EECE 380 Problems Solution Manual, Fall 2011

Chapter 1: Introduction: Waves and Phasors Lesson #1 Chapter — Section: Chapter 1 Topics: EM history and how it relates to other fields Highlights: • EM in Classical era: 1000 BC to 1900 • Examples of Modern Era Technology timelines • Concept of “fields” (gravitational, electric, magnetic) • Static vs. dynamic fields • The EM Spectrum Special Illustrations: • Timelines from CD-ROM Timeline for Electromagnetics in the Classical Era ca. 900 Legend has it that while walking BC across a field in northern Greece, a shepherd named Magnus experiences a pull on the iron nails in his sandals by the black rock he was standing on. The region was later named Magnesia and the rock became known as magnetite [a form of iron with permanent magnetism]. ca. 600 Greek philosopher Thales BC describes how amber, after being rubbed with cat fur, can pick up feathers [static electricity]. ca. 1000 Magnetic compass used as a navigational device. 1752 Benjamin Franklin (American) invents the lightning rod and demonstrates that lightning is electricity. 1785 Charles-Augustin de Coulomb (French) demonstrates that the electrical force between charges is proportional to the inverse of the square of the distance between them. 1800 Alessandro Volta (Italian) develops the first electric battery. 1820 Hans Christian Oersted (Danish) demonstrates the interconnection between electricity and magnetism through his discovery that an electric current in a wire causes a compass needle to orient itself perpendicular to the wire. 2 Lessons #2 and 3 Chapter — Sections: 1-1 to 1-6 Topics: Waves Highlights: • Wave properties • Complex numbers • Phasors Special Illustrations: • CD-ROM Modules 1.1-1.9 • CD-ROM Demos 1.1-1.3 CHAPTER 1 3 Chapter 1 Section 1-3: Traveling Waves Problem 1.1 A 2-kHz sound wave traveling in the x-direction in air was observed to have a differential pressure p � x � t �� 10 N/m2 at x � 0 and t � 50 µs. If the reference phase of p � x � t � is 36 � , find a complete expression for p � x � t � . The velocity of sound in air is 330 m/s. Solution: The general form is given by Eq. (1.17), p � x � t �� Acos � 2πt T � 2πx λ � φ0 � where it is given that φ0 � 36 � . From Eq. (1.26), T � 1 f � 1 � 2 103 �� 0 5 ms. From Eq. (1.27), λ � up f � 330 2 103 � 0 165 m Also, since p � x � 0 � t � 50 µs �� 10 (N/m2) � Acos � 2π 50 10  6 5 10  4 � 36 � π rad 180 � � Acos � 1 26 rad �� 0 31A � it follows that A � 10 0 31 � 32 36 N/m2. So, with t in (s) and x in (m), p � x � t �� 32 36cos  2π 106 t 500 � 2π 103 x 165 � 36 � (N/m2) � 32 36cos � 4π 103t � 12 12πx � 36 � � (N/m2) Problem 1.2 For the pressure wave described in Example 1-1, plot (a) p � x � t � versus x at t � 0, (b) p � x � t � versus t at x � 0. Be sure to use appropriate scales for x and t so that each of your plots covers at least two cycles. Solution: Refer to Fig. P1.2(a) and Fig. P1.2(b). 4 CHAPTER 1 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 -12. -10. -8. -6. -4. -2. 0. 2. 4. 6. 8. 10. 12. Amplitude (N/m2) Distance x (m) p(x,t=0) 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 -12. -10. -8. -6. -4. -2. 0. 2. 4. 6. 8. 10. 12. Amplitude (N/m2) Time t (ms) p(x=0,t) (a) (b) Figure P1.2: (a) Pressure wave as a function of distance at t � 0 and (b) pressure wave as a function of time at x � 0. Problem 1.3 A harmonic wave traveling along a string is generated by an oscillator that completes 180 vibrations per minute. If it is observed that a given crest, or maximum, travels 300 cm in 10 s, what is the wavelength? Solution: f � 180 60 � 3 Hz up � 300 cm 10 s � 0 3 m/s λ � up f � 0 3 3 � 0 1 m � 10 cm Problem 1.4 Two waves, y1 � t � and y2 � t � , have identical amplitudes and oscillate at the same frequency, but y2 � t � leads y1 � t � by a phase angle of 60 � . If y1 � t �� 4cos � 2π 103t � � write down the expression appropriate for y2 � t � and plot both functions over the time span from 0 to 2 ms. Solution: y2 � t �� 4cos � 2π 103t � 60 � � CHAPTER 1 5 0 4 2 -2 -4 0.5 ms 1 ms 1.5 ms 2 ms y (t) 1 y (t) 2 z Figure P1.4: Plots of y1 � t � and y2 � t � . Problem 1.5 The height of an ocean wave is described by the function y � x � t �� 1 5sin � 0 5t � 0 6x � (m) Determine the phase velocity and the wavelength and then sketch y � x � t � at t � 2 s over the range from x � 0 to x � 2λ. Solution: The given wave may be rewritten as a cosine function: y � x � t �� 1 5cos � 0 5t � 0 6x � π 2 � By comparison of this wave with Eq. (1.32), y � x � t �� Acos � ωt � βx � φ0 � � we deduce that ω � 2π f � 0 5 rad/s � β � 2π λ � 0 6 rad/m � up � ω β � 0 5 0 6 � 0 83 m/s � λ � 2π β � 2π 0 6 � 10 47 m 6 CHAPTER 1 2λ x 1.5 1 0.5 0 -0.5 -1 -1.5 y (π/2, t) Figure P1.5: Plot of y � x � 2 � versus x. At t � 2 s, y � x � 2 � � 1 5sin � 1 � 0 6x � (m), with the argument of the cosine function given in radians. Plot is shown in Fig. P1.5. Problem 1.6 A wave traveling along a string in the � x-direction is given by y1 � x � t �� Acos � ωt � βx � � where x � 0 is the end of the string, which is tied rigidly to a wall, as shown in Fig. 1-21 (P1.6). When wave y1 � x � t � arrives at the wall, a reflected wave y2 � x � t � is generated. Hence, at any location on the string, the vertical displacement ys will be the sum of the incident and reflected waves: ys � x � t �� y1 � x � t � � y2 � x � t � (a) Write down an expression for y2 � x � t � , keeping in mind its direction of travel and the fact that the end of the string cannot move. (b) Generate plots of y1 � x � t � , y2 � x � t � and ys � x � t � versus x over the range � 2λ � x � 0 at ωt � π 4 and at ωt � π 2. Solution: (a) Since wave y2 � x � t � was caused by wave y1 � x � t � , the two waves must have the same angular frequency ω, and since y2 � x � t � is traveling on the same string as y1 � x � t � , CHAPTER 1 7 x x = 0 Incident Wave y Figure P1.6: Wave on a string tied to a wall at x � 0 (Problem 1.6). the two waves must have the same phase constant β. Hence, with its direction being in the negative x-direction, y2 � x � t � is given by the general form y2 � x � t �� Bcos � ωt � βx � φ0 � � (1) where B and φ0 are yet-to-be-determined constants. The total displacement is ys � x � t �� y1 � x � t � � y2 � x � t �� Acos � ωt � βx � � Bcos � ωt � βx � φ0 � Since the string cannot move at x � 0, the point at which it is attached to the wall, ys � 0 � t �� 0 for all t. Thus, ys � 0 � t �� Acosωt � Bcos � ωt � φ0 �� 0 (2) (i) Easy Solution: The physics of the problem suggests that a possible solution for (2) is B � � A and φ0 � 0, in which case we have y2 � x � t �� � Acos � ωt � βx � (3) (ii) Rigorous Solution: By expanding the second term in (2), we have Acosωt � B � cosωt cosφ0 � sinωt sinφ0 �� 0 � or � A � Bcosφ0 � cosωt � � Bsinφ0 � sinωt � 0 (4) This equation has to be satisfied for all values of t. At t � 0, it gives A � Bcosφ0 � 0 � (5) 8 CHAPTER 1 and at ωt � π 2, (4) gives Bsinφ0 � 0 (6) Equations (5) and (6) can be satisfied simultaneously only if A � B � 0 (7) or A � � B and φ0 � 0 (8) Clearly (7) is not an acceptable solution because it means that y1 � x � t � � 0, which is contrary to the statement of the problem. The solution given by (8) leads to (3). (b) At ωt � π 4, y1 � x � t �� Acos � π 4 � βx �� Acos � π 4 � 2πx λ � y2 � x � t �� � Acos � ωt � βx �� � Acos � π 4 � 2πx λ Plots of y1, y2, and y3 are shown in Fig. P1.6(b). y (ωt, x) 2 y (ωt, x) s 0 A -A -2λ x 1.5A -1.5A ωt=π/4 y (ωt, x) 1 Figure P1.6: (b) Plots of y1, y2, and ys versus x at ωt � π 4. At ωt � π 2, y1 � x � t �� Acos � π 2 � βx �� Asinβx � Asin 2πx λ � CHAPTER 1 9 y2 � x � t �� � Acos � π 2 � βx �� Asinβx � Asin 2πx λ Plots of y1, y2, and y3 are shown in Fig. P1.6(c). y (ωt, x) 2 y (ωt, x) s 0 A -A -2λ x 2A -2A ωt=π/2 y (ωt, x) 1 Figure P1.6: (c) Plots of y1, y2, and ys versus x at ωt � π 2. Problem 1.7 Two waves on a string are given by the following functions: y1 � x � t �� 4cos � 20t � 30x � (cm) � y2 � x � t �� � 4cos � 20t � 30x � (cm) � where x is in centimeters. The waves are said to interfere constructively when their superposition � ys � �� y1 � y2 � is a maximum and they interfere destructively when � ys � is a minimum. (a) What are the directions of propagation of waves y1 � x � t � and y2 � x � t � ? (b) At t � � π 50 � s, at what location x do the two waves interfere constructively, and what is the corresponding value of � ys � ? (c) At t � � π 50 � s, at what location x do the two waves interfere destructively, and what is the corresponding value of � ys � ? Solution: (a) y1 � x � t � is traveling in positive x-direction. y2 � x � t � is traveling in negative x-direction. 10 CHAPTER 1 (b) At t � � π 50 � s, ys � y1 � y2 � 4 � cos � 0 4π � 30x � � cos � 0 4π � 3x �� . Using the formulas from Appendix C, 2sinxsiny � cos � x � y � � � cosx � y � � we have ys � 8sin � 0 4π � sin30x � 7 61sin30x Hence, � ys � max � 7 61 and it occurs when sin30x � 1, or 30x � π 2 � 2nπ, or x � � π 60 � 2nπ 30 cm, where n � 0 � 1 � 2 � (c) � ys � min � 0 and it occurs when 30x � nπ, or x � nπ 30 cm. Problem 1.8 Give expressions for y � x � t � for a sinusoidal wave traveling along a string in the negative x-direction, given that ymax � 40 cm, λ � 30 cm, f � 10 Hz, and (a) y � x � 0 �� 0 at x � 0, (b) y � x � 0 �� 0 at x � 7 5 cm. Solution: For a wave traveling in the negative x-direction, we use Eq. (1.17) with ω � 2π f � 20π (rad/s), β � 2π λ � 2π 0 3 � 20π 3 (rad/s), A � 40 cm, and x assigned a positive sign: y � x � t �� 40cos � 20πt � 20π 3 x � φ0 (cm) � with x in meters. (a) y � 0 � 0 �� 0 � 40cosφ0. Hence, φ0 �� π 2, and y � x � t �� 40cos � 20πt � 20π 3 x � π 2 � � � 40sin � 20πt � 20π 3 x � (cm), if φ0 � π 2 � 40sin � 20πt � 20π 3 x � (cm), if φ0 � � π 2 (b) At x � 7 5 cm = 7 5 10  2 m, y � 0 � 40cos � π 2 � φ0 � . Hence, φ0 � 0 or π, and y � x � t �� � 40cos � 20πt � 20π 3 x � (cm), if φ0 � 0 � � 40cos � 20πt � 20π 3 x � (cm), if φ0 � π CHAPTER 1 11 Problem 1.9 An oscillator that generates a sinusoidal wave on a string completes 20 vibrations in 50 s. The wave peak is observed to travel a distance of 2.8 m along the string in 50 s. What is the wavelength? Solution: T � 50 20 � 2 5 s � up � 2 8 5 � 0 56 m/s � λ � upT � 0 56 2 5 � 1 4 m Problem 1.10 The vertical displacement of a string is given by the harmonic function: y � x � t �� 6cos � 16πt � 20πx � (m) � where x is the horizontal distance along the string in meters. Suppose a tiny particle were to be attached to the string at x � 5 cm, obtain an expression for the vertical velocity of the particle as a function of time. Solution: y � x � t �� 6cos � 16πt � 20πx � (m) u � 0 05 � t �� dy � x � t � dt � � � � x � 0 � 05 � 96πsin � 16πt � 20πx � � x � 0 � 05 � 96πsin � 16πt � π � � � 96πsin � 16πt � (m/s) Problem 1.11 Given two waves characterized by y1 � t �� 3cosωt � y2 � t �� 3sin � ωt � 36 � � � does y2 � t � lead or lag y1 � t � , and by what phase angle? Solution: We need to express y2 � t � in terms of a cosine function: y2 � t �� 3sin � ωt � 36 � � � 3cos  π 2 � ωt � 36 �  � 3cos � 54 � � ωt �� 3cos � ωt � 54 � � 12 CHAPTER 1 Hence, y2 � t � lags y1 � t � by 54 � . Problem 1.12 The voltage of an electromagnetic wave traveling on a transmission line is given by v � z � t � � 5e  αz sin � 4π 109t � 20πz � (V), where z is the distance in meters from the generator. (a) Find the frequency, wavelength, and phase velocity of the wave. (b) At z � 2 m, the amplitude of the wave was measured to be 1 V. Find α. Solution: (a) This equation is similar to that of Eq. (1.28) with ω � 4π 109 rad/s and β � 20π rad/m. From Eq. (1.29a), f � ω 2π � 2 109 Hz � 2 GHz; from Eq. (1.29b), λ � 2π β � 0 1 m. From Eq. (1.30), up � ω β � 2 108 m/s (b) Using just the amplitude of the wave, 1 � 5e  α2 � α � � 1 2 m ln � 1 5 � 0 81 Np/m. Problem 1.13 A certain electromagnetic wave traveling in sea water was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m and an amplitude of 81.87 (V/m) at a depth of 100 m. What is the attenuation constant of sea water? Solution: The amplitude has the form Aeαz. At z � 10 m, Ae  10α � 98 02 and at z � 100 m, Ae  100α � 81 87 The ratio gives e  10α e  100α � 98 02 81 87 � 1 20 or e  10α � 1 2e  100α Taking the natural log of both sides gives ln � e  10α �� ln � 1 2e  100α � � � 10α � ln � 1 2 � � 100α � 90α � ln � 1 2 �� 0 18 Hence, α � 0 18 90 � 2 10  3 (Np/m) CHAPTER 1 13 Section 1-5: Complex Numbers Problem 1.14 Evaluate each of the following complex numbers and express the result in rectangular form: (a) z1 � 4ejπ � 3, (b) z2 �� 3 ej3π � 4, (c) z3 � 6e  jπ � 2, (d) z4 � j3, (e) z5 � j  4, (f) z6 � � 1 � j � 3, (g) z7 � � 1 � j � 1 � 2. Solution: (Note: In the following solutions, numbers are expressed to only two decimal places, but the final answers are found using a calculator with 10 decimal places.) (a) z1 � 4ejπ � 3 � 4 � cosπ 3 � jsinπ 3 �� 2 0 � j3 46. (b) z2 � � 3ej3π � 4 � � 3 � cos � 3π 4 � jsin � 3π 4 � � � 1 22 � j1 22 � 1 22 � � 1 � j � (c) z3 � 6e  jπ � 2 � 6 � cos � � π 2 � � jsin � � π 2 �� � � j6. (d) z4 � j3 � j � j2 � � j, or z4 � j3 � � ejπ � 2 � 3 � ej3π � 2 � cos � 3π 2 � � jsin � 3π 2 �� � j (e) z5 � j  4 � � ejπ � 2 �  4 � e  j2π � 1. (f) z6 � � 1 � j � 3 � � � 2e  jπ � 4 � 3 � � � 2 � 3e  j3π � 4 � � � 2 � 3 � cos � 3π 4 � � jsin � 3π 4 �� � � 2 � j2 � � 2 � 1 � j � (g) z7 � � 1 � j � 1 � 2 � � � 2e  jπ � 4 � 1 � 2 �� 21 � 4e  jπ � 8 � � 1 19 � 0 92 � j0 38 � � � � 1 10 � j0 45 � Problem 1.15 Complex numbers z1 and z2 are given by z1 � 3 � j2 � z2 � � 4 � j3 14 CHAPTER 1 (a) Express z1 and z2 in polar form. (b) Find � z1 � by applying Eq. (1.41) and again by applying Eq. (1.43). (c) Determine the product z1z2 in polar form. (d) Determine the ratio z1 z2 in polar form. (e) Determine z3 1 in polar form. Solution: (a) Using Eq. (1.41), z1 � 3 � j2 � 3 6e  j33 � 7 � � z2 � � 4 � j3 � 5e j143 � 1 � (b) By Eq. (1.41) and Eq. (1.43), respectively, � z1 � � � 3 � j2 � �� 32 � � � 2 � 2 � � 13 � 3 60 � � z1 � �� � 3 � j2 � � 3 � j2 �� � 13 � 3 60 (c) By applying Eq. (1.47b) to the results of part (a), z1z2 � 3 6e  j33 � 7 � 5ej143 � 1 � � 18ej109 � 4 � (d) By applying Eq. (1.48b) to the results of part (a), z1 z2 � 3 6e  j33 � 7 � 5ej143 � 1 � � 0 72e  j176 � 8 � (e) By applying Eq. (1.49) to the results of part (a), z3 1 � � 3 6e  j33 � 7 � � 3 � � 3 6 � 3e  j3 � 33 � 7 � � 46 66e  j101 � 1 � Problem 1.16 If z � � 2 � j4, determine the following quantities in polar form: (a) 1 z, (b) z3, (c) � z � 2, (d) ��� z , (e) ��� z . Solution: (Note: In the following solutions, numbers are expressed to only two decimal places, but the final answers are found using a calculator with 10 decimal places.) CHAPTER 1 15 (a) 1 z � 1 � 2 � j4 � � � 2 � j4 �  1 � � 4 47ej116 � 6 � �  1 � � 4 47 �  1e  j116 � 6 � � 0 22e  j116 � 6 � (b) z3 � � � 2 � j4 � 3 � � 4 47ej116 � 6 � � 3 � � 4 47 � 3ej350 � 0 � � 89 44e  j10 � . (c) � z � 2 � z � z � � � 2 � j4 � � � 2 � j4 �� 4 � 16 � 20. (d) ��� z � ��� � 2 � j4 � 4. (e) ��� z � �� � � 2 � j4 � � 4 � 4e jπ. Problem 1.17 Find complex numbers t � z1 � z2 and s � z1 � z2, both in polar form, for each of the following pairs: (a) z1 � 2 � j3, z2 � 1 � j3, (b) z1 � 3, z2 � � j3, (c) z1 � 3 � 30 � , z2 � 3 �  30 � , (d) z1 � 3 � 30 � , z2 � 3 �  150 � . Solution: (a) t � z1 � z2 � � 2 � j3 � � � 1 � j3 �� 3 � s � z1 � z2 � � 2 � j3 � � � 1 � j3 �� 1 � j6 � 6 08e j80 � 5 � (b) t � z1 � z2 � 3 � j3 � 4 24e  j45 � � s � z1 � z2 � 3 � j3 � 4 24e j45 � (c) t � z1 � z2 � 3 � 30 � � 3 �  30 � � 3ej30 � � 3e  j30 � � � 2 6 � j1 5 � � � 2 6 � j1 5 � � 5 2 � s � z1 � z2 � 3ej30 � � 3e  j30 � � � 2 6 � j1 5 � � � 2 6 � j1 5 � � j3 � 3e j90 � (d) t � z1 � z2 � 3 � 30 � � 3 �  150 � � � 2 6 � j1 5 � � � � 2 6 � j1 5 �� 0 � s � z1 � z2 � � 2 6 � j1 5 � � � � 2 6 � j1 5 � � 5 2 � j3 � 6e j30 � Problem 1.18 Complex numbers z1 and z2 are given by z1 � 5 �  60 � � z2 � 2 � 45 � 16 CHAPTER 1 (a) Determine the product z1z2 in polar form. (b) Determine the product z1z 2 in polar form. (c) Determine the ratio z1 z2 in polar form. (d) Determine the ratio z 1 z 2 in polar form. (e) Determine � z1 in polar form. Solution: (a) z1z2 � 5e  j60 � 2ej45 � � 10e  j15 � . (b) z1z 2 � 5e  j60 � 2e  j45 � � 10e  j105 � . (c) z1 z2 � 5e  j60 � 2ej45 � � 2 5  j105 � . (d) z 1 z 2 � � z1 z2 � 2 5j105 � . (e) � z1 � � 5e  j60 � � � � 5e  j30 � . Problem 1.19 If z � 3 � j5, find the value of ln � z � . Solution: � z � � � � 32 � 52 � 5 83 � θ � tan  1 � � 5 3 � � 59 � � z � � z � ejθ � 5 83e  j59 � � ln � z �� ln � 5 83e  j59 � � � ln � 5 83 � � ln � e  j59 � � � 1 76 � j59 � � 1 76 � j59 � π 180 � � 1 76 � j1 03 Problem 1.20 If z � 3 � j4, find the value of ez. Solution: ez � e3  j4 � e3 � e  j4 � e3 � cos4 � jsin4 � � e3 � 20 09 � and 4 rad � 4 π 180 � � 229 18 � Hence, ez � 20 08 � cos229 18 � � jsin229 18 � �� � 13 13 � j15 20. CHAPTER 1 17 Section 1-6: Phasors Problem 1.21 A voltage source given by vs � t � � 25cos � 2π 103t � 30 � � (V) is connected to a series RC load as shown in Fig. 1-19. If R � 1 MΩ and C � 200 pF, obtain an expression for vc � t � , the voltage across the capacitor. Solution: In the phasor domain, the circuit is a voltage divider, and � Vc � � Vs 1 jωC R � 1 jωC � � Vs � 1 � jωRC � Now � Vs � 25e  j30 � V with ω � 2π 103 rad/s, so � Vc � 25e  j30 � V 1 � j � � 2π 103 rad/s �� � 106 Ω �� � 200 10  12 F � � � 25e  j30 � V 1 � j2π 5 � 15 57e  j81 � 5 � V. Converting back to an instantaneous value, vc � t �� �� � Vcejωt � �� 15 57e j � ωt  81 � 5 �� V � 15 57cos � 2π 103t � 81 5 � � V � where t is expressed in seconds. Problem 1.22 Find the phasors of the following time functions: (a) v � t �� 3cos � ωt � π 3 � (V), (b) v � t �� 12sin � ωt � π 4 � (V), (c) i � x � t �� 2e  3x sin � ωt � π 6 � (A), (d) i � t �� � 2cos � ωt � 3π 4 � (A), (e) i � t �� 4sin � ωt � π 3 � � 3cos � ωt � π 6 � (A). Solution: (a) � V � 3e  jπ � 3 V. (b) v � t � � 12sin � ωt � π 4 � � 12cos � π 2 � � ωt � π 4 � � � 12cos � ωt � π 4 � V, � V � 12e  jπ � 4 V. (c) i � t �� 2e  3x sin � ωt � π 6 � A � 2e  3x cos � π 2 � � ωt � π 6 � � A � 2e  3x cos � ωt � π 3 � A � � I � 2e  3xe  jπ � 3 A 18 CHAPTER 1 (d) i � t �� � 2cos � ωt � 3π 4 � � � I � � 2ej3π � 4 � 2e  jπej3π � 4 � 2e  jπ � 4 A (e) i � t �� 4sin � ωt � π 3 � � 3cos � ωt � π 6 � � 4cos � π 2 � � ωt � π 3 �� � 3cos � ωt � π 6 � � 4cos � � ωt � π 6 � � 3cos � ωt � π 6 � � 4cos � ωt � π 6 � � 3cos � ωt � π 6 �� 7cos � ωt � π 6 � � � I � 7e  jπ � 6 A Problem 1.23 Find the instantaneous time sinusoidal functions corresponding to the following phasors: (a) � V � � 5ejπ � 3 (V), (b) � V � j6e  jπ � 4 (V), (c) � I � � 6 � j8 � (A), (d) ˜I � � 3 � j2 (A), (e) ˜I � j (A), (f) ˜I � 2ejπ � 6 (A). Solution: (a) � V � � 5ejπ � 3 V � 5ej � π � 3  π � V � 5e  j2π � 3 V � v � t �� 5cos � ωt � 2π 3 � V (b) � V � j6e  jπ � 4 V � 6ej �  π � 4 � π � 2 � V � 6ejπ � 4 V � v � t �� 6cos � ωt � π 4 � V (c) � I � � 6 � j8 � A � 10e j53 � 1 � A � i � t �� 10cos � ωt � 53 1 � � A. (d) � I � � 3 � j2 � 3 61e j146 � 31 � � i � t �� �� � 3 61e j146 � 31 � ejωt � 3 61 cos � ωt � 146 31 � � A CHAPTER 1 19 (e) � I � j � ejπ � 2 � i � t �� �� � ejπ � 2ejωt � cos � ωt � π 2 �� � sinωt A (f) � I � 2ejπ � 6 � i � t �� �� � 2ejπ � 6ejωt � 2cos � ωt � π 6 � A Problem 1.24 A series RLC circuit is connected to a generator with a voltage vs � t �� V0 cos � ωt � π 3 � (V). (a) Write down the voltage loop equation in terms of the current i � t � , R, L, C, and vs � t � . (b) Obtain the corresponding phasor-domain equation. (c) Solve the equation to obtain an expression for the phasor current � I. Vs(t) R L C i Figure P1.24: RLC circuit. Solution: (a) vs � t �� Ri � Ldi dt � 1 C � idt (b) In phasor domain: � Vs � R˜I � jωL˜I � ˜I jωC (c) ˜I � � Vs R � j � ωL � 1 ωC � � V0ejπ � 3 R � j � ωL � 1 ωC � � ωCV0ejπ � 3 ωRC � j � ω2LC � 1 � Problem 1.25 A wave traveling along a string is given by y � x � t �� 2sin � 4πt � 10πx � (cm) 20 CHAPTER 1 where x is the distance along the string in meters and y is the vertical displacement. Determine: (a) the direction of wave travel, (b) the reference phase φ0, (c) the frequency, (d) the wavelength, and (e) the phase velocity. Solution: (a) We start by converting the given expression into a cosine function of the form given by (1.17): y � x � t �� 2cos  4πt � 10πx � π 2  (cm) Since the coefficients of t and x both have the same sign, the wave is traveling in the negative x-direction. (b) From the cosine expression, φ0 � � π 2. (c) ω � 2π f � 4π, f � 4π 2π � 2 Hz (d) 2π λ � 10π, λ � 2π 10π � 0 2 m. (e) up � fλ � 2 0 2 � 0 4 (m/s). Problem 1.26 A laser beam traveling through fog was observed to have an intensity of 1 (µW/m2) at a distance of 2 m from the laser gun and an intensity of 0.2 (µW/m2) at a distance of 3 m. Given that the intensity of an electromagnetic wave is proportional to the square of its electric-field amplitude, find the attenuation constant α of fog. Solution: If the electric field is of the form E � x � t �� E0e  αx cos � ωt � βx � � then the intensity must have a form I � x � t �� � E0e  αx cos � ωt � βx �� 2 � E2 0e  2αx cos2 � ωt � βx � or I � x � t �� I0e  2αx cos2 � ωt � βx � where we define I0 � E2 0. We observe that the magnitude of the intensity varies as I0e  2αx. Hence, at x � 2 m � I0e  4α � 1 10  6 (W/m2) � at x � 3 m � I0e  6α � 0 2 10  6 (W/m2) CHAPTER 1 21 I0e  4α I0e  6α � 10  6 0 2 10  6 � 5 e  4α � e6α � e2α � 5 α � 0 8 (NP/m) Problem 1.27 Complex numbers z1 and z2 are given by z1 � � 3 � j2 z2 � 1 � j2 Determine (a) z1z2, (b) z1 z 2, (c) z2 1, and (d) z1z 1, all all in polar form. Solution: (a) We first convert z1 and z2 to polar form: z1 � � � 3 � j2 �� �  � 32 � 22 e  jtan � 1 2 � 3  � � � 13 e  j33 � 7 � � � 13 ej � 180 �  33 � 7 � � � � 13 ej146 � 3 � z2 � 1 � j2 � � 1 � 4 e  jtan � 1 2 � � 5 e  j63 � 4 � z1z2 � � 13 ej146 � 3 � � 5 e  j63 � 4 � � � 65 ej82 � 9 � (b) z1 z 2 � � 13 ej146 � 3 � � 5 ej63 � 4 � � � 13 5 ej82 � 9 � (c) z2 1 � � � 13 � 2 � ej146 � 3 � � 2 � 13ej292 � 6 � � 13e  j360 � ej292 � 6 � � 13e  j67 � 4 � 22 CHAPTER 1 (d) z1z 1 � � 13 ej146 � 3 � � 13 e  j146 � 3 � � 13 Problem 1.28 If z � 3e jπ � 6, find the value of ez. Solution: z � 3ejπ � 6 � 3cosπ 6 � j3sinπ 6 � 2 6 � j1 5 ez � e2 � 6 � j1 � 5 � e2 � 6 ej1 � 5 � e2 � 6 � cos1 5 � jsin1 5 � � 13 46 � 0 07 � j0 98 � � 0 95 � j13 43 Problem 1.29 The voltage source of the circuit shown in the figure is given by vs � t �� 25cos � 4 104t � 45 � � (V) Obtain an expression for iL � t � , the current flowing through the inductor. vs(t) L i R1 R2 iL A iR2 + - R1 = 20 Ω, R2 = 30 Ω, L = 0.4 mH Solution: Based on the given voltage expression, the phasor source voltage is � Vs � 25e  j45 � (V) (9) The voltage equation for the left-hand side loop is R1i � R2iR2 � vs (10) CHAPTER 1 23 For the right-hand loop, R2iR2 � LdiL dt � (11) and at node A, i � iR2 � iL (12) Next, we convert Eqs. (2)–(4) into phasor form: R1 � I � R2 � IR2 � � Vs (13) R2 � IR2 � jωL � IL (14) � I � � IR2 � � IL (15) Upon combining (6) and (7) to solve for � IR2 in terms of � I, we have: � IR2 � jωL R2 � jωLI (16) Substituting (8) in (5) and then solving for � I leads to: R1 � I � jR2ωL R2 � jωL � I � � Vs � I � R1 � jR2ωL R2 � jωL � � Vs � I � R1R2 � jR1ωL � jR2ωL R2 � jωL � � Vs � I � � R2 � jωL R1R2 � jωL � R1 � R2 � � Vs (17) Combining (6) and (7) to solve for � IL in terms of � I gives � IL � R2 R2 � jωL � I (18) Combining (9) and (10) leads to � IL � � R2 R2 � jωL � R2 � jωL R1R2 � jωL � R1 � R2 � � Vs � R2 R1R2 � � jωL � R1 � R2 � � Vs 24 CHAPTER 1 Using (1) for � Vs and replacing R1, R2, L and ω with their numerical values, we have � IL � 30 20 30 � j4 104 0 4 10  3 � 20 � 30 � 25e  j45 � � 30 25 600 � j800 e  j45 � � 7 5 6 � j8 e  j45 � � 7 5e  j45 � 10ej53 � 1 � � 0 75e  j98 � 1 � (A) Finally, iL � t �� �� � � ILejωt � � 0 75cos � 4 104t � 98 1 � � (A)