ECE Spring Prof. David R. Jackson ECE Dept. Notes 33

Transcription

1 C 6345 Spring 2015 Prof. David R. Jackson C Dept. Notes 33 1

2 Overview In this set of notes we eamine the FSS problem in more detail, using the periodic spectral-domain Green s function. 2

3 FSS Geometry Reflected plane wave Incident plane wave z ( 1, 0) ( 0, 1) ( 0,0) ( 0,1) y ( 1, 0) ( mn, ) h L W ε r b a Dielectric layer Metal patch Transmitted plane wave 3

4 FSS Geometry (cont.) ψ ( ) 0 y0 z 0 = A e ( θ, φ ) 0 0 jk + k y + jk z e = arrival angles k = k sinθ cosφ k = k sinθ sinφ y k = k cosθ z0 0 0 Reflected plane wave Incident plane wave z ( 1, 0) ( 0, 1) ( 0,0) ( 0,1) y ( 1, 0) ( mn, ) h L W ε r b a Dielectric layer Metal patch Transmitted plane wave 4

5 FSS Analysis Assume that the unknown current on the (0,0) patch is of the following form: The FI is then (, ) = (, ) J y A B y s π B ( y, ) = cos, < L/2, y< W/2 L imp A B + = 0, < L/ 2, y < W / 2 Note that the superscript stands for infinite periodic (i.e., the fields due to the infinite periodic array of patch currents). The superscript imp denotes the impressed field (seen by the patches) that eists in the absence of the metal patches. That is, the ident plane-wave field plus that which reflects from the dielectric layer. The FI is enforced on the (0,0) patch; it is then automatically enforced on all patches. 5

6 We have, using Galerkin s method, 0 0 ( ) A B (, y) B ds + B, y ds = 0 imp S S Define Z = B (, y) B ds S 0 S 0 (, ) imp R = B y ds We then have A Z = R 6

7 The (0,0) patch current amplitude is then A = R Z For the patch self reaction we have 1 Z = G ( k, k ) B ( k, k ) 2 dk dk ( 2π ) 2 y y y Single patch 1 Z = G k k B k k ab 2 (, ) (, ) p yq p yq p= q= Periodic patches 7

8 For the RHS term we have S 0 (, ) imp R = B y ds The impressed field as a function of (,y) can be written as imp jk ( 0+ ky0y) y,,0 = e 0,0,0 1+Γ ( ) ( )[ ] where Γ=Γ Γ=Γ T T idence idence This gives us jk ( 0+ ky0y) R = ( 0,0,0)[ 1 +Γ] B (, y) e ds S 0 8

9 Hence, we have ( 0,0,0)[ 1 ] (, ) R = +Γ B k k 0 y0 where L cos k π W B k k LW k 2 2 π L k 2 2 ( ) 2, y = s y 2 2 Γ=Γ Γ=Γ T T idence idence 9

10 For the ident field we have ( ) = θ ( ) θ ( 0,0,0) = ( 0,0,0)( sinφ ) 0,0,0 0,0,0 cos cos φ φ idence T idence In a typical scattering problem we would usually choose θ φ ( ) ( ) 0,0,0 = 1 0,0,0 = 1 idence T idence 10

11 We now calculate the FSS reflection coefficient. The field radiated by the patch currents for z > 0 is 1 ( + ) 0, y, z A G k, k ;0,0 B k, k e e ( ) ( ) ( ) jk k y jk z y z = 2 ( 2π ) Single patch y y 1 pq ( p + yq ) z 0 (, y, z) = A G ( kp, kyq;0,0 ) B ( kp, kyq ) e e ab p= q= jk k y jk z Periodic patches where ( ) 1/2 k = k k k pq z0 0 p yq 11

12 The field of the specular-reflected (0,0) wave radiated by the patches for z > 0 is ref, patch 1 ( 0 + y0 ) z 0 (, y, z) = A G ( k0, ky0;0,0 ) B ( k0, ky0) e e ab jk k y jk z The total specular reflected field is ref, tot ( ) yz = Γe e (,, ) ( 0,0,0) where jk + k y jk z 0 y0 z 0 1 ( ) + A G ( k0, ky0;0,0 ) B ( k0, ky0) e e ab jk + k y jk z 0 y0 z 0 Γ=reflection coefficient from the layer (without the patches) 12

13 The total FSS reflection coefficient is defined from the (0,0) fields as Γ FSS ref, tot ( 0,0,0) ( 0,0,0) Note: The total reflected wave will not in general be perfectly z or T z (unless we are in the pripal planes). The total FSS reflection coefficient is then Γ FSS =Γ+ 1 1 A G k k B k k ab ( 0,0,0) ( 0, 0;0,0 ) ( 0, 0) y y Γ=Γ Γ=Γ T T idence idence ( ) = θ ( ) θ ( 0,0,0) = ( 0,0,0)( sinφ ) 0,0,0 0,0,0 cos cos φ φ idence T idence 13

14 For the layer reflection coefficient we have Γ=Γ Γ=Γ T T idence idence Γ Γ= Z Z in in Z Z 0 0 ( z ) ( z ) Z L + jz1tan k 1h Zin = Z1 Z 1+ jz Ltan k 1 h h Z1 ZL = Z 0 Z 0 kz0 = ωε 0 Z ωµ = T 0 0 kz0 Z 1 k ωµ z1 T 0 = Z1 = ωε 0ε r kz k = k k k k = k k k k1 = k0 ε r z0 0 0 y z1 1 0 y0 14

15 We also have A = R Z ( 0,0,0)[ 1 ] (, ) R = +Γ B k k 0 y0 1 Z = G k k B k k ab 2 (, ) (, ) p yq p yq p= q= ( ) = θ ( ) θ ( 0,0,0) = ( 0,0,0)( sinφ ) 0,0,0 0,0,0 cos cos φ φ idence T idence 15

16 FSS quivalent Circuit Approimate equivalent circuit of patch FSS A periodic array of metal patches forms a capacitive FSS. L C The periodic patch array is modeled as a shunt susceptance in the TN, consisting of an L and a C. Note: There could also be a layer present. 16

17 FSS quivalent Circuit Approimate equivalent circuit of slot FSS A periodic array of slots forms an inductive FSS. L The periodic slot array is modeled as a shunt susceptance in the TN, consisting of an L and a C. C Note: There could also be a layer present. 17

18 At circuit resonance: FSS quivalent Circuit A patch (capacitive) FSS reflects all of the wave. A slot (inductive) FSS transmits all of the wave. Patch FSS L C Slot FSS L C 18

19 Results: FSS Patch Array in Free Space C. C. Chen, Scattering by a two-dimensional array of conducting patches, I Trans. Antennas and Propagation, pp , Sept Note the total reflection at the resonance frequency! 19