1、CHAPTER 1111.1. Show that Exs= Aejk0z+is a solution to the vector Helmholtz equation, Sec. 11.1, Eq. (30),for k0= 00and any and A: We taked2dz2Aejk0z+= (jk0)2Aejk0z+= k20Exs11.2. A 100-MHz uniform plane wave propagates in a lossless medium for which r= 5 and r= 1.Find:a) vp: vp= c/r= 3108/5 = 1.3410
2、8m/s.b) : = /vp= (2108)/(1.34108) = 4.69 m1.c) : = 2/ = 1.34 m.d) Es: Assume real amplitude E0, forward z travel, and x polarization, and writeEs= E0exp(jz)ax= E0exp(j4.69z)axV/m.e) Hs: First, the intrinsic impedance of the medium is = 0/r= 377/5 = 169 .Then Hs= (E0/)exp(jz)ay= (E0/169)exp(j4.69z)ay
3、A/m.f) = (1/2)ReEsHs = (E20/337)azW/m212.3. An H field in free space is given as H(x,t) = 10cos(108tx)ayA/m. Finda) : Since we have a uniform plane wave, = /c, where we identify = 108sec1. Thus = 108/(3108) = 0.33 rad/m.b) : We know = 2/ = 18.9 m.c) E(x,t) at P(0.1,0.2,0.3) at t = 1 ns: Use E(x,t) =
4、 0H(x,t) = (377)(10)cos(108tx) = 3.77103cos(108tx). The vector direction of E will be az, since we requirethat S = EH, where S is x-directed. At the given point, the relevant coordinate isx = 0.1. Using this, along with t = 109sec, we finally obtainE(x,t) = 3.77103cos(108)(109)(0.33)(0.1)az= 3.77103
5、cos(6.7102)az= 3.76103azV/m11.4. Small antennas have low eciencies (as will be seen in Chapter 14) and the eciency increaseswith size up to the point at which a critical dimension of the antenna is an appreciable fractionof a wavelength, say /8.a) An antenna is that is 12cm long is operated in air a
6、t 1 MHz. What fraction of a wavelengthlong is it? The free space wavelength will beair=cf=3.0108m/s106s1= 300 m, so that the fraction =1.2300= 4.0103b) The same antenna is embedded in a ferrite material for which r= 20 and r= 2,000.What fraction of a wavelength is it now?ferrite=airrr=300p(20)(2000)
7、= 1.5m fraction =1.21.5= 0.822111.5. A 150-MHz uniform plane wave in free space is described by Hs= (4+j10)(2ax+jay)ejzA/m.a) Find numerical values for , , and : First, = 2150106= 3108sec1. Second,for a uniform plane wave in free space, = 2c/ = c/f = (3108)/(1.5108) = 2m.Third, = 2/ = rad/m.b) Find
8、H(z,t) at t = 1.5 ns, z = 20 cm: UseH(z,t) = ReHsejt = Re(4 +j10)(2ax+jay)(cos(tz) +jsin(tz)= 8cos(tz)20sin(tz)ax10cos(tz) + 4sin(tz)ay. Now at the given position and time, tz = (3108)(1.5109)(0.20) = /4.And cos(/4) = sin(/4) = 1/2. So finally,H(z = 20cm,t = 1.5ns) = 12(12ax+ 14ay) = 8.5ax9.9ayA/mc)
9、 What is |E|max? Have |E|max= 0|H|max, where|H|max=pHsHs= 4(4+j10)(4j10)+ (j)(j)(4+j10)(4j10)1/2= 24.1A/mThen |E|max= 377(24.1) = 9.08 kV/m.11.6. A uniform plane wave has electric field Es= (Ey0ayEz0az)exejxV/m. The intrinsicimpedance of the medium is given as = |ej, where is a constant phase.a) Des
10、cribe the wave polarization and state the direction of propagation: The wave is linearlypolarized in the y-z plane, and propagates in the forward x direction (from the ejxfactor).b) Find Hs: Each component of Es, when crossed into its companion component of Hs,must give a vector in the positive-x di
11、rection of travel. Using this rule, we findHs=Eyaz+Ezay=Ey0|az+Ez0|ayexejejxA/mc) Find E(x,t) and H(x,t): E(x,t) = ReEsejt= Ey0ayEz0azexcos(tx)H(x,t) = ReHsejt= Ey0az+Ez0ayexcos(tx)where all amplitudes are assumed real.d) Find in W/m2:=12ReEsHs =12E2y0+E2z0e2xcosaxW/m2e) Find the time-average power
12、in watts that is intercepted by an antenna of rectangularcross-section, having width w and height h, suspended parallel to the yz plane, and at adistance d from the wave source. This will beP =Z ZplatedS = | |x=darea =12(wh)E2y0+E2z0e2dcos W22211.7. The phasor magnetic field intensity for a 400-MHz
13、uniform plane wave propagating in acertain lossless material is (2ayj5az)ej25xA/m. Knowing that the maximum amplitudeof E is 1500 V/m, find , , , vp, r, r, and H(x,y,z,t): First, from the phasor expression,we identify = 25 m1from the argument of the exponential function. Next, we evaluateH0= |H| =HH
14、=22+ 52=29. Then = E0/H0= 1500/29 = 278.5 . Then = 2/ = 2/25 = .25 m = 25 cm. Next,vp=240010625= 1.01108m/sNow we note that = 278.5 = 377rrrrr= 0.546Andvp= 1.01108=crr rr= 8.79We solve the above two equations simultaneously to find r= 4.01 and r= 2.19. Finally,H(x,y,z,t) = Re(2ayj5az)ej25xejt= 2cos(
15、2400106t25x)ay+ 5sin(2400106t25x)az= 2cos(8108t25x)ay+ 5sin(8108t25x)azA/m11.8. An electric field in free space is given in spherical coordinates as Es(r) = E0(r)ejkraV/m.a) find Hs(r) assuming uniform plane wave behavior: Knowing that the cross product of Eswith the complex conjugate of the phasor
16、Hsfield must give a vector in the direction ofpropagation, we obtain,Hs(r) =E0(r)0ejkraA/mb) Find : This will be=12ReEsHs =E20(r)20arW/m2c) Express the average outward power in watts through a closed spherical shell of radius r,centered at the origin: The power will be (in this case) just the produc
17、t of the powerdensity magnitude in part b with the sphere area, orP = 4r2E20(r)20Wwhere E0(r) is assumed real.d) Establish the required functional form of E0(r) that will enable the power flow in part cto be independent of radius: Evidently this condition is met when E0(r) 1/r22311.9. A certain loss
18、less materialhasr= 4 and r= 9. A 10-MHzuniformplane wave is propagatingin the aydirection with Ex0= 400 V/m and Ey0= Ez0= 0 at P(0.6,0.6,0.6) at t = 60 ns.a) Find , , vp, and : For a uniform plane wave, = =crr=21073108p(4)(9) = 0.4 rad/mThen = (2)/ = (2)/(0.4) = 5m. Next,vp=21074101= 5107m/sFinally,
19、 =r= 0rrr= 377r49= 251 b) Find E(t) (at P): We are given the amplitude at t = 60 ns and at y = 0.6 m. Let themaximum amplitude be Emax, so that in general, Ex= Emaxcos(ty). At the givenposition and time,Ex= 400 = Emaxcos(2107)(60109)(4101)(0.6) = Emaxcos(0.96)= 0.99EmaxSo Emax= (400)/(0.99) = 403V/m
20、. Thus at P, E(t) = 403cos(2107t) V/m.c) Find H(t): First, we note that if E at a given instant points in the negative x direction,while the wave propagates in the forward y direction, then H at that same position andtime must point in the positive z direction. Since we have a lossless homogeneous m
21、edium, is real, and we are allowed to write H(t) = E(t)/, where is treated as negative andreal. ThusH(t) = Hz(t) =Ex(t)=403251cos(2107t) = 1.61cos(2107t) A/m11.10. In a medium characterized by intrinsic impedance = |ej, a linearly-polarized plane wavepropagates, with magnetic field given as Hs= (H0y
22、ay+H0zaz)exejx. Find:a) Es: Requiring orthogonal components of Esfor each component of Hs, we findEs= |H0zayH0yaz exejxejb) E(x,t) = ReEsejt = |H0zayH0yazexcos(tx+ ).c) H(x,t) = ReHsejt = H0yay+H0zazexcos(tx).d) =12ReEsHs =12|H20y+H20ze2xcosaxW/m222411.11. A 2-GHz uniform plane wave has an amplitude
23、 of Ey0= 1.4 kV/m at (0,0,0,t = 0) and ispropagating in the azdirection in a medium where 00= 1.61011F/m, 0= 3.01011F/m, and = 2.5H/m. Find:a) Eyat P(0,0,1.8cm) at 0.2 ns: To begin, we have the ratio, 00/0= 1.6/3.0 = 0.533. So = r02s1 +000211/2= (22109)r(2.5106)(3.01011)2hp1 + (.533)21i1/2= 28.1Np/m
24、Then = r02s1 +0002+ 11/2= 112rad/mThus in general,Ey(z,t) = 1.4e28.1zcos(4109t112z) kV/mEvaluating this at t = 0.2 ns and z = 1.8 cm, findEy(1.8cm,0.2ns) = 0.74 kV/mb) Hxat P at 0.2 ns: We use the phasor relation, Hxs= Eys/ where =r01p1j(00/0)=r2.51063.010111p1j(.533)= 263+j65.7 = 271614So nowHxs= E
25、ys= (1.4103)e28.1zej112z271ej14= 5.16e28.1zej112zej14A/mThenHx(z,t) = 5.16e28.1zcos(4109t112z14)This, when evaluated at t = 0.2 ns and z = 1.8 cm, yieldsHx(1.8cm,0.2ns) = 3.0A/m22511.12. Describe how the attenuation coecient of a liquid medium, assumed to be a good conductor,could be determined thro
26、ugh measurement of wavelength in the liquid at a known frequency.What restrictions apply? Could this method be used to find the conductivity as well? In agood conductor, we may use the approximation:.= .=r2where =2Therefore, in the good conductor approximation, .= 2/. From the above formula, wecould
27、 also find.=42fwhich would work provided that again, we are certain that we have a good conductor, andthat the permeability is known.11.13. Let jk = 0.2+j1.5m1and = 450+j60 for a uniform plane wave propagating in the azdirection. If = 300 Mrad/s, find , 0, and 00: We begin with =r01p1j(00/0)= 450+j6
28、0andjk = jp0p1j(00/0) = 0.2 +j1.5Then=01p1 + (00/0)2= (450+j60)(450j60) = 2.06105(1)and(jk)(jk)= 20p1 + (00/0)2= (0.2 +j1.5)(0.2j1.5) = 2.29 (2)Taking the ratio of (2) to (1),(jk)(jk)= 2(0)21 + (00/0)2=2.292.06105= 1.11105Then with = 3108,(0)2=1.11105(3108)2(1 + (00/0)2)=1.231022(1 + (00/0)2)(3)Now,
29、 we use Eqs. (35) and (36). Squaring these and taking their ratio gives22=p1 + (00/0)2p1 + (00/0)2=(0.2)2(1.5)2We solve this to find 00/0= 0.271. Substituting this result into (3) gives 0= 1.071011F/m. Since 00/0= 0.271, we then find 00= 2.901012F/m. Finally, using these results ineither (1) or (2)
30、we find = 2.28106H/m. Summary: = 2.28106H/m,0= 1.071011F/m, and 00= 2.901012F/m.22611.14. A certain nonmagnetic material has the material constants 0r= 2 and 00/0= 4104at = 1.5 Grad/s. Find the distance a uniform plane wave can propagate through the materialbefore:a) it is attenuated by 1 Np: First,
31、 00= (4 104)(2)(8.854 1012) = 7.1 1015F/m.Then, since 00/0= (1/2)ReEsJs.a) Show that in a conducting medium, through which a uniform plane wave of amplitude E0propagates in the forward z direction, = (/2)|E0|2e2z: Begin with the phasorexpression for the electric field, assuming complex amplitude E0,
32、 and x-polarization:Es= E0ezejzaxV/m2ThenJs= Es= E0ezejzaxA/m2So that= (1/2)ReE0ezejzaxE0eze+jzax= (/2)|E0|2e2zb) Confirm this result for the special case of a good conductor by using the left hand side ofEq. (70), and consider a very small volume. In a good conductor, the intrinsic impedanceis, fro
33、m Eq. (85), c= (1 + j)/(), where the skin depth, = 1/. The magnetic fieldphasor is thenHs=Escay=(1 +j)E0ezejzayA/mThe time-average Poynting vector is then=12ReEsHs =4|E0|2e2zazW/m2Now, consider a rectangular volume of side lengths, x, y, and z, all of which arevery small. As the wave passes through
34、this volume in the forward z direction, the powerdissipated will be the dierence between the power at entry (at z = 0), and the powerthat exits the volume (at z = z). With small z, we may approximate e2z.= 12z,and the dissipated power in the volume becomesPd= PinPout=h4|E0|2ixyh4|E0|2(12z)ixy =2|E0|
35、2(xyz)This is just the result of part a, evaluated at z = 0 and multiplied by the volume. Therelation becomes exact as z 0, in which case (/2)|E0|2.It is also possible to show the relation by using Eq. (69) (which involves taking thedivergence of ), or by removing the restriction of a small volume a
36、nd evaluatingthe integrals in Eq. (70) without approximations. Either method is straightforward.22811.17. Let = 250 + j30 and jk = 0.2 + j2m1for a uniform plane wave propagating in the azdirection in a dielectric having some finite conductivity. If |Es| = 400 V/m at z = 0, find:a) at z = 0 and z = 6
37、0 cm: Assume x-polarization for the electric field. Then=12ReEsHs =12Re400ezejzax400ezejzay=12(400)2e2zRe1az= 8.0104e2(0.2)zRe1250j30az= 315e2(0.2)zazW/m2Evaluating at z = 0, obtain (z = 0) = 315azW/m2,and at z = 60 cm, Pz,av(z = 0.6) = 315e2(0.2)(0.6)az= 248azW/m2.b) the average ohmic power dissipa
38、tion in watts per cubic meter at z = 60 cm: At this pointa flaw becomes evident in the problem statement, since solving this part in two dierentways gives results that are not the same. I will demonstrate: In the first method, weuse Poyntings theorem in point form Eq.(69), which we modify for the ca
39、se of time-independent fields to read: =, where the right hand side is theaverage power dissipation per volume. Note that the additional right-hand-side terms inPoyntings theorem that describe changes in energy stored in the fields will both be zeroin steady state. We apply our equation to the resul
40、t of part a:= = ddz315e2(0.2)z= (0.4)(315)e2(0.2)z= 126e0.4zW/m3At z = 60 cm, this becomes = 99.1 W/m3. In the second method, we solve forthe conductivity and evaluate = . We use jk = j0p1j(00/0)and =r01p1j(00/0)We take the ratio,jk= j01j000= j0+ 00Identifying = 00, we find = Rejk= Re0.2 +j2250+j30=
41、 1.74103S/mNow we find the dissipated power per volume: = 1.7410312400e0.2z2At z = 60 cm, this evaluates as 109 W/m3. One can show that consistency between thetwo methods requires thatRe1=2This relation does not hold using the numbers as given in the problem statement and thevalue of found above. No
42、te that in Problem 11.13, where all values are worked out, therelation does hold and consistent results are obtained using both methods.22911.18. Given, a 100MHz uniform plane wave in a medium known to be a good dielectric. The phasorelectric field is Es= 4e0.5zej20zaxV/m. Not stated in the problem
43、is the permeability,which we take to be 0. Determine:a) 0: As a first step, it is useful to see just how much of a good dielectric we have. We usethe good dielectric approximations, Eqs. (60a) and (60b), with = 00. Using these, wetake the ratio, /, to find=200.5=01 + (1/8)(00/0)2(00/2)p/0= 2000+1400
44、0This becomes the quadratic equation:0002160000+ 8 = 0The solution to the quadratic is (00/0) = 0.05, which means that we can neglect thesecond term in Eq. (60b), so that .= 0= (/c)p0r. With the given frequency of100 MHz, and with = 0, we findp0r= 20(3/2) = 9.55, so that 0r= 91.3, and finally0= 0r0=
45、 8.11010F/m.b) 00: Using Eq. (60a), the set up is = 0.5 =002r0 00=2(0.5)2108s0=1082(377)91.3 = 4.01011F/mc) : Using Eq. (62b), we find.=r01 +j12000=37791.3(1 +j.025) = (39.5 +j0.99) ohmsd) Hs: This will be a y-directed field, and will beHs=Esay=4(39.5 +j0.99)e0.5zej20zay= 0.101e0.5zej20zej0.025ayA/m
46、e) : Using the given field and the result of part d, obtain=12ReEsHs =(0.101)(4)2e2(0.5)zcos(0.025)az= 0.202ezazW/m2f) the power in watts that is incident on a rectangular surface measuring 20m x 30m atz = 10m: At 10m, the power density is = 0.202e10= 9.2 106W/m2. Theincident power on the given area is then P = 9.2106(20)(30) = 5.5 mW.23011.19. Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. The region betweenthe cylinders is filled with a perfect dielectric for which = 109/4 F/m and r= 1. If E i
