ECE 634 Spring 6 Prof. David R. Jackson ECE Dept. Notes
Fields in a Source-Free Region Example: Radiation from an aperture y PEC E t x Aperture Assume the following choice of vector potentials: A F = = A ˆ ( x, y, ) F ˆ ( x, y, ) A + k A = F+ kf=
Introduce the Fourier transform: jkx ( x ky y ) A + ( k, k, ) = A ( x, y, ) e dxdy x y jkx ( x + ky y ) (,, ) = (,, ) x y x y ( π ) A x y A k k e dk dk A k A k x ky k ( π ) + = ( + + ) = A k k e dk dk jkx ( x + ky y ) ( x, y, ) x y 3
Hence A + ( k kx ky) A = Define k ( k k k ) / x y Correct choice: k k k k, k + k < k = j k + k k, k + k > k x y x y x y x y Then we have A + k A = 4
Solution: A ( k, k, ) = A ( k, k,) e x y x y jk Similarly, F ( k, k, ) = F ( k, k,) e x y x y jk 5
Hence jkx ( x + ky y + k ) (,, ) = (,,) x y x y ( π ) A x y A k k e dk dk jkx ( x + ky y + k ) (,, ) = (,,) x y x y ( π ) F x y F k k e dk dk This is a representation of the potentials as a spectrum of plane waves. 6
Derivative property: A x = ( jk ) A k k e dk dk ( π ) jkx ( x + ky y + k ) x ( x, y,) x y = ( jk ) A k k e dk dk ( π ) jkx ( x + ky y ) x ( x, y, ) x y Also A A jkx ( x + ky y ) = e dk x dk y x ( π ) F x Hence F A x = ( jk ) A ( k, k, ) x x y Similarly for the y and derivatives. 7
Apply B.C. s at = : E E x y A F = jωµε x ε y A F = + jωµε y ε x Hence: E = ( jk )( jk ) A ( jk ) F jωµε ε x x y E = ( jk )( jk ) A + ( jk ) F jωµε ε y y x 8
Hence, we have (A ero subscript denotes =.) kk jk ( ) x y E x kx, ky = A ( kx, ky) + F ( kx, ky) jωµε ε kk ( ) y jkx E y kx, ky = A ( kx, ky) + F ( kx, ky) jωµε ε The left-hand sides are known. We have two equations in two unknowns,. ( A, F ) Hence, we can solve for. ( A, F ) 9
In the space domain, jk ( ) j kxx+ kyy (,, ) = ( x, y) x y ( π ) A x y A k k e e dk dk and similarly for F( xy,, ) We can then find the fields from the TE TM equations.
Example: Radiation from Waveguide b PEC a y x ˆ π x Et ( x, y) = yecos a Find the complex power radiated by the aperture.
kk jk x y E x = A + F = jωµ ε ε E kk = A jk + F y x y jωµ ε ε Hence, from the first equation F A kk x = ωµ k y From the second equation, E kk jk k k = + A y x x y jωµ ε ε ωµ k y
Hence Therefore kk x E y = A kk y + jωµ ε k y k k x = A ky + jωµ ε k y k = A k + k ( ) y x jωµ ε k y k = A k k jωµ ε k y ( ) k y A = E y jωµ ε k k k 3
For the Fourier transform of the aperture field, we have: a b x jkx ( x ky y ) E π + ( k, k ) = E cos e dy dx a y x y a b kb y sin π a cos kx = E b a kb y π a kx 4
Complex power radiated by the aperture: b a ( * P ) c = E H ˆ dxdy b a + + = E H * y x dxdy 5
+ + + + Parseval s theorem: * * f ( x, y) g ( x, y) dxdy = f ( k, k ) g ( k, k ) dk dk ( π ) x y x y x y Hence + + * c = y x x y ( π ) P E H dk dk Also, recall: k y A = E y jωµ ε k k k 6
Now find H : H x x A F = + µ y jωµ ε x H = ( jk ) A + ( jk )( jk ) F x y x µ jωµ ε Hence, or kk x H = ( jk ) + ( jk )( jk ) A x y x µ jωµ ε ωµ ky kk k H = ( jk ) + ( jk )( jk ) jωµ ε E x y x y x y µ jωµ ε ωµ k y k k k 7
Simplifying, kk x H ωε x = E y k y + k k k k Hence, + + * = c ( π ) y x x y P E H dk dk ωε kk P E k k k dk dk + * x c = y( x, y) * * y x y ( π ) + k k k k Notes: k = k * * k = k ( k isreal) * 8
k y Polar coordinates: k ρ φ dk dk x y = k dk d ρ ρ φ k x k k x y = = k k ρ ρ cosφ sinφ k = ρ k + k x y New notation: E ( k, k ) E ( k ρ, φ) y x y y 9
k = k k k =real x y * ( k = k ) k k = k k = k + k = k ρ * x y Hence Also, + + π () dk dk = () k dk d x y ρ ρ φ π ωε k ρ k Pc = Ey ( kρ, φ) sin φ cos φ k dk d * ρ ρ φ π + k k ρ k
π P ( ) c = E y( kρ, φ) F kρ, φ dkρdφ π where ωε k, φ = k sin cos * ρ φ + φ k k ( ρ ) F k ( ) / k = k k ρ k = real, k < k k = imaginary, k > k ρ ρ ( ) ρ φ F k, = real, k < k ( ) ρ φ F k, = imaginary, k > k ρ ρ
k y Imaginary power (vars) Visible-space circle k Real power (watts) k x
Equivalent Circuit y Original problem b E E y s H = Z x Inc Ref E Z s Equivalent problem (Impedance surface) * Pc = ( E H ) ˆ ds S 3
Equivalent Circuit (cont.) so P b a * c = y x b a = + b a * b a s b a cos * s b a E a = ( b) * Zs E H dx dy Z E y E π x = dx dy Z a dx dy Hence Z s ab E = * 4 P c 4
TEN Model I () TEN equivalent circuit Z + V() - Z L TEN modeling equations: π x Ey ( xy,, ) = V ( )cos a π x Hx ( xy,, ) = I ( )cos a 5
TEN Model (cont.) Z + + V ( ) Ey ( xy,, ) = = + + I ( ) H ( xy,, ) ωµ = Z = TE k x where k π = k a Hence Z TE = k ωµ π a 6
TEN Model Therefore Z = Z = TE η π ka Also, Z L V () I() E = = y = H x Z s 7
TEN Model (cont.) Z Z L = / Y L Z = Z = TE η π ka Y L * 4 P c = = = ZL Zs ab E 8
TEN Model (cont.) 4 π Y ( ) L = E y( kρ, φ) f kρ, φ dkρdφ ab E π where ( ) *( ) ωε k f kρ, φ = F kρ, φ = kρ sin φ+ cos φ k k YL = GL + jbl Visible-space region Invisible-space region 9
TEN Model (cont.) Hence 4 π k G ( ) L = E y( kρ, φ) f kρ, φ dkρdφ ab E π 4 π jb ( ) L = E y( kρ, φ) f kρ, φ dkρdφ ab E π k where ωε k, = k sin + cos ( ) ρ f k φ ρ φ φ k k 3
TEN Model (cont.) Explicitly writing out the k term, the result can be written as k 4 π ( ) GL = E y( kρ, φ) F kρ, φ dk d ρ φ ab E π k k ρ 4 π ( ) BL = E y( kρ, φ) F kρ, φ dk d ρ φ ab E π k k k ρ Note: There is a square-root branch-point singularity at k ρ = k. where F ( k ) = k ( f k ) = ( ) k k + ρ, φ ρ, φ ωε ρ sin φ cos φ k 3