ECE 6345ECE 6345Spring 2011Prof. David R. JacksonECE Dept.Notes 101OverviewOverviewIn set set of notes we derive the far-field pattern of a circular patch operating in the dominant TMmodecircular patch operating in the dominant TM11mode. 2Circular Patch: TMCircular Patch: TM11 11 ModeModeyax()()This corresponds to()()1,coszEAJkρφφρ=′This corresponds to a probe on the x axis.111841xkax′=′3111.841x=Circular Patch (cont.)Circular Patch (cont.)Magnetic current model:ˆˆˆˆˆsMnEEρ=−×=−×ρ()ˆ,zsMEaφφ=()1ˆcosAJkaφφ=Choose()11AJka=cossMφφ=4Far Field of Circular PatchFar Field of Circular PatchReciprocity setup:zˆpzθipwHrxhrεysMφxa'φsφ5Far Field of Circular Patch (cont.)Far Field of Circular Patch (cont.)Far-field:()FFEr abθφ=< >(),, ,,ppwaEr abbaHMdSθφ=< >=< >∫()',, cospsSpwHMdSHzdSφρφ φ=− ⋅′′′ ′=−∫∫The primes here denotes source ()()'20'0,, cos,, cosSpwhHzdSHaz adzdφπφρφ φφφφ′′′′′=−∫∫∫coordinates.()0hφφφφ−∫∫6()()'sin cospw pw pwxyHH Hφφφ′′=−+Far Field of Circular Patch (cont.)Far Field of Circular Patch (cont.)()Inside the substrate we have()()()()(),, 11, , 0,0,0 sec cos ( )xyjkx k ypw pwxy xy z zHxyz H e kh k z h′′+′′′ ′⎡⎤=+⎣⎦())(θNkk=())(101θNkkz=The exponent term may be put in cylindrical coordinates as follows:()()()()00sin cos cos sin sin sinxykx ky k a k aθφφ θφφ′′ ′ ′+= +The exponent term may be put in cylindrical coordinates as follows:()()()()()()00sin cos cos sin sin 'sincosykakθφφ φφθφφ′=+′=−7()0scoskθφφFar Field of Circular Patch (cont.)Far Field of Circular Patch (cont.)Hence()()i 'jkθφφ()()()0()sincos'11'sec cossin (0, 0, 0) cos (0, 0, 0)jkapwzzpw pwxyHkhkzheHHθφφφφφ−′=+⎡⎤′′⋅− +⎣⎦Since the horizontal magnetic field components are modeled as current in the TEN, we have()()000 1 ( )pw ipwHHθΓ()(),,0,0,01()pw ipwxy xyHHθ=−Γ:TM :TEppθφ==8:TM,:TEppθφFar Field of Circular Patch (cont.)Far Field of Circular Patch (cont.)TMz()ˆˆpθ=()()00ˆˆ000 iipwEEHφφ⎛⎞⎜⎟()()()()000000ˆ0,0,0sinˆˆ000ipwxipwHxEEHφφηηφφ⎛⎞=−⋅=⎜⎟⎝⎠⎛⎞⎜⎟()()00000,0,0cosipwyHyφφηη=−⋅=−⎜⎟⎝⎠()ˆˆpφ=()()00ˆˆ000ipwEEHθθφ⎛⎞⎜⎟TEz()()()()0000000,0,0cos cosˆˆ000 cos sinipwxipwHxEEHyθθφηηθθφ=⋅=⎜⎟⎝⎠⎛⎞==⎜⎟9()()000,0,0cossinpyHyθθφηη=⋅=⎜⎟⎝⎠Far Field Far Field EEθθTMz()ˆˆpθ=() ( )()0()sincos11sec cosjkapwzzHkhkzheEθφφφ′−′′=+⎛⎞Substituting for Hxand Hy, we have[]()00sin sin cos cos 1 ( )TMEφφφφ θη⎛⎞′′⋅− − −Γ⎜⎟⎝⎠[]()cosφφ′=−−Note:Hence we haveHence, we have() ()()020()sincos01100,, sec cos ( )jkaFFzzhEEr kh kzheπθφφθθφη′−−′=+∫∫10()()1 ( ) cos( ) cosTMadzdθφφ φ φ′′′′⋅−Γ −Far Field Far Field EEθθ(cont.)(cont.)()011 1sec cos()tanc()zz zhkh k z hdz h kh′′+=∫We have that()11 1() ()zz zh−∫()()0()tanc( )1 ()FFTMEEr a h khθφ θ⎛⎞Γ⎜⎟so that()()()()0102cos '0,,()tanc( )1 ()cos ' cos ' 'zjqEr a h khedθπφφθφ θηφφφφ−=−Γ⎜⎟⎝⎠⋅−∫so thatwhere0∫0()sinqkaθ≡φφφ′′ ′=−Let()()()()()22cos coscos cos'cos cosjq jqTMIede dππφφφ φφφ φφ φ φ φφ−′′′−′′′′′′′′≡−= +∫∫11()()()0cos cos cos cosTMede dφφφ φφ φ φ φφ−∫∫Far Field Far Field EEθθ(cont.)(cont.)We have that()cos " cos "cos sin "sinφφφφφφ+= −22φso that()()220ddπφπφφφ−−′′′′=∫∫and()() ( )2cos02cos cosjqTMIe dπφφφφφ′′′′ ′′ ′′≡+∫so that2cos 202coscos cossin sin"cosjqjqededπφπφφφφφφφφ′′′′′′ ′′=′′′′−∫∫220sin sin cosedφφφφ∫equals zero(odd function)1222cos 1 sinφφ′′′′=−Now use()Far Field Far Field EEθθ(cont.)(cont.)22cos cos 200cos cos sinjq jqTMIed e dππφφφφφ φφ′′ ′′′′′′′′=−∫∫Now we use the following identity:112n+⎡⎤⎛⎞Γ⎜⎟12cos 20122sin ( )njq nnnnedJqqπφπφφ+′′⎡⎤⎛⎞Γ+⎜⎟⎢⎥⎝⎠′′ ′′⎢⎥=⎢⎥⎢⎥∫⎢⎥⎣⎦⎛⎞where1231π⎛⎞Γ=⎜⎟⎝⎠⎛⎞133122π⎛⎞Γ=⎜⎟⎝⎠Far Field Far Field EEθθ(cont.)(cont.)Hence2cos "00"2 ()jqed Jqπφφπ=∫2cos " 210()sin " " 2jqJqedqπφφφ π⎛⎞=⎜⎟⎝⎠∫10()cos (2 ) ( )TMJqIJqφπ⎡⎤=−⎢⎥⎣⎦and thusNow useq⎢⎥⎣⎦()()()nJx J x Jx′=−Now use()()()1nn nJx J x Jxx−=−()J14()()()110JxJx Jxx′=−so thatFar Field Far Field EEθθ(cont.)(cont.)Hence12cos()TMIJqπφ′=The far field is then1()TMqφThe far field is then()01,, ( )tanc( ) ()FFEEr ah khQθθφ θ=()1010,, ( )tanc( ) ()2cos ( sin)zEr ah khQJkaθθφ θηπφθ′⋅)(1)(θθTMQΓ15where)(1)(θθTMQΓ−=()ˆˆφTEFar Field Far Field EEφφ()ˆpφ=TEzPerforming similar steps, we have()()()()0( )sin cos11sec cos 1 ( )jkapw TEzzHkhkzh eθφφφθ′−′′=−−Γ⎛⎞[]00sin cos cos cos cos sinEφθφ φθφη⎛⎞′′⋅− +⎜⎟⎝⎠⎛⎞Using reciprocity and performing the integration in z′′, we have()()0102,, ()tanc( )1 ()FF TEzEEr a h khφθφ θη⎛⎞=−Γ⎜⎟⎝⎠16()()2cos0cos sin cosjqedπφφθφφ φφ′−′′′⋅−∫Far Field Far Field EEφφ(cont.)(cont.)Evaluating the integral, we have2cos( )02sin( ) cosjqTEIe dπφφπφφ φφ′−′′′≡−∫∫()[]2cos02cossin cossin c os cos sin sinjqjqededπφπφφφφφφφφφφφ′′′′′′′′ ′′=+′′′′ ′′ ′′=−∫∫[]022cos cos " 200sin c os cos sin sincos sin cos sin sinjq jqedededππφφφφφφφφφφφφ φ φφ′′=′′′′′′ ′′′′=−∫∫∫001()i2Jqφ⎛⎞⎜⎟equals zero(odd function)171()sin2Jqqφπ⎛⎞=−⎜⎟⎝⎠Far Field Far Field EEφφ(cont.)(cont.)Hence()()()100sin() i 2 ()FFJkaEEhkh Pθθφ φ θ⎛⎞⎜⎟()()()100100,,()tanc sin2()sinFFzErahkh Pkaφθφ φπθηθ=−⎜⎟⎝⎠where()()cos 1()TEPθθθ≡−Γ()()