ELECTROMAGNETIC WAVE INTERACTION WITH STRATIFIED NEGATIVE ISOTROPIC MEDIA

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1 Progress In Electromagnetics Research, PIER 35, 1 52, 2002 ELECTROMAGNETIC WAVE INTERACTION WITH STRATIFIED NEGATIVE ISOTROPIC MEDIA J. A. Kong Department of Electrical Engineering and Computer Science Massachusetts Institute of Technology Cambridge, MA 02139, USA Abstract In this paper we provide a general formulation for the electromagnetic wave interaction with stratified media and then specialize to slabs of negative isotropic media. The field solutions are obtained in all regions of the stratified medium. The characteristic waves in negative isotropic media are backward waves. We derive the field solutions and show that a Gaussian beam is laterally shifted by a negative isotropic slab. The amount of beam center shift is calculated for both cases of transmission and reflection. Guided waves in stratified media are studied. Placing a linear antenna and all types of Hertzian dipole antennas in a stratified medium, we obtained solutions in all regions. We demonstrate and locate the positions of the perfect images of the antenna sources in the presence of negative isotropic slabs. 1Introduction 2 Backward Waves in Negative Isotropic Media 3 Reflection and Transmission by a Negative Isotropic Medium 4 Reflection and Transmission by Stratified Media 4.1Reflection Coefficients 4.2 Propagation Matrices and Transmission Coefficients 4.3 Reflection and Transmission by a Slab Medium 5 Guided Waves In Stratified Media 5.1Guidance Conditions 5.2 Fields of Guided Waves 5.3 Coupling of Guided Waves

2 2 Kong 6 Linear Antennas In Stratified Media 6.1Integral Formulation 6.2 Linear Antenna Between Two Negative Isotropic Slabs 6.3 Linear Antenna in Front of a Negative Isotropic Slab 7 Dipole Antennas In Stratified Media 7.1Hertzian Electric and Magnetic Dipoles 7.2 Integral Formulation for Dipoles in Stratified Media 7.3 Dipoles Between Two Negative Isotropic Slabs 7.4 Dipoles in Front of Stratified Isotropic Media 7.5 Dipoles in Front of a Negative Isotropic Slab 7.6 Dipoles In Front of Negative Isotropic Slab Backed by a Perfect Conductor References 1. INTRODUCTION In the constitutive relations for negative isotropic media, both the permittivity ɛ and the permeability µ are negative. Vesolago 1] explored their various electromagnetic properties. He observed that in such media, the wave vector k, the electric field vector E, and the magnetic field vector H form a left-handed system, and called them left-handed subtances. The left-handed materials (LHM) possess negative refractive indeces. Experimental verifications were made with metamaterials and their use as perfect lenses was suggested 2 6]. In light of their potential new applications, negative isotropic media have received wide attention. In this paper we provide a general formulation with complete solution for electromagnetic waves interacting with a stratified medium and then specialize to slabs of negative isotropic medium. After showing that the characteristic waves in negative isotropic meia are backward waves, we consider relection and transmission of TE and TM waves by a negative isotropic medium. The solutions for the reflection and transmission by a stratified medium are then derived and listed by using the propagation matrices 7]. Specializing to a slab of negative isotropic medium, we show how wave beam incident on the slab will be laterally shifted. We then consider guided waves in negative isotropic materials in stratified media. Coupling of guided waves are studied. Next we derive field solutions for a linear antenna situated in a stratified medium. The images and their locations are calculated. Finally we consider Hertzian

3 Stratified negative isotropic media 3 electric and magnetic dipole sources with different orientation inside stratified media. The complete solutions in all regions are derived and listed 8]. Specializing to slabs of negative isotropic medium, we show positions of its images for various configurations, suggesting that perfect images can be made using flat lenzes made of slab negative isotropic slabs. 2. BACKWARD WAVES IN NEGATIVE ISOTROPIC MEDIA The constitutive relations for isotropic media have been written as D = ɛe where ɛ = permittivity (1a) B = µh where µ = permeability (1b) For negative isotropic media, both ɛ and µ are negative. For a plane electromagnetic wave of the form ] ] E(r, t) E = cos(k r ωt) (2) H(r, t) H The Maxwell equations become k E = ωµh (3) k H = ωɛe (4) k E = 0 (5) k H = 0 (6) where k = ω µɛ. The Poynting s vector power density is E H = 1 1 k ωµ (k E) E = ωµ ω 2 (k E) (k H) = E 2 (7) µɛ 1 ωɛ H (k H) = k ωɛ H 2 When µ and ɛ ar both positive, Poynting s power vector is in the same direction as k and so are the group and phase velocities. From (3) (6), we see that the three vectors k, E, and H form a right-handed system. When either µ or ɛ is negative, we have evanescent instead of propgating waves. In negative isotropic media, both µ and ɛ are negative, Poynting s power vector is in the opposite direction of k and so are the group and phase velolcities. The three vectiors k, E, and H form a left-handed Medium (LHM). In this medium, the power propagates in a direction that is opposite to the direction of k. The plane wave in the negative isotropic medium is thus a backward wave.

4 4 Kong H r =0 E r =0 y Region 0 µ 0,ɛ 0 θ r = θ i k r Region t -µ 0, -ɛ 0 z θ i θ t = θ i k k t H i H t E i E t St Figure 1. Reflection and transmission of TM waves by negative isotropic media. 3. REFLECTION AND TRANSMISSION BY A NEGATIVE ISOTROPIC MEDIUM We first study the case of TM wave incidence from free space upon a negative isotropic medium with permittivity ɛ t = ɛ 0 and permeability µ t = µ 0. The incident magnetic field is assumed to have unit amplitude and points in the ˆx direction (Fig. 1). We write, omitting the time convention e iωt, H i =ˆxH ix =ˆxe i(kyy+kzz) 1 E i = ŷk z +ẑk y ] e i(kyy+kzz) ωɛ 0 S i = E H 1 =ŷk y +ẑk z ] e i(k i ki ) r ωɛ 0 The reflected field components for the incident TM wave are H r =ˆxR TM e (kryy+krzz) 1 E r = ŷk rz +ẑk ry ] R TM e (kryy+krzz) ωɛ 0 S r = E H =ŷk ry +ẑk rz ] RTM 2 ωɛ 0 e i(kr k r ) r (8a) (8b) (8c) (9a) (9b) (9c) The incident wave vector k i =ŷk y +ẑk z and the reflected wave vector

5 Stratified negative isotropic media 5 k r =ŷk ry +ẑk rz are governed by the dispersion relations ky 2 + kz 2 = ω 2 µ 0 ɛ 0 = k 2 (10) kry 2 + krz 2 = ω 2 µ 0 ɛ 0 = k 2 (11) The reflection coefficient R TM for the magnetic field component H ix is to be determined by the boundary conditions. In region t, the transmitted TM field components are H t =ˆxT TM e i(ktxx+ktzz) E t = ŷk tz +ẑk ty ] 1 ωɛ t T TM e i(ktyy+ktzz) S t =ŷk ty +ẑk tz ] T TM 2 ωɛ t e i(kt k t ) r (12a) (12b) (12c) where T TM is the transmission coefficient for the magnetic field component H ix. The dispersion relation k 2 ty + k 2 tz = ω 2 µ t ɛ t = k 2 t (13) governs the magnitude k t for the transmitted wave vector k t =ŷk ty + ẑk tz. Let the boundary surface be at z = 0. The boundary condition of continuity of tangential H field gives e ikyy + R TM e ikryy = T TM e ktyy (14) Since (14) must hold for all y and t, it follows that k y = k ry = k ty (15) Eq. (15) is known as the phase matching condition. Eq. (14) then reduces to 1+R TM = T TM (16) From the dispersion relations (10) and (11) by noting that the reflected wave propagates in the negative ẑ direction, we find k rz = k z. The continuity of the tangential components of E x at z = 0 for all y and t gives k z (1 R TM )= k tz T TM (17) ɛ 0 ɛ t Note that the boundary conditions of normal D and normal B components continuous at z = 0 are also satisfied. Solving (16) and

6 6 Kong (17), we find p TM 0t = ɛ 0k tz ɛ t k z (18) R TM = 1 ptm 0t 1+p TM (19) 0t T TM 2 = 1+p TM (20) 0t for the reflection coefficient R TM and the transmission coefficientt TM. In the negative isotropic medium, we have ɛ t = ɛ 0 and µ t = µ 0. From Eq. (18), we find p TM = 1, and from Eqs. (19) and (20), we determine that R TM = 0 and T TM = 1. From the dispersion relation (13), we find k tz = k z. As seen from Fig. 1, the transmitted k t vector is in the same direction as the reflected k r vector. The Poynting power vector density is, from (23c), S t = ŷk y +ẑk z ] T TM 2 ωɛ 0 e i(kt k t ) r (21) such that the transmitted power is directed away from the boundary into the transmitted medium. When the incident wave is an evanescent wave with k i =ŷk y + ẑik Iz and k y >k,wehave H i =ˆxH ix =ˆxe i(kyy+kzz) 1 E i = ŷk z +ẑk y ] e i(kyy+kzz) ωɛ 0 S i = E H 1 =ŷk y +ẑk z ] e i(k i k ωɛ 0 i ) r (22a) (22b) (22c) H t =ˆxT TM e i(ktxx+ktzz) E t = ŷk tz +ẑk ty ] 1 ωɛ t T TM e i(ktyy+ktzz) S t =ŷk ty +ẑk tz ] T TM 2 ωɛ t e i(kt k t ) r (23a) (23b) (23c) For an incident TE wave, the incident wave vector k i =ŷk y +ẑk z, the reflected wave vector k r = ŷk ry +ẑk rz = ŷk y ẑk z, and the transmitted wave vector k t = ŷk y +ẑk tz all satisfy the dispersion

7 Stratified negative isotropic media 7 relations (10), (11), and (13), and the phase match condition (15). The incident electric and magnetic field vectors are E i =ˆxe ik i r 1 H i =ŷk z ẑk y ] e ik i r ωµ 0 S i = E H =ŷk y +ẑkz] 1 e i(k i k ωµ 0 The reflected electric and magnetic field vectors are i ) r (24a) (24b) (24c) E r =ˆxR TE e ikr r (25a) H r = ŷk z ẑk y ] RTE ωµ 0 e ikr r (25b) S r = E H =ŷk y ẑkz] RTE 2 e i(kr k r ) r ωµ 0 In region t, the transmitted TE wave solution takes the form E t =ˆxT TE e ikt r H t =ŷk tz ẑk y ] T TE S t =ŷk y +ẑktz] T TE 2 e i(kt k ωµ t (25c) (26a) ωµ t e ikt r (26b) t ) r (26c) where T TE is the transmission coefficient. From the boundary conditions of continuity of tangential E and H fields, we find p TE 0t = µ 0k tz µ t k z (27) R TE = 1 pte 0t 1+p TE 0t T TE 2 = 1+p TE 0t (28) (29) for the reflection coefficient R TE and the transmission coefficientt TE. In the negative isotropic medium, µ t = µ 0, and k tz = k z such that the transmitted power is directed away from the boundary into the transmitted medium. From Eq. (27), we find p TE = 1. Eqs. (28) and (29) then yield that R TE = 0 and T TE =1.

8 8 Kong y Region 0 µ 0,ɛ 0 Region t -µ 0, -ɛ 0 z θ i θ t = θ i H i k k t H t E i E t S t Figure 2. Reflection and transmission of TE waves by negative isotropic media. Region 0 Region 1 Region l Region n Region t = n + 1 y z µ 0,ɛ 0 µ 1,ɛ 1, µ n,ɛ n µ t,ɛ t µ l ɛ l z = d 1 z = d 2 z = d l z = d l+ 1 z = d n z = d t Figure 3. Stratified medium.

9 Stratified negative isotropic media 9 4. REFLECTION AND TRANSMISSION BY STRATIFIED MEDIA Consider a plane wave incident on a stratified isotropic medium with boundaries at z = d 1, d 2,...,d t (Fig. 3). The (n + 1)th region is semi-infinite and is labeled region t, t = n + 1. The permittivity and permeability in each region are denoted by ɛ l and µ l. The plane wave is incident from region 0 and has the plane of incidence parallel to the y-z plane. All field vectors are dependent on y and z only and independent of x. Since / x =0, the Maxwell equations in any region l can be separated into TE and TM components governed by E lx and H lx.we obtain H ly = 1 iωµ l z E lx (30) H lz = 1 iωµ l y E lx (31) ( 2 y ) z 2 + ω2 µ l ɛ l E lx = 0 (32) E ly = 1 iωɛ l z H lx (33) E lz = 1 iωɛ l y H lx (34) ( 2 y ) z 2 + ω2 µ l ɛ l H lx = 0 (35) The TE waves are completely determined by (30) (32) and the TM waves by (33) (35). The two sets of equations are duals of each other under the replacements E l H l, H l E l, and µ l ɛ l. For a TE plane wave, E x = E 0 e ikzz+ikyy, incident on the stratified medium, the total field in region l can be written as ( E lx = E l + e iklzz + El e ik lzz e ikyy (36) H ly = k ( lz E l + e iklzz El e ik lzz e ikyy ωµ l (37) H lz = k y ωµ l ( E + l e ik lzz + E l e ik lzz ) e ikyy (38)

10 10 Kong Obviously (36) satisfies the Helmholtz wave equation in (32). Substitution of (36) in (32) yields the dispersion relation k 2 lz + k 2 y = ω 2 µ l ɛ l (39) We do not write a subscript l for the y component of k as a consequence of the phase-matching conditions. Truly, there are multiple reflections and transmissions in each layer l. The amplitude E l + thus represents all wave components that have a propagating velocity component along the positive ẑ direction, and El represents those with a velocity component along the negative ẑ direction. We note that in region 0 where l =0, In region t where l = n +1=t, wehave E + 0 = E 0 (40) E 0 = RE 0 (41) E t + = TE 0 (42) Et = 0 (43) because region t is semi-infinite and there is no wave propagating with a velocity component in the positive ẑ direction. We denote the transmitted amplitude by T. The wave amplitudes E + l and E l are related to wave amplitudes in neighboring regions by the boundary conditions. At z = d l+1, boundary conditions require that E x and H y be continuous. We obtain E + l e ik lzd l+1 +E l e ik lzd l+1 = E + l+1 eik (l+1) zd l+1 +E l+1 e ik (l+1) zd l+1 (44) E + l e ik lzd l+1 E l e ik lzd l+1 = p l(l+1) E + l+1 eik (l+1) zd l+1 E l+1 e ik (l+1) zd l+1 ] (45) where p l(l+1) = µ lk (l+1) z (46) µ l+1 k lz for the TE wave. There are n+1boundaries which give rise to (2n+2) equations. In region 0, we have an unknown reflection coefficient R. In region t, we have an unknown transmission coefficient T. There are two unknowns E l + and El in each of the regions l =1,2,..., n. Thus we have a total of (2n + 2) unknowns. To solve for the

11 Stratified negative isotropic media 11 (2n + 2) unknowns from the (2n + 2) linear equations, we can arrange the equations in matrix form with the unknowns forming a (2n +2) column matrix and the coefficients forming a (2n+2) (2n+2) square matrix. The solution is then obtained by inverting the square matrix. This procedure is straightforward but tedious. We shall now describe simpler ways to deal with the problem. For a TM plane wave, H x = H 0 e ikzz+ikyy, incident on the stratified medium, the total field in region l can be written as ( ) H lx = E l + e iklzz + El e ik lzz e ikyy (47) E ly = k ( lz ) E l + e iklzz El e ik lzz e ikyy (48) ωɛ l E lz = k ( y ) E l + e iklzz + El e ik lzz e ikyy (49) ωɛ l Matching boundary conditions at the boundaries, identifying p l(l+1) = ɛ lk (l+1) z ɛ l+1 k lz (50) for TM waves, we obtain the same equations as in (44) and (45) Reflection Coefficients To find a closed form solution for the reflection coefficient R for the stratified medium, we first solve (44) and (45) for A l and B l. E l + e ik lzd l+1 = 1 ) (1+p 2 l(l+1) } {E l+1 + eik (l+1)zd l+1 + R l(l+1) El+1 e ik (l+1)zd l+1 (51) El e ik lzd l+1 = 1 ) (1+p 2 l(l+1) } {R l(l+1) E l+1 + eik (l+1)zd l+1 + El+1 e ik (l+1)zd l+1 (52) where R l(l+1) = 1 p l(l+1) (53) 1+p l(l+1) is the reflection coefficient for waves in region l, caused by the boundary separating regions l and l + 1. We note from (46) that p (l+1) l = 1 p l(l+1) (54)

12 12 Kong which also gives R (l+1) l = R l(l+1) (55) Thus the reflection coefficient in region l +1, R (l+1) l, caused by the boundary separating regions l + 1and l, is equal to the negative of R l(l+1). Forming the ratio of (51) and (52) we obtain ( )] El E l + = ei2k lzd l+1 1 1/Rl(l+1) 2 e i2(k (l+1) z+k lz )d (l+1) + ] R l(l+1) 1/R l(l+1) e i2k (l+1) zd (l+1) +(El+1 /E+ l+1 ) ( )] = ei2k lzd (l+1) 1 1/Rl(l+1) 2 e i2(k (l+1) z+k lz )d (l+1) + ] + R l(l+1) 1/R E l+1 l(l+1) e i2k (l+1) zd (l+1) E l+1 + (56) With the second equality we introduce a notation for writing a continued fraction. Equation (56) expresses (E l + /El ) in terms of El+1 /E+ l+1 and so on, until the transmitted region t, where E t /E+ t = 0, is reached. The reflection coefficient due to the stratified medium is R = B 0 /A 0. Making use of the continued fractions, we obtain R = ei2k 0zd 1 1 (1/R 2 01 ) ] e i2(k 1z+k 0z )d 1 + R 01 (1/R 01 )e i2k 1zd 1 + ei2k1zd2 R 12 1 (1/R 2 12 ) ] e i2(k 2z+k 1z )d 2 + (1/R 12 )e i2k 2zd ei2k(n 1) zdn R (n 1) n ] 1 (1/R(n 1) 2 n ) e i2(knz+k (n 1) z)d n + +R nt e i2knzd (n+1) (57) (1/R (n 1) n )e i2knzdn This is a closed-form solution for the reflection coefficient expressed in continued fractions. Such a solution is very easily programmed for numerical computation Propagation Matrices and Transmission Coefficients For a plane wave incident on a stratified medium, we have obtained the boundary conditions of continuity of tangential electric and magnetic fields at each interface z = d l, with the two equations (44) (45) relating wave amplitudes in regions l and l +1: E l+1 + eik (l+1) zd (l+1) + El+1 e ik (l+1) zd (l+1) = E l + e ik lzd (l+1) + El e ik lzd (l+1) (58)

13 Stratified negative isotropic media 13 E + l+1 eik (l+1) zd (l+1) E l+1 e ik (l+1) zd (l+1) = p (l+1) l E + l e ik lzd (l+1) E l e ik lzd (l+1) ] (59) where for TE waves and for TM waves p (l+1) l = µ l+1k lz µ l k (l+1)z (60) p (l+1) l = ɛ l+1k lz ɛ l k (l+1)z (61) Equations (60) and (61) follow from duality but bear in mind that E l + and El denote amplitudes of tangential electric fields for TE waves and H l + and Hl denote amplitudes of tangential magnetic fields for TM waves. In the last section we determined the reflection coefficients R = E0 /E+ 0 from the (2n+2) boundary conditions. We will now show that the transmission coefficient T = E t + /E+ 0 can be obtained by the use of propagation matrices. We solve for E l+1 + and E l+1 in terms of E+ l and El from (58) (59) and obtain E l+1 + eik (l+1)zd l+1 = 1 ) 2 (1+ p (l+1)l) (E l + e ik lzd l+1 + R (l+1)l El e ik lzd l+1 El+1 e ik (l+1)zd l+1 = 1 ) 2 (1+ p (l+1)l) (R (l+1)l E l + e ik lzd l+1 + El e ik lzd l+1 Expressing in the form of matrix multiplication, we have where E+ l+1 E l+1 = V (l+1) l E+ l E l (62) V (l+1) l = p (l+1) l] e i(k (l+1)z k lz )d l+1 R (l+1) l e i(k (l+1)z+k lz )d l+1 R (l+1) l e i(k (l+1)z+k lz )d l+1 e i(k (l+1)z k lz )d l+1 (63) is called the forward-propagating matrix. In (63), R (l+1) l = 1 p (l+1) l 1+p (l+1) l = R l(l+1) is the reflection coefficient at the boundary separating regions l +1and l, and the first subscript denotes the region with the incident wave.

14 14 Kong It is to be noted for the forward-propagating matrix between layers n and t = n +1, T = V 0 tn E+ n En with V tn = 1 2 (1+ p tn) e i(ktz knz)dt R tn e i(ktz+knz)dt R tn e i(ktz+knz)dt e i(ktz knz)dt By the same token, we may express E l + and El in terms of E l+1 + and El+1 by using (51) (52) and define a backward-propagating matrix. The propagation matrices can be used to determine wave amplitudes in any region in terms of those in any other region. For m>l, we make use of the forward propagation matrix to obtain E+ m Em = V m(m 1) V (m 1)( m 2) V (l+1) l E+ l El Similarly, backward-propagating matrices can be used to express wave amplitudes in any region j in terms of those in region l for l>j. In particular, the transmission coefficient T = Et /E 0 for a stratified medium with t = n + 1layers can be calculated by the multiplication of n + 1propagation matrices. Using the forwardpropagating matrices, we have T = V 0 t0 1 R where V t0 = V tn V n(n 1) V 10 includes all information about the stratified medium. Once V t0 is known, both the reflection and transmission coefficients can be calculated from its matrix elements Reflection and Transmission by a Slab Medium For a slab medium with boundary surfaces at z = d 1 and z = d 2,we find from (57), with t = 2 and n = 1, the reflection coefficient R = ei2k 0zd 1 R (1/R 2 01 ) ] e i2(k 1z+k 0z )d 1 (1/R 01 )e i2k 1zd 1 + R12 e i2k 1zd 2 = R 01 + R 12 e i2k 1z(d 2 d 1 ) 1+R 01 R 12 e i2k 1z(d 2 d 1 ) ei2k 0zd 1 (64)

15 Stratified negative isotropic media 15 Making use of propagation matrix V (l+1) l as shown in (62) E+ 1 E1 = 1 2 (1+ p 10) e i(k 1z k 0z )d 1 R 10 e i(k 1z+k 0z )d 1 R 10 e i(k 1z+k 0z )d 1 e i(k 1z k 0z )d 1 1 R T = (1+ p t1) e i(ktz k 1z)d t R t1 e i(ktz+k 1z)d t R t1 e i(ktz+k 1z)d t e i(ktz k 1z)d t E+ 1 E1 we find the amplitudes inside the slab medium to be E + 1 = 2e i(k 1z k 0z )d 1 (1+ p 01 )(1+ R 01 R 1t e i2k 1z(d 2 d 1 ) ) E1 = 2R 12 e i(k 1z k 0z )d 1 e i2k 1zd 2 (1+ p 01 )(1+ R 01 R 1t e i2k 1z(d 2 d 1 ) ) The transmission coefficient is (65) (66) 4e ik 0zd 1 e ik 1z(d 2 d 1 ) e ik 2zd 2 T = (1+ p 01 )(1+ p 1t )(1+ R 01 R 1t e i2k 1z(d 2 d 1 ) (67) ) For an electromagnetic wave incident on a negative isotropic slab in free space, we let µ 1 = µ 0, ɛ 1 = ɛ 0, and µ t = µ 0, ɛ t = ɛ 0. We find k 1z = k 0z, k 2z = k 0z, p 01 = p 12 =1,R 01 = R 12 = R =0,E1 =0, E 1 + = ei2k 0zd 1, T = e i2k 0z(d 2 d 1 ). Assuming a Gaussian beam with beamwidth g incident on the negative isotropic slab with incident angle θ i and incident vector k i =ŷk iy +ẑk iz, where k iy = k 0 sin θ i,k iz = k 0 cos θ i. For an incident TM wave field, we write H x = Ψ(k y )= g2 k y k iy 2 /4 4π e g2 Then for the transmitted magnetic field, we have H x = dk y e i(kyy+k 0zz) Ψ(k y ) (68) (69) dk y e i(kyy+k 0zz 2(d 2 d 1 )]) Ψ(k y ) (70) Thus on the plane at a constant z in the transmitted region, the Gaussian beam spot is at y out, which is shifted by a distance of 2(d 2 d 1 ) tan θ i in the negative ŷ direction in the presence of the negative isotropic slab as compare to the location of the spot y in in the absence of the slab.

16 16 Kong y y in k Region 0 µ 0,ɛ 0 k Region t µ 0,ɛ 0 z θ i θ t = θ i H i k k 1 H 1 H t k t E i E 1 Region 1 -µ 0, -ɛ 0 St y out E t z = d 1 z = d 2 Figure 4. Beam center shifted by 2(d 2 d 1 ) tan θ i. The reflected beam shift will be twice when we place the negative isotropic slab in front of a perfect conductor. We find µ 1 = µ 0, ɛ 1 = ɛ 0, k 1z = k 0z, T =0,E 1 + = ei2k 0zd 1, E1 = R 12e i2k 0z(d 2 d 1 ), p 01 =1,R 01 =0. For TE waves R12 TE = 1, RTE = e i2k 0z(d 2 d 1 ) e i2k 0zd 1. For TM waves R12 TM =1,R TM = e i2k 0z(d 2 d 1 ) e i2k 0zd 1. In the absence of the negative isotropic slab, the reflection coefficient is R0 TE = e i2k 0zd 2. R0 TM = e i2k 0zd 2. Thus on the plane at a constant z in Region 0, the Gaussian beam spot is at y out, which is shifted by a distance of 4(d 2 d 1 ) tan θ i in the negative ŷ direction in the presence of the negative isotropic slab as compare to the location of the spot y in in the absence of the slab.

17 Stratified negative isotropic media 17 y y in k Region 0 µ 0,ɛ 0 k Region t Perfect Conductor z H i k E i θ i θ t = θ i k 1 H 1 E S 1 1 E1 k 1 S 1 H 1 y out Region 1 -µ 0, -ɛ 0 z = d 1 z = d 2 Figure 5. Beam center shifted by 4(d 2 d 1 ) tan θ i. 5. GUIDED WAVES IN STRATIFIED MEDIA 5.1. Guidance Conditions The geometrical configuration of the problem is shown in Figure 6. There are t layers at z = d 1,d 2,...d t, and s + 1layers at z = d 0,d 1,...,d s. We shall first assume that all regions contain isotropic media. In region l, we denote the permittivity and permeability by ɛ l and µ l. Notice that in region 0, ɛ 0 and µ 0 are not necessarily equal to the free space permittivity ɛ o and permeability µ o.

18 18 Kong Region s Region s+ 1 Region 1 Region 0 Region 1 Region t 1 Region t y 0 z z =d -s + 1 z =d -s + 2 z = d 1 z = d 0 z = d 1 z = d 2 z = d t 1 z = d t Figure 6. Guided Waves in stratified media. For TE waves, the solutions in region l take the following form: ] E lx = E l + e iklzz + El e ik lzz e ikyy (71) H ly = k lz ] E l + e iklzz El e ik lzz e ikyy ωµ l H lz = k y ] E l + e iklzz + El e ik lzz e ikyy (72) ωµ l For TM waves we invoke duality with the replacement E H, H E, µ 0 ɛ 0. The boundary conditions of the interfaces require that tangential electric and magnetic field components be continuous for all x and y. At z = d l+1, we obtain E + l e ik lzd l+1 + E l e ik lzd l+1 = E + l+1 eik (l+1) zd l+1 + E l+1 e ik (l+1) zd l+1 (73) E + l e ik lzd l+1 E l e ik lzd l+1 = p l(l+1) E + l+1 eik (l+1) zd l+1 E l+1 e ik (l+1) zd l+1 ] (74) where p l(l+1) = µ lk (l+1) z µ l+1 k lz = 1 p (l+1)l (75)

19 Stratified negative isotropic media 19 We now determine the wave amplitudes in region l. From (73) (74) we can express E l + and El in terms of E l+1 + and E l+1 or express E+ l+1 and El+1 in terms of E+ l and El. We find E l + e ik lzd l+1 = 1 ) (1+p 2 l(l+1) } {E l+1 + eik (l+1)zd l+1 + R l(l+1) El+1 e ik (l+1)zd l+1 (76) El e ik lzd l+1 = 1 ) (1+p 2 l(l+1) } {R l(l+1) E l+1 + eik (l+1)zd l+1 + El+1 e ik (l+1)zd l+1 (77) for E + l and E l in terms of E + l+1 and E l+1, and E + l+1 eik (l+1)zd l+1 E l+1 e ik (l+1)zd l+1 = 1 2 (1+ p (l+1)l) {E + l e ik lzd l+1 + R (l+1) l E l e ik lzd l+1 } = 1 2 (1+ p (l+1)l) {R (l+1) l E + l e ik lzd l+1 + E l e ik lzd l+1 } (78) (79) for E + l+1 and E l+1 in terms of E+ l and E l. In (76) (??), the Fresnel reflection coefficient R (l+1)l = 1 p (l+1)l 1+p (l+1)l = R l(l+1) (80) where R l(l+1) = 1 p l(l+1) (81) 1+p l(l+1) is the reflection coefficient for waves in region l, caused by the boundary separating regions l and l + 1. The reflection coefficient in region l +1, R (l+1) l, caused by the boundary separating regions l+1and l,is equal to the negative of R l(l+1). There are altogether s + t boundaries which give rise to 2(s + t) equations as shown above. There are altogether s + t + 1regions. In regions t and s we have Et = 0 and E s + = 0 because there are no waves originating from infinity. Thus we have a total of 2(s + t +1) 2=2(s + t) unknowns to be solved from the 2(s + t) equations.

20 20 Kong For z 0, we let l = 0, and obtain the reflection coefficients R 0+ = E0+ /E+ 0+ in the form of continued fractions. We find R 0+ = E 0+ E 0+ + = ei2k 0zd 1 1 (1/R 01 ) 2] e i2(k 0z+k 1z )d 1 + R 01 (1/R 01 )e i2k 1zd 1 +(E 1 /E 1 + ) (82) where E1 /E+ 1 can be expressed in terms of E 2 /E+ 2 and so on until region t where Et /E+ t =0. For z 0, we let l = 0, and obtain the reflection coefficients R 0 = E 0 + /E 0 in the form of continued fractions. We find R 0 = E+ 0 E 0 ) ] 2 = e i2k 0zd 0 1 (1/R 0( 1) e i2(k 0z+k 1z )d 0 + R 0( 1) (1/R 0( 1) )e i2k 1zd 0 +(E + 1 /E 1 ) (83) where E 1 + /E 1 are expressible in terms of E+ 2 /E 2 and so on until region s, where E s/e + s =0. In region 0, the guidance condition is determined from For s = 2 and t = 2, we find R 0+ R 0 = 1(84) R 0+ = R 01 + R 12 ei2k 1z(d 2 d 1 ) 1+R 01 R 12 e i2k 1z(d 2 d 1 ) ei2k 0zd 1 (85) R 0 = R 0( 1) + R ( 1)( 2) e i2k 1z(d 1 d 0 ) 1+R 0( 1) R ( 1)( 2) e i2k 1z(d 1 d 0 ) e i2k 0zd 0 (86) Consider guidance in Region 0 with evanescence in Regions 1and -1 such that k 1z = k1 2 k2 y = iα 1z and k 1z = k 1 2 k2 y = iα 1z.We then have p 01 = i µ 0α 1z ; R 01 = e i2φ 01 ; φ 01 = tan 1 µ 0α 1z µ 1 k 0z µ 1 k 0z (87) p 0( 1) = i µ 0α 1z ; R µ 1 k 0( 1) = e i2φ 0( 1) ; φ 0( 1) = tan 1 µ 0α 1z 0z µ 1 k 0z (88) where φ 01 and φ 0( 1) are Goos-Haschen shifts at the boundaries z = d 1 and z = d 0.

21 Stratified negative isotropic media Fields of Guided Waves For a slab waveguide with s = 1and t =1,wehave The guidance condition becomes R 0+ = R 01 e i2k 0zd 1 (89) R 0 = R 0( 1) e i2k 0zd 0 (90) φ 0( 1) + φ 01 + k 0z d 1 k 0z d 0 = mπ (91) for the mth order guided mode.for an asymmetric slab waveguide, we have E 1 + = 0 (92) E 1 = 1 } (1+p 2 ( 10) ){R ( 1)0 E 0 + ei(k 0z+k 1z )d 0 + E0 e i(k 0z k 1z )d 0 (93) E 1 + = 1 } 2 (1+ p 10) {E 0 + ei(k 0z k 1z )d 1 + R 10 E0 e i(k 0z+k 1z )d 1 (94) E 1 = 0 (95) The field in Region 1is E 1x = E 1 eα 1zz e ikyy (96) H 1y = k 1z E 1 ωµ 1zz e ikyy l (97) H 1z = k y ωµ 1 E 1 eα 1zz ] e ikyy (98) k y < S 1 > =ŷ E 1 2ωµ e 2α 1zz 1 (99) The field in Region 0 is E 0x = E 0 + eik 0zz + E0 e ik 0zz e ikyy (100) H 0y = k 0z E 0 + ωµ E0 e ik 0zz e ikyy 0 (101) H lz = k y ωµ 0 E + 0 eik 0zz + E 0 e ik 0zz ] e ikyy (102) < S 0 > =ŷ k { y }] E 0 + 2ωµ 2 + E0 2 +2Re E 0 + (E 0 ) e i2k 0zz (103) 0

22 22 Kong The field in Region 1is E 1x = E 1 + e α lzz e ikyy (104) H 1y = k 1z E 1 + ωµ 1zz e ikyy 1 (105) H 1z = k y ωµ 1 E + 1 e α 1zz ] e ikyy (106) < S 1 > =ŷ k y ] E 1 + 2ωµ 2 e 2α 1zz (107) 1 When the waveguide is a negative isotropic slab with µ 0 = µ o and ɛ 0 = ɛ o, we observe that the Poynting power inside the wave guide is flowing in the negative ŷ direction while the Poynting power outside the slab flow in the positive ŷ direction Coupling of Guided Waves For the case of s = 2 and t =2,wefind E 2 + = 0 (108) E 2 = 1 ) (1+p 2 2( 1) } {R 2( 1) E 1 + ei(k 1z+k 2z )d 1 + E 1 e i(k 1z k 2z )d 1 (109) E 1 + = 1 } 2 (1+ p 10) {E 0 + ei(k 0z k 1z )d 0 +R 10 E0 e i(k 0z+k 1z )d 0 (110) E 1 = 1 } 2 (1+ p 10) {R 10 E 0 + ei(k 0z+k 1z )d 0 +E0 e i(k 0z k 1z )d 0 (111) E 1 + = 1 } 2 (1+ p 10) {E 0 + ei(k 0z k 1z )d 1 + R 10 E0 e i(k 0z+k 1z )d 1 (112) E1 = 1 } 2 (1+ p 10) {R 10 E 0 + ei(k 0z+k 1z )d 1 + E0 e i(k 0z k 1z )d 1 (113) E 2 + = 1 } 2 (1+ p 21) {E 1 + ei(k 1z k 2z )d 2 + R 21 E1 e i(k 1z+k 2z )d 2 (114) E2 = 0 (115) We let waves be guided in Region 1and Region 1, thus k 0z = iα 0z, k 2z = iα 2z, and k 2z = iα 2z. The electromagnetic fields in all regions take the following forms. In Region 2: ] E 2x = E 2 eα 2zz e ikyy (116)

23 Stratified negative isotropic media 23 H 2y = k 2z ] E 2 ωµ eα 2zz e ikyy 2 H 2z = k y ] E 2 ωµ eα 2zz e ikyy (117) 2 k ] y < S 2 > =ŷ E 2 2ωµ 2 e 2α 2zz (118) 2 In Region 1: ] E 1x = E 1 + eik 1zz + E 1 e ik 1zz e ikyy (119) H 1y = k 1z ] E 1 + ωµ eik 1zz E 1 e ik 1zz e ikyy 1 H 1z = k y ] E 1 + ωµ eik 1zz + E 1 e ik 1zz e ikyy (120) 1 k { }] y < S 1 > =ŷ E 1 + 2ωµ 2 + E 1 2 2Re E 1 + (E 1 ) e i2k1zz (121) 1 In Region 0: ] E 0x = E 0 + eik 0zz + E0 e ik 0zz e ikyy (122) H 0y = k 0z ] E 0 + ωµ eik 0zz E0 e ik 0zz e ikyy 0 H 0z = k y ωµ 0 E + 0 eik 0zz + E 0 e ik 0zz ] e ikyy (123) < S 0 > =ŷ k y 2ωµ 0 E E Re { E + 0 (E 0 ) e i2k 0zz }] (124) In Region 1: ] E 1x = E 1 + eik 1zz + E1 e ik 1zz e ikyy (125) H 1y = k 1z ] E 1 + ωµ eik 1zz E1 e ik 1zz e ikyy 1 H 1z = k y ωµ 1 E + 1 eik 1zz + E 1 e ik 1zz ] e ikyy (126) < S 1 > =ŷ k y E 1 + 2ωµ 2 + E1 2 +2Re 1 In Region 2: E 2x = { }] E 1 + (E 1 ) e i2k1zz (127) E + 2 e α 2zz ] e ikyy (128)

24 24 Kong H 2y = k 2z ωµ 2 E + 2 e α 2zz ] e ikyy H 2z = k y ] E 2 + ωµ e α 2zz e ikyy (129) 2 < S 2 > =ŷ k ] y E 2 + 2ωµ 2 e 2α2zz (130) 2 We observe that if Region 1is a slab medium with positive µ and ɛ, while Region 1is a negative slab, the guided wave in Region 1is propagating in the positive ŷ direction, and the guided wave in Region 1is propagating in the negative ŷ direction. Thus the guided wave direction is reversed through evanescent coupling with Region LINEAR ANTENNAS IN STRATIFIED MEDIA 6.1. Integral Formulation The geometrical configuration of the problem is shown in Figure 7. The origin of the coordinate system is placed in the location of the linear antenna, which is in the ˆx-direction. J(r )=ˆxI δ(y )δ(z ) (131) There are t layers at z = d 1,d 2,...d t and s + 1layers at z = d 0,d 1,...,d s. We shall first assume that all regions contain isotropic media. In region l, we denote the permittivity and permeability by ɛ l and µ l. Notice that in region 0, ɛ 0 and µ 0 are not necessarily equal to the free space permittivity and permeability which we denote by ɛ o and µ o. The solution of the electric field vector for the linear antenna in unbounded medium with permittivity ɛ 0 and µ 0 is E = iωµ 0 dy dz G(r, r ) J(r ) { } i =ˆxiωµ 0 I 4 H(1) 0 (kρ) { i } 1 =ˆxiωµ 0 I dk y e ikyy + ik 0z z 4π k 0z { } ωµ 0 I e =ˆx dk y e ikyy ik 0z z z 0 4πk 0z e ik (132) 0zz z 0 We have from the Maxwell equations in source-free regions H = 1 iωµ 0 E = ii 4π ˆx dk y 1 k 0z e ikyy + ik 0z z

25 Stratified negative isotropic media 25 Region s Region s+ 1 Region 1 Region 0 Region 1 Region t 1 Region t y 0 linear antenna z z =d -s + 1 z =d -s + 2 z = d 1 z = d 0 z = d 1 z = d 2 z = d t 1 z = d t Figure 7. Linear antennas in stratified media. =ŷ I 4π +ẑ I 4π { } e dk y e ikyy ik 0z z e ik 0zz { } e ik 0z z dk y k y k 0z e ikyy e ik 0zz z 0 z 0 z 0 z 0 (133) (134) The solutions in region l take the following form: E lx = H ly = H lz = dk y E + l e ik lzz + E l e ik lzz ] e ikyy (135) dk y k lz ωµ l E + l e ik lzz E l e ik lzz ] e ikyy dk y k y ωµ l E + l e ik lzz + E l e ik lzz ] e ikyy (136) The boundary conditions of the interfaces require that tangential electric and magnetic field components be continuous for all x and y. At z = d l+1, we obtain E + l e ik lzd l+1 + E l e ik lzd l+1 = E + l+1 eik (l+1) zd l+1 + E l+1 e ik (l+1) zd l+1 (137) E + l e ik lzd l+1 E l e ik lzd l+1 = p l(l+1) E + l+1 eik (l+1) zd l+1 E l+1 e ik (l+1) zd l+1 ] (138)

26 26 Kong where p l(l+1) = µ lk (l+1) z = 1 (139) µ l+1 k lz p (l+1)l We now determine the wave amplitudes in region l. From (137) (138) we can express E l + and El in terms of E l+1 + and E l+1 or express E+ l+1 and El+1 in terms of E+ l and El. We find E l + e ik lzd l+1 = 1 ) (1+p 2 l(l+1) } {E l+1 + eik (l+1)zd l+1 + R l(l+1) El+1 e ik (l+1)zd l+1 (140) El e ik lzd l+1 = 1 ) (1+p 2 l(l+1) } {R l(l+1) E l+1 + eik (l+1)zd l+1 + El+1 e ik (l+1)zd l+1 (141) for E + l and E l in terms of E + l+1 and E l+1, and E + l+1 eik (l+1)zd l+1 E l+1 e ik (l+1)zd l+1 = 1 2 (1+ p (l+1)l) {E + l e ik lzd l+1 + R (l+1) l E l e ik lzd l+1 } = 1 2 (1+ p (l+1)l) {R (l+1) l E + l e ik lzd l+1 + E l e ik lzd l+1 } (142) (143) for E + l+1 and E l+1 in terms of E+ l and E l. In (140) (143), the Fresnel reflection coefficient R (l+1)l = 1 p (l+1)l 1+p (l+1)l = R l(l+1) (144) where R l(l+1) = 1 p l(l+1) (145) 1+p l(l+1) is the reflection coefficient for waves in region l, caused by the boundary separating regions l and l + 1. The reflection coefficient in region l +1, R (l+1) l, caused by the boundary separating regions l+1and l,is equal to the negative of R l(l+1). There are altogether s + t boundaries which give rise to 2(s + t) equations as shown above. There are altogether s + t + 1regions. In regions t and s we have Et = 0 and E s + = 0 because there are no waves originating from infinity. Thus we have a total of

27 Stratified negative isotropic media 27 2(s + t +1) 2=2(s + t) unknowns to be solved from the 2(s + t) equations. The wave amplitudes are related to the configurations and the excitation amplitudes of the dipole antenna in region 0. Thus field amplitudes in Region 0 require special attention. For z 0, we notice that Et = 0. Letting l = 0, we obtain the reflection coefficients R 0+ = E0+ /E+ 0+ in the form of continued fractions. We find R 0+ = E 0+ E + 0+ = ei2k 0zd 1 R (1/R 01 ) 2] e i2(k 0z+k 1z )d 1 (1/R 01 )e i2k 1zd 1 +(E 1 /E + 1 ) (146) where E1 /E+ 1 can be expressed in terms of E 2 /E+ 2 and so on until region t where Et /E+ t =0. For z 0, we notice that E s + =0. Letting l =0, we obtain the reflection coefficients R 0 = E 0 + /E 0 in the form of continued fractions. We find ) ] 2 R 0 = E+ 0 E0 = e i2k 0zd 0 1 (1/R 0( 1) e i2(k 0z+k 1z )d 0 + R 0( 1) (1/R 0( 1) )e i2k 1zd 0 +(E + 1 /E 1 ) (147) where E 1 + /E 1 are expressible in terms of E+ 2 /E 2 and so on until region s, where E s/e + s =0. Once the wave amplitudes in region 0 are found, wave amplitudes in other regions can be determined by the use of propagation matrices, and from a set of dual equations for TE waves. In region 0 it becomes necessary that we distinguish the wave amplitudes in region 0 for z 0 from those in region 0 for z<0. For z>0weusee 0+ +,E 0+ ; and for z<0we use E+ 0,E 0. Thus we have E + 0 = E+x 0 E + 0+ = E+x 0 + E lin E0 = E x 0 + E lin E0+ = E x 0 E lin = ωµ 0 I/4πk 0z (148) where E0 x and E 0 +x characterize contributions due to the stratified medium. Let R 0+ = E 0+ E + 0+ R 0 = E+ 0 E 0 = = E0 x E 0 +x + E lin (149) E 0 +x E0 x + E lin (150)

28 28 Kong we find E +x 0 = R 0 (1+ R 0+ ) 1 R 0 R 0+ E lin (151) E x 0 = R 0+(1+ R 0 ) 1 R 0 R 0+ E lin (152) E + 0 = R 0 (1+ R 0+ ) 1 R 0 R 0+ E lin E + 0+ = 1+R 0 1 R 0 R 0+ E lin E 0 = 1+R 0+ 1 R 0 R 0+ E lin E 0+ = R 0+(1+ R 0 ) 1 R 0 R 0+ E lin (153) We have expressed the solution in region 0 where the linear antenna is located in terms of superpositions of the primary excitations in the absence of the stratified medium and the homogeneous solutions of the stratified medium in the absence of the source Linear Antenna Between Two Negative Isotropic Slabs For a linear antenna situated in-between two negative isotropic slabs, we let µ 2 = µ 0, ɛ 2 = ɛ 0 ; µ 1 = µ 0, ɛ 1 = ɛ 0 ; µ 1 = µ 0, ɛ 1 = ɛ 0, and µ t = µ 0, ɛ t = ɛ 0. We find k 2z = k 2z = k 0z, k 1z = k 1z = k 0z, p ( 2)( 1) = p ( 1)0 = p 01 = p 12 =1,R ( 2)( 1) = R ( 1)0 = R 01 = R 12 = R 0 = R 0+ =0. E 0 + =0 E+ 0+ = E } lin E0 = E lin E0+ =0 (154) From (140) to (143), we find E l + = E l+1 + ei(k (l+1)z k lz )d l+1 (155) El = El+1 e i(k (l+1)z k lz )d l+1 (156) It follows that E + 1 =0 E 1 = E line i(k 0z k 1z )d 0 = E lin e i2k 0zd 0 E 2 + =0 E 2 = E line i(k 1z k 2z )d 1 e i2k 0zd 0 = E lin e i2k 0z(d 0 d 1 ) E 1 + = E line i(k 1z k 0z )d 1 = E lin e i2k } 0zd 1 E1 =0 } } (157) (158) (159)

29 Stratified negative isotropic media 29 E 2 + = E line i(k 2z k 1z )d 2 e i2k 0zd 1 = E lin e i2k } 0z(d 2 d 1 ) E2 =0 (160) The field in Region 2 is E 2x = dk y E ] lin e ik 0zz+2(d 0 d 1 )] e ikyy (161) k ] 2z H 2y = dk y E lin e ik 0zz+2(d 0 d 1 )] e ikyy ωµ 2 k ] y H 2z = dk y E lin e ik 0zz+2(d 0 d 1 )] e ikyy (162) ωµ 2 The field in Region 1is E 1x = dk y E ] lin e ik 0zz 2d 0 ] e ikyy (163) k ] 1z H 1y = dk y E lin e ik 0zz 2d 0 ] e ikyy ωµ 1 k ] y H 1z = dk y E lin e ik 0zz 2d 0 ] e ikyy (164) ωµ 1 In Region 0 for z 0 E 0x = dk y E ] lin e ik lzz e ikyy (165) k ] 0z H 0y = dk y E lin e ik lzz e ikyy ωµ 0 k ] y H 0z = dk y E lin e ik lzz e ikyy (166) ωµ 0 In Region 0 for z 0 E 0x = dk y E ] lin e ik lzz e ikyy (167) k ] 0z H 0y = dk y E lin e ik lzz e ikyy ωµ 0 k ] y H 0z = dk y E lin e ik lzz e ikyy (168) ωµ 0 In region 1 E 1x = dk y E lin e ik 0zz 2d 1 ] ] e ikyy (169)

30 30 Kong Region -2 Region -1 Region 0 Region 1 Region 2 µ 0,ɛ 0 -µ 0, -ɛ 0 µ 0,ɛ 0 -µ 0, -ɛ 0 µ 0,ɛ 0 k z Image Linear Antenna Image z = d -1 z = d 0 z = d 1 z = d 2 Figure 8. Images of a linear antenna in Region 0. In region 2 H 1y = H 1z = E 2x = H 2y = H 2z = dk y k 1z ωµ 1 E lin e ik 0zz 2d 1 ] ] e ikyy dk y k y ωµ 1 E lin e ik 0zz 2d 1 ] ] e ikyy (170) dk y E lin e ik 0zz 2(d 2 d 1 )] ] e ikyy (171) dk y k 2z ωµ 2 E lin e ik 0zz 2(d 2 d 1 )] ] e ikyy dk y k y ωµ 2 E lin e ik 0zz 2(d 2 d 1 )] ] e ikyy (172) It is thus seen that the field in Region 2 is due to a linear antenna situated at z =2(d 2 d 1 ), which is a perfect image of the original line source. Likewise a perfect image is formed in Region 2 and located at z =2(d 1 d 0 ) Linear Antenna in Front of a Negative Isotropic Slab For linear antennas in front of stratified media, all regions 1to s are absent, we have E 0 + = 0 and R 0 = 0. It follows that E 0 + =0 E+ 0+ = E } lin E0 =(1+R 0+)E lin E0+ = R0+E (173) lin

31 Stratified negative isotropic media 31 For a linear antenna in front of a slab medium with boundary surfaces at z = d 1 and z = d 2, we find the reflection coefficient R 0+ = R 01 + R 12 ei2k 1z(d 2 d 1 ) 1+R 01 R 12 e i2k 1z(d 2 d 1 ) ei2k 0zd 1 (174) the amplitudes inside the slab medium E + 1 = 2E lin e i(k 1z k 0z )d 1 (1+ p 01 )(1+ R 01 R 1t e i2k 1z(d 2 d 1 ) ) E1 = 2R 12E lin e i(k 1z k 0z )d 1 e i2k 1zd 2 (1+ p 01 )(1+ R 01 R 1t e i2k 1z(d 2 d 1 ) ) and the transmission coefficient T = 4e ik 0zd 1 e ik 1z(d 2 d 1 ) e ik 2zd 2 (1+ p 01 )(1+ p 1t )(1+ R 01 R 1t e i2k 1z(d 2 d 1 ) ) (175) (176) (177) Consider a linear antenna in front of the slab medium, (A) In region 0; z 0 E 0x = H 0y = H 0z = (B) In region 0; z 0 E 0x = H 0y = H 0z = (C) In region 1 E 1x = dk y (1+ R 0+ )E lin e ik lzz ] e ikyy (178) dk y k 0z ωµ 0 (1+ R 0+ )E lin e ik lzz ] e ikyy (179) dk y k y ωµ 0 (1+ R 0+ )E lin e ik lzz ] e ikyy (180) dk y E lin e ik lzz + R 0+ e ik lzz ] e ikyy (181) dk y k 0z ωµ 0 E lin e ik lzz R 0+ e ik lzz ] e ikyy (182) dk y k y ωµ 0 E lin e ik lzz + R 0+ e ik lzz ] e ikyy (183) dk y E + 1 eik lzz + E 1 e ik lzz ] e ikyy (184)

32 32 Kong k ] 1z H 1y = dk y E 1 + ωµ eiklzz E1 e ik lzz e ikyy (185) 1 k ] y H 1z = dk y E 1 + ωµ eik 1zz + E1 e ik 1zz e ikyy (186) 1 (D) In region 2, k 2z = k 0z E 2x = dk y TE ] lin e ik 2zz e ikyy (187) k 2z H 2y = dk y TE lin e ikzz] e ikyy (188) ωµ 2 k ] y H 2z = dk y TE lin e ik 2zz e ikyy (189) ωµ 2 For a linear antenna in front of a negative isotropic slab in free space, we let µ 1 = µ 0, ɛ 1 = ɛ 0, and µ t = µ 0, ɛ t = ɛ 0. We find k 1z = k 0z, k 2z = k 0z, p 01 = p 12 =1,R 01 = R 12 = R 0+ =0,E1 =0, E lin = ωµ 0 I/4πk 0z, E 1 + = E line i2k 0zd 1, T = e i2k 0z(d 2 d 1 ). (A) In region 0; z 0 E 0x = dk y E lin e ] ik 0zz e ikyy (190) k ] 0z H 0y = dk y E lin e ik 0zz e ikyy (191) ωµ 0 k ] y H 0z = dk y E lin e ik 0zz e ikyy (192) ωµ 0 (B) In region 0; z 0 E 0x = H 0y = H 0z = dk y E lin e ik 0zz ] e ikyy (193) dk y k 0z ωµ 0 E lin e ik 0zz ] e ikyy (194) dk y k y ωµ 0 E lin e ik 0zz ] e ikyy (195) H (1) 0 (kρ) = 1 π = 1 π 1 dk y e ikyy + ik 0z z k 0z { 1 e dk y e ikyy ik 0z z k 0z e ik 0zz } z 0 z 0 (196)

33 Stratified negative isotropic media 33 (C) In region 1, d 1 <z<2d 1, k 0z ik y E 1x = = H (1) 0 dk y E lin e ik 0z(z 2d 1 ) e ikyy ( ) k y 2 + z 2d 1 2 (197) H 1 = 1 (ŷ iωµ 0 z ẑ y )E 1x (198) H 1y = H 1z = dk y k 0z ωµ 0 E lin e ik 0z(z 2d 1 ) e ikyy (199) dk y k y ωµ 0 E lin e ik 0z(z 2d 1 ) e ikyy (200) (D) In region 1, 2d 1 <z<d 2, k y k y, k 0z ik y E 1x = H 1y = H 1z = dk y E lin e ik 0z(z 2d 1 ) e ikyy (201) k 0z dk y E lin e ik 0z(z 2d 1 ) e ikyy (202) ωµ 0 dk y k y ωµ 0 E lin e ik 0z(z 2d 1 ) e ikyy (203) (E) In region 2, d 2 <z<2(d 2 d 1 ), k y k y,,k 0z ik y E 2x = H 2y = H 2z = (F) In region 2, 2(d 2 d 1 ) <z, k 0z ik y E 2x = H 2y = H 2z = dk y E lin e ik 0z(z 2(d 2 d 1 )) e ikyy (204) dk y k 0z ωµ 0 E lin e ik 0z(z 2(d 2 d 1 )) e ikyy (205) dk y k y ωµ 0 E lin e ik 0z(z 2(d 2 d 1 ) e ikyy (206) dk y E lin e ik 0z(z 2(d 2 d 1 )) e ikyy (207) dk y k 0z ωµ 0 E lin e ik 0z(z 2(d 2 d 1 )) e ikyy (208) dk y k y ωµ 0 E lin e ik 0z(z 2(d 2 d 1 ) e ikyy (209) It is seen that in the transmitted region, the field originates from a linear antenna located at z =2(d 2 d 1 ), which is a perfect image of the original antenna.

34 34 Kong 7. DIPOLE ANTENNAS IN STRATIFIED MEDIA 7.1. Hertzian Electric and Magnetic Dipoles The most fundamental model for radiating structures is a Hertzian electric dipole which consists of a current-carrying element with an infinitesimal length l. Denoting the current dipole moment with Il, the current density J(r) of a Hertzian dipole pointing in the ẑ direction and located at the origin is J(r )=ẑil δ(r ) (210) The exact expressions for the electric field vector E(r) for the Hertzian dipole is calculated to be E(r) = iωµ I + 1 ] k 2 d r e ik r r 4π r r ) ẑilδ(r = iωµil ẑ + 1 k 2 ] e ikr z 4πr = iωµil { ẑ 1 4π r eikr + 1 k 2 (ˆx x +ŷ y +ẑ z ) ( ikz r 2 + z ]} r 3 )eikr = iωµil { ẑ 1 1+ i 4π r kr 1 ] k 2 r 2 ˆx xz r i kr 3 ] k 2 r 2 ŷ yz r i kr 3 ] k 2 r 2 ẑ zz r i kr 3 ]} k 2 r 2 e ikr = iωµil { ẑ 1+ i 4πr kr 1 ] k 2 r 2 ẑ zz r i kr 3 ] k 2 r 2 ˆρ zρ r i kr 3 ]} k 2 r 2 e ikr { = iωµeikr Il ẑ 1+ i 4πr kr 1 ] k 2 r 2 ˆr z 1+ 3i r kr 3 ]} k 2 r 2 (211) Notice that with g(r) =e ikr /4πr and g(r)/ z =(ik 1/r) cos θg(r). To cast in spherical coordinates, note that ẑ = ˆr cos θ ˆθ sin θ and z = r cos θ. We find from (211) { ( E(r) = iωµeikr i i Il ˆr 4πr kr + kr ) ] 2 2 cos θ+ ˆθ 1+ i ( ) ] } i 2 kr + sin θ kr (212) The magnetic field follows from Faraday s law H(r) = 1 eikr E = ˆφikIl 1+ i ] sin θ (213) iωµ 4πr kr

35 Stratified negative isotropic media 35 The complex Poynting power density is calculated by taking the cross product of E and the complex conjugate of H ] { S = E H kil 2 ( ) ] i 3 ( ) ( ) ] } =η ˆr 1 sin 2 θ 4πr kr ˆθ i i 3 sin 2θ kr kr (214) The time-average Poynting power density is <S>= 1 2 Re{S} =ˆr η ] kil 2 sin 2 θ (215) 2 4πr Let the electric current moment be in a general direction, we write Il =ˆxI x l +ŷi y l +ẑi z l. From (211), (212), and (213) we find E(r) = iωµeikr 4πr H(r) =ˆr Il ikeikr 4πr { ( Il 1+ i 1+ i kr kr 1 k 2 r 2 ] ) ˆr(ˆr Il) 1+ 3i kr 3 ]} k 2 r 2 (216) (217) Notice that ˆr = ẑz/r + ˆρρ/r = ˆxx/r + ŷy/r + ẑz/r. The Poynting vector is ] S = E H k 2 = η 4πr { ˆr(Il) 2 (1+ i k 3 ) ˆr(ˆr Il)2 1+ 2i r3 kr + 3i ] 2i k 3 r 3 +(ˆr Il)Il kr + 2i ]} k 3 r 3 (218) The time-average Poynting power density is <S>= 1 2 Re{S} =ˆrηk2 (Il)2 (ˆr Il) 2 ] 2(4πr) 2 (219) In a negative isotropic medium when k = k the phase velocity of the radiated wave points towards the dipole while the Poynting power is propagating in the direction of increasing r. The dual of a Hertzian electric dipole is a magnetic dipole. A Hertzian magnetic dipole can be realized with the model of a small current loop with area A and carrying current I. The correspondence between the electric and magnetic dipoles can be quantified by letting the Hertzian dipole moment Il to be 7] (Il) e =(ikia) m (220)

36 36 Kong Region s Region s+ 1 Region 1 Region 0 Region 1 Region t 1 Region t ρ z dipole antenna z =d -s + 1 z =d -s + 2 z = d 1 z = d 0 z = d 1 z = d 2 z = d t 1 z = d t Figure 9. Hertzian dipole source in stratified media. E m = ηh e (221) H m = E e η (222) where the subscripts e and m are used to denote the electric dipole and the current loop, respectively. The solutions for a small current loop thus follow from the solutions for electric dipoles Integral Formulation for Dipoles in Stratified Media The geometrical configuration of the problem is shown in Figure 9. The origin of the coordinate system is placed in the location of the dipole which can be a z-directed electric dipole (ZED), a z-directed magnetic dipole (ZMD), an x-directed electric dipole (XED), an y-directed electric dipole (YED), an x-directed magnetic dipole (XMD). or an y- directed magnetic dipole (YMD). There are t layers at z = d 1,d 2,...d t and s+1layers at z = d 0,d 1,...,d s+1. We shall first assume that all regions contain isotropic media. In region l, we denote the permittivity and permeability by ɛ l and µ l. Notice that in region 0,ɛ 0, and µ 0 are not necessarily equal to the free space permittivity and permeability which we denote by ɛ o and µ o. Making use of cylindrical coordinate system, the integrands of transverse field components E s =ˆρE ρ + ˆφE φ and H s =ˆρH ρ + ˆφH φ

37 Stratified negative isotropic media 37 are derived from those of the longitudinal components E z and H z. Let E z = H z = E z ( ) (223) H z ( ) (224) We have from the Maxwell equations in source-free regions E s ( )= 1 ] kρ 2 s z E z( )+iωµ l s H z ( ) (225) H s ( )= 1 ] kρ 2 s z H z( ) iωɛ l s E z ( ) (226) The fields of a dipole radiating in unbounded space with permittivity ɛ 0 and µ 0 can be transformed from spherical coordinates to cylindrical coordinates by using the Sommerfeld identity e ik 0r r = i 2 k 0z H (1) 0 (ρ)e ik 0z z (227) From the previsous section, we find for a Hertzian electric dipole Ilδ(r) situated at the origin, E(r) = iωµ 4π = ωµ 8π H(r) = 1 iωµ E = I + 1 ] k 2 Il eikr r I + 1 ] k 2 Il i 8π Il Noticing that H (1) 0 ( ρ)= H (1) 1 (ρ), we find (A) z-directed electric dipole (ZED): Il =ẑil H z =0; E z = E zed { e ik 0z z e ik 0zz k 0z H (1) 0 (ρ)e ik 0z z k 0z H (1) 0 (ρ)e ik 0z z } H (1) 0 (ρ) (228) (229) z 0 z 0 (230) with E zed = Ilk3 ρ (231) 8πωɛ 0 k 0z where Il is the electric dipole moment.

38 38 Kong (B) x-directed electric dipole (XED): Il =ˆxIl { } e ik 0z z E z = E xed H (1) 1 (ρ) cos φ H z = e ik 0zz } H xed { e ik 0z z e ik 0zz H (1) 1 (ρ) sin φ z 0 z 0 (232) z 0 z 0 (233) with E xed = i Ilk2 ρ 8πωɛ 0 ; H xed = i Ilk2 ρ 8πk 0z (234) (C) y-directed electric dipole (YED): Il =ŷil { Eyed e ik } 0zz E z = E yed e ik 0zz { } e ik 0z z H z = H yed e ik 0zz with E yed = E xed = i Ilk2 ρ 8πωɛ 0 ; H (1) 1 (ρ) sin φ z 0 z 0 (235) H (1) 1 (ρ) cos φ z 0 z 0 (236) H yed = H xed = i Ilk2 ρ 8πk 0z (237) The results of the y-directed electric dipole (YED) can be obtained from that for XED by replacing φ with π/2+φ. Notice that the magnetic dipoles produce fields which are duals of those produced by the corresponding electric dipoles. The results for the magnetic dipoles ZMD, XMD, and YMD, can be obtained by the replacement E H, H E, µ 0 ɛ 0, and Il iωµ 0 IA. We note in particular that at z = 0, the following field components vanish: (A) z-directed electric dipole (ZED) (B) x-directed electric dipole (XED) E ρ = 0 (238) E z = H ρ = H φ = 0 (239) This is seen from (225) (226) and by noting from (227) that e ik 0r =0 at z = 0 (240) z r We now consider dipole sources placed in Region 0 of the stratified isotropic medium (Fig. 9). We assume that all regions contain isotropic

39 Stratified negative isotropic media 39 media. In region l, we denote the permittivity and permeability by ɛ l and µ l. The solutions to the wave equations can be written as superpositions of TE and TM wave components. Let E l + and El denote amplitudes for the TM waves and H l + and Hl denote amplitudes for the TE waves. We find in region l the following solutions: (A) z-directed electric dipole (ZED): Il =ẑil ] E lz = E l + e iklzz + El e ik lzz H (1) 0 (ρ) (241) ik ] lz E lρ = E l + e iklzz El e ik lzz H (1) 0 ( ρ) (242) iωɛ ] l H lφ = E l + e iklzz + El e ik lzz H (1) 0 ( ρ) (243) (B) x-directed electric dipole (XED): Il =ˆxIl E lz = E lρ = E lφ = H lz = H lρ = + H lφ = E + l e ik lzz + E l e ik lzz ] H (1) 1 (ρ) cos φ (244) ik ] lz E l + e iklzz El e ik lzz H (1) 1 ( ρ) cos φ iωµ l kρρ 2 ik lz kρρ 2 H + l e ik lzz + H l e ik lzz ] H (1) 1 (ρ)cos φ] (245) E + l e ik lzz E l e ik lzz ] H (1) 1 (ρ) sin φ] iωµ l H + l e ik lzz + H l e ik lzz ] H (1) 1 ( ρ) sin φ (246) H + l e ik lzz + H l e ik lzz ] H (1) 1 (ρ) sin(φ) (247) ik lz H + l e ik lzz H l e ik lzz ] H (1) 1 ( ρ) sin φ iωɛ l kρρ 2 ik lz kρρ 2 E + l e ik lzz + E l e ik lzz ] H (1) 1 (ρ) sin φ] (248) H + l e ik lzz H l e ik lzz ] H (1) 1 (ρ)cos φ] iωɛ l E + l e ik lzz + E l e ik lzz ] H (1) 1 ( ρ) cos φ (249)

40 40 Kong where H n (1) ( ρ)isthenth order Hankel function of the first kind and H n (1) ( ρ) denotes the derivative of H n (1) (ξ) with respect to its argument ξ. The integrands of transverse field components E s = ˆρE ρ + ˆφE φ and H s = ˆρH ρ + ˆφH φ are derived from those of the longitudinal components E z and H z. In region 0 it becomes necessary that we distinguish the wave amplitudes in region 0 for z 0 from those in region 0 for z < 0. For z>0weusee 0+ +,E 0+,H+ 0+, and H 0+ ; and for z<0weuse E 0 +,E 0,H+ 0, and H 0. (A) z-directed electric dipole (ZED) E + 0 = E+z 0 E + 0+ = E+z 0 + E zed E0 = E z 0 + E zed E0+ = E z 0 H 0 + = H 0 =0 H+ 0+ = H 0+ =0 (250) where E0 z and E 0 +z characterize contributions due to the stratified medium. Let we find R TM 0+ = E 0+ E + 0+ R TM 0 = E+ 0 E 0 = = E0 z E 0 +z + E zed (251) E 0 +z E0 z + E zed (252) E 0 +z = RTM 0 (1+ RTM 0+ ) 1 R0 TM E zed (253) RTM 0+ E0 z = RTM 0+ (1+ RTM 0 ) 1 R0 TM E zed (254) RTM 0+ E 0 + = RTM 0 (1+ RTM 0+ ) 1 R0 TM E zed E + RTM 0+ = 1+RTM R0 TM E zed RTM 0+ E0 = 1+RTM 0+ 1 R0 TM E zed E RTM 0+ = RTM 0+ (1+ RTM 0 ) (255) 0+ 1 R0 TM E zed RTM 0+ H 0 + = H 0 =0 H+ 0+ = H 0+ =0 (B) x-directed electric dipole (XED)