Multislab Notes - Nafl’s 70m“ El e/(th't 0-70 C5047...

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Unformatted text preview: Nafl’s 70m“ El e/(th't 0-70 C5047 (/9347) 8—9 Normal Incidence at Multiple Dielectric Interfaces 402 Reflected a”, In certain practical situations a wave may be incident on several layers of dielecltric media with different constitutive parameters. One such Situatlon IS the use of a diehec- tric coating on glass to reduce glare from sunlight. Another is a radome, Wth is a dome-shaped enclosure designed not only to protect radar installations from in- clement weather but to permit the propagation of electromagnetic waves through 8 Plane Electromagnetic Waves wave " Hr E3 i an} --> "3 Transmitted Incident Wave an6 "1' Medium 1 Medium 3 (53: #3) FIGURE 8—15 z Normal incidence at multiple dielectric interfaces. the enclosure with as little reflection as possible. In both situations, determining the proper dielectric material and its thickness is an important design problem. We now consider the three—region situation depicted in Fig. 8—15. A uniform plane wave traveling in the +z—direction in medium 1 (61, pl) impinges normally at a plane boundary with medium 2 (62, ,uz), at z = 0. Medium 2 has a finite thickness and inter- faces with medium 3 (63, #3) at z = d. Reflection occurs at both 2 = 0 and z = d. Assuming an x-polarized incident field, the total electric field intensity in medium 1 can always be written as the sum of the incident component awafm” and a reflected component axEroejfl‘z: E1 : axwioe—jfi‘z + ErOejfilz)- (8-156) However, owing to the existence of a second discontinuity at z = d, E,0 is no longer related to E0 by Eq. (8438) or Eq. (8—140). Within medium 2, parts of waves bounce back and forth between the two bounding surfaces, some penetrating into media 1 and 3. The reflected field in medium 1 is the sum of (a) the field reflected from the interface at z = 0 as the incident wave impinges on it, (b) the field transmitted back into medium 1 from medium 2 after a first reflection from the interface at z = d, (c) the field transmitted back into medium 1 from medium 2 after a second reflection at z = d, and so on. The total reflected wave is, in fact, the resultant of the initial reflected com- ponent and an infinite sequence of multiply reflected contributions within medium 2 that are transmitted back into medium 1. Since all of the contributions propagate in the —z—direction in medium 1 and contain the propagation factor em 1’, they can be combined into a single term with a coefficient Ero. But how do we determine the relation between E,0 and E0 now? One way to find E,O is to write down the electric and magnetic field intensity vectors in all three regions and apply the boundary conditions. The H1 in region 1 8—9 Normal Incidence at Multiple Dielectric Interfaces 403 that corresponds to the E1 in Eq. (8~156) is, from Eqs. (8—131) and (8433), 1 . . HI = a, ; (Ewe—W — E,0e’”'z). (8—157) 1 The electric and magnetic fields in region 2 can also be represented by combinations of forward and backward waves: E2 = ax(E;e"j”“ + Egeifizz), (8—158) 1 . . H2 = a, — (ER—"’2’ — Egefl’zz). (8—159) ’72 In region 3, only a forward wave traveling in +z-direction exists. Thus, E3 = ange”fi‘3Z, (8—160) E + . H3 = a, —1 e—JI’”. (8—161) ’13 On the right side of Eqs. (8—156) through (8—161) there are a total of four un- known amplitudes: Em, 13;, E;, and E; They can be determined by solving the four boundary—condition equations required by the continuity of the tangential com- ponents of the electric and magnetic fields. ‘ At 2 = 0: E1(0) = E2(0), (8—162) H1(O) = H2(0). (8—163) At 2 = d: E2(d) = E3(d), (8—164) H2(d) = H3(d). (8—165) The procedure is straightforward and purely algebraic (Problem P.8e29). In the fol- lowing subsections we introduce the concept of wave impedance and use it in an alter- native approach for studying the problem of multiple reflections at normal incidence. 8—9.1 WAVE IMPEDANCE OF THE TOTAL FIELD We define the wave impedance of the total field at any plane parallel to the plane boundary as the ratio of the total electric field intensity to the total magnetic field intensity. With a z-dependent uniform plane wave, as was shown in Fig. 8—15, we write, in general, Total Ex(z) 2(2) 2 Total H,(z) ((2). (8-166) For a single wave propagating in the +z-direction in an unbounded medium, the wave impedance equals the intrinsic impedance, r], of the medium; for a single wave traveling in the —z-direction, it is —11 for all z. 404 8 Plane Electromagnetic Waves In the case of a uniform plane wave incident from medium I normally on a plane boundary with an infinite medium 2, such as that illustrated in Fig. 8—14 and discussed in Section 8—8, the magnitudes of the total electric and magnetic field intensities in medium 1 are, from Eqs. (87144) and (8449), E1x(z) = Bide-m” + rem”), (8467) E- . Hug) = _‘°(e~1fiiz _ ’11 refl‘lz). (8-168) Their ratio defines the wave impedance of the total field in medium 1 at a distance 2 from the boundary plane: E1x(z) _ e‘jfl" + FemZ 21(2) _ H1y(z) ” "1 e—fl’lz — reifilz’ (8469) which is obviously a function of z. A distance 2 = ~t to the left of the boundary plane, _ Elx(_/) ew + lee—flirt 21(4)) _ H1y(—/)— "1 eW — refit!" (8—170) Using the definition of F = (112 — r11)/(r/2 + 111) in Eq. (8—170), we obtain { ' ' / Zl(_{) _ n ’12 C05 .61 +1711 5m fir (8_171) 1 111 cos [31f +j112 sin [3%, which correctly reduces to 111 when r12 = 111. In that case there is no discontinuity at z = 0; hence there is no reflected wave and the total-field wave impedance is the same as the intrinsic impedance of the medium. When we study transmission lines in the next chapter, we will find that Eqs. (8—170) and (8—171) are similar to the formulas for the input impedance of a trans— mission line of length 5’ that has a characteristic impedance 111 and terminates in an impedance 112. There is a close similarity between the behavior of the propagation of uniform plane waves at normal incidence and the behavior of transmission lines. If the plane boundary is perfectly conducting, 112 = 0 and F = — 1, and Eq. (8~171) becomes Zl(_{)=j"1tan .315, which is the same as the input impedance of a transmission line of length I that has a characteristic impedance 111 and terminates in a short circuit. (8—172) 8—9.2 IMPEDANCE TRANSFORMATION WITH MULTIPLE DIELECTRICS The concept of total-field wave impedance is very useful in solving problems with multiple dielectric interfaces such as the situation shown in Fig. 8—15. The total field in medium 2 is the result of multiple reflections of the two boundary planes at z = 0 and z = d; but it can be grouped into a wave traveling in the +z-direction and an- other traveling in the —z-directi0n. The wave impedance of the total field in medium 8—9 Normal Incidence at Multiple Dielectric Interfaces 405 2 at the left-hand interface 2 = 0 can be found from the right side of Eq. (8—171) by replacing 112 by 113,111 by 212,31 by [$2, and t’ by d. Thus, 713 COS 32d +j'72 Sin [32d Z 0 = 2( ) 712 :12 cos fizd +jr13 sin [32d (8—173) As far as the wave in medium 1 is concerned, it encounters a discontinuity at z = 0 and the discontinuity can be characterized by an infinite medium with an intrinsic impedance 22(0) as given in Eq. (8—173). The eflective reflection coeflicient at z = O for the incident wave in medium 1 is ErO ___ HrO 22(0) — ’71 — -— = ———————. 8—174 0 Eio Hi0 22(0) + ’71 ( ) We note that F0 diflers from F only in that 112 has been replaced by 22(0). Hence the insertion of a dielectric layer of thickness d and intrinsic impedance r12 in front of medium 3, which has intrinsic impedance 113, has the effect of transforming r13 to Z 2(0). Given 111 and 113, F0 can be adjusted by suitable choices of 112 and d. Once F0 has been found from Eq. (8—174), E,0 of the reflected wave in medium 1 can be calculated: E,0 = FOE“). In many applications, 1‘0 and E,o are the only quantities of interest; hence this impedance-transformation approach is conceptually simple and yields the desired answers in a direct manner. If the fields 15;, E2“, and E, in media 2 and 3 are also desired, they can be determined from the boundary conditions at z = O and z = d, as indicated in Eqs. (8—162) through (8—165). EXAMPLE 8—12 A dielectric layer of thickness d and intrinsic impedance 112 is placed between media 1 and 3 having intrinsic impedances 111 and 173, respectively. Determine d and 112 such that no reflection occurs when a uniform plane wave in medium 1 impinges normally on the interface with medium 2. Solution With the dielectric layer interposed between media 1 and 3 as shown in Fig. 8—15, the condition of no reflection at interface 2 = 0 requires F0 = 0, or Z 2(0) = 111. From Eq. (8—173) we have "2013 COS .Bzd +j’12 Sin 32d) = 771("2 COS [32d +j713 Sin 32‘“- (8—175) Equating the real and imaginary parts separately, we require 113 cos [32d = 111 cos [32d (8—176) and 11% sin [32d = 111113 sin flzd. (8—177) Equation (8—176) is satisfied if either 113 = m ' (8—178) or cos [32d = 0, (8—179) 406 8 Plane Electromagnetic Waves which implies that 0! M = (2n + 1) A d=an+nf, n=QLlu. ohm On the one hand, if condition (8—178) holds, Eq. (8—177) can be satisfied when either (a) 112 = 113 = 7,1, which is the trivial case of no discontinuities at all, or (b) sin [32d = 0, or d = nil/2. and Eq. (8—177) can be satisfied when 112 = On the other hand, if relation (8—179) or (8—180) holds, sin [32d does not vanish, 4mm. We have then two possibilities for the condition of no reflection. 1. When r13 = 111, we require 1 d = —2’ n 2 that is, that the thickness of the dielectric layer be a multiple of a half-wavelength in the dielectric at the operating frequency. Such a dielectric layer is referred to as a half—wave dielectric window. Since 12 = upz/ f = 1/ ftmzez, where f is the operating frequency, a half—wave dielectric window is a narrow-band device. n=arznq @AM) When 173 75 111, we require '72 = V’li’la (8—182a) and v A d=(2n+1)ZZ—, n=0,1,2,.... (8—182b) When media 1 and 3 are different, #12 should be the geometric mean of 111 and n3, and d should be an odd multiple of a quarter wavelength in the dielectric layer at the operating frequency in order to eliminate reflection. Under these conditions the dielectric layer (medium 2) acts like a quarter-wave impedance transformer. We will refer to this term again when we study analogous trans- mission-line problems in Chapter 9. — We see from the above that if a radome is to be constructed around a radar installation (111 = 113 = no), it should be a half-wave window in order to minimize reflection; that is, it should be a multiple of [12/2 (= l/2f2 \mzez) thick at the operating radar frequency f2, where #2 and 62 are the permeability and permittivity, respectively, of the radome material. ...
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Multislab Notes - Nafl’s 70m“ El e/(th't 0-70 C5047...

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