ECE 6345 Spring 2011 Prof David R Jackson ECE Dept Notes 5 1 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 z h 2a er mr x 2b 2 Overview cont The following models are investigated cosine current model gap source model frill model Reference 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 3 Improved Probe Models Cosine current model h z 2a I z er mr x We assume a tube of current as in Notes 4 but with a z variation I z cos k z h Note The derivative of the current is zero at the top conductor PEC Z in 2 Pc I 0 2 Pc complex power radiated by probe current 4 Improved Probe Models cont Gap source model z h 2a 1V er mr x An ideal gap voltage source of height is assumed at the bottom of the probe 1 Z in I D 5 Improved Probe Models cont Frill model h z 2a b er mr x A magnetic frill of radius b is assumed on the mouth of the coax M s z E z E 1 E 1 ln b a TEM mode of coax assuming 1 V M s E 1 Z in I 0 6 Improved Probe Models cont Next we investigate each of the improved probe models in more detail Cosine current model Gap source model Frill model 7 Cosine Current Model z I z 2a h er mr x Assume that Note I z cos k z h I 0 cos kh Represent current as mp z I z I m cos h m 0 8 Cosine Current Model cont Circuit Model I 0 coax feed 1 2 Pc Z in I 0 2 Z in Z in 2 Pc I 0 2 9 Cosine Current Model z h 2a I z er mr x Z in 2 Pc I 0 2 mp z I z I m cos h m 0 Represent the probe current as This will allow us to find the fields and hence the power radiated by the probe current 10 Cosine Current Model cont Using Fourier series theory h h m p z pz mp z m I z cos dz I m cos cos dz h m 0 0 h h 0 The integral is zero unless m m Hence h h mp z I m 1 dmo I z cos dz 0 2 h h 2 mp z Im I z cos dz 0 h 1 dmo h 11 Cosine Current Model cont or 2 Im h 1 dmo mp z cos k z h cos dz 0 h h Result 2 kh I m sin kh 2 2 1 dmo kh mp derivation omitted 12 Cosine Current Model cont Note We have both Ez and E To see this E 0 Time Harmonic Fields Ef Ez 1 1 r Er 0 r r r f z so Er 0 13 Cosine Current Model cont For Ez we represent the field as follows r a mp z E A cos J 0 k r m r h m 0 r a mp z 2 E A cos H 0 k r m r h m 0 z z m m 2 1 2 where kr m 2 mp k h k k 2 2 1 2 zm 14 Cosine Current Model cont At r a Ez Ez so BC 1 Am H 0 2 k r m a Am J 0 k r m a 15 Cosine Current Model cont Also we have H f2 H 1 J sz where BC 2 Er Ez 1 Hf jwm z r To solve for E use H j E 16 Cosine Current Model cont so Hence we have For the mth Fourier term Hf 1 Hz jweEr r f z Hf 1 Er jwe z 2 Hf Ez 1 1 H f 2 jwm jwe z r H f m m 1 1 E 2 m z k zm H jwm jwe r 17 Cosine Current Model cont so that 2 k 2 H f m k zm H m where Hence Ez m jwe r BC 2 2 k 2 k zm k r2m Hf m Ez m jwe 2 kr m r 18 Cosine Current Model cont I Hf2 H 1 J sz 2p a For the mth Fourier term where m Hf2 H Hf m m 1 m J sz Im 2p a Ez m jwe 2 kr m r 19 Cosine Current Model cont Hence jwe Im 2 k A H k a Am J 0 k r m a k 2 r m m 0 r m 2p a rm Using Am H 0 2 k r m a Am J 0 k r m a BC 1 we have 2 H k r m I m 0 k r m a 2 Am H 0 k r m a Am J k a J k a 0 pm 2 p a j we 0 rm 20 Cosine Current Model cont or k r m I m 2 2 A J 0 k r m a H 0 k r m a H 0 k r m a J 0 k r m a J 0 k r m a 2p a jwe m or 2 I k r m m A j J 0 k r m a jwe p k r m a 2p a m Hence 2 k 1 rm Am I m J 0 k r m a 4 we using the Wronskian identity 21 Cosine Current Model cont We now find the complex power radiated by the probe Pc 1 E J s dS 2 s 2p h 1 E a J z sz a dz d f 2 0 0 h p a Ez a J sz dz 0 pa h E a I z dz z 0 2p a 1 h 2 pz mp z m Am H 0 k r m a cos I m cos dz 0 2 m 0 h h m 0 22 Cosine Current Model cont Integrating in z we have Pc 1 h 2 A I H k a 1 dm 0 m m 0 rm 2 m 0 2 Am coefficient 2 k 1 h rm 2 1 dm 0 H 0 k r m a Im J 0 k r m a I m 4 m 0 4 we Hence we have h 1 2 2 2 Pc I k 1 d H m pm m0 0 k r m a J 0 k r m a 16 we m 0 23 Circuit Model cont 2 Pc Z in 2 cos kh Therefore h 1 2 2 2 Z in sec kh I m k r …