Electromagnetics for Engineers

Electromagnetics for Engineers

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Unformatted text preview: Chap'ie? E Section i=3: Eraveiing Waves Probfiem 1.1 A 4-111-12 sound wave traveling in the x—direction in air was observed to have a differential pressure p(x._ z) = 5 "film2 at x = 0 and r z 25 ,us. If the reference phase or" phat) is 42", find a complete expression for p('x,r)_ The velocity of sound in air is 330 mfs. Sedation: The general form is given by Eq- (1.17), 2m 2m: ) p(x,r) = Amos (—3}.— — T +¢o where it is given rhat (90 = 4295:0111 Eq. {1.26), r =1,If= 1/(4 x103): 0.25 ms. From Eq. (1.27}, Also, since 2:: x 25 x 10-‘5 17: rad —— + 42" 2.5 X 10'4 180° : Acos(1.36 rad) = 0.20851, p(x= 0, r : 25 ,us) : 5 (Nlmz) =Acos( it foilows that A = 5/0208 = 24 Nlmz. so. with r in (s) and x in {111), _ 4-) 6L_ - 3 x r o 2 p(x,r)_24cos(mx10 250 275x10 #82-5'1’42) (Wm) = 24cos£81t x 103; — 24.24m+ 42°) (Nlmg). l—P‘mhlem 1.2 For the pressure wave described in Exampie 3-1, plot {a} p(x,r) versus x at I = 0, (b) Mar) verses r at x = 0. Be sure to use appropriate scales for x and: so that each of your piots covers at least two cycies. Semifinal: Refer to Fig. P1.2(a} and Fig. P1.2(b)_ 2 CHAPTER} pix=0.t) "E “E E E g '8 E E i i E E 4 < . . ALL 4:) 41. . 2. ‘ um 515 0.50 {315 mu L25 1.50 LTS 2.00 25 :50 :15 3m 00 u: Lu as 03 :.|:I I: H :0 LB 1:: Dmflm} Tm: Him) (a) (b) Figure P12: (21) Pressure wave as a function of distance at t = O and (b) pressure wave as a fusiction of time at x = 0. Fmbiem 1.3 A hamonic wave traveling afiong a sting is generated by an osciiiator that completes 120 fibrations per minuie. If it is observed that a given crest, or maximum, travels 250 cm in 10 5, what is the wavelength? 303mm: 120 7- ~—— = 2 Hz. f 60 250 cm [4.9— 105 —O.25mfs. 7L PmMem L4 Two waves, y1(r) and 3&0), have identical ampiimdes and osciliate at the same frequency, but yg(r) Eeads no?) by a phase angle of 60°. if 311(23): 4C05(27t X 1031:), write down the expression appropriate for 3520‘) and plot both functions oven: the time span 15mm 0 to 2 ms. Sefiusfiem: mm = 4cos(2:it x103: + 60°). -=_£'3{.§’-‘ Wm? --r-3$§5s'ég_n.- fi‘ . x $3 DJ CHAPTER I Preblem 1.5 The height of an ocean wave is deacfibed by the function y(x,t) = 1.5 Sin(0.5r— 0.63:) {In}- Detemaine the phase velocity and the wavelength and then sketch y(x,r} at r = 2 s over the range from x :— 0 to x = 23L Soiufien: The given wave may be rewritten as a cosine function: y(x, r) z 1.5 035005! ‘— 0.6x —— Tc/Zj. By comparison of this wave with Eq. (1.32), y(xflr) = A casket — Eur-i- $0), wededuce‘ihat 211'. m=2nf=0.5radfs, 13:75:063‘adfm m 0.5 n 21: 21: _ HPZE—E—‘OfiJWSa R—f—Y—§0.47Ifl 4 CW1 Figure P15: Piot ofy(x,2) versus 3:. At t = 2 s, y(x1 2) = 1.5 sin(i -— 0.62:) (In), with the argument of the cosine function given in radians. Plot is shown in Fig- P15. melee: 1.5 A wave traveling along a. string in the +x—direction is given by 3,10%!) 2 Acoskflr _ where x = O is the end of the string, which is tied rigidly to a wail, as shown in Fig. 1-21 (P16). When wave y1(x,r) arrives at the wall, a reflected wave y2(x,z) is generated. Hence, at any iocation on the string, the vertical disPIacement ys will be the sum of the incident and reflected waves: 3,455,?) = yl(x1t) +y2(xit)' (a) Wnlte down an expression for y2(x,r), keeping in mind its direction of travel and the fact that the end of the string cannot move. (33} Generate plots of y1{x,r), yg(x,r) and ys(x,r) versus .7: over the range —2kgx30attor :n/eandathn/z. Sofin‘étion: {3) Since wave y2(x,r) was caused by wave 311 (am), the two waves must have the same angular frequency to, and since y2(x,t) is traveling on the same string as y1(x,z), MRI Incident Wave —§... xy= 0 Figure P1.6: Wave on a string tied to a wdl at x = 0 (Problem 1-6). the two waves must have the same phase constant fi. Hence, with its direction being in the negative x-direction, y; (x, r) is given by the general form MCXJ}=BCOS(W+BI+¢0), (1) where B and $3 are yet-to-be-de‘iennined constants. The total dispiaceinent is y5[x,r] : y1(x,r) +y2(x1£) = Acos[on — 5x} +Bcos(ooz -z- Bx-E— {90). Since the string cannot move at x = 0, the point at which it is attached to the wall, y5(0,z) = 0 fossil I. Thus, y§({},r) :Acosmr+Bcos(mr-1-¢o) : 0. (2) (i) Easy Solutioa: The physics of the probiem suggests the: a possibie solution for (2) is B = —A and $0 = O, in which case we have y2(x,r) = uAcos(cor + (3) (ii) Rigorous Solution: By expanding the second term in (2), we have Aces {or +B(cosmrcos¢o — sine)? same) : 0, (A +Bcos $0) costar — (Bsin¢0)sin (or = 0- (4-) This equation has to be satisfied for 33:11 values of I. At .2 = 0, it gives A+Bcos¢o = 01 (5) 6 CHAPTER} and at wt 2 1r / 2, (4) gives Bsinqao = 0. (6) Equations (5) and (6) can be satisfied simuitaneously Only if A = B = 0 (7) - 01' A:—B and $020. (8) Clearly (7) is not an acceptable solution because it means that y1(x,t) = 0, which is contrary to the staiement of the problem. The solution given by (8) leads to (3). (£3) At (or z 1': f4, y1(x,r) = Acos(1t/4—i3x)= Acos (E — , 3:20:11?) = —Acos(co£+fix) = —Acos (4:- + . Plots of y1, y2, and )3 are shown in Fig. P1.6(b). 83$:th- figure P16: (b) Plots ofyl, 31;, and ys versusx at (or = 12/4. Ate)? = 112/2, y1(x.,r) = A cos(fi/2 +— fix) 2 A siufix = Asin 2%” , CHAPTER I 7 . 2 y;(x,r) = —Acos{7:/2+§3x) =Asinfix=AsmTfix. Plots of y1, ya, and 323 are shown in Fig. P1-6(c). thfiem 1.”? Two waves on a. string are given by the following functioes: y1(x,t) : 3cos(20£ — 30x) (cm), 3:29:11) : —3cos(20r + 303:} (cm), where x is in centimeters. The waves are said to interfere constructiver when their superposition [3:51 = lyi +3221 is a. maximum and they interfere destructiver when lysl is a Izfinimum. {3} Mat are the directions of propagation of waves y;{x1 t) and y3(x,x)‘? (h) At t = (ts/50) s, at what location x do the two waves intesfere constructively, and what is the conesponding value of I ys! ? ‘ (5:) At 2* : (tr/50) s, at what location 5: do the two waves interfere. destmctiveiy, anti what is the corresponding value of [52; | ’3 Somme: {a} y3(x,r) is traveling in positive x—direction. 3720“) is Haveling in negative x—ditection. 8 WERE (b) At 1' = (TE/50) s, y; = y; +313 2 3[cos(0.411t — 30x) — cos{0.4:n:+3x)]. Using the formulas from Appendix C, 23inxsiny = cos(x—y) — (cosx-i—y}, we have 32, : 25in(0.47:] sin30x = 1.9si330x. Hence, lysine; = 1-9 . 7t and it occurs when sm30x = I, or 30x = E + 2am or x = (—— + —) cm, Where 3*: =0,1,2,.-. . . m: {c} [yslmjn = 0 and 1t Occurs when 30:: = ms, or x = — cm. Pmbiem 3.8 Give expressions for y(x,r) for a sinusoidal wave traveling along a string in the negative x—direction, given that ym z 20 cm, k = 30 cm, f : 5 Hz, and (a) y(x,0) : O at x = 0, (b) y(x,0) = 0 atx: 7.5 cm. Solution: For a wave traveling in the negative redirection, we uSe Eq. (1.17) with w : 21tf = 161': (radfs), i3 = 21d?» = Eat/0.3 = 201t/3 (mdfs), A = 20 cm, and x assigned a positive sign: 2011: y(x,:) = 20005 (1075+ Tat-Him) (cm), withx in meters. {2) y(0,0) = 0 = 20¢os¢0. Hence, the = int/2, and 20:: y(x,1‘) = 20605<10flr+ —-3—x:l: 4:) * —203in[10fit+3%"~x) (cm), if¢o = 1:]2, * 205in(10m+ 312%) (cm), if¢o = —1:/2_ (b) Atx = 7-5 cm :75 x10_:’- m, y z 0 = 20C08(fi/2+¢0). Hence, $0 = 0 or R, and (x ‘J _ 20003 (101:: + (cm), ifqm = 0, y ’1 _ —20cos[101rr+ @x) (cm), if¢0=m Problem 1.9 An oscillator that generates a sinusoidal wave on a string completes 20 vibrations in 30 s. The wave peak is observed to travel a distance of 2.8 fit. along the string in 5 3. What is the wavelength? Sofinfion: Probiem 1.3-8 'I‘he vertical displacement of a. suing is given by the harmonic function: y(x,t) = 5 cos( 127E: -— 20H) (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 can obtain an expression for the vertical velocity of the particle as a function of time. Sciatica: y{x, t) = 5 cos(12m‘ — 20m) (in). dytfx, r) 4’3 1:0.05 : 601tsin(12m‘ — 20m)lx=o.as : fifinsinUEttt — it} : —602tsin(12:ztr} (refs). M03354“) = Problem 1.11 Given two waves characterized by mm = Ecostm, mm = 65in(tot + 30° ), does 3120‘) Read or lag yfit), and by what phase angle? Sciatica: We need to express 312(1’) in terms of a cosine function: y2(t) : 63in(cor + 30") = 6cos @— - oar - 30°) :- ocos(60° — EDI) :- 6cos(mt — 60°). 10 CHAPTER 1 Problem 1.12 The voltage of an electromagnetic wave traveling on a transmission line is given by v(z,t) = 39““ sin(21t x 1092‘ — 10m) (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 In, the amplitude of the wave was measured to be 1 V. Find on. Sofiution: (a) This equati0n is similar to that of Eq. (1.28) with o) = 2’11: x 109 1361's and B = 101: rad/m. From Eq. (1.29:1), f = (13/21: :109 Hz = 1 GHz; from Eq. (1.29b), 7L: 212/5 = 0.2 m. From Eq. (1-30), up: m/B=2x103 W5. (b) Using just the amplitude of the wave, —1 1 1 = 33-“2, a = w1n(—) = 0.53 Nplm. 2m 3 ?mb§em 1-13 A certain electromagnetic wave traveling in sea water was observed to have an amplitude of 19-025 (Wm) at a depth of 10 In and an amplitude of 12.13 (Wm) at a. depth of 100 to. What is the attenuation constant of sea water? Sofiution: The amplitude has the form Ae‘”. At 2 = 10 n.1, A240“ = 19.025 and atz = 100 m. Ae*1°°“:12.13. The ratio gives e40“ 19.025 8—100“ 12.13 = 1.568 01' 34°“ = 1.568e“1°°°‘. Taking the natural log of beth sides gives 1n(e*1°°‘) = 11305685100“), —100t = 1n(1.568)- 1000:, 900'. = 111(1568): 0.45. _ 0.45 ——= _3 f . oz 90 5x10 (NPm) 11 CW}? 1 Section 1-5: €®mpiex Numbers mbiem 1.14- Evaluate each of the following complex numbers and express the result in rectangular form: (a) 251 = 364W“, ea) :2 = fi em”, {a} 23 = 2 rim, (d) Z4 = P, {8) ZS 1‘ 1'“, {f} 25 = (1r j)3, {3) Z7 =(1-j}”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) 21: 3am“ = mom/4+ jam/4) = 2.12+f2.12 = 2-12(1+j). (b) 37:: z: = ‘58:?“ = x/i [cos +jsin (3)] = —1-22+j1-22: 1.22(—1+j). is) zs = 22's” = meson/2) + ism—#2)] = —j2. (d) Za=j’=j-j2=—j,or 14 2 i3 = (ejfiflf = ejsfi’rz = cos(31:/2)+jsin(31c/2) = —j. {e} 15 = j“: : bah-mm“Tr = 8'52“ = 1. {i3 zs = (1—33 = (Jig—imp : (fi)3€+;3zg4 = («/5D3[cos(3n/4) * jsin(31:/4)] = *2-1'2 = w2(1-1-j). (g) 27 = (1— a”? = (fie-W4?” = 4_—:2‘"‘ue'fi‘1’8 = i1.19{0.92— j0-38) : :i:(1-10-—j0.45). thaem LE5 Complex numbers 21 and z; are given by z1= Z: I 12 CHAPTER I {3) Express 21 and 2:2 in polar foam. {is} Find izll by applying Eq. (1.41) and again by applying Eq. (1.43). (6:) Determine the product .2122 in polar form. (mi) Determine the ratio Z1/z2 in polar form. (e) Determine z? in polar form. Solution: (2} Using Eq. (1.411), _ - __ z—'*3.7° z1—3—j2w3.oe-’° , 22 = —4 + 1'2 = 4.581514". (b) By Eq. (1.41) and Eq. (1.43), respectively, tzzl = I3 — j2l = 32 + (—2)2 = m: 3.60, IZ1i= W: J13: 3.60- (o) By applying Eq- (1.4713) to the results of part (a), Z122 = 3.653333" x 4.531534" = 16.2ei‘19-7°. (fi) By applying Eq- (1.4833) to the results of part (a), z; 3.6e“133'7° _ — 187.1" z: — 45 M3 40 _0.80e J - . e ' (e) By applying Eq. (1.49) to the results of part (a), z? = (3-66'J33'70f = (3.6)3e"f3x33-7° = 46.66e-i101-1°. Problem 1.16 If 2: : —2 + _;r'31 determine the following quantities in polar form: (a) 1 /z, (W 23, (c) lzll. {fl} 31:1{2}, (e) 3m{z*}. Salmon: (Note: in the following solutions, numbers are expressed to only two decimal places, but flue final anSWEIS are found using a calculaior with EU decimal places-) 13 = I _ —{—2+j3)_1=(3.618123'70r}:(3.61)’1e'j123'7°20.286'j123‘70- (b) 7.3 = (—2+ 3'3)3 = (3.61ei123-7°)3 : (3.61)3ei3“-1° : 46.37ei11-°7°- is) {2:12 = z-z” = (—2+j3)(—2—j3) = 4+9 = 13. (d) 3m{z} = 3m{—2+j3} = 3. {a} 3m{z*} = 3m{—2—j3} = —3 : 3e”. E’mbiem 1.17 Find complex numbers I = 21 +21 and s = z; — 22, both in polar form, fof each of {116 following pairs: {3) Z1: 2+j3, 32:1—1'2. (b) 11: 2. 32 = 'jza (6:) Z: = 34.39:, 22 = 3&, {fl} 21- = 3&1, 22 = 3H. Soiufiom: (a) r = 2: +152 =(2+j3)+{1—j2)= 341'}: 31691343: 5 = 31“ Z2 =(2+j3)*(1-*j2)=1+j5 = 5.103178-69". (h) r = 31412 = 2— i2 = 2.83.3—fl507 S=zx ‘12 = 2+j2:2.83ei45°. » I (c) r=z1+22 = 3fli+3fl91 = 353°“+3g-i30° : (2.6+j1-5)+(2.6—j1.5}= 5.2, s = 211— 2.2 = 3ei3°° - 33—!“ : (2-6+j15) — (2.6—j15) = f3 = 3.4990". _ in?) z = 21 +1: = 3430.2} 3/;150“ = (2.6+ 11.5) + (—2.6— j1.5) = 0, 5: 21—22 = (2.6+j1-5)d(—2-6—j1.5)= 5.2+j3 = 6950’. Probfiom 1.18 Compiex Bombers Z1 and 22 are given by Z1: 51152, Z2 = 215:. 14 CHAPTER 3 (3) Determine the product zlzz in polar form. Cb) Determine the product 47.123 in polar form. (:3) Determine the ratio 21 / 21 in poiar form. (d) Determine the ratio z‘f /z§ in polar form. (e) Determine fl in polar form. Solnfion: o o o (a) 21212 = Se'JfSDo X 2945‘ (I; 108-11? _ o (b) zlz’z‘ = Sea—15° xze-J45 = log—nos _ _ 59° (3) fl = a = 2.5-j105°_ (a) 2:42) 22 2.2 (e) «a = x/Se-fi0° = ifie-i3°°. Fromm 1J9 If 3: = 3 — 1'4, find the value of 111(2)- Sofiufionz —4 121: +\/32+42 = 5, e = tan" (3») = —53.1°, z = Izieja = 58‘j53'10, 111(2) 2 1116645110) = 111(5) + Ink—1511c) 53.1%; = .61—' ,°=1.6 — «—= .6 -—'0- . l j531 1 Jlsoo 11 J 93 Problem 1.281 If z = 3 -— j4, find the value of .22. Sofiufion: 82 = 33—34 = e3 - €_j4 : e3{cos4— jsin4), e3 = 20.99, and 4 rad : E x 180° : 229.18". Hence, 62 = 20.08(cosZZ9-18° — jsin229.18°) = —13-13+j15.20. WWI” .~.. CPLAPIER f 15 Section 1-6: ?hasors fiebfiem 1.21 A voltage source given by v50) : 10cos(2n x 103: — 30°) N) is connected to a series RC load as shown in Fig. 1-19. If R = 1 M9 and C = 100 If, obtain an expression for v42), the voltage across the capacitor. Sofiafiton; lo the phasor domain, the circuit is a voltage divider, and 9 _ g} 1/1735 _ Vs ‘_ SR+1/ij' (i+jcoRC)' Now “275 z 102—3300 V with so : 21tx 103 rat-L's, so i; iOe'i3°° v M “' 1+j([27tx103 radfs) x (1069) x (100x 1041 1.3)) lee-139° v = ——e = 3.5 '3'52-1" v. 1+j1c/5 '9 Converting back to an instantaneous value, yam = mafia” : mes.5eiEW-52-‘°? V = 8.5cos(21t x 103: —62-i°) v, where r is expsessed in seconds. Promotes 1.22 Find the phasors of the following time functions: (a) v(r) = 3cos(cor—n/4) (V), {b} v(r)=123in(mr+it/4) (V), {c} i(x,t‘- = 46‘3‘sin(cor — 1.2/6) (A), {mi} 5(r} = —2cos(o)r+ 311/4) (A), (e) fir) : Zsin(mr + 11/3) + 3 cosfcor — arc/6) (A). Sofinfion: (a) i? = Eve-W“ v. {is} v{r) : 12sin£mr+fi/4) : 12cos(I£/2— (oar+1t/4)j : 12cos (a): —?t/4-) V, fi : 126—4mf4 V. 8:) 5(2) : 42‘3‘sin(e3r—n/6) A = 4e‘3xcos(st/2— (on nation A : 4e'*cos({or — Bit/3) A, F: ée'hfflm3 A. 16 m1 (fl i0: w2a':r.)s((ar+BTU-LL), 483"” = 2;.»‘fi‘emM = 25m“ A- ) T (8) fit) = 23in(mr+n/3)+3cos(mr —‘J‘E/6) = 2cos[TE/2 - (:01 + zit/3]] + 3cos(mt — it/é) = 2cos(~mr +1‘c/6) + 3003(03‘ — 111/6) = 2cos(a)r —1‘E/6) + 3cos(mr — 12/6) : 5cos(m — 31/6), f: 52—33:” A. Pmbiem 1.23 Find the instantaneous time sinusoidal functions corresponding to the following phasors: (a) 17' = —3»ej"’3 (V), (as) ? = jfiefi‘“ (V). (c) I : (3+j4) (A), (d) f: —3+J'2 (A), (e) E = j (A), {f} I = 2&3” (A). Sohfiian: (a) fi = —3ej“{3 V = 331'09’3-10 V = 36—32113 V, “(1): 3cos(cm‘—27t/3) V. (b) E7 = 1'66,ij V : fieflzM-i-n/z} V I 6613,,“ V7 V0) = 6605(f01’+31t/4)V_ (c) ?= (3+j4) A = 5553-1" A, £0) = Scos(mf+53_1°) A {a} 'f: —3+j2 = 3613-5145311 1m : me{3.61ei146-31°eim’} : 3.61 cos(mz +146.31°) A. CHAPTER I 17 {e} f: j = gin/3, i0) : me{ej“flej°x} = (205(0): +1'E/2): — Siflw! A. (f) j": M31132!- i(r) = 9%{2e13fl42j‘9‘} = 2cos(mr + 315/4) A. Pmbiem 1.24 A series RLC circuit is connected to a generator with a voltage 125(1‘) = Vocos(cor +rt/3) (V). (3) Write down the voltage loop equation in terms of the current 1'0), R, L, C, and v50)- (13) Obtain the corresponding phasor-domajn equation. H (c) Solve the equation to obtain an expression for the pnasor current I. Figure P124: RLC circuit. Solution: d_ 1 . I . (a) VS(Z)=RI+LE+E/idf. - ~ ~ ~ I (b) In phasor domain: VS = RI+ijI+ . {is Vgefifl mosh/3 {c} 1=W=W:Wfl- ...'.:.;.;M..«m;-.;.. 18 WK 2 Chapter 2 Seciioes 2% is 2-4: Transmission-flee Mofle‘i fiohiem 2.1 A transmission line of length I connects a load to a sinusoidal voisage source with an oscillasion 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 oransmission fine in the solution of the circuit: (a) i=20cn1,f= lOkHz, (b) I: 50 km, 3“: 601-12, (c) I: 20 cm,f= 300m, (d) I: 1 mm, f=100 GI-Iz. Soiution: A transmission line is negligible when 3/?» g 0-01. z_ E __ (20X10—2m)x(10>< £03 Hz) (a) i — up — W— = 6.67 >< 10-6 (negligible)- (b) % = E; : W2 = 0.0}. (bordefiine). {c} % : if = W = 0.20 (nonncgligible). {é} % = g; 2 WW5)- : 0.33 (nonnegligibie). ?rohim 2.2 Calculate the iine 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 m insulating mateiial where p = no, a, = 2.25, and o = 10—3 Sim. The conductors are made of copper with gt 2 p0 and Ga 2: 5.8 X 107r Sim. The operating frequency is i GHz. Soiufion: Given a : (0.5/2) cm 2 0.25 x 10*2 m, b = (1.0/2) cm = 0.50 ><10'2 m, combining Eqs. (2.5) and {2.6) gives :fi 0‘c a b T i M109 Hz)(41c x 10"“r Him) 1 + i — 275 5.8 X 107 Sim 0.25 X 10—2 m 0.50 X 10": m = 0.788 £21m. ...
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Electromagnetics for Engineers - Chap'ie? E Section i=3:...

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