>Y FOR FE
Study Quids for Fundamentals of Engineering (FE)
Electrical & Computer CST Eicam
Practice over 400 solved problems based on
NCEES® FE CBT Specification Version 9.3
Wasim Asghar
PE, P. Eng, M. Eng
NCEES® is a registered trademark of National Council of Examiners for Engineering and Surveying.
Copyright © 2015 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 publisher.
Printed by CreateSpace, An Amazon.com Company
ISBN-13: 978-1517777920
ISBN-10: 1517777925
Table of Contents
Preface
1
About the author
3
Chapter # 1 - Properties of Electrical Materials
4
Problem Set # 1.1 - Chemical Properties...........................................................................................5
Problem Set # 1.2 - Electrical Properties........................................................................................... 6
Problem Set #1.3 - Mechanical Properties........................................................................................7
Problem Set# 1.4-Thermal Properties........................................................................................... 8
Chapter # 2 - Engineering Sciences
9
Problem Set #2.1 - Work, Energy, Power............................ .......................................................... 10
Problem Set #2.2 - Electrostatics.............................................................. ................................... 12
Problem Set #2.3 - Capacitance................................................................................................... 13
Problem Set #2.4 - Inductance.....................................................................................................14
Chapter # 3 - Circuit Analysis
Problem Set #3.1 - Kirchoffs Laws - KCL, KVL
15
16
Problem Set #3.2 - Series / Parallel Equivalent Circuits.......................................................... .............18
Problem Set #3.3 - Thevenin and Norton Theorems.......................................................................... 20
Problem Set #3.4 - Waveform Analysis...........................................................................................22
Problem Set #3.5 - Phasors........................................................................................................ 23
Problem Set #3.6 - Impedance.....................................................................................................25
Chapter # 4 - Linear Systems
27
Problem Set #4.1 - Frequency / transient response........................................................................... 28
Problem Set #4.2 - Resonance.................................................................................................... 30
Problem Set #4.3 - Laplace Transform........................................................................................... 32
Problem Set #4.4 - Transfer functions........................................................................................... 34
Problem Set #4.5 - Two-Port Theory..............................................................................................36
Chapter # 5 - Signal Processing
38
Problem Set #5.1 - Continuous Time Convolution............................................................................. 39
Problem Set #5.2 - Discrete Time Convolution................................................................................. 42
Problem Set #5.3 - Z Transforms.................................................................................................. 44
Problem Set #5.4 - Sampling.......................................................................................................46
Problem Set #5.5 - Filters.......................................................................................................... 47
Chapter # 6 - Electronics
49
Problem Set #6.1 - Solid-state Fundamentals................................................................................... 50
Problem Set #6.2 - Diodes..........................................................................................................52
Problem Set #6.3 - BJTs..................................... ....................................................................... 54
Problem Set #6.4 - MOSFETs.......................................................................................................57
Problem Set #6.5 - Operational Amplifiers...................................................................................... 60
Problem Set #6.6 - Instrumentation............................................................................................. 63
Problem Set #6.7 - Power Electronics............................................................................................ 65
Chapter # 7 - Power
66
Problem Set #7.1 - Single Phase Power.......................................................................................... 67
Problem Set #7.2 - Three Phase Power/Transmission & Distribution....................................................... 70
Problem Set #7.3 - Voltage Regulation........................................................................................... 72
Problem Set #7.4 - Transformers.................................................................................................. 73
Problem Set #7.5 - Motors & Generators........................................................................................ 75
Problem Set #7.6 - Power Factor..................................................................................................76
Chapter # 8 - Electromagnetics
77
Problem Set #8.1 - Maxwell Equations........................................................................................... 78
Problem Set #8.2 - Electrostatics / Magnetostatics........................................................................... 81
Problem Set #8.3 - Transmission Lines and Wave Propagation..............................................................82
Problem Set #8.4 - Electromagnetic compatibility............................................................................. 84
Chapter # 9 - Control Systems
85
Problem Set #9.1 - Block Diagrams................................................................................................86
Problem Set #9.2 - Bode Plots..................................................................................................... 89
Problem Set #9.3 - Steady Sate Errors............................................................................................92
Problem Set #9.4 - Routh-Hurwitz Criteria & System Stability............................................................... 94
Problem Set #9.5 - Root Locus......................................... ........................................................... 96
Problem Set #9.6 - State Variables............................................................................................. 101
Chapter # 10 - Communications
103
Problem Set #10.1 - Amplitude Modulation................................................................................... 104
Problem Set #10.2 - Angle Modulation.........................................................................................106
Problem Set #10.3 - Pulse Code Modulation (PCM) & Pulse Amplitude Modulation (PAM)...........................107
Problem Set #10.4 - Fourier Transforms....................................................................................... 108
Problem Set # 10.5 - Multiplexing................................................................................................ 110
Chapter # 11 - Computer Networks
111
Problem Set #11.1 - Routing and Switching....................................................................................112
Problem Set #11.2 - Network topologies / Frameworks / Models.........................................................113
Problem Set # 11.3 - Local Area Networks...................................................................................... 114
Chapter # 12 - Digital Systems
115
Problem Set # 12.1 - Number Systems..........................................................................................116
Problem Set # 12.2 - Boolean Logic................................................................................. ............ 117
Problem Set # 12.3 - Logic Gates.................................................................................................118
Problem Set #12.4 - Karnaugh Maps........................................................................................... 120
Problem Set #12.5 - Flip-flops and counters................................................................................... 122
Problem Set # 12.6 - State Machine Design.................................................................................... 125
Chapter # 13 - Computer Systems
127
Problem Set #13.1 - Architecture & Interfacing............................................................................... 128
Problem Set # 13.2- Microprocessor............................................................................................ 130
Problem Set #13.3 - Memory Technology and Systems..................................................................... 131
Chapter # 14- Software Development
132
Problem Set #14.1 - Algorithms................................................................................................. 133
Problem Set #14.2 - Data Structures............................................................................................ 135
Problem Set #14.3 - Software design methods/implementation/testing................................................ 136
Solutions
138
Chapter #1 - Properties of Electrical Materials................................................................................142
Chapter #2 - Engineering Sciences............. ..................................................................................146
Chapter #3
Circuit Analysis..................................................................................................... 151
Chapter #4- Linear Systems..................................................................................................... 160
Chapter #5 - Signal Processing.................................................................................................. 169
Chapter #6 - Electronics.................................. ..................... .................................... ............. 178
Chapter #7 - Power................................................................................................................194
Chapter #8 - Electromagnetics.................................................................................................. 205
Chapter #9 - Control Systems................................................................................................... 214
Chapter # 10 - Communications................................................................................................. 225
Chapter #11- Computer Networks.............................................................................................231
Chapter # 12 - Digital Systems...................................................................................................234
Chapter #13 - Computer Systems............................................................................................... 244
Chapter #14-Software Development..........................................................................................249
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 typical questions may not be very complex. It is therefore important to gain
fundamental understanding of all topics (more on exam taking strategy later). The intended audience
of this book includes final year students, new graduates as well as seasoned professionals who have
been out of school for a while.
Best way to use this book
This book solely focuses on the Electrical and Computer Engineering (ECE) sections of the latest NCEES®
FE Computer Based Testing (CBT) specification. It is centered on the idea of 'problem based learning'.
It is important 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 encourage reader's
familiarity with this reference manual.
Students should conduct multiple reviews of applicable NCEES® FE Reference Handbook sections and
understand the theory behind relevant concepts and formulas. They are also encouraged to use
college/university textbooks because for certain topics there is no option but to consult additional
resources. However it is suggested not to go into too much detail while using textbooks if faced with
time constraints.
It is recommended to attempt problems for each chapter right after studying concepts. You may not be
able to solve all problems in the first attempt. Therefore it is suggested to make note of concepts
requiring review. Once the underlying theory is understood, you should revisit the problems and
attempt 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 the same question therefore an effort has been made to present efficient Solutions.
In cases where you come across questions involving unfamiliar concepts and theories it is
recommended to research such content in order to gain necessary understanding. In fact as part of
exam preparation effort, students should think about different ways in which questions can be asked.
Organization of this book
This book is organized with each section chapter including a reference to applicable NCEES® FE
Reference Handbook topics and page numbers, rating of difficult level by author and content specific
tips for effective studying. Solutions are grouped at the end of this book for ease of review.
1
Copyrighted Material © 2015
FE CBT Exam Taking Tips
It is strongly suggested to purchase NCEES® FE sample exam from www.ncees.org in order to practice
additional problems and simulate exam format on computer. The author suggests adopting either of
the following strategies for taking latest CBT exam:
Strategy # 1 - Three round knock-out
First Round - After reading the question classify it into one of the following categories
'Easy', 'Medium', 'Difficult but solvable' or 'No clue'.
If it is 'Easy' or 'Medium' solve it right away otherwise flag it and move on.
After completing the first round you'll be left with 'Difficult but doable' and 'No clue' questions.
Second Round - Go through the list of flagged questions and try to solve 'Difficult but doable'
questions. Carefully remove the flags from solved questions.
After completing the second round, you'll have only 'No clue' questions left.
Third Round - Depending on the amount of time left in exam either try to solve the remaining
questions or apply elimination method.
Under no circumstances should you leave any question unanswered.
You should use any leftover time to recheck the answers.
Strategy # 2 - Relax, see and conquer
This strategy works best for bolder examinees.
The idea is to go through all questions from get go in order to gain big picture view of the exam.
It is advisable to flag all the difficult questions during the 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 since the fear of
unknown will subside. Now you should start solving non-flagged questions followed by flagged
questions. Any leftover time should be spent rechecking your answers.
The author adopted Strategy # 1 while taking FE CBT - Electrical and Computer Engineering exam in
October 2014 which he passed in first attempt and Strategy # 2 while taking PE - Power Exam in April
2015 which he passed in first attempt as well.
On-demand lectures for FE Electrical and Computer exam - 20% discount!!!
Visit the website www.studvforfe.com to sign-up for the online self-paced step-by-step course tailored
for FE Electrical and Computer Engineering CBT Exam. In order to get a 20% discount coupon for
'Standard life-time course subscription' simply use discount code '200FF' while enrolling.
Reporting errors
This book has undergone multiple 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. Therefore it
would be greatly appreciated if the reader can report any mistakes at comments@studvforfe.com.
2
Copyrighted Material © 2015
About the author
Wasim Asghar is a Licensed Professional Engineer in Texas (PE), Florida (PE) and Ontario (P. Eng) with
consulting experience in power system design, commissioning and plant engineering for leading clients
in Energy, Mining and Infrastructure industries.
He holds Bachelors of Engineering - Electrical with distinction from McMaster University, Hamilton,
Canada (2010) and Master of Engineering - Power Systems from University of Toronto (2013) which
was completed with full-time work.
In 2014, he undertook a two year international work assignment for a major project in Florida and also
decided to pursue PE Licensure in United States. The road to licensure was challenging primarily
because of a lack of useful study material for FE and PE exams.
Wasim passed both exams in first attempts (FE in October 2014 and PE in April 2015). The lessons
learned during exam preparation process inspired him to write this book which is designed to help
aspiring professional engineers better prepare for the latest CBT format of FE Electrical and Computer
Engineering Exam.
Acknowledgements
I am truly thankful to the support offered by these wonderful people:
•
Mother - Farhat, for always believing in my abilities
•
Father-Asghar, for providing me the foundation to realize my goals
•
Wife - Amna, for being part of my dreams
•
Brother - Fahim, for providing continuous assistance during entire lifecycle of this project
•
Uncle - Jawad, for being a great friend and mentor over the years
•
Cousin - Omar, for sharing his book writing experience and offering valuable insights
•
Friend - Emad, for assisting in formatting some text
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 © 2015
Chapter # 1 - Properties of Electrical Materials
Key Knowledge Areas*
Concepts
NCEES® FE Reference Handbook
Section
Page#
Chemical Properties
Electrochemistry
Atomic Bonding
Corrosion
Diffusion
Chemistry
58
Material Science/Structure of Matter
59
Electrical Properties
Capacitance
Conductivity
Material Science/Structure of Matter
Resistivity
Permittivity
Magnetic Permeability
Note: Also review 'Electrical and Computer Engineering7 pages 199-200
59
Mechanical Properties
Material Strength (stress, strain,
Young's modulus etc)
Photoelectric effect
Piezoelectric effect
Material Science/Structure of Matter
5 9 -6 2
Thermal Properties
Expansion
Conductivity
Material Science/Structure of Matter
64
59
Facts about this section
• 4-6 questions can be expected (according to NCEES® FE Specification)
• Difficulty level of this section is rated 'Easy' by the author
Tips for preparing this section
• Understand the concepts and formulas found on above mentioned pages in
NCEES® FE Reference Handbook
• Solve problem sets on next pages and review solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
4
Copyrighted Material © 2015
Problem Set # 1.1 - Chemical Properties
Consult NCEES® Reference Handbook - Pages 58-59 while solving these questions
Problem 1.1 a) __________ will act as sacrificial anode when brought in contact with iron (Fe).
(A) Zn
(B) Cu
(C) Ni
(D) Hg
Problem 1.1 b) Which of the following techniques is not used for corrosion prevention?
(A) Galvanization
(C) Plating
(B) Sacrificial anode
(D) Options A, B and C provide corrosion prevention
Problem 1.1 c)____________ must be present for corrosion to take place.
(A) Anode, cathode and electrolyte
(C) Anode, cathode and sun light
(B) Electrolyte and sun light
(D) Water and sun light
Problem 1.1 d) Calculate the diffusion coefficient for Copper atoms diffusing in Copper if the constant
of proportionality (D0) is 7.8 x 10'5 m2/s and activation energy (Qd) is 250 kJ/mol at 700 °C (973 K)
(A) 7.56 x 10"5 m2/s
(B) 3.78 x 10 5 m2/s
(C) 2.5 x 10 3 m2/s
(D) 1.50 x 10'7 m2/s
Problem 1.1 e) Calculate the activation energy required for Nickel atoms to attain diffusion coefficient
of 1.3 x 10'22 m2/s in host Copper atoms if the constant of proportionality (D0) is 2.7 x 10'5 m2/s at
500 °C (773 K)
(A) 211 kJ/mol
(B) 284 kJ/mol
(C) 256 kJ/mol
(D) 310 kJ/mol
Problem 1.1 f) A corrosion cell consisting of Zinc and Nickel is formed. Which of the following
statement is true?
(A) Nickel will act as Anode
(B) Zinc will act as Anode
(C) Either metal can act as Anode
(D) Corrosion will not occur
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Copyrighted Material © 2015
Problem Set # 1.2 - Electrical Properties
Consult NCEES® Reference Handbook - Pages 59,199 and 200 while solving these questions
Problem 1.2 a) Calculate resistivity of a 100 m long wire with a cross sectional diameter of 2 mm if it
has a resistance of 5 0 at 30° C.
(A) 3.14 x lO '7 Qm
(B) 1.57 x 10'3 Qm
(C) 3.14 x lO '3 Qm
(D) 1.57 x 10'7 Qm
Problem 1.2b) The resistivity of cable 'A' is four times that of cable 'B'. Both cables can offer same
resistance under a given temperature i f _____________
(A) Area is same and A is twice as long as B
(B) Area of B is four times that of A
(C) Area of B is one-fourth of A and same length
(D) Resistance of A and B cannot be same
Problem 1.2c) Capacitance of a parallel plate capacitor can be increased by___________
(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 1.2d) Materials with high electrical conductivity typically have____________
(A) High resistivity
(B) High heat conductivity
(C) Low elasticity
(D) Low ductility
Problem 1.2e) Calculate the magnetic permeability of a medium in which an infinitely long wire
carrying 100A current produces 0.5T magnetic field at a distance of 500cm perpendicular to the wire.
(A) 1 H/m
(B) 0.5 H/m
(C) 0.015 H/m
(D) 2.0 H/m
Problem 1.2f) Photoelectric effect can take place in
under suitable conditions.
(A) Metals
(B) Non-metals
(C) Liquids and gases
(D) Options A, B and C are correct
6
Copyrighted Material © 2015
Problem Set # 1.3 - Mechanical Properties
Consult NCEES® Reference Handbook - Pages 59 - 62 while solving these questions
Problem 1.3 a) A group of researchers have created a new material in laboratory. Its ultimate tensile
strength is found to be 2000 x 108 Pa. Calculate the maximum weight that can be supported by this
material if it is 5 m long and has a diameter of 2 mm.
(A)200 kN
(B)628 kN
(C) 2513 kN
(D) 915 kN
Problem 1.3b) A 100 N weight is suspended from 10 m long wire. It results in a 2 mm length increase.
Calculate its Young's Modulus if the wire has a diameter of 1 mm.
(A) 318 x 109 N/m
(B) 6.36 x 10u N/m
(C) 1273 x 109 N/m
(D) 636 x 103 N/m
Problem 1.3 c) Calculate the 'true' strain in a material if it undergoes 20 mm increase in length (original
length was 200 mm).
(A) 0.1
(B) 0.2
(C) 180
(D) 0.095
Problem 1.3 d)___
is the ability of a material to return back to its original form after
application of force.
(A) Plasticity
(B) Elasticity
(C) Ductility
(D) Malleability
Problem 1.3 e) Piezo-electric effect is associated w ith __________
(A) Photo-electric effect
(C) Electric charge concentration
(B) Magnetic flux
(D) Lightning
Problem 1.3 f) Tensile test curve gives information about all but which of the following mechanical
properties?
(A) Tensile strength
(B) Young's modulus
(C) Ductility
(D) Hardness
7
Copyrighted Material © 2015
Problem Set # 1.4 - Thermal Properties
Consult NCEES® Reference Handbook - Pages 59 and 64 while solving these questions
Problem 1.4 a) Bimetallic strip for thermostat should be made of materials w ith_________
(A) Different lengths
(B) Different coefficients of thermal expansion
(C) Same resistivity
(D) Same density
Problem 1.4 b) A 1 m long material is kept at room temperature (296 K) and constant pressure. Find
the temperature required to cause 6 x 10~3 engineering strain if 7 K temperature rise results in an
engineering strain of 3 x 10'3.
(A)319 K
(B)307 K
(C)310 K
(D)303 K
Problem 1.4 c) A design engineer is calculating space provision required for railway steel track
expansion. The steel being used has a coefficient of thermal expansion 1.2 x 10 5 °C'1. Temperature is
expected to increase from an average of 20 °C to a peak of 45 °C. Calculate the strain that should be
allowed for rail tracks to expand.
(A) 3 X 10'2
(B) 30 x 10 5
(C) 50 x 10'5
(D) 75 x 10'6
Problem 1.4 d) Calculate temperature coefficient of a given metal if its resistance doubles with a 25 K
temperature rise
(A) 0.08 K 1
(B) 0.04 K 1
(C) 25 K 1
(D) 1.5 K 1
Problem 1.4 e) Resistivity of a material is completely independent o f _____________
(A) Resistance
(B) Volume
(C) Temperature
(D) Depends on resistance, temperature & volume
Problem 1.4 f) Students are measuring heat capacities of three different samples of same liquid.
Sample 1 is 1 kg, sample 2 is 2 kg and sample 3 is 3 kg. Which of the following 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
8
Copyrighted Material © 2015
Chapter # 2 - Engineering Sciences
Key Knowledge Areas*
NCEES® FE Reference Handbook
Section
Page#
Concepts
Work, energy, power, heat
Charge, energy, current,
voltage, power
Forces
Work done in electric field
Capacitance
Inductance
Electrical and Computer Engineering
199-201
Facts about this section
• 6 - 9 questions can be expected (according to NCEES® FE Specification)
• Difficulty level of this section is rated 'Easy' by the author
Tips for preparing this section
• Understand the concepts and formulas found on above mentioned pages in
NCEES® Reference Handbook
• Be careful while handling units of measurement and directions for quantities such
as force, electric field strength etc in calculations.
• Work done by an external agent on a charge is considered 'negative7
• Use simplified formulas for electrostatic fields (such as line charge, sheet charge)
wherever applicable.
• Inductors and Resistors behave similarly in series/parallel while capacitors behave
differently.
• Solve problem sets on next pages and review solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
9
Copyrighted Material © 2015
Problem Set # 2.1 - Work, Energy, Power
Consult NCEES® Reference Handbook - Pages 199 - 201 while solving these questions
Problem 2.1 a) Calculate the amount of work necessary to bring a charge Qi = 10~6 C from infinity to Pi
(0, 0, 0 ) while another charge Q2 = 2 x 10"6 C is located at P2 (2, 0, 0 ).
(A) 4.5 x 1CT12 J
(B) 4.5 x 10~3 J
(C) 9 x 10~12 J
(D) 9 x 10 3 J
Problem 2.1 b) Find the potential energy stored by a system of two charges Qi = 5 x 10 6 C & Cb = 10~6 C
located at Pi (1, 0, 0 ) and P2 (0, 1, 0 ) respectively.
(A) 63.6 mJ
(B) 31.9 mJ
(C) 15.9 mJ
(D) 5 mJ
Problem 2.1 c) How many electrons pass through a point on conductor if the conductor carries 1 mA
for 5 s?
(A) 1.56 x 1016
(B) 3.12 x lO 16
(C) 6.24 x 1012
(D) 3.12 x lO 12
Problem 2.1 d) System A contains charge Qi = 50 x 10 6 C and Q2 = 100 x 10'6 C which are 1 m apart.
System B contains charges Q3 = 5 x 10'6 C and Q4 = 10 x 10~6 C which are 1 cm apart. Which of the
following options accurately represent potential energies of System A and System B?
(A) System A = System B = 45 J
(B) System A =4.5 x 1013 J, System B =45 J
(C) System A = 45 J , System B = 4.5 x 1013 J
(D) System A =4.5 J , System B =45 J
Problem 2.1 e) Calculate the amount of work that needs to be done in order to decrease the space
between Qi = 9 x 10"9 C and Q2 = 15 x 10'9 C from 1 m to 1 cm.
(A) 1.215 x 10’6J
(B) 15 x 10 4 J
(C) 1.202 x 10'6J
(D) 1.202 x 10“4 J
Problem 2.1 f) Calculate the energy stored in the electric field of a 200 V capacitor if distance between
the two plates is 0.1 m and each plate has an area of 1 m2 (assume e = 8.85 x 10~12 F/m)
(A) 8.85 x 10 11J
(B) 1.33 x 10"3 J
(C) 3.54 x 10 6J
(D) 1.77 x 10~6 J
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Copyrighted Material © 2015
Problem 2.1 g) Find the amount of work done in moving a charge Qi = 10 x 10"9 C through a distance of
2 cm along y-axis in a 200 V/m a*electric field.
(A) 4 x 10’8 J
(B) 0 J
(C) 4 x 10~9J
(D) 2 x 10"9 J
Problem 2.1 h) Calculate the power dissipated in a 2 Q resistor if 2 C charge passes through it in 1 s.
(A) 4 W
(B) 16 W
(C) 2 W
(D) 8 W
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Copyrighted Material © 2015
Problem Set # 2.2 - Electrostatics
Consult NCEES® Reference Handbook - Pages 199 - 201 while solving these questions
Problem 2.2 a) Calculate the magnitude of force between two point charges Qi = 10 x 10'6 C and Q2 =
100 x 10 6 C if they are located 1 cm apart.
(A) 90 kN
(B) 450 kN
(C) 225 kN
(D) 4.5 kN
Problem 2.2 b) Calculate the potential difference between two parallel plates having electric field
strength of 2000 V/m if they are 1 m apart
(A) 2000 V
(B) 2 V
(C) 0 V
(D) 100 V
Problem 2.2 c) A point charge Qi = 100 x 10 9 C is accelerated over a distance of 200 m in a constant
electric field strength of 1 kV/m. Calculate the potential difference between the two parallel plate.
(A) 22.5 mV
(B) 100 kV
(C) 200 kV
(D) 1 kV
Problem 2.2 d) A charged particle 'Q' with mass 0.01 kg is suspended between two parallel plates
having potential difference of 100 V. Calculate the charge quantity if spacing between the two plates
0.1 m.
(A) 0.98 C
(B) 9.8 x 10'3 C
(C) 4.9 x 10 4 C
(D) 9.8 x 10'5 C
Problem 2.2 e) 0.1 m long conductor carries 5 A in a magnetic field (10 pT). Find the angle between
conductor and magnetic field if the force resulting on conductor is 1 pN
(A) 11.5°
(B) 0°
(C) 90°
(D) 1.145°
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Copyrighted Material © 2015
Problem Set # 2.3 - Capacitance
Consult NCEES® Reference Handbook - Pages 199 - 201 while solving these questions
Problem 2.3 a) Calculate the applied voltage on a capacitor carrying 400 |iC charge with surface area
0.02 m2 and spacing d =0.01 m (assume e =8.85 x 10 12 F/m)
(A) 22.6 x 106 V
(B) 2.2 x 10'14 V
(C) 0 V
(D) 1.5 x 10 10 V
Problem 2.3 b) A parallel plate capacitor has capacitance of 100 |iF. It had an initial voltage of 5 V
across it. Calculate the constant charging current if the voltage across capacitor is recorded as 10 V
after 3 minutes.
(A) 5.4 \xA
(B)
2.7 jiA
(C) 16.6 mA
(D)
8 mA
Problem 2.3 c) A 200 \xf capacitor has voltage v(t) = 240sin(377t) across it. Calculate the energy stored
in the capacitor as a function of time.
(A) 0 J
(B) 5.76sin2(377t) J
(C) 5.76 sin(142129t) J
(D) 1.36sin2(377t) J
Problem 2.3 d) Calculate charging current if voltage across a 100 \xf capacitor increases by 10 V in 5 s.
(A) 1 mA
(B) 0.2 mA
(C) 2 A
(D) 4 A
Problem 2.3 e) Find the equivalent capacitance of the circuit between terminals A-B
l|iF
(A) 1 piF
(C) 0.5 |J.F
(B) 2 nF
(D) 3.5 |iF
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Copyrighted Material © 2015
Problem Set # 2.4 - Inductance
Consult NCEES® Reference Handbook - Pages 199 - 201 while solving these questions
Problem 2.4 a) What is the inductance of a 1 m long coil with 100 turns and a cross sectional area of
0.1 m2 (assume \x=4ti10~7 H/m)?
(A) 1.25 mH
(B)1000 H
(C) 0.625 mH
(D) 2.5 mH
Problem 2.4 b) Calculate the voltage induced in a 5 mH inductor if current is increased in it from 0 to
100 mA in 2 ms.
(A) 5 V
(B) 0.25 V
(C) I V
(D) 0 V
Problem 2.4 c) Find the energy stored in a 100 mH inductor carrying current i(t) =t 2 (for t>0) for 10s
(A) 100J
(B) 10J
(C) 2.5 J
(D) 500 J
Problem 2.4 d) Energy storage capacity of an inductor can be increased by following?
(A) Decreasing voltage across it
(B) Increasing number of turns
(C) Increasing its length
(D) Options A, B, C are correct
Problem 2.4 e) Determine the inductance between terminals A-B in the circuit shown below?
1H
a u r r r r \ ________ -
m
m
2H
2H
1H
b u rrrrx
(A) 6 H
(B) 2 H
(C) 4 H
(D) 8 H
14
Copyrighted Material © 2015
Chapter
# m3
- Girojiiiti:
1 iSr....... ....
......
..............Airnailvsis
............... ......i
....
Key Knowledge Areas*
NCEES® FE Reference Handbook
Section
Page#
Concepts
Kirchoff's Law - KCL, KVL
Series/parallel equivalent
circuits
Thevenin and Norton theorems
Node and loop analysis
Waveform analysis
Phasors
Impedance
Electrical and Computer
Engineering
2 0 0 -2 0 2
Facts about this section
• 1 0 - 15 questions can be expected (according to NCEES® FE Specification). It is the
most heavily weighted section on exam.
• Difficulty level of this section is rated 'Medium' by the author
Tips for preparing this section
• Understand the concepts and formulas found on above mentioned pages in
NCEES® Reference Handbook.
• Learn how to use fundamental circuit equations such as 'Ohm's Law', 'voltage
divider' and 'current divider'.
• Revise basic circuit theory using your university/college electrical textbook. Sound
theoretical understanding is necessary in solving these problems.
• DC circuits can be solved through KCL or KVL but use judgment to decide which
option is best suited for given problem.
• Thevenin Resistance calculation requires 'short circuiting' external voltage
sources and 'open circuiting' current sources.
• Sinusoidal signals require conversion to a standard form (cosine) for phasor
representation.
• Solve problem sets on next pages and review solutions at the end of this book
*
Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
15
Copyrighted Material © 2015
Problem Set # 3.1 - Kirchoffs Laws - KCL, KVL
Consult NCEES® Reference Handbook - Pages 200 - 202 while solving these questions
Problem 3.1 a) Find the voltage magnitude across 10 kQ resistor in the circuit shown below.
(A) 70.5 V
(B) -14.6 V
(C) 1.46 mV
(D) 5V
Problem 3.1 b) Find the current I' in the circuit shown below.
(A) 4 mA
(B) 3 mA
(C) 2 mA
(D) 6 mA
Problem 3.1 c) Find the voltage across 3 Q resistor in the circuit shown below.
4a
(A) 2.5 V
(B) 1.87 V
(C) 7.5 V
(D) 3.75 V
16
Copyrighted Material © 2015
Problem 3.1 d) Determine the current lx in the circuit shown below.
20Q
(A) 4.0 A
(B) 3.5 A
(C) 1.7 A
(D) 1.6 A
Problem 3.1 e) Calculate the current passing through 1 kQ resistor in the circuit shown below.
m
5mA
(A) 3.1mA
(B) 5 mA
(C) 2.5 mA
(D) 1.5 mA
Problem 3.1 f) Calculate the current passing through 5 kQ resistor in the circuit shown below.
ma
2ko.
(A) 1.5 mA
(B)875 nA
(C) 50 nA
(D) 0 A
17
Copyrighted Material © 2015
Problem Set #3.2 - Series / Parallel Equivalent Circuits
Consult NCEES® Reference Handbook - Pages 200 - 202 while solving these questions
Problem 3.2 a) Find the equivalent resistance between terminals A-B of the circuit shown below.
1 0 k fi
AAAr
5k®
1k®
2k®
(A) 5 kQ
(B) 7.5 kQ
(C) 2 kQ
(D) 3.4 kQ
Problem 3.2 b) Find the equivalent resistance between terminals A-B of the circuit shown below.
10kQ
(A) 2 kQ
(B) 3 kQ
(C) 1 kQ
(D) 5 kQ
Problem 3.2 c) Find the equivalent resistance between terminals A-B of the circuit shown below.
1kft
5k®
m
2k®
iokn
(A) 7 kQ
(B) 3.5 kQ
(C) 1.5 kQ
(D) 5 kQ
18
Copyrighted Material © 2015
Problem 3.2 d) Find the equivalent resistance between terminals A-B of the circuit shown below.
5kfl
4kfi
2kn
AM r
A W
2kO
(A) 3 kQ
(B) 2 kQ
(C) 1 kQ
(D) 5 kQ
Problem 3.2 e) Find the equivalent resistance between terminals A-B of the circuit shown below.
ika
...
a
2kO
lOkfl
SkO
AA/V
5kO
2kO
B
(A) 1 kQ
(B) 3 kQ
(C) 5 kQ
(D) 6 kQ
19
Copyrighted Material © 2015
Problem Set # 3.3 - Thevenin and Norton Theorems
Consult NCEES® Reference Handbook - Page 200 - 202 while solving these questions
Problem 3.3 a) Calculate the voltage Voc in circuit shown below using Thevenin Theorem
2kQ
2§cO
(A) 10 V
(B) 5 V
(C)37 V
(D) 100V
Problem 3.3 b) Find Thevenin Resistance of the circuit shown in Problem 3.3a)
(A) 1 kQ
(B) 8 kO
(C) 1.5 kQ
(D) 2 kQ
Problem 3.3 c) Calculate Thevenin Resistance of the circuit shown below
(A) 1.25 kQ
(B) 5 kQ
(C) 2.5 kQ
(D) 3 kQ
20
Copyrighted Material © 2015
Problem 3.3 d) Calculate the short circuit current (lsc) for lx in the network shown below using Norton
theorem
5kfi
lx
(A) 7.5 mA
(B) 15 mA
(C) 5 mA
(D) 12 mA
Problem 3.3 e) Find Req of the circuit shown in Problem 3.3 d)
(A) 1 kQ
(B) 2 kQ
(C) 3 kQ
(D) 4 kQ
21
Copyrighted Material © 2015
Problem Set # 3.4 - Waveform Analysis
Consult NCEES® Reference Handbook - Pages 200 - 202 while solving these questions
Problem 3.4 a) Calculate the maximum voltage Vmax of a full wave rectified sinusoid if its effective value
of is 10 V.
(A) 20.5 V
(B) 5 V
(C) 14.1V
(D)10V
Problem 3.4 b) Findthe sumof following sinusoids:
Vt =
10cos(500t)
(A) 25cos(600t +
&V2 = lScos(100t
45)
(C) 150cos(500t)cos(100t + 45)
+ 45)
(B) 10cos(S00t) - 10.6cos(100t) +
10.6sin(100t)
(D) 10cos(500t) + 10.6cos(100t) —
10.6sin(100t)
Problem 3.4 c) Find the frequency of sinusoidal signal given by 100cos(500t + 50°)
(A) 500 Hz
(B) 250 Hz
(C) 79.5 Hz
(D) 50 Hz
Problem 3.4 d) Calculate Xavefor a halfwave rectified sinusoidal signal given by 15cos(100t + 50°).
(A) 7.5
(B) 4.77
(C) 9.54
(D) 10.6
Problem 3.4 e) Calculate the average value of following periodic current waveform
(A) 4 A
(B) 0.75 A
(C) 1 A
(D) 2 A
22
Copyrighted Material © 2015
Problem Set # 3.5 - Phasors
Consult NCEES® Reference Handbook - Pages 200 - 202 while solving these questions
Problem 3.5 a) Find phasor current for the following circuit
(B) 12.5/-3Q0 A
(A) 100/60° A
(C) 7Q.7/-3Q0 A
(D) 1.875/-12Q0 A
Problem 3.5 b) Express Vt = 212sin(a)t + 50) in phasor form.
(A) 1507-40° V
(B) 212/50° V
(C) 150/50° V
(D) 212/-4Q0 V
Problem 3.5 c) Find the phasor current for circuit shown below
500ft
(A) 0.14/349° A
(B) 0.14/1.4° A
(C) 0.5/Of A
(PI 1.75/50° A
23
Copyrighted Material © 2015
Problem 3.5 d) Find the phasor current passing through capacitor in circuit shown below
(A) 70/10° A
(C) 2665/100° A
(B) 1000/-80° A
(D) 55Q/-1Q0 A
Problem 3.5 e) Find the frequency domain impedance Z of the following circuit.
so
-foja
- W v --------------- 1|----------
z
(A) 5 - 5j n
(C) 5 - 10; £1
20a
(B) 6.2 - 14j (1
(D) 5 n
24
Copyrighted Material © 2015
Problem Set # 3.6 - Impedance
Consult NCEES® Reference Handbook - Pages 200 - 202 while solving these questions
Problem 3.6 a) Determine the equivalent impedance Z for the following circuit.
}2Q
/ r m
2a
\ W
100
i50
-SjO
-10J0
(A) 1 2 - j a
(B) 5 + 2j n
(C) 2 - 3j t t
(D) 0j t t
J
Problem 3.6 b) Determine the equivalent impedance Z for the following circuit
m
ion
|50
*5|Q
Am
(A) 5 - 3 j n
(B) 2 - 3jSl
(C) 1 + 27 n
(D)10j a
25
Copyrighted Material © 2015
Problem 3.6 c) Determine the equivalent impedance for following circuit assuming 60 Hz frequency
100nF
(A) 50 + j 26525 O
(B) 50 - j 26525 a
(C) 50 + ;535 ft
(D) 50 - ;535 O
Problem 3.6 d) Determine the equivalent impedance for following circuit assuming 60 Hz frequency
100
10OyF
2mH
(A) (10 + J2 n) 11( - 7 100 O)
(B) (io + y o .5 n )||(-yioo a)
(C) (10 + j 0.75 n) 11(-726.5 H)
(d ) i o n
26
Copyrighted Material © 2015
Chapter # 4 - Linear Systems
Key Knowledge Areas*
NCEES® FE Reference Handbook
Section
Page #
Concepts
Frequency/transient response
Resonance
Transfer functions
2-port theory
Laplace Transforms
Electrical and Computer
Engineering
202 - 203
Mathematics
33
Facts about this section
• 5 - 8 questions can be expected (according to NCEES® FE Specification).
• Difficulty level of this section is rated 'Easy' by the author
Tips for preparing this section
• Understand the concepts and formulas found on above mentioned pages in
NCEES® Reference Handbook.
• Carefully read the questions involving RC and RL transients in order to establish
initial conditions.
• Use correct formulas for RLC Series and Parallel resonance circuits.
• Transfer functions may require conversion to a standard form in order to
calculate gain, poles and zeros.
• Learn how to do partial fraction expansion for inverse Laplace transforms
• Solve problem sets on next pages and review solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
27
Copyrighted Material © 2015
Problem Set # 4.1 - Frequency / transient response
Consult NCEES® Reference Handbook - Pages 202 - 203 while solving these questions
Problem 4.1 a) The switch shown in circuit below has been in the indicated position for a long time.
Find the voltage across capacitor Vc(t) 5 minutes after the switch changes its position at t =0 s
Switch changes position at t =Os
___________
mm
20V
1mF
(A) 15V
(B) 10V
(C) IV
(D) 3V
Problem 4.1 b) Find an expression for voltage across capacitor Vc(t) for t>0 in the circuit shown below
(A) e't/2 V
(B) 10et/2 V
(C) 10e”t/4 V
(D) 5e~t/2 V
28
Copyrighted Material © 2015
Problem 4.1 c) Find the current i(t) in following circuit if switch is closed at t = 0 s
Switch closes at t s Os
V ——
m
■VSAr
10V
2mH
(B) 1 - e'500000*A
(A) 10 (l-e't/2) A
-sooooot
(C) e
+ 0.01(l-e'500000t) A
-500000t\
(D) 0.01 (1-e
Problem 4.1 d) Find the current i(t) after ten time constants if switch has been in shown position for a
longtime.
Switch opens at t = 0s
(A) 0.5 A
(B) 22.6 [iA
(C) 1 A
(D) 0.18 A
Problem 4.1 e) The switch shown in circuit below has been in the indicated position for a long time.
Find the voltage across capacitor vc(t) after 5 time constants if it changes position at t =0 s.
Switch changes position at t a Os
10V
5mF
lOkft
(A) 9.99 V
(B) 2 V
(C) 67 mA
(D) 35 mA
29
Copyrighted Material © 2015
Problem Set # 4.2 - Resonance
Consult NCEES® Reference Handbook - Pages 202 - 203 while solving these questions
Problem 4.2 a) Calculate the resonant frequency of RLC circuit shown below
10mF
20mH
(A) 1414 rad/s
1Q0Q
(B) 250 rad/s
(C) 500 rad/s
(D) 2236 rad/s
Problem 4.2 b) Find the Band Width (BW) of circuit shown in Problem 4.2a)
(A) 100 rad/s
(B) 5000 rad/s
(C) 250 rad/s
(D) 707 rad/s
Problem 4.2 c) Calculate maximum current for following RLC circuit if it is powered by a 120V source
19|jF
io n
tOmH
(A) 4 A
(B) 12 A
(C) 5 A
(D) 7.5 A
Problem 4.2 d) Calculate the frequency at which maximum current occurs in Problem 4.2c)
(A) 3162 rad/s
(B) 6324 rad/s
(C) 1581 rad/s
(D) 2500 rad/s
30
Copyrighted Material © 2015
Problem 4.2 e) Calculate the current magnification factor for the following RLC circuit
(A) 250
(B) 0.707
(C)500
(D)707
Problem 4.2 f) Calculate the Band Width (BW) of circuit shown in Problem 4.2e)
(A) 377 rad/ s
(B) 707 rad/ s
(C) 500 r ad/ s
(D) 1000 r ad / s
31
Copyrighted Material © 2015
Problem Set # 4.3 - Laplace Transform
Consult NCEES® Reference Handbook - Page 33 while solving these questions
Problem 4.3 a) Find the Laplace Transform of following function
f { t ) = e~bt ( b >
0)
(A) J _
(B)—
x ' s-b
(C) —
1 ; s-b
1 1s+b
(D) —
1 1 s+b
Problem 4.3 b) Find the Laplace Transform of following function
f ( t ) = e~atu ( t — 1 )
1
e—(s+a)
(A) —
(B) ------
c2
(C) —
p-”(S"Ct)
(D) ------
s -fa
s+a
s-a
s-a
Problem 4.3 c) Find the Laplace Transform of following function
/(t) = te~atu ( t — 1 )
...
. . e-(s+a)
s+a
(A) ---- ——
v
(B) -----—
( s + a ) 2+ l
'
(C) e“ (S+0)(7- i TT + 7 ~ t )
(s + a )2
(s -a )2
(D) — ^
(s + a )
(s + a )2
Problem 4.3 d) Find the Laplace Transform of following function
— e " 3^t- 3^)u(t — 3)
/(t) =
(A)
+
s+ 1
(B)
s+ 3
(s + l)(s + 3 )
(C)’ 7( s—
^ :
+ 3 ) ( s —3)
'
'
(D)
e~3s( —---- —)
'
s+ 1
S+3J
Problem 4.3 e) Find the Laplace Transform of following function
f ( t ) = te~aC8 { t -
2)
(A) 2e-2(,+a)
(B)
(C) e _(s+a)
(D) 6 (S+a)
'
_L_
s+a
32
Copyrighted Material © 2015
Problem 4.3 f) Find the inverse Laplace Transform of following function
~
(s + 3 )+ (s + 5 )
(A) 3e~3tu { t ) + 5e~5tu ( t )
(B) e~4tu ( t )
(D) ^e_4tu(t)
(C) 2e~etu ( t )
Problem 4.3 g) Find the inverse Laplace Transform of following function
p ( s) = — — —
y
(s + l)(s + 7 )
J
(A) ( e ~ 8t + e~ l — e~ 7t) u ( t )
(C) - (7 e _t — e~ 7t) u ( t )
(B) (e_t + e ~7t) u ( t )
(D) 5e~7t — 8e~l
6
Problem 4.3 h) Find the inverse Laplace Transform of following function
p
s^+2s+l
s
^
~
(s + 2 ) ( s ) ( s + 3 )
(A)
+ i e-3 t)u (t)
(B) ( e ~ zt + e~3t + ^ u ( t )
(D) 6 (t)(e"2t + e~3t) u ( t )
(C) te~2te~3tu ( t )
Problem 4.3 i) Find the inverse Laplace Transform of following function
(A) 5e~stu ( t ) + - u ( t ) + ” U(t)
(B) \ e ~ stu ( t ) + ^ t u ( t )
(C) ^ w (t) + ^ t u ( t ) ~ ^ ~ e 5tu ( t )(D) 25u ( t ) - e 5tu(t)
25
5
2h
33
Copyrighted Material © 2015
Problem Set # 4.4 - Transfer functions
Consult NCEES® Reference Handbook - Pages 202 - 203 while solving these questions
y
Problem 4.4 a) Determine — for the following circuit
(A)
(C)
WI |(n)
(*+s«||(i)
(R + S L )
( R + s O llfe )
w ife)
(B)
R + (R + SL)||(£)
(D)
R + (R +s L )| | (i)
Problem 4.4 b) Find the voltage transfer function for the following circuit:
r
Vo
(A) (K||sL)
R+
(C)
(B)
sC
(«||s« + ( R + i )
/rt\
/?+
(R\\SL)
R-\—
—
sC \
^
sCJ
(R||SL ) + ( R + i )
34
Copyrighted Material © 2015
Problem 4.4 c) Determine the input impedance transfer function for the following circuit
R
R
( D) R +± + sL\\R
(C )i||s i||R
Problem 4.4 d) Determine the zero, pole and magnitude gain of following transfer function:
20
(S) ~ (s)(10s + 1)
(A) Poles @ 0 and 10, Zero - none, Gain = 20 dB
(B) Poles @ 0 and -0.1, Zero - none, Gain = 6 dB
(C) Pole @ 0, Zero @ 10, Gain = 2 dB
(D) Pole - none, Zeroes @ 0 and 0.1,Gain = 6 dB
Problem 4.4 e) Determine the zero, pole and magnitude gain of following transfer function:
10
“ ( s ) 2(5 s + 1)
(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 = 6 dB
35
Copyrighted Material © 2015
Problem Set # 4.5 - Two-Port Theory
Consult NCEES® Reference Handbook - Pages 202 - 203 while solving these questions
Problem 4.5 a) Find Z parameters for the following circuit
2Q
j-j
£
2fl
a a t
(A) Zu =Z22=4 0, Z12=Z21=2 Q
(B) Zh=Z22=2 0, Z12=Z21=2 0
(C)Zu=Z22=4 0,Z 12=Z21=4n
(D) Zu=Z22=l O, Z12=Z21=4 0
Problem 4.5 b) Find Y parameters for the following circuit
(A) Yh=Y22=S/30/ Y12=Y21=-S/10
(C) Y11=Y22=S/5/ Y12=Y21= -S/10
(B) Y11=Y22=S/5/ Y 12=Y21= -S/20
(D) Y 11=Y22=S/10/ Y12=Y21=-S/10
Problem 4.5 c) Determine Y parameters for the following circuit
2 fi
11
12
2Q
(A) Y11=Y22=S/3/ Y12=Y21=-S/6
(B) Y11=Y22=S/6/ Y 12=Y21=-S/6
(C) Y 11=Y22=S/6/ Y12=Y21=-S/3
(D) Y 11=Y22=S/6/ Y12=Y21=-S/3
36
Copyrighted Material © 2015
Problem 4.5 d) Find Zn for the following circuit
(A) 22 Q
(B) 11 Q
(C) 5 Q
(D) I Q
Problem 4.5 e) Find H12 for the following circuit
100
AAAr
V1
20Q
son
V2
(A) 3
(B) 2
(C) 0.66
(D) 0.33
37
Copyrighted Material © 2015
Chapter # 5 - Signal Processing
Key Knowledge Areas*
NCEES® FE Reference Handbook
Section
Page#
Concepts
Convolution
Difference equations
Z-transforms
Sampling
Analog filters
Digital filters
Electrical and Computer
Engineering
206
205
205
209
2 1 0-211
206
Facts about this section
• 5 - 8 questions can be expected (according to NCEES® FE Specification).
• Difficulty level of this section is rated 'Medium' by the author. Students having a
major in communication engineering may find it easier.
Tips for preparing this section
• Understand the concepts and formulas found on above mentioned pages of
NCEES® FE Reference Handbook.
• Learn how to perform continuous and discrete convolution using graphical
method.
• Develop familiarity with Z transform table and its application in difference
equation.
• Nyquist Theorem shall be applied on signal with highest frequency in cases
involving multiple signals.
• Calculate transfer functions of analog filter circuits given on pages 210-211 of
reference manual in order to gain necessary understanding.
• Solve problem sets on next pages and review solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
38
Copyrighted Material © 2015
Problem Set # 5.1 - Continuous Time Convolution
Consult NCEES® Reference Handbook - Page 206 while solving these questions
Problem 5.1 a) Find the convolved output of following convolution functions
39
Copyrighted Material © 2015
Problem 5.1 b) Find the convolved output of following functions
|f { t ) = efu(—t)
h(t) = 3e tu( t )
(A) ^e tf o r t < 0, - e ^ f o r t > 0
(B) - e 2tf o r t < 0, —^e ~2tf o r t
(C) ^ e ~ lf o r t < 0, ^ e tf o r t > 0
(D) Of or t < 0, 3etf o r t > 0
Problem 5.1 c) Find the convolved output of following functions
x ( t ) = sin(t) (u(t) — u ( t — 7r))
y(t) = u(t) — u (t — 7r)
(A) cos(t) 0 < t < iz, — cos(t) n < t < 2 n
(B) sin(t) 0 < t < n, 1 — sin(t) n < t < 2 n
(C) eos(t) — 1 0 < t < 7r, 1 — cos (t) n < t < 2 n
(D)l — cos (t) 0 < t < n , cos (t) — I n < t < 2 n
40
Copyrighted Material © 2015
Problem 5.1 d) Find the convolved output of following functions
x(t)
yft)
x(t) = 3 (w(t — 3) — u ( t — 5))
(A) 6(1 + t)
(B) 1 + 6 t
(C) 6(1 + t)
(D) 6 (t)
— 1 < t < 1, 6(1 — t ) f o r
for
—1< t
for
/or
y(t) = 2 ( u ( t + 4) — u ( t
for
< 1, 1— 6 t f o r
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 5.1 e) Find the convolved output of following functions for t > 2
y<t)
(B) 2(ef — e f)
(A) 0
(C) 2 ( e ~ t+2 -
e-t)
(D) et — e~ z~2
41
Copyrighted Material © 2015
Problem Set # 5.2 - Discrete Time Convolution
Consult NCEES® Reference Handbook - Page 206 while solving these questions
Problem 5.2 a) Find the convolved output of discrete time functions x[n] and y[n] shown below
3
m
ii
ii
2
yfnj
#
<
1
1
2
1
:5
n
(A) [0 4 10 4 0]
(B) [0 4 10 14 10 4 0]
(C) [0 2 4 8 4 2 0]
(D) [02620]
Problem 5.2 b) Find the convolved output of discrete time functions x[n] and y[n] shown below
3
t)
m
yin]
2
m
1 #
1
1
(A) [ 0 1 4 6 4 1 0 ]
(B) [01410]
(C) [ 0 1 4 8 8 3 0]
(D) [0 1 8 8 1 0]
Problem 5.2 c) Discrete Time Convolution is used to find___________
(A) Zero input solution
(B) Zero state solution
(C) Product of signals
(D) Modulation index
Problem 5.2 d) Find the convolution output of x [ n ] = u [ n] — u [ n — 5] & y [ n ] = 0.2nix[n]
(A) 0.2s
(C)££=S0.2"-fe
(B) —0.25(0.2n - 0.2n“ 6)
(D) n = l 0 . 2 n~k
42
Copyrighted Material © 2015
Problem 5.2 e) Find the convolution output of x [ n ] = u[ n — 2] & y [ n ] = 0.4nu[n]
(A) ~ 0 . 4 n
(B ) - | E ? 0 .4 ” - 1
(C) EfcIJOA n~ku [ n - k ]
(D) 0.42 Y,l=l 0.4~ku[n - k]
43
Copyrighted Material © 2015
Problem Set # 5.3 - Z Transforms
Consult NCEES® Reference Handbook - Page 205 while solving these questions
Problem 5.3 a)_______ __is used to solve a difference equation modeling Discrete Time System.
A) Laplace Transform
(B) Discrete Convolution
C) Z Transform
(D) Integration
Problem 5.3 b) Find the Z Transform of 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 5.3 c) Find the Z Transform of x [ n ] = 0.2ntt[n]
A) (0.2 z " 1) 71
(B) 0.2nz -1
C) 0.2/(l - z - 1)
(D) 1/(1 - 0.2z_1)
Problem 5.3 d) Find the Z Transform of x[n] = [23 105]
A) 2 + 3z -1 + z -2 + 5z -4
(B) 1/(2 + 3z _1 + z -2 + 5z -4)
C) 5(1 — 0.75nz _1)
(D) 2 + 3z
Problem 5.3 e) Find the Z Transform of 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 5.3 f) Find the inverse Z Transform of X ( z ) = ----z —0.5
A) 1/(1 - 0.5z ' 1)
(B) 0.5n“ 1w M
C) 0.5"u[n]
(D) 1 - 0.5’: n
Problem 5.3 g) Find the inverse Z Transform of X (z) =
5z+2
(z -l)(z -4 )
A) ~(4nii[n] — 1nu[n] + 5[n])
(B) cos[n] — 4nu[ n]
C) ~4nu[ n ] + ~ u[ n]
(D)"5[n] - ~ u [ n ] + ~ 4 nu[n]
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Problem 5.3 h) Find the inverse Z Transform of X(z) =
.-^z+1-)
(z -0 .1 )(z -0 .2 )
(A) jO .l" - 2(0.2)nu[n]
(B) -100<5[n] - 209(0.1)"u[n] + 108(0.2)"
(C) jS(n) + 0.1nu[n] + 0.3nu[n]
(D) 0.55[n] + 2(0.1)nw[n]
fz + 0 sl
Problem 5.3 i) Find the inverse Z Transform of X(z) = ----- —1
---V '
'
(A) - — S[n\ + 12(0.1)nu[n] - i( - 0 . 4 ) nu[n]
2
2
(C) 28[n] + 6(0.1)"u[n] — -(0.4)"ii[n]
4
( z —0 . l ) ( z + 0 . 4 )
(B) -0.1nu[n] + -(-0.4)"w [nl
4
2
(D)-(0.1) 7lu|n| + -0 .4 nu[n|
2
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Copyrighted Material © 2015
2
Problem Set # 5.4 - Sampling
Consult NCEES® Reference Handbook - Pages 209 while solving these questions
Problem 5.4 a) Find the minimum sampling frequency required for perfect reconstruction of
sinc(1000nt) + sinc(2000nt )
(A) 1000 Hz
(B) 2000 Hz
(C) 3000 Hz
(D) 500 Hz
Problem 5.4 b) Sampling the continuous time signal x ( t ) = cos (30007rt + 6 ) at fs= 2000 Hz will result
in _____________
(A) Nyquist sampling
(B) Critical sampling
(C) Aliasing
(D) Oversampling
Problem 5.4 c) Determine the aliased frequency of x ( t ) = cos (5007r£ + 6) if it is sample at 500 Hz
(A) 500 Hz
(B) 250 Hz
(C) 0 Hz
(D) There will be no aliasing
Problem 5.4 d) Determine the aliased frequency of x ( t ) = cos (400jit + 6 ) if it is sampled at 300 Hz
(A) 200 Hz
(B) 300 Hz
(C) 100 Hz
(D) 400 Hz
Problem 5.4 e) In order to reconstruct a band limited signal, the signal must b e_____________
(A) Sampled at or above Nyquist rate
(B) Passed through ideal low pass filter
(C) Modulated with high frequency signal
(D) Options A andB are correct
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Problem Set # 5.5 - Filters
Consult NCEES® Reference Handbook - Pages 206, 210 and 211 while solving these questions
Problem 5.5 a) Identify the filter type given by transfer function shown below
,
\
100
H O'") = 11 +, 2nn0j(x )
(A) High Pass Filter
(B) Low Pass Filter
(C) Band Pass Filter
(D) Band Reject Filter
Problem 5.5 b) Identify the filter type given by transfer function shown below
cnn (/*>)
H ( j ( o ) = ------500
( ja ) ) 2 +
200 +
(A) High Pass Filter
(B) Low Pass Filter
(C) Band Pass Filter
(D) Band Reject Filter
1000
Problem 5.5 c) Identify the filter type given by transfer function shown below
^
x
= 1000—
2 0 jco
^—
1 + 20/co
(A) High Pass Filter
(B) Low Pass Filter
(C) Band Pass Filter
(D) Band Reject Filter
Problem 5.5 d) Identify the filter type given by transfer function shown below
300 ( ( j a ) ) 2 +
( / t o ) 2 + 1 0 0 ja) +
(A) High Pass Filter
(B) Low Pass Filter
(C) Band Pass Filter
(D) Band Reject Filter
Problem 5.5 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
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Copyrighted Material © 2015
Problem 5.5 f) _______ is a type of recursive digital filter.
(A) Finite Impulse ResponseFilter
(B) Infinite Impulse Response Filter
(C) Options A and B arecorrect
(D) Recursive filters don't exist
Problem 5.5 g) Digital filtering does not involve___________
(A) Sampling
(B) A/D conversion
(C) D/A conversion
(D) Phase Modulation
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Copyrighted Material © 2015
Chapter # 6 - Electronics
Key Knowledge Areas*
NGiEES FE Refe re nee Hcwclb^j^jk
Section
Page#
212
214-216
Electrical and Computer
214-216
Engineering
212-213
212
Concepts
Solid-state fundamentals
Discrete devices
Bias circuits
Amplifiers
Operational Amplifiers
Instrumentation, Measurement
123-125
and Control
Note: Specific details are not available in NCEES®
FE Reference Handbook
Instrumentation
Power Electronics
Facts about this section
• 7 - 1 1 questions can be expected (according to NCEES® FE Specification).
• Difficulty level of this section is rated 'Hard' by the author. Students having a
major in electronics engineering may find it easier.
Tips for preparing this section
• Understand the concepts and formulas found on above mentioned pages in
NCEES® Reference Handbook.
• Learn how to calculate Q-point of different electronic devices.
• Diode circuits can have different Q-points depending on model (ideal/non-ideal)
• BJT amplification takes place in 'Active Region".
• JFET/MOSFET amplification takes place in 'Saturation Region'
• Derive equations for two-source configuration of an ideal operational amplifier
and CMRR given in NCEES® Reference Handbook (page 212).
• Familiarize yourself with workings of measurement devices such as RTDs,
Thermocouples, Wheatstone Bridge, Ammeters, Voltmeters etc
• Review basic concepts related to power electronics using college/university
textbooks.
• Solve problem sets on next pages and review solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
49
Copyrighted Material © 2015
Problem Set # 6.1 - Solid-state Fundamentals
Consult NCEES® Reference Handbook - Pages 212 while solving these questions
Problem 6.1 a) __________ is typically used as n-type doping material.
(A) Antimony
(B) Silicon
(C) Germanium
(D) Boron
Problem 6.1 b) ______
is typically used as p-type doping material.
(A) Antimony
(B) Silicon
(C) Germanium
(D) Boron
10
Problem 6.1 c) Intrinsic carrier concentration of a semi-conductor wafer (at equilibrium) is nj= 2 x 10
m 3. Its electron and hole mobility are pe=0.20 m V V 1and Ph=0.1 m W 1 respectively. Calculate the
conductivity of this semi-conductor wafer.
(A) 4.8 x l O 10S/m
(B) 9.6 x 10‘10S/m
(C) 1.4 x 10‘9 S/m
(D) 7.2 x 1010S/m
Problem 6.1 d) A p-n junction has acceptor concentration 2 x 1015 m 3, donor concentration of 2 x 1015
m 3 and intrinsic concentration of 1 x 1010irf3. Calculate its contact potential at 300 K.
(A) 0.525 V
(B) 0.026 V
(C) 0.634 V
(D) 0.254 V
Problem 6.1 e) Which of the following 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 between bands
Problem 6.1 f) Drift velocities of charged carriers in semi-conductors (at constant temperature) can be
increased by______________
(A) Increasing applied potential
(B) Decreasing applied potential
(C) Increasing surface area
(D) Decreasing surface area
Problem 6.1 g)________ elements of Periodic Table are typically used as n-type doping materials
(A) Group III
(C) Group V
(B) Group IV
(D) Group VIII
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Problem 6.1 h)
elements of Periodic Table are typically used as p-type doping materials
(A) Group III
(B) Group IV
(C) Group V
(D) Group VIII
Problem 6.1 i).
elements of Periodic Table are typically used as semi-conductor materials
(A) Group III
(B) Group IV
(C) Group V
(D) Group VIII
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Problem Set # 6.2 - Diodes
Consult NCEES® Reference Handbook - Pages 214 while solving these questions
Problem 6.2 a) Find the states of diodes Di and D2 in following circuit using ideal diode model
m
(A) Di On, D2Off
(C) Di Off, D2On
01
(B)
D2
On, D2 On
(D) Dx Off, D2 Off
Problem 6.2 b) Select the graph which accurately approximates Vout for following circuit
Vout
(A)
(C)
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Problem 6.2 c) Find the Q point of diode shown below assuming ideal model
m
(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 6.2 d) Find the diode current in circuit shown below using Von=0.7 V
A/W
10V
10k0
8V
(A) 0.2 mA
(B) 0 mA
(C) 0.13 mA
(D) 0.26 mA
Problem 6.2 e) Find diode currents in following circuit assuming Von = 0.7 V
m
2k®
6k0
(A) ID1= 0, lD2= 0
(B) 101=0, lD2= 2.4 mA
(C) lDi= -2.5 mA, lD2= 2.5 mA
(D) Id i= 0, lD2= -2.5 mA
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Problem Set # 6.3 - BJTs
Consult NCEES® Reference Handbook - Pages 214 - 216 while solving these questions
Problem 6.3 a) Find the emitter current lEand VCefor BJT shown in the circuit below
(A) lE = 0, VCE= 0.5 V
(B) lE= 0, VCE= 1.75 V
(C) lE = 2.09 mA, VCE= 1.68 V
(D) lE = 1. 5mA, VCE= 2.5 V
Problem 6.3 b) Find the collector current lc and VCEfor BJT shown in the circuit below
^0
(A) lc = 0 mA, VCE=0.5 V
(C) lc = 0.75 mA, VCE= 0.1 V
(B) lc = 1.5 mA, VCE= 2 V
(D) lc = 1.33 m A, VCE= 1.3 V
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Problem 6.3 c) Find the transistor state by analyzing circuit shown below
P-100
(A) Saturation
(B) Cut-off
(C) Active region
Problem 6.3 d) Find lEand
1OV0
(D) Transistor state cannot be determined
V Ce for
BJT shown in the circuit below
10V 0
p=too
10kQ
11(0.
(A) lE= 2.4 mA, VCE= 2.7V
(B) lE =0 mA, VCE=0.2V
(C) lE =0 mA, VCE= 10V
(D) lE= 5 mA, VCE=4.5V
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Problem 6.3 e) Find lE and VEc for BJT shown in the circuit below
fk100
(A) lE = 0 mA, VEC= 5 V
(B) lE = 0 mA, VEC= 10 V
(C) lE = 3.2 mA, VEC=6.1 V
(D) lE = 1.5 mA, VEC= 7.5 V
Problem 6.3 f) Solve the circuit shown below to find lEand VEC
2kQ
20ka
u —
y y y — i
p^m
(A)Ie = 0 A ,V ec=0V
(B) lE = 7.75 mA, VEC= 5V
(C) lE = 3.3 mA, VEC= 2 V
(D) lE = 1.5 mA, VEC= 1.88 V
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Problem Set # 6.4 - MOSFETs
Consult NCEES® Reference Handbook - Pages 214 - 216 while solving these questions
Problem 6.4 a) Find the drain current lDs for following circuit (assume k =0.5 mA/V2 & VT = 1 V)
10¥q
im
(A) 0 mA
(B) 1.5 mA
(C) 2.5 mA
(D) 3.2 mA
Problem 6.4 b) Find the operational state of the transistor in circuit shown below
q J Z J L v jo -^ .
5kfi
5V
1V
K
V v so s
O S)
m
AAAr
O 4V
(A) Cut-off
(B) Triode
(C) Saturation
(D) Transistor's state cannot be determined
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la j z _
Problem 6.4 c) Find drain-to-source voltage VDSfor MOSFET circuit shown below (assume k =0.2
mA/V2 & VT = 1 V)
(A) 0. 5 V
(B) 3.28 V
(C)0V
(D) 1.75 V
Problem 6.4 d) Calculate drain current lSDfor following circuit (assume k =0.1 mA/V2 & VT = 1 V)
(A) 5 mA
(C) 0.1 mA
(B) 0.3 mA
(D)0mA
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Problem 6.4 e) Find the transistor operational parameter lDSfor circuit shown below (assume k
mA/V2 & VT = 1 V)
4V
10kQ
10kQ
20kH
10kQ
6.5V
(A) 0 mA
(B) 0.44 mA
(C) 1.5 mA
(D) 0.75 mA
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Problem Set # 6.5 - Operational Amplifiers
Consult NCEES® Reference Handbook - Pages 212 while solving these questions
Problem 6.5 a) Find the output voltage V0 in circuit shown below
10kQ
(A) 5 V
(B) -1 V
(C) 1 V
(D) -5 V
Problem 6.5 b) Find the value of resistance 'R' in the circuit shown below if V0 is 12 V.
R
(A) 1 kQ
(B) 5 kQ
(C) 10 kQ
(D) 3 kQ
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Problem 6.5 c) Find output voltage VQin the circuit shown below
sm
(A) 3 V
(B) 1 V
(C) 2.5 V
(D) 0 V
Problem 6.5 d) Find the Op-Amp output V0 in the circuit shown below
10kQ
(A) -9 V
(B) -35 V
(C) 15 V
(D) -20 V
SOkfi
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Problem 6.5 e) Find the Op-Amp output V0 in the circuit shown below
1Qka
(A) I V
(B) 3 V
(C) 2 V
(D) 0.5 V
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Problem Set # 6.6 - Instrumentation
Consult NCEES® Reference Handbook - Pages 123 -125 while solving these questions
Problem 6.6 a)_________ is not an example of a transducer.
(A) Microphone
(B) Thermocouple
(C) Vernier Calliper
(D) Photo diode
Problem 6.6 b) A 200 Q RTD (@ 20°C) has a temperature coefficient of 0.0039 °C'1. Find the new
resistance if it is now placed in 35°C environment
(A) 188 Q
(B) 211 Q
(C) 200 Q
(D)255 O
Problem 6.6 c) Find Rxif Wheatstone bridge shown below has Ri = 100 Q, R2= 1000 Q & R3= 500 Q
I--------- = = ] _____,
(A)5000 Q
(B) 200 Q
(C)5 0 Q
(D) 100Q
Problem 6.6 d) Find the incremental resistance AR if the voltage difference in the circuit below is 0.
(A) 400 Q
(B) 200 Q
(C) 100 Q
(D) 0 Q
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Problem 6.6 e) Which of the following devices can be used for resistance measurement?
(A) Wheatstone Bridge
(C) RTD
(B) Strain Gage
(D) Options A, B and C are correct
Problem 6.6 f) Find the voltage across 10 kQ resistor if the voltmeter has 100 kQ resistance
10kfi
(A) 2.5 V
(C) 1 V
20kQ
5Ok0
(B) 1.98 V
(D) 5.5 V
Problem 6.6 g) Find the voltage across 10 kQ resistor in Problem 6.6 f) if 500 kQ Voltmeter is used
(A) 2.1 V
(C) 1 V
(B) 1.98 V
(D) 5.5 V
Problem 6.6 h) Find the percentage error in calculating current using 50 Q ammeter as shown in circuit
below?
(A) 1 %
(B) 2 %
(C) 2.5 %
(D) 5 %
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Problem Set # 6.7 - Power Electronics
Problem 6.7 a) A power electronics converter that changes DC to DC is called
(A) Inverter
(B) Chopper
(C) Rectifier
(D) Cycloconverter
Problem 6.7 b) A power electronics converter that changes DC to AC is called
(A) Inverter
(B) Chopper
(C) Rectifier
(D) Cycloconverter
Problem 6.7 c) A power electronics converter that changes AC to DC is called
(A) Inverter
(B) Chopper
(C) Rectifier
(D) Cycloconverter
Problem 6.7 d) A power electronics converter that changes AC to AC is called
(A) Inverter
(B) Chopper
(C) Rectifier
(D) Cycloconverter
Problem 6.7 e) A single phase half-wave controlled rectifier (operating at 110 V) is connected to a 5 Q
resistor. Select the output voltage graph if firing angle is 45°
Vout
Vout
(B)
(A)
Vout
(C)
Vout
k
w
\
(D)
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Copyrighted Material © 2015
Chapter # 7 - Power
Key Knowledge Areas*
NCEES® FE Reference Handbook
Page#
Section
203 - 204
203 - 204
Electrical and Computer
204
Engineering
204
203
Concepts
Single phase and three phase
Transmission and distribution
Transformer
Motors and generators
Power factor
Note: Specific details are not available in NCEES®
FE Reference Handbook
Voltage regulation
Facts about this section
• 8 - 1 2 questions can be expected (according to NCEES® FE Specification).
• Difficulty level of this section is rated 'Medium' by the author. Students having a
major in power engineering may find it easy.
Tips for preparing this section
• Understand the concepts and formulas found on above mentioned pages in
NCEES® Reference Handbook.
• Learn how to convert quantities between A - Y by paying special attention to
phase angle and V3 factor.
• Transformer impedance varies depending on view point (primary or secondary).
• Reactive power required to bring power factor angle from i^to d 2 is given by:
Q = P (ta n
— tan i92 ) where Q = coCV2
• Voltage regulation is given by equation below:
Vs.nl — Vs,fl
V . R = ---- — — J - x 100%
Vs, fl
• Review basic power related concepts using college/university textbooks.
• Solve problem sets on next pages and review solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
66
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Problem Set # 7.1 - Single Phase Power
Consult NCEES® Reference Handbook - Page 203 - 204 while solving these questions
Problem 7.1 a) Find the real power supplied by a generator to single phase load 20 + 5j Q operating at
120 V if the line impedance is 2 + 2j Q.
(A) 592 W
(B) 698 W
(C) 677 W
(D) 225 W
Problem 7.1 b) Find the current passing through circuit shown below
5ft
--------------- w
«
\ --------------- j
s ©
E 1Q
(A) 2/80° A
(B) 1.96/68.7° A
(C) 2/11.3° A
(D) 1Q/-1Q0 A
Problem 7.1 c) Find the apparent power supplied by current source to the circuit shown below
(A) 20/90° VA
(C) 2QQ/-67.30 VA
(B) 100/0? VA
(D) 50/0? VA
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Problem 7.1 d) Calculate the total power absorbed by loads shown in the circuit below
10C2
SJQ
(A) 12 W
(B) 9 W
(C) 5 W
(D) 7 W
Problem 7.1 e) Find the average power absorbed by 5Q resistor shown in the circuit below
(A) 1 W
(B) 5 W
(C) 25 W
(D) 0 W
Problem 7.1 f) Determine the impedance Z required for maximum power transfer in circuit shown
below
(A) 10 O
(C) -j Q
(B) 12 Q
(D) 12-j Q
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Problem 7.1 g) Determine the impedance Z required for maximum power transfer in circuit shown
below
2D
(A) 5 + 2j Q
(B) 5 + 0.6j Q
(C) 5 Q
(D) 1.66j Q
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Problem Set # 7.2 - Three Phase Power/Transmission & Distribution
Consult NCEES® Reference Handbook - Page 203 - 204 while solving these questions
Problem 7.2 a) Calculate the line voltages of a balanced 3 cj>Y- connected system if Van = 120/30° V
(A) 120/30° V, 120/-30° V, 120/150° V
(B) 208/30° V, 120/-30° V, 120/150° V
(C) 208/60° V, 208/-60° V, 208/180° V
(D) 120/60° V, 120/-60° V, 120/180° V
Problem 7.2 b) A balanced 3- <j>Y- connected load with
= 20 + 5j Q is connected to a positive-
sequence balanced 3 cj>Y- connected source Van = 120/30° V. Calculate phase current lan
(A) 5.8/-1Q40 A
(B) 5.8/16° A
(C) 22/30° A
(D) 22/-12Q0 A
Problem 7.2 c) Calculate the line current lL supplied by a balanced positive-sequence 3 cj>Y- connected
source Van =277/0° V to a balanced 3 cj) Y network with Z^, = 5 + 5j Q and Z|ine = 1 + lj O.
(A) 15/0^ A
(B) 195/-150 A
(C) 33/-450 A
(D) 3.3/-450 A
Problem 7.2 d) Calculate the loadvoltage of a positive sequence balanced 3 cj) Y-Y network consisting
of Van = 120/60° V source and Z|ine =2+ lj Q and Z|0ad = 10 + lOj Q.
(A) 104/63° V
(B) 120/30° V
(C) 16.5/44.5° V
(D) 110/53° V
Problem 7.2 e) Calculate the load impedance of a positive sequence balanced 3 cj) Y-Y system with
source voltage Van = 120/0° V, line current lan = 5/-5°A and Zime =0.5 + 0.25j Q.
(A) 10.5/55° Q
(B) 23.4/4.2° Q
(C)7.8/9fQ
(D) 13/50° Q
Problem 7.2 f) Calculate line current la in a network comprising of A source V ab =208/30° V powering a
3 (f>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
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Problem 7.2 g) Calculate the line current la provided by a balanced 3 cj) Y-connected source with Van =
120/0° V to a balanced 3 cj>A connected load having per phase
(A) 20/12° A
(B) 15/45° A
(C) 25.8/-250 A
(D) 35.3/-11.30 A
Z p h a se
= 10 + 2] Q.
Problem 7.2 h) Calculate the equivalent wye load for a load network consisting of balanced 3 cj) Yloads in parallel with balanced 3 <J) A loads if ZphaSe-Y = 10 + 5j Q & ZPhase-A = 6 + 9j Q.
(A) 2.8 Q
(B) 4.2 Q
(C) 8.4 Q
(D) 12.6 Q
Problem 7.2 i) Calculate the equivalent A load in Problem 7.2h).
(A) 2.8 Q
(B) 4.2 Q
(C) 8.4 Q
(D) 12.6 Q
Problem 7.2 j) A positive sequence balanced 3- <
j>Y-Y network has source voltage Van = 120/0° V
feeding load impedance 20/0° Q. Calculate the power generated by source if Z|ine = 0.
(A) 720 VA
(B) 2400 VA
(C) 360 VA
(D) 2160 VA
Problem 7.2 k) A 1 cj>distribution line carries 400 A to a food processing plant consuming 200 kW at
0.83 pf lagging. Calculate the load voltage.
(A)500V
(B)200V
(C) 600 V
(D) 0.5 V
Problem 7.2 I) Calculate 1 <j> distribution line losses if it provides 200 A to a small municipality
(consuming 100 kW ) at 600 V and 0.85 pf lagging.
(A) 2 kW
(B) 20 kW
(C) 10 kW
(D) 0 kW
Problem 7.2 m) Calculate the load voltage at service entrance of a hospital (consuming 125 kW) which
is being fed by a lossless 3 cf>distribution line providing 300 A and 0.694 power factor.
(A)416V
(B)200V
(C)289 V
(D) 600 V
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Problem Set # 7.3 - Voltage Regulation
Problem 7.3 a) A 2.5 MVA 13.8 kV/600 V power transformer has no-load secondary voltage of 650 V.
Calculate its voltage regulation.
(A) -5%
(B) 2.5%
(C) 8.3%
(D) 4.7%
Problem 7.3 b) For an ideal transformer, voltage regulation is _________
(A) >1
(B) > 2.5
(C) = 0
(D) * 0
Problem 7.3 c) A 10 kVA 4000 V/400 V transformer has equivalent series impedance of 1 + 0.5j Q.
Calculate voltage regulation at 0.85 lagging.
(A) 6.3%
(B) 5%
(C) 3.5%
(D) 1.5%
Problem 7.3 d) A 5 MVA 12000 V/240 V power transformer has voltage regulation of 5%. Calculate the
no-load voltage rating.
(A) 276 V
(B) 252 V
(C) 360V
(D)240V
Problem 7.3 e) A leading power factor will result in _______ voltage regulation.
(A) >0
(B) <0
(C) =0
(D) >1
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Problem Set # 7.4 - Transformers
Consult NCEES® Reference Handbook - Page 204 while solving these questions
Problem 7.4 a) A distribution transformer has 50:1 turns ratio. Calculate primary voltage if 1 A current
flows through a 12 Q load connected on the secondary.
(A) 50 V
(B) 600 V
(C) 12 V
(D) 1 V
Problem 7.4 b) Find the rated secondary current for a single phase 600 V / 120 V 15 kVA transformer.
(A) 125 A
(B) 7.2 A
(C) 2.5 A
(D) 15 A
Problem 7.4 c) A 5 kVA transformer has 10:1 turns ratio. Calculate its primary current if rated
secondary voltage is 120 V.
(A) 4.16 A
(B) 8 A
(C) 12 A
(D) 41.6 A
Problem 7.4 d) Calculate the impedance seen by the transformer primary in figure shown below.
50:1
Zp
10Q
(A)2500Q
(B) 10000 Q
(C)50000 Q
(D)25000 Q
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Copyrighted Material © 2015
Problem 7.4 e) Calculate the impedance as seen by transformer secondary in figure shown below
10:1
lOkn
Zs
iokn
(A) 5 kQ
(B) 50 Q
(C) 1 kQ
(D) 10 kQ
74
Copyrighted Material © 2015
Problem Set # 7.5 - Motors & Generators
Consult NCEES® Reference Handbook - Page 204 while solving these questions
Problem 7.5 a) Calculate the number of poles present in a 3-cJ) 480 V induction motor with
synchronous speed of 1800 rpm (operates at 60 Hz).
(A) 1
(B) 2
(C) 4
(D) 8
Problem 7.5 b) Find the synchronous speed of a 3-<J>600 V 2 pole induction motor operating at 60 Hz.
(A) 7200 rpm
(B) 1800 rpm
(C) 3600 rpm
(D) 1200 rpm
Problem 7.5 c) An existing synchronous motor 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 7.5 d) A 2 pole induction motor operating at 60 Hz has a rotational speed of 3400 rpm.
Calculate its slip.
(A) 5.8%
(B) 5%
(C) 4.5%
(D) 5.5%
Problem 7.5 e) Calculate the rotational speed of a 60 Hz, 4 pole 3 phase induction motor operating at
2.3 kV and full load slip of 0.1.
(A) 1800 rpm
(B) 1980 rpm
(C) 1720 rpm
(D) 1620 rpm
75
Copyrighted Material © 2015
Problem Set # 7.6 - Power Factor
Consult NCEES® Reference Handbook - Page 203 while solving these questions
Problem 7.6 a) Find the value of a capacitor bank (at 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 nF
(B) 293 \if
(C) 703 \x¥
(D) 586 |iF
Problem 7.6 b) An automotive plant having a 60 Hz induction motor (480 V) has a lagging power factor
of 0.6 and consumes 100 kW. Calculate the magnitude of reactive power that must be provided in
order to increase power factor to unity.
(A) 33 kVAR
(B) 133 kVAR
(C) 15 kVAR
(D) 0 kVAR
Problem 7.6 cf A 750 |iF capacitor bank is to be connected in parallel to an industrial load drawing 75
kW at 0.80 power factor lagging. Calculate the power factor after connecting capacitor bank.
(A) 0.85 lagging
(B) 0.90 lagging
(C) 0.99 lagging
(D) 0.95 lagging
Problem 7.6 d) A three phase power utility provides 75 kVA at 0.85 lagging power factor and 35 kVA at
lagging power factor at 0.75 lagging power factor to a customer at 600 V and 60 Hz. Calculate the
overall power factor.
(A) 0.75 lagging
(B) 0.82 lagging
(C) 0.85 leading
(D) 1.0
Problem 7.6 e) A balanced 3-phase positive sequence source Van = 120/0° V is connected to a three
phase Y-connected load that consumes 125 kW at 0.85 power factor lagging. Calculate the required per
phase capacitance of a balanced Y-connected capacitor bank if the power factor has to be improved to
0.95 lagging.
(A)10 nF
(B) 8 |iF
(C) 15 mF
(D) 7 mF
76
Copyrighted Material © 2015
Cfaauter # 8 - Electromagnetics
1----------
------
'J l
........................................
•
................................
rr—
-
...................................................... r
■
Key Knowledge Areas*
NCEES® FE Reference Handbook
Section
Page#
205
Electrical and Computer
199-201
Engineering
205
205
Concepts
Maxwell equations
Electrostatics/magnetostatics
Wave propagation
Transmission line
Electromagnetic compatibility
Note: Specific details are not available in NCEES®
FE Reference Handbook.
Facts about this section
• 5 - 8 questions can be expected (according to NCEES® FE Specification).
• Difficulty level of this section is rated 'Easy' by the author.
Tips for preparing this section
• Understand the concepts and formulas found on above mentioned pages in
NCEES® Reference Handbook.
•
_»
V x
E
= ——
—*
• Vx H
is vector form of Faraday's Law
-* dD
= J + —
o t
•
V.D = p
•
V.B =
is vector form of Ampere's Law
is vector form of Gauss' Law for electric fields
0 is vector form of Gauss'
Law for magnetic fields
• Units of measurement and direction for quantities such as force, electric field
strength etc shall be handled carefully in calculations.
• Read "Transmission Line" questions thoroughly to distinguish between
characteristic impedance and load impedance.
• Review basic concepts related to Electromagnetic Compatibility using
college/university textbooks.
•
Solve problem sets on next pages and review solutions atthe end of this book
♦Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
77
Copyrighted Material © 2015
Problem Set # 8.1 - Maxwell Equations
Consult NCEES® Reference Handbook - Pages 34 and 205 while solving these questions
..>
Problem 8.1 a) Calculate the divergence of electric field E given by following equation:
E = 3xi + 2y 2j + xk
(A) 3x + 2y2+ x
(B) 3 + 4y2 + 1
(C) 0
(D) 3 + 4y
Problem 8.1 b) Calculate the divergence of a vector field D given by following equation:
D = xyi + y z j + x z 2k
(A) xy + yz + xz2
(B) y + z + 2xz
(C) x + z +2x
(D) 0
Problem 8.1 c) Calculate net electric flux from a hollow sphere containing four point charges Qa =+1
nC, Qb =+3 nC and Qc =+1 nC and Qd =-2 nC
(A)OVm
(B) 1017 Vm
(C) 339 Vm
(D) 678 Vm
Problem 8.1 d) Calculate net flux from a hollow sphere of radius 5 cm containing two concentric
spheres. Inner sphere has radius of 1 cm with surface charge density of 3 \iC/m2 and the outer sphere
has radius of 3 cm with surface charge density of -5 |iC/m2.
(A) 1250 Vm
(B) -5963 Vm
(C) 3750 Vm
(D) 410 Vm
Problem 8.1 e) Calculate the value of "a" given that B is a magnetic field
B = 3axi + 2yj — 2zk
(A) 3
(B) 1
(C) 0
(D) -1
Problem 8.1 f) Determine which of the following vector fields can be magnetic in nature:
B = 2y 2i + 3xj - 2xyk
A = 2x 2i + y j + 3z 2k
(A) A
(B) B
(C) Both A and B
(D) Neither
nor B
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Copyrighted Material © 2015
Problem 8.1 g) A changing magnetic field is inducing an electric field E given by equation below
E — 2yzi + 3x 2y j + x 2y 2k. Calculate time rate of change of magnetic field
(A)—2y z i + 3x 2y j + x 2k
(B) —2x 2yi — (2y - 2 xy 2) j - ( 6xy — 2z )k
(C) 3x 2j
(D) 0
Problem 8.1 h) Calculate curl of induced electric field produced by a changing magnetic field B given
by J? = —cos2 (3 t)k.
(A) —2cos (3t ) i
(B) sin2 (3 t ) k
(C) —6cos (31) sin (3t ) k
(D) 5sin (3 t)J
Problem 8.1 i) Calculate the voltage induced in a coil with 20 turns if flux passing through it changes
from 0.1 Wb to 1.5 Wb in 2 s.
(A) -28 V
(B) -14 V
(C) -1.4 V
(D)-10V
Problem 8.1 j) A coil of length 20 cm has 50 turns and a cross-sectional area of 2 cm2. Calculate the
induced voltage if current is increased from 50 mA to 100 mA in 1 s (assume \i = 4k x 10'7 H/m).
(A) 0.15 jiV
(C) 10 V
(B) 2.5 mV
(D) 100 mV
Problem 8.1 k) Calculate the magnetic flux density inside a torus with radius 2 cm having 50 turns if 1 A
current is passing through it.
(A) 5 T
(B)O.lmT
(C) 500 nT
(D) 30 T
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Problem 8.11) Two infinitely long parallel wires are placed as shown below. Calculate the magnetic flux
density at point 'A' due to these wires (assume ^ =4n x 10~7 H/m)
Point A
|0,0,§}
A
m m
m
(A) 5 T
(B) 2 mT
(C) 0.4 |iT
(D) 1.75 T
Problem 8.1 m) A current carrying wire in air is generating 0.1 T magnetic flux density at a radial
distance of 20 cm. Calculate the current passing through it.
(A) 20 kA
(B) 100 kA
(C) 100 A
(D) 400 A
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Problem Set # 8.2 - Electrostatics / Magnetostatics
Consult NCEES® Reference Handbook - Page 199 - 201 while solving these questions
Problem 8.2 a) Calculate the magnitude of force on charge Qi = 5 nC located at Pi (0,1, 0) due to
another charge Q2 = 10 nC located at P2 (0, 0, 2).
(A) 50 x 10~6 N
(B) 10 x 10~9N
(C) 20 x 10 6 N
(D) 90 x 10'9N
Problem 8.2 b) Calculate the electric field intensity at origin due to point charges Qi = 50 nC located at
(0,1, 0) and Q2 =-50 nC located at (0, -1, 0).
(A) 0 V/m
(C) 900 V/m
(B) 100 V/m
(D) 50 V/m
Problem 8.2 c) Calculate the electric field at a point P between two infinitely large parallel plates
located in x-y plane having charge densities ps and -p s respectively
(A) 0
(B) Ps/ e
(C) 2ps/e
(D) ps£
Problem 8.2 d) Two infinitely long parallel wires (having line charge densities of + 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 two wires
(A) 3 V/m
(B) 0 V/m
(C) 2 x lO 10V/m
(D) 1.07 x 1011 V/m
Problem 8.2 e) Calculate the force on a current-carrying conductor (2 A) of length 2 m in a uniform
magnetic field of 0.5 T (assume the angle between conductor & field to be 30°)
(A) 2 N
(B) 4 N
(C) ON
(D) 1 N
Problem 8.2 f) Calculate energy stored in magnetic field which has strength of 2 A/m in a 2 m3 volume
(assume |i =4tx x 10~7 H/m).
(A) 10 mJ
(B) 5 nJ
(C) 4 J
(D) 16 J
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Copyrighted Material © 2015
Problem Set # 8.3 - Transmission Lines and Wave Propagation
Consult NCEES® Reference Handbook - Page 205 while solving these questions
Problem 8.3 a) A transmission line with 100 Q characteristic impedance is connected to a 300 + 50j Q
load. Calculate the reflection coefficient T.
(A) 1
(B) 2
(C) 0.5
(D) 0.1
Problem 8.3 b) Calculate the standing wave ratio of a transmission line having characteristic
impedance of 50 Q and load impedance of 500 + 25j Q.
(A) 1
(B) 2
(C) 10
(D) 5
Problem 8.3 c) A transmission line has per unit length inductance of 100 mH and per unit capacitance
of 10 [if. Calculate the magnitude of load impedance that will allow a reflection coefficient I" of 0.5.
(A) 100 Q
(B) 300 Q
(C)200 Q
(D)500 Q
Problem 8.3 d) Wavelength of a transmission line having 200 Q characteristic impedance is 10 m.
Calculate the input impedance at a distance of 100 m if the line is connected to a purely resistive load
of 500 Q.
(A)1000Q
(B)3500Q
(C)200 Q
(D)500 Q
Problem 8.3 e) Calculate the load connected at the end of a transmission line (250 Q characteristic
impedance) if Standing Wave Ratio is 2
(A) 100 Q
(B) 300 Q
(C)1000Q
(D)500 Q
Problem 8.3 f) A transmission line (wavelength 20m) has a characteristic impedance of 100 Q. Find the
expression for voltage at a distance of 100 m.
[ k ) V +ei2n 4-
V ~ e - j27t
(C) V +e j200n + v - e ~ j200n
(B) V +e jl0n +
V ~ e - j1071
(D) ^ ( V +ej2n 4- V ~ e - j2n)
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Problem 8.3 g) In problem 8.3f) find an expression for current at a distance of 100
m
characteristic impedance is 100 Q).
{A) V+ejl0n + V ~ e - jl0n
(C)loo ^V'+e;107r — V ~ e ~ j 10n)
(B) ^ ( V + ejl0n + F " e " ;107r)
(D) Options A, B and C are all incorrect
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(transmission line
Problem Set # 8.4 - Electromagnetic compatibility
Problem 8.4 a) Which of the following is not an example of electromagnetic coupling path?
(A) Inductive
(B) Conductive
(C) Capacitive
(D) Options A, B and C are coupling paths
Problem 8.4 b) Electromagnetic shielding is done in order t o ____________
(A) provide additional physical protection
(C) prevent external electromagneticinterference
(B) enable effective grounding
(D) safe-guard against ultra violet exposure
Problem 8.4 c) A system having cross-talk__________
(A) is immune to external interference
(B) displays electromagnetic interference within itself
(C) facilitates communication
(D) operates optimally
Problem 8.4 d) Potential source(s) of electromagnetic compatibility problems include_______
(A) electric motors
(B) lightning
(C) arc welding
(D) Options A, B and C are all correct
Problem 8.4 e) Negative effects of Electromagnetic Interference can be mitigated by_______
(A) increasing coupling path separation
(B) hardware redundancy
(C) shielding
(D) Options A, B and C are all correct
84
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Chapter # 9 - Control Systems
Key Knowledge Areas*
NCEES® FE Reference Handbook
Section
Page #
Concepts
Block diagrams
Closed-loop I open-loop
response
Steady State Errors
Root Locus
Stability
State variables
Instrumentation, Measurement and
Control
Bode Plots
Electrical and Computer Engineering
126
126
127
128
127
128
207
Facts about this section
• 6 - 9 questions can be expected (according to NCEES® FE Specification).
® Difficulty level of this section is rated "Medium' by the author. Students having a
major in controls engineering may find it easy.
Tips for preparing this section
• Understand the concepts and formulas found on above mentioned pages in
NCEES® Reference Handbook.
® Derive transfer function of classical negative feedback control system model block
diagram given in NCEES® Reference Handbook (page 126) to gain understanding
of block diagrams and closed loop / open loop response.
• Review relevant concepts using college/university textbooks.
• Solve problem sets on next pages and review Solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
85
Copyrighted Material © 2015
Problem Set # 9.1 - Block Diagrams
Consult NCEES® Reference Handbook - Page 126 while solving these questions
Problem 9.1 a) Find the closed loop transfer function for following system
(C)
Gi(s)G2( s )
(D)
1+G1(s)G2( s ) H ( s )
(5)
1 + H (s )
Problem 9.1 b) Find the relationship between Y(s), R(s), N(s) and L(s) for following system.
( , _
1
,U 5 j
“
,
G1(s)G2(s)R(s)
H - G 1 (s )G 2 (s)/V (s) ^
G2(s)L(s)
l+ G ^ N is )
{R\ Y (
1
( *) #( »
-L
G2(S)L(S)
~ 1 + G 1(s )G2(s )N ( s ) t l + G ^ i V O )
(C)
-
_ G 1 ( s ) G 2 (s )/?(s ) - / V ( s ) G 1 ( s ) G 2 ( s ) + L ( s ) G 2 ( 5 )
.
_
i + £
i
(5 ) g 2 (S)
G 2 (5 )/ ? (5 )-iV (S )G i(S )G 2 (5 )+ L (5 )G 2 (s )
.
G 2 (5 )L (5 )
1 + G 1(S)G2(S)
l+ G ^ / V O )
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Copyrighted Material © 2015
Problem 9.1 c) Find the closed loop transfer function for following system.
(A)
(C)
G4(s )(G l ( s )-& 2 ( s ) + & 3 (5))
1 + G 4 ( s ) ( G i ( s ) — G2( s ) + G 3 ( s ) )
^i (5 )^ 2 ( 5 )^ 3 (5 ) ^ 4 (5 )
1 + Gi (5 )G 2 (5 )G 3 ( s )G 4 (s )
5
G 4 ( s ) + G i ( s ) —G 2 ( s ) + G 3 ( )
(B)
1
(D)
+G4(s)+G1(s)—G2(s)+G3(5)
______________________________ ^
4
( 5 ) ______________________________
1+ Gl(s) +G2(s)--G3(s )G4(s)
Problem 9.1 d) Find the closed loop transfer function for following system.
KGt (s)
(A)
(C)
1 +/cg1(5)h(s)
1
l+KG1(s)H(s)
(B)
(D)
1+ H (s)
^ ( 5)+//( 5)
l+ ^ G 1(s)H (s)
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Problem 9.1 e) Calculate closed loop gain of following system
(A)
Gl(s)G2(s)&3 (5)
1+G1(s)G2( s)G3( s ) H
(B)
(s)
______ Gi ( s ) G 2( s) G 3( s )______
(C)
1+ G1( s ) G2( s ) + Gi ( s ) G2( s ) G3( s )
G\ (s)G2(s)+G1(s)G2(s)G3(5)
1+G1($)G2(s)-H(s)+G1(s)G2(s)G3(5)
Gi (s)G2(s)+H(s)G3(s)
(D)
1+G1(s)G2(s)G3( s )
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Problem Set # 9.2 - Bode Plots
Consult NCEES® Reference Handbook - Page 207 while solving these questions
Problem 9.2 a) Select the correct Bode Plot for H ( s ) =
(A)
50
s+1 0
X Axis *Frequency (
§ „
Amplitude Plot
Y Axis •Magnitude <fB
(B)
X Axis
Frequency i
y Axb - Magnitude c
(C)
X Axis. • Frequency t
(D)
X Axis - Frequency <
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Problem 9.2 b) Select the correct Bode Plot (phase response) for H ( s ) = ----
S+20
X Ax?$ -
w
(A)
X Axis-Frequency u
(B)
(C)
X Axis ’ Frequency u
(D)
90
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Problem 9.2 c) Locate the pole(s) for following transfer function:
(50)(s + 2)
(s+ 100)(s + 1000)
(A) s =-100, s =-1000
(B) s =-2
(C) s =-100, s= -1000and s = -2
(D) There are no poles
Problem 9.2 d) Calculate the magnitude gain for following transfer function
s(s + 10)
(A) 10 dB
(B) 100 dB
(C) 20 dB
(D)0dB
Problem 9.2 e) Calculate the magnitude gain for following transfer function
100s
H (s) =
s 2 + 150s + 5000
(A) 100 dB
(B) 50 dB
(C) 1/50 dB
(D) -34 dB
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Copyrighted Material © 2015
Problem Set # 9.3 - Steady Sate Errors
Consult NCEES® Reference Handbook - Page 127 while solving these questions
Problem 9.3 a) Calculate the steady state error for following system if input is 10u(t)
(A)
(B) 0.24
(C) 0.11
(D) 0.33
Problem 9.3 b) Calculate the steady state error for following system if input is 5t u ( t )
(A) 00
(B) 0.24
(C) 0.11
(D) 0.33
Problem 9.3 c) Calculate the steady state error for following system if input is lO t 2u ( t )
(A) 00
(B) 0.24
(C) 0.11
(D) 0.33
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Problem 9.3 d) Calculate the steady error for following system if input is 2 tu(f)
(A) 00
(B) 0.24
(C) 0.11
(D) 0.33
Problem 9.3 e) Calculate the steady state error for following system if input is 3u { t )
(A) 00
(B) 0.24
(C) 0.11
(D) 0.33
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Problem Set # 9.4 - Routh-Hurwitz Criteria & System Stability
Consult NCEES® Reference Handbook - Page 127 while solving these questions
Problem 9.4 a) Find the Routh Array characteristic equation for following system
(A) s + k(s + l)(s + 2) —■0
(B) (s + l) ( s + 2) + s — 0
(C) s 2 + 3s + 2 + ks = 0
(D) s + k = 0
Problem 9.4 b) Compute the entry bi in Routh Array table for following closed loop transfer function
s + I
T ( s ) = -----------------------------------2s5 + 2s3 + 3s4 + 7s2 + 5 + 10
s5
s4
s3
(A)—14/3
(C) 0
2
3
2
7
1
10
bi =?
(B) - 8 / 3
(D) 8/3
Problem 9.4 c) Use Routh Hurwitz criteria to find if the system given by following closed loop transfer
function is stable
T (s) =
2s4 + 3s3 + s 2 + s + 1
(A) Stable
(B) Unstable
(C) Depends on K
(D) Cannot be determined
Problem 9.4 d) Find the range of k for which system given by following closed loop transfer function is
stable
T f .
r(s) =
(fc )O + i o )
3s3 + 5s2 + (k + 10)s + 5k
(A) - 1 < k < 0
(B) - 2 < k < 0
(C) 0 < k < 5
(D) 0 < k
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Problem 9.4 e) Find the range of k for which the following system is stable
(A) k > -17/2
17
(B) k > - 1
(D) k < —1
95
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Problem Set # 9.5 - Root Locus
Consult NCEES® Reference Handbook - Page 128 while solving these questions
Problem 9.5 a) Select the correct root locus for open-loop transfer function G ( s ) H ( s ) =
/
J.O
-2 X )
-13
-10
IP
-0.5
\
(B)
(A)
(C)
-2.5
L5
(D)
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(s )(s + 2 )
Problem 9.5 b) Select the correct root locus for open-loop transfer function G(s)H(s)
frn
hn
-4
-2
(A)
(C)
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_____ l _____
(s) (s 2+75+10)
Problem 9.5 c) Select the correct root locus for open-loop transfer function
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•S“h 5
G(s)H(s) = j 2_5s+4
Problem 9.5 d) Select the correct root locus for open-loop transfer function G (s)H (s ) =
(B)
(C)
(D)
99
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25
(s + l)(s + 2 )
3 5 -2
Problem 9.5 e) Select the correct root locus for open-loop transfer function G (s)H(s ) = S2+ 2S+1
- 30
-2 5
-20
-IS
-IQ
(A)
(C)
(D)
100
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Problem Set # 9.6 - State Variables
Consult NCEES® Reference Handbook - Page 128 while solving these questions
Problem 9.6 a) Find the state equation for following system
y' and y ” represent 1st and 2nd order differentials respectively.
y " + y ' + 2y = 2u ( t )
(A)
*2
w
(C)
-
iie k h o
r o
r3 21 r*il
II
■ SK
■r2l —
(D)
il E l - S ^ 0
L*2
Problem 9.6 b) Find the state equation for following system
2y " + 8y 1+ lOy = 6w(t)
(A)
(C)
IK
y' and y" represent 1st and 2nd order differentials respectively.
3 E W 2 “M
,a -[-° 8 - U
M
™ [a -G
- w
m [ ';l- [ - ° 5
Problem 9.6 c) Find the state equation for following system
3y ' "
6 y "
+
1 2 y '
+ 3y = 9u(t)
y',
y"& y'" represent 1st, 2nd & 3rd order differentials respectively.
-* !*
(A)
&
■ *y
(C) %2 =
*3-
(B)
3 E l + 0 “">
H
' 0
1
0
0
.- 1
—4
0 ' -xr
O'
x2 + 0
1
l
-2 . * 3J
.3.
0
*2 = O
.1
*3 ~Xt
u(t )
'0
(D) X2 = 0
.3
-* 3 -
0
1 O' *ii l
*2 + 3 u(t)
2 3. [x3J
.0.
1
0
2
O' -xr
3'
1 *2 + 0 u(t)
1. L *3.0.
Problem 9.6 d) Given the standard state-variable model for dynamic systems, derive expression for
transfer function
x ( t ) — A x (t ) + Bu(i)
y(t) = Cx(t) + Du(t)
(A)C(s/ — A ) ~ 1B + D
(B) D( sl — A)B + C
(C) B - A + D
(D) Transfer function cannot be determined
101
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Problem 9.6 e) Find the transfer function for following state space model
,a-L°4
y = [ 1
0][^] + [0]u(t)
i f T ( s ) = C ( s I - A ) - 1B + D
(A)7’( s ) = T4 —2
1nS
(C) T ( s ) =
(B) T ( s ) =
1
(D) T ( s ) =
2s—1
1 - 4 s 2- 5 s
2
s 2+5 s +4
Problem 9.6 e) Find the transfer function for following state space model
y [ n + 2] + 3y [ n + 1] + 3y [ n ] = 2u[ n]
(A)
xt [n + 1]
x 2[n + 1]
e s k h o
[-3
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Copyrighted Material © 2015
Chapter # 10 - Communications
Key Knowledge Areas*
NCEES® FE Reference Handbook
Section
Page#
Concepts
Modulation / Demodulation
Digital Communications
Fourier transforms / Fourier
Series
Electrical and Computer Engineering
208-209
206, 209
Mathematics
30-32
Note: Specific details on this topic are not available in
Nc e e S® FE Reference Handbook.
Multiplexing
Facts about this section
• 5 - 8 questions can be expected (accordingto NCEES® FE Specification).
• Difficulty level of this section is rated 'Medium' by the author. Students having a
major in communications engineering may find it easier.
Tips for preparing this section
• Understand the concepts and formulas found on above mentioned pages in
NCEES® Reference Handbook.
• Familiarize yourself with graphical illustrations of different modulation techniques
especially Frequency and Phase Modulation since they appear very similar.
• Study the steps involved in digital communication (PCM, PAM etc).
• Learn how to calculate Fourier Transform/Series using table.
• Develop understanding of the two main multiplexing techniques
-
Time Division Multiplexing
Frequency Division Multiplexing
• Review relevant concepts using college/university textbooks.
• Solve problem sets on next pages and review solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
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Copyrighted Material © 2015
Problem Set # 10.1 - Amplitude Modulation
Consult NCEES® Reference Handbook - Page 208 while solving these questions
Problem 10.1 a) Calculate transmitted signal power if a1000 W carrieris modulated at 80%
(A) 1320 W
(B) 800 W
(C) 390 W
(D)200 W
Problem 10.1 b) 5 kW signal power is transmitted 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 10.1 c) A Costas-loop is used for detecting a signal modulated with which of the following
modulation technique?
(A) Single-Sideband Amplitude Modulation
(B) AngleModulation
(C) Frequency Modulation
(D) Double Sideband Amplitude Modulation
Problem 10.1 d) A message signal Ssin2n(1000t)\s amplitude modulated using a carrier signal of
S0sin2n(40000t). Calculate the modulation index.
(A) 20%
(B) 100%
(C) 10%
(D) 50%
Problem 10.1 e) Calculate the 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%
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Problem 10.1 f) Which of the following options represent amplitude modulated wave for the carrier
and modulating wave shown below?
Carrier Wave
(A)
(B)
(C)
(D) None of the above
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Problem Set # 10.2 - Angle Modulation
Consult NCEES® Reference Handbook - Page 208 - 209 while solving these questions
Problem 10.2 a) Calculate the 98% power bandwidth of a frequency modulated signal with frequency
deviation ratio of 1.25 and message bandwidth of 10 kHz
(A) 20000 Hz
(B) 10000 Hz
(C) 125000 Hz
(D) 45000 Hz
Problem 10.2 b) Angle modulated signals can be demodulated using__________
(A) Costas loop
(B) Phase-lock loop
(C) Envelope detection
(D) Sampling
Problem 10.2 c) In Problem 10.If) which option represents a frequency modulated wave
(A) Option A
(B) Option B
(C) Option C
(D) None
Problem 10.2 d) In Problem 10.If) which option represents a phase modulated wave
(A) Option A
(B) Option B
(C) Option C
(D) None
Problem 10.2 e) Calculate the 98% power bandwidth of 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
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Problem Set # 10.3 - Pulse Code Modulation (PCM) & Pulse
Amplitude Modulation (PAM)
Consult NCEES® Reference Handbook - Page 209 while solving these questions
Problem 10.3 a) ________ quantization levels can be represented by a binary word of length 7 bits
(A) 7
(B) 27
(C) 14
(D) 0
Problem 10.3 b) Calculate the minimum bandwidth required to transmit a pulse code modulated
message m(t), with M(f) = 0 for f >= 100 Hz using 256 quantization levels.
(A) 481 Hz
(C) 25600 Hz
(B) 51200 Hz
(D) 1600 Hz
Problem 10.3 c) A PAM system modulates 15 kHz signal by sampling it using aclock at equal time
intervals. Calculate the minimum clock frequency of this PAM system.
(A) 15 kHz
(B) 30 kHz
(C) 7.5 kHz
(D) 150 kHz
Problem 10.3 d) In Problem 10.3c) calculate the time spacing between adjacent samples of the pulseamplitude modulated wave form.
(A) 66 \xs
(B) 50 [is
(C) 33.3 {is
(D) 75 ^s
Problem 10.3 e) __________ is not part of pulse code modulation process.
(A) Quantization
(C) Sampling
(B) Encoding
(D) Filtering
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Problem Set # 10.4 - Fourier Transforms
Consult NCEES® Reference Handbook - Page 30 - 32 while solving these questions
Problem 10.4 a) Calculate Fourier Transform of function shown in the figure below
Ampltode
5
Time
-4
-3
-2
1
*i
2
3
(A) Ssinc (4/)
(B) 40 sine(8/)
(C) 8 sinc(8/ )
(D) 5 sinc(40/)
4
Problem 10.4 b) Calculate Fourier Transform of function shown in the figure below
Amplitude
3
Time
1
(A) 12 s
i
2
n
3
c
(C) sinc(12/ )
(
4
4
(
B
)
4 sinc(4/)
(D) sinc(4/) e~40‘2^/)
Problem 10.4 c) Calculate Fourier Transform of function given below
x ( t ) = cos(27r(300)t) n ^
(A) 6 sinc(6/ )
(B) 6 sinc(6(/ -- 300))
(C) 3 sinc(6(/ — 300)) + 3 sinc(6(/ + 300))
(D) 6 sinc(6/)e -2 7 T / 3 0 0
Problem 10.4 d) Calculate Fourier Transform of function given below
x ( t ) = e~5t cos(27r(20)t) u ( t )
(A)— -—
(b ) ------ -------1
_____ i_____
' 1 j 2 n f +5
' ' ;2 7 r(/ -2 0 )+ 5
(C) — -— e~40njf
(D)- f ------------ + ------ -------)
j2 n f+ S
2 V ;2 t t (/ - 2 0 ) + 5
j2 n (f+2 0 )+ 5
;2 7 t(/+ 2 0 )+ 5 /
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Problem 10.4 e) Calculate Fourier Transform of function given below
^w =4 II(t
(A) 4 sinc(2/ )
(C) 8 sinc(2 f ) e ~ j4nf
)
(B) 8 sinc(2 ( f — 2))
(D) 2 sinc(2/)
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Problem Set # 10.5 - Multiplexing
Problem 10.5 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 of digital multiplexing
Problem 10.5 b) Three channels are to be frequency multiplexed together
Channel
1 has a Band Width of 50 kHz
Channel
2 has a Band Width of 100 kHz
Channel
3 has a Band Width of 50 kHz
Calculate the 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 10.5 c) Five channels are multiplexed using time division multiplexing. Every channel sends 10
bytes/second. Calculate the frame size if the system can multiplex 1 byte/channel.
(A) 10 bytes
(B) 1 bytes
(C) 5 bytes
(D) 50 bytes
Problem 10.5 d) Calculate the bit rate in Problem 10.5 c)
(A)
100 bps
(C) 500 bps
(B) 400 bps
(D) 1000 bps
Problem 10.5 e) Calculate the bit duration of a time division multiplexer that multiplexes three 50 kbps
channels using 1 bit time slots.
(A) 50000 bps
(B) 150000 bps
(C) 1000 bps
(D) 25000 bps
Problem 10.5 f) Calculate the frame duration in Problem 10.5 e)
(A) 1 \xs
(C) 20
[is
(B) 10[is
(D) lOOpis
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Copyrighted Material © 2015
Chapter # 11 - Computer Networks
Key Knowledge Areas*
NCEES® Reference Handbook
Section
Page#
Concepts
Routing and Switching
Network Topologies
Local area networks
Note: Specific details on this topic are not
available in NCEES® FE Reference Handbook.
F a c ts a b o u t th is s e c tio n
• 3 - 5 questions can be expected (according to NCEES® FE Specification).
• Difficulty level of this section is rated 'Easy' by the author. Students having a
major in computer engineering may find it easier.
T ip s f o r p re p a r in g th is s e c tio n
• Understand basic computer networking concepts while paying special attention
to the topics mentioned above.
• Familiarize yourself with routing/switching process including data forwarding,
routing tables etc.
• Study the differences between major network topologies (bus, ring, star etc).
• Gain understanding of OS I Model.
• Review relevant concepts using college/university textbooks.
• Solve problem sets on next pages and review solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
I ll
Copyrighted Material © 2015
Problem Set # 11.1 - Routing and Switching
Problem 11.1 a) Process of finding efficient paths between nodes based on address is called___
(A) Multiplexing
(B) Buffering
(C) Addressing
(D) Routing
Problem 11.1 b) Which of the following table(s) is maintained by a router?
(A) Time table
(B) Forwarding table
(C) Routing table
(D) Options C & B are correct
Problem 11.1 c) In which of the following OSI Model Layers does routing takes place?
(A) Application Layer
(B) Data Link Layer
(C) Network Layer
(D) Session Layer
Problem 11.1 d )_______ is used to connect two or more networks for transferring data packets
(A) Router
(B) Hub
(C) Switch
(D) Options A, B and C are all correct
Problem 11.1 e)__________ is used to create a network and allow devices within a network to
communicate with each other
(A) Router
(B) Hub
(C) Switch
(D) Options A, B and C are all correct
Problem 11.1 f) A network switch generally operates in which of the following OSI Model Layer?
(A) Application Layer
(B) Data Link Layer
(C) Network Layer
(D) Session Layer
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Problem Set # 11.2 - Network topologies / Frameworks / Models
Problem 11.2 a) ________ is not an example of a common network topology
(A) Bus
(B) Ring
(C) Wireless
(D) All options are examples of common network topologies
Problem 11.2b)_________ network topology has a higher security risk.
(A) Bus
(B) Ring
(C) Star
(D) Wireless
Problem 11.2c) http and email will run in _________ OSI model layer
(A) Network
(B) Physical
(C) Application
(D) Session
Problem 11.2d) Language translation for applications takes place in _______ OSI model layers
(A) Presentation layer
(B) Network layer
(C) Application layer
(D) Transport layer
Problem 11.2e) Session layer is not responsible fo r_________
(A) Modulation/Demodulation
(B) Relation between two end
users
(C) Identifying users
(D) Controlling data exchanged by users
Problem 11.2f) Breaking of a single connection can disrupt an entire network for which of the following
implementations?
(A) Bus
(B) Star
(C) Ring
(D) Options A, B and C arecorrect
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Problem Set # 11.3 - Local Area Networks
Problem 11.3 a) LAN can be implemented using________ topology
A) Ring
(B) Star
C) Bus
(D) Options A, B and C are correct
Problem 11.3 b)________ is a commonly used LAN technology
A) Ethernet
C) Asynchronous Transfer Mode
(B) Wireless
(D) Options A, B and Care correct
Problem 11.3 c) Ethernet is an example o f __________
A) Star topology
(B) Bus topology
C) Ring topology
(D) Mesh topology
Problem 11.3 d) ATM LAN is an example o f__________
A) Star topology
(B) Bus topology
C) Ring topology
(D) Mesh topology
Problem 11.3 e)________ serves the largest geographical region
A) LAN
(B) WAN
C) MAN
(D) PAN
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Chapter # 12 - Digital Systems
Key Knowledge Areas*
NCEES® FE Reference Handbook
Section
Page#
Concepts
Number systems
Boolean logic
Logic gates and circuits
Logic minimization (k-maps)
Flip-flops and counters
State machine design
Programmable logic devices
Data path/controller design
Timing
Electrical and Computer Engineering
217-218
Mathematics
21
Note: NCEES® FE Reference Handbook does not
contain specific details on this topic
Facts about this section
• 7 - 1 1 questions can be expected (according to the NCEES® FE Specification).
• Difficulty level of this section is rated 'Hard' by the author. Students having a
major in computer engineering may find it easier.
Tips for preparing this section
• Understand basic computer networking concepts while paying special attention
to the topics mentioned above.
• Use calculator for converting numbers between different bases where possible.
• Study truth tables of all logic operations especially XOR and XNOR.
• Calculate l's complement and 2's complement carefully.
• Relate graphical symbols with applicable gates, devices etc.
• K-map terms should be carefully grouped by applying correct techniques
• Gain understanding of how different flip-flops and counters work.
• Learn how to navigate between state diagram and state table.
• Get familiar with programmable logic devices, controllers & timing diagrams
• Review relevant concepts using college/university textbooks.
• Solve problem sets on next pages and review solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
115
Copyrighted Material © 2015
Problem Set # 12.1 - Number Systems
Consult NCEES® Reference Handbook - Page 217 while solving these questions
Problem 12.1 a) Convert 8ADHex to binary
(A)1000101011012
(B)IIOIIOIOOIOI2
(C)1010101101002
(D)IOIIIOIOIIOI 2
Problem 12.1 b) Convert 963i0to binary
(A) IOOIOOIOII2
(B) IIIIOOOOII2
(C) IIOOOOOIII2
(D) IOIOOIOOOI2
Problem 12.1 c) Calculate l's complement sum of 00102 and OOII2
(A) IIO I 2
(B) IOIO2
(C)OOIO2
(D)01012
Problem 12.1 d) Calculate l's complement sum of 11012 and 10112
(A) IOIO2
(B) IIIO 2
(C) IOOI2
(D) IOIO2
Problem 12.1 e) Calculate the 2's complement sum of 10112 and 01012
(A) OOOI2
(B) OOOO2
( C ) II II 2
(D)O III 2
Problem 12.1 f) Calculate the 2's complement sum of 11102 and 01012
(A) OOII2
(B) OIOO2
(C) IIOO2
(D) IIIO 2
116
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Problem Set # 12.2 - Boolean Logic
Consult NCEES® Reference Handbook - Page 217 while solving these questions
Problem 12.2 a) Apply DeMorgan's Theorem to the following expression
[A
+ BCD + (F + F)]
(A) { B + C + D)(F)(F) 04)
(C) I + B CD + (F + F)
(B) (A ) ( B C D ) ( E F )
(D) ( A ) ( B + C + D ) ( E ) ( F )
Problem 12.2 b) Apply DeMorgan's Theorem to following expression
[ ( A B C ) ( D + FF)]
(A) I + B + C + D { E + F)
(B) (A B C ) ( D ) ( E F )
{ C )A + B + C + DE
[D)~A~B + DEF
Problem 12.2 c) Simplify the following expression using Boolean logic?
AB + A { B C ) + B ( A + C)
(A) AB + A { B + C ) + B { A C )
(B) AB + C { A + B )
(C) A B + A B C
(D ) A B + A ( B C )
Problem 12.2 d) Simplify the following expression using Boolean algebra?
AB + B ( A + C) + C ( A + B )
(A) AB + AB + BC + AC + BC
(B) AB + BC + AC
{ C ) A B + BC
(D) A B C
Problem 12.2 e) Convert the following expression to Sum of Product form
(A + B ) ( A + B + C )
(A) A + AB + BC
(B) A + B + C
{C) A + B
(D) AB + BC
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Problem Set # 12.3 - Logic Gates
Consult
NCEES® Reference Handbook - Page 217 - 218 while solving these questions
Problem 12.3 a) Find the output expression for logic circuit shown below
(A) AB + CD
(B) A B ( C + D) + ( A + B ) C D
(C) A B ( C + D )
(D) A B ( C + D)
Problem 12.3 b) Find the output expression for logic circuit shown below
(A) 0
(B) A + B
(C) 1
[D )A + B
Problem 12.3 c) Find the output expression for logic circuit shown below
(A) A +
B
(C) A +
BC
+
C
(B) AB + C
(D) AB + BC
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Problem 12.3 d) Find the output expression for logic circuit shown below
[k)AB + C
(B) A B C
('C ) A B C
(D)A + B
+ C
Problem 12.3 e) Find the output expression for logic circuit shown below
(A) 0
(C) A~B
(B) 1
(D) A + B
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Problem Set # 12.4 - Karnaugh Maps
Consult NCEES® Reference Handbook - Page 217 - 218 while solving these questions
Problem 12.4 a) Determine the minimized Sum of Product expression for logic function given by k-map
C
0
1
00
01
AB
11
10
(A) C + AB + BC
(B) A B C +
(C)BC + AB + A B C
(ID)ABC
AB +
+ AB
BC
+BC
Problem 12.4 b) Determine the minimized Sum of Product expression for logic function given by k-map
C
0
AB
1
00
01
11
10
(A) AB
(B) A B C + A B C
AB
(C M
(D M
Problem 12.4 c) Determine minimized Sum of Product expression for logic function given by k-map
CD
00
01
11
10
00
01
AB
11
10
(A) A B 4” AB
(B) B
(C)B
(D) A B + A B
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Problem 12.4 d) Determine the minimized Sum of Product expression for logic function given by k-map
CD
00 01
11
10
00
01
AB
11
10
(A) AC + AB +
A BD
(C) AC + ABC 4-
(B) A + AB + BCD
A BD
(D)
AB +
ABC +
A BD
Problem 12.4 e) Determine the minimized Sum of Product expression for logic function given by k-map
CD
00 01
11
10
00
01
AB
11
10
{A )A B + CD
(B) C D + CD + BC
(C) D + BC
(D) A B C + D
Problem 12.4 f) Determine the minimized Sum of Product expression for logic function given by k-map
C
0
1
00
01
AB
11
10
(A) B + AC
1C) A B C
(B) A C + ABC
+ AC
(D) C +
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AC
Problem Set # 12.5 - Flip-flops and counters
Consult NCEES® Reference Handbook - Page 218 while solving these questions
Problem 12.5 a) Determine the output sequence of Flip Flop # 2 shown in the circuit below
f UP FLOP #1
101 —
111
D
>
fUf»R0i»#2
D
0
Q --------------------1
111
>
(A) 101
(B) 111
(C) 010
(D) 000
0
Q
Problem 12.5 b) Identify the circuit shown below (assume that clock is applied).
(A) JK Flip Flop
(B) RS Flip Flop
(C) D Flip Flop
(D) Multiplexer
Problem 12.5 c) Identify the circuit shown below (assume that clock is applied)
(A) JK Flip Flop
(B) RS Flip Flop
(C) D Flip Flop
(D) Multiplexer
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Problem 12.5 d) Identify the circuit shown below (assume that clock is applied)
(A) JK Flip Flop
(B) RS Flip Flop
(C) T Flip Flop
(D) Multiplexer
Problem 12.5 e) Determine the output sequence (Q) of RS Flip Flop shown in the circuit below
(A) 100
(B) 001
(C)111
(D)110
Problem 12.5 f) Find the final state (Q1Q2Q3) of the counter after three clock cycles. Assume that initial
states of Q1Q2Q3 are zero.
(A) 000
(B) 111
(C) 001
(D) O il
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Problem 12.5 g) Find the final state Q0Q 1Q2Q3 of counter given below after three clock cycles. Assume
that initial states of Q0Q1Q 2Q3 are 1111.
ill
(A)1000
(B)0000
(C)1010
(D)1100
Problem 12.5 h) Find the output sequence of the counter given below. Assume that initial states of
Q0Q1 are 00.
1111
(A) (0 Ojinitial state/ (0 l)cycle 2/
(1 0 ) Cyde 3/ (1 l)cycle4
(B) (0 0) initial state/ ( 10 ) cycle 2/
( o i ) cycle 3/
(i i) cycle 4
(C)(1 1) initial state/
(0 1) cycle 2/
( 1 0 ) cycle 3/
(0 0) cycle 4
(D)
( o i ) cycle 2/ ( 1 0 ) cycle 3/
(0 0) cycle 4
(1 1)
initial state/
Note - Above options are shown in (Q0Q 1) format.
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Problem Set # 12.6 - State Machine Design
Consult NCEES® Reference Handbook - Page 21 while solving these questions
Problem 12.6 a) Complete the state table for Finite State Machine shown below
Next State
Present
State
W=0
W=1
Output
Z
A
C
B
0
B
C
B
1
C
X
X
X
(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 = 1 (A), z = 0
Problem 12.6b) In Problem 12.6a) an input sequence of W = 111 is applied to the Finite State
Machine. Calculate the output sequence if the machine is initially in state A
(A)111
(B)010
(C) 000
(D)011
Problem 12.6 c) In problem 12.6a) an input sequence of W =000 is applied to the Finite State Machine.
Calculate the output sequence if the machine is initially in state A
(A)101
(B)000
(C) 010
(D)111
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Problem 12.6 d) Complete the state table for Finite State Machine shown below
RESET
Next State
Present
State
ab=00
01
10
11
Output
W
A
A
A
C
B
0
B
B
A
C
B
1
C
X
X
C
A
0
D
D
D
C
B
0
(A) 00 - C, 01 —D
(B) 00 —C, 01 - C
(C) 00 - D, 01 - C
(D) 00 —A, 01 - D
Problem 12.6 e) States A, B, C and D of problem 12.6 d) are represented by y2yi - 00, 01,10,11
respectively in the table below. Select the output expression for 'w' as a function of present state.
Present
State
yzVi
Next State
ab=00
01
10
11
Output
w
00
00
00
10
01
1
01
01
00
10
01
1
10
10
11
10
00
0
li
11
11
10
01
0
(A) w = y 2 y 1
(B) w = yx
(C) w -
(D)w = y 2
y 2 y±
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Chapter # 13 - Computer Systems
Key Knowledge Areas*
NCEES® FE Reference Handbook
Section
Page#
Concepts
Architecture
Microprocessors
Memory technology
Interfacing
Note: NCEES® FE Reference Handbook does not
contain specific details on this topic
Facts about this section
• 4 - 6 questions can be expected (according to NCEES® FE Specification).
• Difficulty level of this section is rated 'Easy' by the author. Students having a
major in computer engineering may find it easier.
Tips for preparing this section
• Understand basic computer system concepts while paying special attention to the
topics mentioned above.
• Study computer architecture & identify functional units (ALU, I/O, CU etc).
• Gain fundamental understanding of microprocessors.
• Familiarize yourself with different memory technologies (RAM, ROM, EROM etc).
• Learn how different components of computer system interface with each other.
• Review relevant concepts using college/university textbooks.
• Solve problem sets on next pages and review solutions at the end of this book
♦Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
127
Copyrighted Material © 2015
Problem Set # 13.1 - Architecture & Interfacing
Problem 13.1 a)______________________________is not an example of addressing mode
(A) Register
(C) Indirect
(B) Immediate
(D) Options A, B and C are examplesof addressingmodes
Problem 13.1 b)_is a computer program that converts high level language into machine language
(A) Compiler
(B) Assembler
(C) Interpreter
(D) Translator
Problem 13.1 c) The section of CPU that performs mathematical calculations is called______
(A) Control Unit
(B) I/O Unit
(C) Memory Unit
(D) ALU
Problem 13.1 d) Batch data processing is useful for applications involving______
(A) Quarterly bank statements
(C) Weather monitoring
(B) Stock market quotation
(D) Options A, B and C are examples of batchdataprocessing
Problem 13.1 e) Real time data processing is useful for applications______
(A) Quarterly bank statements
(C) Weather monitoring
(B) High school grade reports
(D) Options A, B and C are examplesof real timedata
processing
Problem 13.1 f) Basic Input Output System (BIOS) is read from ______ during normal startup routine
(A) RAM
(B) ROM
(C) USB
(D) CD
Problem 13.1 g)_______ is the process of transforming data into different format for another system
(A) Encryption
(B) Encoding
(C) Hashing
(D) Decoding
Problem 13.1 h)________ is the process of transforming data for secrecy
(A) Encryption
(B) Encoding
(C) Hashing
(D) Decoding
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Problem 13.1 i) Instruction pipelining results in __________
(A) decreased instruction execution time
(B) increased instruction throughput
(C) allows new types of possible instructions
(D) all of the above
Problem 13.1 j) The control unit of a computer system is responsible for
(A) storage
(B) interpreting program instructions
(C) mathematical operations
(D) input/output operations
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Problem Set # 13.2- Microprocessor
Problem 13.2 a) A microprocessor register that stores address of the last requested program in a
buffer register is called________
(A) Program counter
(B) Stack pointer
(C) Instruction pointer
(D) Accumulator
Problem 13.2 b) A microprocessor register that stores address of current or next instructions in a
buffer register is called________
(A) Program counter
(B) Stack pointer
(C) Accumulator
(D) Options A and B are correct
Problem 13.2 c) A single 1C accepting & executing coded instruction for processing data and controlling
associated circuitry in a computer system is called________
(A) Microprocessor
(B) Microcomputer
(C) Microcontroller
(D) Personal computer
Problem 13.2 d) An interconnected group of ICs, l/Os and memory systems used for data processing
and other application is called________
(A) Microprocessor
(B) Microcomputer
(C) Microcontroller
(D) Personal computer
Problem 13.2 e) An integrated system of a single 1C, I/O circuitry and memory system accepting &
executing code instructions and associated circuitry in computer system is called________
(A) Microprocessor
(B) Microcomputer
(C) Microcontroller
(D) Personal computer
Problem 13.2 f) Bus that transfers information between microprocessor & I/O units is called_______
(A) Address bus
(B) Control bus
(C) Data bus
(D) None of the above
Problem 13.2 g) Bus that selects a location for reading / writing is called________
(A) Address bus
(B) Control bus
(C) Data bus
(D) Main bus
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Problem Set # 13.3 - Memory Technology and Systems
Problem 13.3 a) Which of the following storage devices use random access methods?
(A) Cassette tape
(B) CD
(C) Hard disk
(D) Flash memory
Problem 13.3 b) A Giga Byte contains__ ____ bits
(A) 10003
(B) 2010
(C) 8 x 230
(D) 8 x 2020
Problem 13.3 c) A memory that acts as a buffer between CPU and main memory in order to speed up
processing is called__ _____
(A) DRAM (Dynamic Random Access Memory)
(B) ROM
(C) Cache Memory
(D) EPROM
Problem 13.3 d) Which of the following is an example of secondary memory?
(A) Cache memory
(B) RAM (Random
Access Memory)
(C) DVD
(D) DRAM (Dynamic Random Access Memory)
Problem 13.3 e) Which of the following ROMs can be programmed only once by the user and is non­
erasable?
(A) PROM
(C) EEPROM
(B) EPROM
(D) DRAM
Problem 13.3 f) What 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 13.3 g)________ is a memory unit which varies in size between different computer systems.
(A) Nibble
(C) Bit
(B) Word
(D) Byte
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Chapter # 14 - Software Development
Key Knowledge Areas*
NCEES® FE Reference Handbook
Section
Page#
Concepts
Algorithms
Data structures
Software design methods
Software implementation
Software testing
Note: NCEES® Reference Handbook does not contain
specific details on this topic.
Facts about this section
• 4 - 6 questions can be expected (according to the NCEES® FE Specification).
• Difficulty level of this section is rated 'Easy' by the author. Students having a
major in computer engineering may find it easier.
Tips for preparing this section
• Understand basic software development concepts while paying special attention
to the topics mentioned above.
• Learn how to dry run pseudo codes and simple algorithms including sorting
(bubble, heap etc) and searching (binary, hash etc).
• Familiarize yourself with flow chart (symbols, execution etc).
• Gain fundamental understanding of object-oriented and structured programming
including different types of data structures (static and dynamic).
• Learn key steps involved in software implementation and testing process.
• Review relevant concepts using college/university textbooks.
• Solve problem sets on next pages and review solutions at the end of this book
*Refer to pages 273-275 of NCEES® FE Reference Handbook for Electrical and Computer CBT exam
specification.
132
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Problem Set # 14.1 - Algorithms
Problem 14.1 a) Find the value of V at the end of this code
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
(A) 4
(B) -2
(C) 7
(D)9
Problem 14.1 b) In Problem 14.1a) what is the final value of 'z'?
(A) 4
(B) -2
(C)7
(D)9
Problem 14.1 c) Find the value of 'value' at the end of code given below
int value = 0
for (N = 1; N <= 100; N ++)
{N = 2n
value = N }
print value
(A) 128
(B) 2100
(C) 215
(D) 29
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Problem 14.1 d) Find value of cell A4 in the spreadsheet shown below
1
A
1
B
C
2
5
2
=B$1
3
7
3
=B$2
4
=BlxC2
5
=B$3
5
3
(A) 5
(B) 25
(C) 10
(D) 1
Problem 14.1 e) Find value of cell B4 in the spreadsheet shown below
A
B
C
1
10
=A1 +Cl
=A lx 1
2
15
=A2 +C2
=A2x2
3
20
=A3 +C3
=A3x3
4
25
=A4 +C4
=A4x4
(A) 20
(B) 40
(C)125
(D) 60
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Problem Set # 14.2 - Data Structures
Problem 14.2 a) Array, List, Stacks and Trees are examples o f ________
(A) Algorithms
(B) Programs
(C) Data Structures
(D) Functions
Problem 14.2 b) Performance of a searching algorithm in data structure is based o n ___
(A) Average time
(C) Best-case time
(B) Worst-case time
(D) Performance depends on algorithm length
Problem 14.2 c) Binary tree in which each leaf is at same distance from root is called__
(A) Efficient
(B) Ordered
(C) Complete
(D) Inefficient
Problem 14.2 d) Efficiency of binary search algorithm as a function of 'n' comparisons is
(A) log2n
(B) n/2
(C) n
(D) n2
Problem 14.2 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 14.2 f) Which sorting algorithms has best time performance under stable conditions?
(A) Bubble sort
(B) Heap sort
(C) Insertion sort
(D)
Problem 14.2 g) _
All sorting algorithms have same performance
is an example of static data structure.
(A) Lists
(B) Stacks
(C) Arrays
(D) Trees
135
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Problem Set # 14.3 - Software design
methods/implementation/testing
Problem 14.3 a) Atypical Software Development Lifecycle does not comprise o f ______ phase
(A) Testing
(B) Implementation
(C) Deployment
(D) Marketing
Problem 14.3 b) Software design with a tendency to break due to changes in non-related segments
called_________
(A) Rigid
(B) Fragile
(C) Portable
(D) Immobile
Problem 14.3 c) A good software design will display_____ __
(A) Ability to be extended without requiring modification to original design
(B) Ability to be extended only with extensive modification to original design
(C) Inability to extend
(D) None of the above
Problem 14.3 d) Static software testing involves verification through__ _______
(A) program code review
(B) program code execution
(C) software maintenance
(D) Options A, B and C are all correct
Problem 14.3 e) Dynamic software testing involves verification through___________
(A) program code review
(B) program code execution
(C) software maintenance
(D) Options A, B and C are all correct
Problem 14.3 f) "Top-down" approach is most applicable t o _______ programming
(A) structured
(C) both structured & object oriented
(B) object-oriented
(D) neither structured nor object-oriented
Problem 14.3 g) The concept of "Class" is most applicable to which programming technique
(A) structured
(C) both structured & object oriented
(B) object-oriented
(D) neither structured nor object-oriented
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Problem 14.3 h )___________ language is not used for structured programming
(A) FORTRAN
(B) BASIC
(C) COBOL
Problem 14.3 i) ______
(D) C#
is not used as scripting language
(A) PHP
(C) Python
(B) JavaScript
(D) PHP, JavaScript and Python are scripting languages
137
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Solutions
Problem Answer Problem Answer Problem Answer Problem Answer
1.1a
A
2.1g
B
3.3c
A
4.3c
C
1.1b
D
2.1h
D
3.3d
A
4.3d
D
1.1c
A
2.2a
A
3.3e
A
4.3e
A
l.ld
A
2.2b
A
3.4a
C
4.3f
D
l.le
C
2.2c
C
3.4b
D
4.3g
C
l.lf
B
2.2d
D
3.4c
C
4.3h
A
1.2a
D
2.2e
A
3.4d
B
431
C
1.2b
C
2.3a
A
3.4e
C
4.4a
D
1.2c
D
2.3b
B
3.5a
D
4.4b
C
1.2d
B
2.3c
B
3.5b
A
4.4c
A
1.2e
C
2.3d
B
3.5c
B
4.4d
B
1.2f
D
2.3e
C
3.5d
C
4.4e
D
1.3a
B
2.4a
A
3.5e
B
4.5a
A
1.3b
B
2.4b
B
3.6a
A
4.5b
C
1.3c
D
2.4c
D
3.6b
B
4.5c
A
1.3d
B
2.4d
B
3.6c
B
4.5d
B
1.3e
C
2.4e
C
3.6d
C
4.5e
C
1.3f
D
3.1a
B
4.1a
C
5.1a
C
1.4a
B
3.1b
D
4.1b
B
5.1b
A
1.4b
C
3.1c
B
4.1c
D
5.1c
D
1.4c
B
3.Id
D
4.Id
B
5.Id
C
1.4d
B
3.1e
A
4.1e
C
5.1e
c
1.4e
D
3.I f
B
4.2a
D
5.2a
B
1.4f
B
3.2a
D
4.2b
B
5.2b
C
2. la
D
3.2b
A
4.2c
B
5.2c
B
2.1b
B
3.2c
B
4.2d
A
5.2d
C
2.1c
B
3.2d
€
4.2e
B
5.2e
2.Id
A
3.2e
D
4.2f
D
5.3a
c
c
2.1e
D
3.3a
B
4.3a
A
5.3b .
B
2.I f
D
3.3b
C
4.3b
B
5.3c
D
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Problem Answer
Problem
Answer Problem Answer Problem Answer
53d
A
6.3b
D
7.I f
D
7.6d
B
5.3e
A
6.3c
B
y .ig
A
7.6e
D
5.3f
C
63d
A
7 73
I .Z a
C
8.1a
D
5.3g
D
6.3e
C
7.2b
B
8.1b
B
5.3H
B
6.3f
D
7.2c
C
8.1c
C
5.3i
A
6.4a
D
7.2d
A
8.1d
B
5.4a
B
6.4b
B
7.2e
B
8.1e
C
5.4b
C
6.4c
B
7.2f
B
8. I f
B
5.4c
D
6.4d
C
7.2g
D
8.1g
B
5.4d
C
6.4e
B
7.2h
A
8.1h
C
5.4e
A
6.5a
D
7.2i
C
8.1i
B
5.5a
B
6.5b
B
7.2j
D
A
5.5b
C
6.5c
C
7.2k
C
8-lj
8.1k
5.5c
A
6.5d
B
7.21
A
8.11
C
5.5d
D
6.5e
D
7.2m
B
8.1m
B
5.5e
A
6.6a
C
73a
C
8.2a
D
5.5f
B
6.6b
B
73b
C
8.2b
C
5.5g
D
6.6c
A
73c
A
8.2c
B
6.1a
A
6.6d
B
73d
B
8.2d
D
6.1b
D
6.6e
D
73 e
B
8.2e
D
6.1c
B
6.6f
B
7.4a
B
8.2f
B
6.Id
C
6.6g
A
7.4b
A
83a
C
6.1e
c
6.6h
B
7.4c
A
83b
C
6. I f
A
6.7a
B
7.4d
D
83c
B
6.1g
C
6.7b
A
7.4e
B
83d
D
6.1h
A
6.7c
C
7.5a
C
D
6.1i
B
6.7d
D
7.5b
C
83e
83f
B
6.2a
A
6.7e
B
7.5c
B
83g
C
6.2b
A
7.1a
A
7.5d
D
8.4a
D
6.2c
B
7.1b
B
7.5e
D
8.4b
C
6.2d
C
7.1c
C
7.6a
D
8.4c
B
6.2e
B
7.Id
B
7.6b
B
8.4d
D
6.3a
C
7.1e
B
7.6c
C
8.4e
D
139
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C
Problem
Answer Problem Answer Problem Answer
10.1c
D
11.2c
C
12.5c
C
9.1b
C
c
lO .ld
11.2d
A
12.5d
A
9.1c
A
10.le
C
C
11.2e
A
12.5e
B
9.1d
A
lO .lf
A
11.2f
C
12.5f
B
9.1e
B
10.2a
D
11.3a
D
12.5g
D
9.2a
A
10.2b
B
11.3b
D
12.5h
A
9.2b
D
10.2c
B
11.3c
B
12.6a
C
9.2c
A
10.2d
C
11.3d
A
12.6b
D
9.2d
C
10.2e
D
11.3e
B
12.6c
C
9.2e
D
10.3a
B
12.1a
A
12.6d
A
9.3a
B
10.3b
D
12.1b
B
12.6e
D
9.3b
A
10.3c
B
12.1c
D
13.1a
D
9.3c
A
10.3d
C
12.Id
C
13.1b
A
9.3d
A
10.3e
D
12.le
B
13.1c
D
9.3e
C
10.4a
B
12.I f
A
13.Id
A
9.4a
C
10.4b
A
12.2a
D
13.le
C
9.4b
B
10.4c
C
12.2b
A
13.I f
B
9.4c
B
10.4d
D
12.2c
B
13.Ig
B
9.4d
C
10.4e
C
12.2d
B
13.Ih
A
9.4e
B
10.5a
A
12.2e
C
13.1s
B
9.5a
A
10.5b
f
12.3a
B
13. Ij
B
9.5b
C
10.5c
L
12.3b
C
13.2a
B
9.5c
D
10.5d
B
12.3c
A
13.2b
A
9.5d
B
10.5e
B
12.3d
C
13.2c
A
9.5e
A
10.5f
C
12.3e
B
13.2d
B
9.6a
B
11.1a
D
12.4a
B
13.2e
C
9.6b
D
11.1b
D
12.4b
D
13.2f
C
9.6c
C
11.1c
C
12.4c
C
13.2g
A
9.6d
A
11.Id
A
12.4d
A
13.3a
D
9.6e
D
11.le
C
12.4e
C
13.3b
C
9.6f
A
ll.lf
B
12.4f
13.3c
10.1a
A
11.2a
D
12.5a
c
c
13.3d
c
c
10.1b
D
11.2b
D
12.5b
B
13.3e
A
Problem Answer
9.1a
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Problem Answer
13.3f
13.3g
C
14.1a
C
14.1b
B
14.1c
D
14.Id
B
14. le
C
14.2a
C
14.2b
B
14.2c
C
14.2d
A
14.2e
D
14.2f
B
14.2g
C
14.3a
D
14.3b
B
14.3c
A
14.3d
A
14.3e
B
14.3f
A
14.3g
B
14.3H
D
14.3S
D
B
141
Copyrighted Material © 2015
Chapter #1 - Properties of Electrical Materials
Consult NCEES® Reference Handbook - Pages 58 - 59 for reference
1.1 Chemical Properties - Solutions
l.la) 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 58 of NCEES® Reference Handbook, Fe
has standard oxidation potential of +0.440 V while Zn has a standard oxidation potential of +0.763 V due to
which can act as a sacrificial anode (Cu, Ni and Hg are less electropositive than Fe).
1.1b) 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.
1.1c) CORRECT ANSWER - A
According to NCEES® Reference Handbook (Page 59) 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.
l.ld ) CORRECT ANSWER - A
Relevant formula for this problem:
Qd
D = Dae RT
Substituting values of given quantities results in:
Qd
250
D = D0e RT = 7.8 x 10_5e ''8.314x 973' = 7.56 10“ 5m2s-1
l.le) CORRECT ANSWER - C
Relevant formula for this problem:
Qd
D = Dae RT
Re-arranging the equation results in:
\nD
Qd
= In D0
o —R T
Solving for Qd gives 256.2 kj. mol ~1
l.lf) CORRECT ANSWER - B
As indicated in problem l.la Solution, metals with higher standard oxidation potentials become anode during
corrosion process. According to the table on Page 58 of NCEES® Reference Handbook, Zinc has a standard
oxidation potential of +0.763 V while Nickel has a standard oxidation potential of +0.250 V therefore Zn will act
as the anode in this corrosion cell.
142
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1.2 Electrical Properties - Solutions
Consult NCEES® Reference Handbook - Pages 5 9 , 199 and 200 for reference
1.2a) CORRECT ANSWER - D
Relevant formula for this problem:
RA
p= —
Lt
A = nr2 = 3.141 x 10
m
Solving for p gives 1.57 x 10_70.m
1.2b) CORRECT ANSWER - C
Relevant formula for this problem:
RA
P= T
Given pA = 4pB implies that 5 ^ = 4 ^
lA
H3
Having lA = lB and AB — - Aa results in
8 A
B
B 4 A
= 4 Rb<^Aa^ 4 = UsAa
lAlAlA
Therefore RA = RB will be possible if the cable lengths are equal and area of cable'B' is
one-fourth of cable 'A'
1.2c) CORRECT ANSWER - D
Relevant formula for this problem:
eA
Decreasing d 4s, Increases C 1s
Increasing e i \ Increases C ' t
Therefore decreasing the distance between plates and increasing dielectric strength will result in
higher
capacitance.
1.2d) CORRECT ANSWER - B
Refer to NCEES® Reference Handbook Page 60 for material properties. It can be observed that low electrical
resistivity typically corresponds to high heat conductivity.
1.2e) CORRECT ANSWER - C
B
/
Relevant formula for this problem: H = - = —
(i
Solving for \i =
2nr
results in 0.015 H/m
1.2f) CORRECT ANSWER - D
As indicated on page 59 of NCEES® FE Reference Handbook, photoelectric effect can take place in all forms of
matter.
143
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1.3 Mechanical Properties - Solutions
Consult NCEES® Reference Handbook - Pages 59 - 62 for reference
1.3a) CORRECT ANSWER - B
Relevant formula for this problem:
F
According to the problem details a = 2000 x 108 Pa ,A — nr2 = u x 10 6
Solving for F results in 628.3 kN
1.3b) CORRECT ANSWER - B
Relevant formulas for this problem:
cj
F
AL
E = — ,o = — and e = -—
e
A0
L
Substituting cj and e results in:
E =
Solving for E results in 6.36 x 101:LiVm_1
1.3c) CORRECT ANSWER - D
Relevant formulas for this problem:
AL
£ = — ,£t = ln( l + 8)
L>n
Solving for e results in 0.1 and eT in 0.095
1.3d) CORRECT ANSWER - B
Plasticity involves permanent deformation. Ductility is the ability to deform under stress (for example copper
wires). Malleability is the ability to convert into thin sheets (for example gold, silver).
1.3e) CORRECT ANSWER - C
Lighting 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.
1.3f) CORRECT ANSWER - D
Tensile test curve gives information about tensile strength, ductility and young's modulus. Material hardness is
tested by denting.
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Copyrighted Material © 2015
1.4 Thermal Properties - Solutions
Consult NCEES® Reference Handbook - Pages 59 and 64 for reference
1.4a) CORRECT ANSWER - B
Bimetallic strips are made up of metals having different coefficients of thermal expansion.
1.4b) CORRECT ANSWER - C
Relevant formula for this problem:
£
According to problem details AT = 7 K and £ = 3 x 10 3 . This gives a = —p = 4.28 x 10 4 K 1
Tmitial is 296 K, a is 4.28 x 10-4 K~x and required e is 3 x 10~3
Substituting 6 x 10~3 for strain and 4.28 x 10-4 K~x for a results in AT = 14 K
Treq = ^"initial + A r = 296 K + 14 K = 310 K
1.4c) CORRECT ANSWER - B
Relevant formula for this problem:
£
a =—
AT
According to problem details a = 1.2 x 10“ 5 °C~1 and AT = 25 °C
Solving for e gives 30 x 10~5
1.4d) CORRECT ANSWER - B
Relevant formula for this problem: R = R„[ 1+ a(T — r 0) ]
According to problem details 2R0 = R0 [ 1 + a (25) ]
Solving for a results in 0.04 K~x
1.4e) CORRECT ANSWER - D
RA
Relevant formulas for this problem: p = — and p = pa[ l + a(T — T0) ]
Therefore resistivity depends on resistance, area, length and the temperature of given material.
1.4f) 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 would have been same.
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Chapter #2 - Engineering Sciences
2.1 Work, Energy, Power - Solutions
Consult NCEES® Reference Handbook - Pages 199»201 for reference
2.1a) CORRECT ANSWER - D
Relevant formula for this problem:
Q1Q2
W12 = -— —
4nsRi2
Solving for
given R12 = 2m
results in 9 x 10~3J
2.1b) CORRECT ANSWER - B
Relevant formula for this problem:
ur
— Q1®2
12 ” 4izeR\2
In the above equation R12 = y ( l — 0)2 + (1 — 0)2 = \[2
Solving for W12 gives 31.9 mj
2.1c) CORRECT ANSWER - B
Relevant formula for this problem:
/= q
t
q = I x t = 5 x 10~3C
Electron charge =1.6022 x 10_19C
5 x 10“ 3
= 3.12 x 1016
Number of electrons = -.— — —■
1.6022 X 10
2.Id) CORRECT ANSWER - A
Relevant formula for this problem:
W12 =
Q1Q2
4 tie R \2
Q3Q4
= 45J
ACT
Woa = -— -— = 45/
47T£/?34
Therefore energy of System A =System B
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2.1e) CORRECT ANSWER - D
Relevant formula for this problem:
Q1Q2
W12 = A
, = 1.215 x 10~6J
4ns (Ini)
Energy of system at 1 m spacing is 1.215 x 10~6J
Q1Q2
W12 = --- 7 ;
^= 1.215 x 10~4J
An£(lcm)
Energy of system at 1 cm spacing is 1.215 x 10~4/
The amount of work required is equal to the energy difference between two systems.
A= 1.215 x 10~4 - 1.215 x 10“ 6 = 1.202 x 10“4/
2.1f) CORRECT ANSWER - D
Relevant formula for this problem:
Energy stored in capacitor =
sA
C = —7- = 8.85 x 10_11F and
d
CV2
2
V = 200 V
Therefore energy stored in the capacitor is 1.77 x 10~6J
2.1g) CORRECT ANSWER - B
Relevant formula for this problem:
>P2
W = — 0 ) E.dl
- « /jp
'p i
•p2
/r™
W = —Q I
200 Vm~xax.ay
Jpi
p i
Since the dot product of ax. ay = 0
W = 0
2.1h) CORRECT ANSWER - D
Relevant formula for this problem:
q
I = 1 = 2A
t
P = I 2R
= 22 X 2 = 814/
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2.2 Electrostatics - Solutions
Consult NCEES® Reference Handbook - Pages 199 - 201 for reference
2.2a) CORRECT ANSWER - A
Relevant formula for this problem:
p = Q1 Q2 = 90 kN
4nr/Le
2.2b) CORRECT ANSWER - A
Relevant formulas for this problem:
V
Given, E =2000 V/m and d =1 m.
Solving for V results in 2000 V
2.2c) CORRECT ANSWER - C
Relevant formula for this problem:
V = E x d = 1000 x 200 = 200 kV
2.2d) CORRECT ANSWER - D
According to the problem details:
Gravitations force on electron =Electric force on electron
mg = qE
v
Since E = -
implies
Substituting E in above equation results in Q =
= 9.8 x 10~5C
2.2e) CORRECT ANSWER - A
Relevant formula for this problem:
F = BIL sina
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2.3 Capacitance - Solutions
Consult NCEES® Reference Handbook - Pages 199 - 201 for reference
2.3a) CORRECT ANSWER - A
Relevant formula for this problem:
sA
(8.85 x 10~12)(0.02)
C = — = ------- ————----- = 1.77 X 10-11 F
d
0.01
Since Q = CV this implies that V = ~ = 2.26 106 V
2.3b) CORRECT ANSWER - B
Relevant formula for this problem:
Vc(.t) = Vc( 0 ) + ~ j ic( r ) dr
o
10 = 5+
ic(t ) x 180
c
100 x 10~6
ic( t ) = 2.7 \iA f o r t = 3 mins or 180 s
2.3c) CORRECT ANSWER - B
Relevant formula for this problem:
1
1
Energy Stored = - C v2( t ) = -(200 x 10~6)(2405m377t)2 = 5.76 sin2 377t J
2.3d) CORRECT ANSWER - B
Relevant formula for this problem:
tf
vf ~ vi = l j
ti
1
10 V = — —
100
s)
fiF i(t)(5
wv y
Solving for i(t ) gives 0.2 mA
2.3e) CORRECT ANSWER - C
iF capacitors in parallel will result in 1 \iF+ 1 \iF = 2 \iF. Now we have 1 \iF, 2 \iFand 2 \iFin series with
1
each other. This will result in Cnpt =
.ftt = 0.5 \iF
11
(t)+© +©
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2.4 Inductance - Solutions
Consult NCEES® Reference Handbook - Pages 199 - 201 for reference
2.4a) CORRECT ANSWER - A
Relevant formula for this problem:
N 2uA
i
=
—
Solving for L results in 1.25 mH
2.4b) CORRECT ANSWER - B
Relevant formula for this problem:
di
v = L idt-
According to problem details L = 5 mH, di = 100 mA and dt = 2 ms
Solving for Kresults in 0.25 V
2.4c) CORRECT ANSWER - D
Relevant formula for this problem:
Li2
Energy stored = - y
According to problem details L = 100 mH and i = t2 = 102 = 100 A
Therefore energy stored =500/
2.4d) CORRECT ANSWER - B
Relevant formula for this problem:
Li2
Energy stored = - y
N 2vlA
t = —
Therefore increasing the number of turns increases inductance which will increase energy storage capacity.
2.4e) CORRECT ANSWER - C
2 H branch inductors will result in (2 H + 2 H ) ||(2 H + 2 H ) = 2 H. Now we have 1 H, 2 H and 1 H in series
with each other. This will result in Lnet = l-j-2 + l = 4 H
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Chapter #3 Circuit Analysis
3.1 Kirchoffs Laws - KCL, KVL - Solutions
Consult NCEES® Reference Handbook —Pages 200 - 202 for reference
3.1a) CORRECT ANSWER - B
Let us denote the voltage across 10 kQ resistor with V
KCL can be written as follows:
V -0
V -0
V -0
_____ _|_——— + ———-+ 10 mA — 0
lOkQ
4kQ
3kQ
V
1463 o
= —10 mA
Resulting in V = —14.63 V
3.1b) CORRECT ANSWER - D
KCL can be written as follows (consider V as the voltage across 6 k Q and 2 k Q resistors):
I'
V- 0
10 mA = /' + —+
3
6 ka
V
10 mA = —— +
2
V
+
V- 0
note that I' = — —r
2 k£l
V
kQ.
6 kQ.
6 kQ
Solving for V results in 12 V
Therefore /' =
2 kQ .
= 6 mA
3.1c) CORRECT ANSWER - B
Circuit simplification results in :
2H + 4 0 = 6 Q and 2Q\\2Q = 10
Now we are left with 6 O + 4 0||4 0 = 8 0
10 V
In e t =
-7 7 7 7 =
80
net
1-25 A
1
3X125
Since - I net passes through 3 O (current divider rule), the voltage drop across it is— ^— = 1.875 V
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B.ld) CORRECT ANSWER - D
This circuit can be solved easily using the concept of super position.
Let us first consider 10 A current source only and short 50 V voltage source:
Using current divider rule:
Rt
10
I
---- 1 I =
— — 10 = 3.334
Rx ~
f~Rf
20 + 10
Let us now consider 50 V voltage source only and open 10 A current source
KVL can be written as follows:
50 V = -3 0 tt.Ix
Solving for Ix gives —1.66 A
Summing the two values of Ix gives 1.66 A
3.1e) CORRECT ANSWER - A
The given circuit can be solved easily using current divider rule given below
Rf
Ix = - — — IT
Rx Rf
in the given circuit IT = 5 mA
Rx = 6 ka (since 1 ka and 5 ka are in series) and RT = 10 kfl
10 ka
hka =
mA = 3.125 mA
lkn
6 ka + 10 ka
3.1f) CORRECT ANSWER - B
The given circuit can be solved by use of KCL loop
Let us denote the voltage at node between 10 ka, 5 ka and 2 ka resistors as V
KCL can be written at node V as follows (assuming current is coming into V from voltage sources)
10- V
S-V
V-0
i o ka + 2 ka ~ 5 ka
Multiply the above equation with 10 ka in order to eliminate denominators
10 — F + 5(5 — V ) = 2V
Solving for V gives 4.375 V therefore I5kn =
AQ7C1/
5m = 875 \iA
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3.2 Series/Parallel Equivalent Circuits - Solutions
Consult NCEES® Reference Handbook - Pages 200 - 202 for reference
3.2a) CORRECT ANSWER - D
Rab = 5ka\\(l0ktt + 2kQ.\\lka)
2 m | | l kfl =
tom +
0.66 ktt
2 m ||i ka =
10 m + o.66 m = 10.66 m
5 m | |( i o m + 2 m | 11 m ) = 5 m | |10.66 m = 3.4 m
Therefore RAB = 3.4 m
3.2b) CORRECT ANSWER - A
Rab = 5 m | | 1 0 m | | (2 ka + 4ka\\8kD.)
4ka\\Qkn = 2.66 ka
2k a + 4 /cJQ| |8 ktt = 2 kQ. + 2.66 m = 4.66 m
10 fcn||(2 kn + 4m\\8kti) = 1 0 m | | 4 .6 6 ka = 3.17 8 m
5 m | | io m u (2 m + 4 m ||8 m ) = s m | | 3 .i7 8 m = 2 m
Therefore RAB = 2kQ
3.2c) CORRECT ANSWER - B
rab
= i k a + 4ka\\(s m + 2 m | | io m )
2 m u 10 m
= 1.66 m
5 m + 2 m u 10 m = 5 ka + 1.66 ka = 6.66 m
4 m | | ( 5 m + 2 m | | i o m ) = 4 m | | 6 . 6 6 m = 2.50 m
im
+ 4 m u (5 m + 2 m i l 10 m ) = 1 m + 2.5 m
= 3.50 m
Therefore RAB = 3.50 m
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3.2d) CORRECT ANSWER - C
rab
= i m ||[5 ka\\(2 m +
4 fco) + 2 ka]
2 ktt + 4 ka = 6 fefi
6ka\\$ktt = 2.72 ka
5 feft||(2 kfi + 4fefi) + 2kfl = 2.72 fen + 2 fen = 4.727 fen
1 fen||[5 A:X1|((2 m + 4 fen) + 2 fen] = 1 /cfl||4.727 fen = 0.825 fen ~ 1 kQ
Therefore RAB = 1 kQ.
3.2e) CORRECT ANSWER - D
= 1 fen + [2 fen 4- (5 k a ||5 fen)]||10 fen + 2 fen
2 fen+ (5 fen||5fen) = 4.5 fen
[2 fen + (5 ka\\S fen)]||10 ka = 4.5 ka\\10 ka = 3.1 ka
i fen + [2 ka + (5 wins m ) ] | | io ka + 2ka = 3.ika + 1 fen + 2 fen = 6.1 fen
Therefore RAB = 6 ka
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3.3 Thevenin & Norton Theorem - Solutions
Consult NCEES® Reference Handbook - Page 200 - 202 for reference
3.3a) CORRECT ANSWER - B
Voc is calculated by looking into the circuit through terminals 'a' and 'b' as indicated in the problem
Applying current divider rule, 5 mA current will be divided equally in each branch i.e. 2.5 mA
Voc = 2 kil x 2.5 mA = 5 V
3.3b) CORRECT ANSWER - C
Rth is calculated by looking into the circuit through terminals 'a' and 'b' as indicated in the problem
Current source needs to be open-circuited
Rth = 2 /d!||6 kQ. = 1.5 kfl
3.3c) CORRECT ANSWER - A
Rth is calculated by looking into the circuit through terminals 'a' and 'b' as indicated in the problem
Voltage sources need to be short-circuited
r th
= io m ||2 fcn||5fcn = 1.25 m
3.3d) CORRECT ANSWER - A
Apply Norton Theorem by replacing 5 kSl with a wire to calculate lsc
Let Vx be the node between 2 k£l, 1 kQ. and 2 kQ resistors in the circuit
KCL can be written at Vx as follows
10 V - Vx
2
kti
10 V - Vx
+
Vx
2kaTm’ Vx
sv
“
10 V
10 V —V y
2fefi
2 kQ .
It can be observed that Isr = —— H---- —-- since Vx = 5 V this results in Iqr — 7.5 mA
A
•iC
3.3e) CORRECT ANSWER - A
Req is calculated by looking into the circuit through 5 k£L as shown in the problem
Voltage sources need to be short-circuited
Req = (2 kn ||i ka\ + 2 m)\\2 kn =
1.142 w s i m
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3.4 Waveform Analysis - Solutions
Consult NCEES® Reference Handbook - Pages 200 - 202 for reference
3.4a) CORRECT ANSWER - C
For a full wave rectified sinusoidal wave:
v
_ Xmax
eff ~
^max —
. ^e//
Therefore Vmax = V2 x 10 V = 14.14 V
3.4b) CORRECT ANSWER - D
V\ + V2 = 10 cos(S00t) + 15 cos(100t + 45)
Using trigonometric identity: cos(a + /?) = cos(a). cos(/?) - sin(a) sin (/?)
Therefore V1 + V2 = 10 cos (500t) + 15[cos (lOOt) .cos (45) - sin (lOOt) sin(45)]
Since cos(45) = sin(45) =
15 cos
(lOOt) 15 sin
(lOOt)
V1 + V2 = 10 cos (500t) + ----- ------- - ----- ----------------------- -V1 + V2 = 10cos(500t) + 10.6cos(100t) — 10.6 sin (lOOt)
3.4c) CORRECT ANSWER - C
Comparing 100cos(500t + 50) with standard phasor representation Acos(a)t + $)
a = 2nf = 500
500
/ = — = 79.5 Hz
2tc
3.4d) CORRECT ANSWER - B
For a halfwave rectified sinusoidal signal, average value is given as:
^ » = ^
n
= Vn = 4-77
3.4e) CORRECT ANSWER - C
For a periodic signal, the average value is calculated as following
1 fT
Xa v e = j' J
I f4
1 / f 1
dt = —J x {t) dt = —
f2
r3
r4
\
1
1 dt + J 2 dt + J 1 dt + J 0 d tj = ~ (1 + 2 + 1 + 0) =
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3.5 Phasors - Solutions
Consult NCEES® Reference Handbook - Pages 200 - 202 for reference
3.5a) CORRECT ANSWER - D
Express v ( t ) = 100sin(377t + 60) in standard cosine form
Since sin(i9) = cos (i9 - 90)
Therefore v (t ) = 100 cos(377t + 6 0 - 90) = 100cos(377t - 30°)
Vmax 100
Vrms = —— — —=r = 70.7 V and phase angle is — 30°
V2
V2
Hence the voltage phasor is 70.7/-300 V
XL = jo)L = j x 377 x 100 x 10“ 3n = ;37.7 fl = 37.7/90°Q
V
I = — = 1.875 A with — 120° phase angle
3.5b) CORRECT ANSWER - A
Express v (t ) = 212sin((ot + 50)in standard cosine form
Since sin(^) = cos ($ — 90)
Therefore v (t ) = 212 cos(377t + 50° - 90°) = 212 cos(377t - 40°)
Vrms =
v2
= 150F and phase angle is — 40°
Hence the voltage phasor is 1507-40° V
3.5c) CORRECT ANSWER - B
v (t ) = 100 cos(377t Vrms =
V2
0°)
= 70.7 V and phase angle is 0°
Hence voltage phasor is 70.7/0° V
Z = 500 - y i o o n = 500/-1.14°Q
V
I = — = 0.1414v4 with 1.4° phase angle
I = 0.141/1.14°A
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3.5d) CORRECT ANSWER - C
Express 100sin(377t + 100) in standard cosine form
Since sin($) = cos ($ —90)
Therefore v (t ) = 100cos(377t + 100° —90°)
v ( t ) = 100cos(377t + 10°)
Kms = ~^7=^ = 70.7 V and phase angle is 10°
V2
Hence voltage phasor is 70,7/10° V
Xc = l/ja)C = 0.02657-90° Q
V
I = —- = 2665 A with 100° phase angle
I = 2665/100°A
3.5e) CORRECT ANSWER - B
z = ( 5 n - y io n ) + 2 o n | | -/ 5 n
Z = 6.176 - 14.47; fi s 6.2 - 14;n
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3.6 Impedance - Solutions
Consult NCEES® Reference Handbook - Pages 200 - 202 for reference
3.6a) CORRECT ANSWER - A
Zeq = 10 + ;2 0 + (10- 5/ n)||(2- 5JO)
Solving for Zeq results in 12 - j a
3.6b) CORRECT ANSWER - B
zeq = i o — 5 / n|| 2 - 5j a
Solving for Zeq results in 2.21 - 3.15; a = 2 - 3j a
3.6c) CORRECT ANSWER - B
Z eq
=
R
+
*C
X c = J ^ C = y'377 X 100
X
10“ 9 = ~ j 2652S
Solving for Zeq results in 50 - j26525 a
3.6d) CORRECT ANSWER - C
Zeq
=
(
*
+*
l)
P
c
XL = j(x)L = J377 x 2 x 10~3 = ;0.754n
1
1
Xq ~ JcoC ~ j377 x 100 x 10~6 “ ~ i 26S n
Zeq = 10+ y'0.754 H|| —y'26.5 H
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Chapter # 4 - Linear Systems
4.1 Frequency / transient response - Solutions
Consult NCEES® Reference Handbook - Pages 202 - 203 for reference
4.1a) CORRECT ANSWER - C
vc(0) = 20 V
After the switch changes position, V =0 (since there is no external voltage source anymore)
t = 5 mins = 5 x 60 seconds = 300 s
vc(5 mins) =
+ V ^1 — e~RC ^
R = 100 k£i, C = 1 mF
x?c(5 mins) = 0.995 V = I V
4.1b) CORRECT ANSWER - B
Relevant formula for this problem:
vc(t) — vc(0)e~RC + v ^1 —e~Rc ^ in the given scenario vc(0) = 10 V
Since R = 10 k£l & C = 200 |iF, RC = 2 s
After the switch changes position, V =0 (since there is no external voltage source anymore)
vc(t ) = 10e“ £/2
4.1c) CORRECT ANSWER - D
Relevant formula for this problem:
Rt
Y
+ -(l -
_R t
i(t ) = i(0)e
l
i ( 0) = 0 ,
V = 10 V,R = 1 ka and L = 2 mH= > - = 500000s
Lj
e
l
)
Therefore i(t ) = 0.01 (1 — e“ 500000t)
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4.Id) CORRECT ANSWER - B
Since the switch is closed for a long time, inductor can be assumed as a short wire
5 Xl||10 n = 3.33 n
Therefore the current provided by power source at t = 0:
Using current divider rule, the current through inductor at t =0 is
3 33
i(0) = 0.75 — = 0.5 A
According to the problem details t =
10
t
,
L
where t = — = 66.66 ms given L = 1 H and R = 15 Q
Rt
i(10r) = i(0)e T + 0 = 22.6 \iA
4.1e) CORRECT ANSWER - C
Relevant formula for this problem:
vc(0 ) = 1QV,
5t = SRC = 250s
After the switch changes position, V =0 (since there is no external voltage source anymore)
vc(50) = 10e"5 = 67.3 mV
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4.2 Resonance - Solutions
Consult NCEES® Reference Handbook - Pages 202 - 203 for reference
4.2a) CORRECT ANSWER - D
Relevant formula for this problem:
= -4= and L = 20 mH & C = 10 \iF
s/Tc
Solving for coo results in 2236 rad/s
4.2b) CORRECT ANSWER - B
The given circuit is an example of series resonance for which:
4.2c) CORRECT ANSWER - B
Maximum current will occur at resonance when impedance is purely resistive i.e. Z = R
I
max
=
120 V
______ — 1 ? A
ion
4.2d) CORRECT ANSWER - A
Maximum current will occur at resonance when impedance is purely resistive i.e. Z = R
1
0Jo = —=
= 3162 rad.s 1 given L = 10 mH and C = 10 \iF
yLC
4.2e) CORRECT ANSWER - B
For parallel resonance circuits current magnification factor 'Q' is given by equation below
Q =
R
— -
where
1
0)o = - = = = 707 rad.s 1
yLC
Therefore Q = 500/707 = 0.707
4.2f) CORRECT ANSWER - D
Relevant formula for this problem:
BW =
1000 rad.s-1
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4.3 Laplace Transform - Solutions
Consult NCEES® Reference Handbook —Page 33 for reference
4.3a) CORRECT ANSWER - A
Based on Laplace Transforms table found on page 33 of NCEES® FE Reference Handbook
=
cu m
4.3b) CORRECT ANSWER - B
Based on Laplace Transforms table found on page 33 of NCEES® FE Reference Handbook
L[f(te ~ at
_
t)
u ( t - t)] = e~TS. F ( s )
e ~ a (t+
1 -1 ) _
e - a ( t - l) e ~a
Therefore /(t) = e ~ ^ [ e ~ a^t~1^u(t — 1)]
Since L [ e ~ at1 = —
L
J
s+a
^ p ~ (s )
„ -(s + a )
Implies that L [ f ( t ) 1 = e ^ ----= ------u
J
s+a
s+ a
4.3c) CORRECT ANSWER - C
Based on Laplace Transforms table found on page 33 of NCEES® FE Reference Handbook
L [ f ( t — r )u (t - t)] = e TSF(s) andL(te at) =
1
(s + a)2
/(t) = (t - 1 + l) e -a(t_1+1^u(t - 1) = [(t - l)e _a^
t_1^u(t - 1) + e_a(t_1)u(t - l)]e~a
4.3d) CORRECT ANSWER - D
f ( t ) = [ e~^~3^ — e~3^~3^]u(t — 3)
Based on Laplace Transforms table found on page 33 of NCEES® FE Reference Handbook
4.3e) CORRECT ANSWER - A
f (t ) =
te~atS ( t - 2)
Using Sifting Theorem,
F(s) = 2e~2a e~2s = 2e~2^a+s^
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43f) CORRECT ANSWER - D
F (s ) =
5
(s -I- 3) + (s + 5)
(2s + 8)
2(s + 4)
2 's + 4
Based on Laplace Transforms table found on page 33 of NCEES® FE Reference Handbook
f i t ) = | e " 4tu (0
4.3g) CORRECT ANSWER - C
F (s ) =
s +8
(s + 1)0 + 7)
s +8
A
B
Let ------ —----— = ---- -H----- (s + l)(s + 7)
s+ 1
Multiplying by denominator -> s + 8 = A(s + 7) 4- B(s + 1)
s+ 7
o
1
Lets = - 7 , 1 = B ( —6 ),B = - -
Lets = -1 ,
6
F (s ) =
7
1
(6)(s + l)
(6)(s + 7)
7 = A(6), A = 7/6
Based on Laplace Transforms table found on page 33 of NCEES® FE Reference Handbook
m
(7e~t — e~7t)u (t )
= - ------- 6...... —
4.3h) CORRECT ANSWER - A
F (s ) =
s2 + 2s + 1
(s + 2)(s + 3)(s)
s2 + 2s + 1
A
(s + 2)(s 4- 3)(s)
B
C
s +2+s + 3+ s
s2 + 2s + 1 = j4(s + 3)(s) -I- B(s + 2)(s) + C(s + 2)(s + 3)
1
Lets = 0, 1 = C (2 )(3 ), C = -
6
Lets = - 2,
4 - 4 + 1 = A (l){-2 ),
Lets = - 3,
9 - 6 + 1 = B ( - 1)(-3),
-i
4
1
A = --
4
B =-
1
~ 2(5 + 2) + 3(s + 3) + 6s
4 e -3 t
f(t)=
—+
T +
6
"(
t
)
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j\j\j
4.3i) CORRECT ANSWER - C
s +4
F(s) - (s2)(s + 5)
s +4
_A
^6t (s2)(s + 5)
B
C
s + s2 + s + 5
s + 4 = A (s )(s + 5) + B(s + 5) + C(s2)
Le ts2 = 0,
4
4 = 5 (5 ), F = 1
-1 = C(25), C = - —
25
Lets = -5,
9
1
Comparing coefficients of s , A + C = 0, A = —C, A = —
1
4
1
~ 25(s) + 5 (s2) ~ 25(s + 5)
Based on Laplace Transforms table found on page 33 of NCEES® FE Reference Handbook
/I
=
41
25 ^ "5
e~st\
25~)
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4.4 Transfer Functions - Solutions
Consult NCEES® Reference Handbook - Pages 202 - 203 for reference
4.4a) CORRECT ANSWER - D
Z 0 = ( fl + s i ) | | ( i )
(fi + 5Z') ll( ^ )
vo _ Z o _
Vi
Zt
Z r = R + ( R + SL ) | | ( ^ )
R + (R + s t ) | | (^ r )
4.4b) CORRECT ANSWER - C
Zo = R + i
zt
Vo
Zo
R + JC
Vi
ZT
(R||st) + ( f i + i )
= (r \\sL) + ( r + ± )
4.4c) CORRECT ANSWER - A
The given circuit can be simplified resulting in impedance shown below:
Z = R + ^ - \ \ (R + sL\\R)
4.4d) CORRECT ANSWER - B
H (s ) =
20
s(10s + 1)
Standard form,
20
H (s )
i°s (s + i )
H (s ) = -
2
Poles @ 0 and — 0.1 rad/s
zero - none
\gain\ = 20log (2) = 6 decibel
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4.4e) CORRECT ANSWER - D
His)
10
s2iSs + 1)
Standard form,
His)
His)
10
Ss2is + 1/5)
2
s2(s + 1/5)
Poles @ 0 (2nd order), —0.2 (1st order) rad/s
zero - none
\gain\ = 20log (2) = 6 decibel
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4.5 Two-Port Theory - Solutions
Consult NCEES® Reference Handbook - Pages 202 - 203 for reference
4.5a) CORRECT ANSWER - A
Vi
i2 = o
z n = -± = 4a
h
V2
z21 = — = 2 a
i2 = o
i
vt
Z i2 = -r = 2a
v2
z22= - p = 4 0 ,
/ i=0
*2
*2
4.5b) CORRECT ANSWER - C
1,2 = 0
=
1/1 = °
1,2 = 0
Y* = r 2 = l s -
^ = °
4.5c) CORRECT ANSWER - A
h
l'“
1
/,
^ = 3 s'
^ = °
1/1 = 0
y™ = V2 = - \ s ’
1
^
= v r ~ 6 s-
r-
= t = 4 5'
v*
^ =
4.5d) CORRECT ANSWER - B
zu = ^ = u . n a
i2
h
4.5e) CORRECT ANSWER - C
Vi
20
2
" 12 = i ^ = 30 = 5 = 0-66'
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Chapter # 5 - Signal Processing
5.1 Continuous Time Convolution - Solutions
Consult NCEES® Reference Handbook - Page 206 for reference
5.1a) CORRECT ANSWER - C
Let y(t) = x (t ) * h(t )
It is easier to flip function h (t) into /i(—t)
Region # 1 y(t) = 0 fo r t < 0 no overlap
Region # 2 y(t) = | 8dr = 81fo r 0 < t < 2 partial overlap
jo
Region # 3 y(t) =
[
Jf _2
Region # 4 y(t) = [
8dr = 81 — 8(t — 2) = 16 f o r 2 < t < 4, partial overlap
8dz = 8(4 — t + 2) = 48 — 8t /o r 4 < t < 6, partial overal
Jt_2
Region # 5 y(t) = 0 /o r 6 < t ,no overlap
5.1b) CORRECT ANSWER - A
Let y(t) = /(t) * /i(t)
/t is easier to flip function f i t ) into / ( —t )
f 00
3e^
Region # 1 y(t) = j e^~T^3e^~T^dT =
for t < 0
jo
2
r 00
3e^_t^
Region # 2 y(t) = | e^~T^3e^~T^dT = — -— f o r 0 < t
Jt
I
5.1c) CORRECT ANSWER - D
Let /(t) = x(t) * y(t)
It is easier to flip function y(t) into y(—r)
Region # 1 /(t) = 0 fo r t < 0 no overlap
Region # 2 /(t) =
j
sin (r)dr = 1 —cos (t) fo r 0 < t < n partial overlap
J0
fn
Region # 3 /(t) = J sin(r) dr = cos (t) — 1 / o r n < t < 2n, partial overlap
J t-n
Region # 4 /(t) = 0 /o r n < t ,no overlap
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5.Id) CORRECT ANSWER - C
Let f ( t ) = x (t ) * y(t)
It is easier to flip function y (t ) into y ( - r)
Region # 1 /(t) = 0 fo r t < -1 no overlap
-1+4
J
(2)(3)dr = 6(t + 4 - 3) = 6(t + 1) fo r — 1 < t < 1 partial overlap
3
Region # 3 /(t) = f
6dr = 6(5 - t — 2) = 6(3 — t) /or 1 < t < 3 partial overlap
j t+2
Region # 4 /(t) = 0 fo r 3 < t , no overlap
5.1e) CORRECT ANSWER - C
Let f ( t ) = x (t ) * y(t)
It is easier to flip function y (t ) into y (-r)
Region # 1 /(t) = 0 /o r t < 0 n o overlap
Region # 2 /(t) = | 2e~Tdr = 2 — 2e“ t /o r 0 < t < 2 partial overlap
jo
Region # 3 /(t) = f
j t_ 2
2e~Tdr = 2(e~t+z — e ~t) f o r t > 2
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5.2 Discrete Time Convolution - Solutions
Consult NCEES® Reference Handbook - Page 206 for reference
5.2a) CORRECT ANSWER-B
Flip signal y[n] to make it 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 = 14
Forn =
4, /[ 4]
= 2.3 + 2.2 = 10
For n = 5,
f [ 5] = 2.2 = 4
Forn = 6,
/[ 6] = 0
4
Therefore f [n ] = [0 4 10 14 10 4 0]
5.2b) CORRECT ANSWER - C
Flip signal y[n] to make it y [ —k]
For n < 0
there is no overlap
Forn = 0
a o]
=
Forn = 1
/ [l ]
= 1.1 +2.0
For n
f [2 ]
= 1.2 + 2.1 + 1.0
4
For n
/[ 3]
= 1.3 + 2.2 + 1.1
8
Forn = 4,
f [ 4]
= 1.0 + 2.3 + 1.2
8
o
=
1
For n = 5,/[5] = 1.3 = 3
Forn = 6,
/[ 6] = 0
f [n ] = [0 1 4 8 8 3 0]
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5.2c) 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.
5.2d) CORRECT ANSWER - C
x[n] = u[n] -
u[n — 5]
y[n] = 0.2nu[n]
k = ~ co
k= n
5.2e) CORRECT ANSWER - C
x[n ] = u[n — 2]
y[n] = 0.4nw[n]
k= ~ co
k= n
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5.3 Z Transforms - Solutions
Consult NCEES® Reference Handbook - Page 205 for reference
5.3a) 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.
5.3b) CORRECT ANSWER - B
x[n] = u[n] — u[n — 5]
fc=4
fe = 1 + z 1 +
X[z] = l z
Z~2 +
Z ~ 3 + Z -4
fc=0
5.3c) CORRECT ANSWER - D
x[n] = 0.2nu[n]
k = oo
X[z] =
V
k = oo
0.2k z
0.2ku[k]z~k = ^
fc=0
k=
-k
0
Using geometric series convergence
00
Z
r n = ----- for |r| < 1
1- r
71=0
Therefore
Xlz] =
1
1 - 0.2z " 1
Note: This problem can also be solved used Z Transform table given in NCEES® FE Reference Handbook.
5.3d) CORRECT ANSWER - A
x[n] = [2 3 1 0 5]
k
=oo
X[z] = ^
x[k]z~k = 2z"° + 3z_1 + z " 2 + 5z"4 = 2 + 3z_1 + z “ 2 + 5z“ 4
k=0
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5.3e) CORRECT ANSWER - A
x[n ] = 5(0.75)nu[n]
fe=00
5(0.75) ku[k]z~k = 1 _ Q75z_1
Xlz] =
k=0
5.3f) CORRECT ANSWER - C
X(z) =
z
1
z - 0.5
1 - 0.5z_1
Based on NCEES® FE Reference Handbook table for Z Transform
1
Rk < = > -----------x[n] = (0.5)nu[n]
5.3g) CORRECT ANSWER - D
5z + 2
* (z) " (z — l)(z —4)
Dividing the entire equation by z results in
X(z)
z
5z + 2
Ct
C2
C3
= — + — £- +
z(z — 1)(z - 4 )
z
z — 1 z —4
5z + 2 = ^ (z — 1)(z —4) + C2(z)(z — 4) + C3(z)(z — 1)
Letz = 1,
Let z = 4,
Let z = 0,
7 = C2(—3) => C2 = - ~
22
, ,
11
= C3 (12) => C3 = —
6
1
2 = C1( - l) ( - 4 ) => C± = -
X(z)
1
7
11
z
2z
3(z — 1)
6(z —4)
j (z)=
1
2
7z
llz
...
.
3(z - 1) 6(z - 4)
1
7z
llz
2
3z(l —z x)
6z(l —4z_1)
r 1 S[n] 7 ( l n)u [n] t (U )(4 n)u [n]
x[n\ = — ----------------- 1
--------------
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5.3H) CORRECT ANSWER - B
(z - 2)(z + 1)
*00 =
(z - 0.1)(z - 0.2)
X(z)
(z -2 )(z + l)
Q
—
z
z(z - 0.1)(z - 0.2)
z
C2
t
|-------------------------
_ j_
z - 0.1
C3
z - 0.2
(z - 2)(z + 1) = C±(z - 0.1)(z - 0.2) + C2(z ) ( z - 0.2) + C3( z )(z - 0.1)
Letz = 0.1,
-2.09 = C2(-0.01) => C2 = 209
Let z = 0,
-2 = Ct (0.02) => Ct = -100
Let z = 0.2,
-2.16 = C3(-1 )(-4 ) => C3 = 108
X (z )
-100
z
z
209
108
+ 7----r-rr
(z - 0.1) (z - 0.2)
„ N
209z
108z
X(z) = -100 - ---- —— +
(z - 0.1) 6(z - 0.2)
209z
108z
(1 - O.lz-1)
z ( l - 0.2z_1)
X (z ) = -1 0 0 -7 -— — — - +
x[n] = -100<S[n] - 209(0.1n)w[n] + (108)(0.2n)u[n]
5.3i) CORRECT ANSWER - A
X (z ) =
(z + 0.5)
(z - 0.1) (z + 0.4)
X (z )
(z + 0.5)
z
z(z — 0.1)(z + 0.4)
_Ct
C2
C3
— ------ j----------- -|_
z
z — 0.1 z + 0.4
(z + 0.5) = Ct(z - 0.1)(z + 0.4) + C2(z)(z + 0.4) + C3(z)(z - 0.1)
Letz = 0.1,
0.6 = C2(0.1)(0.5) => C2 = 12
Let z = -0.4,
-25
0.5 = CiC-O.lXO.4) => Cx = —
Let z = 0,
X (z )
1
0.1 = C3 (—0.4)(—0.5) => C3 = -
-25
12
1
+ ---_ ---- ---2z
(z — 0.1) 2(z + 0.4)
-25
X (z )= —
2
+
,
-25
12z
2
(z — 0.1)
X {z ) = — - +
v/
12z
z ( l - O.lz-1)
2z(l + 0.4z_1)
—255[n]
(-0.4n)u[n]
x [n] = ---------- [- 12(0.1)nw[n] —-------------
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2(z + 0.4)
5.4 Sampling - Solutions
Consult NCEES® Reference Handbook - Page 209 for reference
5.4a) CORRECT ANSWER - B
The signal with highest frequency is sinc(2QQ0nt) => sinc[2n(1000)t]
Nyquist sampling rate > highest frequency x 2 = 1000 x 2 = 2000 Hz
5.4b) CORRECT ANSWER - C
Sampling frequency =2000 Hz
Signal frequency = 1500 Hz
Since sampling frequency <2 x signal frequency, aliasing will happen.
5.4c) CORRECT ANSWER - D
Sampling frequency =500 Hz
Signal frequency =250 Hz
Since sampling frequency =2 x 250 Hz =500 Hz, there will be no aliasing.
5.4d) CORRECT ANSWER - C
Sampling frequency =300 Hz
Signal frequency =200 Hz
Sampling frequency <2 x 200 Hz, aliasing will happen.
200 Hz = -100 + 300 Hz
Aliased frequency = 100 Hz
5.4e) CORRECT ANSWER - A
Signal needs to be sampled at or above Nyquist rate for perfect reconstruction.
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5.5 Filters - Solutions
Consult NCEES® Reference Handbook - Pages 206, 210 and 211 for reference
5.5a) 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.
5.5b) 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.
5.5c) 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.
5.5d) 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.
5.5e) CORRECT ANSWER - A
Finite Impulse Response Filter (FIR) is non-recursive because it does not have feedback loop (unity feedback).
5.5f) CORRECT ANSWER - B
Infinite Impulse Response Filter (HR) is recursive because it has a feedback loop.
5.5g) 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.
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Chapter # 6 - Electronics
6.1 Solid-State Fundamentals - Solutions
Consult NCEES® Reference Handbook - Page 212 for reference
6.1a) CORRECT ANSWER - A
According to periodic table found on page 54 of NCEES® FE Reference Handbook, Antimony is a group V element
and has 5 valence electrons due to which it can be used as an n-type doping material.
6.1b) CORRECT ANSWER - D
According to periodic table found on page 54 of NCEES® FE Reference Handbook, Boron is a group III element
and has 3 valence electrons due to which it can be used as a p-type doping material.
6.1c) CORRECT ANSWER - B
Relevant formula for this problem:
(T = rii (ne + \ih)q
Substituting the details given in problem results in:
o = 9.61 x 10-10 S . m " 1
6.1d) CORRECT ANSWER - C
Relevant formula for this problem:
Substituting the details given in problem results in:
V0 = 0.634 V
6.1e) CORRECT ANSWER - C
Insulators have large gap between valence and conduction bands due to which electrons cannot move freely
through the material.
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6.If) 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.
6.1g) CORRECT ANSWER - C
Group V elements of periodic atoms have 5 electrons in their outer orbit which allows them to form 4 covalent
bonds with Si or Ge atoms and lets one electron free making them n-type doping agents. Phosphorous and
Arsenic are examples of n-type doping agents.
6.1h) CORRECT ANSWER - A
Group III elements of periodic atoms have 3 electrons in their outer orbit which allows them to form 3 covalent
bonds with Si or Ge atoms and this creates a hole making them p-type doping agents. Boron and Aluminum are
examples of p-type doping agents.
6.11) CORRECT ANSWER - B
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.
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6.2 Diodes - Solutions
Consult NCEES® Reference Handbook - Page 214 for reference
6.2a) CORRECT ANSWER - A
Let us first assume that both D1 and D2 are ON
SV — I V
^ = ~ T M T = 3mA
iDi + iD2 =
(I V )
= 0.5 mA
iD2 = 0.5 mA — iD1 = 0.5 mA — 3 mA = -2.5 mA
Since iD2< 0, the result is not consistent with assumption.
Let us now assume that D1 is ON and D2 is OFF
SV
im = J r n = 1-66mA
VB2 = 1 V - 1.66 mA X 2 f cn = -2.33 V < 0
Since iD1>0 and vD2<0, the results are consistent with assumptions.
6.2b) CORRECT ANSWER - A
The circuit given in Problem 6.2b) is an example of a full wave rectifier therefore output wave form will be a ful
wave rectified sinusoidal wave.
6.2c) CORRECT ANSWER - B
The given circuit can be rearranged such that
Rthi — 1 kO.\\2 kQ. = 0.66 kQ
VnrA = 2 (?) = 1.33 V
Similarly, Rth2 = 4 kQ\\2 kQ = 1.33 kQ
Vnr? = 4 I - U 2.66 V
Assuming diode is "ON" i.e. forward biased
2.66 V - 1.33 V
lD ~ 1.33 kCl + 0.66 ka = °'66 mA
Since iD>0, Diode is ON. Therefore diode Q point is (OV, 0.66 mA)
Note that voltage drop across diode is assumed 0 V since ideal diode model is being considered.
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6.2d) CORRECT ANSWER - C
Assuming diode is "ON" i.e. forward biased
10 V - (8 + 0.7) V
l° = ---------- w i n ---------- = °-13mA
Since iD>0, Diode is ON
6.2e) CORRECT ANSWER - B
Let us first assume that both DI and D2 are ON
(5 - 0.7) V — (10 - 0.7)7
iD 1
=
------------------------------ 2
l a
------------------------------ =
“
2 '5
m A
Since iD1<0, the result is not consistent with the assumption.
Let us now assume that DI is OFF and D2 is ON
(1 0 -0 .7 )1 /-(-3 1 0
1 0 2
=
-------------------------------5
^
_
,
-------------------------------=
VDi = 5 V - (10 - 0.7)1/ = -4.3 V < 0
Since VD1<0 and iD2>0.7 V the result is consistent with assumptions.
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6.3 BJTs - Solutions
Consult NCEES® Reference Handbook - Pages 214 - 216 for reference
6.3a) CORRECT ANSWER - C
Let us assume that the transistor is operating in active region
Vbe = 0.7 V & lc = plb = 1001b
Base-emitter KVL
5 - 5 Ib - Vbe - 2 I e = 0
Since Vbe = 0.7 V & Ie = (fi + 1)1b = 101 lb
Ib = 20.77 \iA
lc — f$Ib = 2.07 mA
At the collector,
10 “ Vc __ ic
j
_____
Vc = 10 V - 2 ka x 2.07 mA = 5.86 V
Similarly at the emitter,
Ve = 2 ka x 2.09 mA = 4.18 V
Vce =
VC
- Ve = 5.86 V - 4.18 V = 1.68 V
Since Vce > 0.7 V & Ib = 20.7 jjA > 0
Transistor is in active mode
6.3b) CORRECT ANSWER - D
Let us assume that the transistor is operating in active region
Vbe = 0.7V&Ic = (3Ib = 100 Ib
Base-emitter KVL
3 - V be- 21e = 0
= > Ie = 1.35 mA
le = (P + 1)l b = 101 Ib
=> Ib = 1.133 x 10~5,4
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lc = pib = 1,133 mA
At the collector,
7V~VC
3m '
~ l°
Vc = 7 V - 3 ka X 1.133 mA = 3.6 V
Similarly at the emitter,
Vbe = 0.7V&Vb = 3V
3 - Ve = 0.7 V
Ve = 2.3 V
Vce = yc - V e = 3.6 V - 2.3 V = 1.3 V
Since Vce > 0.7 V &Ib > 0
Transistor is in active mode
6.3c) CORRECT ANSWER - B
Let us assume that the transistor is operating in active region
Vbe = 0.7 V 8l Ic = pib = 100 Ib
Base-emitter KVL
0 - Vbe - 3Ie = 0
Since Vbe = 0.7 V
-0.7V
Ie = ~3ka = _0'233 mA
/, = (/? +1) /„
h
=
w
h
) =
_
0
-0 0 2 3
m A
Since Ib< 0, the transistor is in cut off
Ib= Ic = Ie = 0
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6.3d) CORRECT ANSWER - A
The circuit can be rearranged such that
Rth = 20 m il 10 ka = 6.66 ka
/ 10 \
Let us assume that the transistor is operating in active region
Vbe = 0.7V&lc = (3Ib = 100 Ib
Base-emitter KVL
3.3 - 6.6Ib - V be- I e = 0
Since Vbe = 0.7 V & Ie = (J? + 1) I b = 101 Ib
lb = 2.415 x 10“ 5^
lc -
(Jib - 2.41 mA
le = 2.44 mA
At the collector,
10 V - Vc
2 ka
~ lc
Vc = 10 1/ - 2 ka x 2.415 mi4 = 5.17 V
Similarly at the emitter,
Ve - 0
T k a = 'e
Ve = 2.44 V
Vce = vc - v e = 5.17 V - 2.44 V = 2.67 V > 0.7 V
Since Vce > 0.7 V & Ib > 0
Transistor is in active mode
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6.3e) CORRECT ANSWER - C
Let us assume that the transistor is operating in active region
Veb = 0.7 V & Ic = 0Ib = 100 Ib
Base-emitter KVL
5 - 1 x le - Veb - 2lb - 1 = 0
Since le = 101/b
Ib = 32.03 \jlA
lc = pib = 3.2 mA
le = 3.23 mA
At the emitter,
5 -V e
T m = ,‘
Ve = 1.77 V
Similarly at the collector,
Vc - (-5)
soon
kc
= Ic = 3.2 mA
= -^ 4 1 /r
Vec = ve - v c = 1.77 K - (-4.44 V ) = 6.17V > 0.7 V
Since Vec > 0.7 V & Ib > 0
Transistor is in active mode
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6.3f) CORRECT ANSWER - D
Let us assume that the transistor is operating in active region
Veb = 0.7 V & Ic = pib = 100 Ib
Base-emitter KVL
S - V eb- 201b - 4 = 0
Ib = 1.5 x 10~s^4
Ic = pjb = 1.5 mA
Ie = 1.56 mA
At the collector,
Vc - (0)
VC = ^ V
''e c = Ve - V c = 5 V - (3.12 V) = 1.88 K
Since
>
0.7 V
> 0.7 V & Ib > 0
Transistor is in active mode
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6.4 MOSFETs - Solutions
Consult NCEES® Reference Handbook - Pages 214 - 216 for reference
6.4a) CORRECT ANSWER - D
Let us assume that the transistor is operating in saturation mode
Gate-Source loop KVL
0 - V gs- 2Ids + 10 = 0
Ids-
10 - vgs
2
Drain-Source loop KVL
10
-
2 Ids - Vds
2Ids + 10
-
=
0
Vds — 20 — 4lds
Since Ids = k(Vgs - Vtf
= fc fe - V t f
Solving the above given quadratic equations results in following values
Vgs = 3.54 V or Vgs = -2.54 V
Since we assumed saturation, the possible value should be Vgs = 3.54 V making Vgs > V t
As per above given equations
10 - VQS
/ds = --- =
10 - 3.54
---- = 3-23 mA
Vds = 20 — 4/ds = 7.08 V
Checking the assumptions
Vgs = 3.54 V > Vt
Vds = 7.08 V > Vgs — Vt or Vgd < Vt
Transistor is in saturation mode.
6.4b) CORRECT ANSWER - C
Let us assume that the transistor is operating in saturation mode
(SV - 0)
5W1
VS = 4 V
and
Vds = 5 —4 = 1 V,
Vgs = 1 —4 = —3 V
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Therefore transistor is in saturation
6.4c) CORRECT ANSWER - B
The circuit can be rearranged such that
Rth = 60 kD.||40 ka = 24 ka
/ 40 \
V t h
=
4
( l o
o
)
=
1
6
v
Let us assume that the transistor is operating in saturation mode
Vg = 1.6 V,
lgs — 0 A/ Vs = 0 V
Since lds = k(Vgs - Vt) 2 = 0.2 (1 .6 - l ) 2 = 72\iA
Vd = 3.28 V
Vgd = 1.6 - 3.28 = - 1 . 6 8 V < V t
Transistor is in saturation mode.
6.4d) CORRECT ANSWER - C
The given circuit can be rearranged such that
Rth = 20 ka\\20 ka = 10 ka
Let us assume that the transistor is operating in saturation mode
Vg = Vt h = 2 V ,
Isg = 0, Vs g = 4 V - 2 V = 2V
Since Isd = k(Vsg - Vtvf = 0 .1 (2 - l) 2 = 0.1 mA
Writing KVL for source-drain loop:
4 ~ Vsd - Isd X 10 = 0
= > Vsd = 4 - Isd X 10 = 3 V
Kg > Vt
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Vgd = 2 V M
1.5 V
=
0.5 V
< \vtp\
Transistor is in saturation mode.
6.4e) CORRECT ANSWER - B
The circuit can be rearranged such that
Kt/l = 10fcn||20fcfl = 6.6 ka
Let us assume that the transistor is operating in saturation mode
Vg = Vth = 2.6V,
lgs = 0 A
Gate-Source loop KVL
2-6 ~ vg s~ 10lds + 5 = 0
Sinee I ds = k(Vg s- V t) 2
7.6 — Vas
,
x2
- I 0^i = 0 . lf e - 1 )
Solving the above given quadratic equations results in the following
Vgs = 3.12 V or Vgs = -2.1 V
Since we assumed saturation, the possible solution should be
Vgs > Vt implies Vgs = 3.12 V
lds = 0.1(3.12 - l) 2 = 0.44 mA
Vd = 5 - 10/* = 0.6 V
Checking the assumptions
= 3.12 V > Vt
Vdg — 0.6 V — 2.6 V = - 2 V < Vt
Transistor is in saturation mode.
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6.5 Operational Amplifier - Solutions
Consult NCEES® Reference Handbook - Pages 212 for reference
6.5a) CORRECT ANSWER - D
Using the NCEES® FE Reference Handbook equation for inverting amplifier
6.5b) CORRECT ANSWER - B
Using the NCEES® FE Reference Handbook equation for non-inverting amplifier
R = SkQ
6.5c) CORRECT ANSWER - C
Using the NCEES® FE Reference Handbook equation for operational amplifier
-R 2
^
-6
(
6\
+ C1 + R l ) = -3-1 + C1 + 3 ) 15 = 25 ^
(
R2\
6.5d) CORRECT ANSWER - B
This problem can be efficiently solved using principle of superposition as shown below:
6V power source shorted
3 -0
0 - V0
------ = ------ 10 ktt
50 ka
v_ = _15 y
0
3V power source shorted
Summing the two output voltages results in V0 = -15 V — 20 V = —35 V
6.5e) CORRECT ANSWER - D
Using the NCEES® FE Reference Handbook equation for operational amplifier
1 = 0.5 V
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6.6 Instrumentation - Solutions
Consult NCEES® Reference Handbook - Pages 123 - 125 for reference
6.6a) 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.
6.6b) CORRECT ANSWER - B
Relevant formula for this problem:
Rt = Ra[ 1 + a(T - T0) ] = 2 0 0 [1 + 0 .0 0 3 9 (1 5 )] = 21 1 a
6.6c) CORRECT ANSWER - A
Relevant formula for this problem:
R±
Rn
=
R3
rX
=-±R3
Ry
x
R2
= — —
Rt
1000 a
- 50 0 £1 = 50 0 0
3 100 a
0
6.6d) CORRECT ANSWER - B
Relevant formula for this problem:
ARAV x 4R
0.5 V x 4000 O
= = ------- --------------------------= 2 0 0 0
AV = W ) V in
6.6e) CORRECT ANSWER - D
Wheatstone bridge, RTD and Strain Gage can be used to measure resistance.
6.6f) CORRECT ANSWER - B
25 m | 150 ka = 16.6 kSl
&
10 ka\1100 ka = 9.09 ka
9.09
y = -------------------10 = 1.986 V
9.09 + 16.66 + 20
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6.6g) CORRECT ANSWER - A
10 mii5oofcn = 9.8 k a
9.8
V = -------------------10 = 2.10 V
9.8 + 16.66 + 20
6.6h) CORRECT ANSWER - B
ifcn||ifcn = o.5 k a
Without Ammeter,
VFit/i Ammeter,
I = —
= 4 mA
10
I = - — —- — —— =3.92 mA
2 + 0.5 + 0.05
4 - 3.92
% error = ---------100% = 2%
4
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6.7 Power Electronics - Solutions
6.7a) CORRECT ANSWER - B
Chopper converts DC to DC i.e. constant to variable DC or variable DC to constant DC.
6.7b) CORRECT ANSWER - A
Inverter converts DC to AC with required voltage and frequency.
6.7c) CORRECT ANSWER - C
Rectifier converts AC to DC.
6.7d) CORRECT ANSWER - D
Cyclo-converter changes given AC to AC with required frequency and amplitude.
6.7e) CORRECT ANSWER - B
The given circuit is a controlled half wave rectifier with firing angle of 45°.
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Chapter # 7 - Power
7.1 Single Phase Power - Solutions
Consult NCEES® Reference Handbook - Pages 203 and 204 for reference
7.1a) CORRECT ANSWER - A
Relevant formula for this problem:
V
I = — according to the given details V = 120 V and Z = 20 + 5/ SL
120 + jO V
1~ (20 + sj o ) + (2 + 2y n )
/ = 5.19/-17.650A
5 = VI*= 120/0° 5.19/17.65° VA
P = 622cosl7.65° = 592W
7.1b) CORRECT ANSWER - B
Relevant formula for this problem:
V
I — — according to the given details V = 10/80° V and Z = 5 + j H = 5.09/1 1.3°jQ
...
1= 1.96/68.7° A
7.1c) CORRECT ANSWER - C
Z = (5 +; n)||(-2; 0)= 27-67.3° O
V — I x Z =10/0 A x 2/292° O
5= V I*= 2QO/-67.30VA
7.1d) CORRECT ANSWER - B
Z = 10 + Sj - 2j a = 10 + 3j Si = 10.4/16.7°a and V=10/10°V
1/
I = -=
0.96/-6.7 °A
S = VI* = 9.6/16.7° VA
Power absorbed = P =
9.57cos(16.69°) = 9.19
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'
y.le) CORRECT ANSWER - B
Relevant formula for this problem:
(V\2
p=,2fl=y
V2
* = t =sw
7.If) CORRECT ANSWER - D
Maximum power transfer happens when
Z |oad
=
Z * T h e v in in
In order to calculate ZThevininreplace current source with open circuit and voltage source with a short.
^Thevenin — 10 — 5/ + 2 + 6 / = 12 + j Q
Therefore maximum power transfer happens when Z]oa(i =
Z * T h e V in in
=12 - j Q
7.1g) CORRECT ANSWER - A
Maximum power transfer happens when Z|oad=ZThevinin
In order to calculate ZThevinmreplace current source with an open circuit.
^Thevenin = 5 —/ + 3/ O — 5 + 2y O
Therefore maximum power transfer happens when Z\oad= 5 +2j Q
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7.2 Three Phase Power / Transmission & Distribution - Solutions
Consult NCEES® Reference Handbook - Pages 203 and 204 for reference
7.2a) CORRECT ANSWER - C
In a 3-(j>Y-connected system VL.Lleads VL_Nby 30° and is greater by a factor of V3
Vab =V3 x 120/30o+30° V= 208/60° V
Similarly l^c =208/-60°V and Fca =208/180° V
7.2b) CORRECT ANSWER - B
Relevant formula for this problem:
v
.
__ va n
'an ~ y
Substituting the given values in above details, Van=120/30°V and 7.® = 20 + 5j O = 20.6/14°f2
Ian = 5.82/16!A
7.2c) CORRECT ANSWER - C
Relevant formula for this problem:
_
Van
an
Ztj, + ZL
277 + 0; V
S + Sj + l + j O
lan =3 2 .6 /^ A
7.2d) CORRECT ANSWER - A
Relevant formula for this problem:
.
Q Tl
____ Km___
rj
-
rj
Moad + •‘■line
f«„= 120/60° V and Z„„c=2 +j fl
Zhad== 10 + 10 j U
'a n = 7 . 4 / 1 8 ! A
Vn r , = Ian X Z|n ari = 1 0 4 / 6 3 ° V
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7.2e) CORRECT ANSWER - B
Relevant formula for this problem:
CLTL
rs
rj
rj
^•line ' Moad
■
___
rj
Mine ' ^load —
Zload = ~
lan
- Zllne = 23.4 + 1.75; Q = 23.4/4.2°fi
*an
7.2f) CORRECT ANSWER - B
Relevant formula for this problem:
u
i — n
*an ~ ^
According to problem details, Vab = 208/30°^ and Z& = 10 + Sj Q
Since Van lags Vab by 30° and Van is smaller by a factor of V3, this makes Van = 120/0°K
Ian = 10.7/-26.5°A
7.2g) CORRECT ANSWER - D
According to the problem details:
Van = 120/O^V
It is easier to solve this problem by converting ZA tO Zy
ZA = 10 + 2j Q
ZA = 3 X Zy
ZY = 10+j i a = 3.4/11.3°Q
r'an - I 7k
Ly
Solving forIan results in 35.3/-11.3°i4
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7.2h) CORRECT ANSWER - A
According to problem details:
Zy-\ = 10 + 5j Q
ZA_2 = 6 + 9/D
ZA = 3 x ZY
Therefore ZY-2 = 2 + 3j Q
Z eq = Z y - i W Z y - 2 = (10 + 5y) 11(6 + 9j ) S l
= 2.8/48fO
7.2i) CORRECT ANSWER - C
Relevant formula for this problem:
Zdeita = 3
Zwye = 3 x 2.8 = 8.4 Q
X
7.2j) CORRECT ANSWER - D
Relevant formula for this problem:
i
_
* a n ~~ j
Moad
According to problem details:
Van = 120/O2V and Zload = 20/0^0
Ian= 6/0! A
S 3*
=
3 K /*
=
2 1 6 0 /0 ° V A
7.2k) CORRECT ANSWER - C
Relevant formula for this problem:
P = VI cosd
According to problem details:
P = 200 kW,
I = 400 A,
p f = 0.83 lagging
Therefore V
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7.21) CORRECT ANSWER - A
Relevant formula for this problem:
P = VI cosd
According to problem details:
Psource = VI cosd = 600 x 200 x 0.85 = 102 kW
Pload = 100 kW
P lo s s
P so u rce ~
P lo a d
~
2
k W
7.2m) CORRECT ANSWER - B
Relevant formula for this problem:
P = 3VI cosd
According to problem details:
P = 125 kW,
I = 300 A,
p f = 0.694
125 x 1000
V = ------------------= 200 V
3 x 300 x 0.694
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7.3 Voltage Regulation - Solutions
7.3a) CORRECT ANSWER - C
Relevant formula for this problem:
V.R =
650 V — 600 V
m x 100% = ---- — — ------x 100% = 8.33%
Vs,fi
600 V
V s n i- V s fi
7.3b) CORRECT ANSWER - C
An ideal transformer does not have any losses. As such, for an ideal transformer the no-load voltage is equal to
full load voltage and voltage regulation is zero.
7.3c) CORRECT ANSWER - A
Relevant formula for this problem:
V.R = a
V*.fi
10000
h,rated = ~7K?r = 25 A
400
P f = 085 & = 31.78°) phase angle = -31.78
\ = Vs + h,rated f a q + JXeq) = 425 V
VV u
77
“ vs,fi
425-400
V'R = S— — L . = ----- x 100% = 6.3%
400
Vs,fi
7.3d) CORRECT ANSWER - B
Relevant formula for this problem:
V s n l-V s n
V.R = “ -—t— — x 100%
*s,fl
vsnl-
2m v
0.05 = — --------240 V
Vs>nl = 240 V + 240 x 0.05 V = 252 V
7.3e) CORRECT ANSWER - B
For leading power factor Vsj t > Vsni
Therefore V.R < 0
200
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7.4 Transformers - Solutions
Consult NCEES® Reference Handbook —Page 204 for reference
7.4a) CORRECT ANSWER - B
Relevant formula for this problem:
Vs = Is x Rs = 1 x 12 = 12 V
Vp _ 50
V p _S 0
__T
12 “ T
= > Vp = 600 V
7.4b) CORRECT ANSWER - A
Relevant formula for this problem:
= > 15 kVA = K I
S = VI
15000 VA
I' = - u o v - = 1 2 5 A
7.4c) CORRECT ANSWER - A
Relevant formula for this problem:
/ - 5 kVA
s
120 v
/p =
I
4 1 ,6,4
10
T ~ ~ Jr-
A = 4.16^
7.4d) CORRECT ANSWER - D
Relevant formula for this problem:
Z-p
ci
Given Zs = 10 Cl and a = 50
Therefore Zp = 502 x 10fl = 25000 H
7.4e) CORRECT ANSWER - B
Relevant formula for this problem:
Zp = a2Zs
Given Zv = 10 M2||10 kfl = 5 k£l and a = 10
Zp = a2Zs = 102ZS
Zp
SkD.
z = -J L = —
5 100
100
= son
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7.5 Motors & Generators - Solutions
Consult NCEES® Reference Handbook - Page 204 for reference
7.5a) CORRECT ANSWER - C
Relevant formula for this problem:
120/
n = ----p
(120)(60 Hz)
=> 1800 rpm = -------------p
=> p = 4
7.5b) CORRECT ANSWER - C
Relevant formula for this problem:
120/
(120)(60Wz)
ns = ----- = ------- ------- = 3600 rpm
p
2
7.5c) CORRECT ANSWER - B
Relevant formula for this problem:
nc =
120/
V
120/
(120)(50 Hz)
50 Hz & 4 poles, ns = ----- = ------- ------- = 1500 rpm
p
4
120/
(1 2 0 1 0 (6 0 Hz)
p
2
60Hz and 2 poles, ns = ----- = -------- -------- = 3600 rpm
Change in synchronous speed =3600 rpm - 1500 rpm = 2100 rpm
7.5d) CORRECT ANSWER - D
Relevant formula for this problem:
120/
(120)(60)
,
ns- n
3 6 0 0 -3 4 0 0
ns
3600
ns = —- — = ---- ----- = 3600 rpm
Tl
Given
n = 3400 rpm
_ _ n/
Therefore s = —— = --------- = 5.5%
7.5e) CORRECT ANSWER - D •
Relevant formula for this problem:
120/
(120)( 60)
ns = —- — = ---- ----- = 1800 rpm
ns — n
S= - ^ ~
=>
1800 - n
01 = ^ 8 0 0 _
=> n = 1620 rpm
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7.6 Power Factor - Solutions
Consult NCEES® Reference Handbook - Page 204 for reference
7.6a) CORRECT ANSWER - D
Relevant formula for this problem:
^
P(tand1 — tand2)
cosdt = 0.75,
dt = 41.4°
cosd2 = 0.90,
d2 = 25.8°
C_
_____
200000(tan41.4° - tan25.8°)
C = 586 \lF
7.6b) CORRECT ANSWER - B
Relevant formula for this problem:
Qc = P(tan-d1 — tand2)
cos^ = 0.60,
cosd2 = 1.00,
= 53.1°
d2 = 0°
Qc = 100(tan53.1° - tanO0) = 133.1 kVAR
7.6c) CORRECT ANSWER - C
Relevant formula for this problem:
q
—
P(tand1 — tand2)
—
w v rm s
cosd± = 0.8,
dt = 36.86°
Covins = P(tCLnd1 — tand2)
Solving for d2 gives —6.81°, cos($2) = 0.99
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7.6d) CORRECT ANSWER - B
Relevant formula for this problem:
P
Power Factor = —
= 75/cosiO M
S2 = 3S/cos-i0.75
St = St + S2 = 90 + j 62.65 = 109.67/34.84°
P
90
Power Factor = — = —— = 0.82 lagging
S
109
7.6e) CORRECT ANSWER - D
Relevant formula for this problem:
^
P(tand1 — tand2)
^Vrms
cosdt = 0.85,
dl = 31.78°
cosd2 = 0.95,
d2 = 18.19°
125000(tan 31.87 ° - t a n 18.19°)
C~
(2tt)(60)(120)2
C = 6.7 mF = 7 mF
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Chapter # 8 - Electromagnetics
8.1 Maxwell Equations - Solutions
Consult NCEES® Reference Handbook - Pages 34 and 205 for reference
8.1a) CORRECT ANSWER - D
Relevant formula for this problem:
—
—
d
d
_ d
div E — V.E = — 3x + — 2y + — x = 3 + 4y
dx
dy
dz
8.1b) CORRECT ANSWER - B
Relevant formula for this problem:
_
_
d
d
d
div D = V.D — — x y + - r - y z + - j - x z £ = y + z + 2xz
dx '
d y7
dz
8.1c) CORRECT ANSWER - C
Relevant formula for this problem (Gauss' Law):
[
Qenc
+1 nC
TlC — 2 nC + 3 nC + 1 nC
Electric Flux = <b E.ds = ---- = ----- —
£
8.85 x 10~12 C/Vm
Electric Flux = 339 Vm
S.id) CORRECT ANSWER - B
Relevant formula for this problem (Gauss' Law):
Qi
in7
I r > Qenc
Ql + Qi
Electric Flux = (p E.ds — ---- = —— — J
e
8.85 x 10
1 ~12C/Vm
Electric Flux = —5963 Vm
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8.1e) CORRECT ANSWER - C
According to Maxwell's equations, the divergence of a magnetic field is zero.
_
_
d
d
d
div B = V.B = — 3ax + —-2y —— 2z = 3a + 2 — 2 = 0
dx
dy
dz
Therefore a = 0
8.If) CORRECT ANSWER - B
According to Maxwell's equations, the divergence of a magnetic field is zero.
div A — V.A — ——2x2 + — y + — 3z2 = 4x + 1 + 6z
dx
dy
dz
—
—d
„
d
d
div B =V.B = — 2y + — 3x —— 2xy = 0
dx
dy
dz
Since the divergence of B is zero it can be a magnetic field vector
8.1g) CORRECT ANSWER - B
According to Maxwell's equations:
d—
V x E = —— B
dt
Given E = 2yzi + 3x2yj + x 2y 2k
d —_
/dEz
dt**
\ dy
dEy\ . /dEx
dz ) *
v dz
dEz\ . fdEy
dx a
\ dx
dEx\
dy
— B = —2x 2yi — (2y — 2xy2)/ — (6xy — 2z)k
8.1h) CORRECT ANSWER - C
According to Maxwell's equations:
dV x E= - - B
dt
Given B = —cos2(3t ) k
d—
(
d
~dtB = ~ \ d t
0
\
d
t)^i ~~dt
^tSc~ -6 cos (3t) sin (3t)fc
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8.1s) CORRECT ANSWER - B
According to Faraday's Law, induced voltage is given as follow:
d&
v = —N ——
dt
N = 20,
d& = 1.4 Wb,
dt = 2s
20(1.4)
v = ---- = -14 V
8.1j) CORRECT ANSWER - A
Relevant formula for this problem (Faraday's Law):
di
v = L—
dt
N 2\jlA
L = — j— = 3.14 [iH
Given N = SO, A = 2 cm2 and I = 20 cm
Solving for v in the above equation results in 0.15 \iV
8.1k) CORRECT ANSWER - C
Relevant formula for this problem (Ampere's Law):
jH.dl = lenc + fjs %.<ts
lenc — NI = (50)(1 i4) = 50 ^
As per the given details
dD
l
dt
°
B
H = -
B = — !^ £ — =
2n x 2 cm
B f
- f d l = le
j) dl = 2nr = 2n(2 cm)
circumference o f torus
500 jiT
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8.1!) CORRECT ANSWER - C
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
B = - J L = 0.2 \iT
2nr
Magnetic flux density at point A due to first wire is given as:
I\i
B = - J - = 0.2 [iT
2 2nr
*
Summing the two magnetic fields results in:
B -n e t =
B1 + B2 = 0.4 \iT
Note that two magnetic fields in this scenario add due to current direction. If current was flowing in opposite
direction, the magnetic fields would have cancelled each other. Apply right hand rule in such questions.
8.1m) CORRECT ANSWER - B
Relevant formula for this problem (Ampere's Law):
I\i
B =rr~
2nr
2nrB
I = ----- = 100 kA
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8.2 Electrostatics/Magnetostatics - Solutions
Consult NCEES® Reference Handbook - Page 199 - 201 for reference
8.2a) CORRECT ANSWER - D
Relevant formula for this problem:
r = V d 2 + 22) = V5
F = 90 nN
8.2b) CORRECT ANSWER - C
Relevant formula for this problem:
^=
+
r = VT=1
Solving for \E\ results in 900V/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.
8.2c) CORRECT ANSWER - B
Relevant formula for this problem:
—
Ps
E = ^ -k
2e
Using superposition
rp Ps ,
Ps r .
Ps ,
E = — k - —-(-fe) = — k
2s
2eK J
£
Note that if both electric plates 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.
8.2d) CORRECT ANSWER - D
Relevant formula for this problem:
~E = ——— aT
Znsr
I = 2 ^ W ^ + 2 ^ ( b ) (- ^
= 1'07 X
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8.2e) CORRECT ANSWER - D
Relevant formula for this problem:
F = /Z"x B
|F| = 2 x (2 x 0.5) sin30° = 2 sin30° = I N
8.2f) CORRECT ANSWER - B
Relevant formula for this problem:
[iH2
Energy stored in magnetic field = ——- x volume = 5 \if
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8.3 Transmission Lines and Wave Propagation - Solutions
Consult NCEES® Reference Handbook - Page 205 for reference
8.3a) CORRECT ANSWER - C
Relevant formula for this problem:
+ %0
According to the given details Z0 = 100 H,
Zt - 300 4-50/0
r = 0.5 phase 6.91°
8.3b) CORRECT ANSWER - C
Relevant formula for this problem:
r —
~
~Zl +Z0
According to the given details Z0 = 50 H,
Zt = 500 4- 25j H
r = 0.818 phase 0.57°
1 4 |r|
SWR= i ^ w r w
8.3c) CORRECT ANSWER - B
Relevant formula for this problem:
f — ^1~
Z\ 4 Z0
L
- = io o a
z0 =
According to problem details F = 0.5
a c
_
^
~
_
z i ~~
■ “ ZJ + Zo ~Zi 4 l 0 0
zt = 300 a
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8.3d) CORRECT ANSWER - D
Relevant formula for this problem:
2tt _ 2tt
— A ~~10
( a \ - 7 Zl +;'z otan(/?rf)
inW
0 ZQ + j Zl tan (fid)
According to problem details: Z0 = 200 jQ&Zz = 500 Cl
500 + y'200tan(/? x 100)
Zin(100) = 2002 0 0 + ,500tan(/;xl00)
Since tan
(2n
x 100^J = 0
/500\
Zin(100) = 200^— J= 500 a
8.3e) CORRECT ANSWER - D
Relevant formula for this problem:
Zi ZQ
r = -----—
Given Z0 —250 (1
Zi+Z0
i + \r\
l
=>r = 3
SWR= r ^ n = 2
r
= ^Zi——Z0
-2.
Zi+Z0
=> z =
1
soo n
8.3f) CORRECT ANSWER - B
Relevant formula for this problem:
2 tc
2 tt
7T
,a. 0
v(d) = V+e^d + V~e~^d
/? = — = — = —
A
20
10
1/(100) = V+ejl0n + V~e~jl0n
8.3g) CORRECT ANSWER - C
Relevant formula for this problem:
2 n
2 tc
it
^ = T = 20 = lo
/(d) = 16
K
+ re
-
/+= -_&/" = - —
z0
z0
/(100) = - ± ( y +ejl0n - V~e-jl0n)
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8.4 Electromagnetic Compatibility - Solutions
8.4a) CORRECT ANSWER - D
Electromagnetic coupling can happen through induction, conduction and capacitive charging.
8.4b) CORRECT ANSWER - C
Shielding is a method used for protection against electromagnetic interference.
8.4c) CORRECT ANSWER - B
Cross talk happens when both emitter and receiver exist within the same system.
8.4d) CORRECT ANSWER - D
Lightning, arc welding and electric motors are potential sources of electromagnetic interference.
8.4e) CORRECT ANSWER - D
Increasing separation between coupling paths, hardware redundancy and shielding can help mitigate harmful
impacts of electromagnetic interference.
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Chapter # 9 - Control Systems
9.1 Block Diagrams - Solutions
Consult NCEES® Reference Handbook - Page 126 for reference
9.1a) CORRECT ANSWER - C
According to the given control system block diagram:
E(s ) = R(s) - H ( s ) Y ( s )
re s ) = E 0 ) G a(s )G 20 )
r ( s ) = ( R ( s ) - HCs)Y(s)]G1(s)G2(s)
Y(s )
_
f i(s )
CtW C.Q Q
l + H ( s ) C 1( s ) C z ( 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.
9.1b) CORRECT ANSWER - C
According to the given control system block diagram:
E (s ) = «(s) - (Y(s) + JV(s))
y '( s ) = ( £ ( s ) C 1 (5 ) + i ( s ) ) 6 2 ( s )
r ( s ) = ( [ R ( s ) - ( Y ( s ) + JV (s ))]C 1(s ) + i ( s ) ) G2 (s )
yf ^
W C tW C ;W - JVC^C^sjC.Cs) + L( s)G2( s)
^
1 + G1(s )G 2 (s )
9.1c) CORRECT ANSWER - A
According to the given control system block diagram:
E (s ) = R(s ) - Y( s)
r ( s ) = G4 ( s ) ( £ ( s ) G 1 ( s ) - £ ( s ) G 2 ( s ) + E ( s ) G 3 ( s ) )
5 04( 5X 0! ( s )
r(s) = £ ( )
- G2 ( s ) + G3 ( s ) )
/ ( s ) = G4 ( s ) ( K ( s ) - f ( s ) ) ( G 1 ( s ) - G 2 ( 5 ) + G 3 ( s ) )
^
_
S
j
,,
R(s)G4(s)(G1(s ') — C 2 ( 5j + G 3 ( s ) )
~ ( l + G4 ( s ) ( G 1( s ) — G 2 ( s ) + G3 ( s ) ) )
n
_
G4 ( s ) ( G 1 ( s ) - G2 ( s ) + G 3 ( s ) )
“ ( 1 + G4 ( s ) ( G 1 ( s ) - G2 ( s ) + G3 ( s ) ) )
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9.Id) CORRECT ANSWER - A
According to the given control system block diagram:
E(s) = / ? (s ) - K ( s ) « ( s )
K ( s ) = kE(s)Gt (s)
y(s) = kG ^ iR is) - r(sMs))
Y( s) =
r(s) =
kG1(s)R(s')
1 + /cG1 ( s ) / / ( s )
/cGiO?)
1 + /cG1 ( s ) / / ( 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.
9.1e) CORRECT ANSWER - B
According to the given control system block diagram:
£ ( s ) = R(s) - Y(s)
r ( s ) = f O K C - O G j O ) + H ( s ) Y ( s ) + £‘(s)G1(s)G2(s)G3(s)
K ( s ) = ( f l ( s ) - r ( s ) ) G a ( s ) G 2 ( s ) + H ( s ) K ( s ) + ( f i ( s ) - r ( s ) ) G 1( s ) G 2 ( s ) G 3 ( s )
Y(s) = 1
( R ( s) ) ( G 1( s)G2( s) + G1(s)G2( s) G3( s ) )
)
'
,(1 + G 1( s ) G 2 ( s ) - H ( s ) + G 1 ( s ) G 2 ( s ) ) G 3 ( s ) /
T( \ _ ( _____Gl( s)G2(5) + Gl( s)G2(s)G3(s)_____ \
W
“ 1 (1 + G1 ( s ) C 2 ( s ) - H (s) + G 1( s ) G 2 ( s ) ) G 3 ( s ) J
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9.2 Bode Plot - Solutions
Consult NCEES® Reference Handbook - Page 207 for
9.2a) CORRECT ANSWER - A
Transfer function can be rearranged in standard form as shown below:
_
50
_
s + 10
50
5
^+1
lo (-^+l)
j
Gain = 20 log(5) = 13.97 dB. Transfer function has pole at s = —10
9.2b) CORRECT ANSWER - D
Transfer function can be rearranged in standard form as shown below:
//(<>) =
20
1
s + 20
20
i
Gain = 20 log(l) = 0 dB. Transfer function has pole at s = —20
9.2c) CORRECT ANSWER - A
Transfer function can be rearranged into standard form as shown below:
H(s) =
50(5 + 2)
=/ l
(s + 1 0 0 )(s + 1 0 0 0 )
\
( | + 1)
V1000J f_ L _ + i
( s
. i^
V100+ i A l000 + V
Transfer function has pole at s = —100, s = -1000 and zero at s = —2
9.2d) CORRECT ANSWER - C
Transfer function can be rearranged in standard form as shown below:
,
1 0 0 (s + 1)
,
,
(s + 1 )
= (10)
H( s) = ■— ...
s(s + 10)
(S) ^ +1^
Gain = 20 log(10) = 20 dB
9.2e) CORRECT ANSWER - D
Transfer function can be rearranged in standard form as shown below:
H (s ) =
Gain =
100s
100s
(s2 + 1505 + 5000)
0 + 50)(S + 100)
20 log
s
50
+ 1)(_£_ + ^
= —33.9 dB
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9.3 Steady State Errors - Solutions
Consult NCEES® Reference Handbook - Page 127 for reference
9.3a) CORRECT ANSWER - B
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---- — —
'
s -» o
w
s^ol +
G (s )
10
r(t) = 10u(t)
R( s) = —
s
(50)(s + 4)
G(s) =
(s + l)(s + 5)
10
6(00) =
, (50)% + 4)'
(s + 1)(s + 5)
e(CO) = °'24
9.3b) CORRECT ANSWER - A
For a unity feedback system, steady state error can be calculated using final value theorem as shown below:
s i? (s )
e ( o o ) = lim sE(s) = lim---- — -v '
5—>o
v J
s - > o l + G(s)
x
r(t) = St u(t )R(s ) = —r
5
si
100s
G (s ) =
s2 + 11s + 30
^_5_
e(co)=
—
f k —
s2 + 11s + 30
e(0Q) = “
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9.3c) CORRECT ANSWER - A
For a unity feedback system, steady state error can be calculated using final value theorem as shown below:
sjR( s )
e (c o ) = lim sE(s) = lim- ------— —
v
y
5^0
v
'
r ( t ) = 1012 u(t )
G (s ) =
s->o 1 + 6 ( s )
20
R( s) = —
s-3
s2 + 100s + 10
20
s—
e (o o ) = l i m ---------------- ------------------
S-»0 1 _i______ ______
e (o o ) = oo
s2 + 100s + 10
9.3d) CORRECT ANSWER - A
For a unity feedback system, steady state error can be calculated using final value theorem as shown below:
sfl(s)
s-»o 1 + G{s)
e(oo) = lims£Ys) = lim---- — —
v '
s-»o
r(t) = 2t u ( t )
G(s) =
vJ
R(s ) = —r
2
sz
(s + 2)(s + 3)
e (o o ) = lim --------------- -----------------
e (o o ) = oo
s^° i + ____ _____
(s + 2)(s + 3)
9.3e) CORRECT ANSWER - C
For a unity feedback system, steady state error can be calculated using final value theorem as shown below:
, \
sR(s)
e(co) = lims£Ys) = lim ------ —
s->0 v '
s-»0 1 + G(s)
r(t) = 3u( t)
G(s) =
3
R(s ) = s
10(s + 5)
(s + l)(s + 2)
3
s-
e(oo) = lim ------ Txy
—
s->o 1
10(s + 5)
(s + l)(s + 2)
e(oo) = 0.115
v '
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9.4 Routh Hurwitz Criteria & System Stability - Solutions
Consult NCEES® Reference Handbook - Page 127 for reference
9.4a) CORRECT ANSWER - C
Using the classical negative feedback control system diagram given in NCEES® FE Reference Handbook
y
(
s
)=
f
i(
s
)
f
c
(
(
s+
lX
s+
2
)
)
1 + (s + 1)0 + 2)
R(s)ks
Y( s) = ------ — ------s2 + 3s + 2 + ks
The closed loop characteristic equation is s2 + 3s + 2 + ks = 0
9.4b) CORRECT ANSWER - B
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
s4
s3
2
7
2
3
3 x 2 -2 x 7
**=
3
8
=
3
Sign changes for bx therefore system is unstable
9.4c) CORRECT ANSWER - B
The closed loop characteristic equation is 2s4 + 3s3 + s2 + s + l = 0
Routh Array can be developed using NCEES® FE Reference Handbook as shown below:
s4
s3
2
s2
h
s
1
4-3x1
€ 1 = ^ - 1 ---- = - 8
3
1
1
0
1
1
3 -2
—
3
—
1
3
hy —
—
U
2
3-2x0
_
3
—
— 1
JL
-
3
Sign changes for ^therefore system is unstable
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9.4d) CORRECT ANSWER - C
The closed loop characteristic equation is 3s3 + 5s2 + (k + 10)s + 5/c = 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/c
bl "
5k
5
For the system to remain stable
5(k + 10) - 3 X 5/c
> 0 -* k < 5 also 5/c>0-»/c>0
Therefore stability range is 0 < k < 5
9.4e) CORRECT ANSWER - B
The closed loop transfer function of given block diagram is shown below:
Ck + 1)(s + 1)
(s)(s + 2)(s + 3)
T(s) =
1+
T(s) =
' (/c + l)(s + m
<s(s + 2)(s + 3)/
(k + l)(s + 1)
s3 + 5s2 4- s(/c + 7) + /c + 1
The closed loop characteristic equation is s3 + 5s2 + s(k + 7) + /c + 1 = 0
Routh Array can be developed using NCEES® FE Reference Handbook as shown below:
s3
s2
s
5°
b,=
1
5
M
/c + 7
fc + 1
k+1
5(k + 7) - (fc + 1)
34 + 4/c
5
5
For the system to remain stable
34 + 4k
17
b-, > 0 -»--- ---- > 0 -» k > ----1
5
2
Also k + 1 > 0 Therefore k > - 1
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9.5 Root Locus - Solutions
Consult NCEES® Reference Handbook - Page 128 for reference
9.5a) CORRECT ANSWER - A
Refer to page number 128 of NCEES® FE Reference Handbook for root locus.
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 = ---------------n = 2,
n —m
m — 0,
k = 0,1
a = 90°, 270
Asymptote centroids can be calculated using following formula:
<
ja =
Id=iRe(pi) - Z ^ R e i m i )
------------------------------------ ,n = 2,
n —m
m =
0,
(jA
= -1
Locus does not cross imaginary axis.
9.5b) CORRECT ANSWER - C
Refer to page number 128 of NCEES® FE Reference Handbook for root locus.
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:
[(2fc + 1)180°]
a = i! -------i----- n = 3,
n —m
m = 0,
k = 0,1,2
a = 60°, 180°, 300°
Asymptote centroids can be calculated using following formula:
0
I i= 1R e ( P i') - 'E il1Re(mi)
--------------------------- ,71 = 2,
72- 771
1
771 = 0,
oA = —~
3
Locus crosses imaginary axis.
9.5c) CORRECT ANSWER - D
Refer to page number 128 of NCEES® FE Reference Handbook for root locus.
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:
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a
[(2k +1)180°]
„
= --------------------- , n = 2,
m =
71 — 771
1,
,
n
A: = 0
ion0
a = 180
Asymptote centroids can be calculated using following formula:
H = i Re(pi) —Z K i Re(rrii)
Oa = ---------------------------- ,71 = 2,
71 — 771
771 = 1,
oA = 10
Locus crosses imaginary axis.
9.5d) CORRECT ANSWER - B
Refer to page number 128 of NCEES® FE Reference Handbook for root locus.
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:
a
[(2k + 1 ) 1 8 0 ° ]
^
= ------------------------- , n = 2 ,
71 — 771
?
77i = 1 ,
n
k — 0,
_
0
a —18 0 °
Asymptote centroids can be calculated using following formula:
;'Zi=iRe(pi) - I i L 1Re(m i)
aA = ---------------------------- ,71 = 2,
71 — 771
77i=l,
oA — —3
Locus does not cross imaginary axis.
9.5e) CORRECT ANSWER - A
Refer to page number 128 of NCEES® FE Reference Handbook for root locus.
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:
[(2k + 1)180°]
a —------------------------- ,
7 1 — 771
71 = 2,
77i = l ,
k = 0
a
=
180°
Asymptote centroids can be calculated using following formula:
8
Y i =1Re(pi) - Y i L 1Re(mi)
°A = ---------- ---------------- ' n = 2'
m = 1’
°* = - 3
Locus does not cross imaginary axis.
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9.6 State variables - Solutions
Consult NCEES® Reference Handbook - Page 128 for reference
9.6a) 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 x1 = y, x2 = y ' — y
Representing the variables as first order derivatives
xt = x2 and x2 = —2xt — 3x2 + 2u( t)
The state equation is shown below in matrix form
9.6b) 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 x1 = y,x2 = y ' =■ y
Representing the variables as first order derivatives
x± = x2 and x2 = —5x1 — 4x2 + 3u(t)
The state equation is shown below in matrix form
9.6c) CORRECT ANSWER - C
3y " ' + 6y" + 12y' + 3y = 9u( t) -» y ' " + 2y" + 4y' + y = 3u( t)
The given differential equation is of third order therefore it has three state variables.
Let us assume xx = y, x2 = y ' = y, x3 = y " = y
Representing the variables as first order derivatives
xx = x2 , x2 = x3 and x3 = —xt —4x2 — 2x3 + 3u( t)
The state equation is shown below in matrix form
■Xt
' 0
x2 = 0
-1
*3-
1
0
_4
0 ‘ ■Xi
'O'
x2
+ 0 u(t)
1
-2 . X3.
.3.
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9.6d) CORRECT ANSWER - A
Standard state variable model is given as:
y(t) = C x ( t ) + D u ( t )
x(t) = Ax(t) + Bu(t)
Taking Laplace Transforms of the above equations results in:
Y ( s ) = CX(s) + D U (s )
sX(s) = A X (s ) + B U (s )
s I X ( s ) = i4X(s) + B U ( s )
X( s) = (si —A ) ~ 1BU( s)
rearranging the equation results in ( s i — A ) X ( s ) = B U ( s )
substituting X(s)in output equation results in following
7 (5) = [C( sl - A ) ~ %B + D ] U ( s ) where [C( s l - A ) ~ XB + D] is the t ransfer function
9.6e) CORRECT ANSWER - D
As shown in Problem 9.6d), transfer function is related to state equation and output equation as follows:
T(s) = C(sl - A )~ 1B + D
In the given equation C = [1
0], ^4 = J ^
^ J ,B =
and D ~ [0]
9.6f) 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 xt [n] = y{n],x2[n] = y[n + 1]
x^n + 1] = x2[n] and x2[n + 1] = - 3 x t [n] — 3x2[n] + 2u[n]
The matrix form of state equation is shown below
'x1[n + 1]
x2[ n + 1]
0
-3
l] p iW
- 3 n x 2[n]
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Chapter # 10 - Communications
10.1 Amplitude Modulation - Solutions
Consult NCEES® Reference Handbook - Page 208 for reference
10.1a) CORRECT ANSWER - A
Relevant formula for this problem:
According to given details:
Pc = 1000 W & m = 80%
pt = pc ( i + y ) = 1320 w
10.1b) CORRECT ANSWER - D
Relevant formula for this problem:
According to given details:
Pt = SkW &
Pc = 4.75 kW
10.1c) CORRECT ANSWER - D
As per NCEES® FE Reference Handbook, Costas loop is used for detecting Double-Side Modulation (DSB).
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lO.ld) CORRECT ANSWER - C
Relevant formula for this problem:
carrier amplitude
Modulation Index = —---- :----- — —
signal amplitude
According to given details:
m( t ) = 5sin27r(1000t)c(t ) = 50sin27r(4000t)
5
Modulation Index = — = 0.1 = 10%
50
lO.le) CORRECT ANSWER - C
Relevant formula for this problem:
a2 < m2( t ) >
^
1 + a2 < m£(t) >
According to given details:
a = 0.8
< m^(t) > = 0.6
a2 < m j(t) >
0.82 x 0.6
7] = ----- - ^ y;= -— — — — ■
= 0.27 = 27%
1 + a2 < ?nj(t) > 1 + 0.82 x 0.6
lO.lf) CORRECT ANSWER - A
Amplitude modulation results in variation of carrier wave's amplitude in synchronization with that of signal
wave.
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10.2 Angle Modulation - Solutions
Consult NCEES® Reference Handbook - Page 208 - 209 for reference
10.2a) CORRECT ANSWER - D
Relevant formula for this equation:
B = 2 (D + 1) W because D = 1.25 > 1
Therefore B = 2 ( D + 1 ) W = 2 (1.25 + 1) x 10000 = 45000 Hz .
10.2b) CORRECT ANSWER - B
As stated on page 205 of NCEES® FE Reference Handbook, a phase-lock loop can demodulate angle modulated
signals.
10.2c) CORRECT ANSWER - B
Frequency modulation results in variation of carrier wave's frequency by signal wave.
10.2d) 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 and are collectively called as angle modulation.
10.2e) CORRECT ANSWER - D
Relevant formula for this equation:
B = 2 W because D = 0.1 < 1
Therefore B = 2 W = 2 x 1 0 kHz = 20 kHz
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10.3 Pulse Code Modulation (PCM) & Pulse Amplitude Modulation (PAM) - Solutions
Consult NCEES® Reference Handbook - Page 209 for reference
10.3a) CORRECT ANSWER - B
Relevant formula for this problem:
q = 2n where 'q' is the quantization levels and 'n' is the number of bits.
According to the given problem n = 7
Therefore q = 27 = 128 levels
10.3b) CORRECT ANSWER - D
Relevant formula for this problem:
B = 2 W log2 q
q = 28 = 256
B = (2) (100) log2 256
B = 1600 Hz
10.3c) CORRECT ANSWER - B
According to Nyquist Sampling Theorem:
Minimum clock frequency = 2 W = 2 x 15 kHz = 30 kHz
10.3d) CORRECT ANSWER - C
Relevant formula for this problem:
1
1
Ts = — =
= 33.3 fis
fs
30kHz
10.3e) CORRECT ANSWER - D
Quantization, encoding and sampling are part of Pulse Code Modulation (PCM) process.
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10.4 Fourier Transform - Solutions
Consult NCEES® Reference Handbook - Page 30 - 32 for reference
10.4a) CORRECT ANSWER - B
The given function is a rectangular pulse centered at 0 with amplitude of 5 and duration of 8.
This can be represented as 511
According to Fourier Transform pair table given in NCEES® FE Reference Handbook:
f t \ Fourier Transform
511 (—J <
--------------- > 5(8sinc(8/)) = 40 sinc(8/)
10.4b) CORRECT ANSWER - A
The given function is a rectangular pulse centered at 2 with amplitude of 3 and duration of 4.
This can be shown as 30
According to Fourier Transform pair table given in NCEES® FE Reference Handbook:
(
t—
-—
2\
, . ,
F o u r ie r T r a n s fo r m
)<---------------- >3(4sinc(4/))er;27r/(2) = 12 sinc(4/)e“ 4j7r/
10.4c) CORRECT ANSWER - C
According to Fourier Transform pair table given in NCEES® FE Reference Handbook:
II
/ t \ Fourier Transform
---------------- >6 sinc(6/)
)<
/ t \
cos(27r(300)t) n Q
F
o il t ip t
TrnnsfovTYi
10VTierTranSf-°™ 3 sinc(6 {f - 300)) +3 sinc(6( f + 300))
10.4d) CORRECT ANSWER - D
According to Fourier Transform pair table given in NCEES® FE Reference Handbook:
1
Fourier Transform
e~5tu ( t) <
----------- ---- >---------
W
5 +j2 n f
_.
F o u r ie r T r a n s fo r m
1 (
cos (27r(20)t)e_ u(t) ,--------------- g + ^
^
10.4e) CORRECT ANSWER - C
According to Fourier Transform pair table given in NCEES® FE Reference Handbook:
t — 2\ Fourier Transform
(
<
---------------- >4(2 sinc(2/))e-(47r^}
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11\
10.5 Multiplexing - Solutions
10.5a) CORRECT ANSWER - A
Time Division Multiplexing (TDM) is an example of digital multiplexing.
10.5b) CORRECT ANSWER - C
N channels require at least N-l guard bands. Therefore 3 channels will require atleast 2 guard
bands.
Min.BW — Channel #1 BW + Guard Band + Channel# 2 BW + Guard Band +Channel #3 BW
Min. BW = 50 kHz + 5 kHz + 100 kHz + 5 kHz + 50 kHz = 210 kHz
10.5c) CORRECT ANSWER - C
Each frame carries 1 byte per channel.
Since there are 5 channels, the frame size will be:
5 x 1 byte = 5 bytes
10.5d) CORRECT ANSWER - B
Each channel is sending 10 bytes/second and each frame carries 1 byte/channel.
Therefore we have 10 frames/second.
Each frame =5 bytes
Therefore 10 x 5 bytes j second = 50 bytes /second = 400 bits / second = 400 bps.
10.5e) CORRECT ANSWER - B
Each frame carries 1 bit
Since there are 50000 frames per second and 3 bits per frame, total number of bits/second is 150000
10.5f) CORRECT ANSWER - C
There are 50000 frames per second
Frame duration — 1 / 50000 = 20 \is
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Chapter # 11 - Computer Networks
11.1 Routing & Switching - Solutions
11.la) CORRECT ANSWER - D
Routing is the process of finding efficient paths between nodes based on their address.
11.1b) CORRECT ANSWER - D
Router maintains routing table as well as forwarding table.
11.1c) CORRECT ANSWER - C
Network Layer is responsible for performing node management functions such as routing, addressing and traffic
control.
ll.ld ) CORRECT ANSWER - A
Router is responsible for connecting two or more networks for transferring data packets,
ll.le ) CORRECT ANSWER - C
Network switches create a network and allow devices within a network to communicate with each other,
ll. lf) CORRECT ANSWER - B
Network switch typically operates in Data Link Layer. They can also perform routing functionality in Network
Layer if upgraded.
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11.2 Network Topologies / Frameworks / Models - Solutions
11.2a) CORRECT ANSWER - D
Bus, Star and Ring are examples of different network topologies.
11.2b) CORRECT ANSWER - D
Wireless networking has potential security issues associated with data transfer.
11.2c) CORRECT ANSWER - C
Application layer will run http and email.
11.2d) CORRECT ANSWER - A
Presentation layer is responsible for language translations.
11.2e) CORRECT ANSWER - A
Session layer is responsible for managing user interactions.
11.2f) CORRECT ANSWER - C
A single break in connection can disrupt entire network in Ring Topology implementation.
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11.3 Local Area Networks - Solutions
11.3a) CORRECT ANSWER - D
Local Area Networks (LAN) can be implemented using any of the suggested implementations.
11.3b) CORRECT ANSWER - D
LAN can be operated using Ethernet, Wireless or Asynchronous Transfer Mode (ATM) technologies.
11.3c) CORRECT ANSWER - B
Ethernet is an example of Bus topology.
11.3d) CORRECT ANSWER - A
Asynchronous Transfer Mode (ATM) LAN is an example of star topology.
11.3e) 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)
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Chapter # 12 - Digital Systems
12.1 Number Systems - Solutions
Consult NCEES® Reference Handbook - Page 217 for reference
12.1a) CORRECT ANSWER - A
Number system conversion can be efficiently done using calculators.
8ADHEx= IOOOIOIOIIOI2
12.1b) CORRECT ANSWER - B
Number system conversion can be efficiently done using calculators.
96310=11110000112
12.1c) CORRECT ANSWER - D
l's complement addition
(+2) OOIO2
(+3)0011?
(+5) 01012
Note that given numbers are unsigned (MSB of positive numbers is 0)
12.I d ) CORRECT ANSWER - C
l's complement addition
11012=- (0010)2 =- (2)10
10112= - (0100)2= - (4)io
HOI2 (_2)10
+1011,
MU
1 1000
carry ___1
IOOI2
(-6)10
Note that carry is added in l's complement
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12.1e) CORRECT ANSWER - B
2's complement addition
10112 (-5)
+01012
1 0000
(+5)
= 00002
Ignore carry
Note that carry is ignored in 2's complement.
12.1f) CORRECT ANSWER - A
2's complement addition
11102 (- 2 )
+0101z
1 0011
(+5)
(+3) = 00112
Ignore carry
Note that carry is ignored in 2's complement.
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12.2 Boolean Logic - Solutions
Consult NCEES® Reference Handbook - Page 217 for reference
12.2a) CORRECT ANSWER - D
Applying De Morgan's theorem to given logical function results in:
A + BCD + (E + F) = ( a ) ( B C D ) (E + F)
A + BCD + (E + F) = ( I ) ( B + C + D ) ( E ) ( F )
12.2b) CORRECT ANSWER - A
Applying De Morgan's theorem to given logical function results in:
(ABC) (D + EF) = (ABC) + (D + EF)
(ABC) (D + FF) = I + £ + C + D( E + F)
12.2c) CORRECT ANSWER - B
Applying De Morgan's theorem to given logical function results in:
AB + A(BC) + B(A + C ) = A B + A (B +
7P) + B A C
AB + A(BC) + B(A + C) = AB + AB + AC + A C B
AB + A(BC) + B(A + C) = AB + A C + A C B
AB + A(BC) + B(A + C) = AB + C(A + AB)
Since (A + AB) = (A + B)
AB + A (BC) + B (A + C) = AB + C (A + B)
12.2d) CORRECT ANSWER - B
AB + B (A + C) + C (A + B ) = AB + AB + BC + AC + BC
Since A + A = A
AB + B (A + C) + C (A + B ) = AB + BC + AC
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12.2e) CORRECT ANSWER - C
(i4 + f?) (A + B + C) =: A + AB + i4C + AB + B + BC
Since A + A = A
(A +
B) (A+ B
+ C) = A + AB + AC + B + BC
(A +
fi) (y4+B
+ C) = >4(1 4* 5) +
04 +
F)(v4+5
+ C ) = v 4 4 - B 4 - , 4 C = , 4 ( l + C) + £ = ,4 + £
4* B (1 4- C)
Since v4(l + Z?)
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12.3 Logic Gates - Solutions
Consult NCEES® Reference Handbook - Page 217 for reference
12.3a) CORRECT ANSWER - B
(AB)XOR (CD) = AB (CD) + (AB) CD
(AB)XOR (CD) = A B ( C + D) + CD (A + B)
(AB)XOR (CD) = ABC 4- ABD + ACD + BCD
12.3b) CORRECT ANSWER - C
(A B )(A B)
= ( A B ) + (AB)
(A B )(A B) = A + B + A + B
(A B )(A B) = 1
12.3c) CORRECT ANSWER - A
ABC + C = A + B + C + C
ABC + C = A + B + C
12.3d) CORRECT ANSWER - C
(A)(B)(C) = A B C
12.3e) CORRECT ANSWER - B
( A B ) ( A + B) = AAB + A B B
( A B ) ( A + B) = 0 + 0
( AB) ( A + B ) =
1
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12.4 Karnaugh Waps - Solutions
Consult NCEES® Reference Handbook —Page 217 - 218 for reference
12.4a) CORRECT ANSWER - B
C
0
1
12.4b) CORRECT ANSWER - D
C
0
1
12.4c) CORRECT ANSWER - C
CD
00
01
11
10
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12.4d) CORRECT ANSWER - A
CD
00 01
11
10
12.4e) CORRECT ANSWER - C
CD
00 01
11
10
1
1
1
1
1
1
1
V
JL
D + BC
12.4f) CORRECT ANSWER - C
C
0
1
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12.5 Flip Flops and Counters - Solutions
Consult NCEES® Reference Handbook - Page 217 - 218 for reference
12.5a) CORRECT ANSWER - C
Input 101 to Flip Flop #1 appears as 010 at its output#Flip Flop #1 Q is then provided to input of Flip Flop #2 and appears as such on output Q of Flip Flop #2 as 010.
12.5b) 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.
12.5c) 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.
12.5d) 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.
12.5e) 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 (Q from Flip Flop # 1)
R (CV from Flip Flop #1)
a
States
0 0
1
1
0 0
1
0
1
Therefore RS output is 001.
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12.5f) CORRECT ANSWER - B
The given logic circuit is an implementation of Johnson counter.
Clock cycle
0 - initial state
1
2
3
Qi
0
i
i
i
02
0
0
1
1
Qs
0
0
0
1
12.5g) CORRECT ANSWER - D
The given logic circuit is an implementation of 4 bit Asynchronous down counter.
Clock cycle
- initial state
0
O o
Q
i
0 2
O s
1
1
1
1
0
1
1
1
2
1
0
1
1
3
0
0
1
1
1
12.5h) CORRECT ANSWER - A
The given logic circuit is an implementation of a Synchronous up counter.
Clock cycle
0 - initial state
1
2
3
4
Qi
0
1
0
1
0
Qo
0
0
1
1
0
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12.6 State Machine Design - Solutions
Consult NCEES® Reference Handbook - Page 21 for reference
12.6a) CORRECT ANSWER - C
From the state diagram it is clear that if the present state is C and w=0 system transitions to state A and output
is 1. Similarly if the present state is C and w=l system transitions to state B and output is 1.
12.6b) CORRECT ANSWER - D
The table given below summarizes the 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
12.6c) 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
12.6d) CORRECT ANSWER -A
From the state diagram it is clear that if present state is C and ab =00 system stays in state C and output is 0.
Similarly if the present state is C and input ab =01 system transitions to state D and output is 0.
12.6e) CORRECT ANSWER - D
The output expression for w as a function of y2yi can be found using k-map as shown below
Vi
0
1
w = y2
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Chapter #13 - Computer Systems
13.1 Architecture & Interfacing - Solutions
13.1a) CORRECT ANSWER - D
Addressing modes are different ways in which address of an operand is given in an instruction.
Register Addressing Mode - Used when operands are located 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.
13.1b) CORRECT ANSWER - A
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.
Assembler - A language processor that converts assembly language (source code) into machine language (object
code) for execution.
13.1c) CORRECT ANSWER - D
Control Unit-The 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 Unit-The 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.
13.1d) 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.
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13.1e) 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.
13.1f) 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.
13.lg) 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.
13.1H) 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.
13.11) CORRECT ANSWER - B
Instruction pipelining increases instruction throughput. It neither decreases the instruction execution time nor
does it allows new types of instructions.
13.1j) CORRECT ANSWER - B
Control Unit-The 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.
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13.2 Microprocessors - Solutions
13.2a) 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.
13.2b) 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.
13.2c) CORRECT ANSWER - A
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.
13.2d) 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.
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13.2e) 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.
13.2f) CORRECT ANSWER - C
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.
13.2g) CORRECT ANSWER - A
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.
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13.3 Memory Technology & Systems - Solutions
13.3a) 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.
13.3b) CORRECT ANSWER - C
Giga in binary = 230and Byte = 8
Therefore 1 Giga Byte = 8 x 230 bits
13.3c) CORRECT ANSWER - C
Cache memory has low capacity but very high speed. It stores frequently used data for ready access.
13.3d) CORRECT ANSWER - C
DVD is an example of secondary memory storage device. It is non-volatile and cannot be directly accessed by the
computer.
Primary memory systems are volatile and can be accessed directly.
13.3e) 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.
13.3f) 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.
13.3g) CORRECT ANSWER - B
Bit -1 digit, Byte - 8 digits, Nibble - 4 digits and Word -16/32/64 bits depending on system architecture.
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Chapter #14 - Software Development
14.1 Algorithms - Solutions
14.1a) CORRECT ANSWER - C
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.
14.1b) CORRECT ANSWER - B
As shown in above solution z has a final value of -2 at the end of loop #4.
14.1c) CORRECT ANSWER - D
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.
14.1d) CORRECT ANSWER - B
B
1
A
1
5
C
3
2
5
2
=B$1=5
3
7
3
=B$2=2
4
=BlxC2=25
5
=B$3=3
14.1e) CORRECT ANSWER - C
A
B
C
10
=A1+Cl=20
=A lx 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
=A4x 4=100
1
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14.2 Data Structures - Solutions
14.2a) CORRECT ANSWER - C
Array, list, stacks and trees are different implementations of data structures and help organize data efficiently.
14.2b) CORRECT ANSWER - B
Worst-case time is used to measure the performance of searching algorithms. "Chain is only as strong as its
weakest link".
14.2c) CORRECT ANSWER - C
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.
14.2d) CORRECT ANSWER - A
The worst case execution time for a binary search algorithm is log2n.
14.2e) CORRECT ANSWER - D
Bubble, Quick and Heap are examples of sorting algorithms because they sort given data in specific order.
14.2f) CORRECT ANSWER - B
Bubble sort has worst case computational time of n2
Insertion sort has worst case computation time of nlog2n
Heap sort has worst case computation time of nlog2n making it the best performer out of the given options.
14.2g) 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.
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14.3 Software design methods, Implementation and testing - Solutions
14.3a) 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.
14.3b) CORRECT ANSWER - B
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.
14.3c) CORRECT ANSWER - A
A software design that is impossible or very different to extend is generally considered inefficient.
14.3d) CORRECT ANSWER - A
Static software testing involves verification using program code reviews, walk-through & inspections.
14.3e) CORRECT ANSWER - B
Dynamic software testing involves verification through actual program execution.
14.3f) CORRECT ANSWER - A
"Top-down" programming approach is associated with structured programming.
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14.3g) CORRECT ANSWER - B
The concept of "Class" is associated with object-oriented programming.
14.3H) CORRECT ANSWER - D
FORTRAN, BASIC and COBOL are examples of structured programming languages whereas C# is an example of an
object-oriented programming language.
14.3i) CORRECT ANSWER - D
PHP, JavaScript and Python are examples of scripting languages.
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