ECE 6345 Spring 215 Prof. David R. Jackson ECE Dept. Notes 1 1
Overview In this set of notes we derive the far-field pattern of a circular patch operating in the dominant TM 11 mode. We use the magnetic current model. 2
Circular Patch: TM 11 Mode y a x z ( ρφ, = cosφ ( ρ E A J k 1 This corresponds to a probe on the x axis. x 11 k = a x 11 = 1.841 3
Circular Patch (cont. Magnetic current model: M = nˆ Eˆ s = ˆ ρ Eˆ (, cos ( M E a A J ka sφ = z φ = φ 1 Choose 1 A = J ka M cos s φ = 1 ( φ ( V = h at patch edge on axis [V] x 4
Far Field of Circular Patch Reciprocity setup: z θ ε r r M s φ H i ˆp h x y M s φ a φ x 5
Far Field of Circular Patch (cont. Far-field: ( θφ FF E r,, =< ab, > p =< ba, > S S a s = H M ds = = ( ρ φ H,, z cosφ ds φ h ( H a, φ, z cosφ a dz dφ φ The primes here denotes source coordinates. ( ( FF E r, θφ, H a, φ, z cosφ a dz dφ p = h φ ( sin ( cos = x + y Hφ H φ H φ 6
Far Field of Circular Patch (cont. Inside the substrate we have (see Notes 9: ( ( ( x + kyy ( ( jkx xy, xy, z1 z1 H x, y, z = H,, e sec k h cos k ( z + h ( k z = k N ( 1 1 θ The exponent term may be put in cylindrical coordinates as follows: x y ( sinθcosφ( cosφ ( sinθsinφ( sinφ k asinθ( cosφcosφ sinφsin φ' k sinθcos( φ φ kx + ky = k a + k a = + = 7
Far Field of Circular Patch (cont. Hence ( cos ( φ sec z1 z1 H = k h k z + h e ( jka ( sinθcos φ' φ sin φ Hx (,, + cos φ Hy (,, Since the horizontal magnetic field components are modeled as current in the TEN, we have i (,, = (,,( 1 Γ( H H θ xy, xy, p = θ : TM, p= φ: TE 8
Far Field of Circular Patch (cont. TM ( z ˆp = θˆ H H i x i y E (,, = ˆ φ xˆ = ( sinφ η E (,, = ˆ φ yˆ = ( cosφ η E η E η TE z ( ˆp = φˆ H i x E η (,, = ˆ θ xˆ = ( cosθcosφ E η H i y E η (,, = ˆ θ yˆ = ( cosθsinφ E η 9
Far Field E θ TM z ( ˆp = θˆ Substituting for H x and H y, we have ( cos ( φ sec z1 z1 H = k h k z + h e jka ( sin cos E Γ η ( θ φ φ TM [ sinφ sinφ cosφ cosφ]( 1 ( θ Note: [ ] = cos( φ φ Hence, we have FF,, E E r = sec k h cos k ( z + h e θ ( θφ ( ( TM ( ( jka ( sinθcos φ φ h z1 z1 η 1 Γ ( θ cos( φ φ cosφ a dz dφ 1
For the z integral we have Far Field E θ (cont. ( sec k h cos k ( z + h dz = htanc( k h z1 h z1 z1 so that FF E (,, tanc( ( TM E θ r θφ = a h kz1h 1 Γ ( θ I η TM where 2 jq cos( I π e φ φ cos φ φ cosφ dφ TM ( q ( k a sin θ Let φ = φ φ φ jq cos( φ I = e cos φ cos φ + φ dφ TM φ ( ( 11
Far Field E θ (cont. We have that ( cos φ + φ = cosφ cosφ sinφ sinφ and 2 π φ 2 π φ ( dφ = ( dφ so that ( cos( jqcos( φ ITM e cos + d = cos φ φ φ φ jqcosφ 2 φ e cos φ dφ cos sin jq φ e sin cos φ φ φ dφ Now use 2 2 cos φ = 1 sin φ Integrates to zero (odd function 12
Far Field E θ (cont. jqcosφ jqcosφ 2 ITM = cosφ e dφ cosφ e sin φ dφ Now we use the following identity: n+ 1 1 2 π Γ n + 2 sin ( n q jqcosφ 2n e φ dφ = J n q n =,1, 2 where 1 Γ = π 2 3 1 Γ = π 2 2 13
Hence and Far Field E θ (cont. jqcos φ" e dφ = π J q e 2 ( jqcos φ" 2 1 (n = J ( q sin φ dφ = q (n = 1 and thus φ π J1( q ITM = cos (2 J ( q q Next, use so that n J n x = Jn 1 x Jn x x ( ( ( ( ( J x = J x 1 J 1 ( x x 14
Far Field E θ (cont. Hence ITM = cos φ J ( q 1 The far field is then FF E ( Eθ r, θφ, = ( ah tanc( kz1h Q( θ η cos φ J ( kasin θ 1 where TM Q( θ = 1 Γ ( θ 15
TE z ( ˆp = φˆ Far Field E φ Performing similar steps, we have ( ( ( TE jka ( sin θ cos ( φ φ θ H = sec k h cos k z h 1 Γ ( e φ z1 z1 E η [ sinφ cosθcosφ cosφ cosθsinφ] + Using reciprocity and performing the integration in z, we have ( ( TE, θφ, = ( tanc( z1 1 Γ ( θ FF Eφ r a h k h η E ( cos( cos jq φ φ e sin cos θ φ φ φ dφ 16
Far Field E φ (cont. Evaluating the integral, we have jqcos( φ φ ITE e sin( cos d φ φ φ φ ( jqcosφ = e sin cos + φ φ φ dφ [ ] jqcosφ = e sinφ cosφ cosφ sinφ sinφ dφ jqcosφ jqcos φ" 2 = cosφ e sinφ cosφ dφ sinφ e sin φ dφ J ( q q 1 = sinφ integrates to zero (odd function 17
Far Field E φ (cont. Hence E ( ( ( kasinθ FF 1 Eφ r, θφ, = ( ah tanc kz1h sinφ 2 π P( θ η ka sinθ J where ( TE Γ P( θ cosθ 1 ( θ 18