region 0 μ 0, ε 0 d/2 μ 1, ε 1 region 1 d/2 region 2 μ 2, ε 2

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1 W.C.Chew ECE 35 Lecture Note 4. Dielectric Waveguie (Slab). When a wave i incient from a meium with higher ielectric contant at an interface of two ielectric meia, total internal reection occur when the angle of incient i larger than the critical angle. Thi fact can be ue to make wave bouncing between two interface of a ielectric lab to be guie x μ, ε / θ region μ, ε y θ θ z region μ, ε / θ region Since total internal reection occur for both TE an TM wave, guiance i poible for both type of wave I. TE Cae E = ^ye y E y i a olution to the wave equation in each region. In region, we aume a olution of the form E y = E e ;j xx;j zz () x + z =! = : In region, we aume a olution of the form (a) E y =[A e ;j x + B e j x ]e ;jzz () + z =! = : In region, the olution i of the form (a) E y = E e j xx;j zz (3) x + z =! = : (3a)

2 We aume that all the olution in the three region to have the ame z- variation of e ;jzz by the phae matching conition. B In region, we have an up-going wave awell a a own-going wave. The two wave have to be relate by the reection coecient? for the electric el at the bounarie.? i erive earlier in the coure. Therefore at x =, we have e j =? A e ;j (4)? i the reection coecient at the region an interface. At x = ;, we have A e j =? B e ;j (5)? i the reection coecient at the region an interface. Multiplying equation (4) an (5) together, we have, A an B are non-zero only if A B e j =?? A B e ;j : (6) =?? e ;j : (7) The above i known a the guiance conition of a ielectric lab waveguie. If meium 3iequal to meium, then? =?,anthe guiance conition become =? e;j : (8) From before, for a wave incient at an angle,? = co ; co : (9) co + co Since = co, x = co, (9) coul be written a? = ; x = ; x + : () x + x Taking the quare root of (8), we have? e ;j = : () When we chooe the plu ign, B = A from (4), an from () E y =A co( x)e ;jzz ) even in x: () When we chooe the minu ign in () we have B = ;A, an E y = ;ja in( x)e ;jzz ) o in x: (3)

3 Multiplying () by e j an manipulating, we have tan = j x even olution (4) cot = j x o olution: (5) Subtracting (a) from (a) an olving for x, we have x =[! ( ; )+ ] : (6) In orer for (4) an (5) to be atie, x ha to be pure imaginary. In other wor, the wave in region an 3 have to be evanecent an ecay exponentially away from the lab. Hence x = ;j x = ;j[! ( ; ) ; ] (7) an (4) an (5) become tan = x =! ( ; ) even olution (8) ; cot = x =! ( ; ) o olution: (9) We can olve the above graphically by plotting y = tan even olution () y = ; cot o olution () " # y 3 =! ( ; ) = x : () ω(μ ε μ ε ) / y even y y y y 3 o even o μ μ TE π/ TE π 3π β 3

4 y 3 i the equation of a circle the raiu of the circle i given by!( ; ) : (3) The olution to (8) an (9) are given by the interection of y 3 with y an y. We note from (3) that the raiu of the circle can be increae in three way (i) by increaing the frequency, (ii)by increaing the contrat,an (iii) by increaing the thickne of the lab. When x = ;j x, the reection coecient i? = + j x = exp +j tan ; x (4) ; j x an j? j =. Hence there i total internal reection an the wave i guie by total internal reection. Cut-o occur when the total internal reection ceae to occur, i.e. when the frequency ecreae uch that x =. From the iagram, we ee that x =when or!( ; ) = m m = 3 ::: (5)! mc = m m = 3 ::: : (6) ( ; ) The moe that correpon to the m-th cut-o frequency above i labele the TE m moe. TE moe i the moe that ha no cut-o or propagate at all frequencie. At cut-o, x =,an from (a), z =! p (7) for all the moe. Hence, both the group an the phae velocitie are that of the outer region. Thi i becaue when x =, the wave i not evanecent outie, an mot of the energy of the moe i carrie by the exterior el. When!!,! n z = from the iagram for all the moe. From (a), q! ;! p!!: (8) Hence the group an phae velocitie approach that of the ielectric lab. Thi i becaue when!!, x!, an all the el are trappe in the lab an propagating within it. Becaue of thi, the iperion iagram of the ierent moe appear a below. 4

5 β z ω μ ε ΤΕ TE 3 TE ΤΕ ω μ ε ω c ω c ω 3c ω II. TM Cae H = ^yh y For the TM cae, a imilar guiance conition analogou to (7) can be erive = k k e ;j (9) i the reection coecient for the TM el. Similar erivation how that the above guiance conition, for =, =, reuce to tan =! ( ; ) even olution (3) ; cot =! ( ; ) o olution: (3) Note that for equation (7) an (9), when we have two parallel metallic plate, k =,an? =, an the guiance conition become =e ;j which i what we have oberve before. ) = m m= ::: (3) 5