The Spectral Domain Method in Electromagnetics. David R. Jackson University of Houston

Transcription

1 The Spectral Domain Method in Electromagnetics David R. Jackson Universit of Houston 1

2 Contact Information David R. Jackson Dept. of ECE N38 Engineering Building 1 Universit of Houston Houston, TX Phone: Fax: djackson@uh.edu

3 Additional Resources Some basic references are provided at the end of these viewgraphs. You are welcome to visit a website that goes along with a course at the Universit of Houston that discusses (among other things) the spectral domain immittance (SDI) method. ECE 6341: Advanced Electromagnetic Waves Note: You are welcome to use anthing that ou find on this website, as long as ou please acknowledge the source. 3

4 Spectral Domain Immittance Method What is it? It is a powerful, and sstematic method for analzing sources and structures in laered media. Electric or magnetic dipole sources within laered media. Microstrip and printed antennas. Phased arras, FSS structures, planar radomes. Geophsical problems. The method was originall developed b Itoh and Menzel in the earl 198s (references are given at the end). 4

5 Spectral Domain Immittance Method Where does the name come from? S D I The electric and magnetic fields are modeled in the Fourier transform (spectral) domain. The horizontal (transverse) electric and magnetic fields are modeled as voltage and current on a transmission line model of the laered structure (called the transverse equivalent network or N). The N has transmission lines described b characteristic impedances (or admittances). The word immitance means either impedance or admittance. 5

6 Purpose of Short Course To cover the fundamentals of the Spectral Domain Immittance (SDI) method. To explain the phsical principles of the method. To develop the mathematics of the method. To develop a sstematic recipe for appling the method to analzing different tpes of sources and structures in laered media. To illustrate the method with various practical examples. 6

7 Outline Phsical derivation of method for planar electric surface currents. Examples involving planar surface currents: Microstrip line Microstrip patch current General derivation (Fourier transforming Maxwell s equations) that allows for all tpes of sources to be included in one general derivation. Examples: Vertical dipole over the earth (Sommerfeld problem) Slot antenna covered with radome laer (magnetic current) 7

8 Outline Phsical derivation of method for planar electric surface currents. Examples involving planar surface currents: Microstrip line Microstrip patch current General derivation (Fourier transforming Maxwell s equations) that allows for all tpes of sources to be included in one general derivation. Examples: Vertical dipole over the earth (Sommerfeld problem) Slot antenna covered with radome laer (magnetic current) 8

9 Fourier Representation of Planar Currents (cont.) Consider a planar electric surface current in a laered medium z ε, µ i i x ε, µ s s J s ( x, ) Note: The subscript s denotes the source laer 9

10 Fourier Representation of Planar Currents (cont.) A D spatial Fourier transform in (x,) can be used to represent the finite-size current sheet as a collection (superposition) of infinite phased current sheets. + jkx ( x + k) J k k J x e dxd s 1 j( kxx+ k) J x, = J k, k e dk dk x, s(, ) ( π ) s s x x 1

11 Fourier Representation of Planar Currents (cont.) where 1 jkx ( x + k) J x, = J k, k e dk dk p jk ( xmx kn ) J x k k J k k e + s ( π ) p, ;, = s, xm n xm n 1 J k k J k k k k ( π ) s s x x jk ( xmx+ kn ) J k k e k k = n n m m + + (, ) s xm n x p J xk k (, ;, ) s xm n p s xm, n =, s xm n x 11

12 Fourier Representation of Planar Currents (cont.) 1 J k k J k k k k ( π ) p s xm, n =, xm n x k x k (,) k (,) k x (, ) 1

13 Fourier Representation of Planar Currents (cont.) z ε, µ i i x ε, µ s s J s ( x, ) z ε, µ i ε, µ s i s p s xm n J ( xk, ;, k ) x 13

14 Fourier Representation of Planar Currents (cont.) Example A microstrip patch has a dominant (1,) mode surface current that is described b 1 (, ) ˆ π x Js x = x cos W L L W J s h ε r, µ r x 14

15 Fourier Representation of Planar Currents (cont.) Example (cont.) 1 (, ) ˆ π x Js x = x cos W L L/ W/ π jk x 1 jk x J, ˆ s kx k = x cos e dx e d W L L/ W/ L cos kx π L W J (, ) ˆ s kx k = x sinc k π kl x 15

16 Fourier Representation of Planar Currents (cont.) Example (cont.) L cos kx π L W J (, ) ˆ s kx k = x sinc k π kl x Hence 1 J k k J k k k k ( π ) p s xm, n =, xm n x L cos k 1 π L W J k k x k k k ( π ) π kl x x p ˆ s xm, n = sinc x 16

17 Fourier Representation of Planar Currents (cont.) Example (cont.) p jk ( xmx kn ) J x k k J k k e + s p, ;, = s, xm n xm n L cos k 1 π L W J k k x k k k ( π ) π kl x x p ˆ s xm, n = sinc x (, ;, ) J xk k p s x h ε r, µ r k = xk ˆ + k ˆ t x x 17

18 Fourier Representation of Planar Currents (cont.) Consider an infinite phased current sheet having some (k x, k ): p jkx ( x k) J x k k J k k e + s p, ;, = s, x x This phased current sheet launches a pair of plane waves as shown below: (, ;, ) p s x J xk k k + k s k 18

19 Fourier Representation of Planar Currents (cont.) (, ;, ) p s x J xk k k s k k + + k = xk ˆ + k ˆ + zk ˆ x zs k = xk ˆ + k ˆ zk ˆ x zs 1 k = k k zs s t k = k + k t x Note: We choose the phsical choice where k zs is either positive real or negative imaginar. 19

20 Fourier Representation of Planar Currents (cont.) (, ;, ) p s x J xk k k s k k + Goal: Decompose the current into two parts: one that excites onl a pair of z plane waves, and one that excites onl a pair of z plane waves. z and z waves do not couple at boundaries.

21 (u,v) Coordinates Define: k = xˆk + k ˆ t x k t uˆ = vˆ = zˆ uˆ k t Launching of plane wave b source: z p J s k x k t 1

22 (u,v) Coordinates (cont.) ˆv û φ x kx cosφ = sinφ = k t k k t

23 (u,v) Coordinates (cont.) ˆv û x p J s J J p su p sv : : Launches a Launches a z z wave wave 3

24 z - z Properties z E H z t t = ue ˆ = vh ˆ u v φ k k t E H x p J su 4

25 z - z Properties (cont.) z z E H t t = = ve ˆ v uh ˆ u φ k E H x p J sv k t 5

26 z - z Properties (cont.) Proof of launching propert: Consider the z case (u is labeled as x for simplicit here). An infinite medium is considered, since we are onl looking at the launching propert of the waves. p J s ( x, ) z + H t H t k s k s + ( ) nˆ H H = Js H x = H + p s = ˆ J x x Ae x, jk x Exact solution: 1 ( jkxx H ± xz,, =± ˆ )( A) e e jk z z 6

27 z - z Properties (cont.) In the general coordinate sstem: z k s p J s ( x, ) + H t H t k s u jkx ( x k) J x = uae ˆ + p s (, ) ± 1 jkx ( + k ) H ( xz,, ) =± ( vˆ )( A) e e jk z x z z 7

28 z - z Properties (cont.) Two cases of interest: p + + J x, = uj ˆ x, H = vh ˆ p s su t v z p + + J x, = vj ˆ x, H = uh ˆ p s sv t u z 8

29 Decomposition of Current Split general phased current into two parts: p p p J x, = uj ˆ x, + vj ˆ x, s su sv Launches z Launches z p p J x, = J x, uˆ su p p J x, = J x, vˆ sv s s ˆv φ û x p J s 9

30 Transverse Fields z : E = ue ˆ x,, z = ue ˆ z e t u u H = vh ˆ x,, z = vh ˆ z e t v v ( x ) j k x+ k ( x ) j k x+ k ˆv û H E φ x 3

31 Transverse Fields (cont.) z : E = ve ˆ x,, z = ve ˆ z e t v v H = uh ˆ x,, z = uh ˆ z e t u u ( x ) j k x+ k ( x ) j k x+ k ˆv û H E φ x 31

32 Transverse Equivalent Network (N) N modeling equations: : z V z = E z u I z = H z v : V z = E z z v I z = H z u 3

33 N (cont.) z z I + V ε, µ i i Z i Z i = k zi ωε i p Jsu ( x, ) Source 1/ k = k k k zi i x 33

34 N (cont.) z z ε, µ i i I + V Z i Z i ωµ i = k zi p Jsv ( x, ) Source 1/ k = k k k zi i x 34

35 Source Model z Phased current sheet inside laered medium + H t H t p Jsu ( x, ) z = z s ( + ) p ˆ t t s su zˆ H H = J x, = uj x, ( + ) p ˆ v v su zˆ vˆ H H = uj x, ( + ) p v v su uˆ H H = uj ˆ x, 35

36 Source Model (cont.) + H x, H x, = J x, p v v su so H H = J + p v v su I z I z = J + p Hence s s su Also ( + ) t t zˆ E E = ( z uˆ )( E E ) ˆ + = + + u u u u Eu = E u E = E Hence ( + ) ( ) s = s V z V z 36

37 Source Model (cont.) z Model: I + V s p ˆ s I = J u z = z s I s 37

38 Source Model (cont.) z Model: I + V s p ˆ s I = J v z = z s (derivation omitted) I s 38

39 Source Model (cont.) The transverse fields can then be found from u v (,, ) = E xz V ze (,, ) = H xz I ze (,, ) E xz = V ze v (,, ) = H xz I ze u ( x ) j k x+ k ( x ) j k x+ k ( x ) j k x+ k ( x ) j k x+ k 39

40 Source Model (cont.) Introduce normalized voltage and current functions: (following the notation of K. A. Michalski) V z = V z I = 1[ A] / / / i when s I z = I z I = 1[ A] / / / i when s Then we can write: ( p = ˆ) i s V z V z J u ( p = ˆ ) i s I z I z J u ( p = ˆ) i s V z V z J v ( p = ˆ) i s I z I z J v 4

41 Source Model (cont.) The transverse fields are then expressed as ( p ˆ ) j( k x+ k ),, = E xz V z J ue u i s ( p ˆ ) j( k x+ k ),, = H xz I z J ue v i s x x ( p ˆ ) j( k x+ k ),, E xz = V z + J ve v i s ( p ˆ ) j( k x+ k ),, = + H xz I z J ve u i s x x 41

42 Example E ( xz,, ) Find for z due to a surface current at z =, x j ˆ ( kxx k) J = xe + p s z with kx k = k The current is radiating in free space. = k J = p sx 1 kt = 5k x p J s 4

43 Example (cont.) Decompose current into two parts: ( ˆ ˆ) ( ˆ ˆ) E = E u x + E v x x u v k tanφ = =.5 k x o φ = 6.57 kx uˆ xˆ = cosφ = = k 5 k vˆ xˆ = sinφ = = k t t 1 5 ˆv û o φ = 6.57 x p J s ( x, ) 43

44 Example (cont.) ( cosφ ) ( sinφ ) E = E + E x u v k k x = Eu + Ev kt kt 1 = ke x u ke v e k t ( x ) j k x+ k 1 = kv x ( z) + kv ( z) e k t ( x ) j k x+ k 1 = kxvi ( z)( Is ) + kvi ( z)( Is ) e k t ( x ) j k x+ k 1 = kxvi z J s u + kvi z + J s v e k t ( p ) ( p ˆ ˆ) j( k x+ k ) x 44

45 Example (cont.) 1 E = k V z J u + k V z + J v e ( p ) ( p ˆ ˆ) j( k x+ k ) x x i s i s k t 1 p p = kv ( ˆ ˆ) ( ˆ x i z Jsx x u + kv i z + Jsx x v) e k t 1 p p = kv x i z Jsx + kv i z + Jsx e k t ( x ) j( k x+ k ) cosφ sinφ 1 k = kv z J + kv z + J kt kt x p p x i sx i sx k k x t e j k x+ k x ( x ) j k x+ k 45

46 Example (cont.) Hence 1 E xz kv z kv z J e (,, ) = + p x x i i sx k t ( x ) j k x+ k J = p sx 1 kx k = = k k kt = 5k 46

47 Example (cont.) z Model For z > : I i + V i Z Z Vi ( z) = e 1 jk Ii ( z) = e z jk z z z I s = Z 1 z = 47

48 Example (cont.) z Model For z > : I i + V i Z Z Vi ( z) = e 1 jk Ii ( z) = e jk z z z z I s = 1 z = Z 48

49 Example (cont.) Hence 1 E xz kv z kv z J e (,, ) = + p x x i i sx k t ( x ) j k x+ k 1 Z Z p ( x + ) z Ex ( xz,, ) = k x + k Jsx e e kt j k x k jk z where Z = kz ωε Z = ωµ k z ( ) 1/ k = k k k z x 49

50 Example (cont.) Substituting in values, we have: kz = jk Z kx = jη = k k = k Z = j η k = 5k t J = p sx 1 Hence 1 jη Ex ( xz,, ) = 4 k ( jη ) + k e e 5k 4 j 15 Ex ( xz,, ) = e e 5 16 or η jk x+ kz jkx+ k kz j3 Ex ( xz) = η e + e z 16 jk x kz,, 5

51 Finite Source J s ( x, ) For a phased current sheet: p p j( kxx k) J x = J e + s (, ) s ( p ˆ ˆ) j( k x+ k ),, = E x z uv z J u e t i s ( )( p ) j ˆ ( kxx+ k) i s + vˆ V z + J v e x Recall that 1 J = J k k dk dk ( π ) (, ) p s s x x 51

52 Finite Source (cont.) Hence 1 E ( x,, z) = uv z J u vv z J v ( π ) + + { ˆ ˆ ˆ ˆ } t i s i s e ( x ) dk dk j k x+ k x Note: ( x, ), ( x, ) uˆ = uˆ k k vˆ = vˆ k k k ( kx, k) ˆv û k t φ x φ k x Spatial coordinates Wavenumber plane 5

53 N Model for E t We can also write (from the definition of inverse transform) j( kxx+ k) E x,, z = E k, k, z e dk dk ( π ) t t x x Comparing with the previous result, we have E ˆ ˆ ˆ ˆ t kx k z = uvi z J s u vvi z J s v,, Similarl, H ˆ ˆ ˆ ˆ t kx k z = uii z J s v + vii z J s u,, This motivates the following identifications: 53

54 N Model (cont.) (,, ) V z = E k k z uˆ t x (,, ) V z = E k k z vˆ t x (,, ) I z = H k k z uˆ t x (,, ) I z = H k k z vˆ t x I = J k k uˆ (, ) s s x I = + J k k vˆ (, ) s s x 54

55 N Model (cont.) z : I + V I s V = E ˆ t u I = H ˆ t v I = J uˆ s s 55

56 N Model (cont.) z : I + V I s V = E ˆ t v I = H ˆ t u I = J vˆ s s 56

57 Example J = xj ˆ x, s sx Find E( xz,, ) x E x kx k z x u E t u v E t v (,, ) = ˆ ˆ( ˆ) + ˆ( ˆ) ( xuv ˆ ˆ) ( z) ( xv ˆ ˆ) V ( z) = + 57

58 Example (cont.) xˆ uˆ = cosφ = k k xˆ vˆ = sinφ = x t k k t ˆv û φ (, ) k = k k t x x Hence k k E x kx k z V z V z k k x,, = + t t k k x = V ˆ ˆ i z J s u + Vi z J s v k t k t 58

59 Example (cont.) k x k k k x E x ( kx, k, z) = Vi ( z) J sx + Vi ( z) J sx kt kt kt kt 1 = J sx kx Vi ( z) + k Vi ( z) k t 1 1 x (,, ) = sx x i + i ( π ) k t E x z J k V z k V z e ( x ) dk dk j k x+ k x 59

60 Dadic Green s Function where ij G G G G( x x, ; zz, ) G G G G G G (,, ) G E xz xx x xz = x z zx z zz = i due to the unit-amplitude electric dipole at J xz,, = ˆjδ x x δ δ z z ( x,, z ) From superposition: (, ;, ) = (, ;, ) (, ; ) E x z z G x x z z J x z dx d where E E E x = E z J s J J sx s = s J sz We assume here that the currents are located on a planar surface. 6

61 Dadic Green s Function (cont.) (, ;, ) = (, ;, ) (, ; ) E x z z G x x z z J x z dx d s This is recognized as a D convolution: E = G J s Taking the D Fourier transform of both sides, E = G J s where ( x, ;, ) G = Gk k zz 61

62 Dadic Green s Function (cont.) E = G J s Assuming we wish the x component of the electric field due to an x- directed current J sx (x, ), we have E = G J x xx sx In order to indentif G xx, we use 1 1 E( xzz, ;, ) = J k, k kv zz, + kv zz, x sx x x i i ( π ) kt e ( x ) dk dk j k x+ k x 6

63 Dadic Green s Function (cont.) 1 1 Ex ( xzz, ;, ) =,, J sx kv x i zz kv i zz ( π ) k + t j( kxx+ k) e dk dk x E k k zz 1 kv zz kv zz J k k x( x, ;, ) =,,, x i + i sx x k t Recall that E = G J x xx sx Hence 1 G k k zz kv zz kv zz, ;, = (, ) + (, ) xx x x i i kt 63

64 Dadic Green s Function (cont.) We the have G G G G G G G G G G xx x xz = x z zx z zz 1 G k k zz kv zz kv zz, ;, = (, ) + (, ) xx x x i i kt The other eight components could be found in a similar wa. (The sources in the N that correspond to all possible sources are given on the next slide, and from these we can determine an component of the spectral-domain Green s function that we wish, for either an electric current or a magnetic current.) 64

65 Summar of Results for All Sources These results are derived later. V I V I = E u = H u = E = H u v I s = J su k V = M + J ωε t s sv sz V s = M su k I = J + M ωµ t s sv sz Definition of vertical planar currents : = J xz,, J x, δ ( z) z sz = M xz,, M x, δ ( z) z sz 65

66 Sources used in Modeling z : V i = voltage due to 1[A] parallel current source I i = current due to 1[A] parallel current source V v = voltage due to 1[V] series voltage source I v = current due to 1[V] series voltage source I + V I + V I s + - V s = = i s = = v V z V z I I z I z I i s V z V z V v s I z I z V s 66

67 Sources used in Modeling (cont.) z : V i = voltage due to 1[A] parallel current source I i = current due to 1[A] parallel current source V v = voltage due to 1[V] series voltage source I v = current due to 1[V] series voltage source I + V I + V I s + - V s = i s = = i s = V z V z I I z I z I V z V zv v s I z I z V v s 67

68 Outline Phsical derivation of method for planar electric surface currents. Examples involving planar surface currents: Microstrip line Microstrip patch current General derivation (Fourier transforming Maxwell s equations) that allows for all tpes of sources to be included in one general derivation. Examples: Vertical dipole over the earth (Sommerfeld problem) Slot antenna covered with radome laer (magnetic current) 68

69 Microstrip Line Microstrip Line I w Dominant quasi-m mode: sx h r, x (, ) = jk J x B e x x ε µ r I = B / π w We assume a purel x-directed current and a real wavenumber k x. 69

70 Microstrip Line (cont.) Fourier transform of current: w/ 1/ jk jkxx jkxx J π sx kx, k = I e d e e dx w/ w kw x x J, jk x jk x sx kx k = I J e e dx kw J sx kx, k = I J πδ ( kx kx) 7

71 Microstrip Line (cont.) kw J sx ( kx, k ) = I J πδ ( kx kx) 1 1 x (, ;,) =,, sx x i i ( π ) k + t ( x ) dk dk jkx+ k z =, z = Hence we have: x E x J k V k V e I 1 kw Ex( x, ;,) = J πδ ( kx kx) kxvi (,) kvi (,) ( π ) kt + jkx ( x + k) e dk dk x 71

72 Microstrip Line (cont.) I 1 kw Ex( x, ;,) = J δ ( kx kx) kxvi (,) kvi (,) ( π ) kt + jkx ( x + k) e dk dk x Integrating over the δ -function, we have: I 1 kw jkx x jk ( ) E ( x, ;,) = e J k V (,) k V (,) e dk ( π ) + kt x x i i where we now have k = k + k t x = = ( ) 1 1 ( 1 ) k k k k k k z t x k = k k = k k k z t x 7

73 Microstrip Line (cont.) Enforce EFIE using Galerkin s method: w/ w/ E (, ;,) T d = x The EFIE is enforced on the red line. where T( ) = B( ) (testing function = basis function) x Recall that E x e kw J k V k V e dk I jk 1 x x jk x(, ;,) = x i (,) i (,) ( π ) + kt Substituting into the EFIE integral, we have I 1 kw, +, = J T ( k ) kxvi k Vi dk ( π ) kt 73

74 Microstrip Line (cont.) 1 k t kw J T ( k ) kxvi (,) + k Vi (,) dk = Since the testing function is the same as the basis function, 1 k t kw kw J J kxvi (,) + k Vi (,) dk = Since the Bessel function is an even function, 1 k t kw J kxvi (,) + k Vi (,) dk = 74

75 Microstrip Line (cont.) Using smmetr, we have kw J kxvi (,) + k Vi (,) dk = k = k + k t x This is a transcendental equation of the following form: F( k x ) = Note: k < kx < k1 75

76 Microstrip Line (cont.) Branch points: k = k k = k k + k z t x Hence (( ) ) k = k k k z x ( ( )) x = j k k k ( ( ) ) x = j k + k k 1/ 1/ 1/ Note: The wavenumber k z causes branch points to arise. x x 1/ 1/ = j k j k k k + j k k k = ± j k k b x 76

77 Microstrip Line (cont.) Poles (k = k p ): k = k + k = k tp x p k = k k p x or k = ± j k k p x 77

78 Microstrip Line (cont.) Branch points: k = ± j k k b x Poles: k = ± j k k p x k i k p k p C k r 78

79 Microstrip Line (cont.) Note on wavenumber k x ( ) k = ± j k k p x For a real wavenumber k x, we must have that kx > k 1/ Otherwise, there would be poles on the real axis, and this would correspond to leakage into the surface-wave mode of the grounded substrate. The mode would then be a leak mode with a complex wavenumber k x, which contradicts the assumption that the pole is on the real axis. Hence k < k < k < k x 1 79

80 Microstrip Line (cont.) If we wanted to use multiple basis functions, we could consider the following choices: Fourier-Maxwell Basis Function Expansion: M 1 jk 1 x x mπ Jsz ( x, ) = e a cos m w m= w ( n 1) N jkx x w π Jsx ( x, ) = e bn sin n = 1 w Chebshev-Maxwell Basis Function Expansion: jkx x J x e at ( + δ ) M 1 1 m sz (, ) = m m w m= w π w N jkx x sx (, ) = n n 1 n= 1 1 w j4 J x e bu w π w 8

81 Microstrip Line (cont.) We next proceed to calculate the Michalski voltage functions explicitl. 81

82 Microstrip Line (cont.) z V i : (,) k z + V I i i 1 z = x k k k k 1 z1 = 1 x k k k k Z Z1 V i (,) + k z1 1 [A] z = z = h Z k k k η k = = η Z = = z z z1 z1 1 ωε k ωε1 ε r k 8

83 Microstrip Line (cont.) z V i : (,) Z k z I + V i i V i (,) + z = 1 z = x k k k k 1 z1 = 1 x k k k k Z1 k z1 1 [A] z = h Z ωµ η ωµ = = Z = = η µ 1 r 1 kz kz k kz1 kz1 k ( / ) ( / ) 83

84 Microstrip Line (cont.) At z = k z V i (,) = Z in I + V i i = Y = Y + Y 1 + in in in 1 = Y jy1cot kz1h 1 Z V i (,) + 1 [A] z = Hence Z 1 k z1 z = h V i V i (,) (,) = = 1 (, ) D k k m x 1 (, ) D k k e x, = cot D k k Y jy k h m x 1 z1, = cot D k k Y jy k h e x 1 z1 84

85 Microstrip Line (cont.) Effective dielectric constant Low-frequenc results ε h eff r k x = k w k = k ε x eff r eff ε r Dielectric constant of substrate (ε r ) Parameters: w/h =.4 E. J. Denlinger, A frequencdependent solution for microstrip transmission lines, IEEE Trans. Microwave Theor and Techniques, vol. 19, pp. 3-39, Jan

86 Microstrip Line (cont.) Effective dielectric constant ε r Frequenc (GHz) Frequenc variation Parameters: ε r = 11.7, w/h =.96, h =.317 cm ε eff r h k x = k w k = k ε x eff r eff ε r E. J. Denlinger, A frequencdependent solution for microstrip transmission lines, IEEE Trans. Microwave Theor and Techniques, vol. 19, pp. 3-39, Jan

87 Microstrip Line (cont.) Characteristic Impedance 1) Quasi-M Method h w ε r h w eff ε r Original problem Equivalent problem (M) ε eff r k x = k We calculate the characteristic impedance of the equivalent homogeneous medium problem. 87

88 Microstrip Line (cont.) Using the equivalent M problem: Z L = = C L C ε Z eff r = Z air (The zero subscript denotes the value when using an air substrate.) 1 ε eff r h w air Z ε Simple CAD formulas ma be used for the Z of an air line. Z air 8h w w 6 ln + ; for 1 w 4h h = 1π w w ln h h w ; for 1 h 88

89 Microstrip Line (cont.) ) Voltage-Current Method V x V 1 Z = = = Ez (,, z) dz I x I I h z h V + - w ε r 1 E k k z k I z (,, ) = z x t ωε 1 = ωε ( k ) ( ˆ t Ii z J s u) 1 = ωε ( kt ) Ii ( z)( J sx ) (derived later) cosφ 89

90 Microstrip Line (cont.) 1 1 jkx x + k Z = E z( kx, k, z) e dkxdk dz I h ( π ) = x = 1 1 Z = E z( kx, k, z) dkxdk dz I h ( π ) k x Z = ( kt ) Ii ( z)( J sx ) dkxdk dz I h ( π ) ωε k t kw k x Z = ( kt) Ii ( z) IJ πδ ( kx kx) dkxdk dz I ( ) ωε k h π t 9

91 Microstrip Line (cont.) 1 1 kw k x Z = ( kt) Ii ( z) J πδ ( kx kx) dkxdk dz ( ) ωε k h π t 1 1 kw k x Z = ( kt) Ii ( z) J dk dz ( π ) ωε k h t 1 kx kw Z = J Ii ( z) dz dk ( π ) ωε h k = k + k t x 1 kx kw Z = J Ii ( z) dz dk π ωε h 91

92 Microstrip Line (cont.) The final result is 1 kx kw Z = J F ( kt) dk π ωε where k = k + k t x F kt Ii z dz h 9

93 Microstrip Line (cont.) We next calculate the function Ii ( z) Z + V i 1[ A] z Z I 1 i z h I i ( z ) Z 1/ Dm( k in t) tan V = = = = Z Z jz k h in in 1 z1 93

94 Microstrip Line (cont.) Because of the short circuit, I z = Acos k z+ h, h< z< i z1 At z = : ( ) = cos ( z1 ) I A k h i Therefore A = I i cos ( ) ( k h) z1 Hence ( kz1 ( z+ h) ) ( k h) cos Ii ( z) = Ii ( ) cos z1 h< z< 94

95 Microstrip Line (cont.) Hence I i ( z) ( kt) ( kz1 ( z+ h) ) 1/ D cos m = jz1 tan kz1h cos kz1h h< z< or I i ( z) ( kz1 z+ h ) 1 1 cos = Dm( kt) jz 1 sin kz1h 95

96 Microstrip Line (cont.) Hence I i ( z) ( kz1 z+ h ) 1 1 cos = Dm( kt) jz 1 sin kz1h where, = cot D k k Y jy k h m x 1 z1 96

97 Microstrip Line (cont.) We then have F k I z dz t h i ( kz1 z+ h ) 1 1 cos = dz Dm( kt) jz 1 sin k h z1h = cos ( 1 kz z h ) dz Dm( kt) jz + 1 sin ( kz1h) h = sin ( k ( z+ h ) ) D k jz k h k 1 z m t 1 sin ( z1 ) z1 h = ( 1 ) Dm( kt) jz sin kz h 1 sin ( kz1h) kz1 Hence F k t = Dm( kt) jz1 kz1 97

98 Microstrip Line (cont.) 3) Power-Current Method z I P Z = = Sz(,, z) dz d I h z h w ε r Note: It is possible to perform the spatial integrations for the power flow in closed form (details are omitted). 98

99 Microstrip Line (cont.) 4) Power-Voltage Method Z V V = = Pz S,, z dz d h z z h V + - w ε r Note: It is possible to perform the spatial integrations for the power flow in closed form (details are omitted). 99

100 Microstrip Line (cont.) Comparison of methods: At low frequenc all three methods agree well. As frequenc increases, the VI, PI, and PV methods give a Z that increases with frequenc. The Quasi-M method gives a Z that decreases with frequenc. The PI method is usuall regarded as being the best one for high frequenc (agrees better with measurements). 1

101 Microstrip Line (cont.) Quasi-M Method ε r = 15.87, h = 1.16 mm, w/h =.543 E. J. Denlinger, A frequencdependent solution for microstrip transmission lines, IEEE Trans. Microwave Theor and Techniques, vol. 19, pp. 3-39, Jan

102 Microstrip Line (cont.) F. Mesa and D. R. Jackson, A novel approach for calculating the characteristic impedance of printed-circuit lines, IEEE Microwave and Wireless Components Letters, vol. 4, pp , April 5. 1

103 Microstrip Line (cont.) Ω 58 5 (Circles represent results from another paper.) VI PV 46 B Bianco, L. Panini, M. Parodi, and S. Ridella, Some considerations about the frequenc dependence of the characteristic impedance of uniform microstrips, IEEE Trans. Microwave Theor and Techniques, vol. 6, pp , March PI PV e V e I GHz ε r = 1, h =.635 mm, w/h = 1 V e = effective voltage (average taken over different paths). 13

104 Outline Phsical derivation of method for planar electric surface currents. Examples involving planar surface currents: Microstrip line Microstrip patch current General derivation (Fourier transforming Maxwell s equations) that allows for all tpes of sources to be included in one general derivation. Examples: Vertical dipole over the earth (Sommerfeld problem) Slot antenna covered with radome laer (magnetic current) 14

105 Patch Antenna Example Find E( x,,) x z L W h ε r, µ r x Dominant (1,) mode: J sx 1 π x, = cos W L ( x) 15

106 Patch Antenna (cont.) Recall that E = G J x xx sx 1 G k k zz kv zz kv zz, ;, = (, ) + (, ) xx x x i i kt In this problem z = z = 16

107 Patch Antenna (cont.) J sx 1 π x, = cos W L ( x) L/ W/ 1 jkx x jk J π sx kx, k = cos e dx e d W L L/ W/ L cos kx π L W J sx kx, k = sinc k π kl x 17

108 Microstrip Line (cont.) z V i : (,) k z + V I i i 1 z = x k k k k 1 z1 = 1 x k k k k Z Z1 V i (,) + k z1 1 [A] z = z = h Z k k k η k = = η Z = = z z z1 z1 1 ωε k ωε1 ε r k 18

109 Microstrip Line (cont.) z V i : (,) Z k z I + V i i V i (,) + z = 1 z = x k k k k 1 z1 = 1 x k k k k Z1 k z1 1 [A] z = h Z ωµ η ωµ = = Z = = η µ 1 r 1 kz kz k kz1 kz1 k ( / ) ( / ) 19

110 Patch Antenna (cont.) At z = : V i (,) = Z in = Y = Y + Y 1 + in in in 1 = Y jy1cot kz1h 1 Hence V i V i (,) (,) = = 1 (, ) D k k m x 1 (, ) D k k e x (, ) = 1 cot ( 1 ) (, ) = 1 cot ( 1 ) D k k Y jy k h m x z D k k Y jy k h e x z 11

111 Patch Antenna (cont.), = cot D k k Y jy k h m x 1 z1, = cot D k k Y jy k h e x 1 z1 Note: (, ) = D k k D k em, x em, t Hence, we have 1 G k k zz kv zz kv zz (, ;, ) = (, ) + (, ) xx x x i i kt 1 k k x E x kx, k, = J sx kx, k + kt Dm( kt) De( kt) 111

112 Patch Antenna (cont.) Taking the inverse Fourier transform, we have 1 1 E x,, = J k, k ( π ) + + x sx x kt k k x + D k D k m t e t e ( x ) j k x+ k dk dk x 11

113 Example: Vertical Field Find E z (x,,z) inside the substrate for the same patch antenna (-h < z < ). H = jωε E 1 1 H H x Ez = jωε1 x 1 E = jk H + jk H z x x jωε1 1 = + ωε ε r ( kh ) x kh x 113

114 Example (cont.) ( ˆ ˆ) ( ˆ ˆ) H = H u x + H v x x u v u ( cosφ) H v( sinφ) = H + k k x = Hu + Hv kt kt (, ) k = k k t x ˆv û ( ˆ ˆ) ( ˆ ˆ) H = H u + H v u v H u ( sinφ) + H v( cosφ) k k x = H u + H v kt kt φ x 114

115 Example (cont.) Hence cancels 1 k = + kx + k k x + Ez kx Hu Hv k Hu Hv ωε ε r kt kt kt kt or 1 1 E = z Hv ωε ε kx k r kt or E = 1 ( kh ) z t v ωε ε r 115

116 Example (cont.) or 1 E k k z k I z,, = z x t ωε ε r Hence 1 E k, k, z = ( k ) I ( z) J uˆ z x t i s ωε ε r 1 = ( k ) I ( z) J ωε ε r t i sx k k x t The result is then 1 E k k z k J I z,, = z x x sx i ωε ε r 116

117 Example (cont.) Hence, in the space domain we have 1 1 j( kxx+ k) E x z k J I z e dk dk (,, ) = ( π ) z x sx i x ωε ε r Note: Onl z waves contribute to the vertical electric field. On the next page we start calculating the term Ii ( z) 117

118 Example (cont.) We next calculate the function Ii ( z) Z + V i 1[ A] z Z I 1 i z h I i ( z ) Z 1/ Dm( k in t) tan V = = = = Z Z jz k h in in 1 z1 118

119 Example (cont.) Because of the short circuit, I z = Acos k z+ h, h< z< i z1 At z = : ( ) = cos ( z1 ) I A k h i Therefore A = I i cos ( ) ( k h) z1 Hence ( kz1 ( z+ h) ) ( k h) cos Ii ( z) = Ii ( ) cos z1 h< z< 119

120 Example (cont.) Hence I i ( z) ( kt) ( kz1 ( z+ h) ) 1/ D cos m = jz1 tan kz1h cos kz1h h< z< or I i ( z) ( kz1 z+ h ) 1 1 cos = Dm( kt) jz1 tan ( kz1h) cos kz1h or I i ( z) ( kz1 z+ h ) 1 1 cos = Dm( kt) jz 1 sin kz1h 1

121 Example (cont.) Hence I i ( z) ( kz1 z+ h ) 1 1 cos = Dm( kt) jz 1 sin kz1h where, = cot D k k Y jy k h m x 1 z1 11

122 Radiated Power from Patch P SP J sx 1 π x, = cos W L ( x) P SW P SW h ε r Patch Lossless substrate and metal P TOT SP SW e r P = + = P P SP TOT P 1

123 Radiated Power (cont.) P c 1 * = E J s ds 1 = S E J x * sx dxd Use Parseval s Theorem: 1 f x g x dxd f k k g k k dk dk * * (, ) (, ) = (, ) (, ) ( π ) x x x 13

124 Radiated Power (cont.) Hence 1 * = c x sx x 8π P E J dk dk From SDI analsis, E = J G x sx xx G xx where = + k k 1 kx t Dm kt De kt = cot D k Y jy k h m t 1 z1 = cot D k Y jy k h e t 1 z1 L cos kx π L W J sx ( kx, k ) = sinc k π kl x 14

125 Radiated Power (cont.) We then have Using smmetr, 1 1 P G J dk dk = c 8π xx sx x = ( k, k ) ( k, k ) c x x π P G J dk dk xx sx x Polar coordinates: 1 π / P G J kdkdφ = c xx sx t t π C k k t φ ( kx, k) k x 15

126 Radiated Power (cont.) 1 π / P G J kdkdφ = c xx sx t t π C Note that J = Real Function of k, k for real k, k. sx x x Hence π / 1 P = k k π (, φ) (, φ) c t sx t C G J kdkdφ xx t t sx J J sx Note: is analtic but is not. This is a preferable form! 16

127 Radiated Power (cont.) TOT 1 Re π π / C (, ) (, ) sx P G k φ J k φ kdkdφ = xx t t t t Im k t Branch point pole C Re k t k k 1 For real k t we have (proof on next page): G xx = complex, imag, k t k t < k > k Hence, we can neglect the region k t > k, except possibl for the pole. 17

128 Radiated Power (cont.) Proof of complex propert: G xx 1 x kt, φ = + k D ( k ) D ( k ) t k k m t e t Consider the following term (D e is similar): alwas imaginar = cot D k Y jy k h m t 1 z1 Y ωε = k z z 1 t k = k k Y 1 = ωε k z1 1 1 t k = k k z1 1 Y = real, imag, k k t t < > k k 18

129 Space-Wave Power The region kt (, k ) gives the power radiated into space. The residue contribution gives the power launched into the surface wave. Im k t P SP C SW Re k t k k 1 k π / 1 P = Re G k, φ J k, φ kdkdφ SP xx t sx t t t π 19

130 Surface-Wave Power The pole contribution gives the surface-wave power: P G k φ J k φ kdkdφ / 1 π SW = Re xx t, sx t, t t π C From the residue theorem, we have π / SW 1 P = Re π j Res G k, φ J k, φ k dφ { } SW xx t sx t t π π / 1 Re tp sx tp, Res xx t, = π jk J ( k φ) { G ( k φ) } dφ π kt = ktp = π / Re tp sx tp, Res xx t, 1 k j J ( k φ) { G ( k φ) } dφ π kt = ktp k t = k tp 13

131 Surface-Wave Power (cont.) G xx 1 x kt, φ = + k D ( k ) D ( k ) t k cos φ sin φ = + Dm( kt) De( kt) k m t e t The residue of the spectral-domain Green s function at the pole is ( k ) t φ Res G, xx kt = ktp cos φ = D m ktp = pure imaginar Note: This can be calculated in closed form, but the result is omitted here. where = cot D k Y jy k h m t 1 z1 131

132 Surface-Wave Power (cont.) Since the transform of the current is real (assuming that k tp is real), we have π / 1 { } SW tp sx tp, Re Res xx t, π k = k P = k J k φ j G k φ dφ t tp Since the residue is pure imaginar, we then have π / 1 { } SW tp sx tp, Im Res xx t, π k = k P = k J k φ G k φ dφ t tp 13

133 Total Power The total power ma also be calculated directl: P π / 1 Re k, φ k, φ kdkdφ π = G xx J sx TOT t t t t C R Im k t C R Note: h b =.5 k is a good choice for numerical purposes. h b k k1 Re k t 133

134 Surface-Wave Power: Alternative Method The surface-wave power ma then be calculated from: PSW = PTOT PSP This avoids calculating an residues. Im k t C R Total Space h b k k1 Re k t 134

135 EFFICIENCY (%) Results Results: Conductor and dielectric losses are neglected exact CAD h / λ h / λ ε r =. or 1.8 W/L =

136 EFFICIENCY (%) Results (cont.) Results: Accounting for all losses (including conductor and dielectric loss) ε r = 1.8 tanδ = [S/m] σ = exact CAD h / λ h / λ ε r =. or 1.8 W/L =

137 Outline Phsical derivation of method for planar electric surface currents. Examples involving planar surface currents: Microstrip line Microstrip patch current General derivation (Fourier transforming Maxwell s equations) that allows for all tpes of sources to be included in one general derivation. Examples: Vertical dipole over the earth (Sommerfeld problem) Slot antenna covered with radome laer (magnetic current) 137

138 Overview of General SDI Derivation In this part of the notes we derive the SDI formulation using a more mathematical, but more general, approach (we directl Fourier transform Maxwell s equations). This allows for all possible tpes of sources (horizontal, vertical, electric, and magnetic) to be treated in one derivation. 138

139 General SDI Method Start with Ampere s law: = = i H J jωε E = + zˆ t z where ˆ ˆ t = x + x Assume a D spatial transform: ˆ = xˆ jk + jk t x ( ˆ ˆ ) x k = j xk + = = jk t jk uˆ t 139

140 General SDI Method (cont.) i Hence we have jk ˆ t u + zˆ H = J + jωε E z Next, represent the field as Note that uˆ vˆ = zˆ ˆv zˆ uˆ = vˆ zˆ vˆ = uˆ H = uh ˆ + vh ˆ t u v = uˆ H uˆ + ˆ ˆ v( H v) φ û x Take the zˆ, uˆ, vˆ components of the transformed Ampere s equation: 14

141 General SDI Method (cont.) ) i zˆ jk H = J + jωε E t v z z H v i uˆ ) = J u + jωε E u z H u i vˆ ) jk H + = J + jωε E z t z v v Examine the z field: (Ignore equation) ( E,, ) u Hv Ez ˆv i jkth v = J z + jωε E z (1) H v i = J u + jωε E u z () 141

142 z Fields We wish to eliminate E z from Eq. (1). To do this, use Farada s law: = i E M jωµ H i jk ˆ t u + zˆ E = M jωµ H z Take the ˆv component of the transformed Farada s Law: Recall: uˆ vˆ = zˆ zˆ uˆ = vˆ zˆ vˆ = uˆ E jk E + = M jωµ H z u i t z v v (3) 14

143 z Fields (cont.) Substitute E z from (1) into (3) to eliminate E z jk H = J + jωε E i t v z z (1) E jk E + = M jωµ H z u i t z v v (3) E jk J jk H M j H jωε z 1 i u i t z t v + = v ωµ v or E k k z u t i t i + H v + jωµ H v = M v + J z jωε ωε 143

144 z Fields (cont.) Note that kt 1 + jωµ = ω µε jωε jωε 1 = jωε ( kt ) ( kt k ) 1 = jωε kz = jωε ( k kt ) Hence E u k z i kt i H v = M v + J z (4) z jωε ωε 144

145 z Fields (cont.) Equations () and (4) are the final z field modeling equations: H v i = J u jωε E u z E k k = + + z ωε jωε u i t i z M v J z H v 145

146 z Fields (cont.) Define N modeling equations: (,, ) u x V z E k k z (,, ) v x I z H k k z We then have H v i = J u jωε E u z E k k = + + z ωε jωε u i t i z M v J z H v I z = i jωε V Ju k k z jωε ωε V z i t i = I + M v + J z 146

147 Telegrapher s Equation v i - d vs z + d is z L z i + C z + - v + Allow for distributed sources + d v v = vs z L z i + i z t z z so ( d ) s v z = v d s i L t 147

148 Telegrapher s Equation (cont.) Hence, in the phasor domain, V z = jωli + V d s Also, i i = i d z ( C z) z + s so i z = i C d s v t + v t Hence, we have I = jωcv + z I d s 148

149 Telegrapher s Equation (cont.) Compare field equations for z fields with TL equations: I z i jωε V Ju = I = jωcv + z I d s k k z jωε ωε V z i t i = I + M v + J z V z = jωli + V d s 149

150 Telegrapher s Equation (cont.) We then make the following identifications: C ωl = = ε kz ωε Hence TL kz kz = ω LC = ω ε = k ωε Z L k 1 k C ω ε ε ωε TL z z = = = z so k Z TL z TL = = k z kz ωε 15

151 Sources: z For the sources we have, for the z case: I d s = J i u k V = M + J ωε d i t i s v z 151

152 Sources: z (cont.) Special case: planar horizontal surface current sources: J xz,, J x, δ ( z) i i Assume = i = i M xz,, M x, δ ( z) s s Then we have d i i s u su d i i s v sv I = J = J δ z V = M = M δ z These correspond to lumped current and voltage sources: I V s s = J = M i su i sv lumped parallel current source lumped series voltage source 15

153 Sources: z (cont.) For a vertical electric current: Assume i i J xz,, = J ( x, ) δ ( z) z sz planar vertical current distribution z = 153

154 Sources: z (cont.) i i J xz,, = J ( x, ) δ ( z) z d kt i kt i Vs = J z( kx, k) = J sz( kx, k) δ z ωε ωε sz This corresponds to a lumped series voltage source: kt i Vs = J sz( kx, k ) ωε 154

155 Sources: z (cont.) Special case: vertical electric dipole J xz,, = δ xδ δ( z) i z J ( x, ) = δ xδ i sz i J ( k, k ) = 1 sz x Hence V s k ωε t = 155

156 z Fields Use dualit: E H J M i i H E M i J ε µ i H v i = J u jωε E u z E k k = + + z ωε jωε u i t i z M v J z H v E M jωµ H u z H k k =+ + z ωµ jωµ v i = u u i t i z J v M z E v z z 156

157 z (cont.) Define (,, ) v x V z E k k z (,, ) u x I z H k k z V z = i jωµ I Mu I k z k = V + J + M z jωµ ωµ i t i v z V z = jωli + V d s I = jωcv + I z d s We then identif L ωc = = µ kz ωµ k Z TL z = = k z ωµ k z 157

158 z (cont.) For the sources, we have V d s = M i u k I J = + M ωµ d i t i s v z Special case of planar horizontal surface currents: V I s s = M = + J i sv i su lumped series voltage source lumped parallel current source 158

159 Sources: z (cont.) For a vertical magnetic current: Assume i i M xz,, = M ( x, ) δ ( z) z sz This corresponds to a lumped parallel current source: Then we have k I = M k k ωu (, ) t i s sz x 159

160 Sources: z (cont.) Special case: vertical magnetic dipole M xz,, = δ( x) δ δ( z) i z M ( x, ) = δ( x) δ i sz i M k, k = 1 sz x Hence I s k ωu t = 16

161 Summar of Modeling Formulas Results for 3D (volumetric) sources V I V I = E u = H v = E = H u v I V d s d s I V d s d s Horizontal = J i u = M = + J i v = M i v i u V I d s d s Vertical k J ωε t = k M ωµ t = i z i z Distributed sources: either parallel current sources or series voltage sources 161

162 Summar of Modeling Formulas (cont.) Results for D (planar) sources V I V I = E u = H v = E = H u v I V I V Horizontal s s s s = J = M i su = + J = M i sv i sv i su V I s s Vertical k J ωε t = k M ωµ t = i sz i sz Lumped sources: either parallel current sources or series voltage sources 16

163 Summar of Modeling Formulas (cont.) N Models / I / + V / I / + V / I s + - / V s = i = V z V z I / / / s I z I z I / / / i s = v = V z V z V / / / s I z I z V / / / v s 163

164 Summar of Modeling Formulas (cont.) Michalski functions / I i + / V i / I v + / V v 1 [A] [V] 164

165 Outline Phsical derivation of method for planar electric surface currents. Examples involving planar surface currents: Microstrip line Microstrip patch current General derivation (Fourier transforming Maxwell s equations) that allows for all tpes of sources to be included in one general derivation. Examples: Vertical dipole over the earth (Sommerfeld problem) Slot antenna covered with radome laer (magnetic current) 165

166 Sommerfeld Problem In this part of the notes we use SDI theor to solve the classical "Sommerfeld problem" of a vertical dipole over an semi-infinite earth. z Goal: Find E z on the surface of the earth (in the air region). ε Il z = h r x ε = εε 1 rc 166

167 Sommerfeld Problem (cont.) Planar vertical electric current : V s k t = ωε c J where z = sz sz ( J xz,, J x, δ ( z) ) (the impressed source current) For a vertical electric dipole of amplitude I l, we have ( δ δ ) δ J ( x, ) = Il x z z Hence δ J ( x, ) = Il δ x sz Therefore, we have V s k t = ωε c Il 167

168 Sommerfeld Problem (cont.) N: z = h + - Z V s z V s k t = Il ωε Z1 The vertical electric dipole excites z waves onl. 168

169 Sommerfeld Problem (cont.) Find E z (x,,z) inside the air region ( z > ): H = jωε E 1 H H x Ez = jωε x 1 E = jk H + jk H z x x jωε 1 = + ωε ( kh ) x kh x 169

170 Sommerfeld Problem (cont.) ( ˆ ˆ) ( ˆ ˆ) H = H u x + H v x x u v u ( cosφ) H v( sinφ) = H + k k x = Hu + Hv kt kt (, ) k = k k t x ˆv û ( ˆ ˆ) ( ˆ ˆ) H = H u + H v u v H u ( sinφ) + H v( cosφ) k k x = H u + H v kt kt φ x 17

171 Sommerfeld Problem (cont.) Hence cancels 1 k k x k k x E z = kx H u + H v + k H u + H v ωε kt kt kt k t or 1 1 E z = H v kx k ωε kt or E = 1 ( kh ) z t v ωε 171

172 Sommerfeld Problem (cont.) Hence 1 E k k z k I z,, = z x t ωε We use the Michalski normalized current function: k t I ( z) = Iv ( z) Vs = Iv ( z) Il ωε (The v subscript indicates a 1V series source.) We need to calculate the Michalski normalized current function at z = (since we want the field on the surface of the earth). 17

173 Sommerfeld Problem (cont.) Calculation of the Michalski normalized current function Z Z z = h 1V I v + - Z1 z z I v = h + - 1V Z1 z Z Z 1 = = kz ωε k z1 ωε 1 k = k k k z x k = k k k z1 1 x 173

174 Sommerfeld Problem (cont.) I v z = h + - Z1 Z 1V Z z This figure shows how to calculate the Michalski normalized current function: it will be calculated later. 174

175 Sommerfeld Problem (cont.) Return to the calculation of the field: 1 k t E z kx, k, = ( kt) Iv Il ωε ωε Hence we have 1 1 k t j( kxx+ k) E ( x,,) = ( k ) I Il e dk dk ( π ) ωε ωε z t v x or Il 1 j( kxx+ k) E ( x,,) = I e k dk dk ( π ) ( ωε ) z v t x 175

176 Sommerfeld Problem (cont.) Il 1 j( kxx+ k) E ( x,,) = I e k dk dk ( π ) ( ωε ) z v t x Note: I is onl a function of kt v Change to polar coordinates: x = ρcosφ = ρsinφ k k x = = k k t t cosφ sinφ dk dk k dk dφ x t t 176

177 Sommerfeld Problem (cont.) Il 1 j( kxx+ k) E ( x,,) = I e k dk dk ( π ) ( ωε ) z v t x E x I k e d dk π Il 1 3 j( kt cosφρcosφ+ kt sinφρsinφ) z (,,) = v t φ t ( π ) ( ωε ) Switch to polar coordinates dk dk k dk dφ x t t E x I k e d dk π Il 1 3 j( k ) cos (,,) t ρ φ φ z = v t φ t ( π ) ( ωε ) 177

178 Sommerfeld Problem (cont.) Il 1 π 3 j( k ) cos E(,,) t ρ φ φ z x = Iv kt e d ( π ) ( ωε ) φ Use α = φ φ π π φ π jkρ cos φ φ jkρ cosα jkρ cosα φ = α = α t t t e d e d e d φ We see from this result that the vertical field of the vertical electric dipole should not var with angle φ. Integral identit: π jk ( t ρ) cosα I = e d = J k t α π ρ 178

179 Sommerfeld Problem (cont.) Hence we have Il 1 3 E ρ, = J ( k ρ) I k dk ( π ) ( ωε ) z t v t t This is the Sommerfeld form of the field. 179

180 Sommerfeld Problem (cont.) We now return to the calculation of the Michalski normalized current function. I v z = h Z1 Γ Z + - Vv ( z) + - 1V Z z v V z = Ae + Be jk z + jk z z z ( + jk ) zz jkzz e Γ = = B e +Γ Z1 Z < z < h B Iv z e e Z ( + jk ) zz jkzz = +Γ = A B Z + Z 1 18

181 Sommerfeld Problem (cont.) I v z = h Z1 Γ Z + - Vv ( z) + - 1V Z z v z = jk V z Ce z z > h C Iv z e Z z = jk z Here we visualize the transmission line as infinite beond the voltage source. 181

182 Sommerfeld Problem (cont.) I v z = h Z1 Γ Z + - Vv ( z) + - 1V Z z Boundar conditions: v ( + ) ( ) v = 1 V h V h v ( + ) ( = ) v I h I h 18

183 Sommerfeld Problem (cont.) Hence we have v ( + ) ( ) v = 1 V h V h ( ) jk h + jk h jk h Ce B e +Γ e = z z z 1 Z C v ( + ) ( = ) v I h I h B e = e +Γ e ( ) jkzh + jkzh jkzh Z 183

184 Sommerfeld Problem (cont.) Substitute the first of these into the second one: ( ) Ce = 1+ B e +Γ e jk h + jk h jk h z z z Z C B e = e +Γ e ( ) jkzh + jkzh jkzh Z This gives us Z 1 ( + jk ) h jk h + jk h jk h z z z z 1+ B e +Γ e = e +Γ e Z B 184

185 Sommerfeld Problem (cont.) Z 1 B + B e +Γ e = e +Γ e Z ( + jk ) h jk h + jk h jk h 1 z z z z ( + jk ) h jk h + jk h jk h z z z z 1 B e e B e e + +Γ = +Γ ( jk ) z h B e + = 1 cancels Hence we have 1 z B = e jk h 185

186 Sommerfeld Problem (cont.) For the current we then have B Iv z e e Z with Hence ( + jk ) zz jkzz = +Γ 1 z B = e jk h 1 Iv z e e e Z jk zh + jkzz jkzz = +Γ For z = : I v 1 ( ) z = 1 Γ e jk Z h 186

187 Sommerfeld Problem (cont.) We thus have Il 1 3 E ρ, = J ( k ρ) I k dk ( π ) ( ωε ) z t v t t with 1 Iv h e Z z (, ) = ( 1 Γ ) jk Hence Il 1 1 z 3 (,) ( 1 ) jk h E ρ = J k ρ Γ e k dk ( π ) ( ωε ) z t t t Z or Il 1 1 z 3 (,) ( 1 ) jk h E ρ = J k ρ Γ e k dk ( π ) ( ωε ) ωε z t t t kz h 187

188 Sommerfeld Problem (cont.) Il 1 1 z 3 (,) ( 1 ) jk h E ρ = J k ρ Γ e k dk ( π ) ( ωε ) ωε z t t t kz Simplif Il 1 1 ( ) z 3, 1 jk h E ρ = J k ρ Γ e k dk z t t t 4π ωε k z 188

189 Sommerfeld Problem (cont.) Final Result z Il z = h ε r x ε = εε 1 rc ( + ) Il 1 1 ( ) z 3, 1 jk h E ρ = J k ρ Γ e k dk z t t t 4π ωε k z 1 Γ =Γ ( kt ) = Z1 + Z Z Z Z Z `1 kz = ωε kz1 = ωε 1 k = k k z t k = k k z1 1 t 189

190 Sommerfeld Problem (cont.) k ti Note: A box height of about.5k is a good choice. k1 C k k k 1 k tr Zenneck-wave pole ( z ) k tp = k ε ε rc rc + 1 (derivation on next slide) ε rc = complex relative permittivit of the earth (accounting for the conductivit.) 19

191 Sommerfeld Problem (cont.) Zenneck-wave pole ( z ) The TRE is: Z = Z ε k = k rc z z1 ( 1) k ε = ε k k ε tp rc rc rc Z = Z 1 ( k k ) ( k1 k ) ε = rc tp tp k ε k k rc tp = ε rc 1 ε rc k ωε k = ωε z z1 1 ( ε 1) k = ε k k tp rc rc 1 k ε 1 = k ε 1 rc tp rc ε rc Note: Both vertical wavenumbers (k z and k z1 ) are proper. (Power flows downward in both regions.) k tp = k ε ε rc rc

192 Sommerfeld Problem (cont.) Alternative form: The path is extended to the entire real axis. ( + ρ, ) ( ρ) Odd E = J k k dk z t t t Transform the H (1) term: ( 1 ) ( ) Odd kt H ktρ dkt = Odd kt H ( ktρ) dkt We use ( 1 ) ( ( ) ) 1 J( x) = H x + H x ( ) ( ρ ) = Odd k H k dk = = Odd Odd t t t ( ) ( ρ ) k H k dk t t t ( ) ( ρ ) k H k dk t t t 1 H x = H x Use k = k t t 19

193 Sommerfeld Problem (cont.) Hence we have E H k e k dk Il 1 1 ( ) 1 jkz h 3 z ρ, = ( tρ) ( 1 ) t t 4π ωε Γ k z This is a convenient form for deforming the path. 193

194 Sommerfeld Problem (cont.) k ti k 1 k tp C k k k tr k tp k 1 k tp = k ε ε rc rc + 1 The Zenneck-wave is nonphsical because it is a fast wave, but it is proper. It is not captured when deforming to the ESDP (vertical path from k ). 194

195 Sommerfeld Problem (cont.) k ti The Riemann surface has four sheets. k 1 k k tp ESDP Original path C k tp k k 1 k tr Top sheet for both k z and k z1 The ESDP from the branch point at k 1 is usuall not important for a loss earth. Bottom sheet for k z1 Bottom sheet for k z Bottom sheet for both kz and k z1 195

196 Sommerfeld Problem (cont.) k ti k 1 k tp k k k tr k tp k 1 This path deformation makes it appear as if the Zenneck-wave pole is important. 196

197 Sommerfeld Problem (cont.) Throughout much of the th centur, a controvers raged about the realit of the Zenneck wave. Arnold Sommerfeld predicted a surface-wave like field coming from the residue of the Zenneck-wave pole (199). People took measurements and could not find such a wave. Hermann Wel solved the problem in a different wa and did not get the Zenneck wave (1919). Some people (Norton, Niessen) blamed it on a sign error that Sommerfeld had made, though Sommerfeld never admitted to a sign error. Eventuall it was realized that there was no sign error (Collin, 4). The limitation in Sommerfeld s original asmptotic analsis (which shows a Zenneck-wave term) is that the pole must be well separated from the branch point the asmptotic expansion that he used neglects the effects of the pole on the branch point (the saddle point in the steepest-descent plane). When the asmptotic evaluation of the branch-cut integral around k includes the effects of the pole, it turns out that there is no Zenneck-wave term in the total solution (branch-cut integrals + pole-residue term). The easiest wa to explain the fact that the Zenneck wave is not important far awa is that the pole is not captured in deforming to the ESDP paths. 197

198 Sommerfeld Problem (cont.) R. E. Collin, Hertzian Dipole Radiating Over a Loss Earth or Sea: Some Earl and Late th -Centur Controversies, AP-S Magazine, pp , April

199 Sommerfeld Problem (cont.) k ti Alternative path (efficient for large distances ρ) k 1 k tp C k k k tr k tp k 1 This choice of path is convenient because it stas on the top sheet, and et it has fast convergence as the distance ρ increases, due to the Hankel function. 199

200 Outline Phsical derivation of method for planar electric surface currents. Examples involving planar surface currents: Microstrip line Microstrip patch current General derivation (Fourier transforming Maxwell s equations) that allows for all tpes of sources to be included in one general derivation. Examples: Vertical dipole over the earth (Sommerfeld problem) Slot antenna covered with radome laer (magnetic current)

201 Slot Antenna with Radome Find H( x,,) x z L + - V W h ε r, µ r x z V / π (, ) ˆ πx M s x = x cos W L ε r w h 1

202 Slot Antenna with Radome (cont.) (, ) = ( ˆ ˆ) + ( ˆ ˆ) = H u( cosφ ) + H v( sinφ) H k k H u x H v x x x u v I k k I x = + kt kt k k = I M + I M ( x ) ( ) v su v sv kt kt k k = I M u x + I M v x ( x )( ˆ ˆ) ( )( ˆ ˆ) v sx v sx kt kt k k = Iv Msx + Iv kt kt ( x x ) ( M sx ) k k = I M + I M ( x ) ( ) v sx v sx kt kt k k k k t t I V I V V I V I Horizontal s s s s = E u = H v = E = H u = J = M i su = + J = M v i sv i sv i su

203 Slot Antenna with Radome (cont.) Hence (, ) kx k H x kx k = M sx Iv + Iv kt k t For the slot current we have L cos kx kw π L M (, ) ˆ s kx k = xvj π kl x We next calculate the Michalski functions. 3

204 Slot Antenna with Radome (cont.) z Z / I 1V + - / + V Z Z / / 1 I v I v = = Z + jz tan k h = Z jz k h / / / / 1 z1 in Z1 / / 1 + tan z1 h 1 Z in 1 Z in Z Z Z Z 1 1 = = = = kz ωε kz ωε z 1 ωµ k ωµ k z1 4

205 Slot Antenna with Radome (cont.) Input impedance of slot: P c Therefore = P 1 V ( slot Z ) in = 1 * * c sx x slot = 1 * P 1 1 = π M H ds 1 = M H dx d * sx x Z * c sx x slot slot in M H ds = M H dk dk * sx x x V * Pc (Parseval s theorem) 5

206 Slot Antenna with Radome (cont.) Hence, we have * 1 1 k k x P = M I + I dk dk π kt k t c sx v v x so that Z slot in = V 1 k x k M I I dk dk π + k t k t sx v v x 6

207 Slot Antenna with Radome (cont.) Integrating in polar coordinates, and using smmetr to reduce the integration to the first quadrant, we have Z slot in = V / 1 π 4 sx v cos v sin t t π + M I φ I φ kdkdφ with L cos kx kw π L M (, ) ˆ s kx k = xvj π kl x 7

208 References Articles T. Itoh, Spectral Domain Immitance Approach for Dispersion Characteristics of Generalized Printed Transmission Lines, IEEE Trans. Microwave Theor and Techniques, vol. 8, pp , Jul 198. T. Itoh and W. Menzel, A Full-Wave Analsis Method for Open Microstrip Structures, IEEE Trans. Antennas and Propagation, vol. 9, No. 1, Januar 1981, pp

209 References (cont.) Books T. Itoh, Planar Transmission Line Structures, IEEE Press, T. Itoh, Numerical Techniques for Microwave and Millimeter-Wave Passive Structures, Wile,1989. C. Scott, The Spectral Domain Method in Electromagnetics, Artech House, D. Mirshekar-Sahkal, Spectral Domain Method for Microwave Integrated Circuits, Research Studies Press, 199. A. K. Bhattachara, Electromagnetic Fields in Multilaered Structures: Theor and Applications, Artech House, C. Nguen, Analsis Methods for RF, Microwave, and Millimeter-Wave Planar Transmission Line Structures, Wile, 1. 9