ECE Notes 21 Bessel Function Examples. Fall 2017 David R. Jackson. Notes are from D. R. Wilton, Dept. of ECE

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1 ECE 6382 Fall 2017 David R. Jackso Notes 21 Bessel Fuctio Examples Notes are from D. R. Wilto, Dept. of ECE Note: j is used i this set of otes istead of i. 1

2 Impedace of Wire A roud wire made of coductig material is examied. ε 0, µ 0 ε 1, µ 1, σ a k1 = ω µε c ε = ε c1 1 j σ ω The wire has a coductivity of σ. We eglect the variatio of the fields iside the wire ( k << k 1 ). 2

3 Impedace of Wire (cot.) Iside the wire: ( ) E = AJ k e k = ω µε = ω µε c σ ω µε 1 1 j ωε1 = jωµ σ = = ωµ σ 1 jπ /4 ( e ) ωµ 1σ 2 σ j ωε1 jπ /4 ( e ) jk (The field must be fiite o the axis, o φ variatio.) k = k k k ε 0, µ 0 a ε 1, µ 1, σ Note: This assumes that the wire is fed (excited) from the outside. 3

4 Impedace of Wire (cot.) Hece, we have E AJ e π δ /4 = 0 2 j ε 0, µ 0 a ε 1, µ 1, σ where δ = 2 ωµ σ 1 (ski depth) We ca also write the field as 3 /4 3 /4 0 2 j π j π E = AJ e = AJ0 2e δ δ Note : We ca also write E = AJ0 1 j δ ( ) 1 k = β jα = 1 j δ β = α = 1/ δ ( ) 4

5 Impedace of Wire (cot.) Ber Bei υ υ E AJ e π δ j3 /4 = 0 2 Recall: ( ) ( ) ( j3 π /4 x Re J xe ) υ ( ) ( ) ( j3 π /4 x Im J xe ) υ ε 0, µ 0 a ε 1, µ 1, σ Therefore, we ca write E = A Ber0 2 + jbei0 2 δ δ 5

6 Impedace of Wire (cot.) The curret flowig i the wire is I = S J ds ε 0, µ 0 = = = 2π a 0 0 2π a 0 J 2πσ a 0 J ddφ E d d a ε 1, µ 1, σ Hece a j3 π /4 I = 2πσ A J0 2e d δ 0 6

7 Impedace of Wire (cot.) The impedace per uit legth defied as: Hece, Z l = Z a J0 δ 2e a 2πσ δ 2 0 l E ( a) I j3 π /4 j3 π /4 J0 e d ε 0, µ 0 a ε 1, µ 1, σ Note: This assumes that the wire is fed (excited) from the outside. 7

8 Impedace of Wire (cot.) We have the followig helpful itegratio idetity: ( ) = ( ) J x xdx xj x 0 1 ε 0, µ 0 a ε 1, µ 1, σ Hece a 2 L 2 L j3 π/4 1 j3 π/4 a 0 2 = δ 0 0 δ = 0 2 L ( x) 1 0 ( L) L ( ) ( ) J e d e J x xdx J x xdx where a = xj L 2 a = J1 L L a δ 2 j3 /4 e π Use : j3 π /4 x= 2e δ 1 dx = d 2e δ j3 π /4 8

9 Impedace of Wire (cot.) Hece, we have Z l = a j3 π /4 J0 2e δ 1 j3 π/4 a j3 π/4 2πσ aδ e J1 2e 2 δ ε 0, µ 0 where ε 1, µ 1, σ a j3 π /4 a a J0 2e = Ber0 2 + jbei0 2 δ δ δ a j3 π /4 a a J1 2e = Ber1 2 + jbei1 2 δ δ δ a 9

10 Impedace of Wire (cot.) At low frequecy (a << δ): Z l 1 ( 2 ) σ πa ε 0, µ 0 ε 1, µ 1, σ At high frequecy (a >> δ): a Z l Zs 2π a where s s s 1 σδ ( 1 ) Z = R + j R ωµ 2σ 1 = = surface resistace of metal ( ) 10

11 Circular Waveguide The waveguide is homogeeously filled, so we have idepedet TE ad TM modes. a ε r E TM mode: =ψ φ,, ( ) Jυ( k) si( υφ) ψ = e Yυ( k) cos( υφ) jk k k k = 11

12 Circular Waveguide (cot.) (1) φ variatio φ [0, 2 π] ψ( φ, + 2 π, ) = ψ( φ,, ) υ = (uiqueess of solutio) Choose cos( φ ) J ( k ) ψ = Y ( k ) cos( φ) e jk 12

13 Circular Waveguide (cot.) 0, φ, (2) The field should be fiite o the axis ( ) ψ Y ( ) k is ot allowed ψ = cos( φ) J ( k ) e jk k k k 2 =

14 Circular Waveguide (cot.) (3) B.C. s: E a, φ, = 0 ( ) Hece J ( ) 0 ka = 14

15 Circular Waveguide (cot.) J ( ) 0 ka = J (x) Plot show for 0 x 1 x 2 x 3 x ka = x p k = x p a Note: x 0 is ot icluded sice J = a x0 0 (trivial solutio). 15

16 Circular Waveguide (cot.) TM p mode: E cos( ) jk = φ J xp e = 0,1, 2 a 1/2 2 x = = 1, 2,3, a 2 p k k p 16

17 Cutoff Frequecy: TM k = k k k = 0 k = k = x p a 2π f µε = c x p a f TM c c = 2πa ε r x p 17

18 Cutoff Frequecy: TM (cot.) x p values p \ TM 01, TM 11, TM 21, TM 02, 18

19 TE Modes H =ψ φ,, ( ) ψ = cos( φ) J ( k ) e jk I this case we have ψ a, φ, 0 ( ) 19

20 TE Modes (cot.) Set E a,, 0 ( ) φ φ = H = jωε E H Eφ = jωε H 1 At he boudary, the first term o the RHS is ero: H a,, 0 ( ) φ = Hece J ( ) 0 ka = 20

21 TE Modes (cot.) J ( ) 0 ka = J ' (x) Plot show for 1 Recall: 1 J( x) ~ x, 0,1, 2,... = 2! x' 1 x' 2 x' 3 x ka k = x p x p = p = 1,2,3,... a Note: p = 0 is ot icluded (see ext slide). 21

22 TE Modes (cot.) ψ = cos( φ) J jk x p e p= 1, 2, a If p = 0, x = 0 p p = 0 0 J x p = J ( 0) = 0 a = 0 J0 x p = J0( 0) = 1 a ψ jk = e = k = 0 e jk (trivial sol.) This geerates other field compoets that are ero; the resultig field that oly has H violates the magetic Gauss law. 22

23 Cutoff Frequecy: TE k = k k k = 0 k = k = x p a 2π f µε = c x p a f TE c c = x 2πa ε r p 23

24 Cutoff Frequecy:TE x p values p \ TE 11, TE 21, TE 01, TE 31,.. 24

25 TE 11 Mode The domiat mode of circular waveguide is the TE 11 mode. Electric field Magetic field (from Wikipedia) TE 10 mode of rectagular waveguide TE 11 mode of circular waveguide The TE 11 mode ca be thought of as a evolutio of the TE 10 mode of rectagular waveguide as the boudary chages shape. 25

26 Scatterig by Cylider A TM plae wave is icidet o a PEC cylider. y ( φ, ) TM k a x H i θ i x Top view i E i ˆ x H = yh e + y0 jkx ( k) k k x = = k k cosθ siθ i i 26

27 Scatterig by Cylider (cot.) From the plae-wave properties, we have i cos x E = η0h 0 θ e + y i jkx ( k) The total field is writte as the sum of icidet ad scattered parts: For a: E = E + E i s Note: For ay wave of the form exp(-jk ), all field compoets ca be put i terms of E ad H. This is why it is coveiet to work with E. Please see the Appedix. 27

28 Scatterig by Cylider (cot.) We first put i E ito cylidrical form usig the Jacobi-Ager idetity*: 1 E η H cos θe J ( k ) e + i jk = 0 y0 i = j jφ Recall: where 2 2 k cos = kx = k0 k = k0 θi jkx = ( ) ( ) e = j J k e Let k k k x jφ Assume the followig form for the scattered field: 1 ( 2) 0 0cos jk + s E = η Hy θie a H ( k ) e = j jφ *This was derived previously usig the geeratig fuctio. 28

29 Scatterig by Cylider (cot.) At = a E ( a, φ, ) = 0 Hece i (, φ, ) = (, φ, ) s E a E a This yields ( ) ( 2 ) ( ) = J ka ah ka or a = H J (2) ( k ) a ( k a) 29

30 Scatterig by Cylider (cot.) We the have ( ka) + 1 J s jk ( 2) E = η0h 0cos ( ) y θie H ( 2) k e = j H ( ka ) jφ ad s H = 0 (TM ) The other compoets of the scattered field ca be foud from the formulas i the Appedix. 30

31 Curret Lie Source TM : E ( φ),, = ψ( ) I () = I 0 y Coditios: x 1) 2) 3) 4) Allowed agles: Symmetry: = 0 Radiatio coditio: [ ] φ 0, 2π υ = H (2) ( k ) Symmetry: k = 0 ( k = k) E ( ) = AH ( k) Hece (2) 0 31

32 Curret Lie Source (cot.) Our goal is to solve for the costat A: E ( ) = AH ( k) (2) 0 Choose a small circular path: I 0 0 C 32

33 Curret Lie Source (cot.) From Ampere s law ad Stokes theorem: H = J + jωε E i i H d r = J ˆdS + jωε E ˆdS C S S H (2 π ) = I + jωε E ds φ 0 S I 0 0 C Examie the last term (displacemet curret): (2) 0 ( ) E = AH k where (2) 2 j x H0 ( x) ~ γ + l π 2 33

34 Curret Lie Source (cot.) E ( l ) Hece ( ) = O Now use so S E ds C l( ) d dφ 0 Therefore 2π 0 0 Hφ (2 π ) = I 0 H φ = = 1 E jωµ 1 Ak H jωµ ( 2) 0 ( k) (2) E = AH0 ( k) 1 2j 2 1 Ak jωµ π k 2 H H H φ ( 2) 0 ( 2) 0 jωε E jk 1 H = k k k k φ 1 E = jωµ 2 j x ( x) ~ γ l π + 2 2j 2 1 ( x)~ π x 2 34

35 Curret Lie Source (cot.) Hece 1 2j 2 1 Ak (2 π) jωµ π k 2 = I 0 or ( ωµ ) ( 4/ ) I0 A = Thus so ωµ I A = 4 ωµ I = 4 0 (2) E H0 ( k ) 0 35

36 Scatterig From a Lie Curret ωµ I = 4 i 0 (2) E H0 ( kr ) a R = 0 I 0 We use the additio theorem to traslate the Hakel fuctio to the axis. y x (, φ )

37 Scatterig From a Lie Curret (cot.) The additio theorem tells us: H + (2) H ( ) ( ) k J k e < = ( k ) = + ( ) (2) J ( ) k H k e > = (2) 0 0 j( φ φ0 ) 0 0 j( φ φ0 ) 0 0 We use the first form, sice the cylider at = a is iside the circle o which the lie source resides (radius 0 ). 37

38 Scatterig From a Lie Curret (cot.) Icidet field: + i ωµ I0 (2) ( ) ( ) j 0 E H k0 J k e φ = φ 4 = ( valid for < ) 0 ( ) Assume a form for the scattered field: s ψ ωµ I (2) ( ) j( 0 ) ah k e φ φ ( valid for > a) = 4 =

39 Scatterig From a Lie Curret (cot.) Boudary Coditios ( = a): Hece s i E ( a, φ, ) = E ( a, φ, ) a H ( ka) = H ( k ) J ( ka) (2) (2) 0 or H ( 2) 0 a = J ka H ( 2) ( k ) ( ka) ( ) 39

40 Scatterig From a Lie Curret (cot.) Fial result: ( 2) ωµ I H ( k ) E J ka H k e + s 0 0 ( 2 ) j( φ φ0 ) = ( ) (2) ( ) 4 = H ( ka) a I 0 Scattered field y x ( 0, φ0) 40

41 Dielectric Rod 0 = ε 1 r a 1 ε, µ r r Ukow waveumber: k < k < k 0 1 Modes are hybrid* uless: = 0 ( = 0) φ Note: We ca have TE 0p, TM 0p modes *This meas that we eed both E ad H. 41

42 Dielectric Rod (cot.) < a Represetatio of potetials iside the rod: ( ) si ( φ) E = AJ k e ( ) cos( φ) H = BJ k e jk jk where k = k k (k is ukow)

43 Dielectric Rod (cot.) To see choice of si/cos, examie the field compoets (for example E ): From the Appedix: E jωµ 1 H jk E = k k φ k k 43

44 Dielectric Rod (cot.) > a Represetatio of potetials outside the rod: Use ( 2) ( 2) H ( k ) = H ( jα ) 0 0 where ( 2 2) 1/2 k = k k = jα α = k k Note: α 0 is iterpreted as a positive real umber i order to have decay radially i the air regio. 44

45 Dielectric Rod (cot.) Useful idetity: ( ) ( ) 2 ( ) H ( jx) = 1 H ( + jx) Aother useful idetity: ( 1 ) 2 ( + 1) H ( jx) = j K( x) π K (x) = modified Bessel fuctio of the secod kid. 45

46 Dielectric Rod (cot.) The modified Bessel fuctios decay expoetially. 1 1 K0( x) K1( x) K1 ( x) K0 ( x) x x 5 46

47 Dielectric Rod (cot.) Hece, we choose the followig forms i the air regio ( > a): E = CK ( α )si( φ) e 0 0 H = DK ( α )cos( φ) e 0 0 jk jk α = k k

48 Dielectric Rod (cot.) Match E, H, E φ, H φ at = a: Example: 1 0 M11 M12 M13 M14 A 0 M21 M22 M23 M 24 B 0 = M31 M32 M33 M34 C 0 M41 M42 M43 M44 D 0 E = E AJ ( k1a) = CK( α0a) or ( 1 ) ( ) ( α0 ) ( ) ( ) AJ k a + B 0 + C K a + D 0 = 0 so ( ) ( α ) M = J k a, M = K a, M = M =

49 Dielectric Rod (cot.) M11 M12 M13 M14 A 0 M21 M22 M23 M 24 B 0 = M31 M32 M33 M 34 C 0 M41 M42 M43 M44 D 0 To have a o-trivial solutio, we require that Mk ω ( det, ) = 0 k = ukow (for a give frequecy ω) 49

50 Dielectric Rod (cot.) Domiat mode (lowest cutoff frequecy): HE 11 (f c = 0) E This is the mode that is used i fiber-optic guides (sigle-mode fiber). 50

51 Dielectric Rod (cot.) Sketch of ormalied waveumber k / k 0 ε r 1.0 f α = k k At higher frequecies, the fields are more tightly boud to the rod. 51

52 Appedix For ay wave of the form exp(-jk ), all field compoets ca be put i terms of E ad H. E E x y jωµ H jk E = k k y k k x jωµ H jk E = k k x k k y H H x y jωε E jk H = k k y k k x c jωε E jk H = k k x k k y c

53 Appedix (cot.) These may be writte more compactly as E = jωµ H jk E ( ˆ ) ( ) t 2 2 t 2 2 t k k k k H = jωε E jk H ( ˆ ) ( ) t 2 2 t 2 2 t k k k k where Φ Φ ˆ ˆ t x + y x Φ y 53

54 Appedix (cot.) I cylidrical coordiates we have Φ ˆ 1 Φ Φ= ˆ t + φ φ This allows us to calculate the field compoets i terms of E ad H i cylidrical coordiates. 54

55 Appedix (cot.) I cylidrical coordiates we have E E φ jωµ 1 H jk E = k k φ k k jωµ H jk 1 E = k k k k φ H H φ jωε 1 E jk H = k k φ k k jωε E jk 1 H = k k k k φ 55