CHAPTER 11. d 2 dz 2 Aejk 0z+φ = (jk 0 ) 2 Ae jk 0z+φ = k 2 0 E xs

Transcription

1 CHAPTER.. Show that E xs Ae jk 0z+φ is a solution to the vector Helmholtz equation, Sec.., Eq. (6), for k 0 ω µ 0 ɛ 0 and any φ and A: We take d dz Aejk 0z+φ (jk 0 ) Ae jk 0z+φ k 0 E xs.. Let E(z, t) 00 sin 0.z cos 0 8 ta x cos(0.z + 50 ) sin 0 8 ta y V/m. Find: a) E at P(0,, 0.6) at t 5 ns: Obtain E P (t 5) 00 sin [(0.)(0.6)] cos(.5)a x cos [(0.)(0.6) + 50(π)/360] sin(.5)a y b) E at P at t 0 ns: 9.a x + 64a y V/m E P (t 0) 00 sin [(0.)(0.6)] cos(.0)a x cos [(0.)(0.6) + 50(π)/360] sin(.0)a y 9.96a x + 48a y V/m Thus E P (9.96) + (48) 49 V/m. c) E s at P : E s 00 sin 0.za x j500 cos(0.z + 50 )a y. Thus E sp 00 sin [(0.)(0.6)] a x j500 cos [(0.)(0.6) + π(50)/360] a y 3.9a x j73a y V/m.3. An H field in free space is given as H(x, t) 0 cos(0 8 t βx)a y A/m. Find a) β: Since we have a uniform plane wave, β ω/c, where we identify ω 0 8 sec. Thus β 0 8 /(3 0 8 ) 0.33 rad/m. b) λ: We know λ π/β 8.9m. c) E(x, t) at P(0., 0., 0.3) at t ns: Use E(x,t) η 0 H(x,t) (377)(0) cos(0 8 t βx) cos(0 8 t βx). The vector direction of E will be a z, since we require that S E H, where S is x-directed. At the given point, the relevant coordinate is x 0.. Using this, along with t 0 9 sec, we finally obtain E(x, t) cos[(0 8 )(0 9 ) (0.33)(0.)]a z cos(6.7 0 )a z a z V/m.4. In phasor form, the electric field intensity of a uniform plane wave in free space is expressed as E s (40 j30)e j0z a x V/m. Find: a) ω: From the given expression, we identify β 0 rad/m. Then ω cβ (3 0 8 )(0) rad/s. b) β 0 rad/m from part a. 8

2 .4. (continued) c) f ω/π 956 MHz. d) λ π/β π/ m. e) H s : In free space, we find H s by dividing E s by η 0, and assigning vector components such that E s H s gives the required direction of wave travel: We find H s 40 j30 e j0z a y (0. j0.08)e j0z a y A/m 377 f) H(z, t) at P(6,, 0.07), t 7 ps: [ H(z, t) Re H s e jωt] [ ] 0. cos( t 0z) sin( t 0z) a y Then [ [ ] H(.07,t 7ps) 0. cos ( )(7. 0 ) 0(.07) [ ]] +.08 sin ( )(7. 0 ) 0(.07) a y [0.(0.56) 0.08(0.87)]a y a y A/m.5. A 50-MHz uniform plane wave in free space is described by H s (4 + j0)(a x + ja y )e jβz A/m. a) Find numerical values for ω, λ, and β: First, ω π π 0 8 sec. Second, for a uniform plane wave in free space, λ πc/ω c/f (3 0 8 )/( ) m. Third, β π/λ π rad/m. b) Find H(z, t) at t.5 ns, z 0 cm: Use H(z, t) Re{H s e jωt }Re{(4 + j0)(a x + ja y )(cos(ωt βz) + j sin(ωt βz)} [8 cos(ωt βz) 0 sin(ωt βz)] a x [0 cos(ωt βz) + 4 sin(ωt βz)] a y. Now at the given position and time, ωt βz (3π 0 8 )( ) π(0.0) π/4. And cos(π/4) sin(π/4) /. So finally, H(z 0cm,t.5ns) ( ax + 4a y ) 8.5ax 9.9a y A/m c) What is E max?have E max η 0 H max, where H max H s H s [4(4 + j0)(4 j0) + (j)( j)(4 + j0)(4 j0)]/ 4.A/m Then E max 377(4.) 9.08 kv/m. 83

3 .6. Let µ R ɛ R for the field E(z, t) (5a x 30a y ) cos(ωt 50z) V/m. a) Find ω: ω cβ (3 0 8 )(50) s. b) Determine the displacement current density, J d (z, t): J d (z, t) D ɛ 0 ω(5a x 30a y ) sin(ωt 50z) t ( 3.3a x a y ) sin( t 50z) A/m c) Find the total magnetic flux passing through the rectangle defined by 0 <x<, y 0, 0 <z<, at t 0: In free space, the magnetic field of the uniform plane wave can be easily found using the intrinsic impedance: H(z, t) ( 5 a y + 30 ) a x cos(ωt 50z) A/m η 0 η 0 Then B(z, t) µ 0 H(z, t) (/c)(5a y + 30a x ) cos(ωt 50z) Wb/m, where µ 0 /η 0 µ0 ɛ 0 /c. The flux at t 0isnow 0 0 B a y dx dz 0 5 c cos(50z) dz 5 50(3 0 8 sin(50) 0.44 nwb ).7. The phasor magnetic field intensity for a 400-MHz uniform plane wave propagating in a certain lossless material is (a y j5a z )e j5x A/m. Knowing that the maximum amplitude of E is 500 V/m, find β, η, λ, v p, ɛ R, µ R, and H(x,y,z,t): First, from the phasor expression, we identify β 5 m from the argument of the exponential function. Next, we evaluate H 0 H H H Then η E 0 /H 0 500/ Then λ π/β π/5.5 m 5 cm. Next, v p ω β π m/s Now we note that And µr η µ R ɛ R ɛ R v p c µr ɛ R µ R ɛ R 8.79 We solve the above two equations simultaneously to find ɛ R 4.0 and µ R.9. Finally, H(x,y,z,t) Re {(a y j5a z )e j5x e jωt} cos(π t 5x)a y + 5 sin(π t 5x)a z cos(8π 0 8 t 5x)a y + 5 sin(8π 0 8 t 5x)a z A/m 84

4 .8. Let the fields, E(z, t) 800 cos(0 7 πt βz)a x V/m and H(z, t) 3.8 cos(0 7 πt βz)a y A/m, represent a uniform plane wave propagating at a velocity of m/s in a perfect dielectric. Find: a) β ω/v (0 7 π)/( ) 0.4 m. b) λ π/β π/.4 8.0m. c) η E / H 800/ d) µ R : Have two equations in the two unknowns, µ R and ɛ R : η η 0 µr /ɛ R and β ω µ R ɛ R /c. Eliminate ɛ R to find [ ] βcη [ (.4)(3 0 8 ] )(474) µ R ωη 0 ( π)(377) e) ɛ R µ R (η 0 /η) (.69)(377/474) A certain lossless material has µ R 4 and ɛ R 9. A 0-MHz uniform plane wave is propagating in the a y direction with E x0 400 V/m and E y0 E z0 0atP(0.6, 0.6, 0.6) at t 60 ns. a) Find β, λ, v p, and η: For a uniform plane wave, β ω µɛ ω µr ɛ R c π (4)(9) 0.4π rad/m Then λ (π)/β (π)/(0.4π) 5m. Next, v p ω β π 07 4π m/s Finally, η µ ɛ η µr ɛ R 9 b) Find E(t) (at P ): We are given the amplitude at t 60 ns and at y 0.6 m. Let the maximum amplitude be E max, so that in general, E x E max cos(ωt βy). At the given position and time, E x 400 E max cos[(π 0 7 )( ) (4π 0 )(0.6)] E max cos(0.96π) 0.99E max So E max (400)/( 0.99) 403 V/m. Thus at P, E(t) 403 cos(π 0 7 t) 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 and time must point in the positive z direction. Since we have a lossless homogeneous medium, η is real, and we are allowed to write H(t) E(t)/η, where η is treated as negative and real. Thus H(t) H z (t) E x(t) η cos(π 0 7 t).6 cos(π 0 7 t) A/m 85

5 .0. Given a 0MHz uniform plane wave with H s (6a x ja y )e jz A/m, assume propagation in a lossless medium characterized by ɛ R 5 and an unknown µ R. a) Find λ, v p, µ R, and η: First, β, so λ π/β π m. Next, v p ω/β π π 0 7 m/s. Then, µ R (β c )/(ω ɛ R ) (3 0 8 ) /(4π 0 7 ) (5).4. Finally, η η 0 µr /ɛ R 377.4/5 80. b) Determine E at the origin at t 0ns: We use the relation E η H and note that for positive z propagation, a positive x component of H is coupled to a negative y component of E, and a negative y component of H is coupled to a negative x component of E. We obtain E s η(6a y +ja x )e jz. Then E(z, t) Re { E s e jωt} 6η cos(ωt z)a y + η sin(ωt z)a x 360 sin(ωt z)a x 080 cos(ωt z)a y. With ω 4π 0 7 sec, t 0 8 s, and z 0, E evaluates as E(0, 0ns) 360(0.588)a x 080( 0.809)a y a x + 874a y V/m... A -GHz uniform plane wave has an amplitude of E y0.4 kv/m at (0, 0, 0,t 0) and is propagating in the a z direction in a medium where ɛ.6 0 F/m, ɛ F/m, and µ.5 µh/m. Find: a) E y at P(0, 0,.8cm) at 0. ns: To begin, we have the ratio, ɛ /ɛ.6/ So ( ) / µɛ α ω ɛ + ɛ (π 0 9 (.5 0 ) 6 )(3.0 0 ) [ ] / + (.533) 8.Np/m Then ( ) / µɛ β ω ɛ + ɛ + rad/m Thus in general, E y (z, t).4e 8.z cos(4π 0 9 t z) kv/m Evaluating this at t 0. ns and z.8 cm, find E y (.8cm, 0.ns) 0.74 kv/m b) H x at P at 0. ns: We use the phasor relation, H xs E ys /η where µ η ɛ j(ɛ /ɛ ) j j(.533) So now H xs E ys η (.4 03 )e 8.z e jz 7e j4 5.6e 8.z e jz e j4 A/m Then H x (z, t) 5.6e 8.z cos(4π 0 9 t z 4 ) This, when evaluated at t 0. ns and z.8 cm, yields H x (.8cm, 0.ns) 3.0A/m 4 86

6 .. The plane wave E s 300e jkx a y V/m is propagating in a material for which µ.5 µh/m, ɛ 9 pf/m, and ɛ 7.8 pf/m. If ω 64 Mrad/s, find: a) α: We use the general formula, Eq. (35): ( ) / µɛ α ω ɛ + ɛ ( (.5 0 ) 6 )(9 0 ) b) β: Using (36), we write ( ) µɛ β ω ɛ + ɛ + c) v p ω/β ( )/(.3) m/s. d) λ π/β π/(.3) 0. m. e) η: Using (39): η [ ] / + (.867) 0.6 Np/m /.3 rad/m µ ɛ j(ɛ /ɛ ) j e j.36 j(.867) f) H s : With E s in the positive y direction (at a given time) and propagating in the positive x direction, we would have a positive z component of H s, at the same time. We write (with jk α + jβ): H s E s η a z e j.36 e jkx a z 0.69e αx e jβx e j.36 a z 0.69e.6x e j.3x e j.36 a z A/m g) E(3,, 4, 0ns): The real instantaneous form of E will be { E(x,y,z,t) Re E s e jωt} 300e αx cos(ωt βx)a y Therefore E(3,, 4, 0ns) 300e.6(3) cos[( )(0 8 ).3(3)]a y 03 V/m.3. Let jk 0. + j.5m and η j60 for a uniform plane wave propagating in the a z direction. If ω 300 Mrad/s, find µ, ɛ, and ɛ : We begin with µ η ɛ j60 j(ɛ /ɛ ) and jk jω µɛ j(ɛ /ɛ ) 0. + j.5 87

7 .3. (continued) Then ηη µ ɛ + (ɛ /ɛ ) (450 + j60)(450 j60) () and (jk)(jk) ω µɛ + (ɛ /ɛ ) (0. + j.5)(0. j.5).9 () Taking the ratio of () to (), (jk)(jk) ( ηη ω (ɛ ) + (ɛ /ɛ ) ) Then with ω 3 0 8, (ɛ ). 0 5 (3 0 8 ) ( (ɛ /ɛ ) ) ( + (ɛ /ɛ ) ) (3) Now, we use Eqs. (35) and (36). Squaring these and taking their ratio gives α + (ɛ β /ɛ ) + (ɛ /ɛ ) (0.) (.5) We solve this to find ɛ /ɛ 0.7. Substituting this result into (3) gives ɛ.07 0 F/m. Since ɛ /ɛ 0.7, we then find ɛ.90 0 F/m. Finally, using these results in either () or () we find µ H/m. Summary: µ H/m, ɛ.07 0 F/m, and ɛ.90 0 F/m..4. A certain nonmagnetic material has the material constants ɛ R and ɛ /ɛ at ω.5 Grad/s. Find the distance a uniform plane wave can propagate through the material before: a) it is attenuated by Np: First, ɛ (4 0 4 )()( ) F/m. Then, since ɛ /ɛ <<, we use the approximate form for α, given by Eq. (5) (written in terms of ɛ ): α. ωɛ µ ɛ (.5 09 )( ) Np/m The required distance is now z ( ) 706 m b) the power level is reduced by one-half: The governing relation is e αz / /, or z / ln /α ln /(.4 0 3) 44 m. c) the phase shifts 360 : This distance is defined as one wavelength, where λ π/β (πc)/(ω ɛ R ) [π(3 08 )]/[( ) ] 0.89 m..5. A 0 GHz radar signal may be represented as a uniform plane wave in a sufficiently small region. Calculate the wavelength in centimeters and the attenuation in nepers per meter if the wave is propagating in a non-magnetic material for which a) ɛ R and ɛ R 0: In a non-magnetic material, we would have: µ 0 ɛ 0 ɛ R ( ɛ ) / α ω R + ɛ R 88

8 .5. (continued) and β ω µ 0 ɛ 0 ɛ R ( ɛ ) R + + ɛ R / With the given values of ɛ R and ɛ R, it is clear that β ω µ 0 ɛ 0 ω/c, and so λ π/β πc/ω /0 0 3cm. It is also clear that α 0. b) ɛ R.04 and ɛ R : In this case ɛ R /ɛ R <<, and so β. ω ɛ R /c.3 cm. Thus λ π/β.95 cm. Then α. ωɛ µ ɛ ωɛ R µ0 ɛ 0 ɛ R ω c ɛ R ɛ R π 00 ( ) Np/m c) ɛ R.5 and ɛ R 7.: Using the above formulas, we obtain β π 00.5 (3 0 0 ) and so λ π/β.33 cm. Then α π 00.5 (3 0 8 ) + + ( 7..5 ( 7..5 ) + ) / / 4.7 cm 335 Np/m.6. The power factor of a capacitor is defined as the cosine of the impedance phase angle, and its Q is ωcr, where R is the parallel resistance. Assume an idealized parallel plate capacitor having a dielecric characterized by σ, ɛ, and µ R. Find both the power factor and Q in terms of the loss tangent: First, the impedance will be: ( R Z R + ) jωc ( jωc ) R jrωc + (RωC) R jq + Q Now R d/(σ A) and C ɛ A/d, and so Q ωɛ /σ /l.t. Then the power factor is P.F cos[tan ( Q)] / + Q. 89

9 .7. Let η 50 + j30 and jk 0. + jm for a uniform plane wave propagating in the a z direction in a dielectric having some finite conductivity. If E s 400 V/m at z 0, find: a) P z,av at z 0 and z 60 cm: Assume x-polarization for the electric field. Then P z,av Re { E s Hs } (400) e αz Re Re { η 35 e (0.)z a z W/m { 400e αz e jβz a x 400 } a z e (0.)z Re Evaluating at z 0, obtain P z,av (z 0) 35 a z W/m, and at z 60 cm, P z,av (z 0.6) 35e (0.)(0.6) a z 48 a z W/m. } η e αz e jβz a y { 50 j30 b) the average ohmic power dissipation in watts per cubic meter at z 60 cm: At this point a flaw becomes evident in the problem statement, since solving this part in two different ways gives results that are not the same. I will demonstrate: In the first method, we use Poynting s theorem in point form (first equation at the top of p. 366), which we modify for the case of time-average fields to read: P z,av < J E > where the right hand side is the average power dissipation per volume. Note that the additional right-hand-side terms in Poynting s theorem that describe changes in energy stored in the fields will both be zero in steady state. We apply our equation to the result of part a: } a z < J E > P z,av d dz 35 e (0.)z (0.4)(35)e (0.)z 6e 0.4z W/m 3 At z 60 cm, this becomes < J E > 99. W/m 3. In the second method, we solve for the conductivity and evaluate < J E > σ<e >. We use and We take the ratio, Identifying σ ωɛ,wefind σ Re jk jω µɛ j(ɛ /ɛ ) η [ jk η jωɛ j Now we find the dissipated power per volume: µ ɛ j(ɛ /ɛ ) ( ɛ )] jωɛ + ωɛ ɛ { } { } jk 0. + j Re S/m η 50 + j30 σ<e > ( ) ( 400e 0.z) 90

10 .7b. (continued) At z 60 cm, this evaluates as 09 W/m 3. One can show that consistency between the two methods requires that { } Re η σ α This relation does not hold using the numbers as given in the problem statement and the value of σ found above. Note that in Problem.3, where all values are worked out, the relation does hold and consistent results are obtained using both methods..8a. Find P(r,t) if E s 400e jx a y V/m in free space: A positive y component of E requires a positive z component of H for propagation in the forward x direction. Thus H s (400/η 0 )e jx a z.06e jx a z A/m. In real form, the field are E(x, t) 400 cos(ωt x)a y and H(x, t).06 cos(ωt x)a z.nowp(r,t) P(x,t) E(x, t) H(x, t) 44.4 cos (ωt x)a x W/m. b) Find P at t 0 for r (a, 5, 0), where a 0,,, and 3: At t 0, we find from part a, P(a,0) 44.4 cos (a), which leads to the values (in W/m ): 44.4 ata 0, 73.5 ata, 8.3ata, and 39.3ata 3. c) Find P at the origin for T 0, 0.T, 0.4T, and 0.6T, where T is the oscillation period. At the origin, we have P(0,t) 44.4 cos (ωt) 44.4 cos (πt/t). Using this, we obtain the following values (in W/m ): 44.4 att 0, 4.4 att 0.T, 77.8 att 0.4T, and 77.8att 0.6T..9. Perfectly-conducting cylinders with radii of 8 mm and 0 mm are coaxial. The region between the cylinders is filled with a perfect dielectric for which ɛ 0 9 /4π F/m and µ R. If E in this region is (500/ρ) cos(ωt 4z)a ρ V/m, find: a) ω, with the help of Maxwell s equations in cylindrical coordinates: We use the two curl equations, beginning with E B/ t, where in this case, So Then E E ρ z a φ 000 ρ 000 B φ ρ H φ B φ µ 0 sin(ωt 4z)a φ B φ a φ t sin(ωt 4z)dt 000 ωρ We next use H D/ t, where in this case cos(ωt 4z) T 000 (4π 0 7 cos(ωt 4z) A/m )ωρ H H φ z a ρ + ρ (ρh φ ) a z ρ where the second term on the right hand side becomes zero when substituting our H φ.so And D ρ H H φ z a 8000 ρ (4π 0 7 )ωρ sin(ωt 4z)a ρ D ρ t 8000 (4π 0 7 )ωρ sin(ωt 4z)dt 8000 (4π 0 7 )ω cos(ωt 4z) C/m ρ a ρ 9

11 .9a. (continued) Finally, using the given ɛ, E ρ D ρ ɛ 8000 (0 6 )ω cos(ωt 4z) V/m ρ This must be the same as the given field, so we require 8000 (0 6 )ω ρ 500 ρ ω rad/s b) H(ρ,z,t): From part a,wehave H(ρ,z,t) 000 (4π 0 7 )ωρ cos(ωt 4z)a φ 4.0 ρ cos(4 08 t 4z)a φ A/m c) P(ρ,φ,z): This will be P(ρ,φ,z) E H 500 ρ cos(4 08 t 4z)a ρ 4.0 ρ cos(4 08 t 4z)a φ ρ cos (4 0 8 t 4z)a z W/m d) the average power passing through every cross-section 8 <ρ<0 mm, 0 <φ<π. Using the result of part c, we find P avg ( )/ρ a z W/m. The power through the given cross-section is now P π ρdρdφ π 0 3 ln ρ ( ) 0 5.7kW 8.0. If E s (60/r)sin θe jr a θ V/m, and H s (/4πr)sin θe jr a φ A/m in free space, find the average power passing outward through the surface r 0 6,0<θ<π/3, and 0 <φ<π. Then, the requested power will be P avg Re { E s Hs } 5 sin θ πr a r W/m π π/ ( 3 cos θ(sin θ + ) π/3 5 sin θ πr a r a r r sin θdθdφ 5 0 ) π/ W sin 3 θdθ Note that the radial distance at the surface, r 0 6 m, makes no difference, since the power density dimishes as /r. 9

12 .. The cylindrical shell, cm < ρ <. cm, is composed of a conducting material for which σ 0 6 S/m. The external and internal regions are non-conducting. Let H φ 000 A/m at ρ. cm. a) Find H everywhere: Use Ampere s circuital law, which states: H dl πρ(000) π(. 0 )(000) 48π A I encl Then in this case J I Area a z 48 (.44.00) 0 4 a z a z A/m With this result we again use Ampere s circuital law to find H everywhere within the shell as a function of ρ (in meters): H φ (ρ) π ρ πρ 0.0 Outside the shell, we would have ρdρdφ 54.5 ρ (04 ρ ) A/m (.0 <ρ<.0) H φ (ρ) 48π πρ 4/ρ A/m (ρ >.0) Inside the shell (ρ <.0 m), H φ 0 since there is no enclosed current. b) Find E everywhere: We use E J σ a z.09 a z V/m which is valid, presumeably, outside as well as inside the shell. c) Find P everywhere: Use Outside the shell, P E H.09 a z 54.5 ρ (04 ρ ) a φ 59.4 ρ (04 ρ ) a ρ W/m (.0 <ρ<.0 m) P.09 a z 4 ρ a φ 6 ρ a ρ W/m (ρ >.0 m) 93

13 .. The inner and outer dimensions of a copper coaxial transmission line are and 7 mm, respectively. Both conductors have thicknesses much greater than δ. The dielectric is lossless and the operating frequency is 400 MHz. Calculate the resistance per meter length of the: a) inner conductor: First δ πf µσ π(4 0 8 )(4π 0 7 )( ) m 3.3µm Now, using (70) with a unit length, we find R in πaσδ π( 0 3 )( )( ohms/m ) b) outer conductor: Again, (70) applies but with a different conductor radius. Thus R out a b R in (0.4) 0. ohms/m 7 c) transmission line: Since the two resistances found above are in series, the line resistance is their sum, or R R in + R out 0.54 ohms/m..3. A hollow tubular conductor is constructed from a type of brass having a conductivity of. 0 7 S/m. The inner and outer radii are 9 mm and 0 mm respectively. Calculate the resistance per meter length at a frequency of a) dc: In this case the current density is uniform over the entire tube cross-section. We write: R(dc) L σa (. 0 7 )π( ) /m b) 0 MHz: Now the skin effect will limit the effective cross-section. At 0 MHz, the skin depth is δ(0mhz) [πf µ 0 σ ] / [π(0 0 6 )(4π 0 7 )(. 0 7 )] / m This is much less than the outer radius of the tube. Therefore we can approximate the resistance using the formula: R(0MHz) L σa πbδ (. 0 7 )(π(.0))( ) 4. 0 /m c) GHz: Using the same formula as in part b, we find the skin depth at GHz to be δ m. The resistance (using the other formula) is R(GHz) 4. 0 /m. 94

14 .4a. Most microwave ovens operate at.45 GHz. Assume that σ. 0 6 S/m and µ R 500 for the stainless steel interior, and find the depth of penetration: δ πf µσ π( )(4π 0 7 )(. 0 6 ) m 9.8µm b) Let E s 50 0 V/m at the surface of the conductor, and plot a curve of the amplitude of E s vs. the angle of E s as the field propagates into the stainless steel: Since the conductivity is high, we use (6) to write α. β. πf µσ /δ. So, assuming that the direction into the conductor is z, the depth-dependent field is written as E s (z) 50e αz e jβz 50e z/δ e jz/δ 50 exp( z/9.8) exp( j z/9.8) }{{}}{{} amplitude angle where z is in microns. Therefore, the plot of amplitude versus angle is simply a plot of e x versus x, where x z/9.8; the starting amplitude is 50 and the /e amplitude (at z 9.8 µm) is A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0.3 mm and a velocity of m/s. Assuming the conductor is non-magnetic, determine the frequency and the conductivity: First, we use Next, for a good conductor, δ λ π πf µσ σ 4π λ fµ f v λ Hz GHz 4π (9 0 8 )(0 9 )(4π 0 7 ). 05 S/m.6. The dimensions of a certain coaxial transmission line are a 0.8mm and b 4mm. The outer conductor thickness is 0.6mm, and all conductors have σ S/m. a) Find R, the resistance per unit length, at an operating frequency of.4 GHz: First δ πf µσ π( )(4π 0 7 )( ) m.57µm Then, using (70) with a unit length, we find R in πaσδ π( )( )( ohms/m ) The outer conductor resistance is then found from the inner through R out a b R in 0.8 (4.84) 0.97 ohms/m 4 The net resistance per length is then the sum, R R in + R out 5.8 ohms/m. 95

15 .6b. Use information from Secs. 5.0 and 9.0 to find C and L, the capacitance and inductance per unit length, respectively. The coax is air-filled. From those sections, we find (in free space) C πɛ 0 ln(b/a) π( ) F/m ln(4/.8) L µ 0 4π 0 7 ln(b/a) ln(4/.8) H/m π π c) Find α and β if α + jβ jωc(r + jωl): Taking real and imaginary parts of the given expression, we find { } α Re jωc(r + jωl) ω ( ) / LC R + ωl and { } β Im jωc(r + jωl) ω LC + ( R ωl ) + / These can be found by writing out α Re { jωc(r + jωl) } (/) jωc(r + jωl)+c.c., where c.c denotes the complex conjugate. The result is squared, terms collected, and the square root taken. Now, using the values of R, C, and L found in parts a and b, we find α Np/m and β 50.3 rad/m..7. The planar surface at z 0 is a brass-teflon interface. Use data available in Appendix C to evaluate the following ratios for a uniform plane wave having ω rad/s: a) α Tef /α brass : From the appendix we find ɛ /ɛ.0003 for Teflon, making the material a good dielectric. Also, for Teflon, ɛ R.. For brass, we find σ.5 07 S/m, making brass a good conductor at the stated frequency. For a good dielectric (Teflon) we use the approximations: ( ɛ ( ɛ b) α. σ µ ɛ ɛ β. ω µɛ [ + 8 )( ( ɛ ) ω µɛ ɛ ɛ )]. ω µɛ ω c ) ω ɛ R c ɛ R For brass (good conductor) we have α. β. ( ) πf µσ brass π (4 0 π 0 )(4π 0 7 )( ) m Now α Tef α brass / ( ɛ /ɛ ) (ω/c) ɛ R (/)(.0003)(4 00 /3 0 8 ). πf µσbrass λ Tef λ brass (π/β Tef ) (π/β brass ) β brass β Tef c πf µσbrass ω ɛ R Tef (3 08 )( ) (4 0 0 )

16 .7. (continued) c) v Tef v brass (ω/β Tef ) (ω/β brass ) β brass β Tef as before.8. A uniform plane wave in free space has electric field given by E s 0e jβx a z + 5e jβx a y V/m. a) Describe the wave polarization: Since the two components have a fixed phase difference (in this case zero) with respect to time and position, the wave has linear polarization, with the field vector in the yz plane at angle φ tan (0/5) 33.7 to the y axis. b) Find H s : With propagation in forward x, we would have H s e jβx a y e jβx a z A/m 6.5e jβx a y e jβx a z ma/m c) determine the average power density in the wave in W/m : Use P avg Re { [ E s Hs } (0) ] 377 a x + (5) 377 a x 0.43a x W/m or P avg 0.43 W/m.9. Consider a left-circularly polarized wave in free space that propagates in the forward z direction. The electric field is given by the appropriate form of Eq. (80). a) Determine the magnetic field phasor, H s : We begin, using (80), with E s E 0 (a x + ja y )e jβz. We find the two components of H s separately, using the two components of E s. Specifically, the x component of E s is associated with a y component of H s, and the y component of E s is associated with a negative x component of H s. The result is H s E 0 η 0 ( ay ja x ) e jβz b) Determine an expression for the average power density in the wave in W/m by direct application of Eq. (57): We have P z,avg Re(E s Hs ) ( Re E 0 (a x + ja y )e jβz E ) 0 (a y ja x )e +jβz η 0 E 0 η 0 a z W/m (assuming E 0 is real) 97

17 .30. The electric field of a uniform plane wave in free space is given by E s 0(a y + ja z )e j5x. a) Determine the frequency, f : Use f βc π (5)(3 08 ). GHz π b) Find the magnetic field phasor, H s : With the Poynting vector in the positive x direction, a positive y component for E requires a positive z component for H. Similarly, a positive z component for E requires a negative y component for H. Therefore, H s 0 η 0 [ az ja y ] e j5x c) Describe the polarization of the wave: This is most clearly seen by first converting the given field to real instantaneous form: E(x, t) Re {E s e jωt} 0 [ ] cos(ωt 5x)a y sin(ωt 5x)a z At x 0, this becomes, E(0,t) 0 [ ] cos(ωt)a y sin(ωt)a z With the wave traveling in the forward x direction, we recognize the polarization as left circular..3. A linearly-polarized uniform plane wave, propagating in the forward z direction, is input to a lossless anisotropic material, in which the dielectric constant encountered by waves polarized along y(ɛ Ry ) differs from that seen by waves polarized along x(ɛ Rx ). Suppose ɛ Rx.5, ɛ Ry.0, and the wave electric field at input is polarized at 45 to the positive x and y axes. Assume free space wavelength λ. a) Determine the shortest length of the material such that the wave as it emerges from the output end is circularly polarized: With the input field at 45, the x and y components are of equal magnitude, and circular polarization will result if the phase difference between the components is π/. Our requirement over length L is thus β x L β y L π/, or L With the given values, we find, π (β x β y ) πc ω( ɛ Rx ɛ Ry ) L (58.3)πc ω 58.3 λ λ b) Will the output wave be right- or left-circularly-polarized? With the dielectric constant greater for x-polarized waves, the x component will lag the y component in time at the output. The field can thus be written as E E 0 (a y ja x ), which is left circular polarization. 98

18 .3. Suppose that the length of the medium of Problem.3 is made to be twice that as determined in the problem. Describe the polarization of the output wave in this case: With the length doubled, a phase shift of π radians develops between the two components. At the input, we can write the field as E s (0) E 0 (a x + a y ). After propagating through length L, we would have, E s (L) E 0 [e jβ xl a x + e jβ yl a y ] E 0 e jβ xl [a x + e j(β y β x )L a y ] where (β y β x )L π (since β x >β y ), and so E s (L) E 0 e jβ xl [a x a y ]. With the reversal of the y component, the wave polarization is rotated by 90, but is still linear polarization..33. Given a wave for which E s 5e jβz a x + 8e jβz e jφ a y V/m, propagating in a medium characterized by complex intrinsic impedance, η. a) Find H s : With the wave propagating in the forward z direction, we find: H s η [ 8e jφ a x + 5a y ] e jβz A/m b) Determine the average power density in W/m : We find P z,avg Re { { E s Hs } (5) } { } Re η + (8) η 75 Re η W/m.34. Given the general elliptically-polarized wave as per Eq. (73): E s [E x0 a x + E y0 e jφ a y ]e jβz a) Show, using methods similar to those of Example.7, that a linearly polarized wave results when superimposing the given field and a phase-shifted field of the form: E s [E x0 a x + E y0 e jφ a y ]e jβz e jδ where δ is a constant: Adding the two fields gives E s,tot [E x0 ( + e jδ) a x + E y0 (e jφ + e jφ e jδ) ] a y e jβz ( E x0e jδ/ e jδ/ + e jδ/) ( a x + E y0 e jδ/ e jδ/ e jφ + e jφ e jδ/) }{{}}{{} cos(δ/) cos(φ δ/) a y e jβz This simplifies to E s,tot [ E x0 cos(δ/)a x + E y0 cos(φ δ/)a y ] e jδ/ e jβz, which is linearly polarized. b) Find δ in terms of φ such that the resultant wave is polarized along x: By inspecting the part a result, we achieve a zero y component when φ δ π (or odd multiples of π). 99