Fundamentals of Applied Electromagnetics

1 Fundamentals of Applied Electromagnetics 6ebyFawwaz T. Ulaby, Eric Michielssen, and Umberto RavaioliExercise SolutionsFawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallChaptersChapter 1 Introduction: Waves and PhasorsChapter 2 Transmission LinesChapter 3 Vector AnalysisChapter 4 ElectrostaticsChapter 5 MagnetostaticsChapter 6 Maxwell s Equations for Time-Varying FieldsChapter 7 Plane-Wave PropagationChapter 8 Wave Reflection and TransmissionChapter 9 Radiation and AntennasChapter 10 Satellite Communication Systems and Radar SensorsFawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallChapter 1 Exercise SolutionsExercise T.

2 Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise the red wave shown in Fig. What is the wave s (a) amplitude, (b) wavelength, and (c)frequency, given that its phase velocity is 6 m/s?Solution:(a)A=6 V.(b) =4 cm.(c)f=up =64 10 2=150 T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise wave shown in red in Fig. is given by =5 cos 2 t/8. Of the following four equations:(1) =5 cos(2 t/8 /4),(2) =5 cos(2 t/8+ /4),(3) = 5 cos(2 t/8 /4),(4) =5 sin 2 t/8,(a) which equation applies to the green wave? (b) which equation applies to the blue wave?Solution:(a)The green wave has an amplitude of 5 V and a periodT=8 s.

3 Its peak occurs earlier than that of the red wave; hence,its constant phase angle is positive relative to that of the red wave. A full cycle of 8 s corresponds to 2 in phase. The greenwave crosses the time axis 1 s sooner than the red wave. Hence, its phase angle is 0=18 2 = , =5 cos(2 t/T+ 0)=5 cos(2 t/7+ /4),which is given by #2.(b)The blue wave s periodT=8 s. Its phase angle is delayed relative to the red wave by 2 s. Hence, the phase angle isnegative and given by 0= 28 2 = 2,and =5 cos(2 t8 2)=5 sin 2 t/8,which is given by # T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise electric field of a traveling electromagnetic wave is given byE(z,t) =10 cos( 107t+ z/15+ /6)(V/m).

4 Determine (a) the direction of wave propagation, (b) the wave frequencyf, (c) its wavelength , and (d) its phase :(a) z-direction because the signs of the coefficients oftandzare both positive.(b) From the given expression, = 107(rad/s).Hence,f= 2 = 1072 =5 106Hz=5 MHz.(c) From the given expression,2 = =30 m.(d)up=f =5 106 30= 108 T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise the red wave shown in Fig. What is the wave s (a) amplitude (atx=0), (b) wavelength, and(c) attenuation constant?Solution:The wave shown in the figure exhibits a sinusoidal variation inxand its amplitude decreases as a function , it can be described by the general expression =Ae xcos(2 x + 0).

5 From the given coordinates of the first two peaks, we deduce that = , = 5 V and it occurs exactly /2 before the first peak. Hence, the wave amplitude is 5 V, and from 5=5 cos(0+ 0),it follows that 0= .Consequently, =5e xcos(2 + ).In view of the relation cosx= cos(x ), can be expressed as = 5e xcos2 (V).We can describe the amplitude as 5 V for a wave with a constant phase angle of , or as 5 V with a phase angle of cm, (x= ) = 5e cos(2 )=5e .Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallHence,e = ,and = ( )= T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise red wave shown in Fig.

6 Is given by =5 cos 4 x(V). What expression is applicable to (a) theblue wave and (b) the green wave?Solution:Atx=0, all three waves start at their peak value of 5 V. Also, = m for all three waves. Hence, they sharethe general form =Ae xcos2 x =5e xcos 4 x(V).For the red wave, = the blue wave, = the green wave, = T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise electromagnetic wave is propagating in thez-direction in a lossy medium with attenuation constant = Np/m. If the wave s electric-field amplitude is 100 V/m atz=0, how far can the wave travel before its amplitudewill have been reduced to (a) 10 V/m, (b) 1 V/m, (c) 1 V/m?

7 Solution:(a)100e z=10100e m.(b)100e m.(c)100e 6z=ln 10 8 T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise the following complex functions in polar form:z1= (4 j3)2,z2= (4 j3)1 :z1= (4 j3)2=[(42+32)1/2 tan 13/4]2= [5 ]2=25 .z2= (4 j3)1/2=[(42+32)1/2 jtan 13/4]1/2= [5 ]1/2= 5 .Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise that 2j= (1+j).Solution:ej /2=0+jsin( /2) =j 2j= [2ej /2]1/2= 2ej /4= 2(cos /4+jsin /4)= 2(1 2+j1 2)= (1+j).Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise seriesRLcircuit is connected to a voltage source given byvs(t) =150 cos t(V).

8 Find (a) the phasorcurrent Iand (b) the instantaneous currenti(t)forR=400 ,L=3 mH, and =105 :(a) From Example 1 4, I= VsR+j L=150400+j105 3 10 3=150400+j300= (A).(b)i(t) =Re[ Iej t]=Re[ ej105t]= cos(105t )(A).Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise phasor voltage is given by V=j5 V. Findv(t).Solution: V=j5=5ej /2v(t) =Re[ V ej t]=Re[5ej /2ej t]=5 cos( t+ 2)= 5 sin (V).Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallChapter 2 Exercise SolutionsExercise T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise Table 2-1 to compute the line parameters of a two-wire air line whose wires are separated by a distanceof 2 cm, and each is 1 mm in radius.

9 The wires may be treated as perfect conductors with c= .Solution:Two-wire air line : Because medium between wires is air, = 0, = 0and = cm,a=1 mm, c= Rs=[ f c c]1/2=0R =0L = 0 ln (d2a)+ (d2a)2 1 =(4 10 7 )ln (202)+ (202)2 1 =4 10 7ln[10+ 99] = ( H/m).G =0because =0C = 0ln[(d2a)+ (d2a)2 1]= 10 12ln[10+ 99]= (pF/m).Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise the transmission line parameters at 1 MHz for a rigid coaxial air line with an inner conductordiameter of cm and an outer conductor diameter of cm. The conductors are made of copper [see Appendix B for cand cof copper].Solution:Coaxial air line : Because medium between wires is air, = 0, = 0and = cm,b= cm, c= 0, c= 107S/mRs= f c/ c= [ 106 4 10 7/( 107)]1/2= 10 4.

10 R =Rs2 (1a+1b)= 10 42 (13 10 3+16 10 3)= 10 2( /m)L = 02 ln(ba)=4 10 72 ln 2= ( H/m)G =0because =0C =2 ln(b/a)=2 10 12ln 2= (pF/m).Fawwaz T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise that Eq. ( ) is indeed a solution of the wave equation given by Eq. ( ).Solution: V(z) =V+0e z+V 0e zd2 V(z)dz2 2 V(z)?=0d2dz2(V+0e z+V 0e z) 2(V+0e z+V 0e z)?=0 2V+0e z+ 2V 0e z 2V+0e z 2V 0e z= T. Ulaby, Eric Michielssen, and Umberto Ravaioli, Fundamentals of Applied Electromagneticsc 2010 Prentice HallExercise two-wire air line has the following line parameters:R = (m /m),L = ( H/m),G =0, andC = (pF/m). For operation at 5 kHz, determine (a) the attenuation constant , (b) the phase constant , (c) the phasevelocityup, and (d) the characteristic :Given:R = (m /m),G =0,L = ( H/m),C = (pF/m).