Slide 1OverviewDirectivityDirectivity (cont.)Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 161Spring 2011Notes 15ECE 6345ECE 6345Prof. David R. JacksonECE Dept.2OverviewOverviewIn this set of notes we calculate a CAD formula for the directivity and gain of the rectangular patch antenna.3DirectivityDirectivityFor typical substrate thicknesses, we usually have( )( ) ( )22, , 4 , ,,4r rspspS r r S rDPPrq f p q fq fp= =� �� �� �)0,0(maxDD ( )( )24 , 0,00,0rspr S rDPp=Definition of directivity:where4Directivity (cont.)Directivity (cont.)dipsp spP p P=( ) ( )2,0,0 ,0,0 (0,0)hexr rS r S r A=The space-wave radiated power of the patch is (from Notes 12) The radiated power density from the patch in the far field is ( )( )(1,0)2 2cos2, , sinc2 22 2xsx x y yxLkWA J k k WL kLkpq fp� �� �� �� �� � � �� �� �= =� �� �� �� � � �� � � �-� �� � � �� � � �� �%( ) ( ) ( ) ( )(1,0) (1,0)20,0 0,0 ,sx sxpatchSA J J x y dS Il WLp= = = =�%whereso that5Hence,( ) ( )( )22,0,0 ,0,0 (0, 0),0,0 ( )( ,0,0)hexr rhexr patchdiprS r S r AS r IlS r===24 ( ,0, 0)(0,0)diprdipspr S rDpPp=The radiated power density is thenNote: the superscript “hex” denotes a unit-amplitude horizontal electric dipole in the x direction. The superscript “dip” denotes the dipole that has the same dipole moment as the patch.Directivity (cont.)Directivity (cont.)6Hence)0,0(1)0,0(dipDpD 24 ( ,0, 0)(0,0)dipdiprdipspr S rDPp=( )00( , , ) ( ) cos ( )( , , ) ( ) sin ( )E r Il E GE r Il E Fqfq ff qq ff q== -We now calculate the directivity of the dipole:( )2 201, ,2dip dip diprS r E Eq fq fh� �= +� �� �whereandDirectivity (cont.)Directivity (cont.)7Hence( )2 2 2 22 2001, , cos ( ) sin ( )2diprS r Il E G Fq ff q f qh� �= +� �( )( )0 112cos( )1 cot ( )( )rFj k hNNqe qqq=� �-� �� �( )( )10 12cos( )sec1 cot ( )rGNj k hNqqq qqm=� �-� �� �whereDirectivity (cont.)Directivity (cont.)812(0) ( 0)1 cot( )rrG Fj k hem= =-At broadside we have( )2 20 1 0 1 0 1 1sink h N k h n k h n k hq q= - = =Hence1 1( )secrr r rN nq q em m m= =1 1cos( )( )r r rrN ne q e eq m= =Directivity (cont.)Directivity (cont.)9We then have( )( )( )2 2 22 2002202011,0,0 (0) cos sin21 42 41 cotrrrS r Il E GIlrk hffhwmh pem= +� �=� �� �� �+� �� �We can re-write this using:0 0 0kwm h=( )( )2220 0211 2,0, 041 cotrrrS r Il krk hhpem� �� �� �� �=� �� �� �� �+� �� �� �� �� �Directivity (cont.)Directivity (cont.)10From previous calculations in Notes 11:( ) ( )2 22 200 0 16dips rP Il k h k chmp� ��� �� �Directivity (cont.)Directivity (cont.)11To summarize so far, we have( )24 ,0,0(0,0)rdipdipspr S rDPp=( )2 22 200 0 16dipsp rP Il k h k chmp� �=� �� �( )( )2220 0211 2,0,041 cotrrrS r Il krk hhpem� �� �� �� �=� �� �� �� �+� �� �� �� �� �withDirectivity (cont.)Directivity (cont.)12The result isThis may be re-written as 2220 111 1 1(0,0) 31 cot ( )diprrrDk h ck hemm� �� �� �� �� �=� �� �� �� �� �+� �� �Directivity (cont.)Directivity (cont.)221 1221 0 11tan( )1 1(0,0) 3tan ( )diprrrk h kDk h k ck hemm� �� �� �� � � �� �=� �� � � �� �� � � �� �+� �� �13orDirectivity (cont.)Directivity (cont.)2121 11tan( )1(0,0) 3tan ( )diprrrrk hDk h ck heemm� �� �� �� �� �=� �� �� �� �� �+� �� �or( )212111 1(0,0) 3 tanc1 tan ( )diprrD k hck hme� �� �� �� �=� �� �� �+� �� �where( )( )tantancxxx�14Since the substrate is assumed to be thin, we can further approximate this asDirectivity (cont.)Directivity (cont.)13(0,0)dipDc�12 41 11 2 / 51cn n= - +whereNote: for,11n(0,0) 3dipD �15For the patch we have( )2121 111 1 1 3(0, 0) 3tanc1 tan ( )rrD k hp c pck hme� �� �� �� �� �� �� �� �+� �� �Directivity (cont.)Directivity (cont.)( )( )( )( )( ) ( )220422 4 022 02 22 2 0 01103256015170ap k Wa a k Wc k La c k W k L= +� �+ +� �� �� �+� �� �� �+� �� �2420.166050.007610.0914153aac=-==-12 41 11 2 / 51cn n= - +whereandwith16The gain of the patch is related to the directivity as ( )0,0 (0,0 )rG D e=GainGainwhererspQeQ=1 1 1 1 1d c sp swQ Q Q Q Q= + + +CAD formulas for all of the Q factors were presented in Notes 3.