Function generators and oscilloscopes

Gengsheng Lawrence Zeng Megan Zeng Electric Circuits A Concise, Conceptual Tutorial Electric Circuits Gengsheng Lawrence Zeng • Megan Zeng Electric Circuits A Concise, Conceptual Tutorial Gengsheng Lawrence Zeng Utah Valley University Orem, UT, USA Megan Zeng University of California, Berkeley Berkeley, CA, USA ISBN 978-3-030-60514-8 ISBN 978-3-030-60515-5 https://doi.org/10.1007/978-3-030-60515-5 (eBook) # The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed The use of general descriptive names, registered names, trademarks, service marks, etc in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Preface Are you a student who is looking to supplement what you are learning in class? Or are you simply interested in electric circuits? Electric Circuits: A Concise Conceptual Tutorial gives you an opportunity to understand fundamental electrical engineering concepts This book is written in a reader-friendly format like a pictorial dictionary, and you can directly jump to any topic you want to learn more about without having to read the entire book sequentially We hope that this book will help save your time in grasping difficult concepts in electric circuits Good luck and have fun! Orem, UT Berkeley, CA 2020 Gengsheng Lawrence Zeng Megan Zeng vii Contents Voltage, Current, and Resistance Exercise Problems DC Power Supply and Multimeters Exercise Problems 15 Ohm’s Law Exercise Problems 17 19 Kirchhoff’s Voltage Law (KVL) Exercise Problems 23 26 Kirchhoff’s Current Law (KCL) Exercise Problems 31 33 Resistors in Series and in Parallel Exercise Problems 37 40 Voltage Divider and Current Divider Exercise Problems 43 46 Node-Voltage Method Exercise Problems 49 55 Mesh-Current Method Exercise Problems 59 63 10 Circuit Simulation Software Exercise Problems 67 71 11 Superposition Exercise Problems 73 76 12 Thévenin and Norton Equivalent Circuits Exercise Problems 81 90 13 Maximum Power Transfer Exercise Problems 93 95 ix x Contents 14 Operational Amplifiers 97 Exercise Problems 103 15 Inductors 105 Exercise Problems 109 16 Capacitors 111 Exercise Problems 116 17 Analysis of a Circuit by Solving Differential Equations 119 Exercise Problems 122 18 First-Order Circuits 125 Exercise Problems 126 19 Sinusoidal Steady-State (Phasor) 129 Exercise Problems with Solutions 136 20 Function Generators and Oscilloscopes 137 Exercise Problems 148 21 Mutual Inductance and Transformers 149 Exercise Problems 153 22 Fourier Series 157 Exercise Problems 167 23 Laplace Transform in Circuit Analysis 171 Exercise Problems 180 24 Fourier Transform in Circuit Analysis 181 Exercise Problems 185 25 Second-Order Circuits 187 Exercise Problems 198 26 Filters 201 Exercise Problems 207 27 Wrapping Up 209 Exercise Problems 211 Appendix Solutions to Exercise Problems 213 Bibliography (Some Textbooks Used in Colleges) 349 Index 351 Appendix Solutions to Exercise Problems 321 (p) If the secondary side has a load of 100 Ω, the reflected impedance on the primary side is 1000 Ω This is false (q) If the secondary side has a load of 100 Ω, the reflected impedance on the primary side is 10,000 Ω This is true Problem 21.2 This problem is about the dot notation and convention in a transformer Express the induced voltages for each case (a) Fig P21.2a (b) Fig P21.2b (c) Fig P21.2c (d) Fig P21.2d 322 (e) Fig P21.2e (f) Fig P21.2f (g) Fig P21.2g (h) Fig P21.2h Solution (a) v2 ¼ M didt1 and v1 ¼ M didt2 : (b) v2 ¼ ÀM didt1 and v1 ¼ M didt2 : Appendix Solutions to Exercise Problems Appendix Solutions to Exercise Problems (c) (d) (e) (f) (g) (h) v2 v2 v2 v2 v2 v2 323 ¼ ÀM didt1 and v1 ¼ M didt2 : ¼ ÀM didt1 and v1 ¼ M didt2 : ¼ M didt1 and v1 ¼ M didt2 : ¼ M didt1 and v1 ¼ ÀM didt2 : ¼ ÀM didt1 and v1 ¼ ÀM didt2 : ¼ ÀM didt1 and v1 ¼ M didt2 : Chapter 22 Fourier Series Problem 22.1 Match a Fourier series with a periodic function with ω0 ẳ 2/T f t ị ẳ a0 ỵ f t ị ẳ X1 f t ị ¼ a n¼1 n a n¼1 n f ð t ị ẳ a0 ỵ f t ị ẳ X1 cos ðnω0 t Þ X1 X1 cos ðnω0 t Þ b n¼1 n b n¼1 n sin ðnω0 t ị sin n0 t ị X1 n ẳ ẵan cos n0 t ị ỵ bn sin n0 t ị n ẳ odd X1 f t ị ẳ n ¼ an cos ðnω0 t Þ n ¼ odd X1 f t ị ẳ n ẳ bn sin n0 t ị n ẳ odd f t ị ẳ a0 ỵ f t ị ẳ (a) Fig P22.1a X1 X1 nẳ1 nẳ1 ẵan cos n0 t ị ỵ bn sin n0 t ị ẵan cos n0 t ị ỵ bn sin n0 t ị 324 (b) Fig P22.1b (c) Fig P22.1c (d) Fig P22.1d (e) Fig P22.1e (f) Fig P22.1f Appendix Solutions to Exercise Problems Appendix Solutions to Exercise Problems 325 (g) Fig P22.1g (h) Fig P22.1h (i) Fig P22.1i Solution P (a) f t ị ẳ 1n ẳ an cos n0 t Þ n ¼ odd P1 n ¼ bn sin n0 t ị P n ẳ odd (c) f t ị ẳ 1n ẳ ẵan cos n0 t ị ỵ bn sin n0 t ị (b) f t ị ẳ (d) (e) (f) (g) (h) (i) P n ẳ odd f t ị ẳ bn sin n0 t ị nẳ1P b sin n0 t ị f t ị ẳ a0 ỵ Pnẳ1 n f t ị ẳ a0 ỵ nẳ1 an cos n0 t ị P a cos n0 t ị f t ị ẳ Pnẳ1 n f t ị ẳ ẵ a nẳ1Pn cos n0 t ị ỵ bn sin n0 t ị f t ị ẳ a0 ỵ nẳ1 ẵan cos n0 t ị ỵ bn sin n0 t ị: Problem 22.2 Show that the Fourier series has an equivalent exponential form f t ị ẳ X1 ce nẳ1 n jnω0 t : 326 Appendix Solutions to Exercise Problems Solution Using Eulers formula ejn0 t ẳ cos n0 t ị þ j sin ðnω0 t Þ, We have cos ðnω0 t ị ẳ ejn0 t ỵ ejn0 t , sin n0 t ị ẳ ejn0 t ejn0 t : 2j The Fourier series f t ị ẳ a0 ỵ X1 nẳ1 ẵan cos n0 t ị ỵ bn sin ðnω0 t ފ can be rewritten as ! ejn0 t ỵ ejn0 t ejn0 t ejn0 t f t ị ẳ a0 ỵ an ỵ bn , n¼1 2j ! X1 a À jb an þ jbn Àjnω0 t n n jnω0 t þ e e f t ị ẳ a0 ỵ , nẳ1 2 X1 f t ị ẳ c0 ỵ c ejn0 t ỵ cn ejn0 t : n¼1 n X1 where cn ¼ an À bn , c ¼ a0 : Chapter 23 Laplace Transform in Circuit Analysis Problem 23.1 Solve the following differential equation using the Laplace transform method x} t ị ỵ 4x0 t ị ỵ 3xt ị ẳ with initial conditions x0(0) ¼ and x(0) ¼ Appendix Solutions to Exercise Problems 327 Solution Taking the Laplace transform term by term according to the Laplace transform table, we have s2 X sị sx0ị x0 0ị ỵ 4sX sị 4x0ị ỵ 3X sị ẳ , s s2 X sị 2s ỵ 4sX sị ỵ 3X sị ẳ , s X sị ẳ s ỵ ỵ 2s s2 ỵ 4s ỵ ẳ ỵ 9s ỵ 2s2 k k k ẳ 1ỵ ỵ : s sỵ1 sỵ3 ss ỵ 1ịs ỵ 3ị We can use the cover-up method to find the partial fraction expansion k1 ¼ k2 ẳ k3 ẳ ỵ 90ị ỵ 20ị2 ẳ , ỵ 1ị ỵ 3ị ỵ 91ị ỵ 21ị2 ẳ 1, 1ị1 ỵ 3ị ỵ 93ị ỵ 23ị2 ẳ : 3ị3 ỵ 1ị The Laplace transform of the solution is X sị ẳ 5=3 2=3 ỵ ỵ : sỵ1 sỵ3 s The solution can be obtained by nding the inverse Laplace transform (using the Table of Laplace Transform Pairs): x t ị ẳ ỵ et e3t , for t ! 0: 3 Problem 23.2 Use the Laplace transform method to solve for the circuit in Fig P23.2 The initial voltage in the capacitor is V Fig P23.2 328 Appendix Solutions to Exercise Problems Solution We first need to convert the time-domain circuit in Fig P23.2 to its corresponding Laplace-domain circuit in Fig S23.2a The initial condition is converted into a voltage source or a current source We know that if the initial condition is zero, the time-domain relationship for the capacitor i ðt Þ ¼ C dvðt Þ dt has a Laplace-domain counterpart I sị ẳ CsV sị: According to the Laplace transform table, if the initial condition is not zero, the Laplace-domain counterpart is I sị ẳ C ẵsV sị v0ị ẳ CsVsị Cv0ị ẳ Cs V sị ! v ð 0Þ : s There are two ways to handle this initial condition The first way is to use the relationship I sị ẳ CsV sị Cv0ị and treat the initial condition as a current source of value ÀCvð0Þ ¼ À3 μ: The second way is to use the relationship V sị ẳ I sị v0ị ỵ Cs s and treat the initial condition as a voltage source of value v 0ị ẳ : s s Therefore, we can have two Laplace-main circuits, as shown in Fig S23.2a and Fig S23.2b, respectively Appendix Solutions to Exercise Problems Fig S23.2a Fig S23.2b Let us first solve the circuit in Fig S24.1 The node equation is V À 2s V V ẳ , ỵ ỵ k 1k s Vẳ ỵ s 12 k 3s ỵ 2000 k k2 ẳ 1ỵ : ẳ s s ỵ 2000 s2000 ỵ sị 1kỵs Using the cover-up method to find k1 and k2, k1 ¼ k2 ẳ Therefore, 30ị ỵ 2000 ẳ 1, 2000 ỵ 32000ị ỵ 2000 ẳ 2: 2000ị 329 330 Appendix Solutions to Exercise Problems Vẳ ỵ s s ỵ 2000 and vt ị ẳ ỵ 2e2000t V, for t ! 0: Now let us solve the circuit in Fig S23.2b The node equation is V À 2s V 3s V ẳ 0, ỵ ỵ k 1k s Vẳ ỵ3 3s ỵ 2000 : ẳ 2000 þ s sð2000 þ sÞ 2000 s This expression is exactly the same as that for Fig S23.2a Therefore, we have the same solution vt ị ẳ ỵ 2e2000t V, for t ! 0: Problem 23.3 Use the Laplace transform method to solve for the circuit in Fig P23.3 The initial current in the inductor is A Fig P23.3 Solution We first need to convert the time-domain circuit in Fig P23.3 to its corresponding Laplace-domain circuit in Fig S23.3a The initial condition is converted into a voltage source or a current source We know that if the initial condition is zero, the time-domain relationship for the inductor Appendix Solutions to Exercise Problems 331 v t ị ẳ L dit ị dt has a Laplace-domain counterpart V sị ẳ LsI ðsÞ: According to the Laplace transform table, if the initial condition is not zero, the Laplace-domain counterpart is V sị ẳ LẵsI sị i0ị ẳ LsI sị Li0ị ẳ Ls I sị ! i0ị : s There are two ways to handle this initial condition The rst way is to use the relationship V sị ẳ LsI ðsÞ À Lið0Þ and treat the initial condition as a voltage source of value Li0ị ẳ 3: The second way is to use the relationship I sị ẳ V sị i0ị ỵ Ls s and treat the initial condition as a current source of value i ð 0Þ ¼ : s s Therefore, we can have two Laplace-main circuits, as shown in Fig S23.3a and Fig S23.3b, respectively Fig S23.3a 332 Appendix Solutions to Exercise Problems Fig S23.3b Let us first solve the circuit in Fig S23.3a The node equation is V V ỵ ẳ 2V À , s s s Á: V ẳ s ỵ 13 The current through the inductor is i(t) and its Laplace transform is I¼ V 2 ẳ ẳ ỵ : s s s ỵ 13 3s s ỵ After taking the inverse Laplace transform using the Table of Laplace Transform Pairs, it ị ẳ ỵ 2e3t A, for t ! 0: Now let us solve the circuit in Fig S23.3b The node equation is V V ỵ3 þ ¼ À 2V, s s Á: V ẳ s ỵ 13 This expression is exactly the same as that for Fig S23.3a Therefore, we have the same solution Appendix Solutions to Exercise Problems 333 Chapter 24 Fourier Transform in Circuit Analysis Problem 24.1 Use the Fourier transform method to solve for the circuit in Fig P24.1 The initial voltage in the capacitor is V Fig P24.1 Solution This problem is the same as Problem 23.2, in which the Laplace transform method is used The difficult part of solving this problem using the Fourier transform is that the Fourier transform method does not have an explicit way to handle the initial conditions In order to accommodating an initial condition, we need to introduce a source with a step function as follows In Problem 23.2, the initial condition of the capacitor is treated as a voltage source of value in the Laplace domain v ð 0ị ẳ : s s Converting this source into the time domain gives the time-domain source v0ịut ị ẳ uðt Þ, where u(t) is the unit step function The Fourier transform of the unit step function can be found in the Table of Fourier transform Pairs as πδðωÞ þ : jω We can redraw the circuit by introducing the step function sources in Fig S24.1 We now set up a node equation for the circuit in Fig S24.1 in the Fourier domain as follows 334 Appendix Solutions to Exercise Problems Fig S24.1 h i V ị ỵ j 1k ỵ h i V ị ỵ j j ỵ V ẳ 0, 1k ! ! 1 ỵ jmV 3jm ị ỵ ỵ V ẳ 0, j j h i h i 1 ị ỵ j ỵ 3jm ị ỵ j Vẳ , ỵ jm h i h i 1 2000 ị ỵ j ỵ 3j ị ỵ j Vẳ , 2000 ỵ j ! k1 : Vẳ ỵ k2 ị ỵ j 2000 þ jω V À πδðωÞ þ It can be verified that k1 ¼ and k2 ¼ After taking the inverse Fourier transform by using the Table of Fourier Transform Pairs, vt ị ẳ ỵ 2e2000t V, for t ! 0: This answer is the same as that obtained from Problem 23.2, obtained by the Laplace transform The voltage is the capacitor voltage, which is normally denoted by vC, as in Fig P24.1 This voltage includes the everything inside the dotted box in Fig S24.1 We feel that if a circuit problem involves initial conditions or switch actions, the Laplace transform method is easier than the Fourier transform method because the Laplace transform of the unit step function is 1/s, while the Fourier transform of the unit step function is πδ(ω) + 1/( jω) Problem 24.2 The input signal is a signum function Find the inductor current iL Appendix Solutions to Exercise Problems 335 Fig P24.2 Solution This problem cannot be solved by the Laplace transform method because the input signal is not zero when t < This problem cannot be solved by the phasor method either because the input signal is not sinusoidal This problem can be solved either in the time domain or in the Fourier domain We will use the Fourier transform method here The Fourier transform of the signum function is : jω We notice that this circuit is current divider Let us set up the current divider relationship in the Fourier domain as follows IC ¼     1=2 1=2 1 1 ¼ ¼ À : j ỵ jị ỵ j ỵ j j ỵ j Taking the invers Fourier transform yields i C t ị ẳ 1 sgn t Þ À eÀ2t uðt Þ A, 2 where u(t) is the unit step function Problem 24.3 Repeat Problem 24.2 with 10 cos (4t) being the input signal Solution Since the input function is sinusoidal, both the phasor method and the Fourier method can be used The Laplace transform method does not work The Fourier transform of the input function is ... laws and regulations and therefore free for general use The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and. .. 136 20 Function Generators and Oscilloscopes 137 Exercise Problems 148 21 Mutual Inductance and Transformers ... measure voltage, current, and resistance There are two types of multimeters: hand-held digital multimeters and desktop digital multimeters, which can be seen in Figs 2.7 and 2.8 No matter which