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ECE311_W11_Hwk1_Solutions

# ECE311_W11_Hwk1_Solutions - CHAPTER 1 13 Section 1-5...

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Unformatted text preview: 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 4e jπ 3 , 3 e j3π 4 , (b) z2 (c) z3 6e jπ 2 , (d) z4 j3 , (e) z5 j 4 , 1 j 3, (f) z6 1 j 1 2. (g) z7 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.) 2 0 j3 46. (a) z1 4e jπ 3 4 cos π 3 j sin π 3 (b) (g) £ ¤ ECE 311 Homework 1 Solutions Problem 1 Problem 1.15 Complex numbers z1 and z2 are given by  © ¨ £ z2 4 j3 ¡ ¨ £ z1 3 j2 ¢  1 10 j0 45 ¢  ¨  ¨   £ £ £ £  £  £ ¢ ¨ £ z7 1 j 12 2e 21 4 e jπ 4 1 2 jπ 8 1 19 0 92 ¢ © ¨ £ 2 j2 21 j j0 38 ¡ ¢ ¢ ¨ ¢ ¢ 2 3 cos 3π 4 j sin 3π 4  ¢ ¨ ¡ ¨ £¢ £ £ ¢  ¡ £¢ ¨ £ z6 1 j 3 2e 2 3e jπ 4 3 ¡ £  £ ¢ ¡ £ £ (e) z5 (f) j 4 e jπ 2 4 e j2π 1. j3π 4  ¨ £ ¢  © ¢ £ £¢ £ £ z4 j3 23 e jπ e j3π 2 cos 3π 2 j sin 3π 2 ¨ £ j j2 ¨ £ ¢¢  ¡ ¨ © ¢ ¨ £ ¦ £  £ £ (c) z3 (d) z4 6e j3 jπ 2 6 cos π 2 j, or j sin π2 j6. j ¢ © ¨  £  ©  ¨ £ ¤ ¥ © £ £ z2 ¡ 3 e j3π 4 3 cos j sin 1 22 j1 22 1 22 3π 4 3π 4  ©  £ ¤¢  § ©  § £ £ ¢ ¢ ¨ £ £  £ ¨  ¡ ¢£ £ £ £ £ ¡ 1 j 14 (a) (b) (c) (d) (e) CHAPTER 1 Express z1 and z2 in polar form. Find z1 by applying Eq. (1.41) and again by applying Eq. (1.43). Determine the product z1 z2 in polar form. Determine the ratio z1 z2 in polar form. Determine z3 in polar form. 1 Solution: (a) Using Eq. (1.41), ¢ (b) By Eq. (1.41) and Eq. (1.43), respectively, (c) By applying Eq. (1.47b) to the results of part (a), ¢ ¢ ¢ (d) By applying Eq. (1.48b) to the results of part (a), ¢ (e) By applying Eq. (1.49) to the results of part (a), ¢ 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.) © ¨ £ Problem 1.16 If z (a) 1 z, (b) z3 , (c) z 2 , (d) z, (e) z. 2 j4, determine the following quantities in polar form:   ¢  ¢ ¢  3 6e 46 66e ¥ £¥ £ ¢¥ £ z3 1 j33 7 3 6 3e j3 33 7 ¤ 3 ¥ 5e j143 1 ¢  0 72e ¥  £ ¢  £ z1 z2 3 6e ¥  £ z1 z2 3 6e j33 7 j176 8 ¥ £¥ ¥ j33 7 5e j143 1 18e j109 4   £ © ¢ ¨ z1 3 j2 3 j2 13 3 60 j101 1 ¡  £ £¢ ¡ £¦¢ ¨© ¨ z1 3 j2 2 ¡ 32 2 ¢ £ © ¡£ ¨ £ z2 4 j3  £ ¨ £ z1 3 j2 3 6e 5e j143 1 ¥ ¡¥  ¢££ £ j33 7 13 3 60  ©§¥ ¨¦  ©§¥ ¨¦  18 (d) CHAPTER 1 (e) Problem 2 Problem 1.23 Find the instantaneous time sinusoidal functions corresponding to the following phasors: 5e jπ 3 (V), (a) V j6e jπ 4 (V), (b) V (c) I 6 j8 (A), ˜ (d) I 3 j2 (A), ˜ (e) I j (A), ˜ (f) I 2e jπ 6 (A). Solution: (a) (b) 6 cos ωt (c) (d) 3 61 e j146 31 e jωt ¢  ¢¥  ©  £ ¢  £ ¤¢ it 3 61 cos ωt ¡   £ © ¨ £¨¡ ¢ £ I 3 j2 3 61 e j146 31 ¢¥  © £ ¤¢ it 10 cos ωt 53 1 A. ¡ ¢ £ ¢ © £ I 6 j8 A ¥  ¢ © £ ¦¢ vt π4V 10e j53 1 A 146 31 ¡ £ £  £ V j6e V 6e j V ¥ ¤ jπ 4 π4 π2  ¢ ¨ £ ¤¢ vt 5 cos ωt 2π 3 V 6e jπ 4 V ¡  £ £ ¨ £ V 3 V 5e j V 5e ¥  ¤ ¢ £ I 7e jπ 6 A 5e jπ π3 π j2π 3 V ¡¢  ¨ ¨ 4 cos ωt π6 3 cos ωt π6 7 cos ωt £ ¤¢  ¨ ¢ ©   © ¨ ¢ © ¨ ¢ 4 cos ωt π6 3 cos ωt π6 ¢ ¨ © ¡ ¢ ¢ © ¨ 4 cos π 2 ωt π3 3 cos ωt ¢ ¨ © ¢ ©  £ £ £ £ ¤¢ it 4 sin ωt π3 3 cos ωt π6 π6  π6 A  £  £ ¨ £ I 2e 2e e 2e j3π 4 ¡¢  © ¨ £ ¤¢ it 2 cos ωt 3π 4 jπ j3π 4 jπ 4 A  © © ¨£ ¨£ £ £ £ £ CHAPTER 1 (e) e jπ 2 e jωt 19 (f) 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 . R L i Vs(t) C Figure P1.24: RLC circuit. Problem 1.25 A wave traveling along a string is given by ¢ © £ ¢ ¡ yxt 2 sin 4πt 10πx (cm) ¢ ¨ © ¢  ¨ © ¢  R R 1  £ £ ¨ © £ ˜ (c) I Vs j ωL 1 ωC V0 e jπ 3 j ωL 1 ωC  © © £ (b) In phasor domain: Vs ˜ RI  Ri L © © £ ¢ Solution: (a) vs t di dt 1 C i dt ˜ jωLI ˜ I jωC ωCV0 e jπ 3 ωRC j ω2 LC ¢  ¢ © £ £ ¦¢ it  2e jπ 6 e jωt 2 cos ωt ¡ ¨ £¡ ¢ £ I 2e jπ 6 π6A  ¨ £ ¢  © £ ¢ £ ¤¢ it cos ωt  ¡ ¨ £¡ ¢ £ £ I j e jπ 2 © π2 sin ωt A ¢ £ ¤¢ ECE 311 - Homework 2 Solutions CHAPTER 2 33 Chapter 2 Sections 2-1 to 2-4: Transmission-Line Model Problem 2.1 A transmission line of length l connects a load to a sinusoidal voltage source with an oscillation frequency f . Assuming the velocity of wave propagation on the line is c, for which of the following situations is it reasonable to ignore the presence of the transmission line in the solution of the circuit: (a) l 20 cm, f 20 kHz, (b) l 50 km, f 60 Hz, (c) l 20 cm, f 600 MHz, (d) l 1 mm, f 100 GHz. Problem 3 Problem 2.2 Calculate the line parameters R , L , G , and C for a coaxial line with an inner conductor diameter of 0 5 cm and an outer conductor diameter of 1 cm, ﬁlled with an insulating material where µ µ 0 , εr 4 5, and σ 10 3 S/m. The conductors are made of copper with µc µ0 and σc 5 8 107 S/m. The operating frequency is 1 GHz. Solution: Given combining Eqs. (2.5) and (2.6) gives    ©   0 25 2 m 0 50 2 m   ¢ 1 π 109 Hz 4π 10 7 H/m 2π 5 8 107 S/m 0 788 Ω/m §     © ¢   £ £ £ R 1 2π π f µc σc ¡ 1 a 1 b 1 10 ¡    £ b 1 0 2 cm 0 50 10 2 m ¡    £ a 0 5 2 cm 0 25 10 2 m 1 10  0 33 (nonnegligible)  0 40 (nonnegligible)   0 01 (borderline) £     1 33      £ £ £  £ Solution: A transmission line is negligible when l lf 20 10 2 m 20 103 Hz l (a) λ up 3 108 m/s l lf 50 103 m 60 100 Hz (b) 8 m/s λ up 3 10 l lf 20 10 2 m 600 106 Hz (c) λ up 3 108 m/s 3m lf 1 10 100 109 Hz l (d) λ up 3 108 m/s £ ¢ £ ¢ ¢ ¢ £   £   ¢  £ ¢  £   ¦¢  ¦¢  ¦¢  ¦¢           §  £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ λ 0 01. 10 5 (negligible). 34 From Eq. (2.7), CHAPTER 2 From Eq. (2.8), From Eq. (2.9), Problem 2.3 A 1-GHz parallel-plate transmission line consists of 1.2-cm-wide copper strips separated by a 0.15-cm-thick layer of polystyrene. Appendix B gives µc µ0 4π 10 7 (H/m) and σc 5 8 107 (S/m) for copper, and εr 2 6 for polystyrene. Use Table 2-1 to determine the line parameters of the transmission line. Assume µ µ0 and σ 0 for polystyrene. Solution: 3 Problem 2.4 Show that the transmission line model shown in Fig. 2-37 (P2.4) yields the same telegrapher’s equations given by Eqs. (2.14) and (2.16). Solution: The voltage at the central upper node is the same whether it is calculated from the left port or the right port: ¢ ¡ © © ¢¡ © © ¢¡ vz ∆z t 1 2R ∆z i z ∆z t 1 2L ∆z ¢¡ ¨ ¢¡ ¨ ¢¡ © £ £ ¤¢ ¡ vz 1 2 ∆z t vzt 1 2R ∆z i z t 1 2L ∆z ∂ izt ∂t ∂ iz ∂t ∆z t     £     £ © £ £ C ε0 εr 26    εw d w d 10 9 36π ¡ £ G 12 15 10 10 2 1 84 10 ¡    £ £ £ L 10 (F/m) ¡  £        £ £    £ £ R 2Rs 2 π f µc 2 π 109 4π 10 7 w w σc 1 2 10 2 5 8 107 7 3 µd 4π 10 1 5 10 1 57 10 7 (H/m) w 1 2 10 2 0 because σ 0 12 1 38 (Ω/m)   £ £      § £ £ ¢ £  ¢ ¡  £ £ £ C ¢      2πε ln b a 2πεr ε0 ln b a 2π 45 8 854 ln 2 10 12 F/m   £ £ ¢ £ G   2πσ ln b a 2π 10 3 S/m ln 2 9 1 mS/m  362 pF/m £ £ £ L   µ b ln 2π a § 4π 10 7 H/m ln 2 2π 139 nH/m £ cm r44 r4 \$ 14 4 t rL C r4 N ) ‘4 - r C, 4 —C “C:’ cm ‘-‘C_i ‘a -4” c\$à 4 L V ‘p t (N C;;’ C II ‘N -Th ‘4%’ .4. 4 ED \$ N ‘43 .4 44 4%) ...
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