ECE Spring Prof. David R. Jackson ECE Dept. Notes 5

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1 ECE 6345 Sping 15 Pof. David R. Jackson ECE Dept. Notes 5 1

2 Oveview This set of notes discusses impoved models of the pobe inductance of a coaxially-fed patch (accuate fo thicke substates). A paallel-plate waveguide model is initially assumed (at the end of the notes we will also look at the actual finite patch). a h ε, µ b x

3 Oveview (cont.) The following models ae investigated: Cosine-cuent model Gap-souce model Fill model Deivations ae given in the Appendix. Even moe details may be found in the efeence below. Refeence: H. Xu, D. R. Jackson, and J. T. Williams, Compaison of Models fo the Pobe Inductance fo a Paallel Plate Waveguide and a Micostip Patch, IEEE Tans. Antennas and Popagation, vol. 53, pp , Oct. 5. 3

4 Cosine Cuent Model I () a h ε, µ x We assume a tube of cuent (as in Notes 4) but with a vaiation. cos I k h Note: The deivative of the cuent is eo at the top conducto (PEC). Z in I P c P c = complex powe adiated by pobe cuent P 1 = E J ds * c s S p 4

5 Cosine Cuent Model (cont.) Final esult: 1 1 ( ) () Zin = ( kh ) η sec kh ε Im k m(1 m) H ( k ma) J( k ma) 8 ρ + δ ε ρ ρ m= whee I m ( kh) sin( kh) = 1 + δ mo ( kh) ( mπ) k ρm = k1 mπ h 1/ Note: The wavenumbe k ρm is chosen to be a positive eal numbe o a negative imaginay numbe. kρm = k / k ρm δ m 1, m = =, m 5

6 Gap Souce Model a h ε, µ + - 1V x An ideal gap voltage souce of height is assumed at the bottom of the pobe. Z in I 1 6

7 Gap Souce Model (cont.) Final esult: Y in () 1 a H ( k ma) ρ mπ = j4π ε sinc () η h m= (1 + δm) kρmh ( kρma) h whee k ρm = k1 mπ h 1/ Note: The wavenumbe k ρm is chosen to be a positive eal numbe o a negative imaginay numbe. kρm = k / k ρm δ m 1, m = =, m 7

8 Fill Model a h ε, µ b x A magnetic fill of adius b is assumed on the mouth of the coax. M ˆ Eˆ ˆρE s ρ M sφ = E ρ Choose: E ρ 1 1 ρ lnb/a Z in I 1 (TEM mode of coax, assuming 1 V) 8

9 Fill Model (cont.) Final esult: Y in ( ) ( ) H ( kρmb) H ( kρma) = j 4πε ( ) kh ln ( b/ a) η m= ( kρm) (1 + δm) H ( kρma) whee k ρm = k1 mπ h 1/ Note: The wavenumbe k ρm is chosen to be a positive eal numbe o a negative imaginay numbe. kρm = k / k ρm δ m 1, m = =, m 9

10 Compaison of Models Next, we show esults that compae the vaious models, especially as the substate thickness inceases. a h ε, µ b x 1

11 Xin (Ω) Compaison of Models Models ae compaed fo changing substate thickness. a h ε, µ b x Unifom cuent model Fill model Cosine cuent model HFFS simulation h (m) ε =. a =.635 mm f = GH Z = 5 Ω (b =.19 mm) (λ = 15 cm) 11

12 Rin (Ω) Compaison of Models (cont.) Models ae compaed fo vaying substate thickness. a h ε, µ b x Unifom cuent model Fill model Cosine cuent model HFFS simulation ε =. a =.635 mm f = GH Z = 5 Ω (b =.19 mm) 15 (λ = 15 cm) h (m) λ / 4 =.5 [m] d Note: 1

13 Xin (Ω) Compaison of Models (cont.) Fo the gap-souce model, the esults depend on. a h ε, µ + - 1V x Delta=.1mm Delta=.1mm Delta=1mm ε =. a =.635 mm f = GH Z = 5 Ω (b =.19 mm) (λ = 15 cm) h (m) 13

14 Rin o Xin (Ω) Compaison of Models (cont.) The gap-souce model is compaed with the fill model, fo vaying, fo a fixed h. R h = mm Gap souce model eactance Fill model eactance Gap souce model esistance Fill model esistance a h ε, µ + - 1V ε =. a =.635 mm f = GH Z = 5 Ω (b =.19 mm) (λ = 15 cm) x -1 - X (m) b a =.5 mm 3 14

15 Compaison of Models (cont.) These esults suggest the 1/3 ule: The best is chosen as = b a 3 This ule applies fo a coax feed that has a 5 Ω impedance. a h ε, µ b x a h ε, µ + - 1V x 15

16 Xin (Ω) Compaison of Models (cont.) The gap-souce model is compaed with the fill model, using the optimum gap height (1/3 ule). Reactance a h ε, µ b x 1-1 Gap souce model Fill model ε =. a =.635 mm f = GH Z = 5 Ω (b =.19 mm) - (λ = 15 cm) h (m) 16

17 Rin (Ω) Compaison of Models (cont.) The gap-souce model is compaed with the fill model, using the optimum gap height (1/3 ule). 5 Resistance a h ε, µ b x 4 3 Gap souce model Fill model ε =. a =.635 mm f = GH Z = 5 Ω (b =.19 mm) 1 (λ = 15 cm) h (m) 17

18 Pobe in Patch A pobe in a patch does not see an infinite paallel-plate waveguide. y (x, y ) W L x Exact calculation of pobe eactance: Z jx Z cavity in p in X Im Z p in f Z in may be calculated by HFSS o any othe softwae, o it may be measued. f = fequency at which R in is maximum X cavity in 18

19 Pobe in Patch (cont.) Cavity Model Using the cavity model, we can deive an expession fo the pobe eactance (deivation given late) y PMC (x, y ) W e x L e This fomula assumes that thee is no vaiation of the pobe cuent o cavity fields (thinsubstate appoximation), but it does accuately account fo the actual patch dimensions. 19

20 Pobe in Patch (cont.) Final esult: X p ωμ μ h mπx nπy nπw p cos cos sinc L W e e m,n 1, WL e e 1 m 1 mπ nπ n δ δ 4 k L W e e 3/ wp e a a = pobe adius (x, y ) = pobe location

21 Pobe in Patch (cont.) Image Theoy Image theoy can be used to impove the simple paallel-plate waveguide model when the pobe gets close to the patch edge. Using image theoy, we have an infinite set of image pobes. 1

22 Pobe in Patch (cont.) A simple appoximate fomula is obtained by using two tems: the oiginal pobe cuent in a paallel-plate waveguide and one image. This should be an impovement when the pobe is close to an edge. η X =X + X two pobe image in in in ( ) ( ) η ( s) ( a) two 1 1 in 4 4 X = khy ka J ka khy k J k Oiginal k = k ε η = η ε / Image

23 Pobe in Patch (cont.) As shown on the next plot, the best oveall appoximation in obtained by using the following fomula: ( pobe two ) X = max X, X in in in modified CAD fomula 3

24 Xf (Ω) Pobe in Patch (cont.) Results show that the simple fomula ( modified CAD fomula ) woks faily well ( pobe two ) X = max X, X in in in Unifom cuent model Cavity model Unifom cuent model (with one image) Modified CAD fomula HFFS simulation y (x, y ) W 15 L x 1 5 x = x L/ L / cente X edge 4

25 Appendix Next, we investigate each of the impoved pobe models in moe detail: Cosine-cuent model Gap-souce model Fill model 5

26 Cosine Cuent Model I () a h ε, µ x Assume that I ( ) = cos k ( h) Note: I() = cos ( kh) 6

27 Cosine Cuent Model (cont.) Cicuit Model: I() coax feed Z in 1 Pc = Zin I() Z in I P c 7

28 Cosine Cuent Model (cont.) I () a h ε, µ x Z in I P c Repesent the pobe cuent as: mπ I( ) = Im cos m= h This will allow us to find the fields and hence the powe adiated by the pobe cuent. 8

29 Cosine Cuent Model (cont.) Using Fouie-seies theoy: h h m π mπ m π I( )cos d = I cos cos d h h h m m= The integal is eo unless m = m. Hence h mπ h I( )cos d = Im ( 1+ δm ) h h mπ Im = I( )cos d h h ( 1 δ ) + m 9

30 Cosine Cuent Model (cont.) o h mπ Im = cos k( h)cos d h + h ( 1 δ ) mo Result: I m ( kh) sin( kh) = 1 + δ mo ( kh) ( mπ) (deivation omitted) 3

31 Cosine Cuent Model (cont.) Note: We have both E and E ρ To see this: E = (Time-Hamonic Fields) 1 1 Eφ E ( ρe ) ρ + + = ρ ρ ρ φ so E ρ 31

32 Cosine Cuent Model (cont.) Fo E, we epesent the field as follows: ρ < a mπ E A J k = mcos m= h ρm ( ρ ) mπ = ρ > a + + () E A cos H ( k ρ ) m ρm m= h whee k ρm k = mπ h ( k km ) = 1/ 1/ 3

33 Cosine Cuent Model (cont.) At ρ = a E E = (BC 1) + so ( ) = ( ) A H k a A J k a + () m ρm m ρm 33

34 Cosine Cuent Model (cont.) Also we have H H = J φ φ1 s (BC ) whee H φ 1 Eρ E = jωµ ρ To solve fo E ρ, use H = jωε E 34

35 Cosine Cuent Model (cont.) so jωε E E ρ ρ 1 H H = ρ φ = 1 H jωε φ φ Hence we have H φ 1 1 Hφ E = jωµ jωε ρ Fo the m th Fouie tem: 1 1 ( ) E H km H jωµ jωε ρ ( m) ( m) ( m) φ = φ 35

36 Cosine Cuent Model (cont.) so that E ρ ( m) ( m) ( m) φ m φ = ωε kh k H j whee k k = k ρ m m Hence H ( m) φ = jωε E k ρ ρm ( m) 36

37 Cosine Cuent Model (cont.) H H = J = φ φ1 s I π a Fo the m th Fouie tem: ( ) ( ) ( ) H H = J = m m m m φ φ1 s I π a whee ( m) H ( m) φ = jωε E k ρ ρm 37

38 Cosine Cuent Model (cont.) Hence jωε + () ( ) ( ) k ( ) m AmH ρ kρma AmJ kρma = k ρm I m π a ( ) = ( ) A H k a A J k a + () m ρm m ρm Using (BC 1) we have () H () ( k ma) k + ρ Im ρm AH m ( k ma) A + ρ m J ( kpma) = J( kρma) π a jωε 38

39 Cosine Cuent Model (cont.) o k + () () Im m A ρ m J( k ma) H ρ ( kρma) H ( k ma) J ( k ma) ρ ρ = J( kρma) π a jωε o I k m m A m j + ρ = J( kρma) π kρma π a jωε Hence (using the Wonskian identity) 1 k + ρm Am = Im J kρma 4 ωε ( ) 39

40 Cosine Cuent Model (cont.) We now find the complex powe adiated by the pobe: P c 1 * = E J sds = 1 s π h h E ( a) J * s = π a E ( a) J d * s a d dφ π a h * = E ( ) ( ) a I d π a 1 h + () mπ * m π = A mh ( kρma)cos Im cos d m= h m' = h 4

41 Cosine Cuent Model (cont.) Integating in and using othogonality, we have: 1 P AI H k a h + * () c= m m ( ρm ) 1+ m m= ( δ ) = + k A m + coefficient h () 1 ρm * ( 1 δm) H ( kρma) Im J( kρma) Im 4 m= 4 ωε Hence, we have: h 1 () Pc =+ Im kpm(1 + δm) H ( kρma) J( kρma) 16 ωε m= 41

42 Cosine Cuent Model (cont.) Z in = P c cos ( kh) Theefoe, h 1 () Zin = sec ( kh) Im kρm ( 1 + δm) H ( kρma) J( kρma) 8 ωε m= Define: k ρm = k ρm k mπ = εµ kh 4

43 Cosine Cuent Model (cont.) Also, use k = ωε ω µ ωε ε = η ε We then have 1 1 ( ) () Zin = ( kh ) η sec kh ε Im kρm(1 + δm) H ( kρma) J( kρma) 8 ε m= The pobe eactance is: X = Im( Z ) p in 43

44 Cosine Cuent Model (cont.) Thin substate appoximation 1 1 ( ) () Zin = ( kh ) η sec kh ε Im k m(1 m) H ( k ma) J( k ma) 8 ρ + δ ε ρ ρ m= kh<< 1: I m Keep only the m = tem ( kh) sin( kh) = 1 + δ mo ( kh) ( mπ) The esult is 1 () Zin η ( kh) µ J( ka) H ( ka) 4 (same as pevious esult using unifom model) 44

45 Gap Model a h ε, µ + - 1V x ( ( ) ) ( ) mπ E, ρ = Bm H kρmρ cos m= h E ( a, ) 1/, < < =, othewise. Note: It is not clea how best to choose, but this will be e-visited late. 45

46 Gap Model (cont.) a h ε, µ + - 1V x At ρ = a: ( ( ) ) mπ 1/, < < E a, = Bm H ( kρma) cos = m= h, othewise. Fom Fouie seies analysis (details omitted): B m mπ = sinc h + H k a h ( 1 δ ) m ( ) ( ρm ) 46

47 Gap Model (cont.) a h ε, µ + - 1V x Y in = π ajs ( ) whee J ( ) = H ( ) s φ The magnetic field is found fom E, with the help of the magnetic vecto potential A (the field is TM ): 1 A H φ = µ ρ Use: ( ( ) ) mπ A, ρ = Am H ( kρmρ) cos m= h whee 1 jωµε E = + k A Setting ρ = a allows us to solve fo the coefficients A m. 47

48 Gap Model (cont.) a h ε, µ + - 1V x Final esult: () 1 a H ( k ma) ρ mπ Yin = j4π k sinc η () h m= (1 + δ m) kρmh ( kρma) h 48

49 a Fill Model h ε, µ To find the cuent I (), use ecipocity. b x M s φ 1 1 = ρ ln ( b/ a) Z in 1V fill I 1 Intoduce a ing of magnetic cuent K = 1 in the φ diection at (the testing cuent B ). A B S F a b I H M dv A,B B, A V V S F H H b b a M dv M ds s 49

50 Fill Model (cont.) A B S F ( ) b I = H M s ds S = F π b b Hφ (,) Ms φ d d a b a φ ( b a) ( ) b 1 1 b = π Hφ ( ρ,) ρdρ ρ ln ( b/ a a ) π = ln / ρ ρ ρ φ b = π H ρ, M ρdρ b a H b φ sφ ( ρ,) dρ 5

51 Fill Model (cont.) The magnetic cuent ing B may be eplaced by a 1V gap souce of eo height (by the equivalence pinciple). b π gap I( ) = Hφ ( ρ,) dρ ln / ( b a) b a Let : b The field of the gap souce is then calculated as was done in the gapsouce model, using =. 51

52 Fill Model (cont.) a h ε, µ b x Final esult: ( ) ( ) 1 1 H ( kρmb) H ( kρma) Yin = j 4π k ( ) hη ln ( b/ a) m= ( kρm) (1 + δm) H ( kρma) 5

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2 ECE 6345 Sping 05 Po. David R. Jackson ECE Dept. Notes Oveview This set o notes teats cicula polaization, obtained b using a single eed. L W 0 0 W ( 0, 0 ) L Oveview Goals: Find the optimum dimensions