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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition

Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition simplifies even the toughest textbook questions with step-by-step solutions and easy explanations.

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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 1 preview imageSolution Manual for Engineering Electromagnetics and Waves:Advanced TopicsUmran S. Inan, Aziz S. Inan, Ryan K. Said
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 2 preview image
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 3 preview imageContentsATransmission Lines: Advanced Topics1CCylindrical Waveguides9DCavity Resonators37EField-Matter Interactions: Advanced Topics45FElectromagnetic Radiation and Elementary Antennas58
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 4 preview imageSolutions to Addendum ATransmission Lines: Advanced TopicsA.1Nonlinear termination.We start by plotting (see Figure A.1) the current-voltage characteristics of the source and load endsof the 50Ωline given by5=10)s+9sand)L=0:35(1e9L=2)Att=0, a voltage of amplitude9+1=Z0Rs+Z0V0=5010+50(5 V)≃4:17 Vand an accompanying current)+1=9+1=Z0≃0:0833 A are launched at the source end of the line,as indicated by point A which is the intersection of the line9+1=Z0)+1(i.e., line OA) with thesource-end line, defined by 5=10)s+9s. Next, we draw a straight line (line AB) with slope[Z0]1=[50]1passing through point A and find its intersection with the current-voltagecharacteristics of the nonlinear load which is denoted as point B, as shown. The coordinates ofpoint B which correspond to the values of9Land)Latt=t+d=2+ns are approximately givenby9L≃1:072 V and)L≃0:145 A respectively. Using the value of9L, the amplitudes of thereflected voltage and current can be determined as91=9L9+1≃1:0724:167=3:095 Vand)1=91=Z0≃+0:0619 A respectively. We then draw a straight line (line BC) with slope+[Z0]1and passing through point B and find its intersection with the source-end line indicatedas point C, as shown in Figure A.1.This process continues on until steady-state is reached atapproximately point G or att≃8td=16 ns. Note that at steady-state, the source- and load-endvoltages and currents become9s=9L≃2:50 V and)s=)L≃0:25 A which correspond to theintersection point of the source- and load-end voltage-current characteristics as indicated in FigureA.1.
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 5 preview image2Addendum A/ Transmission Lines: Advanced TopicsFigure A.1Figure for Problem A.1.Application of the graphical Bergeron tech-nique to Problem A.1.The points A,B,C,D,E,F, and G correspond respectively to timestd;2td;3td;4td;5td;6td, and 7td.A.2Nonlinear source.(a) We start by plotting the current-voltage characteristics of the source and load ends of the 50Ωline given by)s=8(59s)(59s)2and)L=R1L9L≃0Note that the current-voltage characteristics of the load end of the line is approximately thehorizontal voltage axis sinceRL≃10 MΩ≃ 1. Att=0, an incident voltage of amplitude9+1≃0:798 V and its current counterpart of amplitude)+1=Z019+1≃16:0 mA (which arethe coordinates of point A, the intersection point between line OA and the nonlinear source-endcurrent-voltage characteristics as shown) are launched at the source end of the 50Ωline. Next, wedraw a straight line (line AB) with slope[Z0]1=[50]1passing through point A and find itsintersection with the load-end current-voltage characteristics which is denoted as point B on thehorizontal voltage axis as shown. The coordinates of point B ()B=0 mA and9B≃1:60 V)correspond to the values of)Land9Latt=2+ns from which the amplitudes of the reflectedcurrent and voltage from the load end can be determined as)1=)L)+1=016=16 mA
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 6 preview image3and91=9L9+1≃1:60:798≃0:798 V respectively. We then draw another straight line(line BC) with slope+[Z0]1passing through point B and find its intersection with the nonlinearsource-end current-voltage characteristics indicated as point C as shown in Figure A.2. Using thesecoordinates (9C≃2:31 V;)C≃14:3 mA), the amplitudes of the current and voltage reflectedfrom the source end can be determined as)+2=)C)+1)1≃14:316+16=14:3 mAand9+2=9C9+191≃2:3120:798≃0:714 V respectively. This process continueson. Using the voltage coordinates of points B, D, F, etc., the load-end voltage can be sketched withrespect to time as shown.(b) It can be approximately observed from the Bergeron diagram in Figure A.2 that the steady statecondition is reached int≃12td=24 ns, or at point M.Figure A.2Figure for Problem A.2.Application of the graphical Bergeron techniqueand9Lversust.A.3Effects of source risetime.The one-way time delay along the trace is given bytd= (80 ps-cm1)(25 cm) =2 ns. Att=0,an incident voltage of amplitude
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 7 preview image4Addendum A/ Transmission Lines: Advanced Topics9+1(t) =8><>:Z0V0(Rs+Z0)trt=(4tr)tttrZ0V0Rs+Z0=4 Vttris launched from the source end of the trace. This voltage reaches the short-circuited end of thetrace att=td, when a voltage of amplitude91(t) =8><>:L9+1(ttd) =(4tr)(ttd)td<ttd+tr4 Vttd+tr(wheretd=2 ns) reflects back towards the source sinceL=1. This reflected voltage arrivesat the source end of the line att=2tdand a new voltage of amplitude9+2(t) =8><>:s91(ttd) =(2:4tr)(t2td)2td<t2td+tr2:4 Vt2td+tris launched back towards the load sinces=12:55012:5+50=0:6Att=3td, a new reflected voltage92(t)of amplitude92(t) =8><>:L9+2(ttd) =(2:4tr)(t3td)3td<t3td+tr2:4 Vt3td+tris launched towards the source. This process continues indefinitely, with the voltages and currentsalong the trace gradually reaching their steady-state values. The source-end voltage9s(t)can bewritten in terms of the above voltages as9s(t) =(4tr)[tu(t)(ttr)u(ttr)](A.1)(4tr)[(t2td)u(t2td)(t2tdtr)u(t2tdtr)](A.2)+(2:4tr)[(t2td)u(t2td)(t2tdtr)u(t2tdtr)](A.3)(2:4tr)[(t4td)u(t4td)(t4tdtr)u(t4tdtr)](A.4)+(1:44tr)[(t4td)u(t4td)(t4tdtr)u(t4tdtr)](A.5)+: : :(A.6)
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 8 preview image5Figure A.3Figure for Problem A.3.9sversustfortr=1 ns andtr=0:25 ns.(a) Fortr=1 ns, the source-end voltage9s(t)given by the above expression is sketched as afunction oftin Figure A.3.(b) Fortr=250 ps, the source-end voltage9s(t)is sketched in Figure A.3. We see that all transi-tions (e.g., those starting at 0, 4, or 8 seconds) are sharper fortr=250 ps.A.4Effects of source risetime.The phase velocity along the line can be found asvp= [Z0C]1= [(50Ω)(1 pF-cm1)]1=20 cm-ns1. Att=0, a voltage of amplitude9+1(t) =8><>:Z0V0(Rs+Z0)trt=(2tr)t0<ttrZ0V0Rs+Z0=2 Vttris applied from the source end of the line with source rise timetr=1 ns. This voltage reaches theload end of the linet=td, when a voltage of amplitude91(t) =8><>:L9+1(ttd) =(2tr)(ttd)td<ttd+tr2 Vttd+trreflects back towards the source sinceL≃+1. This voltage arrives at the source end att=2tdwhen a new voltage of amplitude9+2(t) =8><>:s91(ttd) =(23tr)(t2td)2td<t2td+tr23 Vt2td+tris launched back towards the load since the source reflection coefficient iss=255025+50=13Att=3td, a new voltage of amplitude
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 9 preview image6Addendum A/ Transmission Lines: Advanced Topics92(t) =8><>:(23tr)(t3td)3td<t3td+tr23 Vt2td+tris launched back towards the source. This process continues on indefinitely. The source- and theload-end voltage9s(t)and9L(t)can be written in terms of the above voltages as9s(t) =(2tr)[tu(t)(ttr)u(ttr)](A.7)+(2tr)[(t2td)u(t2td)(t2tdtr)u(t2tdtr)](A.8)(23tr)[(t2td)u(t2td)(t2tdtr)u(t2tdtr)](A.9)(23tr)[(t4td)u(t4td)(t4tdtr)u(t4tdtr)](A.10)+(29tr)[(t4td)u(t4td)(t4tdtr)u(t4tdtr)](A.11)+: : :(A.12)and9L(t) =(2tr)[(ttd)u(ttd)(ttdtr)u(ttdtr)](A.13)+(2tr)[(ttd)u(ttd)(ttdtr)u(ttdtr)](A.14)(23tr)[(t3td)u(t3td)(t3tdtr)u(t3tdtr)](A.15)(23tr)[(t3td)u(t3td)(t3tdtr)u(t3tdtr)](A.16)+(29tr)[(t5td)u(t5td)(t5tdtr)u(t5tdtr)](A.17)+: : :(A.18)(a) Forl=5 cm, the one-way time delay istd=l=vp= (5 cm)=(20 cm-ns1) =0:25 ns.Substitutingtd=0:25 ns andtr=1 ns in the above expressions yield the source- and the load-endvoltages9s(t)and9L(t)which are both sketched as a function oftin Figure A.4.(b) Forl=50 cm, the one-way time delay istd=2:5 ns. Again substitutingtd=2:5 ns andtr=1 ns in the above expressions yield the source-end and load-end voltages9s(t)and9L(t)which are both sketched as a function oftin Figure A.4. We note that, as expected, the steady-stateis achieved much quicker fortd=0:25, and with much less ringing.
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 10 preview image7Figure A.4Figure for Problem A.4.9s(t)and9rmL(t)versustfortd=0:25 ns andtd=2:5 ns.A.5Half-wave coaxial line resonator.(a) Using Table 2.2 of the text witha=1 mm,b=4 mm, andf=3 GHz, the line parameters ofthe air-filled coaxial resonator can be evaluated asL=0:2ln4≃0:277H-m1C=55:6ln4≃40:1 pF-m1R=4:151085103p310914106≃2:84Ω-m1andG=0. Since!L≃5226≫2:84≃R, low-loss approximation is valid and the real andimaginary parts of the propagation constantcan be calculated using (3.73) as≃R2√CL≃0:0171 np-m1≃!pLC≃62:9 rad-m1Using these values, the quality factor follows asQair≃2≃1839
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 11 preview image8Addendum A/ Transmission Lines: Advanced Topics(b) For teflon-filled coaxial line, the line parametersCandGhave new values given byCteflon)c2v2pCair≃(3020:7)2(40:1)≃84:2 pF-m1G≃3:3104lnba≃2:38104S-m1Again low-loss approximation is valid since!L≫Rand!C≫G. Using (3.73), we find≃0:0632 and≃91:1 from which theQfactor follows asQteflon≃1442.(c) Using the line parameters in part (a) along with (3.72), the characteristic impedance of the air-filled coaxial resonator can be calculated asZ0≃83:14j0:0226≃83:1Ω. From Section A.2.2,the element values of the equivalent series RLC circuit can be computed asReq=Z02≃0:0710ΩLeq=Z02!≃6:93 nHCeq=2Z0!≃0:406 pFrespectively.
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 12 preview imageSolutions to Addendum CCylindrical WaveguidesC.1Frequency range of the dominant mode.(a) The cutoff frequency of the dominant TE10mode in an air-filled hollow WR-10 rectangularwaveguide witha=0:254 cm andb=0:127 cm is given byfc10=c2a≃310102(0:254)≃59:1 GHz(b) Sincea=2b, the next higher-order modes are TE11, TE20, and TM11. The cutoff frequency ofthese three modes isfc11=fc20=2fc10≃118 GHz(c) Therefore, the frequency band over which only the dominant TE10mode can propagate isfc20fc10≃59:1 GHz.C.2AM waves through a tunnel?The minimum tunnel dimensionaminneeded for the dominant TE10mode to propagate at 1 MHzcan be calculated asfc10≃31082amin=106!amin=150 m!As long asa>amin=150 m, TE10will propagate regardless of the value of the other dimensionb. In practice, since there is no tunnel which has such a large dimension, AM transmission througha tunnel is not feasible.
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 13 preview image10Addendum C/ Cylindrical WaveguidesC.3Propagation experiment in a road tunnel.(a) For the rectangular railway tunnel under consideration (a=17 m andb=4:9 m), the cutofffrequencies of the TE10and TE01modes are given byfc10=c2a≃31082(17)≃8:82 MHzfc01=c2b≃31082(4:9)≃30:6 MHz(b) Using (C.17), the lowest-order modes that do not propagate at the 100 MHz FM frequency areTE92and TM92since the cutoff frequency of these two modes is given byfc92≃31082√(24:9)2+(917)2≃100:3 MHzC.4Radio-wave propagation in railway tunnels.For tunnel 1 (a=8 m andb=4 m), the total number of propagating modes at 100 MHz is 26.These modes are TE10, TE01, (TE and TM)02, (TE and TM)11, (TE and TM)12, (TE and TM)20,(TE and TM)21, (TE and TM)22, (TE and TM)30, (TE and TM)31, (TE and TM)32, (TE and TM)40,(TE and TM)41, and (TE and TM)50. For tunnel 2 (a=6 m andb=4 m), the total number ofpropagating modes at 100 MHz is 18. Unlike tunnel 1, (TE and TM)32, (TE and TM)40, (TE andTM)41, and (TE and TM)50modes do not propagate.C.5Rectangular waveguide modes.(a) For the WR-137 rectangular air waveguide (a=3:484 cm,b=1:58 cm), the cutoff frequenciesof the four lowest-order modes TE10, TE20(and TM20), TE01, and TE11(and TM11) are respectivelycalculated from (C.17) asfc10≃4:31 GHz,fc20≃8:61 GHzfc01≃9:49 GHz, andfc11≃10:4GHz.(b) For the dominant TE10mode, the phase velocityvpand the guide wavelengthat the two endfrequencies (i.e., 5.85 GHz and 8.20 GHz) of the specified frequency range can be calculated using(C.20) and (C.21) asvp10jf=5:85 GHz≃3108√1(4:315:85)2≃4:43108m-s1vp10jf=8:20 GHz≃3108√1(4:318:20)2≃3:52108m-s110f=5:85 GHz≃3108=(5:85109)√1(4:315:85)2≃7:57 cm10f=8:20 GHz≃3108=(8:20109)√1(4:318:20)2≃4:30 cm
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 14 preview image11C.6Dominant mode in a rectangular waveguide.(a) The cutoff frequency of the dominant TE10mode in the WR-137 rectangular air waveguide(a=3:484 cm,b=1:58 cm) can be calculated from (C.17) asfc10=c2a≃310102(3:484)≃4:31 GHzUsing this value, the propagation constant10and wave impedanceZTE10at 3.5 GHz can be foundusing (C.16) and (C.28) as10= 10=√(fc10f)21≃2(3:5109)3108√(4:313:5)21≃52:5 np-m1ZTE10=air√1(fc10=f)2≃377√1(4:31=3:5)2≃j526:3ΩNote that the dominant TE10mode is not propagating at 3.5 GHz as expected sincefc10≃4:31 GHz>f=3:5 GHz.(b) Repeating the same calculations as in part (a) at 7 GHz, we have10=j√1(fc10f)2≃j2(7109)3108√1(4:317)2≃j115:6 rad-m1ZTE10≃377√1(4:31=7)2≃478ΩNote that the dominant TE10mode is a propagating mode at 7 GHz since4:31 GHz<7 GHz.C.7Waveguide wavelength.(a) Using (C.21), the guide wavelength of the dominant TE10mode propagating in a rectangularair waveguide witha=2b=7:214 cm can be written as10=c=f√1(fc10=f)2≃30√[f(GHz)]2(2:08)2cmwherec≃31010cm-s1andfc10=c=(2a)≃2:08 GHz are used. Note that the expressionfound for10above is only valid forf>2:08 GHz (i.e., propagating case). This expression (i.e.,10) is plotted as a function of frequencyfas shown in Figure C.1.(b) If the same waveguide is filled with water (assumeϵr≃81), then,fc10≃2:08=p81≃0:231GHz. Therefore, the guide wavelength for this case is given by10≃30=p81√[f(GHz)]2(0:231)2cm
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 15 preview image12Addendum C/ Cylindrical Waveguides2468510152025f(GHz)λ(cm)TE10TE200.20.40.60.8510152025λ(cm)f(GHz)TE10Air-filledwaveguideWater-filledwaveguideFigure C.1Figure for Problem C.7.Right panel:10and20versusffor air-filled waveg-uide. Left panel:10versusffor water-filled waveguide (i.e.,ϵr≃81).which is valid forf>0:231 GHz. This expression is plotted in Figure C.1 as a function of thefrequencyf.(c) For the propagating TE20mode in the air waveguide of part (a), the20expression is the same asthe10expression except 2.08 should be replaced by 4.16 (i.e.,fc20=2fc10). This expression is validforf>4:16 GHz and is plotted as a function offin Figure C.1. For the propagating TE20modein the water-filled waveguide of part (b),20expression is the same as the10expression foundin part (b) except that 0.231 should be replaced by 0.462. This expression is valid forf>0:462GHz and is plotted in Figure C.1.C.8Rectangular waveguide modes.(a) For the WR-42 rectangular air waveguide (a=1:067 cm,b=0:432 cm), the cutoff frequenciesof the TE10, TE01, TE11, TE20, and TE02modes are given byfc10≃310102(1:067)≃14:06 GHz<40 GHzfc01≃310102(0:432)≃34:72 GHz<40 GHzfc11≃310102√(1:067)2+ (0:432)2≃37:46 GHz<40 GHzfc20≃31010(1:607)≃28:12 GHz<40 GHzfc02≃31010(0:432)≃69:44 GHz>40 GHzNote that all these modes except the TE02mode are propagating at 40 GHz.The propagationconstants and the wave impedances of these modes can be calculated using (C.18) and (C.19) as10≃j2(41010)3108√1(14:0640)2≃j784:3 rad-m1
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Solution Manual for Engineering Electromagnetics and Waves, 2nd Edition - Page 16 preview image13ZTE10≃377√1(14:0640)2≃402:7Ω01≃j2(41010)3108√1(34:7240)2≃j415:9 rad-m1ZTE01≃377√1(34:7240)2≃759:4Ω11≃j2(41010)3108√1(37:4640)2≃j293:8 rad-m1ZTE11≃377√1(37:4640)2≃1075Ω20≃j2(41010)3108√1(28:140)2≃j595:9 rad-m1ZTE20≃377√1(28:140)2≃530Ω02≃2(41010)3108√(69:4440)21≃1189 np-m1ZTE02≃377√1(69:4440)2≃j265:65Ω(b) For the TE10mode, we have from (C.20) and (C.21)vp10≃3108√1(14:0640)2≃3:20108m-s110≃3101041010√1(14:0640)2≃0:801 cmSimilarly, for the TE01mode, we havevp01≃3108√1(34:7240)2≃6:04108m-s101≃3101041010√1(34:7240)2≃1:51 cmIn a similar fashion, for the TE11mode, we obtainvp11≃8:56108m-s1,11≃2:14 cm, andfor the TE20mode, we obtainvp20≃4:22108m-s1, and11≃1:054 cm respectively.
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