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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 =j12 = —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 = (ﬁeWV = (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. l19. 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 1717 ”10°C _ Vs
°“ SR+1/jcoC’(1+jcoRC)' Now 75 = 256—1300 V with (1) = 2n x 103 rad/s, so V __ 25eJ'30o V
° " 1+j<(2n><103 rad/s) >< (106 9) x (200 ><10—‘2 12)) _ = 15.57 ‘1'315" 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,
17:12eI'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 = 10ef531° 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 phasordomain 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 ej33.7°
: mei(130°—33.7°)
_____ mam63°. 22 =1—j2=\/1+4ef“m"2
: 56—1614". 2122 = \/13 em“ x 5 51634" 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 lefthand 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 standingwave 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—0.2eJ‘360" ( 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 standingwave 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 \F2. A necessary condition for a is that its derivative be equal to zero: minimum
_ a 2__ a ‘RLijXL—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 '1340"
‘ 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 6m section of 1500 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 timedomain 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'3054°_
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 win59am = 2.2 cos(81t x 107: — 22.56°) V. W Problem 2.23 Two halfwave 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 antennaterminated 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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