ECE Spring Prof. David R. Jackson ECE Dept. Notes 26

Transcription

1 ECE 6345 Spring 05 Prof. David R. Jackson ECE Dept. Notes 6

2 Overview In this set of notes we use the spectral-domain method to find the mutual impedance between two rectangular patch ennas.

3 z Geometry h L L I r ε I x L y W x x y L W 3

4 Mutual Impedance Formulation Assume two arbitrary ennas, to be general. V + - I V = I + I V = I + I V + - I The two-port system is described by a matrix. 4

5 Mutual Impedance Formulation (cont.) The self impedance is calculated. V + - I V = I + I V = I + I = V V + - I = 0 I I = 0 in (The presence of open-circuited enna does not significly affect the input impedance of enna.) 5

6 Mutual Impedance Formulation (cont.) The mutual impedance is calculated. V = I + I V = I + I V + - I = V I I = 0 V + - Note: = 6

7 Mutual Impedance Formulation (cont.) The open-circuit voltage V is obtained by integrating the electric field produced from current I over the path C. V = E dr C I E = electric field produced by the feed current I, in the presence of enna and enna. V + - C 7

8 Mutual Impedance Formulation (cont.) The open-circuit voltage V is put in the form of a reaction. i J V = E dr C I = I I i + = E J dv V - V J, J i i C i J I 8

9 Mutual Impedance Formulation (cont.) The equivalence principle is used to replace enna with its surface current. I V = J, J i i I = I J, J i i J I J The field produced by current I exciting enna is the same as that produced by the current on enna in free space. I (The enna current J is that excited on enna when it is in the presence of open-circuited enna.) 9

10 Mutual Impedance Formulation (cont.) Reciprocity is invoked, and then the equivalence principle. J I V = J, J i I = I = I J J, J i, J i J I The field produced by current I exciting enna is the same as that produced by the current on enna in free space. J I (The enna current J is that excited on enna when enna is absent.) 0

11 Mutual Impedance Formulation (cont.) Reciprocity is invoked one more time. J V + - I V = J, J I = I J, J J I

12 Mutual Impedance Formulation (cont.) V = J, J I J J V + - I The mutual impedance is then = J, J II I

13 Mutual Impedance Formulation (cont.) Summary J I = J, J II J I J = current on enna, when excited by current I in the presence of open-circuited enna. J = current on enna, when excited by current I in the absence enna. 3

14 Mutual Impedance Between Patches L y W x x y L W = J, J II () () sx sx The two patches are assumed to be identical here. Assume I = I = [A] 4

15 Mutual Impedance Between Patches (cont.) = J, J = A A B, B = A B, B () () () () () () () () sx sx x x x x x x x Denote = B, B, () () xx x x Note : A = A = A () () x x x Then we have = A, x xx Recall that A x = zx xx Note: Formulas for zx and xx were given previously in the analysis of the single patch. Hence zx = xx, xx 5

16 Mutual Impedance Between Patches (cont.) Calculation of reaction, xx E between patch basis functions: = G B x xx x jkx ( x + ky y ) E B = G B e dk dk [ ] ( π ) x x xx x x y Hence, integrating over the surface of patch, we have, ( ( ) ) ( ( ) ) xx =,, xx x y x x y x ( x, y ) x y G k k B k k B k k dk dk ( π ) 6

17 Mutual Impedance Between Patches (cont.), ( ( ) ) ( ( ) ) xx =,, xx x y x x y x ( x, y ) x y G k k B k k B k k dk dk ( π ) From the Fourier shifting theorem, we have B k k B k k e ( ) ( ( ) ) x x, y = x ( x, y) ( x + y ) jk x k y Note: The superscript is dropped henceforth. Hence we have, () () jk ( x x+ ky y) = B, B = G k, k B k, k e dk dk ( π ) ( ) ( ) xx x x xx x y x x y x y 7

18 Mutual Impedance Between Patches (cont.) Converting to polar coordinates, we have, () () jk ( x x+ ky y) = B, B = G k, k B k, k e k dk d ( π ) π ( ) ( ) xx x x xx x y x x y t t 0 0 φ Since the integrand is an even function of k x and k y, we can write π /, () () xx = x, x = xx ( x, y ) x ( x, y ) cos( x ) cos( y ) t t π 0 0 B B G k k B k k k x k y k dk dφ Note: Quadr Quadr Quadr 3 Quadr 4 jkx ( ) jky x 0 ( y 0) + jkx ( ) jky x 0 ( y 0) + jkx ( 0) + jky x ( y 0) jkx ( x 0) + jky ( y 0) e e + e e + e e + e e jky ( y ) = cos kx e + cos kx e ( ) ( ) jky ( y 0) + jky ( y 0) = cos( kx x 0 ) e + e = 4cos x ( kx x 0) cos( ky y 0) ( y ) + jky x 0 8

19 Mutual Impedance Between Patches (cont.) Final form of mutual reaction π /, () () xx = x, x = xx ( x, y ) x ( x, y ) cos( x ) cos( y ) t t π 0 C B B G k k B k k k x k y k dk dφ Im k t LR C h R k k0 β TM 0 Re k t 9

20 Summary zx = xx, xx π /, xx = xx ( x, y ) x ( x, y ) cos( x ) cos( y ) t t π 0 C G k k B k k k x k y k dk dφ π / = G k k B k k k dk dφ (, ) (, ) xx xx x y x x y t t π 0 C π / j h zx = kt Ii h Bx kz h π ωε 0 C ( ) { TM ( ) cosφ sinc } sin ( 0) cos( 0) k x k y dk dφ x y t 0

21 Results D. M. Pozar, "Input Impedance and mutual coupling of rectangular microstrip ennas, IEEE Trans. Antennas Propagat., vol. AP-30. pp. 9-96, Nov. 98.

22 Components of Mutual Impedance We can also consider the various components of π /, xx = xx ( x, y ) x ( x, y ) cos( x ) cos( y ) t t π 0 C G k k B k k k x k y k dk dφ Im k t Total LR C h R k k0 β TM 0 Re k t Surface-wave = Total Space-wave Space-wave (lateral wave)

23 Components of Mutual Impedance (cont.) Results for typical patches Circular patches ε =.94, / = 0.0, =.0 GHz r h λ0 f 3