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HW1 - 22 = Vie/BTW = «3[cos(343 jsin(341 = —1.22 jl.22...

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Unformatted text preview: 22 = Vie/BTW = «3 [cos (343) +jsin (341)] = —1.22+jl.22 =1.22(~1+j). (c) 23 = 6e‘j7t/2 =..6[cos(—1c/2) +jsin(—1t/2)] = ~j6. (d) 24 =13 =j-12 = —j, or 24 =13 = (em/2)3 = 61.3"/2 = cos(31:/2) +jsin(3n:/2) : _j_ (e) ZS =j—4 = (ejn/2)——4 = e—j27r :1. (f) 26 =(1—j)3 = (fie-WV = (fife—1‘3"“ = (fi)3[cos(3n/4) —jsin(3n/4)] = ~2—j2 = —2(1 +j). (g) 27 = (1 —j)1/2 =(\/§e_j"/4)1/2 = :tZI/4e_j"/8 = :hl.l9(0.92—j0.38) = :i:(l.IO—j0.45). \ Problem 1.15 Complex numbers 21 and 22 are given by 21=3~j2, 222—4+j3. CHAPTER 1 _ .- 17 Sectio ' -: Phasors (oblem 1.21 : voltage source given by vs(t) : 25 cos(2n X 103t — 30°) (V) is onnected - a series RC load as shown in Fig. l-19. If R = 1 M9 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 17-17 ”10°C _ Vs °“ SR+1/jcoC’(1+jcoRC)' Now 75 = 256—1300 V with (1) = 2n x 103 rad/s, so V __ 25e-J'30o V ° " 1+j<(2n><103 rad/s) >< (106 9) x (200 ><10—‘2 12)) _ = 15.57 ‘1'31-5" v. 1+ j2n/5 e Converting back to an instantaneous value, vc(t) = mafia/W = m215.57ef(w‘81~5°> v = 15.57 cos (211 x 103t — 815°) v, where t is ex ressed in seconds. roblem 1.22 Find the phasors of the following time functions: a —3cos(cot—1t/3) (V), (b) v(t) = lZsin((ot+1t/4) (V), (c) i(x,t) = 2e‘3xsin(wt+1c/6) (A), (d) i(t) = —2cos(0)t +31t/4) (A), (e) i(t) = 4sin((ot +1t/3) + 3 cos(cot — n/6) (A). Solution: (a) I7=3e~f“/3 v. (b) v(t) = 123in((ot+1t/4)= 12cos(n/2—((1)I+n/4)) = 12cos((ot—n/4) V, 17:12e-I'W/4v. (0) 1(1) = 2e‘3xsin(mt+n/6) A = 2e“3~‘cos (11/2 «(191+ n/6)) A = 2e‘3xcos(cot — n/3) A, 7: 3212—3’Ce—flr/3 A. 18 CHAPTER 1 (d) it( )~ —ZCos((0t + 37t/4), I: _ze]3TE/4=2e—j1tej3TE/4 : ze—jTC/4 A (e) i(t) = 4sin(cot + n/3) + 3 cos((ot — n/6) = 4cos[1t/2 (wt+1c/3)]+3cos( (cot— rc/6) = 4cos(——(ot+1t/6)+ 3cos(a)t— 7t/6) — —4cos(a)t— 1t/6) +3cos(0)t— 715/6)— — 7cos(wt—1t/6), (a)— VI: —5ef"/3 (V), (b) V: j6e‘jn/4 (V), (c) Z=(6+1’8) (A), (d) I = -—3 +12 (A), (e) Z = j (A), (f) I: 2eJ’E/6 (A). Solution: (21) I7: —5ef“/3 V: 5efW3‘“) V = 5e‘f2n/3 V vt()= Scos(a)t— 21t/3)V (b) V = j6e‘jn/4 V = 62i(—“/4+“/2) V = 6ej’t/4 V v(t) = 6003 ((01+1t/4) V (e) = (6 +j8) A = 10ef53-1° A, i(t) = 10005 ((Dt+53.l°) A (d) f: —3 +j2 = 3.61 (#146312 i(t) = 90:43.61 e114é‘31°ef‘°’} = 3.61 cos(cot +146.31°) A ('llAl’TERI 19 (C) I : j : ejn/Z, to) = mean/2W} = cos(u)t+1t/2) = — sinmt A. (f) 7: 2e/“/", i(t) = 9%e{2ej”/6e~j“”} = 2005(0)! + n/6) A. Problem 1.24 A series RLC circuit is connected to a generator with a voltage vs(t) = V0 cos(o)t +1t/3) (V). (3) 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 7. R L i V50) C Figure P1.24: RLC circuit. Solution: a" 1 . l . (a) vs(t)—R1+LE+E/zdt. ~ ~ ~ I (b) In phasor domain: VS = R1+ ijI + W . .l S _ V0 Gin/3 fl (DCVO ejn/ 3 R+j(coL—1/coC) ‘ R+j(ooL #1/(1X')— mRC+j(c02LC— 1) ' (c) i: Problem 1.25 A wave traveling along a string is given by y(x,t) =Zsin(4nl l— l()n\') (cm) CHAPTER 1 Determine (a) 2122, (b) 21 /z’2‘, (c) 2%, and (d) zlz’f, all all in polar form. Solution: 'roblem 1.27 Complex numbers 21 and 22 are given by zl=-—3+j2 22:1—j2 (a) We first convert 21 and 22 to polar form: (1)) (C) 21 = —(3 —12) = — (J32 +22 e—Jtan-‘Z/a) = _ 13 e-j33.7° : mei(130°—33.7°) _____ mam-63°. 22 =1—j2=\/1+4e-f“m"2 : 56—1614". 2122 = \/13 em“ x 5 5163-4" 2 /_65 9132.? _Z_1 = ___._" 13 61.14630 : ‘ IE ej82.9° 2; \/§ ej63.4° 5 ' (fig)2(ejl46.3°)2 ___ ”81292.6“ .L‘L. II ___ 13 @4160“ 6,292.0" : 1335/0791” 21 ... nwmmflkwdv 22 CHAPTER I (d) 212:; = Jig-€146? X 13 e—jl46.3° = 13. Problem 1.28 If 2 = 3ej"/6, find the value of e5. Solution: 2 = 3ej"/6 = 3cos1t/6+j3sin1t/6 =2.6+j1.5 ez = e2.6+j1.5 = 82.6 x 9/15 = e2'6(cos 1.5 +jsin 1.5) = 13.46(0.07 + jO.98) = 0.95 +j13.43. Problem 1.29 The voltage source of the circuit shown in the figure is given by v50) = 25 cos(4 x 104: — 45°) (V). Obtain an expression for iL(t), the current flowing through the inductor. R1=209,R2=3OQ,L=0.4mH Solution: Based on the given voltage expression, the phasor source voltage is Z = zse'j“5° (V). (9) The voltage equation for the left-hand side loop is R1i+R2iR2 = Vs (10) 40 CHAPTER 2 ml» or A, = 24 cm. Accordingly, the first voltage minimum is at 3min = 3 em =* Application of Eq. (2.57) with n = 0 gives 2n7t 9,—2X—K—Xg=—Tt, which gives 6, = —n/2. 5—1 3—1 2 lrl_m_3_1_3_0'5' Hence, I‘ = 0.5 fin/2 = —j0.5. Finally, 1+r 1—j0.5 _ L °[1—r] 50l1+j0.5l (90 J 0) ( Problem 2. . ‘ Using a slotted line, the following results were obtained: distance of - n I um from the load = 4 cm; distance of second minimum from the load = 14 cm, voltage standing-wave ratio = 1.5. If the line is lossless and Z0 = 50 9, find the load impedance. Solution: Following Example 2.5: Given a lossless line with Z0 = 50 Q, S = 15, lmin(0) = 4 cm, 1mm“) = 14 cm. Then X Imin(l) _ [min(0) — E 01' A = 2 X (Imin(1) _ lmin(0)) = 20 cm and 21: Zn rad/cycle B — T _ 20 cm/cycle _ 101: rad/m. From this we obtain 6, = 251mm — (2n+1)1t rad = 2 x 101t rad/m x 0.04 m- 7: rad = —0.21t rad = —36.0°. Also, CHAPTER2 41 S0 1+1“ l+0.2e’j36'0° z :2 x :50 ./ = 67.0— ‘16.4 o. L 0(14‘) (1—0.2e-J‘36-0" ( J ) onnected to a Problem 2.15 A load with impedance ZL = (25 — j50) Q is to be c lossless transmission line with characteristic impedance 20, with 20 chosen such that the standing-wave ratio is the smallest possible. What should Z0 be? Solution: Since S is monotonic with M (Le, a plot of S vs. |F| is always increasing), the value of Zo which gives the minimum possible S also gives the minimum possible lfl, and, for that matter, the minimum possible \F|2. A necessary condition for a is that its derivative be equal to zero: minimum _ a 2__ a ‘RL-i-jXL—Zo‘2 o - wlr _ “/47 8Z0 8Z0 ‘RL +jXL+Z0\ : ,3. (RL —zo)2 +X2 : 4RL(Zg — (Ri+XE)) 320 (RL + ZO)2 +XE ((RL + Z0)2 +X€)2 i V i: g, Therefore, Z3 = Ri +XE or 20 = mm = i/ (252 + (—50%) = 55.9 9. A mathematically precise solution will also demonstrate that this point is a minimum (by calculating the second derivative, for example). Since the endpoints of the range may be local minima or maxima without the derivative being zero there, = O Q and Zo = 0° 9.) should be checked also. the endpoints (namely Z0 ly resistive load has a 0—9 lossless line terminated in a pure ' f 3. Find all possible values of ZL. Solution: S 1 3 1 m=§fi=m=°5 . e load, 9, = O or 1:. For 0, = 0, 1 1“ 1 0.5 ‘ ZL=ZO[I—i?]=5o[ + i=150r2. g S i For a purely resistiv mapfiw ... '. 44 CHAPTER 2 Solution: (a) From Eq. (2.4%), _ ZL—Zo _ (100—1100) —50 : 0.62e—J'29~7°. 1"__ .. 2L +20 (100—1100) + 50 (b) All formulae for Zin require knowledge of B = (0/ up. Since the line is an air line, up = c, and from the expression for vg(t) we conclude 0) = 271: x 109 rad/s. Therefore 27: x 109 rad/s 207: — = —— rad/m. 5‘ 3 x 108 m/s 3 Then, using Eq. (2.63), ZL +jZo tan [3]) z. = ———— ‘“ ZO(Zg+jZLtanBl _ 50 ((100—1100) +j50tan grad) = 12.5—12.7 9. 50+j(100—j100)tan(§rad)) ( J ) An alternative solution to this part involves the solution to part (a) and Eq. (2.61). (c) In phasor domain, Vg = 5 V 6100. From Eq. (2.64), ~ '17me _ 5x(12.5— j12.7) - _ _———— = 1.40 '134-0" ‘ zg+zin 50+(12.5——j12.7) e (V), and also from Eq. (2.64), _ = '11.5° ' ‘ (12.5 — 112.7) 78'4‘31 (mA) 71 = E 1.4e”j34‘°° in Problem 2.22 A 6-m section of 150-0 lossless line is driven by a source with vg(t) = 5005(81: x 10% — 30°) (V) and Zg = 150 Q. If the line, which has a relative permittivity e, = 2.25, is terminated in a load ZL = (150 —j50) 9, find (a) A. on the line, (b) the reflection coefficient at the load, (c) the input impedance, CHAPTER 2 45 (d) the input voltage 12, (e) the time-domain input voltage vi(t). Solution: vg(t) = cos(8nX107t—30°) V, (a) up = jg; = 3 2120: = 2 x 108 (m/s), 2 2 2 l( 8 = 3:; = 8;:11027 = 0.47: (rad/m), .A B1 = 0.411: X 6 = 2.4x (rad). 46 CHAPTER 2 ’ Since this exceeds 21E (rad), we can subtract 211:, which leaves a remainder B1 = 0.41! (rad). (b) r = (C) ZL—ZO _ 150—j50— 150 _ —j50 _ ___._.__ _ __ = 0.16 —J'30-54°_ ZL+ZO 150—j50+ 150 300—150 e Z+‘Ztanl Zin=Zo[ L Jo I3] Zo+jZLtanBl (150— 150) +j150tan(0.41t) = 150 _.————— [150+ j(150— j50) tan(0.41t) = (115.70+j27.42) Q. ((1) ask 2... _ 1W Zg+Zin _ 150+ 115.7 +127,42 = Se—jsoo W 265.7 + j27.42 : 56—1300 X 0_448j7'44° = 2.2e_j22'56° (V). (e) vi(t) = Memefw’] = 2342.2 win-59am = 2.2 cos(81t x 107: — 22.56°) V. W Problem 2.23 Two half-wave dipole antennas, each with impedance of 75 Q, are connected in parallel through a pair of transmission lines, and the combination is connected to a feed transmission line, as shown in Fig. 2.39 (PZ.23(a)). All lines are 50 Q and lossless. (a) Calculate Zim, the input impedance of the antenna-terminated line, at the parallel juncture. (b) Combine Zia. and Zinz in parallel to obtain Z’ , the effective load impedance of the feedline. (c) Calculate Z“ of the feedline. Solution: (3) Z‘ :2 [ZL1+jZOtanBll:| ml 0 Zo +jZL1tanl311 _ 50 { 75 +j50tan[(2n/x)(o.2x)] 50 + j75tan[(21c/?»)(0.2?c)] } = (3520—1862) 9‘ ...
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