ECE Spring Prof. David R. Jackson ECE Dept. Notes 15

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1 ECE 6341 Spring 216 Prof. David R. Jackson ECE Dept. Notes 15 1

2 Arbitrary Line Current TM : A (, ) Introduce Fourier Transform: I I + ( k ) jk = I e d x y 1 I = I ( k ) jk e dk 2π 2

3 Arbitrary Line Current (cont.) View this as a collection of phased line currents:: di = I e jk 1 I = I ( k) dk 2π 1 I = I ( k ) jk e dk 2π Then = µ I 4 j µ 1 4j 2π = ( 2 ) jk da H k e (from Notes 11) ( 2 I ( k ) ) dk H k e jk 3

4 Arbitrary Line Current (cont.) Hence, from superposition, the total magnetic vector potential is µ jk = ( 2 ) ( ) A I k H k e dk 8π j ( 2 2 ) 1/2 k k k = k 2 2 k k, k k = 2 2 j k k, k k 4

5 Example Uniform phased line current I = I e jk I y jk + jk I k I e e d = = j( k k ) I e d = I 2 πδ ( k k ) x δ 1 jkxx x = e dk 2π 1 = 2π Note: e + jk x x dk x x 5

6 Example (cont.) Hence µ A I k H k e dk ( 2 ) jk = 8π j µ ( 2 [ ] ) jk = I2 πδ ( k k) H k e dk 8π j = = µ 8π j µ I 4 j ( 2 ) ( ) I 2πH k e ( 2 ) ( ) H k e jk jk where ( 2 2 ) 1/2 k = k k 6

7 Example (cont.) Hence µ I = 4 j (2) A H ( k ) e jk I where ( 2 2 ) 1/2 k = k k y If k is real, then x k 2 2 k k, k k = 2 2 j k k, k k This is the correct choice of the wavenumber. 7

8 Example (cont.) If k is complex: k = β ( 2 2 ) 1/2 k = k k jα I ( k ) ( k ) < β < k: Im > ( improper) β > k: Im < ( proper) x y This is the physical choice of the wavenumber. Note that the radiation condition at infinity is violated ( - ), so we lose uniqueness. However, we can still talk about what choice of the square root is physical. 8

9 Example (cont.) < β < k: Im k > ( improper) I Region of exponential growth Cone of leakage y x A semi-infinite leaky-wave line source produces a cone of radiation in the near field. 9

10 Example Dipole Il y x I = Il δ I ( k ) = Il 1

11 Example (cont.) Hence µ A I k H k e dk ( 2 ) jk = 8π j µ ( 2) A jk = I H k e dk 8π j Also, A 2 2 k k, k k = 2 2 j k k, k k µ jkr e = I (from ECE 634) 4π r k 11

12 Example (cont.) Hence, e r jkr 1 = 2 j ( 2) jk H ( k ) e dk A spherical wave is thus expressed as a collection of cylindrical waves. 12

13 Example (cont.) Use a change of variables: We then have 2 2 ( ) k = k k 2 2 ( k k ) 1 ( 2) jk 1 k ( 2) H ( k ) e dk H ( k ) e dk 2j = 2j k jk C or CCW 1 k ( 2) jk = H ( k) e dk 2 j k C CW 1/2 k k dk = dk dk 1/2 = k jkr e 1 k ( 2) jk = H ( k) e dk r 2 j k (The path has been deformed back to the real axis, please see the next two slides) 13

14 Example (cont.) Im ( k ) C CCW : counterclockwise around branch cut Mapping equation for C: ( 2 2) 1/2 k = k k k C CCW k Re( k ) k C CCW (, ) k Branch cut Note: The value of k is opposite across the branch cut. The square root is defined so that when k is on the real axis we have: k = j k k, k > k

15 Example (cont.) Im ( k ) C CW : clockwise around branch cut k C CW k Re( k ) Im ( k ) Deform path k k Re( k ) Sommerfeld path Branch cut 15

16 Example (cont.) Alternative form: k = k e r H = H jkr j k ( 2) jk = H ( k) e dk 2 k j k j k = 2 H k e dk H k 2 k e dk j k j k = H k e dk H 2 k 2 k e dk j k ( 1) jk j k ( 2) jk = + H 2 ( k ) e dk H ( k) e dk k 2 k j k ( 1) jk j k ( 2) jk = H ( k) e dk H ( k) e dk 2 k 2 k k = k ( 2 ) jk ( 2) jk ( ) ( ) k ( 2 ) jk ( 2) jk ( ) ( ) k ( ) jk j J k e dk ( 1 ) ( 2 ( ) ) 1 J = H + H 2 16

17 Example (cont.) Hence we have: e 1 k = jkr r j k jk J ( k ) e dk jkr e 1 k ( 2) jk = H ( k) e dk r 2 j k Sommerfeld Identities The integrals are along a Sommerfeld path that stays slightly above or below the branch cuts. 17

18 Far-Field Identity This identity is useful for calculating the far-field of finite (3-D) sources in cylindrical coordinates. r θ I y x Note: We assume that the current decays at = ± fast enough so that a 3-D far field exists. 18

19 Far-Field Identity (cont.) r θ I y x Exact solution: µ A = I k H k e dk 8π j ( 2) jk 19

20 Far-Field Identity (cont.) From ECE 634, as r A µ e 4π r jkr ~ a(, ) θφ ( sinθcosφ + sinθsinφ + cosθ) jk x y a( θφ, ) = J r e dx dy d V µ e = 4π r jkr j k cos ˆ θ I( ) e d Hence A µ e 4π r jkr ~ I ( kcos θ ) 2

21 Far-Field Identity (cont.) Hence, comparing these two, jkr µ ( 2) jk µ e Ik H ( k ) e dk ~ Ik ( cos θ) 8π j 4π r or jkr ( 2) jk e Ik ( ) H ( k ) e dk ~ 2 j Ik ( cos θ) r as r 21

22 Far-Field Identity (cont.) To generalie this identity, use 2 H x e π x nπ π 2 n ~ 2 4 j( x ) Hence ( 2) n ( 2) H ( x)~ j H ( x) n 22

23 Far-Field Identity (cont.) Therefore, we have e I k H k e dk j I k r jkr ( 2) jk n+ 1 ( ) n ( ) ~ 2 ( cos θ) Note: This is valid for Hence, this is valid for r, θ, 18 23

24 Far-Field Identity (cont.) Since the current function is arbitrary, we can write e f k H k e dk j f k r jkr ( 2) jk n+ 1 ( ) n ( ) ~ 2 ( cos θ) for r, θ, 18 24

ECE Spring Prof. David R. Jackson ECE Dept. Notes 6

ECE Spring Prof. David R. Jackson ECE Dept. Notes 6 ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 6 1 Leaky Modes v TM 1 Mode SW 1 v= utan u ε R 2 R kh 0 n1 r = ( ) 1 u Splitting point ISW f = f s f > f s We will examine the solutions as the