ECE 6345ECE 6345ECE 6345ECE 6345Spring 2011Prof. David R. JacksonECE Dept.Notes 2otes1OverviewOverviewThis set of notes treats circular polarization, obtained by using a single feed. yLW≈00yx=W00yxxL(x0, y0)2LOverviewOverviewGoals: Find the optimum dimensions of the CP patchFind the input impedance of the CP patchFind the input impedance of the CP patch Find the pattern (axial-ratio) bandwidth Find the impedance bandwidth of CP patch3Amplitude of Patch CurrentsAmplitude of Patch CurrentsywL00(, )xyWrεxh1[ ]IA=First Step: Find the patch currents (x and y directions), and relate them to the input impedance of the patch. 4pp pAmplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)Lywεx00(, )xyhWrεhxπ⎛⎞1[ ]IA=ˆsinxsxxJxALπ⎛⎞=⎜⎟⎝⎠x-directed current mode (1,0):πy⎛⎞⎟⎜5sinysyπyˆJyAW⎛⎞⎟⎜=⎟⎜⎟⎟⎜⎝⎠y-directed current mode (0,1):Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)ˆˆJnH zH=×=−×x mode:sJnH zH××ysxHJ=soysinxπxˆHyAL⎛⎞⎟⎜=⎟⎜⎟⎟⎜⎝⎠L⎝⎠EjHωε=×∇To find E, use11cosyxHHxEAππ∂⎡⎤∂⎛⎞ ⎛ ⎞==⎜⎟ ⎜ ⎟⎢⎥600coszxrrEAjxyj LLεε εε=−=⎜⎟ ⎜ ⎟⎢⎥∂∂⎝⎠ ⎝ ⎠⎣⎦Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)()xyxyin z z z in inVZVhEhEEZZI= = =− =− + = +I00cosxin xrxjhZALLππωε ε⎛⎞⎛⎞=⎜⎟⎜⎟⎝⎠⎝⎠For the (1,0) mode we have0xiZLAωε ε⎡⎤00cosinrxZLAxjhLωε εππ⎡⎤=⎢⎥⎛⎞⎣⎦⎜⎟⎝⎠or7⎝⎠A similar derivation holds for the y mode.Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)0yin ryZWAjhωε ε⎡⎤=⎢⎥⎛⎞⎣⎦y mode:0cosyyjhWππ⎢⎥⎛⎞⎣⎦⎜⎟⎝⎠()01xrLAωε ε⎡⎤=⎢⎥The patch current amplitudes can then be written as where()()1xxxinyyAAZAAZ==10cosAxjhLππ=⎢⎥⎛⎞⎣⎦⎜⎟⎝⎠1yinAAZ=()0101cosyrWAyjhWωε εππ⎡⎤=⎢⎥⎛⎞⎣⎦⎜⎟⎝⎠8W⎜⎟⎝⎠Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)()0101cosxrLAxjhωε εππ⎡⎤⎢⎥⎛⎞⎣⎦⎜⎟()0101cosyrWAyjhωε εππ⎡⎤⎢⎥⎛⎞⎣⎦⎜⎟cosL⎜⎟⎝⎠cosW⎜⎟⎝⎠LW≈Assumey00LWxy=ssu eWxL(x0, y0)()()111xyAAA≈≡Then9LAmplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)xyRRRBecause of the nearly equal dimensions and the feed along the diagonal, we havexyRRR≈≡Ri= resonant input resistance of the mode i, when excited by itself (e g by a feed along the centerline)2xiAAZ=We then have(e.g., by a feed along the centerline).where22xinyyinAAZAAZ=21AAR=Reminder: The bar denotes impedances that are normalized byR(eitherRorR)10Reminder: The bar denotes impedances that are normalized by R(either Rxor Ry).Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)⎡⎤The A2coefficient can be written as020cosrLRAxjhLωεεππ⎡⎤≈⎢⎥⎛⎞⎣⎦⎜⎟⎝⎠200cosedgerLxRLLπωεε⎝⎠⎛⎞⎜⎟⎡⎤⎝⎠⎢⎥00cosrLxjhLππ⎡⎤⎝⎠≈⎢⎥⎛⎞⎣⎦⎜⎟⎝⎠00cosredgexLRLjhπωεεπ⎡⎤⎛⎞=⎜⎟⎢⎥⎝⎠⎣⎦11Circular Polarization ConditionCircular Polarization Condition()1LW δ=+y()W00yx=x(x0, y0)LThe CP condition is=xAamplitude of x modejAAy±=12=yAamplitude of y modeAxCircular Polarization Condition (cont.)Circular Polarization Condition (cont.)The frequency f0is defined as the frequency for which we get CP at broadside.0ffAxfyfAt f = f0:()212 1xrxAAjQ fA=+−00rx ryffffff≡≡where()212 1yryAAjQ f=+−rx ryxyfffffrx= resonance frequency of (1,0) mode13fry= resonance frequency of (0,1) modeCircular Polarization Condition (cont.)Circular Polarization Condition (cont.)Qfrx211 =−ChooseQfQry2112−=−Q2AA=Then we haveAj421ππAAjAyx=+=1jeeejjAAjjjxy===−+=−2442211ππ14jy−1(LHCP)Circular Polarization Condition (cont.)Circular Polarization Condition (cont.)111ffThe frequency conditions can be written as00011111122211 1111xrxxyfffQfQfQfff−= =+ ≈−0011122 2yryyfffQfQ fQ−=− =− ≈+02xyfff+=so()yxfff +=210or152Circular Polarization Condition (cont.)Circular Polarization Condition (cont.)0011122yxffff Q QQ⎛⎞⎟⎜⎟−= −− =⎜⎟⎜⎟⎜⎝⎠Alsoxyfff−≡ΔLet1fΔ=Then we have0fQ02fQ0f02fQ16xfyffCircular Polarization Condition (cont.)Circular Polarization Condition (cont.)Summary of frequencies00111122xyff ffQQ⎡⎤ ⎡⎤=− =+⎢⎥ ⎢⎥⎣⎦ ⎣⎦(LHCP)0022xyQQ⎢⎥ ⎢⎥⎣⎦ ⎣⎦()00111122xyff ffQQ⎡⎤ ⎡⎤=+ =−⎢⎥ ⎢⎥⎣⎦ ⎣⎦(RHCP)⎣⎦ ⎣⎦f0=frequency for which we get CP at broadside.17f0frequency for which we get CP at broadside.Patch Dimensions for CPPatch Dimensions for CPWLrεLΔLΔrεPMCPMC18()WhfLr,,ε=ΔPhysical Dimensions for CP (cont.)Physical Dimensions for CP (cont.)()WhfLr,,ε=Δ2eLLL=+Δ0erkLεπ=(resonance condition)0002kfπμε≡Let00000022xxyykfkfπμεπμε≡19Physical Dimensions for CP (cont.)Physical Dimensions for CP (cont.)()002exr x rkL k L Lεπεπ=+Δ=()02yrkW Wεπ+Δ=Similarly, we have()0yr()2,,LLhWπε=−ΔHence()()02,,2rxrLLhWkWWhLεεπεΔ=ΔNote: For ΔW, we use the same formula as ΔL , but replace W → L.20()02,,ryrWWhLkεε=−ΔPhysical Dimensions for CP (cont.)Physical Dimensions for CP (cont.)Since the patch is nearly square, the two fringing extensions are nearly equal. Hence we havewhere2LLπ=−Δyq022xrLLkWLεπΔΔ00000022xxyykfkfπμεπμε≡≡02yrWLkε=−Δ000yyNote: For the calculation of ΔL it is probably accurate enough to use the patch dimensions that come from neglecting fringing.21the patch dimensions that come from neglecting fringing.Hammerstad’s FormulaHammerstad’s Formula0.2620 300Wε⎡⎤+⎢⎥⎡⎤+0.3000.4120.2580.813rerehLhWhεε⎢⎥⎡⎤+Δ=⎢⎥⎢⎥−⎣⎦⎢⎥+⎣⎦⎣⎦111rrεεε⎛⎞⎛⎞+−⎟⎟⎜⎜=+⎟⎟⎜⎜22112reεhW=+⎟⎟⎜⎜⎟⎟⎜⎜⎝⎠⎝⎠+22Input Impedance of CP PatchInput Impedance of CP Patch() () ()()()12 112 1xyin in inrxryRRZf Zf ZfjQ fjQ f=+= ++−+−1 1/(2) 11/(2)rx ryfQf Q−=− − =andAt f0(LHCP)()011inRRZfjj=+−+so(1)(1) (1)(1)(1 )(1 ) 2jjRjRjRjRjjj++− ++−==−+RZin=orThe CP frequency f0is also the resonance frequency where the 23ininput impedance is real (if we neglect the probe inductance).Input Impedance of CP PatchInput Impedance of CP Patch⎛⎞Hence, at the resonance (CP) frequency f0we have20cosin edgexZRLπ⎛⎞=⎜⎟⎝⎠Note: We have a CAD formula forRedge.Note: We have a CAD formula