Slide Number 1OverviewOverview (cont.)Improved Probe ModelsImproved Probe Models (cont.)Slide Number 6Slide Number 7Cosine Current ModelSlide Number 9Cosine Current ModelSlide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Circuit Model (cont.)Circuit Model (cont.)Circuit Model (cont.)Circuit Model (cont.)Gap ModelGap Model (cont.)Gap Model (cont.)Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Slide Number 40Slide Number 41Slide Number 42Slide Number 43Slide Number 44Slide Number 45Slide Number 46Slide Number 47Slide Number 48Slide Number 491 Spring 2011 Notes 5 ECE 6345 Prof. David R. Jackson ECE Dept.2 Overview This set of notes discusses improved models of the probe inductance of a coaxially-fed patch (accurate for thicker substrates). A parallel-plate waveguide model is initially assumed (at the end of the notes we will also look at the actual finite patch). zh,rrεµ2ax2b3 Overview (cont.) The following models are investigated:  cosine-current model  gap-source model  frill model “Comparison of Models for the Probe Inductance for a Parallel Plate Waveguide and a Microstrip Patch,” H. Xu, D. R. Jackson, and J. T. Williams, IEEE Trans. Antennas and Propagation, Vol. 53, pp. 3229-3235, Oct. 2005. Reference:4 Improved Probe Models Cosine-current model:    cosIz kz hNote: The derivative of the current is zero at the top conductor (PEC).  220cinPZIPc = complex power radiated by probe current zh,rrεµ2axI (z) We assume a tube of current (as in Notes 4) but with a z variation.5 Improved Probe Models (cont.) Gap-source model: An ideal gap voltage source of height ∆ is assumed at the bottom of the probe. zh,rrεµ2ax1V + - ∆  1inZI6 Frill model: A magnetic frill of radius b is assumed on the mouth of the coax.  s ρˆˆˆM zE z ρE   zh,rrεµ2axb  11lnρEρ b/asφρME(TEM mode of coax, assuming 1 V)  10inZIImproved Probe Models (cont.)7 Next, we investigate each of the improved probe models in more detail: Improved Probe Models (cont.)  Cosine-current model  Gap-source model  Frill model8 Cosine Current Model Represent current as: Assume that 0( ) cosmmmzIz Ihπ∞==∑( ) cos ( )Iz kz h= −Note: (0) cos( )I kh=zh,rrεµ2axI (z)9 21(0)2c inP ZI=Cosine Current Model (cont.) Circuit Model: (0)Icoax feed inZ 220cinPZI10 Cosine Current Model Represent the probe current as: 0( ) cosmmmzIz Ihπ∞==∑ 220cinPZIThis will allow us to find the fields and hence the power radiated by the probe current. zh,rrεµ2axI (z)11 Hence 000( )cos cos coshhmmmz mz mzI z dz I dzh hhπ ππ∞=′′ =  ∑∫∫( )( )001 ( )cos22( )cos1hm mohmmoh mzI I z dzhmzI I z dzhhπδπδ  +=  =+∫∫Using Fourier-series theory: Cosine Current Model (cont.) The integral is zero unless m = m′.12 Result: ( )02cos ( )cos1hmmomzI k z h dzhhπδ= −+∫or 222 ()sin( )1 () ( )mmokhI khkh mδπ=+−(derivation omitted) Cosine Current Model (cont.)13 (Time-Harmonic Fields) Note: We have both Ez and Eρ To see this: so 0E∇⋅ =11() 0zEEEzφρρρρ ρ φ∂∂∂+ +=∂ ∂∂0Eρ≠Cosine Current Model (cont.)14 For Ez, we represent the field as follows: where ( )00coszm mmmzE A Jkhρπρ∞−−==∑aρ<aρ>( )(2)00coszm mmmzE A Hkhρπρ∞++==∑( )1/2221/222mzmmkkhkkρπ= −= −Cosine Current Model (cont.)15 At so aρ=zzEE+−=(BC 1) ( ) ( )(2)00m m mmAH ka AJkaρρ+−=Cosine Current Model (cont.)16 Also we have To solve for Eρ , use (BC 2) 21szHHJφφ−=1zEEHjzρφωµ ρ∂∂−= −∂∂EjHωε=×∇Cosine Current Model (cont.) where17 so 11zHHjEzHEjzφρφρωερφωε∂∂= −∂∂∂= −∂Hence we have 2211zHEHj jzφφωµ ωε ρ∂∂=−− −∂∂For the mth Fourier term: ( )()() 2 ()11mmmzzmEH kHjjφφωµ ωε ρ∂=−−− −∂Cosine Current Model (cont.)18 Hence ()2 () 2 ()mmmzzmEkH k H jφφωερ∂−=−∂22 2zm mkk kρ−=so that ()()2mmzmEjHkφρωερ∂= −∂where Cosine Current Model (cont.) (BC 2)19 where ()()2mmzmEjHkφρωερ∂= −∂212szIHHJaφφπ−==( ) ( ) ( )212m mmmszIHHJaφφπ−==For the mth Fourier term: Cosine Current Model (cont.)20 Hence we have ( ) ( )(2)002()2mmm m m mmIjk AH ka AJ kakaρρρρωεπ+−−′′−=(2)0(2)000()() ()() 2mmmm m m pmmH ka kIAH k a A J k aJka a jρρρρπ ωε++′′−=−Cosine Current Model (cont.) ( ) ( )(2)00m m mmAH ka AJkaρρ+−=Using (BC 1)21 (2) (2)00 0 0 0()() ()() ()2mmmm m m m mkIA JkaH ka H kaJ ka Jkaajρρ ρ ρρ ρπ ωε+′′−=−02()2mmmmmkIA j Jkaka a jρρρπ π ωε+−=−( )2014mmm mkA I Jkaρρωε+= −Hence or or (using the Wronskian identity) Cosine Current Model (cont.)22 *2*00*0*0(2) *000 '0121()2()() ()21( )cos cos2scshz szhz szhzhmm mmmP E J dSE a J adz da E a J dzaE a I z dzamz m zA H k a I dzhhπρφπππππ∞∞+′= =−= ⋅−== −= −′    =−⋅        ∫∫∫∫∫∑∑∫We now find the complex power radiated by the probe: Cosine Current Model (cont.)23 ()( )* (2)0002(2) *00 001() 12211 () ()44c mm m mmmm m m mmmhP AIH k akhH ka I JkaIρρρρδδωε∞+=∞==−+  =−+ −    ∑∑Integrating in z we have Hence we have 22 (2)00 001(1 )()()16c m pm m m mmhP I k H kaJkaρρδωε∞==++∑Am+ coefficient Cosine Current Model (cont.)24 Circuit Model (cont.) Therefore, 22cos ( )cinPZkh=( )22 2 (2)00 001sec() 1 ()()8in m m m m mmhZ kh I k H