ECE b Applied Electromagnetics Notes Set 4b Instructor: Prof. Vitali Lomain Department of Electrical and Computer Engineering Universit of California, San Diego 1
Uniform Waveguide (1) Wave propagation in the + direction: [ ˆ ] E= E + E ˆ = e (, ) + e (, ) e t t j [ ˆ ] H = H + H ˆ = h (, ) + h (, ) e t t Mawell s equations: + t ˆ ( Et + E ˆ ) = jωµ ( Ht + H ˆ ) j ε, µ H = jωε E ˆ E + j ˆ E = jωµ H t t t E = jωµ Hˆ t t ˆ H + j ˆ H = jωε E t t t H = jωε Eˆ t t
Uniform Waveguide () Transverse field components: E H 1 = ( jωµ ˆ H j E ) t t t t 1 = ( jωε ˆ E j H ) t t t t We onl have to deal with the longitudinal components! = t = ω µε Equations for the longitudinal components: 1 ( ˆ ) ˆ t ωε te + th = ωε E t 1 t ( ωµ ˆ ) ˆ th te = ωµ H t 3
Uniform Waveguide (3) Consider two cases: Case A: Homogeneousl filled waveguide: = is independent of and t E+ E= in Ω t t H + H = in Ω t t What is the implication??? E and H can eist independentl!!! E H =, TM modes E = H, TE modes 4
Uniform Waveguide (4) Wh TM or TE? Wh not TEM? TEM can onl eist in waveguides made of two separate conductors. Eample: coaial waveguide. What s the consequence of have either E or H? 1 S= E H is not directl in the direction. Zigag propagation. Phase velocit is greater than c. Energ velocit is smaller than c. What happens at cutoff? 5
Uniform Waveguide (5) Case B: Inhomogeneousl filled waveguide: = is a function of and t 1 ( ˆ ) ˆ t ωε te + th = ωε E t 1 t ( ωµ ˆ ) ˆ th te = ωµ H t Consequence: E and H must co-eist: Hbrid modes. What if the filling is piecewise homogeneous? E and H are coupled b the discontinuous interface in order to satisf the boundar conditions. 6
Uniform Waveguide (6) To anale TM modes: j Solve te + E t = in Ω then Et = te t jωε E = on Γ H = ˆ E To anale TE modes: t t H n = on Γ t t t jωµ Solve H + H = in Ω then E = ˆ H H t t t j = H t t t 7
Uniform Waveguide (7) General characteristics: Propagation constant: ω µε t ω µε > t = ω µε = t j t ω µε ω µε < t Propagation Cutoff c = Attenuation t Phase, group, and energ velocities: v p ω c = = 1 ( ) c v g 1 d c = = c 1 dω v e P c = = c 1 W + W e m 8
Uniform Waveguide (8) Guided wavelength: π λ λ g = = 1 ( ) c Wave impedance in the -direction: Resistive f > f TM c Zw = = η 1 = f = f ωε Capacitive f < f c c c Resistive f > f TE ωµ η Zw = = = f = f 1 ( c ) Inductive f < f c c c 9
Uniform Waveguide (9) c What s the meaning of 1? E E H H c sinθ = c θ c cosθ 1 = Mode orthogonalit: 1. All TM modes are orthogonal to each other. All TE modes are orthogonal to each other 3. TM and TE modes are mutuall orthogonal 1
Rectangular Waveguide (1) An empt, infinitel long waveguide: 1. TE modes H + H = t t b ε, µ a or Let H H H + + t = H(,, ) = f( ) g ( ) e j f g g f fg + + t = 11
Rectangular Waveguide () 1 f 1 g + + t = f g Consider 1 f 1 g f g =, = 1 f f = + f = f + = t j Two independent solutions:, e j e or sin, cos Similarl: f( ) = Ccos + Dsin 1 1 g( ) = C cos + D sin 1
Rectangular Waveguide (3) General solution: ( )( ) H(,, ) = Ccos + Dsin C cos + Dsin e j 1 1 Boundar condition: n H = E = E = E = H = = D 1 = E = a H = a = sin a= a = mπ mπ = m =,1,,... 13 a
Rectangular Waveguide (4) H = = D = H = b = sin b= b = nπ nπ = n= b,1,,... jωµ nπ mπ nπ jmn E = Hmn cos sin e b a b tmn jωµ mπ mπ nπ jmn E = Hmn sin cos e a a b tmn E = 14
Rectangular Waveguide (5) jmn mπ mπ nπ jmn H = Hmn sin cos e a a b tmn jmn nπ mπ nπ jmn H = Hmn cos sin e b a b tmn mπ nπ jmn H = Hmn cos cos e a b Consider : tmn mπ nπ = + a b mn mπ nπ = a b m, n =,1,, ecept for m = n = 15
Rectangular Waveguide (6) TE modal field distribution: TE TE 1 1 TE TE TE TE3 11 1 16
Rectangular Waveguide (7). TM modes General solution: E+ E= t t ( )( ) E(,, ) = Ccos + Dsin C cos + Dsin e j 1 1 Boundar condition: E = E = E = E = E = C = = 1 E = a 17
Rectangular Waveguide (8) E = = sin a= a E = C = = a = mπ mπ = m= 1,,... a E = = sin b= b b = nπ nπ = n= 1,,... b mπ nπ jmn E = Emn sin sin e a b 18
Rectangular Waveguide (9) jmn mπ mπ nπ jmn E = Emn cos sin e a a b tmn jmn nπ mπ nπ jmn E = Emn sin cos e b a b tmn jωε nπ mπ nπ jmn H = Emn sin cos e b a b tmn jωε mπ mπ nπ jmn H = Emn cos sin e a a b tmn H = 19
Rectangular Waveguide (1) TM modal field distribution: TM TM 11 1 TM31 TM TM 1 41 TM
Rectangular Waveguide (11) > + mπ nπ mπ nπ ( a ) ( b ) ( a ) ( b ) mπ nπ mn = = ( a ) + ( b ) j ( ) + ( ) < ( ) + ( ) mπ nπ mπ nπ a b a b Define mπ nπ = ( a ) + ( b ) cutoff wavenumber 1 = ( ) + ( ) cutoff frequenc π µε mπ nπ f a b π π λ = = mπ nπ ( a ) + ( b ) cutoff wavelength 1
Rectangular Waveguide (1) Guide wavelength: mn = π λ gmn π π π λ λ λ gmn = = = = = mn λ 1 1 1 f ( ) ( λ ) ( f ) Phase velocit: v pmn Energ velocit: v emn ω ω c = = = > mn λ 1 P = = c 1 W + W e m ( λ ) λ ( ) λ c
Rectangular Waveguide (13) Group velocit: v gmn 1 1 1 1 = = = = d mn d d 1 dω d dω ω 1 ( ) ( λ ) λ = c 1 = c 1 < c ( ) µε 3
Rectangular Waveguide (14) Wave impedance: Z w E H = = Real f > fc or λ > λc TE η η Zwmn = = = f = f or ( ) ( ) c λ = λc λ f 1 λ 1 Imaginar or f f < fc λ < λc E H Real f > fc or λ > λc ( ) ( ) TM λ f Zwmn = η 1 λ = η 1 or f = f = fc λ = λc Imaginar f < fc or λ < λc 4
Rectangular Waveguide (15) Consider f 1 mπ nπ = + π µε a b m,n =,1,, ecept for m = n = TE 1 TE,TE 1 Cutoff Single mode 1.4.83 TE 11,TM 11 TE 1 f f c1 Define the dominant mode as the mode having the lowest cutoff frequenc. Assume a > b, the dominant mode: TE m= 1, n= 1 1 π f 1 = 1 = λ 1 = a c a µε c a c 5
Rectangular Waveguide (16) TE 1 modal field: jωµ a π j1 E = H1 sin e π a j a π 1 sin π a 1 j H = H e cos π j1 H = H1 e a 1 J s = nˆ H Surface current distribution 6
Rectangular Waveguide (17) Epress the modal field in terms of plane waves: H E = ˆ e e η 1 j( ) j( + ) 1 1 1 1 1 Superposition of two plane waves! E H E H 7