ECE 6345 Sprng 2015 Prof. Davd R. Jackson ECE Dept. Notes 25 1
Overvew In ths set of notes we use the spectral-doman method to fnd the nput mpedance of a rectangular patch antenna. Ths method uses the eact spectral-doman Green s functon, so all radaton physcs, ncludng surface-wave ectaton, s automatcally ncluded (no need for an effectve permttvty). It does not account for the probe nductance (the way t s formulated here), so the CAD formula for probe nductance s added on at the end. D. M. Pozar, Input mpedance and mutual couplng of rectangular mcrostrp antennas, IEEE Trans. Antennas and Propagaton, vol. 30, pp. 1291-1196, Nov. 1982. 2
Spectral Doman Method z h J z J s (, y ) 0 0 L ε r I0 = 1[ A] The probe s vewed as an mpressed current. Set = 0 E ( y) S, S s the patch surface [ ] [ ] J + E J = 0 E (, y) S s z Ths s the Electrc Feld Integral Equaton (EFIE) 3
Spectral Doman Method (cont.) Let J ( y, ) = AB( y, ) B s π, = cos L ( y) The EFIE s then AE B E J [ ] + = 0 z Pck a testng functon T(,y): S [ ] { } z T (, y) A E B + E J ds = 0 [ ] A T (, y) E B ds + T (, y) E J z ds = 0 S S 4
Spectral Doman Method (cont.) Galerkn s Method: T,y B,y (The testng functon s the same as the bass functon.) Hence [ ] A B(, y) E B ds + B, y E J z ds = 0 S S The soluton for the unknown ampltude coeffcent A s then A B(, y) E J z ds S = = B(, y) E[ B] ds S J B z, B, B = [ ] J z, B E J z B, y ds S B, B E B B (, y) ds S 5
Spectral Doman Method (cont.) The nput mpedance s calculated as: n = 2P I 0 0 n 2 2 1 = E 2 zj I 2 = V V E J dv z z P n = comple power comng from mpressed probe current (n the presence of the patch). * z dv V = volume of probe current (The probe current s real and equal to 1.0 [A].) The total feld comes from the patch and the probe: E = E J + E J [ ] z z z z s 6
Spectral Doman Method (cont.) Hence = J E J dv J E J dv [ ] n z z z z z s V V Defne: J E J dv = J, J probe z z z z z V Then we have or [ ] J E J dv = n probe z z s V [ ], n = probe A Jz Ez B dv = probe A B Jz V 7
Spectral Doman Method (cont.) = A B, J n probe z where A = J B z, B, B We have from recprocty that J, B = B, J z z Note: z s easer to calculate than z. so that n = + probe B B, J z, B 2 8
Spectral Doman Method (cont.) Defne: B, B B, J z z Note: The mnus sgn s added to agree wth typcal MoM conventon. Note: The subscrpt notaton on j follows the usual MoM conventon. We then have: n probe 2 z 9
Spectral Doman Method (cont.) Note: The probe mpedance may be appromately calculated by usng a CAD formula: probe jx p X ηµ h 1 ln γ ln π ln µε = p 0 r λ0 a / λ r r 0 γ 0.57722 ( Euler's constant) Ths result comes from a probe nsde of an nfnte parallel-plate wavegude. Nlote: Calculatng probe eactly from the spectral-doman method can be done, but ths would be a lot of work, and the mprovement would be small. 10
Spectral Doman Method (cont.) The net goal s to calculate the reactons and z n closed form. For the patch-patch reacton we have: [ ] = B, B = E B B ds S From prevous SDI theory, we have E = G B so 1 jk ( + ky y ) E B = G B e dk dk [ ] ( 2π ) 2 y Hence, ntegratng over the patch surface, we have 1 = G ( k, k ) B ( k, k ) B ( k, k ) dk dk ( 2π ) + 2 y y y y 11
Spectral Doman Method (cont.) Snce the Fourer transform of the bass functon (cosne functon) s an even functon of k and k y, we can wrte: 1 = G k k B k k dk dk ( 2π ) (, ) 2 (, ) 2 y y y or 1 = G k k B k k dk dk (, ) 2 (, ) 2 y y y π 0 0 Note: z = z = 0 n the spectral-doman Green s functon here. 12
Spectral Doman Method (cont.) Convertng to polar coordnates, we have π /2 1 = G k φ B k φ k dk dφ 2, (, ) 2 t t t t π 0 C Im k t LR C h R β 0 k k 1 0 Re k t Note: The path must etend to nfnty. 13
Spectral Doman Method (cont.) From prevous calculatons, we have: G k k 1 1 φ D D 2 2,,0 cos φ + sn y TE ( z ) D ( k ) = Y jy cot k h t 0 1 1 D ( k ) = Y jy cot k h TE TE TE t 0 1 z1 L cos k π W 2 B ( k, ky) = LW snc k y 2 2 2 2 π L k 2 2 14
Spectral Doman Method (cont.) For the patch-probe reacton we have 0 [ ](,, ) = E B y z J dv z z z V = h z [ ](,, ) E B y z dz 0 0 Note: z s the voltage drop at the feed locaton due to the current B. E = G J z z s so 1 jk ( 0+ ky y 0) E y z = G z B e dk dk (,, ) ( 2π ) + + z 0 0 2 z y 15
Spectral Doman Method (cont.) To calculate G ~ z use: E z H H = jωε1 y 1 y so that 1 E = jk H + jk H z y y jωε1 We need the transforms of the transverse magnetc feld components. 16
Spectral Doman Method (cont.) Usng spectral-doman theory, we have cosφ ( snφ) cosφ I ( snφ) ( ) cosφ ( )( snφ) ( snφ) cosφ ( cosφ)( snφ) H = H + H u v TE = I + = I J + I J TE sv su = I J + I J TE s s TE = I J snφcosφ+ I J cosφsnφ s s ˆ t y = ˆ t y ˆ s V z u E k,k,z I z v H k,k,z I z u J k,k s y = ˆ t y = ˆ t y = ˆ s TE V z v E k,k,z TE I z u H k,k,z I z v J k,k TE s y k y ( k, ky) y ˆv kt k t û φ k φ 17
Spectral Doman Method (cont.) We also have V z uˆ E t k,k y,z H y = H usnφ+ H vcosφ I z = vˆ H t k,k y,z TE = I snφ + I cosφ Is z uˆ J s k,ky TE = I ( J sv ) snφ + I ( J su ) cosφ TE V z = vˆ E t k,k y,z TE = I ( J s ( snφ) ) snφ+ I ( J s cosφ) cosφ TE I z = uˆ H t k,k y,z TE 2 2 = I sn φ I J s cos φ TE Is z = vˆ J s k,ky k y ( k, ky) y ˆv kt k t û φ k φ 18
Spectral Doman Method (cont.) Recall: 1 E = jk H + jk H z y y jωε1 k k y We then have TE 2 I J s sn φ jkh y + jk yh = jktcosφ 2 I J s cos φ TE I J s snφcosφ + jkt snφ + I J Note: s snφcosφ = kt cosφ = k snφ t ( 2 2 ) = jk I J cosφ cos φ + sn φ = t s jk J cosφ I t s 19
Spectral Doman Method (cont.) Hence we have 1 E = jk J I ( cosφ ) z t s jωε1 Usng E = G J z z s we then dentfy that k G z z I z ωε t = cosφ 1 20
Spectral Doman Method (cont.) From TL theory, we have the property that I ( z) = I ( h)cos k ( z+ h) z1 (The short crcut at z = -h causes the current to have a zero dervatve there.) Hence k G z = cos I ( h)cos k ( z+ h) t φ 1 z z ωε1 21
Spectral Doman Method (cont.) For the feld due to the patch bass functon, we then have 1 jk ( 0+ kyy0) E y z = E z e dk dk (,, ) ( 2π ) ( 2π ) z 0 0 2 z y = 1 2 + + + + jk ( 0+ kyy0) G z B e dk dk z y + + 1 k t jk ( 0+ kyy0) = cos φ 2 ( I ( h) ) cos k 1( z + h) B e dk dk ( 2π ) ωε 1 z y z Recall that 0 (,, ) E y z dz = h z 0 0 0 h Note that h cos k ( z + h) dz = cos k z dz = hsnc k h z1 z1 z1 0 z = z+ h 22
Spectral Doman Method (cont.) Hence we have 1 kt jk ( 0+ kyy0) = I h B h k h e dk dk ( 2π ) + + ( ) cosφ snc( 1 ) z 2 z y ωε 1 where cosφ = k k t 23
Spectral Doman Method (cont.) The ntegrand s an even functon of k y and an odd functon of k (due to the cosne term). Hence we use the followng combnatons to reduce the ntegraton to one over the frst quadrant: Quadrant 1 Quadrant 2 Quadrant 3 Quadrant 4 jk jky 0 ( y 0) + jk jky 0 ( y 0) + jk ( 0) + jky ( y 0) jk ( 0) + jky ( y 0) e e e e e e + e e jky ( y 0) + jky ( y 0) = 2 jsn k e 2 jsn k e 0 0 jky ( y 0) + jky ( y 0) = 2 jsn ( k0) e + e = 2 jsn ( k 0) 2cos( ky y 0) = 4 jsn k cos ky ( 0) ( y 0) The result s then π /2 j h z =+ 2 ki t hb kh z π ωε 1 0 0 { cosφ snc( 1 ) } sn ( 0) cos( 0) k ky kdkdφ y t t 24
Spectral Doman Method (cont.) The fnal result s then: π /2 j h z = 2 kt I h B kz h π ωε1 0 C Im k t { 2 cosφ snc } 1 sn cos k k y dk dφ 0 y 0 t LR C h R β 0 k k 1 0 Re k t Note: The path must etend to nfnty. 25
Spectral Doman Method (cont.) Note on materal loss: The spectral-doman method already accounts for radaton nto space and nto surface waves, and accounts for delectrc loss by usng an comple permttvty. In order to account for conductor loss, we can use 1 1 1 tanδ eff = = + Qloss Qd Qc ( 1 j tan ) ε = ε δ eff r r eff where Qd = tanδ Q η ( kh) = µ 0 0 c r ave 2 Rs It s also possble to account for conductor loss by usng a mpedance boundary condton on the patch, but usng an effectve loss tangent s a smpler approach (no need to modfy the code smply ncrease the loss tangent to account for conductor loss). 26
Spectral Doman Method (cont.) We now calculate the needed current functon I ( h) 0 + V (0) z 1[ A] I ( h) 1 z = h V (0) = (1) 0 1 z1 1 D n 1 = Y jy cot k h = 27
Spectral Doman Method (cont.) so I (0 ) V (0) I (0 ) = n = j tan k h n + V (0) 1 z1 n 1 I (0 ) j tan k h D 1 = 1 z1 z From last slde: (0) 1 D V = 28
Spectral Doman Method (cont.) Also, I ( z) = I ( h)cos k z+ h h z < 0 z1 so I (0 ) = I ( h)cos k h z1 And therefore Hence I ( h) = I (0 )sec k h z1 1 = j1 tan k z1h kz1h D 1 sec 1 1 I ( h) = sec kz1h Y0 jy1 cot ( kz1h) j1 tan ( kz1h) 29
Results D. M. Pozar, Input mpedance and mutual couplng of rectangular mcrostrp antennas, IEEE Trans. Antennas Propagat., vol. AP-30. pp. 1191-1196, Nov. 1982. [6] E. H. Newman and P. Tulyathan, Analyss of mcrostrp antennas usng moment methods, IEEE Trans. Antennas Propagat., vol. AP-29. pp. 47-53, Jan. 1981. 30
Note: Usng two bass functons s mportant for crcular polarzaton or for dual-polarzed patches. Two Bass Functons W S J sy (, ) = (, ) + (, ) J y AB y AB y s y y y L J s (, y ) 0 0 EFIE: [ ] E : AE B + AE y B y + E J z = 0 on S [ ] Ey : AE y B + AE y y B y + E y J z = 0 on S 31
Two Bass Functons (cont.) Galerkn testng: S E B ds = 0 S E B ds y y = 0 A B, B + A B, B = J, B y y z A B, B + A B, B = J, B y y y y z y Defne: B, B j j B, J z z 32
Two Bass Functons (cont.) A + A = y y z A + A = y y yy yz y = y (recprocty) By symmetry, y = = y 0 Ths follows snce y = B, By = By ye, y B y, ds S and y [ ](, ) = Odd B( y, ) = Even ( y) E B y y y (from symmetry) 33
Two Bass Functons (cont.) Hence, the two testng equatons reduce to: A = z A = y yy yz The soluton s: Usng recprocty: A A y = = z yz yy A A y = = z zy yy 34
Two Bass Functons (cont.) As before, = [ ] ( J J ) J E J dv n probe z z s V = J, J n probe s z = A B, J A B, J probe z y y z = A + A probe z y zy s s Hence, usng the prevous results for A and A y, we have n 2 z = probe 2 zy yy 35
Two Bass Functons (cont.) From our dervaton: π /2 1 = G k φ B k φ k dk dφ 2, (, ) 2 t t t t π 0 C G = + k k 2 2 1 k y 2 t Dm kt De kt Smlarly, π /2 1 2 yy =, 2 yy t y t, t t π 0 C 2 2 1 k y k Gyy = + 2 kt Dm( kt) De( kt) G k φ B k φ k dk dφ 36
Two Bass Functons (cont.) From our dervaton: π /2 j h z = 2 kt I h B kz h π ωε 1 0 C { 2 cosφ snc } 1 sn ( 0) cos( 0) k k y dk dφ y t Smlarly, π /2 j h zy = 2 kt I h By kz h π ωε 1 0 C { 2 snφ snc } 1 sn ( 0) cos( 0) k y k dk dφ y t 37
Two Bass Functons (cont.) The transforms of the bass functons are: L cos k π W 2 B ( k, ky) = LW snc k y 2 2 2 2 π L k 2 2 W cos k y π L 2 B y( k, k y) = WL snc k 2 2 2 2 π W k y 2 2 38