c o s x ) 2 — 1
(A) 0
(C) sin 2%
(B) 2
(D) sin x cos x
Problem 1.1 h) In the triangle given below, length o f side 'a' = _
6
Copyrighted Material © 2018
Problem Set # 1.2 - Complex Numbers
Consult NCEES® FE Reference Handbook - Page 23 while solving these questions
Problem 1.2 a) W hich o f the follow ing options represent (8 + 4j ) — (5 + 0j ) in polar form ?
(A) 3 + 4j
(B) 5/5 3°
(C) 5 / 3 T
(D) 8 /7°
Problem 1.2 b) W hich of the follow ing options represent (4 + 4y)/(2 + 3j ) in polar form ?
(A) 3 + 4j
(B) 5/53°
(C) 1.5/ —11°
(D) 8 /7°
Problem 1.2 c) Which o f the follow ing options represent 2/30° x 4/15° in rectangular form ?
(A) 8/4 5°
(B) 8 + V 2 /
(D) 8
(C) 5.7 + 5.7j
Problem 1.2 d) Which o f the follow ing options represent 4/30° + 6/30° in rectangular form ?
(A) 10/60°
(B) 24/90°
(C) 8.6 + Sj
(D) 5.4 + 3.8;
Problem 1.2 e) 3 + 4j can be expressed a s __________
(A) 5 (cos 53° + j sin 53°)
(B)
(C) 5/5 3°
(D) All of the above
Problem 1.2 f) The complex conjugate of Z = (2 + 67) x (3 + 3j ) is _
7
Copyrighted Material © 2018
Problem Set # 1.3 - Discrete Mathematics and Progressions
Consult NCEES® FE Reference Handbook - Pages 22 and 30 while solving these questions
Problem 1.3 a) W hich o f the follow ing sets is a proper subset o f set {2, 4, 6, 8, 10,12}?
(A) {2, 3, 4, 5}
(B) {2, 4, 6, 8, 10,12, 14}
(C) {2, 4, 6}
(D) {2, 4, 6, 8, 10,12}
Problem 1.3b) W hich o f the follow ing sets is a subset of set {a, b, c, d, e}?
(A) {x, y, z}
(B) {a, b, c, d, e}
(C) {a, f, h, e}
(D) None of the above
Problem 1.3 c) {1, 2, 3, 4, 5} and {a, b, c, d, e} are examples o f _________ sets.
(A) Disjoint
(B) Proper
(C) Sub
(D) Overlapping
Problem 1.3 d) W hat is the cartesian product of {1, 2} x {a, b, c, d}?
(A) {(l,a),(l,b),(2,a),(2,b)}
(B) {(l,a),(l,b),(l,c),(l,d),(2,a),(2,b),(2,c),(2,d)}
(C) {(a,l),(b,l),(c,l),(d,l),(a,2),(b,2),(c,2),(d,2)}
(D) None of the above
Problem 1.3 e) Which o f the follow ing relation(s) is not an example of a function?
(A) {(l,b ),(l,c),(l,d )}
(B) {(1,a),(2,a),(3,a ) }
(C) {(l,a),(2,b),(3,c)}
(D) None of the above
Problem 1.3 f) {(a,l),(b,l),(c,2),(d,2)} is an example o f _______ function.
(A) Injective
(B) Surjective
(C) Bijective
(D) It's not a function
Problem 1.3 g) The 35th term of progression given below is ________ .
2, 4, 6, 8, 10, 12 Problem 1.3 h)_______ is the sum of an arithm etic progression with 120 term s for which first term is 1
and last term is 259.
8
Copyrighted Material © 2018
Problem 1.3 i) The 12th term of progression given below is _______ .
3, 9, 27, 81
Problem 1.3 j)______ is the sum of a geom etric progression for which first term is 1, last term is 19683
and common ratio is 3.
Problem 1.3 k) Select appropriate region(s) representing A n B.
U
R4
9
Copyrighted Material © 2018
Problem Set # 1.4 - Analytic Geometry
Consult NCEES® FE Reference Handbook - Pages 22, 24 - 27 while solving these questions
Problem 1.4 a) Calculate the angle between tw o lines given by follow ing equation.
y1 = x 1 + 4
y2 = Sx2 + 6
(A) 78.6°
(B) 11.3°
(C) 38.6°
(D) 33.7°
Problem 1.4 b) Find the equation o f a straight line passing through points (2,10) and (3,12).
(A) y = 3x + 12
(B) y = 2 x + 10
(C) y = 2x + 6
(D) y = 3x + 10
Problem 1.4 c) Equation o f a straight line with a slope o f 2 and intercept o f -5 is ______ .
Problem 1.4 d) Select the option(s) representing a straight line perpendicular to line given below.
y = 4x + 4
(A)y = —x/4
(B) y = —x/4 + 4
(C) y = —4x
(D) y = —4x + 4
Problem 1.4 e) A conic section equation containing just x 2 term or y 2 term represents_______ .
(A) Parabola
(C) Ellipse
(B) Hyperbola
(D) Circle
Problem 1.4 f) A conic section given by equation 2(x — 1 0 )2 + 8 (y — 6 )2 = 200 represents____
(A) Parabola
(C) Ellipse
(B) Hyperbola
(D) Circle
Problem 1.4 g) The eccentricity o f conic section given in problem 1.4 f) is __________ .
Problem 1.4 h) A conic section given by equation 2(x — 1 0 )2 — 8 ( y — 6 )2 = 200 represents___
(A) Parabola
(C) Ellipse
(B) Hyperbola
(D) Circle
Problem 1.4 i) The eccentricity of conic section given in problem 1.4 h) is __________ .
10
Copyrighted Material © 2018
Problem 1.4 j) A conic section given by equation (y — 8 ) 2 = 4(x — 2) represents_______ .
(A) Parabola
(B) Hyperbola
(C) Ellipse
(D) Circle
Problem 1.4 k) The directrix of conic section given in problem 1.4 j) is a t _________ .
Problem 1.41) The volum e of a right circular cylinder with radius lm and height 2m is _______ .
(A) 2.1 m3
(B) 4.2 m3
(C) 6.3 m3
(D) 3.7 m3
Problem 1.4 m) The area o f a right circular cone with radius 2 m and height 4 m is ______ .
(A) 40 m2
(B) 10 m2
(C) 20 m2
(D) 8 m2
Problem 1.4 n) Which of the follow ing geom etric shapes will hold largest volum e of liquid provided
they have the same radius and height?
(A) Right circular cone
(C) Paraboloid of Revolution
(B) Right circular cylinder
(D) All of them will hold same volum e
11
Copyrighted Material © 2018
Problem Set # 1.5 - Calculus
Consult NCEES® FE Reference Handbook - Pages 27 - 29 while solving these questions
Problem 1.5 a) Calculate the derivative o f follow ing function.
f i x ') = 2 ta n 2 x + s in 2 x
(A) 4 tan x + 2 sin x
(B) 2 sec2 x + co s2 x
(C) 4 ta n x sec2 x + 2 sin x cos x
(D) 0
Problem 1.5 b) Calculate the derivative of follow ing function.
/ ( x ) = 4 x 2 + 6x + 2 y 2
(A) 8 x + 6
(B) 8 x + 6 +
(C) 4 x + 6 + 2y
4y
(D) 0
Problem 1.5 c) Calculate the derivative o f follow ing function.
/ ( x ) = 2 tan x sec x
(A) 2 s in x / co s2 x
(B) 2(tan 2x sec x + sec3 x)
(C) 2 tan x sec3 x
(D) 0
Problem 1.5 d) Calculate the derivative of follow ing function.
fix )
= 2 s in ” 1 x + 2 c o s-1 x
(A) 2/Vl - x 2
(B) -2 / V l - x 2
( C ) 4 s in x c o s x
(D) 0
Problem 1.5 e) Calculate local minimum and maximum points o f function given below.
fix ) = 4x3
+ x 2 — 2x + 8
— 1 < x <1
(A) There is no m inimum/maximum
1
1
(B) x = - (m in), x = — - (m ax)
1
1
(C) x = - (m ax), x =
1
(D) x = - (m in), no m ax
— - (m in)
12
Copyrighted Material © 2018
Problem 1.5 f) Calculate local minimum and maximum points of function given below.
/ ( x ) = 3x 3 + 3x 2 — 3x + 3
—2 < x < 1
(A) There is no m inim um /m aximum
(B) x = - (m in ),x = —l(m a x )
1
(C) x = —(m a x ),x = —l( m in )
1
(D) x = - (m in), no m ax
Problem 1.5 g) The point of inflection for function given in problem 1.5 f) is at x = __
Problem 1.5 h) Evaluate the follow ing limit:
3 x 2 — 2x — 1
lim
* ^ i4 x 2 + 6 x — 10
(A) 0
(B) oo
(C) 2 /7
(D) 4
Problem 1.5 i) Evaluate the follow ing limit:
lim
4cosx
x^n/2 2 — 2 sin x
(A) n
(B) 0
(C) oo
(D) 3n/2
Problem 1.5 j) Evaluate the follow ing indefinite integral:
4
/x + 3 dx
1 . _ !/
IT
(A)^=tan 1 1 x ^ -1
(B) 2 In |2x + 6|
(C) 4Vx + 3
(D) 4 In |x + 3 1
Problem 1.5 k) Evaluate the follow ing indefinite integral:
J
(sin 2 x + co s2 x)d x
(A) 1
(B) x / 2 — s in 2 x / 4
(C) x
(D) x / 2 + sin 2 x / 4
13
Copyrighted Material © 2018
Problem 1.5 I) The definite integral of function given below is _________ .
I x e 2xdx
Jo
Problem 1.5 m) Select the point on graph at which derivative of given function is highest.
Problem 1.5 n) Select the point on graph at which derivative o f given function is low est.
14
Copyrighted Material © 2018
Problem Set # 1.6 - Differential Equations
Consult NCEES® FE Reference Handbook - Pages 30 - 31 while solving these questions
Problem 1.6 a) Solve the follow ing 1st order linear differential equation with given initial values.
2 y' + 4 y = 0
y (0 ) = 6
(A )y = 3e~4t
(B) y = 6e~4t
(C)y = 3e~2t
(D)y = 6e~2t
Problem 1.6 b) W hat is the solution o f 1st order linear differential equation given below ?
y' + 2 x 2y = x 2
Problem 1.6 c) W hat is the solution o f 2nd order differential equation given below?
y " + 6y' + 9y = 0
Problem 1.6 d) Solve the follow ing 2nd order differential equation with initial conditions given below.
y" + 2y' - 8y = 0
y (0 ) = 6,
y '( 0) = 0
Problem 1.6 e) The 2nd order differential equation given below i s ________ damped.
2
y " + 4y ’ + 8 y = 0
Problem 1.6 f) Match the given differential equation with correct type.
1st order linear non-homogeneous differential equation
y " + 8y f + 12y = 0
1st order linear homogeneous differential equation
2nd order linear homogeneous differential equation
15
Copyrighted Material © 2018
Problem Set # 1.7 - Matrix and Vector analysis
Consult NCEES® FE Reference Handbook - Pages 34 - 35 while solving these questions
Problem 1.7 a) Calculate the sum o f matrices A and B given below.
1
A = 0
.0
0
1
0
0
0
1.
0
B = 0
.2
0
2
0
2~
0
0.
1
(A) 0
.0
0
1
0
0
0
1.
0
(B) 0
.2
1
(C) 0
0
3
0
2'
0
(D) A and B cannot be added
2
0
2
0
2'
0
1.
1.
Problem 1.7 b) Calculate the sum of matrices A and B given below.
A =
0
0
1
0
.0
0
1.
2
0
0"
0
2
0
.0
0
2.
'2
(A) 0
.0
0
1
0
0‘
0
2.
CD
1
0
'2
(C) 0
.0
0
2
0
O'
0
1.
(D)
B =P
LO
°1
IJ
A and B cannot be added
Problem 1.7 c) Calculate the product of matrices A x B of matrices A and B given below.
A =
(A)
(C)
4
1
8
L12
2
3
2
8
0
2
B =
1
2
0
1
(B)
(D) A and B cannot be multiplied
Problem 1.7 d) The determ inant of matrix given below is
4
2
5
6
Problem 1.7 e) The inverse o f matrix given in problem 1.7 d) is
16
Copyrighted Material © 2018
Problem 1.7 f) Calculate the product of matrices A x B of matrices A and B given below.
'2
A = 4
.6
14
(A) 32
.54
(C)
20'
22
36.
7
B = 8
.9
1’
2
3.
'7
(B) 16
.27
lo ­
ll
12.
20'
44
72.
(D) A and B cannot be m ultiplied
l-«J JLi
Problem 1.7 g) Which of the following option(s) Is not true for all matrices?
(A) ,4 x B = f i x A
(B)A + B = B + A
(C) m A ]-1 = [ A l- ^ A ]
(D) A - 1 = ad j(A )/ d et(A )
Problem 1.7 h) The determ inant of matrix given below is _______________ .
"2 4
6'
8 10
.1 3
12
5.
Problem 1.7 i) The dot product of vectors A and B is _________
A = 2i +j +3k
B=
=i + 2/ + 4k
Problem 1.7 j) The cross product of vectors A and B is _________ .
A = 3i + 2j + k
B = i + 4j + Ok
Problem 1.7 k) Select the correct resultant vector from operations given below.
X =f
Y = ----►
X + ¥ = ___________
O ption - A
--------► O ption - B
O ption - C
<------
17
Copyrighted Material © 2018
O ption - D
Chapter # 2 - Probability and Statistics
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
Engineering Probability and
Statistics
3 7 -5 3
Measures o f central tendencies and dispersions
Probability distributions
Expected value in decision making
Estimation for a single mean
Facts about this section
•
4 - 6 questions can be expected on the exam (according to NCEES® FE Specification).
•
Difficulty level o f this section is rated 'M e d iu m ' by the author.
Tips for preparing this section
•
Understand concepts found on above mentioned pages of NCEES® FE Reference Handbook.
•
Some o f the most im portant concepts and equations of this section and sub-sections include:
■
Measures of central tendencies and dispersion - Calculation of mean, mode, median,
standard deviation, sample range and sample variance.
■
Probability distributions - Normal distribution, Binomial distribution, continuous
distribution, discrete distribution.
■
Laws o f probability - Probability definition, Law of total probability, Law of joint
probability, Bayes' Theorem.
■
Expected v alu e- Probability density functions, Probability mass functions, Cumulative
distribution function.
■
•
Estimation for a single mean - Confidence interval, norm al-distribution, t-distribution.
Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification details can be found on pages 265-267 of NCEES® FE Reference Handbook.
18
Copyrighted Material © 2018
Problem Set # 2.1 - Measures of central tendencies
Consult NCEES® FE Reference Handbook - Page 37 while solving these questions
Problem 2.1 a) Arithm etic mean of follow ing observations is ________ .
2, 4 , 10, 8, 4, 8
Problem 2.1 b) The average annual summer tem perature o f a small town is recorded as follows.
2017(90°F)
2016(84°F)
2015(88°F)
2014(92°F)
W hat is the weighted average of summer tem perature if 50% weight is assigned to 2017, 30% weight
to 2016, 10% to 2015 and 10% to 2014?
Problem 2.1 c) Sample standard deviation o f follow ing data set is ________ .
2, 4, 6, 8
Problem 2.1 d) Sample geom etric mean o f follow ing data set is ________ .
1, 2, 3, 4
Problem 2.1 e) Sample root mean square o f follow ing data set is __________ .
3, 5, 6,1 1
Problem 2.1 f) M edian value o f data set given below is ________ .
2, 3, 7,1, 4, 9, 0
Problem 2.1 g) M edian value o f data set given below is ___________.
90, 60, 70,110, 50, 40, 200, 210
Problem 2.1 h) M ode o f the data set given below is ___________.
1, 3, 3, 4, 9, 7, 3, 4
Problem 2.1 i) Sample range o f data set given below is ________ .
50, 20, 10, 50, 70, 80,100, 120
Problem 2.1 j) Select the correct property of given statistical measurement.
Influenced by outliers
M ean
Not a measure o f central tendency
M arginally influenced by outlier
19
Copyrighted Material © 2018
Problem Set # 2.2 - Probability distributions
Consult NCEES® FE Reference Handbook - Pages 38 - 39 while solving these questions
Problem 2.2 a) According to latest weather forecast, the probability that weekend will be sunny is
0.25, cloudy is 0.35 and both sunny and cloudy is 0.15. W hat is the probability that weekend will be
sunny, cloudy or both?
(A) 0.25
(B) 0.30
(C) 0.45
(D) 0.75
Problem 2.2 b) According to the data collected by a superstore chain, 20% of custom ers prefer online
shopping whereas remaining customers prefer in-store shopping. 50% of the customers preferring
online shopping are under 30 years o f age and 40% o f the customers preferring in-store shopping are
under 30 years o f age.
W hat is the probability o f a random ly selected custom er having preference for online shopping given
that person is under 30 years o f age?
Problem 2.2 c) The probability of an engineering firm hiring a new graduate after an on-campus
interview is 0.10. The probability of hiring 2 new graduates after 30 on-campus interviews is ______ .
Problem 2.2 d) A top-seed tennis player is confident that she has 40% chance o f winning each of the
four grand slam tournam ents next year. W hat is the probability that she will win 2 to 4 grand slams?
Problem 2.2 e) The average GMAT® score of successful applicants at a well-reputed business school is
740 with a standard deviation o f 20. Calculate the probability o f admission for students with scores less
than 700. Assume that acceptance based on GMAT® score follow s normal distribution.
Problem 2.2 f) The average useful life of a typical dishwasher is 10 years with standard deviation o f 2
years. Calculate the probability of this dishwasher lasting between 12 to 14 years. Assume that useful
life o f dishwasher follow s normal distribution.
Problem 2.2 g) The total probability of events A and B based on Venn diagram given below is _______ .
U
P(A n B) =o.i
20
Copyrighted Material © 2018
Problem Set # 2.3 - Expected values & Estimation for a single mean
Consult NCEES® FE Reference Handbook - Pages 38 - 39,44, 45,47 while solving these questions
Problem 2.3 a) The expected value o f a discrete random variable X w ith probability mass function
/ ( * * ) given below is __________ .
f (%
k)
1
22
^
^3
^
Problem 2.3 b) The variance of X in problem 2.3a) is ____________.
Problem 2.3 c) According to the guidance provided by a fund manager at an investment bank, the fund
is forecasted to perform in accordance with following probability distribution.
Gain
-5%
Probability
0.10
0%
0.20
5%
0.30
10%
0.30
15%
0.10
Investors can expect fund perform ance to be approxim ately______ %.
Problem 2.3 d) A typical college student spends X proportion o f study tim e on com pleting assignments.
Probability density function of X is given as follows:
/( x ) = 4x — 1
0 < x <1
/(x ) = 0
otherwise
The expected value of study tim e spent by student in doing assignments is _______ .
Problem 2.3 e) According to the data collected by an autom obile insurance company, the probability
distribution function of num ber of traffic violations per driver (denoted by X) can be given as follows:
f i x ) = 2 x -3
x > 1
fix ) = 0
otherwise
The expected number of traffic violations per driver is _________ .
Problem 2.3 f) The total area under probability density function and probability mass function is always
equal t o ____________.
Problem 2.3 g) The average income o f 60 households based on a recent census in a small tow n is found
to be $55,000. The household income is assumed to be normally distributed with a standard deviation
of a = $5,000. Determine 99% confidence interval for the population mean of household income.
(A) $55,000 ± $5000
(B) $55,000 ± $2500
(C) $55,000 ± $1662
(D) $55,000 ± $3325
21
Copyrighted Material © 2018
Problem 2.3 h) In problem 2.3 g), the standard error of mean (SEM) is _________ .
Problem 2.3 i) According to a survey involving 10 participants, the average tim e dedicated to low
intensity exercise by retirees was recorded as 100 minutes per week with a sample standard deviation
o f 20 minutes. Calculate 90% confidence interval for the population mean o f tim e dedicated to low
intensity exercise by retirees assuming it follow s t-distribution.
(A) 100 ± 20
(B) 100 ± 5.5
(C) 100 ±11.5
(D) 100 ±18
Problem 2.3 j) In problem 2.3 i), the margin of error is ________ .
22
Copyrighted Material © 2018
Chapter # 3 - Ethics and Professional Practice
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
Ethics
3 -5
Code of Ethic
NCEES M odel Law and M odel Rules
Intellectual property
Facts about this section
•
3 - 5 questions can be expected on the exam (according to NCEES® FE Specification).
•
Ethics and Professional Practice is one o f the most lightly weighted section on exam.
•
Difficulty level o f this section is rated 'Easy' by the author.
Tips for preparing this section
•
Understand concepts found on above m entioned pages o f NCEES® FE Reference Handbook.
•
Some o f the most im portant concepts o f this section and sub-sections include:
■
Code of ethics - Background and definitions.
■
NCEES M odel Law and M odel Rules - M odel Rules, Section 240.15, Rules o f Professional
Conduct, Licensee's Obligation to the Public, Licensee's Obligation to Employer and
Client, Licensee's Obligation to other Licensees.
■
Intellectual property - Trademark, Copyright, Patent, Industrial design, Trade secret.
• Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification details can be found on pages 265-267 o f NCEES® FE Reference Handbook.
23
Copyrighted Material © 2018
Problem Set # 3.1 - Codes of Ethics & NCEES® Model Law and Rules
Consult NCEES® FE Reference Handbook - Pages 3 - 4 while solving these questions
Problem 3.1 a) A recently licensed professional engineer is working under the supervision of a senior
engineer. The junior engineer brings a potential design issue to senior engineer's attention which can
result in a safety hazard. The senior engineer tells her that design is based on standard industry
practice, construction is underway and the probability of safety incident occurring is practically zero.
W hat should be junior engineer's response after learning these facts?
(A) Junior engineer is not required to do anything further because senior engineer is competent.
(B) Junior engineer is not required to do anything further because he's already expressed concern.
(C) Junior engineer shall escalate his concerns to a higher level because he's still worried about safety.
(D) Junior engineer shall escalate his concerns to a higher level because it will raise his profile.
Problem 3.1 b) John is a well-respected engineer with a specialization in pow er system engineering
and some experience in control systems. John has an excellent track record of delivering successful
projects on tim e and under budget. He is asked by his new supervisor to take an assignment involving
control systems scope. John's new supervisor does not know his past work experience very well, but he
has heard wonderful things about John from other managers.
Is it ethically acceptable for John to accept this assignment?
(A) Yes, accepting the assignment will give him an opportunity to enhance skills in control systems.
(B) Yes, it is his professional obligation to work diligently for the employer.
(C) No, John is a power systems specialist and should only work in that area.
(D) John should review assignment requirem ents and his expertise with his supervisor and then decide.
Problem 3.1 c) Sarah is a professional engineer and she is brought on a project as a discipline lead. The
project is in its final design stages. Project manager has asked her to quickly review near com plete
engineering docum ents and seal them. These documents were prepared by other engineers who are
not licensed. However, she knows that all of them are com petent based on her experience.
Should Sarah sign and seal these documents?
(A) Yes, she is fam iliar with quality of team.
(B) Yes, she is the new lead and can review docum ents even though tim eline is tight.
(C) No, documents were not developed by licensed engineers.
(D) No, documents were not developed under her supervision and review tim e may be insufficient.
24
Copyrighted Material © 2018
Problem 3.1 d) A professional engineer is reviewing equipm ent drawings for compliance on behalf of a
client. The engineer notices that vendor is providing a technically acceptable alternative to one o f the
specification requirements listed in original bid. Upon further research, the engineer finds that
proposed alternative is cheaper than the one required by client's specification.
W hat should be included in the engineer's recommendation report to the client?
(A) Rejection, because proposed alternative is cheaper than original requirem ent
(B) Rejection, because proposed alternative is not the same as one required by specification
(C) Acceptance, because proposed alternative is technically acceptable.
(D) Acceptance, because proposed alternative is acceptable but negotiate a price credit from supplier.
Problem 3.1 e) M ark is a licensed professional engineer who independently provides engineering
consultancy services to various clients. He recently came across a proprietary solution w hile working
for one of his regular clients which can be deployed in sim ilar scenarios faced by many other clients.
Can M ark use this unique solution in sim ilar applications for other clients?
(A) Yes, M ark is obligated to provide solutions to his clients to the best o f his knowledge
(B) Yes, M ark's clients hire him because of his wide experience and expect innovative solutions
(C) No, M ark cannot use confidential details w ithout obtaining consent from relevant clients
(D) No, M ark is obligated to produce innovative solutions for each client.
Problem 3.1 f) Drag and drop correct NCEES® M odel Rules, Section 240.15, Rule o f Professional
Conduct reference in front of given description.
Respect of fellow licensees________ .
A-4
A -l
C-3
C-4
B -l
B-4
Problem 3.1 g) Which o f the follow ing statement(s) is correct?
(A) Only licensed engineers are expected to conduct them selves professionally.
(B) Ethical problems are always straight forward.
(C) Engineers possess special knowledge which is not common in public domain.
(D) Engineer is not liable for violating an ethical principle if he/she wasn't aware o f it.
25
Copyrighted Material © 2018
Problem Set # 3.2 - Intellectual Property
Consult NCEES® FE Reference Handbook - Page 5 while solving these questions
Problem 3.2 a) An advertising agency interested in protecting creative marketing slogans for its clients
should co n sid e r__________registration.
(A) Trademark
(B) Copyright
(C) Patent
(D) Industrial design
Problem 3.2 b) A m anufacturing plant interested in protecting its innovative processing techniques
should co n sid er__________ registration.
(A) Trademark
(B) Copyright
(C) Patent
(D) Industrial design
Problem 3.2 c) A publishing com pany interested in protecting content of its publications should
co n sid e r________ registration.
(A) Trademark
(B) Copyright
(C) Patent
(D) Industrial design
Problem 3.2 d) A toy m anufacturer interested in preventing com petitors from copying its packaging
styles should co n sid er________ registration.
(A) Trademark
(B) Copyright
(C) Patent
(D) Industrial design
Problem 3.2 e) Drag and drop correct symbol against given intellectual property registration products.
1 - Tradem ark_____
2 - Copyright_______
3 - Registered Trademark
A =©
B=®
C
D
=
T
E=Qe
F=™
Problem 3.2 f) Which of the follow ing option(s) is an example o f intellectual property?
(A) Stocks
(C) Software program
(B) Bonds
(D) Real estate
26
Copyrighted Material © 2018
Chapter # 4 - Engineering Economics
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
Engineering Economics
1 3 1 -1 3 7
Time value of money
Cost estimation
Risk Identification
Analysis
Facts about this section
• 3 - 5 questions can be expected on the exam (according to NCEES® FE Specification).
• Engineering Economics is one o f the most lightly weighted section on exam.
• Difficulty level o f this section is rated 'Easy' by the author.
Tips for preparing this section
• Understand concepts found on above mentioned pages of NCEES® FE Reference Handbook.
• Some of the most im portant concepts and equations of this section and sub-sections include:
■
Time value of money - Single payment compound amount, Single payment present
worth, Uniform series sinking fund, Capital recovery, Uniform series compound amount,
Uniform series present worth, Uniform gradient worth, Uniform gradient future worth,
Uniform gradient uniform series.
■
Cost estimation - Inflation, Depreciation, Book value, Capitalized costs.
■
Risk identification - Definition, Rate-of-Return.
■
Analysis - Break-even analysis, Benefit-cost analysis.
• Learn how to use interest rate tables given in NCEES® FE Reference Handbook.
• Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification details can be found on pages 265-267 of NCEES® FE Reference Handbook.
27
Copyrighted Material © 2018
Problem Set # 4.1 - Time value of money
Consult NCEES® FE Reference Handbook - Page 131 while solving these questions
Problem 4.1 a) Calculate the future worth o f $25,000 investment, 20 years from now. 12% annual
interest rate can be assumed for this calculation.
Problem 4.1 b) Calculate the present worth of a retirem ent fund if it is expected to have $750,000 in it
approxim ately 30 years from now. 4% annual interest rate can be assumed for this calculation.
Problem 4.1 c)_______ converts an annuity to a future amount.
(A) Capital Recovery
(B) Uniform Series Compound Am ount
(C) Uniform Series Present W orth
(D) Uniform Series Sinking Fund
Problem 4.1 d )_______ converts a present value to an annuity.
(A) Capital Recovery
(B) Uniform Series Compound Am ount
(C) Uniform Series Present W orth
(D) Uniform Series Sinking Fund
Problem 4.1 e) A retired couple is considering 20-year term annuity with their $200,000 cash savings at
6% annual interest rate. The yearly annuity am ount they can expect to receive is _______ .
Problem 4.1 f) Amanda needs to have $________ saved today if she plans to get a 20-year term
annuity that will pay her $40,000 annually for the duration of its term ? 8% annual interest rate can be
assumed for this calculation.
Problem 4.1 g) Calculate the future worth o f an investm ent plan in which $15,000 are invested each
year for 30 years. Investments earns interest annually at a rate o f 8%.
Problem 4.1 h) A nominal interest rate o f 10% compounded m onthly will result in an annual effective
interest rate o f _____ %.
Problem 4.1 i) Sam is considering enrollm ent into a 6-year degree program at a college that com bines
undergraduate and graduate course work. He can expect to pay $40,000 in tuition for 1st year, $42,000
for 2nd year, $44,000 for 3rd year and so on (i.e. tuition will increase by $2,000 each year).
Assuming 2% annual interest rate, Sam's total tuition in today's dollars will b e _______ .
28
Copyrighted Material © 2018
Problem Set # 4.2 - Cost estimation
Consult NCEES® FE Reference Handbook - Pages 131 - 132 while solving these questions
Problem 4.2 a) Calculate the annual inflation adjusted interest rate if annual inflation is 4% and annual
interest rate over the same period is 6%.
(A) 6.5%
(B) 24%
(C) 12.7%
(D) 10.2%
Problem 4.2 b) W hat is the accumulated depreciation o f a car at year 4 if its useful life is 30 years,
initial cost was $32,000 and end of life salvage value is $2,000?
Problem 4.2 c) Calculate the accumulated depreciation o f office printer in year 3 using MACRS factors
table if it has a useful life o f 10 years. The initial cost was $7,000 and salvage value will be $500.
Problem 4.2 d) Calculate the accumulated depreciation of a transport truck in year 2 using MACRS
factors table if its useful life is 5 years. The initial cost was $50,000 and salvage value will be $2000.
Problem 4.2 e) Calculate 2nd year book value o f transport truck given in problem 4.2 d) using MACRS
depreciation table.
Problem 4.2 f) The capitalized cost o f operating a restaurant with annual operating expenses of
$20,000 at 8% interest rate is __________ .
Problem 4.2 g)_____________depreciation method results in same value decline each year.
29
Copyrighted Material © 2018
Problem Set # 4.3 - Risk identification and analysis
Consult NCEES® FE Reference Handbook - Pages 131 -132 while solving these questions
Problem 4.3 a) Two production lines have different cost structures for manufacturing same item.
Product line A has a fixed cost of $100,000 and variable cost of $5per item. Product line Bdoes
not
have any fixed costs, but it has a higher variable cost o f $10 per item. The break-even point of both
product lines is ________ .
Problem 4.3 b) W hich o f the follow ing statement(s) are true about problem 4.3 a)?
(A) Product line A is more econom ical below break even point
(B) Product line B is more economical below break even point
(C) Both product lines are equally economical at break even point
(D) None o f the above
Problem 4.3 c) It costs a chair manufacturing com pany $50,000 in fixed cost and $2 per chair in
variable cost to m anufacture a chair. How many chairs need to be sold to break-even if each selling
price of each chair is $27?
Problem 4.3 d) A shipping company has short-listed tw o contractors for a multi-year project.
Assuming identical scope o f work, service quality and annual interest rate o f 8%, which of the follow ing
contractors is offering a com petitive quote?
Contactor A - Down paym ent of $20,000 and final payment of $40,000 in year 6.
Contractor B - $10,000 payment each year for six years.
Problem 4.3 e) Perform a cost-benefit analysis to determ ine which of the follow ing investment options
is more profitable (assume 10% annual interest rate).
Option A - $100,000 purchase of stocks, expected to sell for $200,000 in year 10.
Option B - $100,000 real estate investment that pays $5000 in rent each year and is expected to sell for
$160,000 in year 10.
Option C - $100,000 deposit in a savings account earning 10% annual interest for 10 years.
Problem 4.3 f ) ________ is a risk analysis technique that graphically organizes decision making process
by using probability distributions.
30
Copyrighted Material © 2018
Chapter # 5 - Properties of Electrical Materials
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
Chemical Properties
Electrochemistry
Chemistry
59
Atom ic Bonding
Corrosion
M aterial Science/Structure of M atter
60
Diffusion
Electrical Properties
Capacitance
Conductivity
Resistivity
60
M aterial Science/Structure of M atter
Perm ittivity
Magnetic Perm eability
Note: Electrical Properties are also discussed on pages 199-200 NCEES® FE Reference Handbook.
Mechanical Properties
Material Strength (stress, strain,
Young's modulus)
Photoelectric effect
M aterial Science/Structure of M atter
6 0 -6 2
Piezoelectric effect
Thermal Properties
Expansion
M aterial Science/Structure of M atter
Conductivity
65
60
Facts about this section
• 4-6 questions can be expected on the exam (according to NCEES® FE Specification).
• Difficulty level of this section is rated 'Easy' by the author.
Tips for preparing this section
•
Understand concepts found on above m entioned pages o f NCEES® FE Reference Handbook.
• Handle units o f measurement carefully.
•
Some of the im portant equations o f this section relate to corrosion reaction, diffusion
coefficient, resistivity, resistance, capacitance, magnetic field strength, stress, strain, Young's
modulus and tem perature coefficient.
• Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification details can be found on pages 265-267 of NCEES® FE Reference Handbook.
31
Copyrighted Material © 2018
Problem Set # 5.1 - Chemical Properties
Consult NCEES® FE Reference Handbook - Pages 59-60 while solving these questions
Problem 5.1 a)
will act as a sacrificial anode when brought in contact with iron (Fe).
(A) Zn
(B) Cu
(C) Ni
(D) Hg
Problem 5.1 b) W hich o f the follow ing techniques are used for corrosion prevention?
(A) Galvanization
(B) Sacrificial anode
(C) Plating
(D) All o f the above
Problem 5.1 c)
must be present for corrosion to take place.
(A) Anode, cathode and electrolyte
(B) Electrolyte and sun light
(C) Anode, cathode and sun light
(D) W ater and sun light
Problem 5.1 d) Calculate the diffusion coefficient for Copper atoms diffusing in Copper if constant of
proportionality (D0) is 7.8 x 10~5 m2/s and activation energy (Qd) is 250 J/mol at 700 °C (973 K).
Problem 5.1 e) Calculate the activation energy required by Nickel atoms to attain diffusion coefficient
o f 1.3 x 10~22 m2/s in host Copper atoms if the constant of proportionality (Do) is 2.7 x 10~5 m2/s at
500 °C (773 K).
Problem 5.1 f) Which of the follow ing statements is true about Zinc-Nickel corrosion cell?
(A) Nickel will act as Anode
(C) Either metal can act as Anode
(B) Zinc will act as Anode
(D) Corrosion will not occur
32
Copyrighted Material © 2018
Problem Set # 5.2 - Electrical Properties
Consult NCEES® FE Reference Handbook - Pages 60,200 and 201 while solving these questions
Problem 5.2 a) Calculate resistivity of a 100 m long wire with a cross sectional diam eter o f 2 mm if its
resistance is 5 Q at 30° C.
(A) 3.14 x 10~7 Qm
(B) 1.57 x 10~3 Qm
(C) 3.14 x 10~3 Qm
(D) 1.57 x 10 7 Qm
Problem 5.2 b) The resistivity of cable 'A' is four tim es that of cable 'B'. Both cables can offer same
resistance under a given tem perature i f _______________ .
(A) Area is same and A is twice as long as B
(B) Area of B is four tim es that
(C) Area of B is one-fourth of A and same length
of
(D) Resistance of A and B cannot
A
be same
Problem 5.2 c) Capacitance of a parallel plate capacitor can be increased b y _____________
(A) Decreasing the cross-sectional area of plates
(B) Decreasing the distance between plates
(C) Inserting an insulating material with a higher dielectric constant
(D) Options B & C are correct
Problem 5.2 d) M aterials with high electrical conductivity typically h a ve ______________ .
(A) High resistivity
(B) High heat conductivity
(C) Low elasticity
(D) Low ductility
Problem 5.2 e) Calculate the magnetic perm eability of a medium in which an infinitely long wire
carrying 100 A current produces 0.5 T magnetic field 50 cm perpendicular to the wire.
Problem 5.2 f) Photoelectric effect can take place i n ____________under suitable conditions.
(A) M etals
(B) Non-metals
(C) Liquids and gases
(D) All of the above
33
Copyrighted Material © 2018
Problem Set # 5.3 - Mechanical Properties
Consult NCEES® FE Reference Handbook - Pages 60 - 62 while solving these questions
Problem 5.3 a) A group o f researchers have created a new material in laboratory. Its ultimate tensile
strength is found to be 2000 x 108Pa. Calculate the maximum weight that can be supported by this
m aterial if it is 5m long and has a diam eter of 2mm.
(A) 200 kN
(B) 628 kN
(C) 2513 kN
(D) 915 kN
Problem 5.3 b) Suspending 100 N weight from a 10m long w ire results in 2mm length increase.
Calculate Young's modulus of this wire if its diam eter is 1 mm.
(A) 318 x 109 N/m 2
(B) 6.36 x 1011 N/m 2
(C) 1273 x 109 N /m 2
(D) 636 x 103
N/m 2
Problem 5.3 c) Calculate 'true' strain o f a material that undergoes 20mm increase in length if its
original length was 200 mm.
Problem 5.3 d )____________is the ability o f a material to return to its original form once applied force
is removed.
(A) Plasticity
(B) Elasticity
(C) Ductility
(D) M alleability
Problem 5.3 e) Piezo-electric effect is associated w it h ____________
(A) Photo-electric effect
(B) Magnetic flux
(C) Electric charge concentration
(D) Lightning
Problem 5.3 f) Tensile test curve gives inform ation about all but which o f the follow ing mechanical
properties?
(A) Tensile strength
(B) Young's modulus
(C) Ductility
(D) Hardness
34
Copyrighted Material © 2018
Problem Set # 5.4 - Thermal Properties
Consult NCEES® FE Reference Handbook - Pages 60, 65 and 201 while solving these questions
Problem 5.4 a) Bim etallic strip for a therm ostat should be made of materials w it h ___________
(A) Different lengths
(B) Different coefficients o f therm al expansion
(C) Same resistivity
(D) Same density
Problem 5.4 b) A material o f 1 m length is kept at room tem perature (296 K) and constant pressure. It
is known that a 7 K tem perature rise results in an engineering strain of 3 x 10~3. Find the tem perature
required to cause 6 x 10~3 engineering strain for this material.
(A )319 K
(B) 307 K
(C)310 K
(D )303 K
Problem 5.4 c) A design engineer is calculating space allowance required for railway steel track
expansion. Thermal expansion coefficient of steel being is 1.2 x 10-5OC 1. Tem perature is expected to
increase from an average o f 20 °C to a peak of 45 °C. Calculate the strain that can be experienced by
tracks rail tracks due to therm al expansion.
(A) 3 x 10-2
(B) 30 x 10 s
(C) 50 x 10~5
(D) 75 x 10~6
Problem 5.4 d) W hat is the tem perature coefficient o f a given metal specimen if its resistance doubles
with a 25 K tem perature rise?
Problem 5.4 e) Resistivity o f a material is com pletely independent o f _______________ .
(A) Resistance
(B) Volum e
(C) Temperature
(D) Depends on resistance, tem perature & volum e
Problem 5.4 f) Students are measuring heat capacities of three different samples o f same liquid.
Sample 1 is 1 kg, sample 2 is 2 kg and sample 3 is 3 kg. W hich o f the follow ing options correctly
indicates their heat capacities?
(A) Sample 1 > Sample 2 > Sample 3
(B) Sample 1 < Sample 2 < Sample 3
(C)Sample 1= Sample 2 = Sample 3
(D) It cannot be determined
35
Copyrighted Material © 2018
Chapter # 6 - Engineering Sciences
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
Electrical and Com puter Engineering
200-202
Work, energy, power, heat
Charge, energy, current, voltage, power
Electrostatic forces
W ork done in electric field
Capacitance
Inductance
Facts about this section
•
6 - 9 questions can be expected on the exam (according to NCEES® FE Specification).
•
Difficulty level o f this section is rated 'Easy' by the author.
Tips for preparing this section
•
Understand concepts found on above m entioned pages of NCEES® FE Reference Handbook.
•
It is im portant to correctly handle direction o f vector quantities such as force, electric field
strength and magnetic field strength in calculations.
•
Learn how to use right hand rule for determ ining vector direction.
•
Understand the difference between work, force, energy and power.
•
W ork done by an external agent on a charge is considered 'negative'.
•
Learn how to calculate dot product and cross product. Review situations in which dot product
and cross products are different and apply them to vector quantities such as work and force.
•
Identify cases in which simplified form ulas for electrostatic fields involving line charges and
sheet charges can be applied.
•
Inductors and resistors behave sim ilarly in series and parallel arrangement whereas capacitors
behave differently.
•
Perm ittivity of medium s = 8.85 x 10~12 F m ~1
•
Perm eability o f medium \i = AnlO ~7 HmT 1
•
Some of the im portant equations of this section relate to work, power, current, potential
difference, electrostatic forces, electric field strength, capacitance and inductance.
•
Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification details can be found on pages 265-267 of NCEES® FE Reference Handbook.
36
Copyrighted Material © 2018
Problem Set # 6.1 - Work, Energy, Power
Consult NCEES® FE Reference Handbook - Pages 200 - 202 while solving these questions
Problem 6.1 a) Calculate the am ount o f work necessary to bring a charge Q i = 10~6 C from infinity to P i
(0, 0, 0) in the presence o f another charge Ch = 2 x 10 ~6 C located at P 2 (2, 0, 0).
(A) 4.5 x 1 0 12 J
(B) 4.5 x IO -3 J
(C) 9 x 1 0 12 J
(D) 9 x 10"3 J
Problem 6.1 b) Calculate potential energy stored by system o f two charges Q i = 5 x 10'6 C & Q 2 = 10 ~6 C
located at P i (1 , 0, 0 ) and P 2 (0 , 1 , 0 ) respectively.
(A) 63.6 mJ
(B) 31.9 mJ
(C) 15.9 mJ
(D) 5 mJ
Problem 6.1 c) How many electrons pass through a point on conductor if it carries 1 mA for 5 s?
(A) 1.56 x 10 16
(B) 3.12 x 10 16
(C) 6.24 x 10 12
(D) 3.12 x l O 12
Problem 6.1 d) System A contains charges Q i = 50 x 10"6 C and Ch = 100 x 10~6 C that are 1 m apart.
System B contains charges Q 3 = 5 x 10 ~6 C and 0.4 = 10 x 10 ‘6 C that are 1 cm apart. W hich o f the
follow ing options accurately represent potential energies of System A and System B?
(B) System A = 4.5x 10 13 J,System
(A) System A = System B = 45 J
(C) System A = 45 J
, System B = 4.5 x 10 13 J
B = 45 J
(D) System A = 4.5 J, System B = 45
J
Problem 6.1 e) Calculate the am ount of work that needs to be done to decrease the space between
Q i = 9 x 10 9 C and Q 2 = 15 x 10 9 C from 1 m to 1 cm.
(A) 1.215 x 10~6 J
(B) 15 x 10 4 J
(C) 1.202 x 10 6 J
(D) 1.202 x 10 4 J
Problem 6.1 f) Calculate the energy stored in the electric field o f a parallel plate capacitor that has
potential difference of 200 V if the distance between tw o plates is 0.1 m and each area of each plate is
1 m 2 (assume 8 = 8.85 x 1 0 12 F/m).
(A) 8.85 x 1 0 11 J
(B) 1.33 x 10 3 J
(C) 3.54 x 10~6 J
(D) 1.77 x 10~6 J
37
Copyrighted Material © 2018
Problem 6.1 g) Calculate the am ount of work done in moving a charge Q i = 10 x 10~9 C 2 cm along yaxis in a 200 V/m ax electric field.
(A) 4 x 10'8 J
(B) 0 J
(C) 4 x 10~9 J
(D) 2 x 10'9 J
Problem 6.1 h) Calculate real power dissipated in a 2 Q resistor if 2 C charge passes through it in 1 s.
38
Copyrighted Material © 2018
Problem Set # 6.2 - Electrostatics
Consult NCEES® FE Reference Handbook - Pages 200 - 202 while solving these questions
Problem 6.2 a) Calculate the magnitude o f force between tw o point charges Q i = 10 x 10~6 C and Ch =
100 x 10~6 C that are located at a distance of 1 cm.
(A) 90 kN
(B) 450 kN
(C) 225 kN
(D) 4.5 kN
Problem 6.2 b) Calculate the potential difference between tw o parallel plates that are 1 m apart and
have an electric field strength o f 2000 V/m.
(A) 2000 V
(B) 2 V
(C) 0 V
(D) 100 V
Problem 6.2 c) A point charge Q i = 100 x 10 9 C is accelerated 200 m between tw o parallel plates in a
constant electric field strength of 1 kV/m. Calculate the potential difference.
(A) 22.5 mV
(B) 100 kV
(C) 200 kV
(D) 1 kV
Problem 6.2 d) A charged particle 'Q' with mass 0.01 kg is suspended between two parallel plates with
potential difference of 100 V. Calculate the charge quantity if spacing between plates is 0.1 m.
Problem 6.2 e) 0.1 m long conductor carries 5 A in a 10 jiT magnetic field. Find the angle between
conductor and magnetic field if the force resulting on conductor is 1 1xH.
(A) 11.5°
(C) 90°
(B) 0°
(D) 1.145°
39
Copyrighted Material © 2018
Problem Set # 6.3 - Capacitance
Consult NCEES® FE Reference Handbook - Pages 200 - 202 while solving these questions
Problem 6.3 a) Calculate voltage applied across parallel plate capacitor carrying 400 [xC charge w ith a
0.02 m2 plate surface area and 0.01 m plate spacing (assume £ = 8.85 x 1 0 12 F/m).
(A) 22.6 x 106 V
(B) 2.2 x 1 0 14 V
(C )0 V
(D) 1.5 x 10 10 V
Problem 6.3 b) The capacitance o f parallel plate capacitor is 100 piF. Initial voltage across capacitor was
5 V. Calculate constant charging current if voltage across capacitor is recorded as 10 V after 3 minutes.
(A) 5.4 |JiA
(B) 2.7 nA
(C) 16.6 mA
(D) 8 mA
Problem 6.3 c) A 200 \if capacitor has voltage v(t) = 240sin(377t)V across it. Calculate the energy
stored in this capacitor as a function o f time.
(A) 0 J
(B) 5.76sin2(377t) J
(C) 5.76 sin(142129t) i
(D) 1.36sin2(377t) J
Problem 6.3 d) Calculate charging current if voltage across a 100 \xf capacitor increases by 10 V in 5 s.
(A) 1 m A
(B) 0.2 mA
(C) 2 A
(D) 4 A
Problem 6.3 e) Calculate the equivalent capacitance o f circuit shown below.
1pF
(A) 1 HF
(B) 2 nF
(C) 0.5 nF
(D) 3.5 \i¥
40
Copyrighted Material © 2018
Problem Set # 6.4 - Inductance
Consult NCEES® FE Reference Handbook - Pages 200 - 202 while solving these questions
Problem 6.4 a) W hat is the inductance o f a 1 m long coil with 100 turns and a cross sectional area of
0.1 m2 (assume \x = 4 n l0 '7 H/m)?
(A) 1.25 mH
(B )1000 H
(C) 0.625 mH
(D) 2.5 mH
Problem 6.4 b) Calculate the voltage induced in a 5 mH inductor if current in the inductor is increased
from 0 to 100 m A in 2 ms.
(A) 5 V
(B) 0.25 V
(C) I V
(D) 0 V
Problem 6.4 c} Calculate energy stored in a 100 mH inductor carrying current i(t) = t 2 (for t >0) for 10 s.
(A )100 J
(B) 10 J
(C) 2.5 J
(D) 500 J
Problem 6.4 d) Energy storage capacity of an inductor can be increased b y _________ .
(A) Decreasing voltage across it
(B) Increasing num ber o f turns
(C) Increasing its length
(D) Options A, B, C are correct
Problem 6.4 e) Calculate the equivalent inductance of circuit shown below.
1H
2H
m
(A) 6 H
(C) 4 H
(D) 8 H
41
Copyrighted Material © 2018
Chapter # 7 - Circuit Analysis
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
Electrical and Com puter Engineering
201 - 203
Kirchoff's Laws - KCL, KVL
Series/parallel equivalent circuits
Thevenin and Norton theorem s
Node and loop analysis
W aveform analysis
Phasors
Impedance
Facts about this section
• 1 0 - 1 5 questions can be expected on the exam (according to NCEES® FE Specification).
•
Circuit analysis is the most heavily weighted section on exam.
•
Difficulty level o f this section is rated 'M edium ' by the author.
Tips for preparing this section
• Understand concepts found on above mentioned pages of NCEES® FE Reference Handbook.
• Some o f the im portant equations relevant to this section include Ohm's Law, KCL, KVL, voltage
divider, current divider, Thevenin, Norton, source transform ation, average value, effective
value and RMS value.
•
Revisit basic circuit theory using your university/college electrical textbook.
•
Circuits can be solved using different methods such as KCL, KVL, Thevenin, Norton,
superposition but certain techniques may be better suited for a given problem.
•
Thevenin resistance calculation requires 'short circuiting' external voltage sources and 'open
circuiting' current sources.
•
Equivalent resistance in series Rs = R± + R 2 + R3 ..... + R n
•
Equivalent resistance in parallel Rp = 1 /(1 /Rt + 1/R2 + I/R 3 ..... + 1 /Rn)
•
Equivalent inductance in series Ls = Lt + L 2 + L 3 ..... + L n
•
Equivalent inductance in parallel Lp = 1/(1/Lt + 1 / L 2 + 1 / L 3 ..... + 1 / L n)
•
Equivalent capacitance in parallel Cp = Ct + C2 + C3 .....+Cn
•
Equivalent capacitance in series Cs = 1 / ( 1 / ^ + 1/C2 + 1 /C 3 .....+1 /Cn)
•
Sinusoidal signals require conversion to a standard form (cosine) for phasor representation.
•
Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification details can be found on pages 265-267 of NCEES® FE Reference Handbook.
42
Copyrighted Material © 2018
Problem Set # 7.1 - Kirchoffs Laws - KCL, KVL
Consult NCEES® FE Reference Handbook - Pages 201 - 203 while solving these questions
Problem 7.1 a) Calculate voltage across 10 kQ resistor in the circuit shown below.
(A) 70.5 V
(B) -14.6 V
(C) 1.46 mV
(D) 5V
Problem 7.1 b) Calculate current passing through 2 kQ resistor in the circuit shown below.
2ku ?
6ka
10mA
(A) 4 mA
(B) 3 mA
(C) 2 mA
(D) 6 mA
r/3
Problem 7.1 c) Calculate voltage across 3 Q resistor in the circuit shown below.
4Q
(A) 2.5 V
(B) 1.87 V
(C) 7.5 V
(D) 3.75 V
43
Copyrighted Material © 2018
Problem 7.1 d) Calculate current lx in the circuit shown below.
m
(A) 4 A
(B) 3.5 A
(C) 7 A
(D) 1.7 A
Problem 7.1 e) Calculate current passing through 1 kO resistor in the circuit shown below.
sm
(A) 3.1 mA
(B) 5 mA
(C) 2.5 mA
(D) 1.5 mA
Problem 7.1 f) Calculate current passing through 5 kQ resistor in the circuit shown below.
10k0
2ko
44
Copyrighted Material © 2018
Problem Set # 7.2 - Series / Parallel Equivalent Circuits
Consult NCEES® FE Reference Handbook - Pages 201 - 203 while solving these questions
Problem 7.2 a) Find the equivalent resistance between term inals A-B o f the circuit shown below.
iok0
AA/V
m
B
(A) 5 kQ
(B) 7.5 kQ
(C) 2 kQ
(D) 3.4 kQ
Problem 7.2 b) Find the equivalent resistance between term inals A-B o f the circuit shown below.
(A) 2 kQ
(B) 3 kQ
(C) 1 kQ
(D) 5 kQ
Problem 7.2 c) Find the equivalent resistance between term inals A-B o f the circuit shown below.
ikfl
5kQ
AAAr
m
(A) 7 kQ
(B) 3.5 kQ
(C) 1.5 kQ
(D) 5 kQ
10kQ
45
Copyrighted Material © 2018
Problem 7.2 d) Find the equivalent resistance between term inals A-B of the circuit shown below.
5kfl
4kfl
AAAr
2kfi
AAAr
2kft
1kO
(A) 3.0 kQ
(B) 2.75 kQ
(C) 0.825 kQ
(D) 5.50 kQ
Problem 7.2 e) The equivalent resistance between term inals A-B of the circuit shown below is
1*0
2k£l
iska
ma
•AAAr
sm
5k£i
46
Copyrighted Material © 2018
Problem Set # 7.3 - Thevenin and Norton Theorems
Consult NCEES® FE Reference Handbook - Page 201 - 203 while solving these questions
Problem 7.3 a) Calculate voltage Voc in the circuit shown below using Thevenin theorem.
2m
zm
(A) 10 V
(B) 5 V
(C )37 V
(D) 100 V
Problem 7.3 b) Thevenin resistance of the circuit given in problem 7.3 a) is
Problem 7.3 c) Calculate Thevenin resistance of the circuit shown below.
(A) 1.25 kQ
(B) 5 kQ
(C) 2.5 kQ
(D) 3 kQ
47
Copyrighted Material © 2018
Problem 7.3 d) Calculate short circuit current (lsc) for lx in the follow ing network using Norton theorem .
(A) 7.5 mA
(B) 15 mA
(C) 5 mA
(D) 12 mA
Problem 7.3 e) Req of the circuit given in problem 7.3 d) is
48
Copyrighted Material © 2018
Problem Set # 7.4 - Waveform Analysis
Consult NCEES® FE Reference Handbook - Pages 201 - 203 while solving these questions
Problem 7.4 a) Calculate maximum voltage o f a full wave rectified sinusoid if its effective value is 10
(A) 20.5 V
(B) 5 V
(C) 1 4.1V
(D) 10V
Problem 7.4 b) Calculate sum of follow ing sinusoids:
V i = 10cos(500t)
(A) 25cos(600t
+ 45)
V2 = 15cos(100t + 45°)
(B) 10cos(500t) -
(C) 150cos(500t)cos(100t + 45)
10.6cos(100t) + 1 0 .6 sin (1 0 0 t)
(D) 1 0 c o s(5 0 0 t) + 10.6cos(100t)
-
10.6sin(100t)
Problem 7.4 c) Calculate frequency o f sinusoidal signal given by 1 0 0 c o s (5 0 0 t + 50°).
(A) 500 Hz
(B) 250 Hz
(C) 79.5 Hz
(D) 50 Hz
Problem 7.4 d) Calculate average value o f a h alfw ave rectified signal given by 15cos(100t + 50°).
(A) 7.5
(B) 4.77
(C) 9.54
(D) 10.6
Problem 7.4 e) The average value of periodic current waveform shown below is ____ .
Am ps
2
1
2
3
4
Time
49
Copyrighted Material © 2018
Problem Set # 7.5 - Phasors
Consult NCEES® FE Reference Handbook - Pages 201 - 203 while solving these questions
Problem 7.5 a) Calculate phasor current o f circuit shown below.
(A) 100/60° A
(B) 12.5/-3Q0 A
(C) 7Q.7/-3Q0 A
(D) 1.87-120° A
Problem 7.5 b) Express v ( t ) = 2 1 2 sin(a)t + 50)1/ in phasor form.
(A) 1507-40° V
(B) 212/50° V
(C) 150/50° V
(D) 212/-4Q0 V
Problem 7.5 c) Calculate phasor current of circuit shown below.
5000
(A) 0.14/349° A
(C) 0.5/0^ A
(B) 0.14/11.3° A
(D) 1.75/50° A
50
Copyrighted Material © 2018
Problem 7.5 d) Calculate phasor current passing through the capacitor in circuit shown below.
(A) 70/10° A
(B) 10007-80° A
(C) 2665/100° A
(D) 55Q/-1Q0 A
Problem 7.5 e) The frequency domain impedance 'Z' of circuit given below is
50
*1OJ0
2m
51
Copyrighted Material © 2018
-p a
Problem Set # 7.6 - Impedance
Consult NCEES® FE Reference Handbook - Pages 201 - 203 while solving these questions
Problem 7.6 a) Determine equivalent impedance of circuit shown below.
(a) 1 2 - j d,
(B) 5 + 2y n
(C) 2 - 3j S l
(D) Oj Cl
Problem 7.6 b) Determine equivalent impedance o f circuit shown below.
20
(A) 5 - 3jS l
(B) 2 - 3j £ l
(C) 1 + 2jS l
(D) 10 j a
52
Copyrighted Material © 2018
Problem 7.6 c) Determine equivalent impedance of circuit shown below (assume 60 Hz frequency).
100nF
X
50ft
(A) 50 + j 26525 SI
(B) 50 - ;2 6 5 2 5 O
(C) 50 + ;5 3 5 f l
(D) 50 - ;5 3 5 £1
Problem 7.6 d) Determine equivalent impedance o f circuit shown below (assume 60 Hz frequency).
10a
100{j F
2mH
(A) (10+7*2 a ) | | ( - ; 1 0 0 ft)
(B) (10 + y'0.5 n )| |( —;1 0 0 ft)
(C) (10 + J0.75 a)\\(-j26.5 n )
(d) i o a
53
Copyrighted Material © 2018
Chapter # 8 - Linear Systems
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
Electrical and Com puter Engineering
203 - 204
M athem atics
34
Frequency / transient response
Resonance
Transfer functions
2-port theory
Laplace Transforms
Facts about this section
•
5 - 8 questions can be expected on the exam (according to NCEES® FE Specification).
•
Difficulty level of this section is rated 'Easy' by the author.
Tips for preparing this section
•
Understand concepts found on above m entioned pages of NCEES® FE Reference Handbook.
•
Some o f the im portant equations relevant to this section include RC and RL transient circuits,
RLC series and parallel resonance circuits, Laplace Transforms pairs and Two-port parameters.
•
RLC series and parallel resonance circuits are very sim ilar therefore students need to be careful.
•
Initial conditions o f RC and RL transient circuit shall be established carefully.
•
Transfer functions shall be converted to a standard form to calculate gain, poles and zeros.
•
Develop fam iliarity with Laplace Transform pairs.
•
Two-port parameters require proper enforcem ent of voltage and current conditions.
•
Learn how to perform partial fraction expansion w hile calculating inverse Laplace transforms.
•
Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification details can be found on pages 265-267 o f NCEES® FE Reference Handbook.
54
Copyrighted Material © 2018
Problem Set # 8.1 - Frequency / transient response
Consult NCEES® FE Reference Handbook - Pages 203 - 204 while solving these questions
Problem 8.1 a) Assume that switch shown in circuit below has been in indicated position for a long
time. Calculate voltage across capacitor vc( t ) 5 m inutes after switch changes position at t = 0 s.
Switch ch a n g e s position at t = Os
^
— ,------- -
100KQ
ImF
(A) 15V
(B) 10V
(C) IV
(D) 3V
Problem 8.1 b) Calculate voltage across capacitor vc( t ) for t >0 in the circuit shown below.
(A) e~t/2V
(B) 1 0 e -^ 2V
(C) 10e_t/4V
(D) Se~t/2V
55
Copyrighted Material © 2018
Problem 8.1 c) Calculate the current i ( t ) in follow ing circuit if switch closes at t = 0 s.
Switch closes at I * Os
—
1k£l
#«*
10V
2mH
(A) 10(1 - e ~ ) A
(B) 1 - g - 500000^
(C) e~500000t + 0.01(1 - e - 500000t) A
(D) 0.01 (1 - e- 500000t) A
Problem 8.1 d) Calculate current i( t ) ten tim e constants after opening the switch. Assume that switch
has been in shown position for a long time.
Switch©pansit i~©$
(A) 0.5 A
(B) 22.6 \iA
(C) 1 A
(D) 0.18 A
Problem 8.1 e) Assume that switch shown in circuit below has been in indicated position for a long
time. Find the voltage across capacitor vc( t ) after 5 tim e constants if it changes position at t = 0 s.
Switch changes position at t *■0s
10V
5mF
lOkO
(A) 9.99 V
(B) 2 V
(C) 67 mV
(D) 35 mV
56
Copyrighted Material © 2018
Problem Set # 8.2 - Resonance
Consult NCEES® FE Reference Handbook - Pages 203 - 204 while solving these questions
Problem 8.2 a) Calculate resonant frequency of RLC circuit shown below
lOpF
20mH
1.00Q
(A) 1414 rad/s
(B) 250 rad/s
(C) 500 rad/s
(D) 2236 rad/s
Problem 8.2 b) The band W idth (BW) of circuit shown in problem 8.2a) is
Problem 8.2 c) Calculate maximum current of RLC circuit shown below.
IQjjF
tOQ
10mH
(A) 4 A
(B) 12 A
(C) 5 A
(D) 7.5 A
Problem 8.2 d) The frequency at which maximum current occurs in problem 8.2c) is
57
Copyrighted Material © 2018
Problem 8.2 e) Calculate the current magnification factor o f RLC circuit shown below.
(A) 250
(B) 0.707
(C )500
(D )707
Problem 8.2 f) The band width (BW) o f circuit shown in problem 8.2 e) is
58
Copyrighted Material © 2018
Problem Set # 8.3 - Laplace Transform
Consult NCEES® FE Reference Handbook - Page 34 while solving these questions
Problem 8.3 a) Find the Laplace transform o f follow ing function.
f { t ) = e~bt ( b > 0)
(A) —
(B) ——
(C) —
(D) —
' s+ b
'
s -b
'
1 s+ b
1 s -b
Problem 8.3 b) Find the Laplace transform of follow ing function.
f ( t ) = e~~atu ( t — 1)
1
p -(s+ a )
(A) —
(B) ---------
'
' s+a
s+ a
e2
p - is - a )
(C) —
(D) --------
' s -a
s -a
Problem 8.3 c) Find the Laplace transform o f follow ing function.
/ ( t ) = te~atu ( t — 1)
. .
s+ a
' * (s+a)2+ l
e-(s+a)
(
-(s+a)/( --------1
1 ^
(C) e~ts+a)
1--------)
1 j
\ s + a )2
} (s -a )2
(D)
5
K ’ (s+a)2
(s + a Y
Problem 8.3 d) Find the Laplace transform o f follow ing function.
f(t) =
(A)e3S^
+^
— e_3('t“ 3'))u (t — 3)
(B)
(S+lf(S+3)
Problem 8.3 e) Find the Laplace transform of follow ing function.
f i t ) = te~ac8 (t - 2)
(A) 2<T*»+»>
,
.
(C) e “ (5+a)
'
'
(B)
p -is+ a )
(D)
s+ a
—
59
Copyrighted Material © 2018
Problem 8.3 f) Find the inverse Laplace transform o f follow ing function:
F (s ) =
(A) 3 e -3t + Se~5t
5
(s + 3) + (s + 5)
(B) e~4t
(C) 2e~et
( D ) je “ 4t
Problem 8.3 g) Find the inverse Laplace transform o f follow ing function:
s+8
F (s )
(A) e~8t +
(s + l) ( s + 7)
— e~7t
(C) - (7 e - t — e~7t)
6
(B) e~z + e~7t
(D) Se~7t — 8 e _t
Problem 8.3 h) Find the inverse Laplace transform o f follow ing function:
F{s) =
S 2 + 25 + 1
(s + 2 ) ( s ) ( s + 3)
(A )-^ - — - e 2t + - e 3t
(B) e 2t + e 3 t+ -
(C) te~2te~3t
(D) 6{t)(e-2t + e " 3t)
6
2
3
3
Problem 8 .3 1) Find the inverse Laplace transform o f follow ing function:
F (s ) =
(A)5e~ 5t + i + j
(C) ~ u { t ) + ^ t ~ ^ e ~ 5t
s+ 4
( s 2) (s + 5)
(B) ^e~5 t-h~t
(D) 25 u ( t ) - e~5t
60
Copyrighted Material © 2018
Problem Set # 8.4 - Transfer functions
Consult NCEES® FE Reference Handbook - Pages 203 - 204 while solving these questions
Problem 8.4 a) Calculate voltage transfer function for the circuit shown below.
R
(A)
(C)
(B)ll(sL)
(R+si.)ll(^)
w if e )
(B)
R + (R + S «I| (£ )
(R + sL)
(D)
(R + s « l| (i)
R+ (R + St ) | | ( i )
(R + s lO llfe )
Problem 8.4 b) Calculate voltage transfer function for the circuit shown below.
Ft
Vo
R+
(A) (R\\sL)
(B)
(K+s )
(R U s L )
k(R+£)
(D) m sl)+(R+±)
R+
(R||SL ) + ( R + i )
61
Copyrighted Material © 2018
Problem 8.4 c) Calculate input impedance transfer function for the circuit shown below.
r
n
(A) R + ^ \ K R + sLl\R)
(B) fi+i||si||fl
(C)i||sZ,||R
( D ) R + ± + sL\\R
Problem 8.4 d) Calculate zero(s), pole(s) and magnitude gain o f follow ing transfer function:
20
H
~ (s)(1 0 s + 1 )
(A) Poles @ 0 and 10, Zero - none, Gain = 20 dB
(B) Poles @ 0 and 0.1,Zero - none, Gain = 26 dB
(C) Pole @ 0, Zero @ 10, Gain = 2 dB
(D) Pole - none, Zeroes @ 0and 0.1, Gain = 6 dB
Problem 8.4 e) Calculate zero(s), pole(s) and magnitude gain o f follow ing transfer function:
10
0 ) 2(5s + l )
(A) Poles @ 0 & 5, Zero - none, Gain = 10 dB
(B) Poles 1st order @ 5, 2nd order @ 0, Zero - none, Gain = 10 dB
(C) Pole 2nd order @ 0, Zero - none, Gain = 10 dB
(D)Poles 1st order @ 0.2, 2nd order @ 0, Zero - none, Gain = 20 dB
62
Copyrighted Material © 2018
Problem Set # 8.5 - Two-Port Theory
Consult NCEES® FE Reference Handbook - Pages 203 - 204 while solving these questions
Problem 8.5 a) Calculate z-parameters of tw o-port network shown below.
2Q.
—
j-j
.... •^►n*
V1
io
20
V2
20
(A) Zll=Z22=4 Q, Zl2=Z21=2 Q
(B) Zi i =Z22=2 Q, Zl2=Z21=2 Q
(C) Zll=Z22=4 Q, Zl2=Z21=4 Q
(D) Zll=Z22=l Q, Zl2=Z21=4 Q
Problem 8.5 b) Calculate y-parameters of tw o-port network shown below.
(A) yn=y22=S/30, y i 2=y2i= -S/10
(B) yn=y 22=S/5 , y i 2=y2i= -S/20
(C) yn=y 22=S/5 , y12=^21= -S/10
(D) yn=y 22=S/10 / y12=^21= -S/10
Problem 8.5 c) Calculate y-parameters o f tw o-port network shown below.
28
11
12
2D
V2
(A) yn=y 22=S/3 / y i 2=y2i= -S/6
(B) yn=y 22=S/6 , y12=^21= -S/6
(C) yn=y 22=S/6/ y12=^21- -S/3
(D) yu=y 22=S/6 / y i2=y2i= -S/3
63
Copyrighted Material © 2018
Problem 8.5 d) Z-param eter z n of tw o-port network shown below is
11
VI
100
20fl
...... W V ................. V W
^
200
12
¥2
15S
Problem 8.5 e) H-parameter h i 2 of tw o-port network shown below is
¥t
200
aon
64
Copyrighted Material © 2018
V2
Chanter # 9 - Signal Processing
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
Convolution
206
Difference equations
Z-transforms
Electrical and Com puter Engineering
Sampling
Analog filters
209
2 1 0 -2 1 1
207
Digital filters
Facts about this section
•
5 - 8 questions can be expected on the exam (according to NCEES® FE Specification).
•
Difficulty level of this section is rated 'M edium ' by the author.
•
Students with a major in com m unication engineering may find this section
easier.
Tips for preparing this section
•
Understand concepts found on above mentioned pages of NCEES® FE Reference Handbook.
•
Some o f the im portant equations relevant to this section include continuous tim e convolution,
discrete tim e convolution, z-transform, and Nyquist theorem .
•
Graphical convolution of continuous tim e functions involves flipping sim pler function about yaxis, determ ining regions o f overlap, identifying limits o f integration and correctly calculating
integrals.
•
Graphical convolution of discrete tim e functions involves flipping sim pler function about y-axis,
determ ining regions o f overlap and calculating integrals.
•
Develop fam iliarity with Z transform table and its application in difference equation.
•
Identify signal with highest frequency to correctly apply Nyquist theorem .
•
Calculate transfer functions of analog filter circuits given on pages 210-211 o f NCEES® FE
Reference Handbook to develop understanding.
•
Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification details can be found on pages 265-267 of NCEES® FE Reference Handbook.
65
Copyrighted Material © 2018
Problem Set # 9.1 - Continuous Time Convolution
Consult NCEES® FE Reference Handbook - Page 206 while solving these questions
Problem 9.1 a) W hich o f the follow ing options correctly represents convolved output o f x(t) and h(t)?
m
h(t)
x(t) = 4 (u(t) - u (t — 4))
y(t) = 2(u(t) - u ( t - 2))
66
Copyrighted Material © 2018
Problem 9.1 b) W hich of the follow ing options correctly represents convolved output of f(t) and h(t)?
|f ( t ) = etu{—t)
h(t ) = 3e tu(t )
(A) ^etf o r t < 0, ^e~ff o r t > 0
(B) - e2tf o r t < 0 , —^e~2tf o r t > 0
(C) ^ e_t/ o r t < 0, j etf o r t > 0
(D) Of or t < 0, 3 e lf o r t > 0
Problem 9.1 c) W hich of the follow ing options correctly represents convolved output of x(t) and y(t)?
x ( t ) = s in (t) ( u ( t) — u (t — 7r))
y ( t ) = u ( t ) — u ( t — ri)
(A) co s(t) 0 < t < n, —co s ( t ) n < t < 2 n
(B) s in (t) 0 < t < n, 1 — sin (t) tc < t < 2 n
(C) co s(t) — 1 0 < t < 7 r , 1 — co s(t) n < t < 2n
(D )l — co s(t) 0 < t < n , 1 — co s(t) n < t < 2 n
67
Copyrighted Material © 2018
Problem 9.1 d) W hich o f the follow ing options correctly represents convolved output of x(t) and y(t)?
x(t)
3
yft)
■2
3
5
t
4
x ( t) = 3 {u (t — 3) — u ( t — 5))
(A) 6(1 + t)
(B) 1 + 6 t
(C) 6(1 + t)
(D) 6 (t)
— 1 < t < 1, 1 — 6 1 f o r
fo r
/o r
fo r
t
y ( t ) = 2 ( u ( t + 4) — u ( t + 2))
- 1 < t < 1, 6(1 - t) f o r
fo r
“2
1 < t < 3
— 1 < t < 1, 6(3 — t) f o r
— 1 < t < 1, 3 — t f o r
1< t < 3
1 < t < 3
1 < t <3
Problem 9.1 e) Find the convolved output of follow ing functions for t > 2.
yW
y ( t ) =u(t) - u ( t - 2)
(B) 2{el -
(A) 0
(C) 2(e~t+2 -
e~l )
t _
(D )e. 1
e - fc)
g —t —2
68
Copyrighted Material © 2018
Problem Set # 9.2 - Discrete Time Convolution
Consult NCEES® FE Reference Handbook - Page 206 while solving these questions
Problem 9.2 a) Find convolved output of discrete tim e functions x[n] and y[n] shown below.
xfn ]
3
1
I1
2
*
1
3
11
2
1
M
1
f
1
2
;I
i
n
(A) [0 4 10 4 0]
(B) [0 4 10 14 10 4 0]
(C) [0 2 4 8 4 2 0]
(D) [ 0 2 6 2 0 ]
2
Problem 9.2 b) Find convolved output of discrete tim e functions x[n] and y[n] shown below.
3
ir
x{n]
2
3
yW
2
l1
1
1
ii
|
ii
I
i!
1
?
i i ..... ............
1
2
:1
n
(A) [ 0 1 4 6 4 10]
(B) [ 0 1 4 10]
(C) [0 1 4 8 8 3 0]
(D) [0 1 8 8 1 0]
Problem 9.2 c) Discrete tim e convolution is used to fin d _____________
(A) Zero input solution
(B) Zero state solution
(C) Product o f signals
(D) M odulation index
Problem 9.2 d) Find convolved output of x[n] = u[n] — u[n — 5] and y[n ] = 0.2nu[n].
(A) 0.25
{cm k==00-2n" fe -
(B) —0.25(0.2n - 0.2n~6)
0.2n~k
(D)
0.2"-*
69
Copyrighted Material © 2018
3
n
Problem 9.2 e) Find convolved output of x[n] = u[n — 2] and y[n] = 0Anu[n].
(A) ~0.4n
(C) E K ? 0 .4 n- '£
;
:
(D) 0.42 E ^ 0 . 4 ~ fc
70
Copyrighted Material © 2018
Problem Set # 9.3 - Z Transforms
Consult NCEES® FE Reference Handbook - Page 206 while solving these questions
Problem 9.3 a)____________is used to solve a difference equation modeling discrete tim e system.
(A) Laplace Transform
(B) Discrete Convolution
(C) Z Transform
(D) Integration
Problem 9.3 b) Find the z-transform o f x[n] = u[n ] — u[n — 5].
(A) z _1 + z -2 + z -3 + z -4
(B) 1 + z _1 + z~2 + z~3
+ z~4
(C) 5z “ 5
(D) 1 + z -1 + z~2 + z -3 + z " 4 + z -5
Problem 9.3 c) Find the z-transform of x [n] = 0.2nu[n].
(A) (0.2 z - 1) 71
(B)
0.2nz _1
(C) 0 . 2 / ( l - z " 1)
(D)
1 /(1 - 0 .2 Z "1)
Problem 9.3 d) Find the z-transform o f x[n] = [2 3 1 0 5 ] .
(A) 2 + 3 z _1 + z~2 + 5z~4
(B) 1 /(2 + 3 z -1 + z~2 + 5z~4)
(C) 5(1 — 0.75nz -1 )
(D) 2 + 3 z
Problem 9.3 e) Find the z-transform o f x [n] = 5(0.75n)u[n].
(A) 5 /(1 - 0.75Z"1)
(B) 5(0.75nz " n)
(C) 5(1 - 0.75nz _1)
(D) 0.5nz “ n
Problem 9.3 f) Find the inverse z-transform o f X (z ) =
'
K
J
2-0.5
(A) 1 /(1 - 0 .5 z-1 )
(B) 0.5n_1u[n]
(C) 0.5nw[n]
(D) 1 - 0.5n
Problem 9.3 g) Find the inverse z-transform o f X (z ) =
Sz+2
(A) ^ (4 nu[n] — 1nu[n] -1- £[n])
(B)
(C) ^4 nu[n] + i u[n ]
(D)^S[n] - ^u[ n] + ^ 4 nw[n]
71
Copyrighted Material © 2018
cos[n] — 4 nu[n]
Problem 9.3 h) Find the inverse z-transform o f follow ing function
, ,
(z - 2) ( z + 1)
(z - 0.1) (z - 0.2)
(A)
i o . l " - 2(0.2 ) nu[n]
(B) - 1 0 0 £ [ n ] - 2 0 9 (0 .1 )nU [n] - 108(0.2 ) nu[n]
(C)
i5 ( n ) +
(D) 0.55[n] + 2(0.1)nu[n]
0.1"u [n ] + 0.3 nu [n]
Problem 9.3 i) Find the inverse z-transform o f follow ing functions
(z + 0.5)
(z — 0.1) (z + 0.4)
(A) - — 8[n] + 12(0.1)nu[n] + - ( —0.4)"u[n]
2
2
(B) - 0 .1 nu[n] + - ( - 0 . 4 ) nu [n l
4
(C) 2S[n] + 6(0.1)nu[n] — ~(0.4)nu[n]
2
(D) -(0 .1 )nw[n] + -0 .4 nu[n]
4
2
72
Copyrighted Material © 2018
2
Problem Set # 9.4 - Sampling
Consult NCEES® FE Reference Handbook - Pages 209 while solving these questions
Problem 9.4 a) The minimum sampling frequency required for perfect reconstruction of signal given
below is ___________.
sinc(1000nt) + sinc(2QQ0nt )
Problem 9.4 b) Sampling continuous tim e signal cos(30007zt + 6) at 2000 Hz will result in ________
(A) Nyquist sampling
(B) Critical sampling
(C) Aliasing
(D) Oversampling
Problem 9.4 c) Determine aliased frequency o f x ( t ) = cos(5007rt + 6) if it is sample at 500 Hz.
(A) 500 Hz
(B) 250 Hz
(C) 0 Hz
(D) There will be no aliasing
Problem 9.4 d) Determine aliased frequency of x ( t) = cos(4007Zt + O') if it is sampled at 300 Hz.
(A) 200 Hz
(B) 300 Hz
(C) 100 Hz
(D) 400 Hz
Problem 9.4 e) For com plete reconstruction o f a band limited signal, it must b e _______________
(A) Sampled at or above
Nyquist rate
(B) Passed through ideal low pass filter
(C) M odulated with highfrequency signal
(D) Options A and B are
73
Copyrighted Material © 2018
correct
Problem Set # 9.5 - Filters
Consult NCEES® FE Reference Handbook - Pages 207, 210 and 211 while solving these questions
Problem 9.5 a) Identify the filter type given by transfer function shown below:
H (jw ) =
100
1 + 20 jco
(A) High Pass Filter
(B) Low Pass Filter
(C) Band Pass Filter
(D) Band Reject Filter
Problem 9.5 b) Identify the filter type given by transfer function shown below.
H 0 w ) = ------(ja>y + £00 + 1000
(A) High Pass Filter
(B) Low Pass Filter
(C) Band Pass Filter
(D) Band Reject Filter
Problem 9.5 c) Identify the filter type given by transfer function shown below
, ,
20 j<o
H(ja)) = 1 0 0 0 — — —
1
+ 20 jo)
(A) High Pass Filter
(B) Low Pass Filter
(C) Band Pass Filter
(D) Band Reject Filter
Problem 9.5 d) Identify the filter type given by transfer function shown below
3 0 0 (0 a ))2 + ^ )
H (jw ) =
O’to)2 + 100/6) + i
50
(A) High Pass Filter
(B) Low Pass Filter
(C) Band Pass Filter
(D) Band Reject Filter
Problem 9.5 e)________ is a type of non-recursive digital filter.
(A) Finite Impulse Response Filter
(B) Infinite Impulse Response Filter
(C) Options A and B are correct
(D) Non-recursive filters don't exist
74
Copyrighted Material © 2018
Problem 9.5 f) ________ is a type o f recursive digital filter.
(A) Finite Impulse Response Filter
(B) Infinite Impulse Response Filter
(C) Options A and B are correct
(D) Recursive filters don't exist
Problem 9.5 g) Digital filtering does not in v o lv e _____________ .
(A) Sampling
(B) A/D conversion
(C) D/A conversion
(D) Phase M odulation
75
Copyrighted Material © 2018
Chapter # 10 - Electronics
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
Solid-state fundam entals
212
Discrete devices
Bias circuits
2 1 4 -2 1 7
Electrical and Com puter Engineering
Am plifiers
Operational Am plifiers
2 1 2 -2 1 3
212
Instrumentation
Power Electronics
2 1 4 -2 1 7
Instrumentation, M easurem ent and Control
124 - 1 2 6
Note: Specific details are not found in NCEES® FE Reference Handbook
Facts about this section
•
7 - 1 1 questions can be expected on the exam (according to NCEES® FE Specification).
• Difficulty level o f this section is rated 'Hard' by the author.
© Students with a major in electronics engineering may find this section easier.
Tips for preparing this section
•
Understand concepts found on above m entioned pages of NCEES® FE Reference Handbook.
•
Some of the im portant equations relevant to this section include conductivity of sem i­
conductors, contact potential, diode Shockley equation, BJT mathematical relationships, JFETs
and MOSFETs mathematical relationships.
• Learn how to calculate Q-point of diodes, BJTs, JFETs and MOSFETs.
• BJT am plification takes place in 'active region'.
•
JFET/MOSFET amplification takes place in 'saturation region'.
•
Derive equations for two-source configuration o f an ideal Operational Am plifier and CMRR
given in NCEES® FE Reference Handbook (page 212).
•
Familiarize yourself with workings o f measurem ent devices such as RTDs, thermocouples,
W heatstone bridge, ammeters, voltm eters etc.
•
Review basic concepts related to pow er electronics using your college/university textbooks.
•
Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification can be found on pages 265-267 o f NCEES® FE Reference Handbook.
76
Copyrighted Material © 2018
Problem Set # 10.1 - Solid-State Fundamentals
Consult NCEES® FE Reference Handbook - Pages 212 while solving these questions
Problem 10.1 a)
is typically used as n-type doping material.
(A) Antim ony
(B) Silicon
(C) Germanium
(D) Boron
Problem 10.1 b)
is typically used as p-type doping material.
(A) Antim ony
(B) Silicon
(C) Germanium
(D) Boron
Problem 10.1 c) Intrinsic carrier concentration of a sem i-conductor wafer at equilibrium is ni = 2 x 1010
rrr3. Electron and hole m obilities are \xn= 0.20 m2V ^ s1 and ^p=0.1 m2V ^ s1 respectively. Calculate
conductivity of this sem i-conductor wafer.
(A) 4.8 x 1 0 10 S/m
(B) 9.6 x 1 0 10 S/m
(C) 1.4 x 1 0 9 S/m
(D)7.2 x 1 0 10 S/m
Problem 10.1 d) A p-n junction has acceptor concentration 2 x 1015 rrr3, donor concentration o f 2 x
1015rrv3 and intrinsic concentration o f 1 x 1010rrr3. Calculate contact potential of this p-n junction at
300 K.
Problem 10.1 e) Which o f the follow ing class of materials has largest energy gap between valence band
and conduction band?
(A) Semi-conductors
(B) Conductors
(C) Insulators
(D) All materials have same energy gaps
Problem 10.1 f) Drift velocities of charged carriers in sem i-conductors (at constant tem perature) can
be increased b y _________________ .
(A) Increasing applied potential
(B) Decreasing applied potential
(C) Increasing surface area
(D) Decreasing surface area
Problem 10.1 g) G ro u p ______ elem ents of periodic table are typically used as n-type doping materials.
Problem 10.1 h) G ro u p _____ elements of periodic table are typically used as p-type doping materials.
Problem 10.1 i) G ro u p _____ elements of periodic table are typically used as sem i-conductor materials.
77
Copyrighted Material © 2018
Problem Set # 10.2 - Diodes
Consult NCEES® FE Reference Handbook - Pages 214 while solving these questions
Problem 10.2 a) Find operating states of diodes D i and D2 in follow ing circuit assuming ideal model.
nu
01
m
5V
(A) D i On, D2 O ff
(B) D i On, D2 On
(C) D i Off, D2 On
(D) D i Off, D2 Off
Problem 10.2 b) Select the graph that accurately approxim ates V o u t for follow ing circuit.
78
Copyrighted Material © 2018
Problem 10.2 c) Find the Q point of diode shown below assuming ideal model
4V
(A) (0.7 V, 0.5 mA)
(B) (0 V, 0.66 mA)
(C) (0 V, 1 mA)
(D) (0 V, 1.33 mA)
Problem 10.2 d) Calculate the diode current in follow ing circuit assuming V 0n = 0.7 >
o10V
AAAr
-o
10k0
8V
Problem 10.2 e) Find diode currents in follow ing circuit assuming Von = 0.7 V.
D1
2M
5m
(A) Id i = 0, ID2 - 0
(B) Idi - 0, Id2= 2.4 mA
(C) Id i = -2.5 mA, Id2= 2.5 mA
(D) Idi - 0, Id2= ~2.5 mA
79
Copyrighted Material © 2018
Problem Set # 10.3 - BJTs
Consult NCEES® FE Reference Handbook - Page 215 while solving these questions
Problem 10.3 a) Calculate em itter current Ie and voltage V c e for BJT shown in the circuit given below.
10V
2§Cfi
.ate
(A) lE = 0, V ce = 0.5 V
(B) lE =0, V CE = 1.75 V
(C) lE = 2.09 mA, V CE = 1.68 V
(D) lE = 1. 5mA, V ce = 2.5 V
Problem 10.3 b) Calculate collector current lc and voltage V ce for BJT shown in the circuit given below.
TV
m
p«too
mo,
(A) lc = 0 mA, V CE= 0.5 V
(B) lc = 1.5 mA, VCE= 2 V
(C) lc = 0.75 mA, V CE= 0.1 V
(D) lc = 1.13 mA, VCE= 1.3 V
80
Copyrighted Material © 2018
Problem 10.3 c) Determine the state o f transistor shown in the circuit given below.
Problem 10.3 d) Calculate em itter current Ie and voltage V ce for BJT shown in the circuit given below.
10V 0
W q
(A) lE = 2.4 mA, VCE= 2.7V
(B) lE =0 mA, V Ce = 0.2V
(C) lE =0 mA, V Ce= 10V
(D) Ie = 5 mA, VCE = 4.5V
81
Copyrighted Material © 2018
Problem 10.3 e) Calculate em itter current Ie and voltage V Ec for BJT shown in the circuit given below.
(A) lE = 0 mA, V EC= 5 V
(B) lE = 0 mA, V EC = 10 V
(C) lE = 3.2 mA, V EC= 5.1 V
(D) lE = 1.5 mA, V EC= 7.5 V
Problem 10.3 f) Calculate em itter current band voltage Vec for BJT shown in the circuit given below.
(A) |E = 0 A, Vec = 0 V
(B) lE = 7.75 mA, V EC= 5V
(C) lE = 3.3 mA, Vec = 2 V
(D) lE = 1.5 mA, V EC= 2 V
82
Copyrighted Material © 2018
Problem Set # 10.4 - MOSFETs
Consult NCEES® FE Reference Handbook - Page 217 while solving these questions
Problem 10.4 a) Calculate drain current ip for follow ing circuit (assume K = 0.5 m A/V2 & Vt = 1 V).
(A) 0 mA
(B) 1.5 mA
(C) 2.5 mA
(D) 3.2 mA
Problem 10.4 b) Find the operational state of transistor shown below (assume K = 0.5m A/V2 V j =
m
sv
IV
8m
AA/V
Q 6V
83
Copyrighted Material © 2018
Problem 10.4 c) Calculate V d s for MOSFET circuit shown below (assume K = 0.2 m A/V2 & V t = 1 V).
(A) 0. 5 V
(B) 3.28 V
( C )0 V
(D) 1.75 V
Problem 10.4 d) Calculate drain current ip for follow ing circuit (assume K = 0.1 m A/V2 & V j = 1 V).
(A) 5 mA
(B) 0.3 mA
(C) 0 .1 m A
(D )0 m A
84
Copyrighted Material © 2018
Problem 10.4 e) Calculate drain current iD for circuit shown below (assume K = 0.1 m A/V2 & V t = 1 V).
4V r\
10V
(A) 0 mA
(B) 0.44 mA
(C) 1.5 mA
(D) 0.75 mA
85
Copyrighted Material © 2018
Problem Set # 10.5 - Operational Amplifiers
Consult NCEES® FE Reference Handbook - Page 212 while solving these questions
Problem 10.5 a) Find the output voltage Vo in the circuit shown below.
mm
Problem 10.5 b) Find the value o f resistance 'R' in the circuit shown below if Vo is 12 V.
R
Problem 10.5 c) Find the output voltage Vo in the circuit shown below.
I te
86
Copyrighted Material © 2018
Problem 10.5 d) Find the output voltage V0 in the circuit shown below.
sow*
nm
Problem 10.5 e) Find the output voltage V0 in the circuit shown below.
1ma
§ua
87
Copyrighted Material © 2018
Problem Set # 10.6 - Instrumentation
Consult NCEES® FE Reference Handbook - Pages 124 -126 while solving these questions
Problem 10.6 a)___________is not an example o f a transducer.
(A) M icrophone
(B) Therm ocouple
(C) Vernier Calliper
(D) Photo diode
Problem 10.6 b) A 200 Q RTD has a tem perature coefficient o f 0.0039 ° C 1 at room tem perature (20°C).
Find the new resistance o f this RTD if it is placed in a 35°C environment.
(A) 188 Q
(B) 211 Q
(C )200 Q
(D )255 Q
Problem 10.6 c) Determine Rx if W heatstone bridge is balanced and Ri= 100Q, R2 = 1000Q, R3 = 500Q.
Problem 10.6 d) Calculate incremental resistance R if the voltage difference in the circuit below is 0.5V.
Problem 10.6 e) W hich of the follow ing devices can be used for resistance measurem ent?
(A) W heatstone Bridge
(B) Strain Gage
(C) RTD
(D) Options A, B and C are correct
88
Copyrighted Material © 2018
Problem 10.6 f) Calculate voltage across 10 kQ resistor measured by a voltm eter with 100 kQ
resistance.
10kQ
20kQ
(A) 2.5 V
(B) 1.98 V
(C) 1 V
(D)
SOkO
5.5 V
Problem 10.6 g) Calculate voltage across 10 kQ resistor in Problem 10.6 f) if 500 kQ voltm eter is used.
(A) 2 .1 V
(B) 1.98 V
(C) 1 V
(D)
5.5 V
Problem 10.6 h) W hat is the percentage error in calculating current using 50 Q am m eter shown in the
circuit given below?
(A) 1%
(B) 2 %
(C) 2.5 %
(D) 5 %
89
Copyrighted Material © 2018
Problem Set # 10.7 - Power Electronics
Problem 10.7 a) Power electronics converter that changes DC to DC is c a lle d _________
(A) Inverter
(B) Chopper
(C) Rectifier
(D) Cycloconverter
Problem 10.7 b) Power electronics converter that changes DC to AC is c a lle d _________
(A) Inverter
(B) Chopper
(C) Rectifier
(D) Cycloconverter
Problem 10.7 c) Power electronics converter that changes AC to DC is c a lle d _________
(A) Inverter
(B) Chopper
(C) Rectifier
(D) Cycloconverter
Problem 10.7 d) Power electronics converter that changes AC to AC is c a lle d _________
(A) Inverter
(B) Chopper
(C) Rectifier
(D) Cycloconverter
Problem 10.7 e) A single phase half-wave controlled rectifier is powered by a 110 VAC source and
connected to a 5 Q resistor. Select the output voltage graph across resistor if firing angle is 45°
Vout
(A)
(B)
Vout
(C)
(D)
90
Copyrighted Material © 2018
Chapter # 11 - Power
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Page #
Section
203 - 205
Single phase & three phase
Transmission & distribution
203 - 205
Electrical and Com puter Engineering
Transformer
205
M otors and generators
205
Power factor
203
Voltage regulation
Note: Specific details are not found in NCEES® FE Rel :erence Handbook
Facts about this section
•
8 - 1 2 questions can be expected on the exam (according to NCEES® FE Specification).
© Difficulty level of this section is rated 'M edium ' by the author.
•
Students with a major in power engineering may find this section easier.
Tips for preparing this section
•
Understand concepts found on above m entioned pages o f NCEES® FE Reference Handbook.
•
Some o f the im portant equations relevant to this section include single phase power, three
phase power, conversion between line and phase quantities, conversion between A - Y
quantities, maximum power transfer, transform er turns ratio, pow er factor, voltage regulation
and synchronous speed.
•
Reactive pow er required to bring power factor angle from i^ to d2 is given by:
Q = P (ta n i9x - tan 0 2)
•
Q = coCV2
Voltage regulation is given by equation shown below:
V.R =
Til
fl
x 100%
* S, f l
•
Maxim um pow er transfer occurs when Z x = Z?h.
• Transform er impedance varies depending on view point (primary or secondary).
•
Understand the difference between leading/lagging pow er factor and real/reactive power.
•
Review basic power related concepts using your college/university textbooks.
•
Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification can be found on pages 265-267 o f NCEES® FE Reference Handbook.
91
Copyrighted Material © 2018
Problem Set # 11.1 - Single Phase Power
Consult NCEES® FE Reference Handbook - Pages 203 - 205 while solving these questions
Problem 11.1 a)______ W real power is supplied by a generator to single phase load 20 + 5j Q
operating at 120 V if line impedance is 2 + 2j Q.
Problem 11.1 b) Calculate current passing through the circuit shown below.
50
(A) 2/80° A
(B) 1.96/68.7° A
(C) 2/11.3° A
(D) 1Q/-1Q0 A
Problem 11.1 c) Calculate apparent power supplied by current source to the circuit shown below.
(A) 20/90° VA
(B) 100/0^ VA
(C) 2QQ/-67.30 V A
(D) 50/0^ VA
Problem 11.1 d) Calculate total power (in Watts) absorbed by the circuit shown below.
100
1 )0
Copyrighted Material © 2018
Problem 11.1 e) Calculate average pow er absorbed by 5 0 resistor shown in the circuit given below.
m
(A) 1 W
(B) 5 W
(C) 25 W
(D) 0 W
Problem 11.1 f) Determine impedance 'Z' required for maximum power transfer.
(A) 10 Q
(B) 12 Q
(C) - j Q
(D) 12-j Q
Problem 11.1 g) Determine impedance 'Z' required for maximum power transfer.
20
-jft
1/QPA
(A) 5 - 2j 0
(B) 5 + 0.6j Q
(C) 5 0
(D) 1.66j Q
93
Copyrighted Material © 2018
50
Problem Set # 11.2 - Three Phase Power/Transmission &
Distribution
Consult NCEES® FE Reference Handbook - Pages 203 - 205 while solving these questions
Problem 11.2 a) Calculate line voltages of a balanced 3 4>Y- connected system if Van = 120/30° V.
(A) 120/30° V, 12Q/-3Q0 V, 120/150° V
(B) 208/30° V, 12Q/-3Q0 V, 120/150° V
(C) 208/60° V, 208/-60° V, 208/180° V
(D) 120/60° V, 12Q/-6Q0 V, 120/180° V
Problem 11.2 b) A balanced 3- c|>Y- connected load with Zcj, = 20 + 5j Q is connected to a positivesequence balanced 3 cf>Y- connected source Van = 120/30° V. Calculate the phase current lan.
(A) 5.8/-1Q40 A
(B) 5.8/16° A
(C) 22/30° A
(D) 22/-12Q0 A
Problem 11.2 c) Calculate line current L supplied by a balanced positive-sequence 3 4» Y- connected
source Van = 277/0° V to a balanced 3 cj>Y network with
(A) 15/0^ A
(B) 195/-150 A
(C) 33/-450 A
(D) 3.3/-450 A
= 5 + 5j Q and Zime = 1 + l j Q.
Problem 11.2 d) Calculate load voltage of a positive sequence balanced 3 cj>Y-Y network consisting o f
Van = 120/60° V source w ith Zr,ne = 2 + l j Q and Zioad = 10 + lO j Q.
(A) 104/63° V
(B) 120/30° V
(C) 16.5/44.5° V
(D) 110/53° V
Problem 11.2 e) Calculate load impedance of a positive sequence balanced 3 cf>Y-Y system with source
voltage Van = 120/0° V, line current lan = 5/-5°A and Zrme = 0.5 + 0.25j Q.
(A) 10.5/55° Q
(B) 23.4/4.2° Q
(C) 7.8/9^ Q
(D) 13/50° Q
Problem 11.2 f) Calculate line current la in a network com prising of A source V 3b = 208/30° V powering
a 3 cj>balanced Y connected load bank with ZPhaSe = 10 + 5j Q (assume lossless line).
(A) 18.5/30° A
(B) 1Q.7/-26.50 A
(C) 18.5/-26.50 A
(D) 10.7/30° A
94
Copyrighted Material © 2018
Problem 11.2 g) Calculate line current la provided by a balanced 3 c|) Y-connected source with V an =
120/0° V to a balanced 3 4>A connected load having per phase Zphase = 10 + 2j Q.
(A) 20/12° A
(C) 25.8/-250 A
(B) 15/45° A
(D) 35.3/-11.30 A
Problem 11.2 h) Calculate equivalent wye load for a load network consisting of balanced 3 cj>Y- loads in
parallel with balanced 3 4>A loads if Z Phase-Y = 10 + 5j Q & Z Phase-A = 6 + 9j Q.
(A) 2.8 O
(B) 4.2 Q
(C) 8.4 Q
(D) 12.6 Q
Problem 11.2 i) Calculate equivalent A load o f Problem 11.2h).
(A) 2.8 Q
(B) 4.2 Q
(C) 8.4 Q
(D) 12.6 Q
Problem 11.2 j) A positive sequence balanced 3- <
j>Y-Y network has source voltage Van = 120/0° V
feeding load impedance 20/0° Q. Calculate pow er generated by source if Znne = 0.
(A) 720 VA
(B) 2400 VA
(C) 360 VA
(D) 2160 VA
Problem 11.2 k) A food processing plant consumes 200 kW at 0.83 pf lagging. It is serviced by a 1 cj>
distribution line carrying 400 A current. W hat is the voltage across load?
(A) 500 V
(B) 200 V
(C) 600 V
(D) 0.5 V
Problem 11.2 I) Calculate losses o f a l c[>distribution line if it provides 200 A to a small m unicipality that
consumes 100 kW at 600 V and 0.85 pf lagging.
(A) 2 kW
(B) 20 kW
(C) 10 kW
(D) 0 kW
Problem 11.2 m) Calculate phase voltage at service entrance of a hospital (consuming 125 kW) which is
being fed by a lossless 3 4> balanced Y network providing 300 A line current at 0.694 pow er factor.
(A) 416 V
(B) 200 V
(C )289 V
(D) 600 V
95
Copyrighted Material © 2018
Problem Set # 11.3 - Voltage Regulation
Problem 11.3 a) A power transform er is rated 2.5 M V A 13.8 kV/600 V with no-load secondary voltage
o f 650 V. Calculate voltage regulation o f this transformer.
(A) -5%
(B)
2.5%
(C) 8.3%
(D)
4.7%
Problem 11.3 b) For an ideal transformer, voltage regulation is ___________
(A) >1
(B) > 2.5
(C) = 0
(D) * 0
Problem 11.3 c) A distribution transform er is rated 10 kVA 4000 V/400 V with equivalent series
impedance of 1 + 0.5j Q. Calculate voltage regulation at 0.85 lagging.
Problem 11.3 d) A power transform er is rated 5 M V A 12000 V/240 V with voltage regulation o f 5%.
Calculate no-load voltage rating of this pow er transformer.
(A) 276 V
(B)
252 V
(C )360 V
(D )240 V
Problem 11.3 e) A leading pow er factor will result in ________ voltage regulation.
(A) >0
(B) < 0
(C) = 0
(D) > 1
96
Copyrighted Material © 2018
Problem Set # 11.4 - Transformers
Consult NCEES® FE Reference Handbook - Page 205 while solving these questions
Problem 11.4 a) A distribution transform er has 50:1 turns ratio. Calculate primary voltage if 1 A current
flows through a 12 Q load connected on transform er secondary.
(A) 50 V
(B) 600 V
(C) 12 V
(D) 1 V
Problem 11.4 b) Calculate rated secondary current of 1-phase transform er rated 15 kVA 600 V / 120 V.
(A) 125 A
(B) 7.2 A
(C) 2.5 A
(D) 15 A
Problem 11.4 c) A 5 kVA distribution transform er has 10:1 turns ratio. Calculate primary current if
rated secondary voltage is 120 V.
Problem 11.4 d)
Q impedance is seen by transform er primary in the figure shown below.
zp
io n
Problem 11.4 e)_____ Q impedance is seen by transform er secondary in the figure shown below.
10:1
97
Copyrighted Material © 2018
Problem Set # 11.5 - Motors & Generators
Consult NCEES® FE Reference Handbook - Page 205 while solving these questions
Problem 11.5 a) Calculate the number of poles present in a 3-cJ> 480 V induction m otor with
synchronous speed of 1800 rpm (operates at 60 Hz).
(A) 1
(B) 2
(C) 4
(D) 8
Problem 11.5 b) Find the synchronous speed o f a 3-4> 600 V 2 pole induction m otor operating at 60 Hz.
(A) 7200 rpm
(B) 1800 rpm
(C) 3600 rpm
(D) 1200 rpm
Problem 11.5 c) An existing synchronous m otor is retrofitted from 50 Hz, 4 pole construction to a 60 Hz
2 pole construction. Calculate the change in its synchronous speed.
(A) 1500 rpm
(B) 2100 rpm
(C) 3600 rpm
(D) 0 rpm
Problem 11.5 d) Calculate slip o f a 2 pole 600 V induction m otor operating at 60 Hz with a rotational
speed of 3400 rpm.
Problem 11.5 e) Calculate the rotational speed o f a 60 Hz, 4 pole 3 phase induction m otor operating at
2.3 kV and full load slip o f 0.1.
(A) 1800 rpm
(B) 1980 rpm
(C) 1720 rpm
(D) 1620 rpm
98
Copyrighted Material © 2018
Problem Set # 11.6 - Power Factor
Consult NCEES® FE Reference Handbook - Page 203 while solving these questions
Problem 11.6 a) Calculate the size o f a capacitor bank (assume 60 Hz and 600 V) required to increase
the power factor of an industrial load consuming 200 kW from 0.75 pf lagging to 0.9 lagging.
(A) 400 |iF
(B) 293 \if
(C) 703 |iF
(D) 586 nF
Problem 11.6 b) An autom otive plant having a 60 Hz induction m otor (480 V) has a lagging pow er
factor of 0.6 and consumes 100 kW. Calculate the magnitude o f reactive power that must be provided
to increase pow er factor to unity.
(A) 33 kVAR
(B) 133 kVAR
(C) 15 kVAR
(D) 0 kVAR
Problem 11.6 c) A 750 |iF capacitor bank operating at V rms = 400V, 60 Hz is to be connected in parallel
to an industrial load consuming 75 kW at 0.80 power factor lagging. The power factor after connecting
capacitor bank is _______ .
Problem 11.6 d) A three phase pow er utility provides 75 kVA at 0.85 lagging power factor and 35 kVA
at 0.75 lagging power factor to a custom er at 600 V and 60 Hz. The overall power factor at custom er's
end is _______ .
Problem 11.6 e) A balanced 3-phase positive sequence source Van = 120/0°V, 60Hz is connected to a
three phase Y-connected load that consumes 375 kW at 0.85 power factor lagging. Calculate the
required per phase capacitance o f a balanced Y-connected capacitor bank if power factor needs to be
improved to 0.95 lagging.
(A)10 [if
(B) 8 [if
(C) 15 mF
(D) 7 mF
99
Copyrighted Material © 2018
Chapter # 12 - Electromagnetics
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
M axw ell equations
Electrostatics/magnetostatics
W ave propagation
Transmission line
Electromagnetic com patibility
205
Electrical and Com puter Engineering
200
205 - 206
205 - 206
Note: Specific details are not found in NCEES® FE Reference
Handbook on this topic.
F a c ts a b o u t t h is s e c t io n
•
5 - 8 questions can be expected on the exam (according to NCEES® FE Specification).
•
Difficulty level of this section is rated 'Easy7 by the author.
T ip s f o r p r e p a r in g t h is s e c t io n
•
Understand concepts found on above mentioned pages of NCEES® FE Reference Handbook.
•
Some o f the im portant equations relevant to this section include M axw ell's equations,
transmission line reflection coefficient equation, standing wave ratio equation, wave equation.
•
Review vector calculus and learn how to calculate divergence and curl o f a given vector.
•
Divergence of a magnetic field is zero.
•
V X E = —dB/dt is the vector form of Faraday's Law.
•
V X H = J + dD/dt is the vector form o f Am pere's Law.
•
V. D = p is the vector form o f Gauss' Law for electric field.
•
V.B = 0 is the vector form o f Gauss' Law for magnetic field.
•
Read transmission line questions carefully to distinguish between characteristic impedance and
load impedance.
•
Review basic concepts related to electrom agnetic compatibility.
•
Solve problem sets on next pages and review solutions at the end of this book.
*Exam specification can be found on pages 265-267 of NCEES® FE Reference Handbook.
100
Copyrighted Material © 2018
Problem Set # 12.1 - Maxwell Equations
Consult NCEES® FE Reference Handbook - Pages 35 and 205 while solving these questions
Problem 12.1 a) Calculate divergence of electric field E given by follow ing equation:
E = 3 xi + 2y2j + xk
(A) 3x + 2y 2 + x
(B) 3 + 4y 2 + 1
(C) 0
(D) 3 + 4y
Problem 12.1 b) Calculate divergence of a vector field D given by follow ing equation:
D = xyi + yzj + x z 2k
(A) xy + yz + xz 2
(B) y + z + 2xz
(C) x + z +2x
(D) 0
Problem 12.1 c) Calculate net electric flux of a hollow sphere containing four point charges Qa = +1 nC,
Qb = +3 nC and Qc = +1 nC and Qd = -2 nC
(A) 0 Vm
(B) 1017 Vm
(C) 339 Vm
(D) 678 Vm
Problem 12.1 d) Calculate net flux of a hollow sphere of radius 5 cm containing tw o concentric
spheres. Inner sphere has radius of 1 cm with surface charge density of 3 |iC/m 2 and outer sphere has
radius o f 3 cm with surface charge density of -5 |iC/m2.
(A) 1250 Vm
(B) -5963 Vm
(C) 3750 Vm
(D) 410 Vm
Problem 12.1 e) Calculate the value o f "a" given that f? is a magnetic field.
B = 3 axi + 2yj — 2zk
(A) 3
(B) 1
(C)0
(D) -1
Problem 12.1 f) W hich of the follow ing vector fields can be magnetic in nature?
A = 2x 2i + y j + 3 z 2k B = 2y 2i + 3 x j - 2xyk
(A) A
(B ) B
(C) Both A and B
(D) Neither Anor B
101
Copyrighted Material © 2018
Problem 12.1 g) A changing magnetic field is inducing an electric field E given by equation below
E = 2yzi + 3 x 2yj + x 2y 2k. Calculate tim e rate of change of the magnetic field
(B) —2 x 2yi — (2 y — 2 x y 2)/' - (6 xy — 2 z) k
(A) —2 yz i + 3 x 2yj + x 2k
(C) 3 x 2j
(D) 0
Problem 12.1 h) Calculate curl o f induced electric field produced by a changing magnetic field B given
by £ = —cos2 (3 t)k.
(A) —2 cos (3 t ) i
(B) sin2 (3 t ) k
(C) —6 cos (3 t) sin (3 t ) k
(D) 5 sin (3 t ) j
Problem 12.1 i) Calculate the voltage induced in a coil with 20 turns if flux passing through it changes
from 0.1 W b to 1.5 W b in 2 s.
(A) -28 V
(B) -14 V
(C) -1.4 V
(D )-1 0 V
Problem 12.1 j) A coil o f length 20 cm has 50 turns and a cross-sectional area o f 2 cm 2. Calculate
induced voltage if current is increased from 50 m A to 100 mA in 1 s (assume \i = 4n x 10 '7 H/m).
Problem 12.1 k) Calculate magnetic flux density inside a torus of radius 2 cm having 50 turns if 1 A
current is passing through it.
Problem 12.11) Two infinitely long parallel wires are placed as shown below. The magnetic flux density
at point 'A' due to these wires (assume [i = 4h x 10 7 H/m) is _______ .
P o in t A
CO,0 , 0 )
1A
Problem 12.1 m) A current carrying wire in air is generating 0.1 T magnetic flux density at a radial
distance of 20 cm. The total current passing through this w ire is ______ .
102
Copyrighted Material © 2018
Problem Set # 12.2 - Electrostatics / Magnetostatics
Consult NCEES® FE Reference Handbook - Page 200 while solving these questions
Problem 12.2 a) Calculate magnitude of force on charge Q i = 5 nC located at P i( 0 , 1, 0) due to another
charge Ch = 10 nC located at P2 (0 , 0 , 2 ).
(A) 50 x 10~6 N
(B) 10 x 10~9 N
(C) 20 x 1 0 6 N
(D) 90 x 10'9 N
Problem 12.2 b) Calculate electric field intensity at origin due to point charges Q i = 50 nC located at (0,
1 , 0 ) and Ch = -50 nC located at (0 , - 1 , 0 ).
(A) 0 V/m
(B) 100 V/m
(C) 900 V/m
(D) 50 V/m
Problem 12.2 c) Calculate electric field at a point P between tw o infinitely large parallel plates located
in x-y plane having charge densities ps and - p s respectively
(A) 0
(B) ps/£
(C) 2ps/e
(D) ps£
Problem 12.2 d) Two infinitely long parallel wires (having line charge densities o f + 2 C/m and -1 C/m
respectively) located along z axis axes have 1 m space between them. Calculate the electric field mid­
way between the tw o wires
(A) 3 V/m
(B) 0 V/m
(C) 2 x 10 10 V/m
(D) 1 x 10 11 V/m
Problem 12.2 e) Calculate force on a 2 A current-carrying conductor of length 2 m in a uniform 0.5 T
magnetic field (assume that angle between conductor and field is 30°).
(A) 2 N
(B) 4 N
(C) ON
(D) 1 N
Problem 12.2 f) Calculate the energy stored in a magnetic field that has a strength of 2 A/m in a 2 m 3
volum e (assume |i = 4tt x 10~7 H/m).
103
Copyrighted Material © 2018
Problem Set # 12.3 - Transmission Lines and Wave Propagation
Consult NCEES® FE Reference Handbook - Pages 205 - 206 while solving these questions
Problem 12.3 a) A transmission line with 100 Q characteristic impedance is connected to a 300 + 50j Q
load. The reflection coefficient I" at the load is ______ .
Problem 12.3 b) Calculate standing wave ratio o f a transmission line having characteristic impedance
o f 50 Q and load impedance o f 500 + 25j Q.
(A) 1
(B) 2
(C) 10
(D) 5
Problem 12.3 c) A transm ission line has per unit length inductance o f 100 mH and per unit capacitance
o f 10 [if. Calculate magnitude o f load impedance that will allow a reflection coefficient T of 0.5.
( A )100 Q
(B )300 Q
(C )200 Q
(D )500 Q
Problem 12.3 d) W avelength of a transmission line having 200 Q characteristic impedance is 10 m.
Calculate input impedance at 100 m if the line is connected to a purely resistive load o f 500 Q.
(A )1 0 0 0 Q
(B )3500 Q
(C )200 Q
(D )500 Q
Problem 12.3 e) Calculate load connected at the end o f a transmission line (250 Q characteristic
impedance) if standing wave ratio is 2 .
( A )100 Q
(B )300 Q
(C )1000 Q
(D )500 Q
Problem 12.3 f) A transm ission line (wavelength 20m) has a characteristic impedance of 100 Q. Find
the expression for voltage at 100 m.
{A)V+ej2n + V~e~j2n
(B) V+eJ10n + V ~ e - jl0n
(C) V+ej200n + y - e ~j200n
(D) ^
( V+ej2n + V ' e ~j2n)
Problem 12.3 g) In problem 12.3 f), find an expression for current at 100 m.
{A)V+ejl0n + V - e ~ il0TC
(B) - ~ ( V +ei1071 +
(C)ioo ^^+eJl()7r ~ V~e~jl0n)
(D) None of the above
104
Copyrighted Material © 2018
Problem Set # 12.4 - Electromagnetic compatibility
Problem 12.4 a) W hich of the follow ing is not an example o f electrom agnetic coupling path?
(A) Inductive
(B) Conductive
(C) Capacitive
(D) Options A, B and C are coupling paths
Problem 12.4 b) Electromagnetic shielding is done t o ______________
(A) provide additional physical protection
(B) enable effective grounding
(C) prevent external electromagnetic interference
(D) safe-guard against ultra violet exposure
Problem 12.4 c) A system having cross-talk____________
(A) is im m une to external interference
(B) displays electrom agnetic interference within itself
(C) facilitates communication
(D) operates optim ally
Problem 12.4 d) Potential source(s) o f electrom agnetic com patibility problems in c lu d e _______
(A) electric motors
(B) lightning
(C) arc welding
(D) Options A, B and
C are all correct
Problem 12.4 e) Negative effects of electrom agnetic interference can be mitigated b y _______
(A) increasing coupling path separation
(B) hardware redundancy
(C) shielding
(D) All of the above
105
Copyrighted Material © 2018
Chapter # 13 - Control Systems
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Section
Page #
Block diagrams
127
Closed-loop / open-loop response
127
Steady State Errors
128
Instrumentation, M easurem ent and
Control
Root Locus
129
Stability
128
State variables
129 - 1 3 0
Bode Plots
Electrical and Com puter Engineering
208
Facts about this section
•
6 - 9 questions can be expected on the exam (accordingto NCEES® FE Specification).
© Difficulty level of this section is rated 'M edium ' by the author.
•
Students with a major in controls engineering may find this section easier.
Tips for preparing this section
•
Understand concepts found on above mentioned pages of NCEES® FE Reference Handbook.
•
Some of the im portant equations relevant to this section include steady state error - final value
theorem, root locus parameters, Routh-Hurwitz parameters, state equations, Bode Plot.
•
Derive transfer function o f classical negative feedback control system model block diagram
given in NCEES® Reference Handbook (page 127) to gain understanding of block diagrams and
closed loop / open loop response.
•
Review steps involved in developing Root Locus sketch.
•
Learn how to determ ine stability o f a given system using Routh Hurwitz criteria.
•
Review relevant concepts using college/university textbooks.
•
Solve problem sets on next pages and review solutions at the end of this book.
*Exam sp ecification can be found on pages 265-267 o f NCEES® FE Reference Handbook.
106
Copyrighted Material © 2018
Problem Set # 13.1 - Block Diagrams
Consult NCEES® FE Reference Handbook - Page 127 while so lv in g these questions
Problem 13.1 a) Find the closed loop tra n sfer fu n ction fo r fo llo w in g system .
(A)
(C)
G±( s)G2( s)
^2 (s)
(B)
1- H ( s )
Gi ( s)G2(s)
1 + G1(s )G 2 ( s ) t f( s )
(D)
1 + G 1 (s ) G 2 (s )//(s )
^2 ( 5 )
1 + jF / ( s )
Problem 13.1 b) Find the relationship between Y(s), R(s), N(s) and L(s) for follow ing system.
/ a\ y ( ^ _
[R) y
-
mi
-
,
G^ (S)G2(S)R(S)
G2(s)L( s)
1 + G l( s ) G2(s)tf (s) ^
1 + G 1(s)iV(s)
4.
g 2& L ( s)
l+ G ^ t ffs )
Gi ^ R( s)
W Y ^ S ) - i + g i(5)g 2(5)jv(5) t
_ G1(s)G2(s)/?(s)-iV(s)G1(5)G2(s)+L(s)G2(s)
(C)
1+ G 1 (s)G2(5)
”
,m „ r . _ G 2
1 D )U S J-
^
—7V(s)<71
C 2Cs)+ Z,Cs)(72C^) ,
"r
1 + G 1(s )G2(s )
G2(s)L(s)
1 + G 1(s)JV(s)
107
Copyrighted Material © 2018
Problem 13.1 c) Find the closed loop transfer function for follow ing system.
G4(s ) ( G1(5 ) —G2(s ) + G3(5))
(A)
1+G4(s)(G 1(s )-G 2 (s)+ G 3 (s))
(B)
G4 (s)+ G 1 ( 5) — G2 ( 5) + G3 (5)
1+G4 (s)+ G1( 5 ) - G 2 (s )+ G 3 (s)
G4( 5 )
G i (s) G2( s)G3(s) G4( s)
(C)
1 + G i (s ) G 2 (s ) G 3 ( s ) G 4 (s )
(D)
'
7 1 + G 1 (s )+ G 2 (s ) - G 3 (s )G 4 (s )
Problem 13.1 d) Find the closed loop transfer function for follow ing system.
kG±(s )
(A)
l + k G ±(s)H(s)
(B)
1
(C)
l+ fc G 1( s ) H ( s )
(D)
1 + H (s )
fcGiCsjH- H(s )
l+ /c G 1( s ) H ( s )
108
Copyrighted Material © 2018
Problem 13.1 e) Calculate closed loop gain of follow ing system.
G 1 (s)G 2 (s)+ G 1 (s)G 2 (s) G3(s )
G±(s)G2(s)G3( s )
(A)
1+G 1 (s)G 2 (s)G 3 (s)H (s)
____________________ G
(C)
l ( s ) G
2
( s ) ^
3
( s ) _____________________
1 + G i(s )G 2 (s)+ G 1 (s)G 2 (s)G 3 (s)
(B)
1+G1( s)G2(s) - H ( s ) + G 1( s )G2(s )G3(5 )
G1(js)G2(s)+H(s)G3i's)
(D)
1+G1(s)G2(s)G3( s )
109
Copyrighted Material © 2018
Problem Set # 13.2 - Bode Plots
Consult NCEES® FE Reference Handbook - Page 208 while solving these questions
Problem 13.2 a) Select the correct Bode Plot for H(s ) =
50
s+10
(B)
(C)
(D)
110
Copyrighted Material © 2018
Problem 13.2 b) Select the correct Bode Plot (phase response) for H(s ) =
aii
m-
t
'»
m
®
is#.
(A)
x A m - fmifsme-f u
(B)
KAm~Fmmmss>ia
g.ss
0.1
’i
(C)
XAxi* m
(D)
111
Copyrighted Material © 2018
Problem 13.2 c) Determine pole(s) of following transfer function:
H(s) =
(50) (s + 2)
(s + 1 0 0 ) (s + 1 0 0 0 )
(A) s = 100, s = 1000
(B) s = 2
(C) s = 100, s= 1000 and s = 2
(D) There are no poles
Problem 13.2 d) The magnitude gain o f follow ing transfer function is ____
i ir,.) = 1 0 0 (s + ^
(
s ( s + 1 0)
Problem 13.2 e) The magnitude gain o f follow ing transfer function is ____
H(s) =
100s
s 2 + 150s + 5000
112
Copyrighted Material © 2018
Problem Set # 13.3 - Steady Sate Errors
Consult NCEES® FE Reference Handbook - Page 128 while solving these questions
Problem 13.3 a) Calculate steady state error of follow ing system if input is 10 u(t ).
(A) 00
(B) 0.24
(C) 0.11
(D) 0.33
Problem 13.3 b) Calculate steady state error of follow ing system if input is 5 tu(t).
(A) 00
(B) 0.24
(C) 0.11
(D) 0.33
Problem 13.3 c) Calculate steady state error o f follow ing system if input is 10 t 2u(t).
(A) OO
(B) 0.24
(C) 0.11
(D) 0.33
113
Copyrighted Material © 2018
Problem 13.3 d) Calculate steady error o f follow ing system if input is 2 tu(t).
(A) 00
(B) 0.24
(C) 0.11
(D) 0.33
Problem 13.3 e) Calculate steady state error o f follow ing system if input is 3 u(t).
(A) 00
(B) 0.24
(C) 0.11
(D) 0.33
114
Copyrighted Material © 2018
Problem Set # 13.4 - Routh-Hurwitz Criteria & System Stability
Consult NCEES® FE Reference Handbook - Page 128 while solving these questions
Problem 13.4 a) Routh Array characteristic equation of follow ing system is _______ .
Problem 13.4 b) Compute the entry b i in Routh Array table o f follow ing closed loop transfer function
s+ 1
T(s) =
+ 2 s 3 + 3 s 4 + 7s2 + s + 10
s5
S'4
2
2
3
7
1
10
s3 b i= ?
Problem 13.4 c) Determine the stability o f system given fo lowing closed loop transfer function using
Routh Hurwitz criteria o f system stability.
T(s) =
2 s 4 + 3 5 s + s2 + s + 1
(A) Stable
(B) Unstable
(C) Depends on K
(D) Cannot be determined
Problem 13.4 d) Determine the range of k for which system given by follow ing closed loop transfer
function is stable.
T( s) =
(fe)(s + 1 0 )
3 s 3 + 5 s 2 + (k + 1 0 )s + 5 k
Problem 13.4 e) Determine the range of k for which system given by follow ing block diagram is stable.
115
Copyrighted Material © 2018
Problem Set # 13.5 - Root Locus
Consult NCEES® FE Reference Handbook - Page 129 while solving these questions
Problem 13.5 a) Select the correct root locus for open-loop transfer function G(s)H(s ) =
(A)
(C)
(D)
116
Copyrighted Material © 2018
(s)(s+2)
1
Problem 13.5 b) Select the correct root locus for open-loop transfer function G(s)H(s ) = — — +y—1Q^
(A)
(B)
(C)
(D)
117
Copyrighted Material © 2018
Problem 13.5 c) Select the correct root locus for open-loop transfer function G(s)H(s)
(A)
(B)
(C)
118
Copyrighted Material © 2018
5+5
s2—55+4
Problem 13.5 d) Select the correct root locus for open-loop transfer function G(s)H(s)
(B)
(D)
119
Copyrighted Material © 2018
2S
( s + l) ( s + 2 )
35—2
Problem 13.5 e) Select the correct root locus for open-loop transfer function G (s)//(s ) = —-------
S +2S+1
im
!
(C)
I
(D)
120
Copyrighted Material © 2018
Problem Set # 13.6 - State Variables
Consult NCEES® FE Reference Handbook - Pages 129 -130 while solving these questions
Problem 13.6 a) Find state equation for follow ing system.
y " + 3 y' + 2y = 2u ( t )
y’ and y” represent 1 st and 2 nd order differentials respectively.
2]GM 3“«
» K H - 2 - J E M ”!-®
Problem 13.6 b) Find state equation for follow ing system.
2y " + 8y' + 10 y = 6u ( t )
(A)
(C)
* i]
x2\
ro
Li
* il
r o
L—8
x2\
y' and y " represent 1 st and 2 nd order differentials respectively.
l
o-
(B)
EK
(D)
Problem 13.6 c) Find state equation for follow ing system.
o
0
1
1
K>
1
0
1
*3-
O
'Xi
(C) x2 =
(B) *2 = 0
.1
-*3.
* i-
O'
1 1
*2 + 3 u ( t )
2 3. Lx 3J
.0.
0
1
(D) *2 =
0
0
0' -Xl"3
1 *2 + 0 u ( t )
-*3-
.3
2
1. Lx 3 J
* !-
O'
u(i)
+
0
*2
u 3J
.3.
o
’* 1*
EH” ae,K]«®
j---------
(A)
y', y "& y "' represent 1 st, 2 nd & 3rd order differentials respectively.
O
3 y " ' + 6y " + 1 2 y ' + 3 y = 9u ( t )
.0.
Problem 13.6 d) Derive expression for transfer function given standard state-variable model.
x(t) = Ax(t) + Bu(t)
y{t) = Cx(t) + Du(t)
(A j C i s I - A ^ B + D
(B) D ( s l — A )B + C
(C) B - A + D
(D Transfer function cannot be determined
121
Copyrighted Material © 2018
Problem 13.6 e) Find transfer function for follow ing state space model.
[|]-L .y&GH)
"4
y = [1
(a ) r ( s ) =
(c) r ( s ) =
'
'
v
J
0 ] [**] + [0 ]u (t)
ifT{s) = C i s I - A ^ B + D
(B) r ( s )
4 —2 s
1
1 - 4 s 2- 5 s
(D) T (^s ) = ,
2s—1
'
7
v y
2
s 2+5s+4
Problem 13.6 f) Find transfer function for following state space model.
y[n + 2] + 3y[n + 1] + 3 y[n] = 2 u[n]
(A)
(C)
x j n + 1 ]‘
] =[—3 - s i E K H
x 2[n + l]_
■*i[n + i ] i ro
[x2[n + 1]J L l
II r^i t«]]
OJ x2[n]\
roi
L3J K)
(B)
x±[ n + 1 ]
[x2[n + 1 ]
(D)
xi l n + i ] l _ r i
x2\n + 1]J
1-3
122
Copyrighted Material © 2018
] - e a x [n]J
2
-* zM
Chapter # 14 - Communications
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
M odulation / Demodulation
Digital Com munications
Fourier transform s / Fourier Series
M ultiplexing
Section
Page #
Electrical and Com puter Engineering
209-210
206, 210
M athem atics
3 1 -3 3
Note: Specific details on this topic are not available in NCEES®
FE Reference Handbook.
Facts about this section
•
5 - 8 questions can be expected on the exam (accordingto NCEES® FE Specification).
•
Difficulty level o f this section is rated 'M edium ' by the author.
•
Students with a major in com m unications engineering may find this section easier.
Tips for preparing this section
•
Understand concepts found on above mentioned pages of NCEES® FE Reference Handbook.
•
Some of the im portant equations relevant to this section include efficiency o f amplitude
modulation, pow er transm itted in a message, modulation index, 98% bandwidth power of FM
signal, PCM, PAM, Fourier Transform pairs.
•
Learn how to identify graphical illustrations of AM , FM and PM signals.
•
Frequency and Phase M odulated signals appear very similar.
•
Understand the steps involved in digital signal processing.
•
Learn how to calculate Fourier Transform/Series using table.
•
Develop understanding of the two main multiplexing techniques
Time Division M ultiplexing
Frequency Division M ultiplexing
•
Review concepts relevant to communication using your college/university textbooks.
•
Solve problem sets on next pages and review solutions at the end of this book.
*Exam specification can be found on pages 265-267 of NCEES® FE Reference Handbook.
123
Copyrighted Material © 2018
Problem Set # 14.1 - Amplitude Modulation
Consult NCEES® FE Reference Handbook - Page 209 while solving these questions
Problem 14.1 a) Calculate transm itted signal power if a 1000 W carrier is modulated at 80%.
(A) 1320 W
(B) 800 W
(C) 390 W
(D) 200 W
Problem 14.1 b) 5 kW signal power is transm itted by a 4.75 kW carrier signal using sine wave. Find the
modulation index for this transmission.
(A) 10%
(B) 105%
(C) 75%
(D) 32%
Problem 14.1 c) Costas-loop is used for detecting a signal modulated with which o f the following
modulation technique?
(A) Single-Sideband Am plitude M odulation
(B) Angle Modulation
(C) Frequency M odulation
(D) Double Sideband Am plitude M odulation
Problem 14.1 d) Calculate modulation index of an am plitude modulated message signal
Ssin2n(1000t) using a carrier signal o f S0sin2n{4000t).
(A) 20%
(B) 100%
(C) 10%
(D) 50%
Problem 14.1 e) Calculate efficiency of an amplitude modulated wave for which modulation index is
0.8 and normalized average power is 0 .6 .
(A) 32%
(B) 18%
(C) 27%
(D) 44%
124
Copyrighted Material © 2018
Problem 14.1 f) W hich of the follow ing options represents amplitude modulated wave?
Carrier Wave
(A)
(B)
(C)
(D) None of the above
125
Copyrighted Material © 2018
Problem Set # 14.2 - Angle Modulation
Consult NCEES® FE Reference Handbook - Page 209 while solving these questions
Problem 14.2 a) Calculate 98% pow er bandwidth o f a frequency modulated signal with frequency
deviation ratio of 1.25 and message bandwidth o f 10 kHz.
(A) 20000 Hz
(B) 10000 Hz
(C) 125000 Hz
(D) 45000 Hz
Problem 14.2 b) Angle modulated signals can be dem odulated u sin g ____________
(A) Costas loop
(B) Phase-lock loop
(C) Envelope detection
(D) Sampling
Problem 14.2 c) In Problem 14.If) which option represents a frequency modulated wave?
(A) Option A
(B) Option B
(C) Option C
(D) None
Problem 14.2 d) In Problem 14.If) which option represents a phase modulated wave?
(A) Option A
(B) Option B
(C) Option C
(D) None
Problem 14.2 e) Calculate 98% power bandwidth o f a frequency modulated signal with frequency
deviation ratio of 0.1 and message bandwidth of 10 kHz.
(A) 10000 Hz
(B) 5000 Hz
(C) 1000 Hz
(D) 20000 Hz
126
Copyrighted Material © 2018
Problem Set # 14.3 - Pulse Code Modulation (PCM) & Pulse
Amplitude Modulation (PAM)
Consult NCEES® FE Reference Handbook - Page 210 while solving these questions
Problem 14.3 a)_________ quantization levels can be represented by a binary word of length 7 bits.
Problem 14.3 b) Calculate minimum bandwidth required to transm it a pulse code m odulated message
m(t), with M(f) = 0 for f >= 100 Hz using 256 quantization levels.
(A) 481 Hz
(B) 51200 Hz
(C) 25600 Hz
(D) 1600 Hz
Problem 14.3 c) A PAM system modulates 15 kHz signal by sampling it using a clock at equal tim e
intervals. Calculate minimum clock frequency of this PAM system.
(A) 15 kHz
(B) 30 kHz
(C) 7.5 kHz
(D) 150 kHz
Problem 14.3 d) In Problem 14.3c) calculate tim e spacing between adjacent samples o f the pulseam plitude modulated wave form.
(A) 66 |is
(B) 50 us
(C) 33.3 |is
(D) 75 \is
Problem 14.3 e ) ____________is not part of pulse code modulation process.
(A) Quantization
(B) Encoding
(C) Sampling
(D) Filtering
127
Copyrighted Material © 2018
Problem Set # 14.4 - Fourier Transforms
Consult NCEES® FE Reference Handbook - Page 31 - 33 while solving these questions
Problem 14.4 a) Fourier transform o f function shown in the figure given below is ______ .
Ampllteide
s
Time
*4
-3
1
-1
-2
2
3
4
Problem 14.4 b) Fourier transform o f function shown in the figure given below is
Jtospiiude
3
Problem 14.4 c) Calculate Fourier transform of function given below:
x ( t ) = cos(27r(300)t) n ( - )
\6 /
(A) 6 sin c ( 6 / )
(B) 6 s in c ( 6 ( / — 300))
.
(C) 3 s in c ( 6 ( f — 3 0 0 )) + 3 s in c ( 6 ( / + 3 0 0 ))
(D) 6 s in c ( 6/ ) e “ 27^ 300
Problem 14.4 d) Calculate Fourier transform of function given below
x ( t ) = e~5t cos(2Tc(20)t) u ( t )
(A) 7
;27t(/-20)+5
j2rtf+5
1
( Q -j 2 n f + 5
e -4 0n jf
(D)- ( —
v
+
2 V ;'27
j 2 nt(/-20)+5
( f -
Problem 14.4 e) Fourier transform o f function given below is
't — 2
x(t) = 4n
128
Copyrighted Material © 2018
j 2tt(/+20)+5
1
;27t(/+20)+5>
Problem Set # 14.5 - Multiplexing
Problem 14.5 a)__________ is an example of digital multiplexing.
(A) Time-division multiplexing
(B) Frequency-division
(C) Wavelength division
(D) Options A, B and C are examples o f digital multiplexing
Problem 14.5 b) Three channels are to be frequency multiplexed together
Channel 1 has a Band W idth o f 50 kHz
Channel 2 has a Band W idth of 100 kHz
Channel 3 has a Band W idth o f 50 kHz
Calculate minimum link bandwidth if a guard band of 5 kHz is required between channels.
(A) 200 kHz
(B) 215 kHz
(C) 210 kHz
(D) 220 kHz
Problem 14.5 c) Five channels are multiplexed using tim e division multiplexing. Every channel sends 10
bytes/second. Calculate frame size if the system can multiplex 1 byte/channel.
(A) 10 bytes
(B) 1 bytes
(C) 5 bytes
(D) 50 bytes
Problem 14.5 d) The bit rate in problem 14.5 c) is ______ bps.
Problem 14.5 e) Calculate bit duration o f a tim e division m ultiplexer that multiplexes three 50 kbps
channels using 1 bit tim e slots.
(A) 50000 bps
(B) 150000 bps
(C) 1000 bps
(D) 25000 bps
Problem 14.5 f) The frame duration in problem 14.5 e) is ______.
129
Copyrighted Material © 2018
Chapter # 15 - Computer Networks
Key Knowledge Areas*
NCEES® Reference Handbook
Concepts
Section
Page #
Routing and Switching
Network Topologies
Local area networks
Electrical and Com puter Engineering
219 - 222
Facts about this section
•
3 - 5 questions can be expected on the exam (accordingto NCEES® FE Specification).
• Difficulty level o f this section is rated 'Easy' by the author.
• Students with a major in com puter engineering may find this section easier.
Tips for preparing this section
•
Understand concepts found in above mentioned pages of NCEES® FE Reference Handbook.
•
Learn routing/switching process including data forwarding, routing tables etc.
•
Review differences between major network topologies (bus, ring, star etc.).
• Understand OSI and TCP/IP model.
• Review relevant concepts using your college/university textbooks.
• Solve problem sets on next pages and review solutions at the end o f this book.
:Exam specification can be found on pages 265-267 of NCEES® FE Reference Handbook.
130
Copyrighted Material © 2018
Problem Set # 15.1 - Routing and Switching
Consult NCEES® FE Reference Handbook - Page 219 - 222 while solving these questions
Problem 15.1 a) Process of finding efficient paths between nodes based on address is called____
Problem 15.1 b) Which o f the follow ing table(s) is maintained by a router?
(A) Time table
(B) Forwarding table
(C) Routing table
(D) Options C & B are correct
Problem 15.1 c) In which o f the following OSI M odel Layers does routing takes place?
(A) Application Layer
(B) Data Link Layer
(C) Network Layer
(D) Session Layer
Problem 15.1 d)_______ is used to connect tw o or more networks for transferring data packets.
(A) Router
(B) Hub
(C) Switch
(D) Options A, B and C are all correct
Problem 15.1 e)_______ is used to create a network and allow devices within a network to
com m unicate with each other.
(A) Router
(B) Hub
(C) Switch
(D) Options A, B and C are all correct
Problem 15.1 f) A network switch generally operates in which of the follow ing OSI M odel Layer?
(A) Application Layer
(B) Data Link Layer
(C) Network Layer
(D) Session Layer
131
Copyrighted Material © 2018
Problem Set # 15.2 - Network topologies / Frameworks / Models
Consult NCEES® FE Reference Handbook - Page 219 - 222 while solving these questions
Problem 15.2 a)_________ is not an example of a common network topology
(A) Bus
(B) Ring
(C) Star
(D) All options are examples o f common network topologies
Problem 15.2b)__________ network topology has a higher security risk.
(A) Bus
(B) Ring
(C) Star
(D) W ireless
Problem 15.2c) http and email will run in __________ OS I model layer.
Problem 15.2d) Language translation for applications takes place in ________ OSI model layer.
Problem 15.2e) Session layer is not responsible f o r __________
(A) M odulation/Dem odulation
(B) Relation between tw o end users
(C) Identifying users
(D) Controlling data exchanged by users
Problem 15.2f) Breaking a single connection can disrupt an entire network in _______ implementation.
(A) Bus
(B) Star
(C) Ring
(D) Options A, B and C are correct
132
Copyrighted Material © 2018
Problem Set # 15.3 - Local Area Networks
Consult NCEES® FE Reference Handbook - Page 219 - 222 while solving these questions
Problem 15.3 a) LAN can be implemented u sing_________ topology.
A) Ring
(B) Star
C) Bus
(D) Options A, B and C are correct
Problem 15.3 b)_________ is a com m only used LAN technology.
A) Ethernet
(B) W ireless
C) Asynchronous Transfer M ode
(D) Options A, B and C are correct
Problem 15.3 c) Ethernet is an example o f ____________.
A) Star topology
(B) Bus topology
C) Ring topology
(D) Mesh topology
Problem 15.3 d) ATM LAN is an example o f ____________.
A) Star topology
(B) Bus topology
C) Ring topology
(D) Mesh topology
Problem 15.3 e)_________ serves the largest geographical region.
A) LAN
(B) W AN
C) M AN
(D) PAN
133
Copyrighted Material © 2018
Chapter # 16 - Digital Systems
Key Knowledge Areas*
Concepts
NCEES® FE Reference Handbook
Section
Page #
Electrical and Com puter Engineering
2 1 8 -2 1 9
M athem atics
22
Num ber systems
Boolean logic
Logic gates and circuits
Logic m inimization (k-maps)
Flip-flops and counters
State machine design
Programmable logic devices
Data path/controller design
Timing
Note: NCEES® FE Reference Handbook does not contain
specific details on these topics.
Facts about this section
• 7 - 1 1 questions can be expected on the exam (according to the NCEES® FE Specification).
• Difficulty level of this section is rated 'Hard' by the author.
•
Students with a major in com puter engineering may find this section easier.
Tips for preparing this section
• Understand concepts found on above mentioned pages of NCEES® FE Reference Handbook.
• Learn how to use calculator for converting numbers between different bases.
•
Study truth tables o f all logic operators especially XOR and XNOR.
•
Understand the difference between l 's com plem ent and 2's com plem ent calculations.
•
Review graphical symbols o f logic gates, devices etc.
• Understand rules regarding grouping of K-map terms for function minimization.
• Gain understanding of how different flip-flops and counters work.
• Learn how to navigate between state diagram and state table.
•
Get fam iliar with programmable logic devices, controllers and tim ing diagrams.
•
Review relevant concepts using college/university textbooks.
•
Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification can be found on pages 265-267 of NCEES® FE Reference Handbook.
134
Copyrighted Material © 2018
Problem Set # 16.1 - Number Systems
Consult NCEES® FE Reference Handbook - Page 218 while solving these questions
Problem 16.1 a) 8 A D hex is equal t o _______ in binary num ber system.
Problem 16.1 b) 963io is equal t o _______ in binary num ber system.
Problem 16.1 c) Calculate l's com plem ent sum of OOIO2 and OOII 2.
(A )IIO I2
(B) IOIO 2
(C)OOIO 2
(D) OIOI 2
Problem 16.1 d) Calculate l's com plem ent sum of I I O I 2 and I O I I 2.
(A) IOIO 2
(B) I I I O 2
(C)IOO I 2
(D) IOIO 2
Problem 16.1 e) Calculate the 2's com plem ent sum of I O I I 2 and OIOI 2.
(A)OOOI2
(B)OOOO2
(C) I I H 2
(D)0111 2
Problem 16.1 f) Calculate the 2's com plem ent sum o f I I I O 2 and OIOI 2.
(A)O O II 2
(B) OIOO2
(C) IIOO 2
(D) I I I O 2
135
Copyrighted Material © 2018
Problem Set # 16.2 - Boolean Logic
Consult NCEES® FE Reference Handbook - Page 218 while solving these questions
Problem 16.2 a) Apply DeMorgan's Theorem to follow ing expression.
[A + BCD + (E + F )]
(A) (B + C + D ) ( I ) ( F ) (A)
(C) I + B CD + ( F + F )
(B) (A ) { B C D ) ( E F )
(D) ( Z ) ( B + C + D ) ( F ) ( F )
Problem 16.2 b) Apply DeMorgan's Theorem to follow ing expression.
[(ABC)(D + £ F ) ]
( A ) I + B + C + D(E + F )
(B) ( A B C ) { D ) { E F )
( Q A + B + C + DE
(D) A B + D F F
Problem 16.2 c) Simplify follow ing expression using Boolean logic.
AB + A(BC) + B(A + C)
(A) AB + A(B + C) + B ( J C )
(B) AB + C(.4 + 5 )
(C)AB + A B C
{D )A B +A (B Q
Problem 16.2 d) Simplify following expression using Boolean logic.
AB + B(A + C) + C(A + B)
(A) AB + AB +BC + AC + BC
(C) AB + BC
(B) AB + BC + AC
(D )ABC
Problem 16.2 e) Convert follow ing expression to Sum of Product form.
(A + B) (A + B + C)
(A) A + AB + BC
(B) A + B +
(C M + B
(D )AB + BC
136
Copyrighted Material © 2018
C
Problem Set # 16.3 - Logic Gates
Consult NCEES® FE Reference Handbook - Page 218 while solving these questions
Problem 16.3 a) Find output expression for logic circuit shown below.
(A ) AB + CD
(B ) AB ( C + D) + (A + B)CD
(C) AB( C + D )
(D)AB(C + D)
Problem 16.3 b) Find output expression for logic circuit shown below.
(A) 0
{B)A+B
(C) 1
(D ) A + B
Problem 16.3 c) Find output expression for logic circuit shown below.
(A M + B + C
(B ) A B + C
(CM + bc
(D) AB + BC
137
Copyrighted Material © 2018
Problem 16.3 d) Find output expression for logic circuit shown below.
(A) AB + C
{C)ABC
(B) ABC
(D ) A + B + C
Problem 16.3 e) Find output expression for logic circuit shown below.
(A) 0
(B) 1
{C)AB
{D)A + B
138
Copyrighted Material © 2018
Problem Set # 16.4 - Karnaugh Maps
Consult NCEES® FE Reference Handbook - Page 219 while solving these questions
Problem 16.4 a) Determine minimized 'Sum of Product' expression for logic function given by k-map.
C
0
1
00
AB
01
11
10
(A) C + A B + BC
(B) A B C + AB + BC
{ Q B C + AB + A B C
(D)ABC + A B
+BC
Problem 16.4 b) Determine minimized 'Sum of Product' expression for logic function given by k-map.
C
0
1
00
AB
01
11
10
(A )AB + A B
(B) ABC + A BC
(CM
(DM
Problem 16.4 c) Determine minimized 'Sum of Product' expression for logic function given by k-map.
CD
00
AB
01
11
10
00
01
11
10
(A) A B + AB
(B) B
(C)B
(D) AB + A
139
Copyrighted Material © 2018
Problem 16.4 d) Determine minimized 'Sum of Product' expression for logic function given by k-map.
CD
00
01
11
10
00
AB
01
11
10
(A) AC + AB + A BD
(B) A + AB + BCD
(C) AC + ABC + A B D
(D) AB + ABC + A B D
Problem 16.4 e) Determine minimized 'Sum of Product' expression for logic function given by k-map.
CD
00
AB
01
11
10
00
01
11
10
(A) A B + C D
(B ) C D + CD + BC
(C) D + BC
(D) ABC + D
Problem 16.4 f) Determine minimized 'Sum of Product' expression for logic function given by k-map.
C
0
AB
1
00
01
11
10
(A) B + AC
(B) A C + ABC
{ C ) A B C + AC
(D) C + A C
140
Copyrighted Material © 2018
Problem Set # 16.5 - Flip-flops and counters
Consult NCEES® FE Reference Handbook - Page 219 while solving these questions
Problem 16.5 a) Determine output sequence of Flip Flop # 2 shown in the circuit given below.
FLPFiO §>*1
tot
0
FUPR.QP#2
0
Gi
11t — >
0
G
m
(A) 101
(B) 111
(C) 010
(D) 000
Problem 16.5 b) Identify the circuit shown below (assume that clock is applied).
(A) JK Flip Flop
(B) RS Flip Flop
(C) D Flip Flop
(D) M ultiplexer
Problem 16.5 c) Identify the circuit shown below (assume that clock is applied).
(A) JK Flip Flop
(B) RS Flip Flop
(C) D Flip Flop
(D) M ultiplexer
141
Copyrighted Material © 2018
Problem 16.5 d) Identify the circuit shown below (assume that clock is applied).
(A) JK Flip Flop
(B) RS Flip Flop
(C) T Flip Flop
(D) M ultiplexer
Problem 16.5 e) Determine the output sequence (Q) of RS Flip Flop shown in the circuit given below.
m i «m i : i}
ci
o
5
(A) 100
(B )001
(C) 111
(D )110
Problem 16.5 f) Find final state (Q1Q 2Q3) of the counter after three clock cycles. Assume that initial
states o f Q 1O 2Q 3 are zero.
(A) 000
(B)111
(C) 001
(D) O i l
142
Copyrighted Material © 2018
Problem 16.5 g) Find final state Q 0O 1Q 2Q 3 o f counter given below after three clock cycles. Assume that
initial states of Q 0Q 1Q 2Q 3 are 1 1 1 1 .
Ill
( A ) 1000
(B)0000
(C )1010
(D )1100
Note - Above given options are shown in (Q3Q 2Q 1Q 0) format.
Problem 16.5 h) Find output sequence of counter given below. Assume that Q dQ i initial states are 00.
1111
( A ) ( 0 0 ) initial state/ ( 0 l)cy cle 2/ ( 1 0)cyde3» ( 1 l)c y c le 4
(B ) ( 0 0 ) initial state/ ( 1 0 ) cycle 2/ ( 0 1 ) cycle 3/ ( i i ) cycle 4
(C ) ( 1 1 ) initial state/ ( 0 1) cycle 2/ ( 1 0 ) cycle 3/ ( 0 0 ) cycle 4
( D ) ( 1 1 ) initial state/ ( o i ) cycle 2/ ( 1 0 ) cycle 3/ ( 0 0 ) cycle 4
Note - Above given options are shown in (Q1Q 0) format.
143
Copyrighted Material © 2018
Problem Set # 16.6 - State Machine Design
Consult NCEES® FE Reference Handbook - Page 22 while solving these questions
Problem 16.6 a) Com plete state table for 'Finite State M achine' shown below.
Next State
o
II
Present
State
i
i
A
C
B
C
?
C
w= 1
B
B
?
Output
Z
0
1
?
(A) w = 0 (B), w = 1 (A), z = 1
(B) w = 0 (A), w =
1 (B), z = 0
(C) w = 0 (A), w = 1 (B), z = 1
(D) w = 0 (C), w =
l (A), z = 0
Problem 16.6 b) In problem 16.6 a) an input sequence of w = 111 is applied to the 'Finite State
M achine'. Calculate output sequence if machine is initially in state A.
(A) 111
(B) 010
(C)000
(D )011
Problem 16.6 c) In problem 16.6 a) an input sequence of w = 000 is applied to 'Finite State Machine'.
Calculate output sequence if machine is initially in state A.
(A) 101
(B) 000
(C)010
(D)111
144
Copyrighted Material © 2018
Problem 16.6 d) Complete state table for 'Finite State M achine' shown below.
8ES-EF
Next State
Present
State
ab=00
A
A
B
Output
Z
10
C
C
C
11
B
B
C
B
?
01
A
A
?
A
1
1
0
D
D
D
C
B
0
(A) 00 - Q 01 - D
(B) 0 0 - Q 0 1 - C
(C) 00 — D, 0 1 - C
(D) 0 0 - A, 0 1 -D
Problem 16.6 e) States A, B, C and D of problem 12.6 d) are represented by y 2y i - 00, 0 1 ,1 0 ,1 1
respectively in the table below. Select simplified output expression for 'Z' as a function o f present
state.
Next State
Present State
Y2yi
00
01
10
11
ab=00
01
10
11
00
01
10
11
00
00
11
11
10
10
10
10
01
01
00
(A) z = y2 y±
(B) z = y±
(C )z = T i T\
( ° ) z = y2
145
Copyrighted Material © 2018
01
Output Z
1
1
0
0
Chapter # 17 - Computer Systems
Key Knowledge Areas*
Concepts
NCEES® FE Reference Handbook
Section
Page #
Electrical and Com puter Engineering
223 - 224
Architecture
Microprocessors
M em ory technology
Interfacing
Facts about this section
• 4 - 6 questions can be expected on the exam (according to NCEES® FE Specification).
• Difficulty level o f this section is rated 'Easy' by the author.
• Students with a major in com puter engineering may find this section easier.
Tips for preparing this section
•
Understand concepts found in above mentioned pages of NCEES® FE Reference Handbook.
• Review com puter architecture and identify functional units (ALU, I/O, CU etc.).
• Learn fundam ental concepts related to microprocessors.
• Familiarize yourself with different m emory technologies (RAM, ROM, EROM etc.).
• Learn how different components o f com puter system interface with each other.
• Review relevant concepts using college/university textbooks.
• Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification can be found on pages 265-267 of NCEES® FE Reference Handbook.
146
Copyrighted Material © 2018
Problem Set # 17.1 - Architecture & Interfacing
Consult NCEES® FE Reference Handbook - Pages 223 - 224 while solving these questions
Problem 17.1 a)_____ is not an example o f addressing mode
(A) Register
(B) Immediate
(C) Indirect
(D) Options A, B and Care examples of addressing
modes
Problem 17.1 b)____ is a com puter program that converts high-level language into machine language.
Problem 17.1 c) The section of CPU that performs mathematical calculations is c a lle d _______ .
Problem 17.1 d) Batch data processing is useful for applications involving_______ .
(A) Quarterly bank statements
(B) Stock market quotation
(C) W eather m onitoring
(D) Options A, B and C are examples o f batch data
processing
Problem 17.1 e) Real tim e data processing is useful f o r _______ applications.
(A) Quarterly bank statements
(B) High school grade reports
(C) W eather m onitoring
(D) Options A, B and Care examples o f real tim e data processing
Problem 17.1 f) Basic Input Output System (BIOS) is read fr o m _______ during normal start-up routine.
(A) RAM
(B) ROM
(C) USB
(D) CD
Problem 17.1 g)_______ is the process o f transforming data into different form at for another system.
(A) Encryption
(B) Encoding
(C) Hashing
(D) Decoding
Problem 17.1 h)_________ is the process of transform ing data for secrecy.
(A) Encryption
(B) Encoding
(C) Hashing
(D) Decoding
147
Copyrighted Material © 2018
Problem 17.1 i) Instruction pipelining results in ____________.
(A) decreased instruction execution tim e
(B) increased instruction throughput
(C) allows new types o f possible instructions
(D) All of the above
Problem 17.1 j) The control unit of a com puter system is responsible for
(A)storage
(B)
interpreting program instructions
(C) m athematical operations
(D) input/output operations
148
Copyrighted Material © 2018
Problem Set # 17.2- Microprocessor
Consult NCEES® FE Reference Handbook - Pages 223 - 224 while solving these questions
Problem 17.2 a) A microprocessor register that stores address of the last requested program in a
buffer register is c a lle d __________ .
(A) Program counter
(B) Stack pointer
(C) Instruction pointer
(D) Accum ulator
Problem 17.2 b) A m icroprocessor register that stores address o f current or next instructions in a
buffer register is c a lle d __________ .
(A) Program counter
(C) Accum ulator
(B) Stack pointer
(D) Options A and B are correct
Problem 17.2 c) A single 1C accepting & executing coded instruction for processing data and controlling
associated circuitry in a com puter system is c a lle d _________ .
(A) M icroprocessor
(B) M icrocom puter
(C) M icrocontroller
(D) Personal com puter
Problem 17.2 d) An interconnected group of ICs, l/Os and memory systems used for data processing
and other application is ca lle d _________ .
(A) M icroprocessor
(B) M icrocom puter
(C) M icrocontroller
(D) Personal com puter
Problem 17.2 e) An integrated system of a single 1C, I/O circuitry and memory system accepting &
executing code instructions and associated circuitry in com puter system is c a lle d _________ .
(A) M icroprocessor
(B) M icrocom puter
(C) M icrocontroller
(D) Personal com puter
Problem 17.2 f) Bus that transfers information between microprocessor & I/O units is c a lle d ________ .
Problem 17.2 g) Bus that selects a location for reading / w riting is ca lle d _________ .
149
Copyrighted Material © 2018
Problem Set # 17.3 - Memory Technology and Systems
Consult NCEES® FE Reference Handbook - Pages 223 - 224 while solving these questions
Problem 17.3 a) W hich of the follow ing storage device uses random access methods?
(A) Cassette tape
(B) CD
(C) Hard disk
(D) Flash memory
Problem 17.3 b) A Giga Byte co n tain s________ bits.
Problem 17.3 c) A m emory that acts as a buffer between CPU and main memory to speed up
processing is ca lle d _______ .
(A) DRAM (Dynamic Random Access Mem ory)
(C) Cache M em ory
(B) ROM
(D) EPROM
Problem 17.3 d) Which of the follow ing is an example o f secondary memory?
(A) Cache memory
(B) RAM (Random Access Memory)
(C) DVD
(D) DRAM (Dynamic Random
Access Memory)
Problem 17.3 e) W hich of the follow ing ROMs can be programmed only once and is non-erasable?
(A) PROM
(B) EPROM
(C) EEPROM
(D) DRAM
Problem 17.3 f) W hat is the difference between EPROM & EEPROM?
(A) EPROM and EEPROM are same
(B) EPROM is not erasable
(C) EEPROM can be erased electrically, EPROM can be erased ultra violet light only
(D) EPROM is volatile
Problem 17.3 g)_________ is a memory unit which varies in size between different com puter systems.
(A) Nibble
(B) W ord
(C) Bit
(D) Byte
150
Copyrighted Material © 2018
Chapter # 18 - Software Development
Key Knowledge Areas*
NCEES® FE Reference Handbook
Concepts
Algorithm s
Data structures
Software design methods
Page #
Section
2 2 4 -2 2 7
Electrical and Com puter Engineering
Software implementation
Software testing
Facts about this section
•
4 - 6 questions can be expected on the exam (accordingto the NCEES® FE Specification).
•
Difficulty level of this section is rated 'Easy' by the author.
•
Students with a major in com puter engineering may find this section
easier.
Tips for preparing this section
•
Understand concepts found in abovementioned pages of NCEES® FE Reference Handbook.
•
Learn how to dry run pseudo codes and simple algorithms including sorting (bubble, heap etc.)
and searching (binary, hash etc.).
•
Review flow chart developm ent and its symbols.
•
Gain fundamental understanding of object-oriented and structured programming including
different types of data structures (static and dynamic).
•
Learn key steps involved in software im plem entation and testing process.
•
Review relevant concepts using your college/university textbooks.
•
Solve problem sets on next pages and review solutions at the end o f this book.
*Exam specification can be found on pages 265-267 of NCEES® FE Reference Handbook.
151
Copyrighted Material © 2018
Problem Set # 18.1 - Algorithms
Consult NCEES® FE Reference Handbook - Page 224 - 227 while solving these questions
Problem 18.1 a) The value o f 'x' after execution o f follow ing code will b e ______ .
int x = 0 , y = 0 , z = 10
do while z > 0
{
y =y+i
x = 2y - 1
z =z - 3
}
end while
print x, print z
Problem 18.1 b) In problem 18.1a) the final value of 'z' is ______ .
Problem 18.1 c) The value o f Value 7 after execution of follow ing code will b e _______.
int value = 0
for (N = 1; N <= 100; N ++)
{ N = 2N
value = N }
print value
Problem 18.1 d) Determine value of cell A4 in the spreadsheet section shown below.
A
1
2
3
4
1
5
7
=BlxC2
B
5
C
3
2
3
5
=B$1
=B$2
=B$3
Problem 18.1 e) Determine value of cell B4 in the spreadsheet section shown below.
A
1
2
3
4
B
10
15
20
25
C
=A1 + C l
=A2 + C2
=A l x l
= A2 x 2
=A3 + C3
= A3 x 3
=A4 + C4
=A 4 x 4
152
Copyrighted Material © 2018
Problem Set # 18.2 - Data Structures
Consult NCEES® FE Reference Handbook - Page 224 - 227 while solving these questions
Problem 18.2 a) Array, List, Stacks and Trees are examples o f _________ .
(A) Algorithm s
(B) Programs
(C) Data Structures
(D) Functions
Problem 18.2 b) Performance of a searching algorithm in data structure is based o n _______ .
(A) Average tim e
(B) Worst-case time
(C) Best-case tim e
(D) Performance depends on algorithm length
Problem 18.2 c) Binary tree in which each leaf is at same distance from root is c a lle d _________ .
Problem 18.2 d) Efficiency o f binary search algorithm as a function of 'n' comparisons is
(A) log 2n
(B) n/2
(C) n
(D) n 2
Problem 18.2 e)
is an example of a sorting algorithm.
(A) Bubble
(B) Heap
(C) Quick
(D) Options A, B and C are examples of sorting algorithms
Problem 18.2 f) Which sorting algorithms has best tim e performance under stable conditions?
(A) Bubble sort
(B) Heap sort
(C) Insertion sort
(D)
Problem 18.2 g) _
is an example of static data structure.
(A) Lists
(B) Stacks
(C) Arrays
(D) Trees
153
Copyrighted Material © 2018
All sorting algorithms have same p
Problem Set # 18.3 - Software design
methods /implementation/testing
Consult NCEES® FE Reference Handbook - Page 224 - 227 while solving these questions
Problem 18.3 a) A typical Software Development Lifecycle does not comprise o f _______ phase.
(A) Testing
(C) Deployment
(B) Implementation
(D) M arketing
Problem 18.3 b) Software design with a tendency to break due to changes in non-related segments
ca lle d __________ .
Problem 18.3 c) A good software design will d isplay ________ .
(A) Ability to be extended w ithout requiring modification to original design
(B) A bility to be extended only with extensive m odification to original design
(C) Inability to extend
(D) None of the above
Problem 18.3 d) Static software testing involves verification th ro u g h __________ .
(A) program code review
(B) program code execution
(C) software maintenance
(D) Options A, B and C are all correct
Problem 18.3 e) Dynamic software testing involves verification th ro u g h _____________.
(A) program code review
(B) program code execution
(C) software maintenance
(D) Options A, B and C are all correct
Problem 18.3 f) "Top-down" approach is most applicable t o ________ programming.
(A) structured
(B) object-oriented
(C) both structured & object oriented
(D) neither structured nor object-oriented
Problem 18.3 g) The concept of "Class" is most applicable to which programming technique.
(A) structured
(B) object-oriented
(C) both structured & object oriented
(D) neither structured nor object-oriented
154
Copyrighted Material © 2018
Problem 18.3 h)
(A) FORTRAN
(C) COBOL
Problem 18.3 i)
(A) PHP
___language is not used for structured programming.
(B) BASIC
(D) C#
is not used as scripting language.
(B)JavaScript
(D) PHP, JavaScript and Python are scripting languages
(C) Python
155
Copyrighted Material © 2018
Solutions
Problem
Answer
Problem
Answer
Problem
Answer
Problem
Answer
1.1a
B
1.4f
C
1.7b
D
2.3d
5/6
1.1b
B, C
1.4g
0.87
1.7c
A
2.3e
2
1.1c
C
1.4h
B
1.7d
14
2.3f
1
l.ld
See sol
1.4i
1.1
1.7e
See sol.
2.3g
C
l.le
B
1.4j
A
1.7f
D
2.3h
$645
l.lf
B
1.4k
x = -1
1.7g
A
2.3i
C
l.lg
C
1.41
C
1.7h
0
2.3j
11.5
l.lh
2m
1.4m
A
1.7i
16
3.1a
C
1.2a
B
1.4n
B
1.7j
See sol.
3.1b
D
1.2b
C
1.5a
C
1.7k
B
3.1c
D
1.2c
C
1.5b
A
2.1a
6
3.Id
D
1.2d
C
1.5c
B
2.1b
88.2°F
3.1e
C
l.le
l.lf
D
1.5d
D
2.1c
V20/3
3.If
C3
-12-24j
1.5e
B
2.Id
2.21
3.1g
C
1.3a
C
1.5f
B
2.1e
6.9
3.2a
A
1.3b
B
1.5g
x=-l/3
2.If
3
3.2b
C
1.3c
A
1.5h
C
2.1g
80
3.2c
B
1.3d
B
1.5i
C
2.1h
3
3.2d
D
1.3e
A
1.5j
D
2.1i
110
3.2e
See sol.
1.3f
B
1.5k
C
2.1j
See sol.
3.2f
C
1.3g
70
1.51
See sol.
2.2a
C
4.1a
$241,157
1.3h
15600
1.5m
C
2.2b
0.24
4.1b
$231,225
1.3i
531441
1.5n
B
2.2c
0.228
4.1c
B
l-3j
29524
1.6a
D
2.2d
0.51
4.1d
A
1.3k
R2
1.6b
See sol.
0.0228
4.1e
$17440
1.4a
D
1.6c
See sol.
0.1359
4.If
$392,724
1.4b
C
1.6d
See sol.
l.le
l .l f
l.lg
0.50
4.1g
$1.69mn
1.4c
y=2x-5
1.6e
Under-
2.3a
3.18
4.1h
10.47%
1.4d
A, B
1.6f
See sol.
2.3b
0.66
4.1i
251416
1.4e
A
1.7a
C
2.3c
5.5%
4.2a
D
156
Copyrighted Material © 2018
Problem
Answer
Problem
Answer
Problem
Answer
Problem
Answer
4.2b
$4,000
5.4d
0.04K1
7.2b
A
8.2f
See sol.
4.2c
$2,968
5,4e
D
7.2c
B
8.3a
A
4.2d
$26,000
5.4f
B
7.2d
C
8.3b
B
4.2e
$24,000
6.1a
D
7.2e
6kO
8.3c
C
4.2f
$250,000
6.1b
B
7.3a
B
8.3d
D
4.2g
See sol.
6.1c
B
7.3b
1.5kQ
8.3e
A
4.3a
20,000
6.1d
A
7.3c
A
8.3f
D
4.3b
B, C
6.1e
D
7.3d
A
8.3g
C
4.3c
2,000
6. If
D
7.3e
lkQ
8.3h
A
4.3d
A
6.1g
B
7.4a
C
8.3i
C
4.3e
C
6.1h
8W
7.4b
D
8.4a
D
4.3f
See sol.
6.2a
A
7.4c
C
8.4b
C
5.1a
A
6.2b
A
7.4d
B
8.4c
A
5.1b
D
6.2c
C
7.4e
1A
8.4d
B
5.1c
A
6.2d
See sol.
7.5a
D
8.4e
D
5.Id
See sol.
6.2e
A
7.5b
A
8.5a
A
5.1e
See sol.
6.3a
A
7.5c
B
8.5b
C
5.If
B
6.3b
B
7.5d
C
8.5c
A
5.2a
D
6.3c
B
7.5e
See sol.
8.5d
11.11Q
5.2b
C
6.3d
B
7.6a
A
8.5e
0.66
5.2c
D
6.3e
C
7.6b
B
9.1a
C
5.2d
B
6.4a
A
7.6c
B
9.1b
A
5.2e
0.015H/m
6.4b
B
7.6d
C
9.1c
D
5.2f
D
6.4c
D
8.1a
C
9.Id
C
5.3a
B
6.4d
B
8.1b
B
9.1e
C
5.3b
B
6.4e
C
8.1c
D
9.2a
B
5.3c
0.095
7.1a
B
8.Id
B
9.2b
C
5.3d
B
7.1b
D
8.1e
C
9.2c
B
5.3e
C
7.1c
B
8.2a
D
9.2d
C
5.3f
D
7.Id
D
8.2b
See sol.
9.2e
C
5.4a
B
7.1e
A
8.2c
B
9.3a
C
5.4b
C
7.If
875|iA
8.2d
See sol.
9.3b
B
5.4c
B
7.2a
D
8.2e
B
9.3c
D
157
Copyrighted Material © 2018
Problem
Answer
Problem
Answer
Problem
Answer
Problem
Answer
9.3d
A
10.3b
D
ll.lf
D
11.6d
0.82
9.3e
A
10.3c
Cut-off
ll.lg
A
11.6e
D
9.3f
C
10.3d
A
11.2a
C
12.1a
D
9.3g
D
10.3e
C
11.2b
B
12.1b
B
9.3h
B
10.3f
D
11.2c
C
12.1c
C
9.3i
A
10.4a
D
11.2d
A
12.Id
B
9.4a
2000Hz
10.4b
Triode
11.2e
B
12.le
C
9.4b
C
10.4c
B
11.2f
B
12.If
B
9.4c
D
10.4d
C
11.2g
D
12.lg
B
9.4d
C
10.4e
B
11.2h
A
12.Ih
C
9.4e
A
10.5a
-5V
11.2i
C
12.li
B
9.5a
B
10.5b
5kQ
11.2j
D
12. lj
0.15\iV
9.5b
C
10.5c
2.5V
11.2k
C
12.1k
500|iT
9.5c
A
10.5d
-35V
11.21
A
12.11
0.4piT
9.5d
D
10.5e
0.5V
11.2m
B
12.1m
lOOkA
9.5e
A
10.6a
C
11.3a
C
12.2a
D
9.5f
B
10.6b
B
11.3b
C
12.2b
C
9.5g
D
10.6c
5000Q
11.3c
7%
12.2c
B
10.1a
A
10.6d
200Q
11.3d
B
12.2d
D
10.1b
D
10.6e
D
11.3e
B
12.2e
D
10.1c
B
10.6f
B
11.4a
B
12.2f
5|xJ
10.Id
0.634V
10.6g
A
11.4b
A
12.3a
See sol.
lO .le
C
10.6h
B
11.4c
4.16A
12.3b
C
10. If
A
10.7a
B
11.4d
25kQ
12.3c
B
10. lg
Group V
10.7b
A
11.4e
50Q
12.3d
D
10.Ih
Group III
10.7c
C
11.5a
C
12.3e
D
10.li
Group IV
10.7d
D
11.5b
C
12.3f
B
10.2a
A
10.7e
B
11.5c
B
12.3g
C
10.2b
A
11.1a
592 W
11.5d
5.5%
12.4a
D
10.2c
B
11.1b
B
11.5e
D
12.4b
C
10.2d
0.13mA
11.1c
C
11.6a
D
12.4c
B
10.2e
B
11.Id
9.19 W
11.6b
B
12.4d
D
10.3a
C
11.le
B
11.6c
0.98
12.4e
D
158
Copyrighted Material © 2018
Problem
Answer
Problem
Answer
Problem
Answer
Problem
Answer
13.1a
C
14.1c
D
15.2c
App.
16.5c
C
13.1b
C
14. Id
C
15.2d
Present.
16.5d
A
13.1c
A
14.le
C
15.2e
A
16.5e
B
13.Id
A
14. If
A
15.2f
C
16.5f
B
13. le
B
14.2a
D
15.3a
D
16.5g
D
13.2a
A
14.2b
B
15.3b
D
16.5H
A
13.2b
D
14.2c
B
15.3c
B
16.6a
C
13.2c
A
14.2d
C
15.3d
A
16.6b
D
13.2d
20dB
14.2e
D
15.3e
B
16.6c
C
13.2e
-33.9dB
14.3a
128
16.1a
See sol.
16.6d
A
13.3a
B
14.3b
D
16.1b
See sol.
16.6e
D
13.3b
A
14.3c
B
16.1c
D
17.1a
D
13.3c
A
14.3d
C
16.1d
C
17.1b
Compiler
13.3d
A
14.3e
D
16.1e
B
17.1c
ALU
13.3e
C
14.4a
See sol.
16.If
A
17.Id
A
13.4a
See sol.
14.4b
See sol.
16.2a
D
17.le
C
13.4b
-8/3
14.4c
C
16.2b
A
17.If
B
13.4c
B
14.4d
D
16.2c
B
17. lg
B
13.4d
0 < k <5
14.4e
See sol.
16.2d
B
17.Ih
A
13.4e
k > -1
14.5a
A
16.2e
C
17.li
B
13.5a
A
14.5b
C
16.3a
B
17. Ij
B
13.5b
C
14.5c
C
16.3b
C
17.2a
B
13.5c
D
14.5d
400 bps
16.3c
A
17.2b
A
13.5d
B
14.5e
B
16.3d
C
17.2c
A
13.5e
A
14.5f
20|is
16.3e
B
17.2d
B
13.6a
B
15.1a
Routing
16.4a
B
17.2e
C
13.6b
D
15.1b
D
16.4b
D
17.2f
Data bus
13.6c
C
15.1c
C
16.4c
C
17.2g
Address
13.6d
A
15.Id
A
16.4d
A
17.3a
D
13.6e
D
15.le
C
16.4e
C
17.3b
See sol.
13.6f
A
15.If
B
16.4f
C
17.3c
C
14.1a
A
15.2a
D
16.5a
C
17.3d
C
14.1b
D
15.2b
D
16.5b
B
17.3e
A
159
Copyrighted Material © 2018
Problem
Answer
17.3f
C
17.3g
B
18.1a
7
18.1b
-2
18.1c
29
18.Id
25
18.16
125
18.2a
C
18.2b
B
18.2c
Complete
18.2d
A
18.2e
D
18.2f
B
18.2g
C
18.3a
D
18.3b
Fragile
18.3c
A
18.3d
A
18.3e
B
18.3f
A
18.3g
B
18.3h
D
18.3i
D
160
Copyrighted Material © 2018
Chapter #1 - Mathematics
1.1 Algebra and Trigonometry
Consult NCEES® FE Reference Handbook - Pages 23 - 24 for reference
1.1a) CORRECT ANSWER - B
logs(3x - 12) - log3(x) = 2
According to logarithmic identities given in NCEES® FE Reference Handbook:
x
( 3x - 12)
log* - lo g y = log - -» logs (3x - 12) - log3(x) = log3---- — ---The given logarithmic function can be rearranged as follows:
(3x - 12)
l o g 3
o
o
=
2
(3x - 12)
---- — ---- = 32 -> 3x — 12 = 9x
(x)
x = —2
1.1b) CORRECT ANSWERS - B, C
ln x + ln(x + 8) = ln(x — 12)
According to logarithmic identities given in NCEES® FE Reference Handbook:
log* + lo g y = lo g xy -> ln x + ln(x + 8) = ln x ( x + 8)
The given logarithmic function can be rearranged as follows:
ln x (x + 8) = ln(x — 12)
Taking anti-log on both sides of equations results in following:
x{x + 8) = x — 12 -> x 2 + 8x = x — 12
x 2 + Ix + 12 = 0
x = —4, —3 solved using quadratic equation.
Helpful tip - Review logarithmic identities given in NCEES® FE Reference Handbook.
Note: According to latest exam format, it is possible to see a question with multiple correct answers.
1.1c) CORRECT ANSWER - C
log4(x2 -f 8x + 17) = 0
x 2 -I- 8x + 17 = 4° = 1
x 2 + 8x + 16 = 0
x = —4 can be obtained by solving above given quadratic equation.
Helpful tip - Learn how to solve quadratic equations using calculator.
161
Copyrighted Material © 2018
I . Id) CORRECT ANSWER - log10 7 / log10 5
This problem falls under the category of 'Logarithms'.
According to logarithmic details given in NCEES® FE Reference Handbook:
log&x = loga x / loga b
log5 7 can be expressed in log-base 10 as follows:
log5 7 = log10 7 / lo g 10 5
l.le ) CORRECT ANSWER - B
This problem falls under the category of Trigonometry'.
According to the trigonometric identities given in NCEES® FE Reference Handbook:
1
cos2 X
cot2 x = — 5— = —t -z—
tan2 x
sin 2 x
sin 2 x + cos2 x = 1
Substituting this identity in given trigonometric expression results in:
cos2 X
\
—r—x---- 1 - 1 = sin 2 x + cos2 x = 1
(
sin 2 x
)
l. lf ) CORRECT ANSWER - B
This problem falls under the category of 'Trigonometry'.
According to the trigonometric identities given in NCEES® FE Reference Handbook:
1
cos2 X
cot2 X = — r— = — -—
tan2 x
sin 2 x
sin 2 X + cos2 X = 1
Substituting this identity in given trigonometric expression results in:
cos2 x
cos2 x
cot
x
sin2 x
(cot2 x + 1 ) = ---------------/ cos2 x
A — ----/(
---------------— — — = --- ---- — cos^ x
x
o
s
2
x
+
sin
2
x^
\sin 2 x
/
V
sir# x
l.lg ) CORRECT ANSWER - C
This problem falls under the category of 'Trigonometry'.
(sin x + cos x )2 — 1 = (sin2 x + cos2 x + 2 sin x cos x) — 1
According to the trigonometric identities given in NCEES® FE Reference Handbook:
sin2 x + cos2 x = 1
s in 2 x = 2 sin x cos x
Therefore, (sinx + cosx)2 - 1 = (1) + (sin2x) - 1 = sin2x
162
Copyrighted Material © 2018
cos2 x
l. lh ) CORRECT ANSWER - 2m
This problem falls under the category of 'Trigonometry'.
According to the 'Law of Sines' given in NCEES® FE Reference Handbook:
a
b
c
sinv4
sin B
sinC
a
sin 60°
a=
2m
sin 60°
(2m) (sin 60°)
sin 60°
= 2m
Note - Equilateral triangle has equal sides and angles.
163
Copyrighted Material © 2018
1.2 Complex Numbers - Solutions
Consult NCEES® FE Reference Handbook - Page 23 for reference
1.2a) CORRECT ANSWER - B
This problem falls under the category of 'Algebra of Complex Numbers'.
cl — 8 + 4j
b — 5 + 0;
a — b = (8 + 4/) - (5 + Oj)
a — 6 = 3 + 4)
Polar coordinate representation can be found as shown below:
r = |3 + 4;| = ^ 3 2 + 42 = 5
Therefore, a — b = 5/53° in polar format.
Helpful tip - Complex number arithmetic including conversions can be performed quickly using calculators.
1.2b) CORRECT ANSWER - C
This problem falls under the category of 'Algebra of Complex Numbers'.
a = 4 + 4/ = 5.66/45° in polar format.
b = 2 + 3j -3 .6 1 /56.3° in polar format.
a/b = (5.66/45°)/(3.61/56.3°)
a/b = (1.56/45°-56.3°) = (1.56/-110)
Helpful tip - Complex number division can be performed easily in polar form.
1.2c) CORRECT ANSWER - C
This problem falls under the category of 'Algebra of Complex Numbers'.
a = 2/30°
b = 4/15°
a x b = (2/30°) (4/15°) = 8/45°
a x b = 8 cos 45° + j 8 sin 45° = 5.7 + 5.7j
Helpful tip - Complex number multiplication is easier in polar form.
164
Copyrighted Material © 2018
1.2d) CORRECT ANSWER - C
This problem falls under the category of 'Algebra of Complex Numbers'.
a —4/30° = 4 cos 30° + ;4 sin 30° = 3.46 + 2j
b = 6/30° = 6 cos 30° + ;6 sin 30° = 5.2 + 3;
a + b = (3.46 + 2j) + (5.2 + 3;) = 8.66 + 5;
a + b — 8.66 + 5;
Helpful tip - Complex number additions and subtraction are easier in rectangular form.
1.2e) CORRECT ANSWER - D
This problem falls under the category of 'Algebra of Complex Numbers'.
5(cos 53 + ; sin 53) is rectangular form of 3 + 4;.
SeJ'53 is Euler's form of 3 + 4j.
5/53° is polar form of 3 -I- 4j.
Therefore, all given options are accurate representation of 3 + 4j.
1.2f) CORRECT ANSWER - -1 2 - 24;
This problem falls under the category of 'Algebra of Complex Numbers'.
Z = ( 2 + 6j ) x (3 + 3;)
Z = 6+
6j + 18j + 18j2
Z = 6+
24; + 18(—1)
Z = 6 +24; - 18 = -12 + 24j
Z* = -12 - 24j
165
Copyrighted Material © 2018
1.3 Discrete Mathematics and Progressions - Solutions
Consult NCEES® FE Reference Handbook - Page 22 for reference
1.3a) CORRECT ANSWER - C
This problem falls under the category of 'Discrete Math - Set Theory'.
Set A is defined as the proper subset of Set B if every element in Set A is also present in Set B, and there exists at
least one element in Set B which is not present in Set A.
According to above given definition only {2,4, 6} qualifies as a proper subset of {2,4, 6, 8,10,12}.
1.3b) CORRECT ANSWER - B
This problem falls under the category of 'Discrete Math - Set Theory'.
Set A is defined as a subset of Set B if every element in Set A is also present in Set B. Equal sets are also subsets.
According to above given definition {a, b, c, d, e} qualifies as a subset of {a, b, c, d, e}.
1.3c) CORRECT ANSWER - A
This problem falls under the category of 'Discrete Math - Set Theory'.
{1, 2, 3 , 4, 5} and {a, b, c, d, e} are examples of disjoint sets because they don't share any common element.
1.3d) CORRECT ANSWER - B
This problem falls under the category of 'Discrete Math - Set Theory'.
Cartesian product of A x B contains ordered pairs in the format (a, b).
Therefore, {(l,a),(l,b),(l,c),(l,d),(2,a),(2,b),(2,c),(2,d)} is the cartesian product of {1,2} x {a, b, c, d}.
1.3e) CORRECT ANSWER - A
This problem falls under the category of 'Discrete Math - Function Characteristics'.
Function is defined as a set of relations between inputs (domain - x) and outputs (range - y) such that each
input is related to only one output.
According to above given definition, {(1, a), (1, b), (1, c)} is not a function because input - 1 is related to multiple
outputs a, b and c.
1.3f) CORRECT ANSWER - B
This problem falls under the category of 'Discrete Math - Function Characteristics'.
{(a, 1), (b, 1), (c, 2), (d, 2)} is a surjective function because each output is linked to at least one input.
Helpful tip - Understand the difference between injective, surjective and bijective functions.
166
Copyrighted Material © 2018
1.3g) CORRECT ANSWER - 70
According to the problem statement, progression is 2,4, 6, 8 ,1 0 ,1 2 .....
It can be observed that given progression is Arithmetic such that:
First term value 'a! = 2, Difference between terms rdr = 2, # of terms V = 35, last term 'V ■
According to the arithmetic progression formula given in NCEES® FE Reference Handbook:
I = a + ( n — l)d
l 35 = 2 + (35 - 1)2 = 70
Therefore, 35th term of the given arithmetic progression is 70.
Helpful tip - Review arithmetic progression formulas given in NCEES® FE Reference Handbook.
1.3h) CORRECT ANSWER - 15600
This problem falls under the category of 'Progressions and Series'.
According to the problem statement, we are given arithmetic progression such that:
First term value 'a' = 1, # of terms ln' = 120, last term T = 259
According to the arithmetic progression formula given in NCEES® FE Reference Handbook:
S = n(a + Z)/2
120(1 + 259)
5 = ----- ----------- L = i 5 6oo
Therefore, sum of given arithmetic progression is 15600.
1.31) CORRECT ANSWER - 531441
This problem falls under the category of 'Progressions and Series'.
According to the problem statement, progression is 3, 9, 27, 8 1 .....
It can be observed that given progression is geometric such that:
First term value 'a' = 3, common ratio 'r' = 3, # of terms 'n' = 12, last term 'V =?
According to the geometric progression formula given in NCEES® FE Reference Handbook:
I = a r n_1
l12 = (3)(3)12-1 = 531441
Therefore, 12th term of the given geometric progression is 531441.
Helpful tip - Review geometric progression formulas given in NCEES® FE Reference Handbook.
167
Copyrighted Material © 2018
l.Bj) CORRECT ANSWER - 29524
This problem falls under the category of 'Progressions and Series'.
According to the problem statement, progression is geometric such that:
First term value V = 1, common ratio Y = 3, last term T = 19683
According to the geometric progression formula given in NCEES® FE Reference Handbook:
S = (a - r/)/( 1 - r)
5r _
_ 1 - (3 x 19683) = 29524
_
_
Therefore, sum of given geometric progression is 29524.
1.3k) CORRECT ANSWER - R2
A n B is the intersection of Sets A and B.
R2 represents overlapping areas of Sets A and B.
U
R4
168
Copyrighted Material © 2018
1.4 Analytic Geometry - Solutions
Consult NCEES® FE Reference Handbook - Pages 22,24 - 27 for reference
1.4a) CORRECT ANSWER - D
Angle a between two straight lines is given by following equation:
„
a = tan - 1
m 7 —rri-]
=-------
1 + m 1m2
m1 and m2 are slopes of lines L± and L2 respectively.
According to problem statement:
yt = xt + 4 = (1)*! + 4
m1 = 1
y2 = Sx2 + 6 = ( 5 ) x 2 + 6 m2 = 5
5 -1
4
= tan- 1 - = 33.7°
a = tan- 1 ----- —
1 + (5)(1)
6
Helpful tip - Review straight line equations given in NCEES® FE Reference Handbook.
1.4b) CORRECT ANSWER - C
The standard form of straight line equation is given below.
y —mx 4- b
According to problem statement, straight line passes through (2, 10) and (3,12).
y2 - V i
1 2 -1 0
m = --------- = —-— — = 2
x2 —Xi
3 —2
y = 2x + b
Line passes through (2,10), therefore, 10 = 2(2 ) + b
b = 1 0 -4 = 6
Therefore, equation of a straight line passing through (2,10) and (3,12) is y = 2x + 6
1.4c) CORRECT ANSWER -y = 2 x - S
The standard form of straight line equation is given below.
y = mx + b
According to problem statement, slope is 2 and intercept is -5.
y = (2)x + ( - 5 )
Therefore, equation of a straight line with slope 2 and intercept -5 is y = 2x —5.
169
Copyrighted Material © 2018
1.4d) CORRECT ANSWERS - A, B
The equation of straight line given in the problem statement is y = 4 x + 4 according to which slope is 4.
Therefore, perpendicular line will have a slope of —1/4 and its general equation will be y = —x/4 + c
Straight line equations given below meet above mentioned requirements:
x
y = -4
y = —x /4 + 4
Note: According to latest exam format, it is possible to see a question with multiple correct answers.
1.4e) CORRECT ANSWER - A
This problem falls under the category of 'Conic Sections'.
Standard forms of parabola equations are given below.
(x — h)2 = 2p (y — k)
(y “ k) 2 = 2p(x — h)
It can be observed that conic equation containing either x 2 o r y 2 represents a parabola.
1.4f) CORRECT ANSWER - C
This problem falls under the category of 'Conic Sections'.
2(x — 10)2 + 8(y — 6)2 = 200
Let us first convert the given conic section equation into standard form as shown below:
2(x - 10)2
200
8 (y - 6)2 __ 200
+
200
“ 200
(x - 10)2
100
+
(y - 6)2 _
(x - 10)2
25
(10)2
“ 1
(y - 6)2
+
(5)2
1
It can be observed that given conic section is an ellipse (standard form of ellipse given below for reference).
( x - h )2 ( (y - k)2 _ ^
In our case, a = 10, b = 5, h = 10, k = 6.
1.4g) CORRECT ANSWER - 0.87
For the given ellipse, a = 10, b = 5, h = 10, k = 6.
Eccentricity of an ellipse is given by following equation:
e = V 1 - b2/a2 = V l - 25/100 = 0.866
Helpful tip - Understand the difference between equations of ellipse about x-axis and y-axis.
170
Copyrighted Material © 2018
1.4h) CORRECT ANSWER - B
This problem falls under the category of 'Conic Sections'.
2(x - 10)2 - 8 (y - 6)2 = 200
Let us first convert the given conic section equation into standard form as shown below:
2(x - 10)2
200
8(y - 6)2 _ 200
(x - 10)2
( y - 6)2 _
200
(x - 10)2 (y - 6)2 _
~ 200 10025“ 1(10)2
(5)2
1
It can be observed that given conic section is hyperbola (standard form of hyperbola given below for reference).
(x - h)2
(y - k)2
a2
b2
~
1.4i) CORRECT ANSWER - 1.1
For the given hyperbola, a = 10, b = 5, h = 10, /c = 6.
Eccentricity of hyperbola is given by following equation:
e = V 1 + b2/a2 = V l + 25/100 = 1.1
1.4j) CORRECT ANSWER - A
This problem falls under the category of 'Conic Sections'.
(y — 8)2 = 4(x - 2) = 2(2)(x - 2)
It can be observed that given conic section is a parabola (standard form of parabola given below for reference).
(y — k) 2 = 2p(x — h)
1.4k) CORRECT ANSWER - X = - 1
For the given parabola, p = 2.
Directrix of parabola is given by following equation:
1.41) CORRECT ANSWER - C
According to the formulas given in NCEES® FE Reference Handbook, volume (V) of a right circular cylinder is:
V = nr2h = 7r(lm )2(2m) = 6.28 m 3
Helpful tip - Review formulas of different geometric shapes given in NCEES® FE Reference Handbook.
171
Copyrighted Material © 2018
1.4m) CORRECT ANSWER - A
This problem falls under the category of 'Mensuration of Areas and Volumes'.
According to the formulas given in NCEES® FE Reference Handbook, area (A) of a right circular cone is:
A = nr ( r + ^ r 2 + h2^
According problem details: A = n r(r + V r 2 + h2) - A - n(2m) (2m + ^ (2m)2 + (4m)2) = 40.6m2
1.4n) CORRECT ANSWER - B
This problem falls under the category of 'Mensuration of Areas and Volumes'.
According to the formulas given in NCEES® FE Reference Handbook:
nr2h
^right circular cone =
^
brig h t circular cylinder ~
h
nd2h
^paraboloid o f rev ~
g
nr2h
”
^2
It can be observed that for a given radius and height, right circular cylinder can hold more volume of liquid than
right circular cone and paraboloid of revolution.
172
Copyrighted Material © 2018
1.5 Calculus - Solutions
Consult NCEES® FE Reference Handbook - Pages 27 - 29 for reference
1.5a) CORRECT ANSWER - C
This problem falls under the category of 'Differential Calculus'.
fipc) = 2 tan2 x + sin2 x
.
d
d
f (x) = 2 — tanz x -I- — s n r x
dx
dx
/
d
\
d
f (x) = 2 2 tan x — tan x -I- 2 sin x — sin x
\
dx
J
dx
According to the derivatives given in NCEES® FE Reference Handbook:
d
dx
d
ta n x = secz x
— sin x = cos*
dx
Substituting these formulas results in given indefinite integral results in:
f i x ) = 4 tan x sec2x + 2 sin x cos x
1.5b) CORRECT ANSWER - A
This problem falls under the category of 'Differential Calculus'.
f i x ) = 4x2 + 6x + 2y 2
d ,
d
d
/ (x) = 4 — x 2 + 6 — x + 2 — y 2
dx
dx
_
dx
According to thederivatives given
in NCEES® FE Reference Handbook:
d_
xn = n x n 1
dx
f i x ) = 4{2x) + 6(1) + 2(0)
f i x ) = 8x + 6
1.5c) CORRECT ANSWER - B
This problem falls under the category of 'Differential Calculus'.
f i x ) = 2 tan x sec x
d
f (x) = 2 —- (tan x sec x)
dx
d
d
173
Copyrighted Material © 2018
According to the derivatives given in NCEES® FE Reference Handbook:
d
ax
— ta n x = secz x
d
— secx = secx tanx
ax
Substituting these formulas results in given indefinite integral results in:
f '( x ) = 2(tan x (secx tanx) + se cx(sec2x))
f '( x ) = 2(tan2 x s e c x + sec3 x)
1.5d) CORRECT ANSWER - D
This problem falls under the category of 'Differential Calculus'.
/ ( x ) = 2 sin-1 x + 2 cos-1 x
f r(x) = 2 -— (sin_1x) + 2 — (cos- 1 x)
cLoc
CLOC
According to the derivatives given in NCEES® FE Reference Handbook:
d
_1
— Sin 1 X =
dx
...
1
V l- X 2
d
—1
dx
Vl -x 2
— COS 1 X = - 7 =
Substituting these formulas results in given indefinite integral results in:
f(x ) = 0
1.5e) CORRECT ANSWER - B
This problem falls under the category of 'Differential Calculus'.
According to the problem statement:
/ ( x ) = 4 x 3 + x 2 — 2x + 8
—1 < x < 1
To calculate the maximum and minimum of a function, we calculate first order derivative as shown below:
f ' ( x ) = 12x 2 + 2x — 2
The roots of f ' ( x ) can be calculated using factorization, quadratic equation or calculator.
1
12x + 2 x — 2 ^ x = — ,x = —1/2
Now, we need to calculate second order derivative f " ( x ) = 24x + 2
Substituting x = —1/2 in second derivative equation results in negative value. Therefore, it is a local maximum.
Substituting x = 1/3 in second derivative equation results in positive value. Therefore, it is a local minimum.
174
Copyrighted Material © 2018
1.5f) CORRECT ANSWER - B
This problem falls under the category of 'Differential Calculus'.
According to the problem statement:
f { x) = 3x3 + 3x2 — 3x + 3
—2 < x < 1
To calculate the maximum and minimum of a function, we calculate first order derivative as shown below:
f i x ' ) = 9x2 + 6x —3
The roots of f i x ' ) can be calculated using factorization, quadratic equation or calculator.
1
9x2 + 6 x - 3 = 0 - * x = - , x = —1
Now, we need to calculate second order derivative f " ( x ) = 18x + 6
Substituting x = —1 in second derivative equation results in negative value. Therefore, it is a local maximum.
Substituting x = 1/3 in second derivative equation results in positive value. Therefore, it is a local minimum.
1.5g) CORRECT ANSWER - X = -1/3
In problem 1.5f) f " ( x ) = 18% + 6
Point of inflection is calculated as shown below:
f !(x) = 18* + 6 = 0 -» x = —1/3
The above given value of x is a point of inflection because at this point second derivative is zero and second
derivative also changes sign as x increases through it (test by substituting incremental values).
1.5h) CORRECT ANSWER - C
This problem falls under the category of 'Differential Calculus - L'Hospital's Rule'.
According to the problem statement:
3x2 —2x —l _ 3(1)2 - 2 ( 1 ) - 1 _ 0
4 x 2 + 6x —10 4(1)2 + 6(1) — 10 0
0/0 is an indeterminate form (co/co is the other indeterminate form).
Therefore, we need to use L'Hospital's rule to evaluate this limit.
3 x2 - 2 x - l
jim — -—
7E (3x2 ~ 2x ~ !)
6 x — 2 _ 6 (1 ) - 2
2
(4x2 + 6x — 10)
i ™ 8 x + 6 - 8 (l) + 6
7
---- — = lim -^r--------------------
175
Copyrighted Material © 2018
1.51) CORRECT ANSWER - C
This problem falls under the category of 'Differential Calculus - L'Hospital's Rule'.
According to the problem statement:
x->n /2
4 cos*
4 cos 7t/2
4(0)
0
2 — 2 s in x
2 — 2sin7r/2
2 — 2(1)
0
0/0 is an indeterminate form.
Therefore, we need to use L'Hospital's rule to evaluate this limit.
n
= 2 tan — = oo
2
dx
1.5j) CORRECT ANSWER - D
This problem falls under the category of 'Integral Calculus'.
f
I
4
-d x = 4
J x+3
f
1
------- dx
J x+3
According to the indefinite integrals given in NCEES® FE Reference Handbook:
1
-d x = - I n |ax + b |
J ax + b
a
f
I
1
Substituting this formula in given indefinite integral results in:
1
4 - I n \x + 3| = 41n \x + 3|
Helpful tip - Review tables of derivatives and indefinite integrals given in NCEES® FE Reference Handbook.
1.5k) CORRECT ANSWER - C
This problem falls under the category of 'Integral Calculus'.
j (sin2 x + cos2 x) dx
According to the trigonometric identity given in NCEES® FE Reference Handbook:
sin2 x 4- cos2 x — 1
Substituting this formula in given indefinite integral results in:
j (sin2 x + cos2 x) dx = J 1 dx = x
176
Copyrighted Material © 2018
1.51) CORRECT ANSWER - ^
4
\
4
This problem falls under the category of 'Integral Calculus'.
I xe2x dx
Jo
According to the trigonometric identity given in NCEES® FE Reference Handbook:
f
e ax
J
az
I xeax dx = —7T (ax — 1)
Substituting this formula in given indefinite integral results in:
1.5m) CORRECT ANSWER - C
Derivative is highest when rate of rise over run is highest. In other words, it's a measure of steepness.
It can be observed that the given function is steepest at point C.
1.5n) CORRECT ANSWER - B
Derivative is lowest when rate of rise over run is lowest.
It can be observed that the given function has negative slope at point B.
177
Copyrighted Material © 2018
1.6 Differential Equations - Solutions
Consult NCEES® FE Reference Handbook - Pages 30 - 31 for reference
1.6a) CORRECT ANSWER - D
This problem falls under the category of 'Differential Equations'.
2y' + 4y = 0
y(0) = 6
Standard form of 1st order differential equation with constant coefficient is given below:
yf + ay = Q
Solution of standard equation is y = Ce~at
Divide given equation by 2 to convert into standard form y' + 2y = 0
In our case, a = 2. Therefore, general solution is y = Ce~2t.
We can calculate coefficient using initial conditions as shown below:
y(0) = 6
6 = C e(-2X°)
C= 6
Therefore, y = 6e~2t
-~x3
1.6b) CORRECT ANSWER - y = .........
This problem falls under the category of 'Differential Equations'.
y r + 2x 2y = x 2
We can use method of separation of variables to solve this differential equation as shown below.
— + 2 x 2y = x 2 -» — = x 2 — 2 x2y -» — = x 2( l — 2y)
CvA1
CvtA>
•
Taking integrals on both side of equations results in following:
1
2
x
ln | l - 2y| = y
2
ln | l - 2y| = - - x 3
Helpful tip - Learn separation of variables and integrating factor methods for solving 1st order differential
equations.
178
Copyrighted Material © 2018
1.6c) CORRECT ANSWER - y = (C1 + C2x)e~3x
This problem falls under the category of 'Differential Equations'.
y " + 6y' + 9y = 0
We can use method of undetermined coefficients, as explained in NCEES® FE Reference Handbook.
The solution will be of form y = Cerx with characteristic equation: r 2 + 6r + 9 = 0
Comparing this equation with standard form r 2 + ar + b = 0 shows that a = 6, b = 9.
Solving r 2 + 6r + 9 = 0 results in r1 = r 2 = —3
Since a2 = 4b = 36, solution is critically damped and can be represented as follows.
y — (.Ci + C2x)eriX -> y — (Ci + C2x)e~3x
1.6d) CORRECT ANSWER - y = 4e2x + 2e~4x
This problem falls under the category of 'Differential Equations'.
y " + 2y' - 8y = 0
y(0) = 6
y '(0 ) = 0
We can use method of undetermined coefficients, as explained in NCEES® FE Reference Handbook.
The solution will be of form y = Cerx with characteristic equation: r 2 + 2r — 8 = 0
Comparing this equation with standard form r 2 + ar + b = 0 shows that a = 2, b = —8.
Solving r 2 + 2 r — 8 = 0 results in rt = 2,r 2 = —4
Since a2 = 4 > 4b = —32, solution is overdamped and can be represented as shown below:
y = CiLer'lX + C2erzX
y' = C1r1eriX + C2r2er2X
We can calculate coefficients using initial conditions as shown below:
y(0) = 6
6=
+ C2e ^ m = Cx + C2 -> = 6 - C2
y ,(0) = 0 -*
0=
+ C2( - 4 ) e (" 4)(0) = 2 ^ - 4C2 = 0
2
2(6 -
C2) - 4C2 = 0
-
4C2 = 0
12 - 2C2 - 4C2 = 0
Solving these equations results in
= 4 and C2 = 2. Therefore, y = 4e2* + 2e~4x
179
Copyrighted Material © 2018
1.6e) CORRECT ANSWER - Underdamped
This problem falls under the category of 'Differential Equations'.
2 y " + 4y' + 8y = 0
Divide the given equation by 2 to convert into standard form y " + 2y ' + 4 y = 0
We can use method of undetermined coefficients, as explained in NCEES® FE Reference Handbook.
The solution will be of form y = Cerx with characteristic equation: r 2 + 2 r + 4 = 0
Comparing this equation with standard form r 2 + ar + b = 0 shows that a = 2,b = 4.
Since a2 = 4 < 4b = 16, solution is underdamped.
1.6f) CORRECT ANSWER - 2ndorder linear homogeneous differential equation
y " + 8y' + 12y = 0 is an example of 2nd order linear homogeneous differential equation.
Helpful tip - Review differences between several types of differential equations.
180
Copyrighted Material © 2018
1.7 Matrix and Vector Analysis - Solutions
Consult NCEES® FE Reference Handbook - Pages 34 - 35 for reference
1.7a) CORRECT ANSWER - C
This problem falls under the category of 'Matrices'.
According to the problem statement:
O'
0
1.
0
B= 0
.2
0
2
0
1 +0
A+ B = 0+0
.0 + 2
0+0
1+2
0+0
0 + 2'
1
=
0+0
0
.2
1 + 0.
1
A= 0
.0
0
1
0
2
0
0.
0
3
0
2'
0
1.
Helpful tip - Matrix operations (addition, subtraction, multiplication, inverse) using calculator can save time and
reduce chances of error. Alternatively, you may perform conversion by hand and cross-check it with calculator.
1.7b) CORRECT ANSWER - D
This problem falls under the category of 'Matrices'.
According to the problem statement:
1
A= 0
0
0 0
1 0
0 1
B
rl
0‘
Lo
1-
Matrix addition requires both matrices to have same number of rows and columns. In this case, the two
matrices do not have same dimensions. Therefore, given matrices cannot be added.
1.7c) CORRECT ANSWER - A
This problem falls under the category of 'Matrices'.
According to the problem statement:
2
A= 4
L6
1
2
3J
B
_ rl
0
L2
1
'2 x 1 + 1 x 2
AxB = 4 x 1+ 2 x 2
.6 x 1 + 3 x 2
2 x 0 + 1 x 1'
’4
4x 0+2x 1 = 8
6 x 0 + 3 x 1.
.12
V
2
3.
1.7d) CORRECT ANSWER - 14
This problem falls under the category of 'Matrices'.
4
det J
2
51
6
(4 x 6) - (2 x 5) = 14
Helpful tip - Learn how to find to find determinant of a 2 x 2 matrix using calculator.
181
Copyrighted Material © 2018
1.7e) CORRECT ANSWER -
1
7
_5_
14
2
7
Matrix inverse is calculated using following equation
[4
[2
A- i = adj(A) _ ad) b
5]
r 6
6] _ L—2
[
6J
MM
-5
4
14
]
5
6/14
-2/14
-5 /1 4
4/14 .
14
2
7 ■
Helpful tip - Learn how to find inverse of a 2 x 2 matrix using calculator.
1.7f) CORRECT ANSWER - D
According to the problem statement:
2
A= 4
.6
r
2
3.
7
B= 8
.9
10
11
12.
Matrix multiplication requires columns of first matrix to be equal to the number of rows in second matrix. In this
case, the two matrices do not meet this requirement.
1.7g) CORRECT ANSWER - A
This problem fails under the category of 'Matrices'.
A x B is not always equal to B x A.
1.7h) CORRECT ANSWER - 0
This problem falls under the category of 'Matrices'.
'2
det 8
.1
4
10
3
6'
12 = 2[(10)(5) - (3)(12)] - 4[(8)(5) - (12)(1)] + 6[(8)(3) - (10)(1)]
5.
‘2
det 8
.1
4
10
3
6‘
12 = 2(14) - 4(28) + 6(14) = 0
5.
Helpful tip - Learn how to find determinant and inverse of a 3 x 3 matrix using calculator.
1.7i) CORRECT ANSWER -16
A = 2 i+ j + 3k
B = i + 2j
4k
A.B = (2 i +j + 3k). (i + 2j + 4fc)
1 1 = 2.1 + 1.2 + 3.4 = 16
Helpful tip - Learn how to calculate angle between vectors ^4 and B using dot product.
182
Copyrighted Material © 2018
1.7j) CORRECT ANSWER - - 4 i + ; + 10k
A = 3 i + 2j + k
B = i + 4/ 4- Ok
i
2 x S = (3£ + 2; + fc) x (£ + 4/ + Ok) = 3
j
2
k
1
1 4
0
A x B = (2 x 0 - 1 x 4 )i - (3 x 0 - 1x 1)/ + (3 x 4 - 2 x l)fc
A x B - —4i + j + 10 k
Helpful tip - Learn how to use calculator to find vectorproduct.
1.7k) CORRECT ANSWER - B
We can use head-tail rule to confirm that resultant vector will be Option B
x + r=
Helpful tip - Review head-tail rule for vector addition and subtraction.
183
Copyrighted Material © 2018
Chapter #2 - Probability and Statistics
2.1 Measures of Central Tendencies and Dispersions
Consult NCEES® FE Reference Handbook - Page 37 for reference
2.1a) CORRECT ANSWER - 6
Arithmetic mean of given data set can be calculated as shown below.
_
1
36
X = - (2 + 4 + 1 0 + 8 + 4 + 8 ) = — =6
6
6
2.1b) CORRECT ANSWER - 88.2°F
Weighted average of given data set can be calculated as shown below.
— _ d WiXi) _ (0.50 x 90°F) + (0.30 x 84°F) + (0.10 x 88°F) + (0.10 x 92°F)
Xw ~
Zwi
~
0.50 + 0.30 + 0.10 + 0.10
~~88'2°F
Helpful tip - Arithmetic mean X is a special case of weighted average in which all weights are equal.
2.1c) CORRECT ANSWER - J 20/3
Sample standard deviation is calculated by using the equation given below:
ZU V i-xy
s =
N
n —1
1
X = —(2 + 4 + 6 + 8) = 5,
4
_
n= 4
(2 - 5)2 + (4 - 5) 2 + (6 - 5)2 + (8 - 5)2
s =
M
(3)2 + ( l ) 2 + ( l ) 2 + (3)2
4 -1
s = V 20/3
Helpful tip - It is important to understand the difference between 'sample' and 'population' because formulas
are different for sample and population.
2.Id) CORRECT ANSWER - 2.21
Sample geometric mean is calculated by using the equation given below:
s = y x xx 2x 3 ~ X n
s = V (l)(2 )(3 )(4 )
s = V24 = 2.21
Helpful tip - Review measure of central tendency/dispersion formulas given in NCEES® FE Reference Handbook.
184
Copyrighted Material © 2018
2.1e) CORRECT ANSWER - 6.9
Sample root mean square is calculated by using the equation given below:
s =
N
s =
s =
\ Y *
—(32 + 52 + 62 + l l 2)
4
M
1
- ( 9 + 25 + 36 + 121) = 6.9
4
M
2.If) CORRECT ANSWER - 3
Let us first arrange the given data in ascending order.
2 ,3 ,7 ,1 ,4 ,9 ,0 ^ 0 ,1 ,2 ,3 ,4 ,7 ,9
It can be observed that the median of given data set is 3 because it bisects the data set.
2.1g) CORRECT ANSWER - 80
Let us first arrange the given data in ascending order.
90,60,70,110,50,40,200,210
40,50,60,70,90,110,200,210
( 7l\ ^
It can be observed that we have even number of terms (n = 8) due to which median will be average of ( j ) and
^ + lj
terms. Therefore, median = (70 + 90)/2 = 80
2.1h) CORRECT ANSWER - 3
Mode of a given data set is the value that occurs with greatest frequency. It can be observed that in the given
data set 1, 3, 3,4, 9, 7, 3,4, the value 3 occurs with highest frequency.
Helpful tip - There can be more than one modes in a data set. For instance, in this example if we replace 9 with
4, then it becomes a bimodal data set with two modes i.e. 3 and 4.
2.1i) CORRECT ANSWER -110
According to the definition given NCEES® FE Reference Handbook, sample range is the difference between
largest and smallest sample value.
In our case, largest and smallest sample values are 120 and 10 respectively. Sample range = 120 - 1 0 = 110.
2.1j) CORRECT ANSWER - Influenced by outliers
Mean is highly influenced by outliers. Standard deviation is a measure of dispersion. Median is marginally
influenced by outliers.
185
Copyrighted Material © 2018
2.2 Probability Distributions - Solutions
Consult NCEES® FE Reference Handbook - Pages 38 - 39 for reference
2.2a) CORRECT ANSWER - C
This problem falls under category of 'Law of Total Probability' which is given by equation below:
P(A + B) = P(A) + P ( B ) ~ P(A, B)
Let us identify the probabilities as shown below:
Probability of sunny weather = P(A) = 0.25
Probability of cloudy weather = P{B) = 0.35
Probability of both sunny and cloudy weather = P(A, B) = 0.15
Probability of sunny weather, cloudy weather or both = P(A + B) = 0.25 + 0.35 — 0.15 = 0.45
2.2b) CORRECT ANSWER - 0.24
This problem falls under the category of 'Bayes' Theorem'.
According to NCEES® FE Reference Handbook, Bayes' Theorem is given by equation shown below:
p(B LQ P(Aj ) is the probability of event Aj in population of A.
P[ Bj ) is the probability of event Bj in population of B.
Let X and Y indicate online and in-store customers respectively.
Probability of a customer preferring online shopping X = P(X) = 0.20
Probability of a customer preferring in-store shopping Y = P(Y) = 0.80
Probability of online customer X under 30 years of age = P(S |X) = 0.50
Probability of in-store customer Y under 30 years of age = P(S |Y) = 0.40
The probability of randomly selected customer under 30 years of age preferring online shopping (X).
P(X\S)
P (J)P (5 |Z )
P(S\X)P(X) + P(S\Y)P(T)
(0.20)(0.50)
P ( XS) =
^
^ = 0.238 = 0.24
v
'
(0.20)(0.50) + (0.80)(0.40)
186
Copyrighted Material © 2018
2.2c) CORRECT ANSWER - 0.228
This problem falls under the category of 'Binomial Distribution' which is given by equation below:
Pn(x) = C(n,x)pxqn~x
n = No. of on-campus interviews.
x = No. of hiring.
q = Failure probability or rejection probability.
p = Success probability or hiring probability.
According to given scenario:
n = 30 x = 2 p = 0.10 q = 0.90
P30( 2) = C(30,2)0.120.930-2 = 0.228 = 22.8%
Therefore, the engineering firm has 22.8% chance of hiring 2 new graduates after 30 interviews.
Helpful tip - Binomial distribution involves binary outcomes such as yes/no, pass/fail, head/tail etc. To gain
better understanding, it is recommended to review Binomial distribution examples and practice problems.
2.2d) CORRECT ANSWER - 0.51
This problem falls under the category of 'Binomial Distribution' which is given by equation below:
Pn(x) = C{n,x')pxqu~x
n = No. of grad slam tournaments.
x = No. of tournament successes.
q —Failure probability.
p = Success probability.
According to given scenario:
n = 4 x = 2,3,4 p — 0.40 q = 0.60
P4(2) = C{4,2)0.420.64~2 = 0.34 = 34%
P4 ( 3 ) = C(4, 3)0.430.64~3 = 0.15 = 15%
P4(4) = C(4, 4)0.440.64~4 = 0.025 = 2.5%
PT = p4( 2) + P4(3) + P4(4) = 0.34 + 0.15 + 0.025 = 0.51
Therefore, tennis player has 51% chance of winning 2 to 4 grand slam tournaments out of 4.
187
Copyrighted Material © 2018
2.2e) CORRECT ANSWER - 0.0228
This problem falls under the category of 'Gaussian distribution'.
According to the problem statement* = 700, (i = 740, a = 20.
Since,
& 0 & a =£ 1 this distribution should be standardized as shown below:
x — \i
700 — 740
Z = ------ - = ------ — ------
Note that, Z < —2 is the same as Z > 2.
According to the unit normal distribution table given in NCEES® FE Reference Handbook:
P{Z > 2) = R{ 2) - 0.0228
Helpful tip - Learn how to use normal distribution table given in NCEES® FE Reference Handbook.
2.2f) CORRECT ANSWER - 0.1359
According to the problem statement xt = 12 x 2 = 14 \i = 10, a = 2.
Since, \i =£ 0 & a =£ 1 this distribution should be standardized as shown below:
1 2 -1 0
2
1 4 -1 0
4
Z 2 = ---- ----- = - = 2
2
2
2
Z± < Z < Z 2
P{Z± < Z < Z 2) = F( 2) - F ( 1) = 0.9772 - 0.8413 = 0.1359
2.2g) CORRECT ANSWER - 0.50
This problem falls under category of 'Law of Total Probability' which is given by equation below:
P(A + B) = P(A ) + P(B) - P(A , B)
P(A + B) = 0.20 + 0.40 - 0.10 = 0.50
U
p (a
nB) = o.i
P(A) = 0.2
P(B) = 0.4
188
Copyrighted Material © 2018
2.3 Expected values and Estimation for a single mean - Solutions
Consult NCEES® FE Reference Handbook - Pages 38 - 39,44 and 47 for reference
2.3a) CORRECT ANSWER - 3.18
This problem falls under the category of 'Expected Values'.
According to formula given in NCEES® FE Reference Handbook:
11
e [x ] = ^ Xfc/Ok) = * i / O i ) + X 2/ O 2) + X3/ O 3)
k ~ l
£[X] = (2) ( i ) ( 3 x 2 + 2) + (3) ( i ) ( 3 x 3 + 2) + (4) ( i ) ( 3 x 4 + 2) = 3.18
Helpful tip - Review probability density function, cumulative distribution function and expected values formulas
given in NCEES® FE Reference Handbook.
2.3b) CORRECT ANSWER - 0.66
This problem falls under the category of 'Expected Values'.
According to formula given in NCEES® FE Reference Handbook:
a 2 = V[X] = E L iO f c - iU)2/ ( x fe) = (*! - M)2/ ^ ) + (x 2 - v ) 2f ( x 2) + (x 3 - v ) 2f ( x 3)
M=
2
+3+4
=3
1
cr2 = V[X] = — [(2 - 3 )2(3 x 2 + 2) + (3 - 3 )2(3 x 3 + 2) + (4 - 3 )2(3 x 4 + 2)] = 0.66
33
2.3c) CORRECT ANSWER - 5. 5%
This problem falls under the category of 'Expected Values'.
Let X be the 'expected performance of fund'. According to formula given in NCEES® FE Reference Handbook:
E[X] = Y t= i xkf ( x k) = (—5%) (0.10) + (0)(0.20) + (5% )(0.30) + (10% )(0.30) + (15% )(0.10) = 5.5%
2.3d) CORRECT ANSWER - 5/6
This problem falls under the category of 'Expected Values'.
According to formula given in NCEES® FE Reference Handbook:
E[X] = r x f ( x ) d x = C x(4 x - 1)dx
E[X] = j (4x2 - x ) d x = ^4x3 ~^x2]l =
5
- 0 = - = 0.83
6
Therefore, expected value of study time spent by student in doing assignment is 0.83 or 83%.
189
Copyrighted Material © 2018
2.3e) CORRECT ANSWER - 2
This problem falls under the category of 'Expected Values'.
According to formula given in NCEES® FE Reference Handbook:
E[X] =
— J T x ( 2x ~3 )dx
f 00
2
E[X\ = J 2x~2dx = (_ 2 + 1 )^ 2+1] f = - 2 x - ' ] ? =
2
/
2\
j ) = [0 - ( -2 )] = 2
Therefore, expected number of traffic violations per driver are 2.
231) CORRECT ANSWER -1
The total area under probability density function and probability mass function is always equal to 1.
2.3g) CORRECT ANSWER - C
This problem falls under the category of 'Estimation for a single mean'.
According to the problem statement, we have a normal distribution with following details:
X = $55,000 n =
60
=
$5,000
C.I = 99%
According to the formula given in NCEES® FE Reference Handbook:
X — Za
< /I < X + Za
2 Vn
2 y/n
Za = 2.5758 for C.I = 99%
2
$5,000
^
$5,000
$55,000 - (2.5758)— — - < [x < $55,000 + (2.5758)- ^ V60
V60
$56,662 < n < $53,338
Therefore, 99% confidence interval for the mean household income is ($53,338, $56,662).
2.3h) CORRECT ANSWER - $645
This problem falls under the category of 'Estimation for a single mean'.
Standard error of mean (SEM) of problem 2.3g) can be calculated as shown below:
a
Vn
=
$5,000
- = $645
V60
Therefore, SEM is $645.
190
Copyrighted Material © 2018
2.3i) CORRECT ANSWER - C
This problem falls under the category of 'Estimation for a single
mean'.
According to the problem statement, we have at-distribution with the following details:
*=100
n = 10
s = 20
C.I = 90%
According to the formula given in NCEES® FE Reference Handbook:
s
X — ta —
<
2 -sJn
_
s
fl < X + ta —
2 Vn
Degrees of freedom v = n — 1 = 9
a 0.1
a = (1 - Cl) = 1 - 0.9 = 0.1 -* - = — = 0.05
ta = 1.83 for C. I = 90% ( a = 0.05) v = 9
2
20
20
100 - 1 . 8 3 - = < m < 100 + 1 . 8 3 - =
V lO
V Io
Therefore, 90% confidence interval for the mean low intensity exercise time is 100 ± 11.5.
2.3j) CORRECT ANSWER - 11.5
This problem falls under the category of 'Estimation for a single mean'.
Margin of error in problem 2.3i) can be calculated as shown below:
s
20
2y n
VIO
ta —
= — 1.83
= 11.5
Therefore, margin of error is 11.5 minutes.
191
Copyrighted Material © 2018
Chapter #3 - Ethics and Professional Practice
3.1 Codes of Ethics & NCEES® Model Law and Rules - Solutions
Consult NCEES® FE Reference Handbook - Pages 3 - 4 for reference
3.1a) CORRECT ANSWER - C
This problem falls under the category of 'Code of Ethics'.
According to rule A3 of Model Rules Section 240.15, Rule of Professional Conduct:
"Licensees shall notify their employer or client and such other authority as may be appropriate when their
professional judgment is overruled under circumstances in which the health, safety, or welfare of the public is
endangered."
Therefore, junior engineer shall escalate his concerns to a higher level.
3.1b) CORRECT ANSWER - D
This problem falls under the category of 'Code of Ethics'.
According to rule B1 of Model Rules Section 240.15, Rule of Professional Conduct:
"Licensees shall undertake assignments only when qualified by education or experience in the specific
technical fields of engineering or surveying involved."
John should review assignment requirements and his skills with new supervisor and then decide accordingly.
3.1c) CORRECT ANSWER - D
This problem falls under the category of 'Code of Ethics'.
According to rule B2 of Model Rules Section 240.15, Rule of Professional Conduct:
"Licensees shall not affix their signatures or seals to any plans or documents dealing with subject matter in
which they lack competence, nor to any such plan or document not prepared under their responsible charge."
Therefore, Sarah should not sign and seal these documents because they were not prepared under her
responsible charge and she may not have sufficient time to review them.
3.Id) CORRECT ANSWER - D
According to rule A4 of Model Rules Section 240.15, Rule of Professional Conduct:
"Licensees shall, to the best of their knowledge, include all relevant and pertinent information in an objective
and truthful manner within all professional documents, statements, and testimony."
Therefore, professional engineer should share all details in his recommendation and act in client's best interest.
192
Copyrighted Material © 2018
3.1e) CORRECT ANSWER - C
This problem falls under the category of 'Code of Ethics7.
According to rule B9 of Model Rules Section 240.15, Rule of Professional Conduct:
"Licensees shall not use confidential information received in the course of their assignments as a means
of making personal profit without the consent of the party from whom the information was obtained."
Therefore, Mark should first obtain consent from client prior to using their unique design on other projects.
3.If) CORRECT ANSWER - C3
This problem falls under the category of 'Code of Ethics'.
Obligation to respect of fellow licensees is discussed in rule C3 of Model Rules Section 240.15, Rule of
Professional Conduct.
3.1g) CORRECT ANSWER - C
This problem falls under the category of 'Code of Ethics'.
Engineers both licensed and non-licensed are expected to conduct themselves professionally.
Ethical problems are not always straight forward.
Engineers possess special knowledge which is not common in public domain like other professions.
It is responsibility of engineers to educate themselves about ethical practices.
193
Copyrighted Material © 2018
3.2 Intellectual Property - Solutions
Consult NCEES® FE Reference Handbook - Page 5 for reference
3.2a) CORRECT ANSWER - A
This problem falls under the category of 'Intellectual Property'.
According to the definitions for trademark, patent, copyright and industrial design provided in NCEES® FE
Reference Handbook, trademark is the most applicable product for advertising agency under the given scenario.
3.2b) CORRECT ANSWER - C
This problem falls under the category of 'Intellectual Property'.
According to the definitions for trademark, patent, copyright and industrial design provided in NCEES® FE
Reference Handbook, patent is the most applicable product for this manufacturer under given scenario.
3.2c) CORRECT ANSWER - B
This problem falls under the category of 'Intellectual Property'.
According to the definitions for trademark, patent, copyright and industrial design provided in NCEES® FE
Reference Handbook, copyright is the most applicable product for the publisher under given scenario.
3.2d) CORRECT ANSWER - D
This problem falls under the category of 'Intellectual Property'.
According to the definitions for trademark, patent, copyright and industrial design provided in NCEES® FE
Reference Handbook, industrial design is the most applicable product for packaging firm under given scenario.
3.2e) CORRECT ANSWER - 1-F, 2-A, 3-B
Trademark - ™
Copyright - ©
Registered Trademark -
®
3.2f) CORRECT ANSWER - C
This problem falls under the category of 'Intellectual Property'.
Stocks, bonds and real estate are examples of tangible assets whereas software program is an example of
intellectual property.
194
Copyrighted Material © 2018
Chapter # 4 - Engineering Economics
4.1 Time value of money - Solutions
Consult NCEES® FE Reference Handbook - Page 131 for reference
4.1a) CORRECT ANSWER -$241,157
i% = 12% P = $25,000, n = 20 years
F =?
F = $25,000 x (~,12%, 20 years^j = $25,000 x 9.6463 = $241,157
Therefore, future worth of given investment will be $241,157.
4.1b) CORRECT ANSWER -$231,225
i% = 4% F = $750,000, n = 30 years
P =?
P = $750,000 x (^,4%, 30 years'^
P = $750,000 x 0.3083 = $231,225
Therefore, present worth of retirement fund is be $231,225.
4.1c) CORRECT ANSWER - B
Uniform Series Compound Amount factor converts an annuity to a future amount.
4.1d) CORRECT ANSWER - A
Capital recovery factor converts a present value to an annuity.
4.1e) CORRECT ANSWER - $17,440
i% = 6% P = $200,000 n = 20 years A =?
A = $200,000 x ^—,6%, 20 years^
A = $200,000 X 0.0872 = $17,440
Therefore, expected yearly payment of given annuity is $17,440.
4.If) CORRECT ANSWER - $392,724
i% = 8% A = $40,000 7i = 20 years P =?
P = $40,000 x |jj-,8%, 20 years^
P = $40,000 x 9.8181 = $392,724
Therefore, Amanda needs to have $392,724 saved today to purchase annuity described in problem.
195
Copyrighted Material © 2018
4.1g) CORRECT ANSWER - $1,699,248
i% = 8% A — $15,000 n = 30 years F =?
F = $15,000 x
, 8%, 30 years^
F = $15,000 x 113.282 = $1,699,248
Therefore, expected future worth of this investment will be $1,699,248.
4.1h) CORRECT ANSWER - 10.47%
r = 10% m = 12 ie =?
/
r\ m (
0.1\12
ie = ( H - — J = \1 + — J - 1 = 0.1047 = 10.47%
Therefore, annual effective interest rate is 10.47%.
4.1i) CORRECT ANSWER - $251,416
We are asked to calculate present value of Sam's overall tuition cost.
It involves annuity and gradient which can be converted to present value as shown below:
P = $40,000 x
2%, 6^ + $2000 x
2%, 6^ = $40,000 x 5.6014 + $2000 x (13.6801) = $251,416
196
Copyrighted Material © 2018
4.2 Cost estimation - Solutions
Consult NCEES® FE Reference Handbook - Pages 131 -132 for reference
4.2a) CORRECT ANSWER - D
Note that we are given inflation rate (f) in problem statement hence it also needs to be taken into consideration.
Inflation adjusted interest rate per interest period (d) can be calculated as shown below.
d = i + / + i x / = 6% + 4% + (6% x 4%) = 10.24%
4.2b) CORRECT ANSWER - $4,000
This problem falls under the category of depreciation analysis using straight line method.
We can calculate accumulated depreciation using the formula shown below.
C - Sn $32,000 - $2,000
Dj = -------- = ----------— ----------= $1,000
J
n
30
Accumulated depreciation in year 4 = 4 x $1,000 = $4,000
4.2c) CORRECT ANSWER - $2,968
This problem falls under the category of depreciation analysis using MACRS.
As per the given details, we need to use 10-year recovery period and following formula.
Dj = (factor) C
Accumulated depreciation in year 3 =
actor) C = 10% x $7,000 + 18% x $7,000 + 14.4% x $7,000
Accumulated depreciation in year 3 = $2968
Helpful tip - Salvage value is not relevant in MACRS.
4.2d) CORRECT ANSWER - $26,000
This problem falls under the category of depreciation analysis using MACRS.
As per the given details, we need to use 5-year recovery period and following formula.
Dj = ( f actor) C
Accumulated depreciation in year 2 = J^(factor)C = 20% x $50,000 + 32% x $50,000 = $26,000
4.2e) CORRECT ANSWER - $24,000
This problem falls under the category of depreciation analysis using MACRS.
Book value of the truck in year 2 can be calculated as shown below.
BV = G - ^ D j = $50,000 - $26,000 = $24,000
197
Copyrighted Material © 2018
4.2f) CORRECT ANSWER - $250,000
This problem falls under the category of capitalized costs.
Capitalized costs can be calculated using the formula given below.
A $20,000
Capitalized Cost = P = — = —— —— = $250,000
i
0.08
4.2g) CORRECT ANSWER - Straight line
Straight line depreciation method results in same value decline each year.
198
Copyrighted Material © 2018
4.3 Risk Identification and analysis - Solutions
Consult NCEES® FE Reference Handbook - Pages 131 -132 for reference
4.3a) CORRECT ANSWER - 20,000
This problem falls under the category of break-even analysis.
Cost structure o f Product Line A = $100,000 + $5x
Cost structure o f Product Line B = $10*
At break-even production point, the cost structure of both product lines will be same.
$100,000 + Sx = $10*
Solving for %results in 20,000 items.
4.3b) CORRECT ANSWER - B, C
This problem falls under the category of break-even analysis.
At x = 20,000 both lines are equally economical.
For x > 20,000, Production line A is more economical because it has lower variable cost.
For x < 20,000, Production line B is more economical because it has no fixed cost.
4.3c) CORRECT ANSWER - 2000
This problem falls under the category of break-even analysis.
Chair cost = $50,000 + %2x
Chair sale price = $27x
At break-even production point, the chair sales will cover cost of production.
$50,000 + 2x = %27x
Solving for x results in 2000. Therefore, at least 2000 chairs need to be sold before making any profit.
4.3d) CORRECT ANSWER - Contractor A
This problem falls under the category of 'Benefit-Cost Analysis'.
We can compare given options by looking at their present worth of their costs. The one with lower cost is better.
^Contractor a = $20,000 +
$40,000 X ^ ,8 % , 6^ = $20,000 + $40,000 x 0.6302 = $45,208
^Contractor b = $10,000 x (^, 8%, 6^ = $10,000 x 4.6229 = $46,229
Therefore, contractor A is offering better price.
199
Copyrighted Material © 2018
4.3e) CORRECT ANSWER - Investment Option C
This problem falls under the category of 'Benefit-Cost Analysis'.
We can compare given options by looking at their future worth.
Option resulting in highest future worth should be recommended.
^Option a =
$200,000
F O p t io n B =
$5,000 x
Fovtionc = $100,000 x
10%, l o ) + $160,000 = $5,000 x 15.9374 + $160,000 = $239,687
10%, 10) = $100,000 x 2.5937 = $259,370
It can be observed that Option C offers the best return on investment.
4.3f) CORRECT ANSWER - Decision tree
Decision tree is a risk analysis technique that graphically organizes decision making process by using probability
distributions.
200
Copyrighted Material © 2018
Chapter #5 - Properties of Electrical Materials
5.1Chemical Properties - Solutions
Consult NCEES® FE Reference Handbook - Pages 59 - 60 for reference
5.1a) CORRECT ANSWER - A
Metals with higher standard oxidation potentials become anode during corrosion process. Sacrificial anodes are
used to protect metals against corrosion. According to the table on page 59 of NCEES® Reference Handbook, Fe
has standard oxidation potential of + 0.440 V whereas Zn has a standard oxidation potential of +0.763 V.
Therefore, Zn will act as a sacrificial anode (Cu, Ni and Hg are less electropositive than Fe).
5.1b) CORRECT ANSWER - D
Galvanization prevents corrosion by application of a protective zinc layer to steel. Plating inhibits corrosion by
preventing contact with atmosphere using layer of tin, nickel or chromium. Sacrificial anode prevents corrosion
at expense of more electro positive metal which is purposefully introduced.
5.1c) CORRECT ANSWER - A
According to NCEES® Reference Handbook (Page 60) anode, cathode and an electrolyte are required for
corrosion to take place. Absence of any one of these necessary components can prevent corrosion. This concept
forms basis of all corrosion prevention techniques.
5.Id) CORRECT ANSWER - 7. 56 10~5m2s -1
Diffusion coefficient can be calculated using following equation.
Qa
D = D0e RT
D0 = 7.80 x 10_57722s _1, Qd = 250j . m o r 1 ,T = 973 K,R = 8.314/.m o i-1! -1
Qd
250
D = D0e RT = (7.8 x 10-5)e 8.314x 973 = 7.56 10_5m 2s -1
5.1e) CORRECT ANSWER -256.2 kj. m ol'1
Activation energy Qd is related to diffusion coefficient through equation given below.
Qd
D = D0e RT
Taking natural logarithm of entire equation results in:
In D = In D0 —
Qd
Qd = RTQnD - In D0) = (8.314)(773)(ln 2.7 x 10“ 5 - In 1.3 x 10~22) = 256.2 kJ.mol~1
5.If) CORRECT ANSWER - B
Metals with higher standard oxidation potentials become anode during corrosion process. According to the
table on page 59 of NCEES® Reference Handbook, Zinc has a standard oxidation potential o f +0.763 V whereas
Nickel has a standard oxidation potential o f+0.250 V. Therefore, Zn will act as the anode in this corrosion cell.
201
Copyrighted Material © 2018
5.2 Electrical Properties - Solutions
Consult NCEES® FE Reference Handbook - Pages 60, 200 and 201 for reference
5.2a) CORRECT ANSWER - D
Resistivity is related to resistance, area and length through following equation:
p =—
A = nr 2 = n ^—- — J = 3.141 x 10 ~6m 2 I = 100 m R = 5ft.
RA (5ft)(3.141 x 10~6m 2)
p = — = - — —— — ------------ = 1.57 x 10
H
I
100772
ft.m
5.2b) CORRECT ANSWER - C
Resistivity can be calculated using p = RA/l
According to problem statement, pA = 4pB implies that ^7 ^- = 4 ^ ^
lA
If lA = lB and Ab = -A a then ^
4
M
lB
= 4 MW*)/* =
‘■A
lA
Therefore, RA = RB if cable lengths are equal and area of cable 'B' is one-fourth of cable TV.
5.2c) CORRECT ANSWER - D
Capacitance of a parallel plate capacitor is given by C = s A/d
It can be observed that: Decreasing d 4/, increases C ^
Increasing e i \ increases C i '
Therefore, decreasing distance between plates and increasing dielectric strength will increase capacitance.
5.2d) CORRECT ANSWER - B
By referring to Properties of Materials table on page 61 of NCEES® Reference Handbook, it can be observed that
low electrical resistivity typically corresponds to high heat conductivity.
5.2e) CORRECT ANSWER - 0.015 H / m
Magnetic permeability can be calculated using following equation.
B
I
2nrB
H = - = - ------> [l = — —
|i 2nr
I
According to problem statement, I = 100.4 B = 0.57 r = 50 cm
^
=
2nrB
I
27r(50cm)(0.5T')
= — >..— £>----- = 0.015 H/m
100 ^
'
5.2f) CORRECT ANSWER - D
Photoelectric effect can take place in all forms of matter.
202
Copyrighted Material © 2018
5.3 Mechanical Properties - Solutions
Consult NCEES® FE Reference Handbook - Pages 60 - 62 for reference
5.3a) CORRECT ANSWER - B
Force is related to stress and area through following equation:
F
cj = ---- > F =
A
ox A
According to the problem details, o = 2000 x IQ8 Pa , A = n r 2 = n
—n x 10~6 m 2
F = a x A = (2000 x 108 Pa)(n x 1G“ 6 m 2) = 628.3 kN
5.3b) CORRECT ANSWER - B
Young's modulus (E) can be calculated using following equation:
o
F
AL
5.3c) CORRECT ANSWER - 0.095
True strain eT can be calculated using following equations.
AL
20 mm
e = — = ----------= 0.1
L0 200 mm
sT = ln ( l + e) = ln ( l + 0.1) = 0.095
5.3d) CORRECT ANSWER - B
Plasticity involves permanent deformation. Ductility is the ability to deform under stress (such as copper wires).
Malleability is the ability to convert into thin sheets (such as gold, silver and aluminum).
5.3e) CORRECT ANSWER - C
Lightning is a natural phenomenon due to static electricity. Magnetic flux is the number of magnetic field lines
through a surface. Photoelectric effect is a phenomenon involving electron emission from matter due to energy
absorption from electromagnetic radiation.
5.3f) CORRECT ANSWER - D
Tensile test curve provides information about tensile strength, ductility and young's modulus of a given material.
Material hardness is tested by denting.
203
Copyrighted Material © 2018
5.4 Thermal Properties - Solutions
Consult NCEES® FE Reference Handbook - Pages 60,65 and 201 for reference
5.4a) CORRECT ANSWER - B
Bimetallic strips are fabricated using metals with different coefficients of thermal expansion.
5.4b) CORRECT ANSWER - C
Thermal expansion coefficient can be calculated using following equation.
e
3 x 1 0 -3
_.
a —— = — — — = 4 .2 8 x1 0 K
AT
7K
Tinitiai = 296 K,a = 4.28 x 1Q~4 K ' 1, zreq = 3 x 10~3
6 x 10~3
Erpo
Ar = l T = 4 .2 8 x lO - « * - i = 14*
Treq = Tinitiai + AT =
296 K + 14 K = 310 K
5.4c) CORRECT ANSWER - B
Thermal expansion coefficient is given by following equation:
£ = a x AT = (1.2 x 10"5 °C“ 1) x (25 °C) = 30 x 10~5
a
5.4d) CORRECT ANSWER - 0.04 K 1
Resistance is related to temperature coefficient through following formula:
R = R0 [ l + a(T —T0) ]
According to problem details:
1
2R0 = R0 [ 1 + a ( 25) ] ^ a = — = 0.04 K~x
5.4e) CORRECT ANSWER - D
RA
p
=
—
p =
p 0 [ l +
a(T - T^]
Therefore, resistivity depends on resistance, area, length and the temperature of given material.
5.4f) CORRECT ANSWER - B
Heat capacity is directly proportional amount of material. Sample # 3 will have the highest heat capacity because
it contains the largest about of substance being tested. Sample # 1 has the least amount of substance therefore
it will have the lowest heat capacity. Specific heat capacity of all three samples will be equal.
204
Copyrighted Material © 2018
Chapter #6 - Engineering Sciences
6.1 Work, Energy, Power - Solutions
Consult NCEES® FE Reference Handbook - Pages 200 - 202 for reference
6.1a) CORRECT ANSWER - D
It is important to understand the difference between work, energy and force.
Work done on a point charge to in moving it from p i to p2 in the presence of an electrical field is given by:
W12 = -Q
fPzE. dl = - Q x CVz Qi
JPl
JPl 4 ^
Q2V fPzar21.dl = - Q ± Q2\ r21
4ner?JPl
Vl4
21
.----------------------------------------
Q.Q..
IW12I
ar21-dl = —Q1
r 21 = V ( 2 - ° ) 2 + (° - ° ) 2 + (0 - 0 )z = 2 m
47Tfr2i
M/12 = 9 x 1(T3/
6.1b) CORRECT ANSWER - B
Work and energy have same units of measurements.
Potential energy of given system is equal to the amount of work required to keep charges apart as shown below.
Wyi =
r±2 = V(1 - 0)2+ (0 - l)2+ (0 - 0)2 = V2
W12 = 31.9 m/
6.1c) CORRECT ANSWER - B
d{i) — dq(t)/d(t) -» dq(t) = i(t) x d(t) = 1mA x 5s = SmC
Electron charge = 1.6022 x 10-19C
Number of electrons = 5mC/1.6022 x 10-19C = 3.12 x 1016
6.Id) CORRECT ANSWER - A
Potential energy of System A can be calculated as follows:
W12 =
4
Q1Q2— = 45/
neri2
Potential energy of System B can be calculated as follows:
T.r
Q3Q4
= 4^
Arj
= 4S}
Therefore, potential energies of both systems are equal.
205
Copyrighted Material © 2018
6.1e) CORRECT ANSWER - D
The amount of work required to decrease spacing will be equal to difference system energy at lm and 1cm.
System energy at 1 m charge spacing can be calculated as shown below.
w
=
Q1Q2
4ne(lrri)
= 1215 x 10 ~6J
System energy at 1 cm charge spacing can be calculated as shown below.
W
12
QlQl
A
— = 1.215 x 10-4;
4ns (lcm )
J
- —
The amount of work required is equal to energy difference between two systems.
A = (1.215 x 10“ 4/) - (1.215 x 10“ 67) = 1.202 x 10“ 47
6.If) CORRECT ANSWER - D
Energy stored in a capacitor is given by CV2f 2
sA (8.85 x 10~12F/m )(lm 2)
C = — = ------------- ...... -...... ....... . = 8.85 x 10_11F
d
0.1 m
CV2 (8.85 x 10-11F )(2 0 0 7 )2
= 1--------------- — --------— = 1.77 x 10 ~6J
2
2
J
6.1g) CORRECT ANSWER - B
Work done by an external agent in moving charge Q from PI to P2 in an electric field is given by:
fP2
W = - Q \ E.dl
Jpi
Note that charge is moved along y-axis and electric field is along x-axis.
r P2
W = —Q I
200 Km
1ax. ay
Jp i
Dot product ax. ay = 0 -» W = 0
6.1h) CORRECT ANSWER - 8 W
Electric current passing through a surface at tim e't' is given by following equation:
q 2C
I = - = — = 2A
t
Is
P = I 2R = (2A)2 x 2H = 8 W
206
Copyrighted Material © 2018
6.2 Electrostatics - Solutions
Consult NCEES® FE Reference Handbook - Pages 200 - 202 for reference
6.2a) CORRECT ANSWER - A
Electrostatic force between point charges can be calculated using formula given below.
F =
Qt Q2
(10 x 10“ 6C)(100 x 10“ 6C)
—= -------------------------- — ----------------------------s 90 kN
4nr2e 4n(0.01m)2 x (8.85 x 10 12Fm *)
6.2b) CORRECT ANSWER - A
Potential difference (V) can be calculated using formula given below.
V
E = - ^ v = E x d = 2000 VmT1 x lm = 2000V
a
6.2c) CORRECT ANSWER - C
Potential difference (V) can be calculated using formula given below.
E
V
V = E x d = 1000 VwT1 x 200m = 200 kV
6.2d) CORRECT ANSWER - 9.8 X 10~5C
According to problem details, weight of charge is being balanced by electrostatic force.
Gravitation force on electron = Electrostatic force on electron
mg = qE
((V
V \'
V
V
£=r ms=,b)
mgd
(0.01) x (9.8ms-2) x (0.1m)
V
100F
9.8 x 10“ 5C
6.2e) CORRECT ANSWER - A
Force on a current-carrying conductor can be calculated using following formula:
F = IL x B = lLBsina
a=sin_1Gfe)
J
1 x 10 ~6N
\
a = sin 1 ( — — — ------ — — ~r 7 = 11.5
\SAx 0.1m x 10 x 10~6 /
207
Copyrighted Material © 2018
6.3 Capacitance - Solutions
Consult NCEES® FE Reference Handbook - Pages 200 - 202 fo r reference
6.3a) CORRECT ANSWER - A
Capacitance of a parallel plate capacitor can be calculated as shown below.
eA (8.85 x 10_12F m -1)(0.02m2)
C = — = ----------------— ----- - --------- - = 1.77 x 1 0 "11 F
d
0.01 m
Q
400 fiC
VC = TT =
— L 11 r- = 22.6 106 V
c C 1.77 x 10-11 F
6.3b) CORRECT ANSWER - B
The current-voltage relationship of a capacitor is given by following equation.
1
v. (t) = pc (0 ) + - J
i c ( r ) dx
ic(t) x t
,
(10V - 510(100 x 1 0 -6F)
107 = SV + - - - - - -» £c(180s) = ------------- -------------------- 100 x 10-6F
c
180s
ic(3min) = 2.7 \lA
6.3c) CORRECT ANSWER - B
Energy stored in a capacitor is given by ^Cv 2(t).
1
1
- C v 2{t) = - ( 2 0 0 x 10~6 F){240sin377tV)2 = 5.76 sin2 377t J
6.3d) CORRECT ANSWER - B
The current-voltage relationship of a capacitor is given by following equation.
1
vc(t) = vc( 0) + - J
1
ic(r )d T -> v c( t ) - v c( 0) = - J
ic(r)d r
1
10V = ■„ i(5s)(5s) -» i(5s) = 0.2mA
100 fiF K
J
y '
6.3e) CORRECT ANSWER - C
1 \iF branch capacitors are in parallel. They will result in 1 \iF + 1 \iF = 2 |iF.
1 \iF, 2 \iF, 2 [iF will be in series with each other.
C
’^
= (t ) + © + © = °'5 " F
Helpful tip - Addition of capacitance in series and parallel is different from resistance and inductance.
208
Copyrighted Material © 2018
6.4 Inductance - Solutions
Consult NCEES® FE Reference Handbook - Pages 200 - 202 for reference
6.4a) CORRECT ANSWER - A
Inductance can be calculated using following formula.
L=
N2[iA
I
1002 X 4n x 1Q~7HmT1 x 0.1m2
1m
= 1.25 mH
6.4b) CORRECT ANSWER - B
Voltage-current relationship of inductor is given by Faraday's Law as shown below.
di
V = L—
dt
100mA
V = 5mH x — ------= 0.257
2ms
6.4c) CORRECT ANSWER - D
Energy stored in an inductor can be calculated as shown below.
hi2
-
100 x 10 ~SH x (100v4)2
= ---------------2---------------= 5 0 °;
6.4d) CORRECT ANSWER - B
Energy stored in an inductor can be calculated as shown below.
,
N2]iA -2
Li
~ ~ l
2
2
Therefore, increasing number of turns increases inductance which will increase energy storage capacity.
6.4e) CORRECT ANSWER - C
2 H branch inductors will result in (2 H + 2 H)||(2 H + 2 //) = 2 H.
1 H, 2 H, 1 H are in series with each other.
Leqv = 1H + 2H + 1H = 4 H
Helpful tip - Addition of inductance in series and parallel is like addition of resistance in series and parallel
209
Copyrighted Material © 2018
Chapter #7 - Circuit Analysis
7.1 Kirchoff's Laws - KCL, KVL - Solutions
Consult NCEES® FE Reference Handbook - Pages 201 - 203 for reference
7.1a) CORRECT ANSWER - B
Let us denote the voltage across 10 kQ. resistor as V10kn-
y
We can write KCL at V10kn as follows:
Viokn — 0
10 fcO
+
V^okjQ — 0
4 k£l
+
V10kn — 0
3 kD.
+ 10mA = 0
Vioka
= —10mA -» V10kn = -1 4 .6 3 7
14630
7.1b) CORRECT ANSWER - D
_
Let us denote the voltage across 2 kO resistor as V2kn- We can write KCL at V2kn as follows:
I'
V2kn - 0
10mA = ,’ + 3 + T
I' =
^2ka ~ Q
2 ko
A
173
V2kn ■ ^2kn ■
™
1 0 m A = 2 k fl + 6 id T + 6 k n " V2kfi = 12V
„
V2kn
12V
^ „
I - TTTTT = X7T7 = 6 mA
2kO
2kO
7.1c) CORRECT ANSWER - B
40
Rcircuit = (20 + 40) + 4 0 11(30 + 20||20) = 80
h
^source
10 V
^circuit
^ ^
= 1.25 A
We can use current divider rule to calculate I3n.
Rt
Ix = - ---- — h
R x "I- R f
Ix = Isn, R t = 4 0 , R x = 30 + 2 0||2 O , IT = ISOurce
40
30 + 2 0| 12 O -H 40
1.25A = 0.625A
V3n = ha X 30 = 0.625A x 30 = 1.875 V
210
Copyrighted Material © 2018
20 a
7.Id) CORRECT ANSWER - D
v w
s
We can use principle of super-position to solve this circuit.
\ / fx
Let us first consider 10A current source only.
10Q
1GA
Using current divider rule:
0
ion
Ixl — Rx -j- R f It — 200 + io n 10A = 3.33A
rt
Let us now consider 50V voltage source only.
200
KVL can be written as follows:
- 5 0 V = 300 x Ix2
50V
1*2 = -1 .6 6 A
Ix = I x i + 1x2 = 1-66A = 1.7A
7.1e) CORRECT ANSWER - A
The given circuit can be solved using current divider rule as shown below:
Rt
Ix = - — — lT
R x 4* R x
IT = 5 tylA ,Rx = 6 kVL (since 1 kQ. and 5 kO. are in series), RT = 10 kQ.
10 kO
lkfl
6 kO + 10 kO
5 mA
Iikfl = 3.125 mA
7.If) CORRECT ANSWER - 875 |iA
We can solve this circuit using KCL.
toko v 2kQ
AAAr
AAAr
Let us denote the voltage across 5 kO resistor as. V5kn
IQ -V s k n
10 kO
5 — V5kQ _ V5kn — 0
2 kO
5 kO
10V
Multiply the above equation with lO/cO for simplification.
10 - V5kn + 5(5 - Vskn) = 2V5kn -> V5kn = 4.375K
V5kfi
Iskn — 5kO
4.375 V
5kO
= 875 ^iA
211
Copyrighted Material © 2018
ska
5V
7.2 Series/Parallel Equivalent Circuits - Solutions
Consult NCEES® FE Reference Handbook - Pages 201 - 203 for reference
10kfl
7.2a) CORRECT ANSWER - D
tOka
AAA
....^
rab = ska\\(io m + 2 m | | i m )
OJ6kO
2 kQ\\l kQ = 0.66 m
10 m + 2 m||i m = 10.66 m
Rab = 5 m i l 10.66 kQ = 3.4 kQ.
Therefore, RAB = 3.4 kQ..
*10.66kQ
5kS2
7.2b) CORRECT ANSWER - A
toko
-A A A r
Rab = 5 kQ\\10 kQ\\(2 kQ + 4kQ\\8kQ)
2HCS
A
"AAAr
...'"•AAAr
"' Y V V — -y*
4 m ||8 m = 2.66 m
2 /c Q + 4 m ||8 kQ = 2 kQ + 2.66 m = 4.66 m
4ks
B
10 kQ 11(2 m + 4 fcft||8 m ) = 10 m ||4.66 kQ = 3.178 m
5 m |110 kQ\I(2 m + 4 m ||8 fcft) = SkQ\\3.178 kQ = 2 kQ *
-A A A r
I ^
■AAAr
3J.78fc0
Therefore, RAB = 2 kQ.
7.2c) CORRECT ANSWER - B
rab = i m
2 m | | io m
im
+ 4m||(5 m + 2 m ||io m )
= i. 6 6 m
skQ + 2 mu io m = 5 m + 1.66 m = 6.66 m
4m||(5 m + 2m ||iom ) = 4m||6.66m= 2.50 m
i m + 4m||(5m + 2m||iom) = 1m + 2 .5m = 3 .50 /
Therefore, #45 = 3.50 m .
212
Copyrighted Material © 2018
5kQ
7.2d) CORRECT ANSWER - C
rab = i fen ||(5 k£i||(2 fen + 4 fen) + 2 fen)
2 fen + 4 kVL = 6 fen
6fen||5fen = 2.72 fen
5 ka \\(2 fen + 4 fen) + 2 fen = 2.72 fen + 2fen = 4.
= 1 fen||4.727 fen = 0.825 fen
Therefore,
= 0.825 fen.
7.2e) CORRECT ANSWER - 6 M l
1W*
A
Given circuit can be simplified as shown on the right.
ska
10ka
-W V —
----5k<Qs
rab = i kn + [2 fen + (5 fen||5 fen)]||io fen + 2 fen
2 fen+ (5 fen||5fen) = 4.5 fen
2kfi
[2fen + (5fen||5fen)]||iofen = 4.5fen||iofen = 3.1 fen
B- W
V
1 fen + [2 fen + (5 fen||5 fen)]||io fen + 2 fen = 3.1 fen + 1 fen + 2 fen = 6.1 fen
Therefore, RAB = 6 fen.
213
Copyrighted Material © 2018
■AAA/—
7.3 Thevenin & Norton Theorem - Solutions
Consult NCEES® FE Reference Handbook - Page 201 - 203 for reference
7.3a) CORRECT ANSWER - B
Voc is the voltage drop across terminals 'a' and 'b' as indicated in given problem.
According to current divider rule, 5 mA current will be divided equally in each branch i.e. 2.5 mA
Voc = 2 kQ x 2.5 mA = 5 V
2*0
2m
7.3b) CORRECT ANSWER -1.5 kQ
Rth is calculated by looking into terminals 'a' and 'b'.
Current source is open-circuited as shown.
Rth = 2 kQ||(2 kQ + 2 kQ + 2 kQ) = 1.5 kQ.
7.3c) CORRECT ANSWER - A
Rth is calculated by looking into terminals ‘a’ and 'b'.
Voltage sources need to be short-circuited.
Rth = 10 kft\\2 k{l\\5k£l = 1.25 kQ.
7.3d) CORRECT ANSWER - A
_
Apply Norton Theorem by shorting 5 kQ with a wire to calculate lsc
KCL can be written at Vx as follows:
10 V - Vx
1 0 V -V X
Vx
X + — — - A = -2 L -^
2 kQ
2 kQ
1 kQ
Lsc
VX = S V
10 V 1 0 V -V X 10 V 1 0 V -5 V
+—
— = t^ +
2 kQ
2 kQ
2 kQ
2 kQ
Isc = 7.5 mA
7.3e) CORRECT ANSWER -1 kQ
veq
Req is calculated by looking into circuit through 5 kQ.
Voltage source need to be short-circuited as shown.
w
AAAr
“AAAr
2M
Req = (2 kQ\\lkQ + 2kQ)\\2kQ
ate
1m
Req = 1.142 kQ = I kQ
214
Copyrighted Material © 2018
m
7.4 Waveform Analysis - Solutions
Consult NCEES® FE Reference Handbook - Pages 201 - 203 for reference
7.4a) CORRECT ANSWER - C
For a full wave rectified sinusoidal wave, maximum and effective values are related as follows.
Xeff =
Xmax = ^ 2 x Xe„
Kna* = V2 x v eff = V2 X 10 K = 14.14 V
7.4b) CORRECT ANSWER - D
V1 + V2 = 10 cos(500t) + 15 cos(100t + 45)
cos(a + (3) = cos(a) cos(/?) - sin(a) sin(/?)
+ ^2 — 10 cos(500t) + 15[cos (lOOt) cos (45) — sin (lOOt) sin(45)]
1
cos(45°) = sin(45°) = —
15 cos(100t) 15 sin(100t)
V1 + V2 = 10 cos(500t) + ------- p ----- - ---------- P ------
V2
V2
V1 + V2 = 10 cos(500t) + 10.6cos(100t) - 10.6 sin(100t)
7.4c) CORRECT ANSWER - C
Compare 100 cos(500t + 50) with standard phasor representation Acos(a)t + $).
0) = 2nf = 500
500
/ = — = 79.5 Hz
2n
7.4d) CORRECT ANSWER - B
For a halfwave rectified sinusoidal signal, average value is given as:
xm = —
n
= ■ - = 4.77
n
7.4e) CORRECT ANSWER - 1A
Average value of the given periodic signal can be calculated as follows:
I f 7
U ve = f j
I f 4
x (t ) dt =
lave = ^1 dt +
4 j dt
j 2 dt + J 1 dt J 0
= —(1 + 2 + 1 + 0 ) ~
215
Copyrighted Material © 2018
1^
7.5 Phasors - Solutions
Consult NCEES® FE Reference Handbook - Pages 201 - 203 for reference
7.5a) CORRECT ANSWER - D
v (t ) = 100sin(377t + 60°) requires conversion to standard cosine form as shown below.
sin(i9) = cos(d — 90°). Therefore, v (t ) = 100 cos(377t + 60° — 90°) = 100cos(377t - 30°)
Vrms =
v
1 on
= 70.7 V and phase angle is —30°.
Therefore, voltage phasor is 70.7/-300 V
XL = jcoL = j x 377 x 100 x 10“ 3n = J37.7 Q = 37.7/90°Q
j —JL= 1.875 A with phase angle -1 2 0 °
XL
Therefore, current phasor I = 1.875/-120°A.
7.5b) CORRECT ANSWER - A
v (t ) = 212 sin(a)t + 50) requires conversion to standard cosine form as shown below.
sin ($) = cos(i9 — 90). Therefore, v (t) = 212 cos(377t + 50° — 90°) = 212 cos(377t - 40°)
Vrms — j = = 150V and phase angle is —40°.
Therefore, voltage phasor is 1507-40° V.
7.5c) CORRECT ANSWER - B
v (t) = 100 cos(377t - 0°)
Vrms = -jljr- = 70.7 V and phase angle is 0°.
Therefore, voltage phasor is 70.7/0° V.
z = 5 0 0 - y i o o n = 5oo/-ii.3 ° q
I _v_
z
7o.7v_ _ q
sooon
with phase angle 11.3°.
r
a
Therefore, current phasor I = 0.14/11.3°A.
216
Copyrighted Material © 2018
7.5d) CORRECT ANSWER - C
v (t) = 100sin(377t + 100) requires conversion to standard cosine form as shown below.
sin($) = cos($ — 90). Therefore, v (t) = 100cos(377t + 100° — 90°) = 100cos(377t + 10°)
Vrms =
= 70.7 V and phase angle is 10°.
Therefore, voltage phasor is 70.7/10° V.
Xc = 1/j(oC = 0.0265/-900 Q
/ = — = 2665 A and phase angle is 100°
Therefore, current phasor is / = 2665/100°A.
7.5e) CORRECT ANSWER -6 .2 - 14/0
Z = (5 0 - ; 1 0 n ) + (20 0 )||(-;5 0)
[(1000;)(20n + 50;)]
Z = ( 5 n - ; 1 0 0 ) + [( 2 o n - 5o /)( 2 on + 5n/)]
Z = (5 0 - 7 1 0 0 ) +
Z = (5 0 -
10 O) +
[20000; - 5000]
[4000 + 250]
[20000; - 5000]
[4000 + 250]
2000 .
500
Z = 50 + — 0 - 1 0 ; ' 0 Z = 6.176 - 14.47; a = 6 .2 - 14;0
217
Copyrighted Material © 2018
7.6 Impedance - Solutions
Consult NCEES® FE Reference Handbook - Pages 201 - 203 for reference
7.6a) CORRECT ANSWER - A
Zeq = 10 + ;2 a + (10 - 5j n)||(2 - Sjtt)
Zeq — 10 + ;2 fl +
540 - 770j
244
Zeq = 10 + j2 a + (2 .2 1 -3 .1 5 jO )
Zeq = 12 —j Cl
7.6b) CORRECT ANSWER - B
Zeq = 1 0 -5 ; n il 2 - 5jCl
_ 540 - 770;
244
Zeq ~
Zeq = 2.21 - 3.15yft
Zeq = 2 - 3jCl
7.6c) CORRECT ANSWER - B
IQOnF Z
%eq ~ R + XC
Xr =
c
1
1
7 377 x 100 x 10 “ 9
= —/2 6 5 2 5 n
z
Zeq = 50 - 726525 Cl
500 <
7.6d) CORRECT ANSWER - C
Zeq = ( R + X l )\\Xc
100 <
XL = jo)L = /377 x 2 x 10“ 3 = y0.754n
Xr =
1
ja)C
1
j 377 x 100 x 10
100pF Z
2mH jE
'eq
1 0 + /0.754 n|| — jf'26.5 n
218
Copyrighted Material © 2018
Chapter # 8 - Linear Systems
8.1 Frequency / transient response - Solutions
Consult NCEES® FE Reference Handbook - Pages 203 - 204 for reference
8.1a) CORRECT ANSWER - C
Relevant formula for this problem:
vc( t ) = vc(Q)e~l?c+ v(l-e~ T tc^ J
vc(0) = 20 V
After switch changes position, V = 0 (there is no external voltage source anymore)
t = 5 x 60 s = 300 s,
tmF
t = RC = 100s
i?c(300s) = vc(0)e~RC + V ^1 — e~RC ^
R =
100 kSl, C — 1 mF
C irc u it after switch closes
300
i7c(300s) = 20e~iooV = 0.995 V = 1 V
Helpful tip - Review and understand RC transient equations given in NCEES® FE Reference Handbook.
8.1b) CORRECT ANSWER - B
Relevant formula for this problem:
vc(t) = vc(0)e~Rc + F ^ l - e~RC ^
vc(0) = 10 V
ma
R = 10 ka,C = 200 \iF,RC = 2 s
After switch changes position, V = 0 (there is no external voltage source anymore)
v ^
,
C irc u it after switch opens
vc(t) = lO e -^ V
8.1c) CORRECT ANSWER - D
1kQ
A/S/Sr
Relevant formula for this problem:
10V
_Rt
V
_Rt
i(t ) = /(0)e l + - ( 1 - e l )
2mH
R
i( 0) = 0,V = 10 V,R = 1 ka,L = 2mH ^ - = 500000s
JLr
Circuit after switch closes
i(t) = 0.01 (1 — e -500000t)i4
Helpful tip - Review and understand RL transient equations given in NCEES® FE Reference Handbook.
219
Copyrighted Material © 2018
8 . I d ) CORRECT ANSWER - B
Relevant formula for this problem:
i( t )
_ Rt
v
l
+ - (1
= i(0 )e
__El
- e i )
Inductor can be considered a short wire because switch has been closed for a long time.
Current provided by power source before switch opens at t = 0:
10 V
10V
/ = ______________ == ____________ = 0 75,4
i o n + 3.330
io n + 5 n | | io a
m
Voltage across 5 fl||10 H before switch opens can be calculated as shown below:
17(0) = 0.75,4 x 5 n||10 a = 2.5V
m
Current through inductor at t = 0 can be calculated using current divider rule.
2.5 V
i( 0 ) = -Z— = 0.5 4
Circuit after switch opens
According to problem details t = 10t
L
t = — = 66.66 ms
L- 1H
R = 15 H
Note that 7 = 0 after switch opens.
_Rt
i( 10r) = i(0)e l + 0 = 22.6 \iA
8.1e) CORRECT ANSWER - C
Relevant formula for this problem:
5mF
vc(t) = vc( 0)e~Rc + v ( l - e~Rc ^
vc( 0 ) =
1 07 ,
5t
tm
= SRC = 2 5 0 s
After switch changes position, V = 0 (there is no external voltage source anymore)
t?c(50) = 10e~5 = 67.3 mV
220
Copyrighted Material © 2018
Circuit after switch
changes position.
8.2 Resonance - Solutions
Consult NCEES® FE Reference Handbook - Pages 203 - 204 for reference
8.2a) CORRECT ANSWER - D
The given circuit is an example of series resonance circuit for which relevant formula is given below.
1
1
...... — = 2 2 3 6 Ydd/s
C00 = —= = .
VLC
yJlOmH x 10juF
8.2b) CORRECT ANSWER - 5000 rad/s
Bandwidth of series resonance circuit can be calculated using formula given below.
0)o R 100X2
BW = - £ = - = — — - = 5000 rad/s
Q L 20 mH
'
8.2c) CORRECT ANSWER - B
Maximum current occurs at resonant frequency because its impedance is lowest and purely resistive i.e.
1201/
Ijnax ~ T o r T " 12A
8.2d) CORRECT ANSWER - 3162 rad/s
Maximum current occurs at resonant frequency because impedance is lowest and purely resistive i.e. Z
1
0)o = - =
V ic
1
.= 3162 Tdd/S
^JlOmH x 10 p.F
= ...
8.2e) CORRECT ANSWER - B
The given circuit is an example of parallel resonance circuit for which current magnification factor 'Q' is
calculated as shown below:
Q = —R <o0L
, -11
(jon = —=1 = 707 rad.s
-JhC
ion
Q
_________________________________
v
(20m//)(707 r a d .s -1)
_
n 7Q7
8.2f) CORRECT ANSWER -1000 rad. s_1
Bandwidth of parallel resonance circuits can be calculated using formula given below.
BW =
(X),o
707
BW — — — rad.s
0.707
1 = 1000 rad. s
221
Copyrighted Material © 2018
8.3 Laplace Transform - Solutions
Consult NCEES® FE Reference Handbook — Page 34 for reference
8.3a) CORRECT ANSWER - A
According to Laplace Transform pairs provided in NCEES® FE Reference Handbook
L [e ~a t ] = 7 T -a -* m
t ) ] = JTb
8.3b) CORRECT ANSWER - B
/(t) = e~at = e-a(~t+1-^ = e-a(t-l)e-a = e-(a)[g-a(t-1)M(t _ !)]
According to Laplace Transform pairs provided in NCEES® FE Reference Handbook:
£ [ / ( t - t ) i i(t - t)] = e~tsF(s)
L[e~at] = — -—
s+ a
It can be observed that in given case, t = 1.
U
J
p-(s)
p -(s + a )
s+ a
s+ a
8.3c) CORRECT ANSWER - C
f i t ) = i t - 1 + 1) e - a^ - 1+1M t ~ 1) = [(t - l)e~ a^t_1^u(t - 1) + e -^ -^ u C t “ l) ] e “ a
According to Laplace Transform pairs provided in NCEES® FE Reference Handbook:
£ [ / ( t - T)u(t - t)] = e~zsF{s)
Lite~at) =
1
(s + a)2
It can be observed that in given case, t = 1.
X [/ ( t ) ] = F ( S) =
e-(s+a ) ( _ J _ _ + _ L )
8.3d) CORRECT ANSWER - D
f i t ) = [ e~^~3^— e~3^~3^]uit — 3) = e~^~3^uit — 3) — e~3^~3^uit — 3)
As explained in above given solutions, Laplace Transform of this function will be as follows:
£^
=
e' 3s { j T - i - j h )
8.3e) CORRECT ANSWER - A
f i t ) = te~at5it —2)
According to sifting property of delta function / / ( t ) 5 ( t — T)dt = f(T ).
Fis) = i2 )e ~ ^ a e~(2)5 = 2e~2^a+s^
222
Copyrighted Material © 2018
8.3f) CORRECT ANSWER - D
F(s) =
5
(s + 3) + (s + 5)
(2s + 8)
2(s + 4)
2 s+4
According to Laplace Transform pairs provided in NCEES® FE Reference Handbook
£ te“ at] = 7
s +T a^ / M = I! e' 4t
8.3g) CORRECT ANSWER - C
F(s) =
s+8
(s + 1)(s + 7)
s+8
.4
(s + l) ( s + 7)
5+1
B
s+7
_l-------Multiplying by denominator -»
1
Lets = -1 , 1 = B ( - 6),B = - 6
7
1
(6) (s + 1)
(6)(s + 7)
1 = A{ 6), A = 7/6
Lets = - 1 ,
1/7
^ 0 0 = T T T T T T T T “ 7>T77~ ■ ^
s 4- 8 = ^4(s + 7
1
= 7l
6 Vs + 1
s + 7/
According to Laplace Transform pairs provided in NCEES® FE Reference Handbook
L ie~at] = — j— ^ / ( O = ^ (7 e~t - e ~ 7t)
s+a
6
8.3H) CORRECT ANSWER - A
F(s)
s2 + 2s + 1
(s + 2)(s + 3)(s)
s2 + 2s + 1
(s + 2)(s + 3)(s)
A
B
C
_l------- — H—
s+2 s+3 s
s2 + 2s + 1 = ^4(s + 3)(s) + B(s + 2)(s) + C(s + 2)(s + 3)
1
Lets = 0, 1 = C(2)(3) ,C = 6
1
^ = - -
Lets = -2 ,
4 - 4 + 1 = A( 1)(—2),
Lets = - 3,
9 - 6 + 1 = f l( - l) ( - 3 ) ,
4
B =-
-1
4
1
F fc) = ------------ 1----------------------------[- —
^;
2(s + 2) 3(s + 3) 6s
According to Laplace Transform pairs provided in NCEES® FE Reference Handbook
1
1
s+a
s
1
e~2t
6
4e~3t
z
3
223
Copyrighted Material © 2018
+ B(s + 1)
8.3i) CORRECT ANSWER - C
s+4
F ° ° ~ (s2)(s + 5)
5
+4
(s2)(s + 5)
_A
B
C
S + S2 + S + 5
s + 4 = A(s~)(s + 5) + B(s + 5) + C(s2)
Lets 2 = 0,
4
4 = 5 (5 ), £ = -
Lets = - 5 ,
- 1 = C( 25), C = - —
1
9
1
Comparing coefficients of s , A + C = 0, A = —C, A = —
1
4
W “ 25(s) + 5 (s2)
.
1
1
25(s + 5)
1
u (t)
41
e~st
Helpful tip - Review partial fraction expansion for calculating inverse Laplace Transforms.
224
Copyrighted Material © 2018
8.4 Transfer Functions - Solutions
Consult NCEES® FE Reference Handbook - Pages 203 - 204 for reference
8.4a) CORRECT ANSWER - D
—A/Sfo"
Z0 = (fi+ s£ .)||(^ :)
ZT = K + ( f i + SZ ,)||(^ )
Vo
Vi
k,
z„
( « + sm i ( ^ )
Vi
Zr
R + (R + sL)|| ( i )
R
8.4b) CORRECT ANSWER - C
Zn — R +
rAAAr
u n n rrL
sC
L
Vo
R + sC
Vo _ z0
n ~ z
X
8.4c) CORRECT ANSWER - A
R
Equivalent impedance can be calculated as shown below.
- v w
Z = R + — \\(R + sL\\R)
sC
8.4d) CORRECT ANSWER - B
«(S) =
20
s(10s + 1)
Transfer function can be expressed in standard form as shown below.
H(s) =
20
------ r
s (oT+1)
It can be observed that given transfer function has two poles and no zeros.
p± = 0 rad.s -1 p2 = 0.1 rad/s
\gain\ = 20log (20) = 26 dB
225
Copyrighted Material © 2018
R
A /W
8.4e) CORRECT ANSWER - D
H(s) =
10
s2(Ss + 1)
Transfer function can be expressed in standard form as shown below.
H(s) =
10
( o +1)
It can be observed that given transfer function has two poles and no zeros.
= 0 rad. s -1 (2ndorder) p2 = 0.2 rad. s ~1(1storder)
\gain\ = 20log(10) = 20 dB
226
Copyrighted Material © 2018
8.5 Two-Port Theory - Solutions
Consult NCEES® FE Reference Handbook - Pages 203 - 204 for reference
8.5a) CORRECT ANSWER - A
Vi
I2 — 0
z-ii — —- — 4 O,
k
¥1
V2
Z21 ~ ~r~ — 2 0,
I2 — 0
Vi
zi 2 = — = 2 0 ,
/i = 0
k
¥2
V2
i?
z 22 — — — 4 0 ,
I1 — 0
8.5b) CORRECT ANSWER - C
m
= r r i s’
h
= -L
y
-721
V±
=—
10G
V2 = 0
-
1
V2 =
5,
10
VI
0
2
10fl
■>~AAAr
1
k
¥2=0
108
y' 2 = v2 = ~ w s’
Vl = 0
y22=|=J5'
¥1 = 0
•100
¥2
20
Kl = 0
¥1
1
yi1 = i
= s s’
k
k2
1
y2i —
1,2 = 0
6
- 6 S-
h
1
y 22 = t t = ^ s ,
^i — 0
y-12 — — — — — 5,
>
^ = °
2a
2®
11
■ ^ ^ 1.—
¥1 = 0
■AAAr
12 = 0
%
0
V1
k
280
t$ii
V2
10Q
8.5e) CORRECT ANSWER - 0.66
h 12
=
V±
V?
20
= —
30
2
= - = 0.66,
3
2Q
¥2 = 0
12
..■<”"
20
v, = 0
3
8.5d) CORRECT ANSWER -11.11 O
^1
= - = 11.110,
12
.... ^ ...........A / V \ r "
8.5c) CORRECT ANSWER - A
k
II
/x = 0
227
Copyrighted Material © 2018
2C
¥2
Chapter # 9 - Signal Processing
9.1 Continuous Time Convolution - Solutions
Consult NCEES® FE Reference Handbook - Page 206 for reference
9.1a) CORRECT ANSWER - C
x (t )
y (t) = x (t) * h(t)
It is easier to flip h (t) into h{—t).
<-
Region # 1 y (t) = 0 f o r t < 0 no overlap
t-2
t
•>
4
T
Region # 2 y (t) = I 8dr = St fo r 0 < t < 2 partial overlap
Jo
Region # 3 y (t) = f 8d r = 8t —8(t — 2) = 16 / o r 2 < t < 4, partial overlap
h -2
Region # 4 y (t) = f 8dr = 8(4 — t + 2) = 48 — 8t f o r 4 < t < 6, partial overal
h -2
Region # 5 y (t) = 0 fo r 6 < t , no overlap
9.1b) CORRECT ANSWER - A
y (t) = / ( t ) * h(t )
It is easier to flip / ( t ) into / ( —t ).
J ruu
e^~T^3e^~r^dT =
36®
— /o r t < 0
o
Region# 2 y (t) = j
^
e^ T^3e^ T^dr
/o r 0 < t
9.1c) CORRECT ANSWER - D
f i t ) = x (t) * y (t)
It is easier to flip y(t) into y ( —r).
Region # 1 f ( t ) = 0 / o r £ < 0 no overlap
Region # 2 f i t ) = [ si
sin(r)dT = 1 — cos(t) fo r 0 < t < n partial overlap
Jo
rU
Region # 3 f i t ) = J sin(r) d r = 1 — cos(t) f o r n < t <2n, partial overlap
J t -n
Region # 4 f ( t ) = 0 fo r n < t ,no overlap
228
Copyrighted Material © 2018
9.Id) CORRECT ANSWER - C
4^
Let f { t ) = x (t ) * y(t)
X (T )
y(t-
It is easier to flip y(t) into y (-x ).
Region # 1 f ( t ) = 0 f o r t < —1 no overlap
■>
i+ 2
3
t+ 4
5
•t+4
(2)(3 )dr = 6(t + 4 — 3) = 6(t + 1) / o r - 1 < t < 1 partial overlap
3
Region # 3 f ( t ) = f
6dx — 6(5 — t — 2) = 6(3 - t) / o r 1 < t < 3
partial overlap
Jt+7.
Region # 4 f ( t ) = 0 / o r 3 < t , no overlap
9.1e) CORRECT ANSWER - C
Let f { t ) = x(t) * y(t)
It is easier to flip y(t) into y ( —t ).
Region # 1 / ( t ) = 0 / o r t < 0 no overlap
Region # 2 f ( t ) = [ 2e~Tdr = 2 — 2e-t / o r 0 < t < 2 partial overlap
jo
Region #3 f ( t ) = [
2e~Tdr = 2(e ~t+2 - e~c) fo r t > 2
Jt_ 2
229
Copyrighted Material © 2018
9.2 Discrete Time Convolution - Solutions
Consult NCEES® FE Reference Handbook - Page 206 for reference
9.2a) CORRECT ANSWER - B
f[n ] = x[n] * y[n] =
x[k]y[n - k].
It is easier to flip y[n] into y [—k].
For n < 0 ,
there is no overlap
Forn =
0,
/ [ 0] = 2.0 =
0
Forn =
1,
/ [ l] = 2.2 +
2.0 =
Forn =
2,
f [ 2] = 2.3 + 2.2 + 2.0 = 4 + 6 = 10
Forn =
3,
/ [ 3] = 2.2 +
2.3 + 2.2 =
Forn =
4,
f[4 ] — 2.3 +
2.2 = 10
For n = 5,
/[5] = 2.2 = 4
Forn = 6,
/ [ 6] = 0
4
14
Therefore, f[n ] = [0 4 10 14 10 4 0]
9.2b) CORRECT ANSWER - C
f[n ] = x[n\ * y[n] =
x[k]y[n - k].
It is easier to flip y[n ] into y [—k].
For n < 0
there is no overlap
Forn =
0,
/[0] = 0
Forn =
1,
/ [ l] = 1.1 + 2.0 =
Forn =
2,
f[2 ] = 1.2 + 2.1 + 1.0 = 4
F orn =
3,
/ [ 3] = 1.3 + 2.2 + 1.1 = 8
F orn =
4,
f[4 ] = 1.0 + 2.3 + 1.2 = 8
For n = 5,
/[5] = 1.3 = 3
Forn = 6,
/ [ 6] = 0
1
Therefore, f[n ] = [0 1 4 8 8 3 0]
230
Copyrighted Material © 2018
9.2c) CORRECT ANSWER - B
Zero input solution is found using characteristic polynomial roots.
Zero state response is the output of a system to a specific input when system has zero initial conditions and it
found using D-T convolution.
9.2d) CORRECT ANSWER - C
x[n] = u[n] — u[n — 5]
y[n] = 0.2nu[n]
x[n] * y[n ] =
> x[k]y[n —k]
k=-oo
k - oo
x[n] * y[n] =
^
(u[k] —u[k —5])(0.2n ku[n —k ])
x[n ]*y[n ]=
^
u [k ](0 .2 - * u [n - k ]) -
^
u[ k -
5 ](0 .2 - M n - * ])
9.2e) CORRECT ANSWER - C
x[n] = u[n —2]
y[n] = 0.4nu[n]
k = ~ oo
fc=oo
fc = -00
fc=n
k=n
231
Copyrighted Material © 2018
9.3 Z Transforms - Solutions
Consult NCEES® FE Reference Handbook - Page 206 for reference
9.3a) CORRECT ANSWER - C
Z transforms are Laplace transform equivalent in difference equations. Laplace transforms are used in
differential equations. Discrete convolution provides zero state solution of a Discrete Linear Time Invariant
system.
9.3b) CORRECT ANSWER - B
x[n] = u[n ] — u[n —5]
fc=4
k =oo
X[z] =
z k — 1 + z 1 + z 2 + z 3 + r—4
(u[k] — u[k —5)z k =
k=0
fc=0
9.3c) CORRECT ANSWER - D
x[n] = 0.2nu[n]
fc=oo
J^fz] = ^
k =oo
0.2ku[/c]z fe =
fc=0
k =oo
0.2fez k =
fc=0
(0.2z 1) /
fc=0
According to geometric series convergence formula:
CO
yT r n = - — - / o r |r| < 1
71=0
k=co
X[z] = £
0. 2kz~k =
1 - 0 .2 Z-1
k=0
Note: This problem can also be solved used Z Transform table given in NCEES® FE Reference Handbook.
9.3d) CORRECT ANSWER - A
x[n] = [ 2
3 1 0 5]
k =oo
x[k\z k = 2z ° + 3z
X[z] =
1 + z 2 4- 5z 4 = 2 + 3z 1 + z 2 + 5z'
k= 0
232
Copyrighted Material © 2018
9.3e) CORRECT ANSWER - A
x[n] = 5(0.75)nu[n]
k = 00
fe = 00
A 'W =
^
5 ( 0 .7 5 ) k u [ k ] z ~ k = 5
/c=0
( 0 .7 5 Z - 1 ) '1 = 1 _ Q 7 5 z _
fc=0
9 .3 f) CORRECT ANSWER - C
z
* 0 0 = z - 0.5
z
1
z ( l - 0.5z-1)
1 - 0.5z_1
According to z-transform pairs given in NCEES® FE Reference Handbook:
1
1 — /?z 1
x[n] = (0.5)n
9.3g) CORRECT ANSWER - D
5z + 2
X(Z) ~ ( z - l ) ( z - 4 )
Inverse z-transform can be calculated using partial fraction expansion.
X(z)
z
5z + 2
_ q
C2
z(z - 1)(z — 4)
C3
z
z — 1z - 4
5z + 2 = Cx(z - 1)(z - 4) + C2(z)(z - 4) + C3(z)(z - 1)
Letz = 1 ,
7 = C2(—3) -» C2 =
Let z = 4,
22 = C3 (12) -> C3 = —
L e t z = 0,
2 = C1(—1)(—4)
11
6
X (z)
1
7
z
2z
3(z — 1)
+
1
Cj = -
11
6 (z — 4)
1
7z
llz
^ ) =2
o - ^3(z
7 —- 1)
TT + 6 (z - 4)
1
7z
llz
2
3 z ( l - z x)
6 z ( l - 4z x)
r n S[n] 7(1") , (11)(4n)
x[n\ = — --------------- -1---- -----
233
Copyrighted Material © 2018
9.3h) CORRECT ANSWER - B
X(z) =
X(z)
z
( z - 2 ) ( z + l)
(z —0.1 )(z — 0.2)
C3
z — 0.2
( z - 2 ) ( z + l)
Ci
C2
— ---- 1---- r—r +
z(z — 0.1)(z — 0.2)
z
z — 0.1
(z - 2)(z + 1) = Ct (z - 0.1)(z - 0.2) + C2(z)(z - 0.2) + C3(z)(z - 0.1)
L e tz = 0.1,
-2.0 9 = C2(-0.01) -> C2 = 209
Let z = 0,
-2 =
-2 .1 6 = C3(0.2)(0.1) -> C3 = - 1 0 8
L e t z = 0.2,
X (z)
(0 - 0.1)(0 - 0.2) -♦ ^ = - 1 0 0
-1 0 0
X(z) = - 1 0 0
209
108
(z - 0.1)
(z - 0.2)
209z
X(z) = - 1 0 0 w
(z — 0.1)
->
209z
108z
(1 - O .lz-1)
z ( l - 0.2z_1)
108z
6(z - 0.2)
x[n] = —1005[n] - 209(0.1n) - (108)(0.2n)
9.31) CORRECT ANSWER - A
X(z)
(z + 0.5)
(z — 0.1) (z + 0.4)
(z + 0.5)
X(z)
z
z(z — 0.1)(z + 0.4)
C2
C±
C3
— ------ 1-----------— r +
z
z — 0.1
z + 0.4
(z + 0.5) = C i(z - 0.1)(z + 0.4) + C2(z )(z + 0.4) + C3(z )(z - 0.1)
L e tz = 0.1,
0.6 = C2(0.1)(0.5) -» C2 =
Let z = -0.4,
0.1 = C3 (—0.4)(—0.5) ^ C 3 = -
Let z = 0,
1
-2 5
0.5 = C iC -O .l)^ ^ ) -> C± = —
X (z) __ - 2 5
12
1
”T ~ ~ ~2z~ + (z - 0.1) + 2(z + 0.4)
-2 5
*0) =
12
_ -2 5
12z
z
* (z) " T " + (z - 0.1) + 2(z + 0.4)
12z
z
+ ~r
,—
rr
+
z ( l — O .lz x) 2 z ( l + 0.4z_1)
-2 5
(—0.4n)
x[n] = — S[n] + 12(0.l ) n + — - —
Helpful tip - Review partial fraction expansion for calculating inverse Z-Transform.
234
Copyrighted Material © 2018
9.4 Sampling - Solutions
Consult NCEES® FE Reference Handbook - Page 209 for reference
9.4a) CORRECT ANSWER - 2000 Hz
The given signal is a summation of two sine functions sinc(1000nt) and sinc(2000nt).
sinc(2000nt ) = sinc[2n (1000)t] has highest frequency component.
According to Nyquist Theory, perfect reconstruction requires sampling rate to be > 2 x highest frequency.
Therefore, required sampling rate shall be > 2 x 1000 Hz = 2000 Hz.
9.4b) CORRECT ANSWER - C
According to problem statement x(t) = cos(27r(1500)t + 0)
Signal frequency = 1500 Hz and sampling frequency = 2000 Hz.
Aliasing will occur because sampling frequency < 2 x signal frequency.
9.4c) CORRECT ANSWER - D
According to problem statement x (t) = cos(27r(250)t + 0)
Signal frequency = 250 Hz and sampling frequency = 500 Hz.
Aliasing will not occur because sampling frequency = 2 x 250 Hz = 500 Hz.
9.4d) CORRECT ANSWER - C
According to problem statement x (t) = cos(27r(200)t 4- 9)
Signal frequency = 200 Hz and sampling frequency = 300 Hz.
Aliasing will occur because sampling frequency < 2 x 200 Hz.
Alias frequency = |signal frequency - n x sampling frequency | 'n' is an integer thatbrings n x sampling frequency
closest to signal frequency. In our case, n = 1 as shown below.
Alias frequency = |signal frequency - n x sampling frequency |=1200 Hz - 1 x 300Hz |=
100 Hz.
9.4e) CORRECT ANSWER - A
According to Nyquist Theorem, signal needs to be sampled at or above Nyquist rate for perfect reconstruction.
235
Copyrighted Material © 2018
9.5 Filters - Solutions
Consult NCEES® FE Reference Handbook - Pages 207, 210 and 211 for reference
9.5a) CORRECT ANSWER - B
Comparing the given transfer function to First-Order Low-Pass Filter transfer function on page # 210 of NCEES
Reference Handbook shows that it is analog implementation of First-Order Low-Pass Filter.
9.5b) CORRECT ANSWER - C
Comparing the given transfer function to Band-Pass Filter transfer function on page # 211 of NCEES Reference
Handbook shows that it is analog implementation of Band-Pass Filter.
9.5c) CORRECT ANSWER - A
Comparing the given transfer function to First-Order High-Pass Filter transfer function on page # 210 of NCEES
Reference Handbook shows that it is analog implementation of First-Order High-Pass Filter.
9.5d) CORRECT ANSWER - D
Comparing the given transfer function to Band Reject Filter transfer function on page # 211 of NCEES Reference
Handbook shows that it is analog implementation of Band Reject Filter.
9.5e) CORRECT ANSWER - A
Finite Impulse Response Filter (FIR) is non-recursive because it does not have feedback loop (unity feedback).
9.5f) CORRECT ANSWER - B
Infinite Impulse Response Filter (HR) is recursive because it has a feedback loop.
9.5g) CORRECT ANSWER - D
Sampling, A/D conversion and D/A conversion are important processes involved in digital filtering where as
Phase Modulation applies to analog domain.
236
Copyrighted Material © 2018
Chapter # 10 - Electronics
10.1 Solkl-State Fundamentals - Solutions
Consult NCEES® FE Reference Handbook - Page 212 for reference
10.1a) CORRECT ANSWER - A
According to periodic table found in NCEES® FE Reference Handbook, Antimony is a group V element. Members
of this group have 5 valance electrons which makes them suitable for n-type doping.
10.1b) CORRECT ANSWER - D
According to periodic table found in NCEES® FE Reference Handbook, Boron is a group III element. Members of
this group have 3 valance electrons which makes them suitable for p-type doping.
10.1c) CORRECT ANSWER - B
According to the formula given in NCEES® FE Reference Handbook, conductivity of a semiconductor is given by:
a = q (nil,, + p\ip)
rii = n = p = 2 x 10lo m " 3,
q = 1.6 x 10-19C,
|4.n = 0.20m2V~1s~1,\ip = 0.10m2F “ 1s “ 1
a = (1.6 X 10~19)((2 x 10lo )(0.20) + (2 x 10lo )(0.10)) = 9.61 X 1 0 "10 S.rrT 1
lO.ld) CORRECT ANSWER - 0.634V
According to the formula given in NCEES® FE Reference Handbook, built-in potential of p-n junction is given by:
0
nf )
_ (kT\ (, W A
U H
_ rn
(, (2 x 1015)(2 X 1015)\ _
(°-026)^ ln
(1 x 1010) 2
J
°-6UV
10.le) CORRECT ANSWER - C
Insulators have large gap between valence and conduction bands due to which electrons cannot move freely
through the material.
237
Copyrighted Material © 2018
lO.lf) CORRECT ANSWER - A
The average velocity at which electrons move under the influence of an electrical field is called the drift velocity
and it is directly proportional to the applied electric field.
lO.lg) CORRECT ANSWER - Group V
Group V elements of periodic table have 5 electrons in their outer orbit this allows them to form 4 covalent
bonds with Si or Ge atoms. It leaves one electron free which makes them n-type doping agents. Phosphorous
and Arsenic are examples of n-type doping agents.
lO.lh) CORRECT ANSWER - Group III
Group III elements of periodic table have 3 electrons in their outer orbit which allows them to form 3 covalent
bonds with Si or Ge atoms. It creates a hole which makes them p-type doping agents. Boron and Aluminum are
examples of p-type doping agents.
lO.li) CORRECT ANSWER - Group IV
Group IV elements of periodic atoms have 4 electrons in their outer orbit which allows them to form 4 covalent
bonds. The electrical properties of semi-conductors such as Silicon and Germanium can be modified by doping.
238
Copyrighted Material © 2018
10.2 Diodes - Solutions
Consult NCEES® FE Reference Handbook - Page 214 for reference
10.2a) CORRECT ANSWER - A
1kft
Let us first assume that both D1 and D2 are ON
S V -IV
i m
= 3 mA
(i v)
iDl + lD2
2 ka
= 0.5 mA
iD2 = 0.5 mA —iD1 = 0.5 mA — 3 mA = —2.5 mA
iD2< 0, result is not consistent with assumption.
Let us now assume that D1 is ON and D2 is OFF.
5V
lD 1
3 k£l
= 1.66 mA
vD2 = l V - 1.66 mA x 2 ktt = -2 .3 3 V < 0
Since iD1>0 and vD2<0, the results are consistent with assumptions.
10.2b) CORRECT ANSWER - A
The circuit given in problem 10.2b) is an example of a full wave rectifier. Therefore, its output wave form will be
a full-wave rectified sinusoidal wave.
10.2c) CORRECT ANSWER - B
The given circuit can be rearranged such that
0„66kH
03W 1
Rthi = 1 kft||2 kO = 0.66 kO
V,o c i
f2'
1.33 V
Similarly, Rth2 = 4 kO||2 kQ. = 1.33 kCl
Voc2 = * { i ) = 2.66V
Assuming diode is "ON" i.e. forward biased
2.66 V - 1.33 V
1d —
1.33/cO + 0.66 m
= 0.66 mA
Since iD>0, Diode is ON. Therefore, diode Q point is (0V, 0.66 mA)
Note that voltage drop across diode is assumed 0 V since ideal diode model is being considered.
239
Copyrighted Material © 2018
10.2d) CORRECT ANSWER - 0.13 mA
Assuming diode is "ON" i.e. forward biased
i
o--------- VW
1 0 V - 8 V - 0.77
l D~ --------- 1 0 m --------- - 0 1 3 m j 4
10V
lO k a
Diode is ON because iD>0.
10.2e) CORRECT ANSWER - B
Let us first assume that both DI and D2 are ON.
(5 - 0.7) 7 — (10 — 0.7)7
lD 1
2 kCl
= —2.5 mA
Since iD1 < 0, the result is not consistent with the assumption.
Let us now assume that DI is OFF and D2 is ON.
(107 - 0.77) - ( - 3 7)
Ska
= 2.46 mA
VD1 = 5 7 — (107 - 0.77) = -4 .3 7 < 0
Since iD2 > 0 and VD1 < 0.77 the result is consistent with assumptions.
‘D1
Q
5V
2kn
X
|
5kQ
\ i\ i\
I
0.7V
D I an d D2 ON
-A A A r
-3¥
0.7V
!D2
10V
w
D1
2kfi
5kO
SV
■3V
q
D l O FF a n d D2 ON
yy
‘ 02 / V
10V
240
Copyrighted Material © 2018
10.3 BJTs - Solutions
Consult NCEES® FE Reference Handbook - Pages 215 for reference
10.3a) CORRECT ANSWER - C
To analyze a BJT circuit we assume a state of operation, enforce conditions and then verify assumptions.
Let us assume that transistor is operating in active region.
In active region: Vbe = 0.7 V Ib > 0mA
Vce > 0.7V
Base-emitter KVL can be written as shown below:
5F - (5 k a )(Ib) - Vbe - (2kO)0e) = 0
SV - (5 kO )(lb) - 0.7V - (2fcn)(101/b) = 0
Ib = 20.77 [iA
m
Ic = p ib = (100)(20.77 \iA) = 2.07 mA
jy y V —
KCL can be written at the collector as shown below:
TS/St
_10F
_ _ -_ V
_c = h
e
Vc = 10 V - (2 fcn)(2.07 mA) = 5.86 V
KCL can be written at the emitter as shown below:
Ve ~ 0 _ r
2 ka ~ le
Ve = (2 kD.)(2.09 mA) = 4.18 V
Ke = VC- V e = 5.86 V - 4.18 V = 1.68 V
Our assumptions are correct because Vce > 0.7 V, Ib > 0mA.
Therefore, transistor is in operating in active mode.
10.3b) CORRECT ANSWER - D
To analyze a BJT circuit we assume a state of operation, enforce conditions and then verify assumptions.
Let us assume that transistor is operating in active region.
In active region: Vbe = 0.7 V Ib > 0mA
Vce > 0.7V
Base-emitter KVL can be written as shown below:
3V - 0.7V
3F - Vbe - (2 k n )(Ie) = 0 -* Ie = ............... = 1.15 mA
241
Copyrighted Material © 2018
le = (p + 1)/f, = 101 Ib - * I b = 1.13 x 10“ 5^
N/
lc = p ib = 1.138 mA
KCL can be written at collector as shown below:
IV
Vr
3ka
3k£i
3V
Ic
vc = 7V - (3 fcO)(1.138 mA) = 3.58 V
Vbe = Vb ~ Ve = 3K - ve = 0.7V ->Ve = 2.3V
Vce = VC- V e = 3.6 V - 2.3 V = 1.3 V
ana
L
Our assumptions are correct because Vce > 0.7 V, Ib > 0mA.
Therefore, transistor is operating in active mode.
10.3c) CORRECT ANSWER - Cut-off
To analyze a BJT circuit we assume a state of operation, enforce conditions and then verify assumptions.
Let us assume that transistor is operating in active region.
In active region: Vbe = 0.7 V Ib > 0mA
Vce > 0.7V
Base-emitter KVL can be written as shown below:
0 - Vbe - (3fcn)(/e) = 0
0
- 0.7V
Ie = - ^ M T = - ° 233mA
le = iP + 1)4
L
= —0.0023 mA
(/? + !)
b
Transistor is operating in cut-off region because Ib < 0.
Ih
—
lr
—
—
0.
242
Copyrighted Material © 2018
10.3d) CORRECT ANSWER - A
The given circuit can be simplified using Thevenin theorem.
Rth = 20 /cO|| 10 ka = 6.66 kSl
Vth can be calculated using voltage divider rule as shown below.
/10 kQ.\
ka) = 333V
Let us assume that transistor is operating in active region.
In active region: Vbe = 0.7 V lb > 0mA
Vce > 0.7V
Base-emitter KVL can be written as shown below:
3.3V - (6.6k a )(Ib) ~ Vbe ~ (1 kD.)(Ie) = 0
3.3F - (6.6kO)(Ib) - 0.7V - (lk Q .)(W l Ib) = 0
Ib = 2.44 x 1 0 "5^
lc = pib = (100)(2.44 x 10~5i4) = 2.44 mA
Ie = 101 Ib = 2.46 mA
KCL can be written at the collector as shown below:
10 V - Vc _ r
2 Mi
~ Ic
Vc = 10 V - (2 M!)(2.44 mA) = 5.12 V
KCL can be written at the emitter as shown below:
Ve - 0
_____
Ve = (1 kSi)Qe) = 2A6V
Vce =
VC- V e = 5.12 V - 2.46 V = 2.66 V > 0.7 V
Our assumptions are correct because Vce > 0.7 V, Ib > 0mA
Therefore, transistor is operating in active mode.
243
Copyrighted Material © 2018
10V
10.3e) CORRECT ANSWER - C
To analyze a BJT circuit we assume a state of operation, enforce conditions and then verify assumptions.
Let us assume that transistor is operating in active region.
In active region: Veb = 0.7 V lb > 0mA
Vec > 0.7V
Emitter - base KVL can be written as shown below;
SV - (1 kO)(Ie) - Veb - (2 kO)(Ib) - I V = 0
4V - ( lka)(101fb) - 0.7V - (2 kO)(Ib) = 0
Ib = 32.03 \iA
soqji
lc = pib = 3.2 mA
•jy
Ie = 3.23 mA
2to
lu
KCL can be written at the emitter as shown below:
1ko
sv -v e
Ve = SV - (1 ka)(3.23mA) = 1.77 7
KCL can be written at the collector as shown below:
Vc - (-5 1 0
■.—^.- ..- = lc = 3.2 mA
500 a
c
Vc = (3.2mA)(S00 O) - SV = -3.4V
Vec = Ve - V c = 1.77 V - (-3 .4 V) = S.17V > 0.7 V
Our assumptions are correct because Vec > 0.7 V, lb > 0mA.
Therefore, transistor is operating in active mode.
244
Copyrighted Material © 2018
10.3f) CORRECT ANSWER - D
To analyze a BJT circuit we assume a state of operation, enforce conditions and then verify assumptions.
Let us assume that transistor is operating in active region.
In active region: Veb = 0.7 V Ib > 0mA Vec > 0.7V
jgL
Emitter - base KVL can be written as shown below:
>v j
5 F - Veb ~ (20 kO)0b) - 4 F = 0
IV - 0.7V - (20m ) ( / b) = 0
Ib
20 ka
= 1.5 x 10~5,4
Ic = pjb = 1.5 mA
L = 1.51 mA
KCL can be written at the collector as shown below:
O v e
5V
Vc _- _ (0V)
___
= Ic = 1-50 mA
Vc = (/c)( 2 k£i) = 3V
Vec = Ve - V c = 5 V — (3 V) = 2 V > 0.7 V
Our assumptions are correct because Vec > 0.7 V, Ib > 0mA
Therefore, transistor is operating in active mode.
245
Copyrighted Material © 2018
10.4 MOSFETs - Solutions
Consult NCEES® FE Reference Handbook - Page 217 for reference
10.4a) CORRECT ANSWER - D
To analyze a MOSFET circuit we assume state of operation, enforce conditions and then verify assumptions.
Let us assume that transistor is operating in saturation region.
Gate-Source KVL can be written as shown below:
0 - v a s -(2 fc n )(/d) + 107 = 0 - » i d =
10V-
10V —vgs
2ka
Drain-Source KVL can be written as shown below:
- (2 k m d) + 10V = 0 ^ vds = 20V - (4/cO)(id)
10V - (2 W l)(id) -
According to MOSFET mathematical relationships: id = K(vgs — Ft) 2
10V - v,gs
2m
= k {vbs - vty
Solving above given quadratic equations results in following values:
vgs = 3.54 V or vgs = -2 .5 4 V
Since we assumed saturation, value should be vgs = 3.54 V -» vgs > Vt
10V - vas 1 Q V - 3.541/
id = — = ----- — ------= 3.23 mA
2k£l
2kfL
vds = 20V - (4fcfi)(id) = 7.08 V
gs
3.54 V > V t
vds = 7.08 V > vqs - Vt or vqd < Vt
Our assumptions are correct because vgs > Vt and vgd < Vt. Therefore, transistor is in saturation mode.
10.4b) CORRECT ANSWER - Triode
According to details provided in the question:
vg = I V
vs = 6 V
(57-0)
id
5kCL
vd = SV
ig = 0
= 1 mA
vsd = vs - v d = 6V - S V = l V
v sg ~ v sd = v dg ~
— IV
vsg = vs - vg = 6V - IV = 5F -> vsg > Vt
> Vt
Therefore, transistor is operating in triode region.
246
Copyrighted Material © 2018
10.4c) CORRECT ANSWER - B
The given circuit can be simplified using Thevenin theorem.
Rth = 60 /en||40 kil = 24 kSl
v
=
4 0 k£L \
w l( ---------= 16V
\iooka)
Let us assume that transistor is operating in saturation region.
vg = 1.6 V, ig = 0 A, vs = 0 V
id = K(vgs - Vtf = 0.2mAV~~2(1.6V - IV )2 = 72 \iA
vd = 3.28 V ,
vds = 3.28F - OF = 3.28V
vgs = 1.6V > Vt
vgd = 1.6V - 3.28V = -1 .6 8 V < V t
Our assumptions are correct because vgs > Vt and vgd < Vt.
Therefore, transistor is operating in saturation mode.
10.4d) CORRECT ANSWER - C
The given circuit can be simplified using Thevenin theorem.
Rth = 20ka\\20ka= 10 ka
/20k£L\
v^ = 4v ( 2m i ) = 2 v
Let us assume that transistor is operating in saturation region.
vg = Vth = 2V,
ig = Q, vsg = 4 V - 2 V = 2V
k = K(vsg - Vtpf = 0.lmAV~2(2V - IV )2 = 0.1 mA
Source-Drain KVL can be written as follows:
4V —vsd —id x 10k£l = 0
vsd = 4V — id x 10kQ. = 3 V
vsg = 2V > V t
Vdg = vsg - vsd = 2 V - 3 V = - 1 V < Vt
Our assumptions are correct because vsg > Vt and vdg < Vt.
Therefore, transistor is operating in saturation mode.
247
Copyrighted Material © 2018
10.4e) CORRECT ANSWER - B
The given circuit can be simplified using Thevenin theorem.
Rth = 10 kQ1120 kQ = 6.6 kQ
To analyze a MOSFET circuit we assume state of operation, enforce conditions and then verify assumptions.
Let us assume that transistor is operating in saturation region.
Vg — Vfa — 2.6 V, igs — 0 A
Gate-Source KVL can be written as shown below:
2.6V - vgs - (10 kQ)(id) + SV = 0
Solving the above given quadratic equations results in following values:
vgs = 3.12 V vgs = -2 .1 V
Since we assumed saturation, the possible solution should be
Vgs > V f ^ Vgs = 3-12 V
id = Q.lmAV~2(3.12V - IV )2 = 0.44 mA
vd = 10V - (10 kQ)(id) = 5.6 V
vgs = 3.12 V > V t
vgd = 2.6 7 - 5.6 V = -3 V < Vt
Our assumptions are correct because vgs > Vt and vgd < Vt .
Therefore, transistor is operating in saturation mode.
248
Copyrighted Material © 2018
10.5 Operational Amplifier - Solutions
Consult NCEES® FE Reference Handbook - Page 212 for reference
10.5a) CORRECT ANSWER - -5 V
According to NCEES® FE Reference Handbook, inverting amplifier gain is given by:
R2
W k tt, N
v° = - £ v‘ = - w i W = - 5 v
10.5b) CORRECT ANSWER - 5 kQ
According to NCEES® FE Reference Handbook, non-inverting amplifier gain is given by:
R?\
Vr
=
(
1
+
i
(
h
=
{
R
1
+
m
)
2 V
=
1
2
V
-
R
=
5
k
a
10.5c) CORRECT ANSWER - 2.5 V
According to NCEES® FE Reference Handbook, non-inverting amplifier gain is given by:
R2
^ = - i ^
(
6 kQ
R2\
+ { 1 + i d Vb = - m
(
6kQ\
lv + { 1 + m
) 1-5v = 2-5 v
10.5d) CORRECT ANSWER - -35 V
We can solve this problem using principle of superposition as shown below:
6V power source shorted results in following output.
3V - 0
0 — v0
10 kQ ~ 50 kQ
Vq1 ~ ~ 15 V
3V power source shorted results in following output.
67-0
0 - F0
15 kQ ~ 50 kQ
V°2 ~ ~ 2° V
Vq = v01 + v 02 = —15 V —20 V = - 3 5 V
10.5e) CORRECT ANSWER - 0.5 V
According to NCEES® FE Reference Handbook, non-inverting amplifier gain is given by:
R2
(
R2\
v° = - i v“ + { 1 + T h
5 kQ
(
5 Mi \
= ~ T ^ 2V + { 1 + T o m ) l v = °-5 v
249
Copyrighted Material © 2018
10.6 Instrumentation - Solutions
Consult NCEES® FE Reference Handbook - Pages 124 -126 for reference
10.6a) CORRECT ANSWER - C
Transducer converts a physical quantity/value into electrical signal. Microphone (voice), thermocouple (heat)
and photodiode (light) convert different physical quantities into electrical signals.
10.6b) CORRECT ANSWER - B
Resistance at a given temperature can be calculated using formula given below:
Rt = R0[ 1 + a(T - T0)] = 2 0 0 n [l + O .O O S^ C -H lS0*:)] = 211 fl
10.6c) CORRECT ANSWER - 5000 0
Wheatstone bridge is balanced therefore, unknown resistance can be calculated as shown below:
R±
R3
r2
Rx
- i =
r
R2
= -± r
Rt
1000 a
=——
500 n = 5000 n
100 0
10.6d) CORRECT ANSWER - 200 O
AR of Wheatstone bridge can be calculated using formula given below:
AR
^ ~ 4(R) ^in
AV x 4R 0.5 V x 4000 a
AR = ---- -------= -------- ------------ = 200 a
Vin
10 7
10.6e) CORRECT ANSWER - D
Wheatstone bridge, RTD and Strain Gage can be used to measure resistance.
10.6f) CORRECT ANSWER - B
Voltage across 10 kQ. can be calculated as shown below:
10 m | | 100 k£i
Vlom ~ io m u io o ka + 25 m || 5 o m + 20 m 10V ~ 1-986 v
250
Copyrighted Material © 2018
10.6g) CORRECT ANSWER - A
Voltage across 10 ka can be calculated as shown below:
10 m | | 5°0 ka
Vioka - 10 /dl||500 m + 25 m| |so m + 20 k a 10V ~ 2-10 v
10.6h) CORRECT ANSWER - B
Actual current without ammeter can be calculated as shown below:
107
2ka + 1 k a \11 ka
10V
2ka + 0.5 ka = 4 mA
Current measured by ammeter is shown below:
2ka + 1 ka\\i ka + 0.05 ka = 3.92 mA
% error =
4 mA — 3.92 mA
251
Copyrighted Material © 2018
10.7 Power Electronics - Solutions
10.7a) CORRECT ANSWER - B
Chopper converts DC to DC i.e. constant to variable DC or variable DC to constant DC.
10.7b) CORRECT ANSWER - A
Inverter converts DC to AC with required voltage and frequency.
10.7c) CORRECT ANSWER - C
Rectifier converts AC to DC.
10.7d) CORRECT ANSWER - D
Cyclo-converter changes given AC to AC with required frequency and amplitude.
10.7e) CORRECT ANSWER - B
According to problem statement, we have a single phase half-wave rectifier circuit with a resistive load of 50
and firing angle 45° as shown below. Thyristor is an electronic switch and works as explained below.
•
It turns ON/starts conducting if gate current (pulse) is applied when Anode-Cathode is forward biased.
•
It turns OFF when Anode-Cathode becomes reverse biased or anode current falls below threshold.
f1
Vout
ON
OFF
ON
OFF
Thyristor will turn on at 45° firing angle and it will stay on until voltage polarity changes as shown above.
252
Copyrighted Material © 2018
Chapter # 11 - Power
11.1 Single Phase Power - Solutions
Consult NCEES® FE Reference Handbook - Pages 203 - 205 for reference
11.1a) CORRECT ANSWER - 592 W
$l<p = Vl<pll<p = P +JQV = 120V Zune = 2 + 2j£l
__
V
Zline + Zload
Zioad = 20 + 5j£l
120 V
(2 + 2/0) + (20 + 5/a)
/ = 5.19/-17.650 A
S*p = VicpKcp = (120/0°) (5.19/17^65°)VA
P = ScosO = 622 cosl7.65° = 592 W
11.1b) CORRECT ANSWER - B
According to given details, V = 10/8 0 °V and Z = 5 + j a = 5.09/11.3° a.
I = V/Z = 1.96/68.7° A
11.1c) CORRECT ANSWER - C
^i<p = Vi(pli<p
Z = (5 + j n )||(-2 j tt)= 2 /-67.30 a
Vi<P = h i p * Z =10/0 A x 2/-67.30 f l =20/-67.3° V
Sicp = Vicpllcp = 200/d67,3? VA
ll.ld ) CORRECT ANSWER - 9.19 W
s*p = Vicpiicp = P +JQZ = 10 + 5; - 2j a = 10 + 3; a = 10.4/16.7°a
V= 10/1 0 ° V
v
I = - = 0.96/ —6.7 >4
^i«p =
= 9.6/16.7° V A
P = 9.57cos(16.69°) = 9.19 14/
253
Copyrighted Material © 2018
ll.le ) CORRECT ANSWER - B
It can be observed that the voltage across 5 Q resistor is 5/0° V. Power absorbed by 5 Q resistor can be
calculated as shown below.
l l . l f ) CORRECT ANSWER - D
Maximum power transfer occurs when Ztoad = Z^hevenin
To calculate ZThevenin, open circuit current source and short circuit voltage source.
Z Thevenin = 10 — 5/ + 2 + 6j = 12 + j Q
Therefore, maximum power transfer happens when Zioad = Z*Thevenin = 12 —jVL
l l. l g ) CORRECT ANSWER - A
Maximum power transfer occurs when Ztoad = Z^hevenin
To calculate Z Thevenin, open circuit current source.
Z Thevenin — 5 — / + 3/ O = 5 + 2/ O
Therefore, maximum power transfer happens when Ztoad = Z jhevenin = 5 — 2jQ.
254
Copyrighted Material © 2018
11.2 Three Phase Power / Transmission & Distribution - Solutions
Consult NCEES® FE Reference Handbook - Pages 203 - 205 for reference
11.2a) CORRECT ANSWER - C
According to problem statement, Van = 120/30° V.
In the given scenario, Vbn = 120/30o-120° V and Vcn = 120/30o+120° V
Line voltages lead phase voltages by 30° and are greater by a factor of V3.
V ab = V3 X 120/30o+30° V= 208/60° V
Vbc = 208/-60°V and Vca = 208/180° V
11.2b) CORRECT ANSWER - B
In a balanced 3cf>-Y network, phase current can be calculated as shown below.
j
_ Van
*an ~
^(p
Van = 120/30° V and Zv = 20 + 5j Q =20.6/14° P.
Ian = 5.82Z16!A
11.2c) CORRECT ANSWER - C
In a balanced 3cf)-Y network, phase current equals line current.
7an
an
=
Z(p + Z[
277 + 0j V
(5 + 5jd ) + (1 + jff)
Ian = h = 32.6/245° A
11.2d) CORRECT ANSWER - A
In a balanced 3c|>-Y network, phase current and load voltage can be calculated as shown below.
j
________________
(X T l
rp
Van___________
j
r j
z load ' L line
Van= 120/60^ V and ZUne = 2 + ; 0 Zload = 10 + 10;0
I an = 7.4/18! A
Vload = I an X Zload = 104/63! V
255
Copyrighted Material © 2018
11.2e) CORRECT ANSWER - B
In a balanced 3<|>-Y network, line current, voltage and impedance are related as shown below.
_
Ij CLTl -ry
*7
^ load
Van
ry
^load
• ^lin e
I
__
' ^line ~
*an
Zioad = V
- T ~ Zline = 23.4 + 1.75j Q = 23.4/4,2° «
inn
11.2f) CORRECT ANSWER - B
It is important to note that given network is A-Y.
j
_ ^an
'a n ~
L< p
According to problem details, Vab = 208/30°7 and Z& = 10 + 5j O
Van lags Vai) by 30° and it is smaller by a factor of V3, this makes Van = 120/0°F
Ian = 10.7/-26.5°A
11.2g) CORRECT ANSWER - D
It is important to note that impedance is connected in a A arrangement and source is Y connected.
Van = 1 2 0 / W
This problem can be solved easily by converting ZA to ZY as shown below:
ZA = 10 + 2jO
Z& — 3 x ZY -* ZY —
10 + 2 / Q
~
ZY = 3.4/11.3°D
256
Copyrighted Material © 2018
11.2h) CORRECT ANSWER - A
It is important to note that one load network is Y-connected and second load network is A-connected.
Zy_ i = 10 + 5jO
ZA_2 = 6 + 9jQ
ZA = 3 X Zy
Z &—
2 = 2 + 3; Q
Z Y- 2 = ~
z eq.y = Zr^ ltZy., = (10 + 5j)||(2 + 3j)(l
Z eq. Y =
2.8/48°fl
11.2i) CORRECT ANSWER - C
Equivalent delta load can be calculated using this formula:
Z eq_A — 3 x Zeq__Y — 3 x 2.8 == 8.4/5
11.2j) CORRECT ANSWER - 0
3~4» complex power of a balanced system can be calculated using following formula.
s 3p = 3vpi;
,
*an
_ ^an
rj
■Moad
Van = 120/0y
Zload = 201&Q
Ian=6/0! A
S3p = 3VpI* = 2160 /0£VA
11.2k) CORRECT ANSWER - C
Real power is related to voltage, current and power factor as shown below.
PI
200 kW
P = VI cos'd -> V = ------ =
— ... N = 602V
cosd (400,4)(0.83)
257
Copyrighted Material © 2018
11.21) CORRECT ANSWER - A
Line losses will be equal to the difference between power supplied by source and power consumed by load.
Pload =
100 kW
Psource
= V I cosd = 6 0 0 F x
x
20(L4
Ploss = Psource ~ Pload = 1 0 2 k W ~
0.85 = 102 kW
lOOkW = 2 kW
11.2m) CORRECT ANSWER - B
3-c|> real power is related to voltage, current and power factor as shown below.
^34>= 3 ^ / 4, cosd -> 7$ =
= 125 kW,
^34,
31$ cosd
1$ = IL = 300 A,
p f = 0.694
125 kW
V* “ 3 x 300.4 x 0.694 " 200 V
258
Copyrighted Material © 2018
11.3 Voltage Regulation - Solutions
11.3a) CORRECT ANSWER - C
Voltage regulation can be calculated using formula given below.
V sn i-V s fi
6 5 0 V — 6 00 V
V . R = — ------ --------- -----------------------= 8 .3 3 %
VS if l
60 0 V
11.3b) CORRECT ANSWER - C
Ideal transformers are lossless. Therefore, for an ideal transformer the no-load voltage is equal to full load
voltage and voltage regulation is zero.
11.3c) CORRECT ANSWER - 7%
Voltage regulation can be calculated using formula given below.
Vp
V R = ^S,nl ~ ^S’f l =
^S’f l
VStfl
V s ,fi
lO k V A
Is,rated =
= 25 A
p f = 0.85 ( 0 = 3 1 .7 8 °)
phase angle =
-3 1 .7 8 °
\ = vs + Is,rated 0Req + j X eq) = 427.8 V
Vv
u
4 2 7 .8 F — 4 0 0 F
— — = ---------------------------------------------------- r —
a~ V s.fi
V.R =
V S tfi
400V
-- x 1 0 0 % = 6 .9 8 % = 7 %
11.3d) CORRECT ANSWER - B
No-load voltage rating can be calculated using voltage regulation formula given below.
T/ n _ V s,n l ~ V s .fl
vs ,fl
0.05 =
Vs>nl - 240 V
240 V
VsM = 240 V + 240 V x 0.05 = 252 V
11.3e) CORRECT ANSWER - B
Voltage regulation can be calculated using formula given below.
rr n __ Vs.Tll ~ VS,fl
-
v
v s,fl
In cases of leading power factor Vsji > Vsni.Therefore, V.R < 0
259
Copyrighted Material © 2018
11.4 Transformers - Solutions
Consult NCEES® FE Reference Handbook - Page 205 for reference
11.4a) CORRECT ANSWER - B
VS = IS x RS = 1A x 12 a = 12V
Vv
50
Vp
■j| = a = y - > ^ = T
50
->7p = 1 2 7 x 5 0 = 6 0 0 7
11.4b) CORRECT ANSWER - A
Secondary current can be calculated using the formula given below:
S<p
I
SSec VsecIsec —>15 kVA
Sprtm
Is
15kVA
=
.= 125 A
s 120 V
11.4c) CORRECT ANSWER - 4.16 A
Secondary current can be calculated using the formula given below:
Sa 5 kVA
Is = 7T = T^TTTT = 41.6 A
5
Vs 120 V
/ - —=
p
a
10
Is
10
-i = a = —
Ip
1
= 4_i6 A
11.4d) CORRECT ANSWER - 25000 fl
Transformer primary and secondary impedances are related by formula shown below.
Zv = a2Zs = 502 x lO H = 25000H
11.4e) CORRECT ANSWER - 50 H
Transformer primary and secondary impedances are related by formula shown below.
Z
£ jp
K
= a2Z -»
Z =2—
/
U
js
a*
zp = io fcn||io m = sk n
a = 10
_ _ zp _ 5 fcO
7
s
a2 102
100
260
Copyrighted Material © 2018
11.5 Motors & Generators - Solutions
Consult NCEES® FE Reference Handbook - Page 205 for
11.5a) CORRECT ANSWER - C
AC motor poles can be calculated using the formula given below.
Yl
—
5
120/
120 /
(120)(60 Hz)
p
ns
(1800 rpm)
----------» r ) — ------- zz ------------------- =
4
11.5b) CORRECT ANSWER - C
Synchronous speed of AC motor can be calculated using the formula given below.
120/
(120)(60 Hz)
ns = ------ = ------------------ = 3600 rpm
p
2
11.5c) CORRECT ANSWER - B
Synchronous speed of AC motor can be calculated using the formula given below.
120/
n = ------V
120/
(120)(50 Hz)
ns-50Hz = — — = -------- 7-------- = 1500 rpm
p
4
120/
(120 7)(60 Hz)
ns-60Hz = ------ = ---------- ^--------- = 3600 rpm
p
L
ns-60Hz ~ ns-50Hz - 3600 rpm - 1500 rpm = 2100 rpm
11.5d) CORRECT ANSWER - 5.5%
Synchronous speed and slip of AC motor are given by following formula.
120/
(120)(60)
ns = -------= ---------------- = 3600 rpm
p
2
n = 3400 rpm
ns —n 3600 — 3400
s = — ----- = ------ — ------- = 5.5%
ns
3600
11.5e) CORRECT ANSWER - D
Synchronous speed and slip of AC motor are given by following formula.
120/
(120)(60)
n = ------ = ------- ------ = 1800 rpm
p
4
ns —n
1800 — n
s = -----------> 0.1 = — ..— ------ >n = n, —ns = 1620 rpm
ns
1800
5
F
261
Copyrighted Material © 2018
11.6 Power Factor - Solutions
Consult NCEES® FE Reference Handbook - Page 203 for reference
11.6a) CORRECT ANSWER - D
Capacitance required to improve power factor angle from 6± to 92 can be calculated as follows.
P(tand 1 —tand2)
^Vrms
cosdt = 0.75,
=
41.4°
cosd2 = 0.90,
d2 =
25.8°
- tan25.8°)s ^ no
_ 200kW(tan41A°
v
(2 tt)(60H z )(6 0 0 F )2
“
T?
^
11.6b) CORRECT ANSWER - B
Reactive power required to improve power factor angle from 9t to 02 can be calculated as follows.
Qc = P(tand1 —tan$2)
c o s = 0.60,
d1 = 53.1°
cosd2 = 1.00,
d2 = 0°
Qc = 100(tan53.1° - tanO0) = 133.1 kVAR
11.6c) CORRECT ANSWER - 0.98 lag
Capacitance required to improve power factor angle from 6t to 02 can be calculated as follows.
P(tand 1 —tand2)
^Vrms
cosd± = 0.8,
d1 =
36.86°
Co) Vrms = P(tand1 —tand2)
d2 = tan- 1 1 tand1 ------ — 8 .3 3 0
cos(i92) = 0.989
262
Copyrighted Material © 2018
11.6d) CORRECT ANSWER - 0.82 lag
S± = 7 5 kVA/cos^OBS
Pt = 7SkVA x pft = 75 kVA x 0.85 = 63.75 kW
Qt = 7SkVA x sin(cos_10.85) = 39.5kVAR
S2 = 35kVA/cos-iQ.7S
P2 = 35 kVA x pf 2 = 3SkVA x 0.75= 26.25kW
Q2 = 35 kVA x sin(cos- 1 0.75) =23.1kVAR
St =
+ S2 = 90 kW + j62.6SkVAR = 109.67WA 734.84°
Pt
90 kW
P f t = T t = W k V A = °-82 la"
inS
11.6e) CORRECT ANSWER - D
Capacitance required to improve power factor angle from 0± to 02 can be calculated as follows.
P(tand 1 —tand2)
Marins
cosdt = 0.85,
01 = 31.78°
cosd2 = 0.95,
d2 = 18.19°
P34) = 3 75 kW
P14, = 125 kW
125/cW(tan 31.87 0 - tan 18.19 °)
Ci^ _
(2n) (60 Hz) (120V)2
Cl4) = 6.7 mF = 7 m F
263
Copyrighted Material © 2018
Chapter # 12 - Electromagnetics
12.1 Maxwell Equations - Solutions
Consult NCEES® F E Reference Handbook - Pages 35 and 205 for reference
12.1a) CORRECT ANSWER - D
Divergence of electric field can be calculated using Del operator as shown below.
_
_
/d
d
d \ ,
_
d
d
_
d
div E = V.E = l — i + — / + — & ) . ( 3 xi + 2 y2j + xk) = — 3x + — 2 v + — x = 3 + 4y
\dx
dy
dz)
'
dx
dy
dz
J
Helpful tip - Review vector operators i.e. gradient, curl and divergence given in NCEES® FE Reference Handbook.
12.1b) CORRECT ANSWER - B
Divergence of given vector can be calculated using Del operator as shown below.
_
_
( d
d
d \
d
d
d
div D = V.D = \— i + — i + — k )(x y i + yzj + xzzk)==— xy + — yz + -—xzz = y + z + 2xz
\dx
dy
dz J
7'
dx J dyJ
dz
*
12.1c) CORRECT ANSWER - C
According to Gauss' Law, electric flux over a closed surface can be calculated as shown below.
iQenc +1 nC —2 nC + 3 nC + 1 nC
(p= $E.ds =
= ------ — -----— ..- .- .............. = 339 Vm
T
e
8.85 x 1 0 -12 C/Vm
Helpful tip - Review vector and integral form of Gauss' Law given in NCEES® FE Reference Handbook.
12.1d) CORRECT ANSWER - B
According to Gauss' Law, electric flux over a closed surface can be calculated as shown below.
cp = <b E.ds =
Qenc
C
£
Ql
( 1 \2
31*^2 x 47r( l M j
Q2 =
V2
C
m2
—5|i— r x
Qz
8.85 x 10~12C/Vm
q
= 3 -76 X l °
C
( 3 \2
4n —
= —56.5 x 1 0 _ 9 C
V100 /
(p = —5963 Vm
264
Copyrighted Material © 2018
12.1e) CORRECT ANSWER - C
We can calculate the value of "a" by taking divergence of magnetic field B.
_
_
/d
d
d \
d
d
d
div B = V.B = [ — i + — j + -T~k) (3 axi + 2yi —2zk) = — 3 ax + — 2y —— 2z = 3 a + 2 — 2 = 0
\dx
dy
dz)
dx
dy
dz
3 a = 0 -> a = 0
Helpful tip - Divergence of a magnetic field is always equal to zero.
12.If) CORRECT ANSWER - B
According to Maxwell's equations, divergence of a magnetic field is zero.
— 2x2+ — y + — 3z 2= 4x + 1 + 6z
div A =
V.A=
—
div B =
—d
n
d
d
V.B=
— 2y + — 3x — — 2xy = 0
dx
dy
dz
dx
dy
dz
Divergence of B is zero.
Therefore, B can be a magnetic field vector.
12.lg) CORRECT ANSWER - B
The rate of change of magnetic field B is equal to the curl of electric field strength vector E.
d—
V x E= —r B
dt
E = 2yzi + 3 x 2yj + x 2y2k
d—
/dEz
dEy\ m /dEx
dEz\ m /dEy dEx\ ^
dt
\ dy
dz ) l
dx\ dx
\ dz
dy)
d_
„
— B = —2x yi - (2y — 2xy2) j —(6 xy —2z)k
12.Ih) CORRECT ANSWER - C
The rate of change of magnetic field B is equal to the curl of electric field strength vector E.
d—
V x E = — —B
dt
B =
- c o s 2(3 t ) k
_
d—
V x E = ~—B =
/ d
— ^— —
„
\
d
„
c o s 3 t j k = — cos23 t k =
—6 c o s (3 t ) s in (3 t )fc
265
Copyrighted Material © 2018
12.li) CORRECT ANSWER - B
According to Faraday's Law, voltage induced in a coil can be calculated as shown below.
d&
1.4 Wb
v = - N — = - 2 0 — -----= - 1 4 V
dt
2s
Helpful tip - Review vector and integral form of Faraday's Law given in NCEES® FE Reference Handbook.
12.1j) CORRECT ANSWER - 0 .15 \iV
According to Faraday's Law, voltage induced in a coil can be calculated as shown below.
di N 2\iAdi
(50)2 x 4n x 10~7Hm~1 x 2cm 2 100mA —50mA
------------ — ----------------------x — ------------------= 0.15 \iV
v = L— = — — — =
dt
I dt
20 cm
Is
12.1k) CORRECT ANSWER - 500 \iT
Magnetic flux density can be calculated using Ampere's Law as shown below.
dD
H. d l
lenc "I" 11
I
^
/enc = W/ = (5 0 )(li4 ) = 5 0 /l
According to problem statement:
dD
//, dt ~" °
B
H=—
—j) dl = lenc
j) dl = 2nr = 2n (2 cm)
circumference o f torus
B = --- --------- zz: 500 \iT
2n x 2 cm
12.11) CORRECT ANSWER - 0 .4 \iT
This problem can be solved using Ampere's Law and principle of superposition
Magnetic flux density at point A due to first wire is given as:
I\i
B1 = - ^ - = 0 . 2 [ i T
2nr
Magnetic flux density at point A due to first wire is given as:
fMB2 = ~ ~ = 0.2 \iT
2nr
Adding the two magnetic fields result in Bnet = Bt + B2 — 0.4 |xT
Note that in this case magnetic fields add due to current direction. If current was flowing in opposite direction,
magnetic fields would have cancelled each other. Apply right hand rule to determine magnetic field direction.
266
Copyrighted Material © 2018
12.1m) CORRECT ANSWER - 100 kA
Current passing through wire can be calculated using Ampere's Law.
B=
I =
I\i
2nr
2nrB
PAn x 10 7H/m
Helpful tip —Review vector and integral form of Ampere's Law given in NCEES® FE Reference Handbook.
267
Copyrighted Material © 2018
12.2 Electrostatics/Magnetostatics - Solutions
Consult NCEES® FE Reference Handbook - Page 200 for reference
12.2a) CORRECT ANSWER - D
Electrostatic forces on charges can be calculated using the formula given below.
p
— Q 1® 2
4n£r2
(5n C )(1 0n C )
P
_
4tt x 8.85 x 1 0 -12Fm - 1 x (V5)
F = 90 nN
12.2b) CORRECT ANSWER - C
Electric field intensity can be calculated as shown below.
~F -
I=
* , @2 ,
4nsr21 4ner 2
Qi
Qi
.
4nsr2
.
4ner2
E = ( — i _____ — ') i
\4ner2 4ner2)
50 nC
—50 nC
47Tf(Vl)2
4 n e { 4 lf
|I| = 900 V/m
Note that if both charges were of same polarity, the total would have been zero. It is important to understand
that electric field lines diverge from positive charges and converge on to negative charges.
12.2c) CORRECT ANSWER - B
Electric field intensity of a sheet charge with density ps can be calculated as shown below.
We can use principle of superposition as shown below.
7;
Ps .
Ps S
, N
Ps ,
E = — k -~ -(-U ) = —k
2s
2e
e
Note that if both electric plates were of samepolarity, thetotal would have
been zero.It is important to
understand that electric field lines diverge from positivechargesand converge on tonegative
268
Copyrighted Material © 2018
charges.
12.2d) CORRECT ANSWER - D
Electric field intensity of a line charge with density p can be calculated as shown below,
p -
9 ~
2ner r
We can use principle of superposition as shown below.
1 =
+2 ^ 5 )
^
X 1011 ^
Helpful tip - Pay special attention to direction of vectors while performing addition and subtraction.
12.2e) CORRECT ANSWER - D
Force on a current carrying conductor in uniform magnetic field can be calculated using following equation.
F = IL x B
Since we are only interested in the magnitude of force, we can take modulus as shown below.
\F\ = ILBsind
\F\ = 2 A x 2 m x 0.5 T sin30° = I N
12.2f) CORRECT ANSWER - 5fij
Energy stored in a magnetic field can be calculated using following equation.
IxH2
Energy stored = — —x Vol
4n x 10~7Hm-1
Energy stored = ---------------------x ( 2Am
) x 2m 6
Energy stored = 5/i/
269
Copyrighted Material © 2018
12.3 Transmission Lines and Wave Propagation - Solutions
Consult NCEES® FE Reference Handbook - Pages 205- 206 for reference
12.3a) CORRECT ANSWER - 0. 5 /6.91°
Reflection coefficient can be calculated using following formula.
n _ Zt - Z0 _ ((300 + 50; H) - 100 fl)
r ~ z t + z 0 ~ (300 + soj a) + ioo a
r = 0.5 phase 6.91°
Helpful tip - Review transmission line reflection coefficient given in NCEES® FE Reference Handbook.
12.3b) CORRECT ANSWER - C
Reflection coefficient can be calculated using following formula.
Zi —Z0
(500 + 25j O) — 50 n
Z[ + Z 0
(500 + 25j 0 ) + 50 fl
r = 0.818 phase 0.57°
l + |f|
1 + 0.818
^ * = T H rf = 130515= “
Helpful tip - Review transmission line standing wave ratio equation given in NCEES® FE Reference Handbook.
12.3c) CORRECT ANSWER - B
Reflection coefficient is given by following formula.
p —
Zi + Z 0
~
Characteristic impedance can be calculated using following formula.
Z0 =
L
- = 100 n
According to problem details, r = 0.5
Qc _
■ ~ Z i + Z 0 ~ Z t + 100
~
~
z t = 300 n
270
Copyrighted Material © 2018
12.3d) CORRECT ANSWER - D
2n
Zt + iZ0 tan(j8d)
= 10
Z in id ) = ZoZ 0 + j Z l ten(fid)
2n
^=T
z o = 200a z z = 500 a
Zin (100) = 200
ta n g x
500 + /200tan(/? x 100)
200 + y'500tan(/? x 100)
lo o ) = 0
/500H\
zin(ioo) = 200 a — — = 500 a
inK J
\200ay
12.3e) CORRECT ANSWER - D
Standing wave ratio is given by following equation.
l + \r\
1
W fi = T H T i = 2
" r =3
Reflection coefficient can be calculated using following formula.
Z,-Z„
1
Zt + Z 0
3
r = -i — 2. -»_ =
Zj - 2 5 0 a
----------- >z, = 500 a
Zj + 2 5 0 a
1
12.3f) CORRECT ANSWER - B
2n
P
2n
n
20 “ 10
V(d) = V+eJPd + V~e~ifid
7(100) = V+eJ10n + V~e~jl0n
Helpful tip - Review voltage across transmission line equation given in NCEES® FE Reference Handbook.
12.3g) CORRECT ANSWER - C
2n
^=T
2n
n
“ 20~ l0
1(d) = i+ ejPd + r e -JPd
+ _ F+
Zo
- _
V~
Zo
/(100) = - ~ ( V+eJ10n - V~e-jl0n)
Helpful tip - Review current across transmission line equation given in NCEES® FE Reference Handbook.
271
Copyrighted Material © 2018
12.4 Electromagnetic Compatibility - Solutions
12.4a) CORRECT ANSWER - D
Electromagnetic coupling can take place through induction, conduction and capacitive charging.
Helpful tip - Review electromagnetic coupling theory.
12.4b) CORRECT ANSWER - C
Shielding is a method used for protection against electromagnetic interference.
Helpful tip - Learn different methods that can be used to mitigate electromagnetic interference.
12.4c) CORRECT ANSWER - B
Cross talk happens when both emitter and receiver exist within the same system.
Helpful tip - Understand cross talk and learn how it can be reduced.
12.4d) CORRECT ANSWER - D
Lightning, arc welding and electric motors are potential sources of electromagnetic interference.
Helpful tip - Learn about various sources of electromagnetic interference.
12.4e) CORRECT ANSWER - D
Increasing separation between coupling paths, hardware redundancy and shielding can help mitigate harmful
impacts of electromagnetic interference.
272
Copyrighted Material © 2018
Chapter # 13 - Control Systems
13.1 Block Diagrams - Solutions
Consult NCEES® FE Reference Handbook - Page 127 for reference
13.1a) CORRECT ANSWER - C
Let us define error E(s) as E(s ) = R(s ) — H(s)Y(s )
Y(s) = E (s )G 1(s)G 2(s) = (R (s) - H ( s ) r ( s ) ) G 1(s)G 2(s) = ff(s)G 1(s)G 2(s) - H (s )K (s )G 1(s)G 2(s)
r ( s ) + H(s)Y{s')G1(,s)G2(s) = R (s)G 1(s)G 2(s) -> 7 0 0 ( 1 + /f(s)G 1(s)G 2(s)) = B (s )G 1(s)G 2(s)
Y d s ) ..
G ^ G ^ s)
R (s )
1 + //(*)&! (s)G 2(s)
Note: Closed loop transfer function can also be calculated using the classical negative feedback control system
model relations given on page 126 of NCEES® FE Reference Handbook.
13.1b) CORRECT ANSWER - C
Let us define error E(s) as E(s ) = R(s) —( Y(s) + N(s ))
r ( s ) = ( E ^ G , (5) + L O ) ) G 2( s ) = [ff(s) - ( y ( s ) + W(s ))]G 1(s )G 2(s ) + L ( s )G 2(s )
r ( s ) = R ( s )G 1( s )G 2( s ) - F ( s )G 1(s )G2(s ) - W(s )G 1( s )G 2( s ) + L(s)G 2(s)
r ( s ) + r ( s ) G 1(s)G 2(5) = fi(s )G 1(s)G 2(s) - W(s )G 1( s )G 2( s ) + L ( s )G2(s )
« M G 1(s)G 2(s) - N ( s )G 1(s )G 2(s ) + L (s)G 2(s)
W
1 + G1(s )G2(s )
13.1c) CORRECT ANSWER - A
Let us define error E(s) as E(s ) = R(s) — F(s)
r( s ) = £’(s)(G1(s) - C2(s) + G3(s))G 4(s)
Y(s) = (R (s ) - F ( s )) ( G 1(s) - G2(s ) + G3(s))G 40 )
F (s ) + F (s ) ^G4(s )(C 1(s ) — G2(s ) + G 3 ( s ))^ = R ( s )G 4 ( s ) ( G 1( s ) — C2(s ) + G3( s ) )
F (s ) ( l + G4 ( s ) ( G 1( s ) — G 2 ( s ) + G 3( s ) ) J = R ( s ) G 4 ( s ) ( G 1( s ) — G2(s ) + G3( s ) )
YCs) = R( j )
G4^
Gl^ ~
+
( l + G4( s )(G 1(s ) - C2(s) + G3(s)))
n f) =
R(s)
- g2(^) + g 3(s))
1 + G4(s)(G1(s) — G2(s) + G3(s))
273
Copyrighted Material © 2018
13.Id) CORRECT ANSWER - A
Let us define error E(s) as E(s ) = R(s ) — H(s)Y(s)
Y(s) = kE(s)Gt (s)
-> F(s) = kR(s)G1(s) - kY(s)H(s)G 1(s)
r( s ) =
F(s) + /cF(s)//(s)G1 (s) = k R & G ^ s ) -> F ( s ) ( l + k H ^ G ^ s ) ) = /c/?(s)G1 (s)
y(s) = k (s )
F(s)
i?(s)
/cG^s)
1 + kG1(s)H(s )
= T(s ) =
w
kG±(s)
l + fcGiCs)//^)
Note: Closed loop transfer function can also be calculated using the classical negative feedback control system
model relations given on page 126 of NCEES® FE Reference Handbook.
13.1e) CORRECT ANSWER - B
Let us define error E(s) as E(s ) = R(s) — F(s)
r( s ) = E(s)G1(s)G2(s) + H(s)Y(s) + E(s)G1(s)G2(s)G3(s)
y (s ) = (fi(s) - r ( s ) ) c 1( s ) c 2(s) + t f ( s ) n s ) + (fi(s ) - y ( s ) ) c 1( s ) c 2( s ) c 3(s)
r( s ) = fi(s)C 1(s)G2(s) - K ( s ) C 1( s ) G 2( s ) + H (s)Y (s) + fi(s)C 1(s)C2(s)C3(s) - r ( s ) C 1(s)C2(s)C3(s)
y (s ) + y (s )C 1(s)C 2(s) - H (s )Y (s) + F (s )C 1(s)G 2(s)G 3(s) = R (s)(G 1(s)G 2(s) + G1(s)G 2(s)G 3(5))
K ( s ) ( l + G1(s )G 2( s ) - « (s ) + G1(s)G2(s)G3(s)) = R(s)(Gt (s)G2(s) + G1(s)G2(s)G3(s))
K(s) = R(s)
C i (s )G 2( s ) + G1( s )G 2( s )G 3( s )
1 + G1( s )G 2( s ) - H (s) + G1(s )G2( s )G 3( s )
y (s )
fi( s )
G1(S)G2(S) + G1(S)G2(S)G3(S)
= T (S )
=
1 + G1(s )G 2( s ) - H (s) + G1(s )G 2( s )G 3( s )
Note: Closed loop transfer function can also be calculated using the classical negative feedback control system
model relations given on page 126 of NCEES® FE Reference Handbook.
274
Copyrighted Material © 2018
13.2 Bode Plot - Solutions
Consult NCEES® FE Reference Handbook - Page 208 for reference
13.2a) CORRECT ANSWER - A
Transfer function can be rearranged in standard form as shown below:
50
+
50
1 0
1
( t
0
o
5
+
1
T
)
o
+
1
Gain = 20 log(5) = 13.97 dB. Transfer function has pole at 5 = 10 rad. s~1
13.2b) CORRECT ANSWER - D
Transfer function can be rearranged in standard form as shown below:
H(s) =
20
1
s + 20
JL + i
20
Gain = 201og(l) = 0 dB. Transfer function has pole a ts = 20 r a d .s -1
13.2c) CORRECT ANSWER - A
Transfer function can be rearranged into standard form as shown below:
50(s + 2)
_/
i
\
( f +1)
^ 10Q^ S + 1000) - VioooJ/ s
\/
s
U 0 0 + 1A lO O O + V
Transfer function has pole at s = 100 rad. s _1, s = 1000 rad. s -1 and zero at s = 2 rad. s -1
13.2d) CORRECT ANSWER - 20 dB
Transfer function can be rearranged in standard form as shown below:
H(s) =
100(s + 1)
(s + 1)
,
.-.......... = (10)
s(s + 10)
( s) ( t o + i )
Gain = 201og(10) = 20 dB
13.2e) CORRECT ANSWER - -3 3 .9 dB
Transfer function can be rearranged in standard form as shown below:
H(s) =
100s
100s
s
(s2 + 150s + 5000)
(s + 50)(s + 100)
50 (JL + 1 ) ( ^ + l )
Gain = 20 log
= —33.9 dB
275
Copyrighted Material © 2018
13.3 Steady State Errors - Solutions
Consult NCEES® FE Reference Handbook - Page 128 for reference
13.3a) CORRECT ANSWER - B
For a unity feedback system, steady state error can be calculated using final value theorem as shown below:
sR(s)
s^o 1 + GO)
e(co) = lim sE(s) = lim ----- ——
v
s-»o
w
10
r( t) = 10u(t) -> R(s) = —
s
G(s) =
(50)(s + 4)
(s + l) ( s + 5)
10
e (o o )
5—
= lim
(s + l) ( s + 5)
10
e(co) = lim ------- — r r r ~ = -------------- ;
(50) (5 + 4)
(s + l) ( s + 5)
s->o
10
= °-24
(50)(0 + 4)
+ (0 + 1)(0 + 5)
Helpful tip - Understand steady state error calculation formula and learn how to use it for different inputs.
13.3b) CORRECT ANSWER - A
For a unity feedback system, steady state error can be calculated using final value theorem as shown below:
e (o o )
v
si? (5 )
= lim sE(s) = lim ----- ——s->o v J s-»0 1 + G(s)
5
R(s) = —
sz
r( t) = 5 t u (t )
G(s) =
100s
s2 + 11s + 30
5
s—
e(oo) = lim
,
1+
s->o 1
e(oo) = lim
s-* 0 1 +
s
100s
s 2 + 11s + 30
5
s
10° S
s 2 + l l s + 30
276
Copyrighted Material © 2018
13.3c) CORRECT ANSWER - A
For a unity feedback system, steady state error can be calculated using final value theorem as shown below:
sR(s)
e(co) = lim sE(s) = lim ----- --v
y
s->o
w
s->o 1 + G ( s )
20
r( t) = 10 1 u (t) -» R(s ) = —r
s-5
G(s) =
s 2 + 100 s + 10
20
20
20
e(oo) = lim ------------ -------------= lim ------------ ^
s-»0 1 _i_______ _______
s->0 -1 _i_______ _______
s2 + 100 s + 10
s 2 + 100 s + 10
------------- = oo
1 ,_______ 0______
0 + 100 ( 0) + 10
13.3d) CORRECT ANSWER - A
For a unity feedback system, steady state error can be calculated using final value theorem as shown below:
sR(s)
e(oo) = lim sE(s) = lim - ---- — —
v '
s-»o w
s->o 1 + G(s)
2
r ( t ) = 2 t u ( t ) -> R(s) = —
G(s) =
(s + 2)(s + 3)
_2_
e(co) = lim --------- ^
2
---------= lim
(s + 2)(s + 3)
s
(s + 2)(s + 3)
2
0
_
1 (0 + 2)(0 + 3)
IB.Be) CORRECT ANSWER - C
For a unity feedback system, steady state error can be calculated using final value theorem as shown below:
sR{s)
e(oo) = lim sE(s) = lim ----- - 7- s-»o
s->o 1 + G(s)
3
r( t) = 3u(t) -* R(s) = s
,
G(s)
10(s + 5)
(s + l) ( s + 2 )
3
St
e(oo) = lim -------- 7777— —=r—
s->°.
10(s + 5)
(s + l) ( s + 2)
3
= lim ------ 7777— —=r—
s->o 1
10(s + 5)
(s + l) ( s + 2)
3
= ...... ........................- 77:...... -........... 7
10(0 + 5)
(0 + 1)(0 + 2)
277
Copyrighted Material © 2018
13.4 Routh Hurwitz Criteria & System Stability - Solutions
Consult NCEES® FE Reference Handbook - Page 128 for reference
13.4a) CORRECT ANSWER - s2 + 3s + 2 + ks = 0
According to classical negative feedback control system diagram given in NCEES® FE Reference Handbook:
R(s)k
Y(s) =
1+
Y(s) =
s
0 + l) ( s + 2)
ks
(s + l) ( s + 2)
R(s)ks
s 2 + 3s + 2 + ks
The closed loop characteristic equation is s 2 + 3s + 2 + ks = 0
13.4b) CORRECT ANSWER - -8/3
The closed loop characteristic equation is 2s5 + 3s4 + 7s2 + s + 10 = 0
Routh Array can be developed using formulas given in NCEES® FE Reference Handbook as shown below:
s5
CM
s 4
3
s3
2
7
bi =
3 x 2 -2
X
7
8
..1................3...................3..
The given system is unstable because bt is negative.
13.4c) CORRECT ANSWER - B
The closed loop characteristic equation of given system is 2s4 + 3s3 + s 2 + s + l = 0
Routh Array can be developed using NCEES® FE Reference Handbook as shown below:
s4
s3
s2
s
2
3
b±~
3 -2
1
3
~ 3
1
1
1
0
3-2x0
b2 = ----- ------ = 1
2
3
-
i —3 x l
ci = 6 t
= 8
The given system is unstable because ct is negative.
278
Copyrighted Material © 2018
-
13.4d) CORRECT ANSWER - 0 < k < 5
The closed loop characteristic equation of given system is 3s3 + 5s2 + (k + 10)s + Sk = 0
Routh Array can be developed using NCEES® FE Reference Handbook as shown below:
s3
s2
s
s°
k+10
5k
3
5
5 (k + 10) - 3 x 5 k
5
h=
Sk
For the system to remain stable all entries in first column should have same sign (positive).
S(k + 10) - 3 x 5 k
-------------> 0 - »
5
—
k< 5
Sk>0^k>0
By combining both conditions we obtain stability range as 0 < k < 5
13.4e) CORRECT ANSWER - k > - 1
The closed loop characteristic equation of given system is shown below:
(k + l) ( s + 1)
= _ 0 ) 0 + 2)0 + 3)
_
(k + l) ( s + 1)
s3 + 5s2 + s(k + 7) + k + 1
1 , ( (fc + 1)(5 + 1) ^
\s(s + 2 )(s + 3)/
The closed loop characteristic equation is s 3 + 5s2 + s(k + 7) + fc + l = 0
Routh Array can be developed using NCEES® FE Reference Handbook as shown below:
s3
s2
s
s°
5(k + 7) — (k + 1)
34 + 4 k
5
5
1
5
k+7
k+1
bl
k+1
— ----------------------------------------- — -------------------
1
For the system to remain stable all entries in first column should have same sign (positive).
34 + 4k
17
bt > 0 -» ---- ----- > 0 -> k > — —
/c + l > 0 - > / e > —1
By combining both conditions we obtain stability range as k > —1
279
Copyrighted Material © 2018
13.5 Root Locus - Solutions
Consult NCEES® FE Reference Handbook - Page 129 for reference
13.5a) CORRECT ANSWER - A
Refer to page number 128 of NCEES® FE Reference Handbook for root locus equations.
For the given system it can be observed that open-loop poles exist at s =0, s = -2 and there are no zeros. Hence
n = 2 and m = 0.
Locus originates at open-loop poles and terminates at zeros. However, since there is no zero (m < n), (n - m = 2)
branches will terminate at infinity at asymptote angles a as shown below:
[ ( 2k + 1)180°]
a = ±±-------- 1------ n = 2,
n —m
m = 0,
k = 0,1
a = 90°,270
Asymptote centroids can be calculated using following formula:
£F=i Re(pt) - £ £ i Re(mi)
oA = ---------------------------------- ,n = 2,
n —m
m = 0,
o& = —1
Locus does not cross imaginary axis.
13.5b) CORRECT ANSWER - C
Refer to page number 128 of NCEES® FE Reference Handbook for root locus equations.
For the given system it can be observed that open-loop poles exist at s =0, s = -2 and s = -5 also there are no
zeros. Hence n = 3 and m = 0.
Locus originates at open-loop poles and terminates at zeros. However, since there is no zero (m < n), (n - m = 3)
branches will terminate at infinity at asymptote angles a as shown below:
[ ( 2k + 1)180°]
a = - -------- ------- n = 3,
n —m
m = 0,
k = 0,1,2
a = 60°, 180°, 300°
Asymptote centroids can be calculated using following formula:
Z?=1Ret e ) - S “ iR e (m i)
„
<jA = --------------— --------------- ,n = 3,
n —m
„
m - 0,
o',, "
7
::
5
Locus crosses imaginary axis.
13.5c) CORRECT ANSWER - D
Refer to page number 128 of NCEES® FE Reference Handbook for root locus equations.
For the given system it can be observed that open-loop poles exist at s = 1 and s = 4 also there is a zero at s = -5.
Hence n = 2 and m = 1.
Locus originates at open-loop poles and terminates at zeros. However, since m < n (n - m = 1) branches will
terminate at infinity at asymptote angles a as shown below:
280
Copyrighted Material © 2018
Asymptote centroids can be calculated using following formula:
Z?=1« e ( p ,) - Z E 1/te(mi) ,71 = 2,
(Ja = ---------------------------------n —m
771 = 1,
<JA = 10
Locus crosses imaginary axis.
13.5d) CORRECT ANSWER - B
Refer to page number 128 of NCEES® FE Reference Handbook for root locus equations.
For the given system it can be observed that open-loop poles exist at s = -1 and s = -2 also there is a zero at s = 0
Hence n = 2 and m = 1.
Locus originates at open-loop poles and terminates at zeros. However, since m < n, (n - m = 1) branches will
terminate at infinity at asymptote angles a as shown below:
m + i) i8 o ° ]
„
a = --------------------, n = 2,
n —m
,
n
k = 0,
m = l,
a = 180u
Asymptote centroids can be calculated using following formula:
(J —---------------------------------- ,71 = 2,
n —m
771=1,
(Ja = —3
Locus does not cross imaginary axis.
13.5e) CORRECT ANSWER - A
Refer to page number 128 of NCEES® FE Reference Handbook for root locus equations.
For the given system it can be observed that two open-loop poles exist at s = 0 and there is a zero at s = 2/3.
Hence n = 2 and m = 1.
Locus originates at open-loop poles and terminates at zeros. However, since there is no zero (m < n), (n - m = 2)
branches will terminate at infinity at asymptote angles a as shown below:
m + i) i8 o ° ]
„
a —--------------------f 7i = 2,
n —m
771 = 1,
,
n
k= 0
a = 180u
Asymptote centroids can be calculated using following formula:
< * = ------------- ^
------------- ’ n = 2’
m = 1'
<* = - 3
Locus does not cross imaginary axis.
281
Copyrighted Material © 2018
13.6 State variables - Solutions
Consult NCEES® FE Reference Handbook - Pages 129 -1 30 for reference
13.6a) CORRECT ANSWER - B
y " + 3y' + 2y = 2u(t)
The given differential equation is of second order therefore it has two state variables.
Let us assume x 1 = y,x 2 = y' = y
Representing the variables as first order derivatives
x± = x2 and x 2 = —2x1 —3x2 4- 2u(t)
The state equation is shown below in matrix form
x i] _
0
-—2
1
—3
13.6b) CORRECT ANSWER - D
2y " + 8y' + 10y = 6u(t) -» y " + 4y ' + 5y = 3u(t)
The given differential equation is of second order therefore it has two state variables.
Let us assume x± = y,x 2 = y' = y
Representing the variables as first order derivatives
x\ = x 2 and x 2 = —5 ^ — 4x2 + 3u(t)
The state equation is shown below in matrix form
x±\ _
0
x 2\ ~ 1 -5
1
-4
13.6c) CORRECT ANSWER - C
3 y " ' + 6y " + 12y' + 3y = 9u (t) -> ym + 2y" + 4 y ' + y = 3 u(t)
The given differential equation is of third order therefore it has three state variables.
Let us assume x± = y,x 2 = y' = y,x 3 = y " = y
Representing the variables as first order derivatives
xt = x2 , x 2 = x3 and x 3 = —xt —4x2 —2x3 + 3u(t)
The state equation is shown below in matrix form.
0
0
o ■ -X-LX2 +
1
.- 1
-4
-2 .
■0
^2
*3-
=
1
.*3.
O'
.3.
0
282
Copyrighted Material © 2018
13.6d) CORRECT ANSWER - A
Standard state variable model is given as follows.
x ( t) = A x ( t ) + B u (t)
y ( t ) = C x(t ) + D u (t)
Taking Laplace Transform of above equations results in following.
Y(s) = CX(s ) + D U (s)
s X ( s ) = A X (s) + BU(s)
slX (s) = A X (s ) + BU(s) -> (si - A )X (s ) = BU(s)
X(s) = (si - A ) ~ 1BU(s)
Y (s) = [C(sl - A)~XB + D] U( s)
[C(sl - A )^ B + D] is the transfer function.
13.6e) CORRECT ANSWER - D
As shown in above given solution, transfer function is related to state equation as follows.
T(s) = C(sl - A)~1B + D
c =[1 0], =[_°4 X} B = Q o = m
rl
r(s) = [i o](.g
n * ) = [i
T(s) =
T(s) =
o](R
01
r0
? ]-[_ °4
1 i . _ i [01
+
- 15] ) - 1 [“ ]
as
*
(s)(s + 5) + 4
2
s2 + 5s + 4
13.6f) CORRECT ANSWER - A
y[n + 2] + 3y[n + 1] + 3y[n] = 2u[n]
This is a second order difference equation therefore it has two state variables.
Let us assume x j n ] = y[n],x 2[n] = y[n + 1]
x±[n + 1] = x2[n] and x2[n + 1] = S x ^ n ] — 3x2[n] + 2u[n]
The matrix form of state equation is shown below
'xx[n + 1]'
0
-3
[ * iM
W in ]
+
| u[n]
283
Copyrighted Material © 2018
Chapter # 1 4 - Communications
14.1 Am plitude Modulation - Solutions
Consult NCEES® FE Reference Handbook - Page 209 for reference
14.1a) CORRECT ANSWER - A
Power transmitted in a signal can be expressed as follows:
(
m2\
pt = pc ( i + T ]
According to problem statement:
Pc = 1000 W m — 80% = 0.80
(
m2\
(
(0.80)2\
Pt = Pc (1 + — j = (1000 W) ( 1 +
.- ..........-.......... j =
1320
W
Note: Power transmitted in a signal formula is not given in NCEES® FE Reference Handbook.
14.1b) CORRECT ANSWER - D
Power transmitted in a signal can be expressed as follows:
According to problem statement:
Pt = 5 kW
Pc = 4.75 kW
14.1c) CORRECT ANSWER - D
As per NCEES® FE Reference Handbook, Costas loop can be used for detecting Double-Side Modulation (DSB).
14.Id) CORRECT ANSWER - C
Modulation index can be calculated using following formula:
signal amplitude
YYl —-----------------------carrier amplitude
According to problem statement:
m (t) = 5sin27r(1000t)
c(t) = 50sin27r(4000t)
5
m = — = 0.1 = 10%
284
Copyrighted Material © 2018
14.le) CORRECT ANSWER - C
Efficiency of an amplitude modulated wave can be calculated using following formula
a2 < m |(t) >
^
1 + a2 < ra^(t) >
According to problem statement:
a = 0.8
< m 2(t) > = 0.6
a2 < rrinit) >
^
1 + a2 < m j( t) >
0.82 x 0.6
77 “ 1 + 0.82 x 0.6
rj = 0.27 = 27%
14.If) CORRECT ANSWER - A
Amplitude modulation results in variation of carrier wave's amplitude by signal wave.
Helpful tip - Learn how to identify different forms of modulation graphically.
285
Copyrighted Material © 2018
14.2 Angle Modulation - Solutions
Consult NCEES® FE Reference Handbook ~ Page 209 for reference
14.2a) CORRECT ANSWER - D
98% power bandwidth can be calculated using following formula:
B = 2 (D + 1) W
According to problem statement, D = 1.25 > 1
W = 10kHz
B = 2 (D + 1)W
B = 2 (1.25 + 1) x 10000Hz
B = 45000 Hz
Helpful tip - Review Angle Modulation formulas given in NCEES® FE Reference Handbook.
14.2b) CORRECT ANSWER - B
As per NCEES® FE Reference Handbook, phase-lock loop can demodulate angle modulated signals.
14.2c) CORRECT ANSWER - B
Frequency modulation results in variation of carrier wave's frequency by signal wave.
14.2d) CORRECT ANSWER - C
Phase modulation results in variation of carrier wave's phase by signal wave. Note that phase and frequency
modulation are very similar. They are collectively called as angle modulation.
14.2e) CORRECT ANSWER - D
98% power bandwidth can be calculated using following formula:
B = 2W
According to problem statement, D = 0.1 < 1 W = 10 kHz
B = 2W
B = 2 X 10 kHz
B = 20 kHz
286
Copyrighted Material © 2018
14.3 Pulse Code Modulation (PCM) & Pulse Amplitude Modulation (PAM) - Solutions
Consult NCEES® FE Reference Handbook - Page 210 for reference
14.3a) CORRECT ANSWER - 128
Quantization levels can be calculated using following formula.
q = 2n where 'q' is the quantization levels and 'n' is the number of bits.
According to the given problem n = 7
q =
27 = 128 levels
Helpful tip - Review PCM and PAM formulas given in NCEES® FE Reference Handbook.
14.3b) CORRECT ANSWER - D
Minimum bandwidth for PCM message transmission can be calculated using following formula.
B = 2 W lo g2 q
q = 2n = 28 = 256
B = (2) (100 Hz) log2 256
B = 1600 Hz
14.3c) CORRECT ANSWER - B
According to Nyquist sampling theorem, fs > 2W
Therefore, f s>min = 2 W
fs,min
2 X 15 kHz
fs,min
30 kHz
14.3d) CORRECT ANSWER - C
Time spacing between adjacent samples can be calculated using following formula.
1
Ts ~ 30 kHz
Ts = 33.3 |xs
14.3e) CORRECT ANSWER - D
Quantization, encoding and sampling are part of Pulse Code Modulation (PCM) process.
287
Copyrighted Material © 2018
14.4 Fourier Transform - Solutions
Consult NCEES® FE Reference Handbook - Pages 31 - 33 for reference
14.4a) CORRECT ANSWER - 40 sinc(8 /)
The given function is a rectangular pulse centered at t = 0, amplitude of 5 and duration of 8 i.e. 511
Fourier Transform can be calculated using pairs given in NCEES® FE Reference Handbook as shown below.
/ 1\ Fourier Transform
5n ( - ) <
------------------ » 5(8sinc(8/)) = 40sinc(8/)
\8 /
14.4b) CORRECT ANSWER -12 sinc(4/)e“w
The given function is a rectangular pulse centered at t = 2, amplitude of 3 and duration of 4 i.e. 311
Fourier Transform time shifting property needs to be used as shown below.
/t — 2\ Fourier Transform
3 n f—— J <
----------------- > 3(4sinc(4/))e
= 12sinc(4f ) e
4jnf
14.4c) CORRECT ANSWER - C
Fourier Transform modulation property needs to be used as shown below.
t \ Fourier Transform
Transform
©
<----- > 6sinc(6/ )
cos(27r(300)t) II
/ t \ Fourier Transform
N
,
<------------------ » 3 sinc(6 ( f —300)) +3 sinc(6 ( f + 300))
14.4d) CORRECT ANSWER - D
Fourier Transform can be calculated using pairs given in NCEES® FE Reference Handbook as shown below.
1
Fourier Transform
e~5tu(t)< ----------------
5 + j2 n f
Fourier Transform 1
/
cos(2*(20)t)e- u(t)«---------------- +
1
1
\
, 20) ' r>+ /2,r(/
20))
14.4e) CORRECT ANSWER - 8 sin c(2 /)e _(47rj/)
Fourier Transform can be calculated using pairs given in NCEES® FE Reference Handbook as shown below.
/t —2\ Fourier Trans form
4 n f - r - J <-— >4(2 sinc(2
Helpful tip - Learn how to use Fourier Transform table given in NCEES® FE Reference Handbook.
288
Copyrighted Material © 2018
14.5 Multiplexing - Solutions
14.5a) CORRECT ANSWER - A
Time Division Multiplexing (TDM) is an example of digital multiplexing.
14.5b) CORRECT ANSWER - C
N channels require at least N -l guard bands. Therefore 3 channels will require at least 2 guard bands.
BWmin = BWch ± + Guard Band + BWch 2 + Guard Band + BWch 3
BWmin = 50 kHz + 5 kHz + 100 kHz + 5 kHz + 50 kHz
BWmin= 210 kHz
14.5c) CORRECT ANSWER - C
Each frame carries 1 byte per channel.
There are 5 channels.
Therefore, frame size will be:
5 x 1 byte = 5 bytes
14.5d) CORRECT ANSWER - 400 bps
Each channel is sending 10 bytes/second and each frame carries 1 byte/channel.
Frame rate = 10 frames/second.
Frame size = 5 bytes.
10 frame/second x 5 bytes/frame = 50 bytes /second = 400 bits / second = 400 bps.
14.5e) CORRECT ANSWER - B
Each frame carries 1 bit.
There are 50000 frames per second and 3 bits per frame. Therefore, total number of bits/second is 150000.
14.5f) CORRECT ANSWER - 20 |is
There are 50000 frames per second.
Frame duration = 1/50000
Frame duration = 20 (is
289
Copyrighted Material © 2018
Chapter # 15 - Computer Networks
15.1 Routing & Switching - Solutions
Consult NCEES® FE Reference Handbook - Pages 219 - 222 for reference
15.1a) CORRECT ANSWER - Routing
Routing is the process of finding efficient paths between nodes based on their address.
15.1b) CORRECT ANSWER - D
Router maintains routing table as well as forwarding table.
15.1c) CORRECT ANSWER - C
Network Layer performs node management functions such as routing, addressing and traffic control.
Helpful tip - Review OSI Model and TCP/IP Model layers given in NCEES® FE Reference Handbook.
15.Id) CORRECT ANSWER - A
Router is responsible for connecting two or more networks for transferring data packets.
15.1e) CORRECT ANSWER - C
Network switches create a network and allow devices within a network to communicate with each other.
15.If) CORRECT ANSWER - B
Network switch typically operates in Data Link Layer. They can also perform routing functionality in Network
Layer, if upgraded.
290
Copyrighted Material © 2018
15.2 Network Topologies / Frameworks / Models - Solutions
Consult NCEES® FE Reference Handbook - Pages 219 - 222 for reference
15.2a) CORRECT ANSWER - D
Bus, Star and Ring are examples of different network topologies.
Helpful tip - Review network topologies given in NCEES® FE Reference Handbook.
15.2b) CORRECT ANSWER - D
Wireless networking has potential security issues associated with data transfer.
15.2c) CORRECT ANSWER - Application layer
Application layer will run http and email.
Helpful tip - Review OSI Model and TCP/IP Model layers given in NCEES® FE Reference Handbook.
15.2d) CORRECT ANSWER - Presentation layer
Presentation layer is responsible for language translations.
15.2e) CORRECT ANSWER - A
Session layer is responsible for managing user interactions.
15.2f) CORRECT ANSWER - C
A single break in connection can disrupt entire network in Ring Topology implementation.
291
Copyrighted Material © 2018
15.3 Local Area Networks - Solutions
Consult NCEES® FE Reference Handbook - Pages 219 - 222 for reference
15.3a) CORRECT ANSWER - D
Local Area Networks (LAN) can be implemented using any of the suggested implementations.
15.3b) CORRECT ANSWER - D
LAN can be operated using Ethernet, Wireless or Asynchronous Transfer Mode (ATM) technologies.
15.3c) CORRECT ANSWER - B
Ethernet is an example of Bus topology.
15.3d) CORRECT ANSWER - A
Asynchronous Transfer Mode (ATM) LAN is an example of star topology.
15.3e) CORRECT ANSWER - B
Local Area Network (LAN) can cover small areas (office, hospital etc.).
Wide Area Network (WAN) can cover large geographical regions (states, countries etc.).
Metropolitan Area Network (MAN) can cover municipalities.
Personal Area Network (PAN) can cover distance within range of a person (typically 10 m).
292
Copyrighted Material © 2018
Chapter # 16 - Digital Systems
16.1 Number Systems - Solutions
Consult NCEES® FE Reference Handbook - Page 218 for reference
16.1a) CORRECT ANSWER - 1000101011012
Number system conversion can be efficiently done using calculators.
8ADHex= 1 0 0 0101 011012
16.1b) CORRECT ANSWER - 11110000112
Number system conversion can be efficiently done using calculators.
9 63 10= I I I I O O O O I I 2
16.1c) CORRECT ANSWER - D
l's complement addition can be performed as shown below.
(+ 2 ) OOIO2
(+ 3 ) 0011?
(+ 5 ) O IO I 2
Note that given numbers are unsigned (MSB of positive numbers is 0)
16. Id) CORRECT ANSWER - C
l's complement addition can be performed as shown below.
The given binary numbers are negative.
I I O I 2 = - (0010)2 = - (2)io
I O I I 2 = - (0100)2 = - (4)io
1
IIO I2
(-2)10
+1011?
(-4 )m
1000
carry ___ 1
IOOI2
(-6)10
Note that carry is added in l's complement.
293
Copyrighted Material © 2018
16.Ie) CORRECT ANSWER - B
2's complement addition can be performed as shown below.
IO II 2 (-5)
+OIOI7
1 0000
(+5)
= OOOO2
Ignore carry
Note that carry is ignored in 2's complement.
16.If) CORRECT ANSWER - A
2's complement addition can be performed as shown below.
IIIO 2 (-2)
+0101?
(+5)
1 0011
(+3)
= OOII2
Ignore carry
Note that carry is ignored in 2's complement.
294
Copyrighted Material © 2018
16.2 Boolean Logic - Solutions
Consult NCEES® FE Reference Handbook - Page 218 for reference
16.2a) CORRECT ANSWER - D
Applying De Morgan's theorem to given logical function results in:
A + BCD + (E + F) = (A)(BCD) ( I + F) = ( I ) ( I + C + D )(F )(F )
Helpful tip - Review De Morgan's theorem given in NCEES® FE Reference Handbook.
16.2b) CORRECT ANSWER - A
Applying De Morgan's theorem to given logical function results in:
CABC){D + EF) = (ABC) + (D + E F ) = A + B + C + D(E + F)
16.2c) CORRECT ANSWER - B
Applying De Morgan's theorem to given logical function results in:
AB + A(BC) + B(A + C) = AB + A(B + C) + BAC
AB + A(BC) + B(A + C) = AB + AB + AC + A C B
AB + A(BC) + B(A + C) = A B + A C + A C B
AB + A(BC) + B(A + C) = AB + C(A + AB)
According to Redundancy Law: A + AB = A + B
AB + A (BC) + B (.A + C) = AB + C (A + B)
Helpful tip - Review laws and identities of Boolean Logic.
16.2d) CORRECT ANSWER - B
AB + B (A + C) + C (A + B) = AB + AB + BC + AC + BC
According to Idempotent Law: AB + AB = AB
BC + BC = BC
AB + B (A + C) + C (A + B) = AB + BC + AC
295
Copyrighted Material © 2018
16.2e) CORRECT ANSWER - C
{A + B) (A + B +
C) = A + AB + AC +AB + B +
BC
According to Idempotent Law: AB + AB = AB
(A + B) {A+ B +
C) = A + AB + AC + B + BC
(A + B) (A + B +
C) = A (1 + B) + AC + B (1 + C)
According to Identity Law: (1 + B) = 1.
{A + B) (A + B + C ) = A + B + AC = A ( I + C) + B = A + B
296
Copyrighted Material © 2018
16.3 Logic Gates - Solutions
Consult NCEES® FE Reference Handbook - Page 218 for reference
16.3a) CORRECT ANSWER - B
(.AB)XOR (CD) = AB (CD) + (AB) CD
(AB)XOR (CD) = AB (C + D) + CD (A + B)
(AB)XOR (CD) = ABC + ABD + ACD + BCD
Helpful tip - Review logical operations and gates given in NCEES® FE Reference Handbook.
16.3b) CORRECT ANSWER - C
(A B )(Ib)
= (AB) + (AB)
(AB)(AB) = A + B + A + B = (A + A ) + (B + 1 ) = 1 + 1 = 1
(AB)(AB) = 1
16.3c) CORRECT ANSWER - A
AB C + C = (~A + B + C) + C
According to Idempotent Law: C + C = C
ABC + C = A + B + C
16.3d) CORRECT ANSWER - C
(A )(1 )(C ) = A B C
16.3e) CORRECT ANSWER - B
( AB) ( A + B) = AAB + ABB
( AB) ( A + B ) = 0 + 0
( AB) ( A + B) = 1
297
Copyrighted Material © 2018
16.4 Karnaugh Maps - Solutions
Consult NCEES® FE Reference Handbook - Page 219 for reference
16.4a) CORRECT ANSWER - B
C
0
1
Helpful tip - Review rules related to function minimization using K-maps.
16.4b) CORRECT ANSWER - D
C
0
1
16.4c) CORRECT ANSWER - C
CD
00
01
11
10
298
Copyrighted Material © 2018
16.4d) CORRECT ANSWER - A
CD
00
01
11
10
16.4e) CORRECT ANSWER - C
CD
16.4f) CORRECT ANSWER - C
C
0
1
299
Copyrighted Material © 2018
16.5 Flip Flops and Counters - Solutions
Consult NCEES® FE Reference Handbook - Page 219 for reference
16.5a) CORRECT ANSWER - C
Input 101 to Flip Flop # 1 appears as 010 at its output^.
Q of Flip Flop # 1 acts as input of Flip Flop # 2 which appears as 010 on the output Q of Flip Flop # 2.
16.5b) CORRECT ANSWER - B
Comparing the input output relation of given circuit with RS Flip Flop truth table reveals that it is an
implementation of RS Flip Flop. Refer to the truth table given in NCEES® FE Reference Handbook.
16.5c) CORRECT ANSWER - C
Comparing the input output relation of given circuit with D Flip Flop truth table reveals that it is an
implementation of D Flip Flop. Refer to the truth table given in NCEES® FE Reference Handbook.
16.5d) CORRECT ANSWER - A
Comparing the input output relation of given circuit with JK Flip Flop truth table reveals that it is an
implementation of JK Flip Flop. Refer to the truth table given in NCEES® FE Reference Handbook.
16.5e) CORRECT ANSWER - B
Input 001 to Flip Flop # 1 appears as 001 at its output Q.
Flip Flop # 1 output Q is then provided to Flip Flop # 2 set input (S) and Flip Flop # 1 output Q' is provided to Flip
Flop # 2 reset input (R).
Following table summarizes the input output relation
Input / Output
S (Qfrom Flip Flop # 1)
R (Q' from Flip Flop # 1)
Q
States
0
1
0
1
1
0
0
0
1
Therefore, RS output is 001.
300
Copyrighted Material © 2018
16.5f) CORRECT ANSWER - B
The given logic circuit is an implementation of Johnson counter. It is important to note that without a positive
clock cycle (which I've indicated with 1), Flip Flops will not turn on. In this problem, Flip Flop # 1 has clock input
of 001, Flip Flop # 2 has 010 and Flip Flop # 3 has 100. It means that at the proverbial drop of hat, Flip Flop # 1
will be on for first clock cycle and off for next two cycles (remember 001), Flip Flop # 2 will be on during second
clock cycle and off during first and last (remember 010) whereas Flip Flop # 3 will be on for last clock cycle only
(100). Initially, Q3 was D-input of Flip Flop # 1. Q3 was 0 initially, hence Q3 was 1. So, when the first clock cycle
arrived at Flip Flop # 1, the 1 (Q3) sitting at D will result Q± from 0 to 1. These transitions are summarized below,
Clock cycle
0
1
2
3
Qi
0
1
1
1
Q2
0
0
1
1
Q3
0
0
0
1
16.5g) CORRECT ANSWER - D
The given logic circuit is an implementation of 4-bit Asynchronous down counter. Note that flip-flops are
'positive edge triggered1meaning that outputs change on a rising clock edge. The rising edge of the Q output of
each flip-flop triggers the clock input of the next flip-flop. Initial output states are 1111. The rising edge of first
clock pulse applied will cause the output Qo to go to logic 0, and the next CK pulse will make Qo output return to
logic 1. At the same time Q i will go from 1 to 0. As Qo and clock input of first flip flop go high, it makes Q i low
because Q i = 0 was sitting at second flip flop's input waiting for positive edge trigger. The next (third) CK pulse
will cause Qo to go low again (because Qo = 0 was sitting at first flip flop's input waiting for the positive edge
trigger), now both Qo and Q i will be low. This process can be extended to illustrate that it's a down counter.
Clock cycle
0
1
2
3
Qo
1
0
1
0
Qi
1
1
0
0
Q2
1
1
1
1
Qs
1
1
1
1
16.5H) CORRECT ANSWER - A
The given logic circuit is an implementation of a 2-bit synchronous up counter. Initial states are Qo Q i = 0 0 .1st
clock cycle - Flip Flop # 1 will toggle and Qo will change from 0 to 1. Flip Flop # 2 will not change because when
the clock is applied at Flip Flop # 1, Qo = J2 = K2 = 0, implies Q i = 0. 2ndclock cycle - Flip Flop # 1 toggles again and
Qo will change from 1 to 0 because J = K = 1 results in toggle. Now, Q i will also change state to 1 because at this
time J2=K2 =1 which causes Flip Flop #2 to toggle. 3rdclock cycle - Flip Flop # 1 toggles again and Qo will change
from 0 to 1 but there will be no change of state for Flip Flop # 2. At the end of 3rd clock cycle, Qo Q i = 11.4th
clock cycle - Flip Flop # 1 toggles again and Qo will change from 1 to 0 because J = K = 1 results in toggle. Now, Q i
will also change state to 0 because at this time J2 = K2 = 1 which causes Flip Flop # 2 to toggle.
Clock cycle
Initial state
1
2
3
4
Qo
0
1
0
1
0
Qi
0
0
1
1
0
301
Copyrighted Material © 2018
16.6 State Machine Design - Solutions
Consult NCEES® FE Reference Handbook - Page 22 for reference
16.6a) CORRECT ANSWER - C
According to the state diagram, if present state is C and w =0, system will transition to state A and output will
be 1. Similarly, if present state is C and w = 1 system will transition to state B and output will be 1.
Helpful tip - Learn how to navigate from state table to state diagram and vice versa.
16.6b) CORRECT ANSWER - D
The table given below summarizes transitions from initial state A considering inputs 111.
Current State and Input
A, 1
B, 1
B, 1
Next State
B
B
B
Output
0
1
1
16.6c) CORRECT ANSWER - C
The table given below summarizes the transitions from initial state A considering inputs 000.
Current State and Input
A, 0
C,0
A, 0
Next State
C
A
C
Output
0
1
0
16.6d) CORRECT ANSWER - A
According to the state diagram, if present state is C and ab = 00 system will stay in state C and output will be 0.
Similarly, if present state is C and input ab = 01 system will transition to state D and output will be 0.
16.6e) CORRECT ANSWER - D
The output expression for 'z' as a function of y2yi can be found using k-map as shown below.
72
302
Copyrighted Material © 2018
Chapter #17 - Computer Systems
17.1 Architecture & Interfacing - Solutions
Consult NCEES® FE Reference Handbook - Pages 223 - 224 for reference
17.1a) CORRECT ANSWER - D
Addressing modes are unique ways in which address of an operand is given in an instruction.
Register Addressing Mode - Used when operands are in registers.
Immediate Addressing Mode - Used when operands are stored part of instructions i.e. constants.
Direct Addressing Mode - Used when operands are provided in memory addressing modes.
Other examples include indirect addressing mode, memory deferred addressing mode, scaled addressing mode.
17.1b) CORRECT ANSWER - Compiler
Compiler - A language processor that converts high-level language (source code - understandable by humans)
into machine language (object code - understandable by computer) for execution.
Interpreter - A language processor that converts high-level language (source code - understandable by humans)
into an intermediate (low-level language) for execution.
The difference between an interpreter and compiler is that as opposed to an interpreter a compiler translates
the program into machine language before execution.
Assem bler-A language processor that converts assembly language (source code) into machine language (object
code) for execution.
17.1c) CORRECT ANSWER - ALU
Control U n it-T he Control Unit is part of computer's Central Processing Unit (CPU). It directs the operation of
memory unit, arithmetic logic unit and I/O units by interpreting instructions.
I/O U nit-T he I/O Unit comprises of devices used to enter and extract details to and from a computer.
ALU - The ALU (Arithmetic Logic Unit) is responsible for making mathematical and logical operations.
17.1d) CORRECT ANSWER - A
Batch data processing - It involves processing high volume data in groups/batches. It is an efficient way of
processing quarterly bank statements, payroll, school reports etc.
Real time data processing - It involves continuous data processing. It is an efficient way of processing stock
quotations, customer service etc.
303
Copyrighted Material © 2018
17.1e) CORRECT ANSWER - C
Batch data processing - It involves processing high volume data in groups/batches. It is an efficient way of
processing quarterly bank statements, payroll, school reports etc.
Real time data processing - It involves continuous data processing. It is an efficient way of processing stock
quotations, customer service, weather monitoring etc.
17.If) CORRECT ANSWER - B
RAM - Random Access Memory (RAM) is a type of computer storage memory. Typically it is volatile and BIOS is
loaded into RAM only after computer boots.
ROM - Read Only Memory (ROM) is a type of computer storage memory. It is non-volatile and BIOS is generally
stored in ROM which is used to boot the computer.
USB Mass Storage Device - It is typically used as secondary memory storage device for portable data storage
options.
17.1g) CORRECT ANSWER - B
Encryption - A data conversion process which prevents unauthorized personnel from accessing it.
Encoding - A data transformation process involving changing data format for another system.
Hashing - Transformation of string into shorter fixed length value representing original string.
Decoding - It is the opposite of encoding and converts encoded data to its original format.
17.1H) CORRECT ANSWER - A
Encryption - A data conversion process which prevents unauthorized personnel from accessing it.
Encoding - A data transformation process involving changing data format for another system.
Hashing-Transformation of string into shorter fixed length value representing original string.
Decoding - It is the opposite of encoding and converts encoded data to its original format.
17. li) CORRECT ANSWER - B
Instruction pipelining increases instruction throughput. It neither decreases the instruction execution time nor
does it allow new types of instructions.
17. lj) CORRECT ANSWER - B
Control U n it-T he Control Unit is part of computer's Central Processing Unit (CPU). It directs the operation of
memory unit, arithmetic logic unit and I/O units by interpreting instructions.
304
Copyrighted Material © 2018
17.2 Microprocessors - Solutions
Consult NCEES® FE Reference Handbook - Pages 223 - 224 for reference
17.2a) CORRECT ANSWER - B
Program counter register - contains address of next instruction to be executed.
Stack pointer register - contains address of last executed instruction.
Instruction pointer register - contains address of the current instruction being executed.
Accumulator register - contains results of arithmetic and logic operations.
17.2b) CORRECT ANSWER - A
Program counter register - contains address of next instruction to be executed.
Stack pointer register - contains address of last executed instruction.
Instruction pointer register - contains address of the current instruction being executed.
Accumulator register - contains results of arithmetic and logic operations.
17.2c) CORRECT ANSWER - A
M icroprocessor-A single Integrated Circuit (1C) accepting & executing code instructions for processing data and
controlling associated circuitry in a computer system
Microcomputer - An interconnected group of Integrated Circuits (ICs), l/Os and memory systems used for data
processing and other applications.
Microcontroller - An integrated system of a single 1C, l/Os, memory system and associated circuitry accepting &
executing coded instructions in computer system.
17.2d) CORRECT ANSWER - B
Microprocessor - A single Integrated Circuit (1C) accepting & executing code instructions for processing data and
controlling associated circuitry in a computer system
Microcomputer - An interconnected group of Integrated Circuits (ICs), l/Os and memory systems used for data
processing and other applications.
Microcontroller - An integrated system of a single 1C, l/Os, memory system and associated circuitry accepting &
executing coded instructions in computer system.
305
Copyrighted Material © 2018
17.2e) CORRECT ANSWER - C
Microprocessor - A single Integrated Circuit (1C) accepting & executing code instructions for processing data and
controlling associated circuitry in a computer system
Microcomputer - An interconnected group of Integrated Circuits (ICs), l/Os and memory systems used for data
processing and other applications.
Microcontroller - An integrated system of a single 1C, l/Os, memory system and associated circuitry accepting &
executing coded instructions in computer system.
17.2f) CORRECT ANSWER - Data bus
Address Bus - carries physical address for reading/writing.
Data Bus - carries data between different units of a computer system.
Control Bus - connects CPU with other components.
17.2g) CORRECT ANSWER - Address bus
Address Bus - carries physical address for reading/writing.
Data Bus - carries data between different units of a computer system.
Control Bus - connects CPU with other components.
306
Copyrighted Material © 2018
17.3 Memory Technology & Systems - Solutions
Consult NCEES® FE Reference Handbook - Pages 223 - 224 for reference
17.3a) CORRECT ANSWER - D
Cassette tapes, CDs, DVDs and Hard Disks are examples of Sequential Access storage devices.
Flash memory is an example of Random Access storage device.
Helpful tip - Review memory technology and systems given in NCEES® FE Reference Handbook.
17.3b) CORRECT ANSWER - 8 X 230 bits
Giga in binary = 230and Byte = 8. Therefore, 1 Giga Byte = 8 x 230 bits
Helpful tip - Review binary prefixes and unit conversions.
17.3c) CORRECT ANSWER - C
Cache memory has low capacity and very high speed. It stores frequently used data for ready access.
17.3d) CORRECT ANSWER - C
DVD is a secondary memory storage device. It is non-volatile and cannot be directly accessed by computer.
Primary memory systems are volatile and can be accessed directly.
Helpful tip - Understand the difference between primary memory storage and secondary memory storage.
17.3e) CORRECT ANSWER - A
PROM - Programmable Read-Only Memory can be programmed only once.
EPROM - Erasable Programmable Read-Only Memory can be erased using ultra violet light.
EEPROM - Electrically Erasable Programmable Read-Only Memory can be erased electrically.
Helpful tip - Learn the difference between various types of ROMs.
17.3f) CORRECT ANSWER - C
EPROM - Erasable Programmable Read-Only Memory can be erased using ultra violet light.
EEPROM - Electrically Erasable Programmable Read-Only Memory can be erased electrically.
17.3g) CORRECT ANSWER - B
Bit -1 digit, Byte - 8 digits, Nibble - 4 digits and Word -16/32/64 bits depending on system architecture.
307
Copyrighted Material © 2018
Chapter #18 - Software Development
18.1Aigortthms - Solutions
Consult NCEES® FE Reference Handbook - Pages 224 - 227 for reference
18.1a) CORRECT ANSWER - 7
Values at the end of loop # 1, y = 1, x = 1, z = 7
Values at the end of loop # 2, y = 2, x = 3, z = 4
Values at the end of loop # 3, y = 3, x = 5, z = 1
Values at the end of loop # 4, y = 4, x = 7, z = -2
z < 0 at the end of loop # 4 therefore program will not run loop # 5 and values at the end of loop # 4 are final.
18.1b) CORRECT ANSWER - -2
As shown in above solution z has a final value of -2 at the end of loop # 4.
18.1c) CORRECT ANSWER - 29
Values at the end of loop # 1, N = 21+ 1 = 3, value = 2
Values at the end of loop # 2, N = 23 + 1 = 9, value = 23
Values at the end of loop # 3, N = 29 + 1 > 100, value = 29
N >100 at the end of loop # 3 therefore program will not run loop # 4 and values at the end of loop # 3 are final.
18.1d) CORRECT ANSWER - 25
1
A
1
B
5
C
3
2
5
2
=B$1=5
3
7
3
=B$2=2
4
=BlxC2=25
5
=B$3=3
18.1e) CORRECT ANSWER - 125
A
B
C
1
10
=A1 + Cl=20
= A1 x 1=10
2
15
=A2 + C2=45
= A2 x 2=30
3
20
=A3 + C3=80
= A3 x 3=60
4
25
=A4 + C4=125
= A4 x 4=100
308
Copyrighted Material © 2018
18.2 Data Structures - Solutions
Consult NCEES® FE Reference Handbook - Pages 224 - 227 for reference
18.2a) CORRECT ANSWER - C
Array, list, stacks and trees are different implementations of data structures and help organize data efficiently.
Helpful tip - Review the similarities and differences between arrays, lists, stacks and trees.
18.2b) CORRECT ANSWER - B
Worst-case time is used to measure the performance of searching algorithms. "Chain is only as strong as its
weakest link".
Helpful tip - Understand algorithm performance measurement method especially for searching and sorting.
18.2c) CORRECT ANSWER - Complete
Nodes of a 'full' binary tree can either be leaves or each node can possess exactly two children.
A binary tree is complete if all of its levels except possibly the last level are full. Consequently, each leaf will be
at the same distance from root.
Helpful tip - Review workings of binary tree, lists and arrays. Also understand associated terminology.
18.2d) CORRECT ANSWER - A
The worst-case execution time for a binary search algorithm is log2 n.
Helpful tip - Review workings of all searching algorithms and understand their performance.
18.2e) CORRECT ANSWER - D
Bubble, Quick and Heap are examples of sorting algorithms because they sort given data in specific order.
Helpful tip - Review workings of all sorting algorithms and understand their performance.
18.2f) CORRECT ANSWER - B
Bubble sort has worst case computational time of n 2
Insertion sort has worst case computation time of n 2
Heap sort has worst case computation time of nlog2n making it the best performer out of the given options.
18.2g) CORRECT ANSWER - C
Static data structure is one in which the memory size is fixed. Dynamic data structure is one in which memory is
allowed to expand and contract dynamically. Lists & Stacks are dynamic data structures since their memory
allocation can be changed. Array is an example of static data structure since its memory requirement/size is
fixed.
309
Copyrighted Material © 2018
18.3 Software design methods, implementation and testing - Solutions
Consult NCEES® FE Reference Handbook - Pages 224 - 227 for reference
18.3a) CORRECT ANSWER - D
A typical Software Development Lifecycle comprises of following steps:
•
System requirements
•
Design
•
Implementation
•
Testing
•
Deployment
•
Operation and Maintenance
Marketing is not a standard phase of Software Development Lifecycle.
18.3b) CORRECT ANSWER - Fragile
Rigid - A software design that is difficult to change.
Fragile - A software design that is prone to breaking in multiple places whenever a change is made.
Portable - A software design that ca n be used in different environments.
Immobile - A software design that is difficult to reuse for different projects.
18.3c) CORRECT ANSWER - A
A software design that is impossible or very different to extend is generally considered inefficient.
18.3d) CORRECT ANSWER - A
Static software testing involves verification using program code reviews, walk-through & inspections.
18.3e) CORRECT ANSWER - B
Dynamic software testing involves verification through actual program execution.
18.3f) CORRECT ANSWER - A
"Top-down" programming approach is associated with structured programming.
310
Copyrighted Material © 2018
18.3g) CORRECT ANSWER - B
The concept of "Class" is associated with object-oriented programming.
18.3H) CORRECT ANSWER - D
FORTRAN, BASIC and COBOL are examples of structured programming languages whereas C# is an example of an
object-oriented programming language.
18.3i) CORRECT ANSWER - D
PHP, JavaScript and Python are examples of scripting languages.
311
Copyrighted Material © 2018
A lso b y W a s im A s g h a r p .e ., p .en g , m .eng
Fundamentals of Engineering (FE) Electrical and Computer - Practice Exam # 1
Fundamentals of Engineering (FE) Electrical and Computer - Practice Exam # 2
Fundamentals of Engineering (FE) Electrical and Computer - Practice Exam # 3
312
Copyrighted Material © 2018