ECE Spring Prof. David R. Jackson ECE Dept. Notes 24

Transcription

1 ECE 6345 Sprng 015 Prof. Dav R. Jackon ECE Dept. Note 4 1

2 Overvew In th et of note we erve the SDI formlaton ng a more mathematcal, bt general, approach (we rectly Forer tranform Maxwell eqaton). Th allow for all poble type of orce to be treate n one ervaton.

3 General SDI Metho Start wth Ampere law: where = = H J E = + ˆ ˆ t = x + y x t y ˆ Ame a D patal tranform: ( ) ˆ( ) = xˆ jk + y jk t x y ( ˆ ˆ ) x yky = j xk + = = jk t jk ˆ t 3

4 General SDI Metho (cont.) Hence we have jk ˆ t + ˆ H = J + E Next, repreent the fel a H = ˆH + vˆh + H ˆ v ( ) ( ) ( ) = ˆ H ˆ + vˆ H vˆ ++ ˆ H ˆ Note that ˆ vˆ = ˆ ˆ ˆ = vˆ ˆ vˆ = ˆ Take the v ˆ, ˆ, ˆ component of the tranforme Ampere eqaton 4

5 General SDI Metho (cont.) ) ˆ jk H = J + E t v H v ˆ ) = J + E H vˆ ) jk H + = J + E t v v Examne fel: Ignore eqaton ( E,, ) Hv E ˆv jkth v = J + E (1) H v = J + E () 5

6 We wh to elmnate E Fel. To o th, e Faraay law: = E M jωµ H jk ˆ t + ˆ E = M jωµ H Take the ˆv component of the tranforme Faraay Law: E jk E + = M jωµ H t v v (3) 6

7 Fel (cont.) Sbttte E from (1) nto (3) to obtan ( ) E jk J jk H M j H 1 t t v + = v ωµ v Pttng all the orce on the RHS: E k k t t + H + jωµ H = M + J ωε v v v Note that kt 1 + jωµ = ω µε 1 = = 1 k ( kt ) ( kt k ) 7

8 Fel (cont.) Hence E k kt H v = M v + J (4) ωε 8

9 Fel (cont.) Eqaton () an (4) are rewrtten a H v = J E E k k = + + ωε t M v J H v 9

10 Fel (cont.) Defne: ( ) = (,, ) V E kx ky ( ) = (,, ) I Hv kx ky We then have: I = V J k k ωε V t = I + M v + J 10

11 Telegrapher Eqaton v v + L + C + - v + Allow for trbte orce + ( ) v v = L + v t o v = L + t v 11

12 Telegrapher Eqaton (cont.) Hence, n the phaor oman, V = jωli + V v = C + t + Alo, ( ) o Hence, n the phaor oman, v = C + t I = jωcv + I 1

13 Telegrapher Eqaton (cont.) Compare fel eqaton for fel wth TL eqaton: I ( ) ( ) = jω ε V + J I = jωcv + I k k ω ω ε ωε V t = j I + M v + J V = jωli + V 13

14 Telegrapher Eqaton (cont.) We then make the followng entfcaton: C L = = ε k ωε Hence TL k k = ω LC = ω ε = k ωε Z L C k ω ε k ωε TL 0 = = = or k Z TL TL 0 = = k k ωε 14

15 Sorce: For the orce we have, for the cae: I = J k V = M + J ωε t v 15

16 Sorce: (cont.) Specal cae: planar rface-crrent orce Ame J xy,, = J xy, δ ( ) ( ) ( ) M xy,, = M xy, δ ( ) ( ) ( ) ( ) I = J δ ( ) V = M δ v Then we have I = J Th a lmpe parallel crrent generator. Smlarly, V = M v Th a lmpe ere voltage generator. 16

17 Sorce: (cont.) Specal cae: vertcal planar electrc crrent If ( ) J xy,, = f( xy, ) δ ( ) Then we have k V = f k k ωε (, ) t x y = 0 Example: f( xy, ) δ( x) δ( y) ( x y) f k, k = 1 = (nt-amplte vertcal electrc pole) 17

18 Fel Ue alty: E H J M H E M J ε µ H v = J E E k k = + + ωε t M v J H v E M jωµ H H k k =+ + ωµ jωµ v = t J v M E v 18

19 (cont.) Defne ( ) = (,, ) V Ev kx ky ( ) = (,, ) I H kx ky V ( ) = + jω µ I M I k kt = jω V + J v + M ω µ ωµ We then entfy: L C = = µ k ωµ k Z TL 0 = = k ωµ k 19

20 (cont.) For the orce, we have V = M k I J = + M ωµ t v Specal cae of horontal rface crrent: V I = M = + J v Specal cae of vertcal planar crrent: M = gxy (, ) δ ( ) I k g ωµ t = ( ) 0

21 Smmary V I V I = E = H v = E = H v I = J k V = M + J ωε t v V = M k I J = + M ωµ t v Specal cae of horontal rface crrent: I V = J = M v I V = M = + J v 1

22 Smmary (cont.) Specal cae of vertcal planar crrent: V k ωε t = f ( J = f( xy, ) δ ( ) ) I k t = ωµ g ( M = gxy (, ) δ ( ) )

23 Example Calclate G To calclate G H = J + E e: We then have E 1 H y H x 1 = J x y o that 1 ( ) 1 E = jk H + jk H J x y y x ob ob 3

24 Example (cont.) We have H = H coφ+ H x v = I coφ+ I ( nφ) ( nφ) H = H nφ+ H y v = I nφ + I coφ coφ The part cancel when we bttte thee expreon nto the expreon for the tranform of E, o we have k ( co n ) 1 t E = φ + φ I J ωεob ob 4

25 Example (cont.) or kt 1 E = I J ωε ob ob From the trength of the voltage generator e to a ntamplte vertcal pole at, we then have I k I t = ωε rc v We alo have J = J = f( xy, ) δ ( ) where f( xy, ) = δ( x) δ( y) f ( k, k ) = 1 x y 5

26 Example (cont.) Hence, we have 1 kt 1 E = Iv δ ωεob ωε rc ob ( ) Becae of the elta fncton, we can replace the obervaton bcrpt wth the orce bcrpt, o that 1 kt 1 E = Iv δ ωεob ωε rc rc ( ) Note: If the vertcal pole at an nterface between two fferent materal, then the elta-fncton term not well efne. In th cae, we nterpret the orce a reng nfntemally on one e of the nterface (the orce e) 6

27 Example (cont.) In the pace oman we have (after takng the D nvere Forer tranform): 1 1 k t jkx ( x + kyy) E = I e k k ( π ) ωεob ωε rc v x y 1 rc ( ) ( ) ( x) ( y) δ δ δ where we have e ( ) = δ ( ) δ ( ) F 1 1 x y 7

28 Example (cont.) Convertng to polar coornate, we have (for any fncton F): π jkx ( x + kyy) jkx ( x + kyy) F k e k k = F k e k φ k ( ) ( ) t x y t t t 0 0 = = = = π jk ( tρ)( coφ coφ+ nφ nφ) F ( kt) kt e φ kt 0 0 π jk ( tρ) co( φ φ) F ( kt) kt e φ kt 0 0 π jk ( tρ) coφ F ( kt) kt e φ kt ( 0 ) ( ) π ( ρ) F k k J k k t t t t 8

29 In the pace oman we then have Example (cont.) o that k t E = Iv J0 ( ktρ) ρ π ωε 0 ob ωε rc 1 δ δ δ rc ( ) ( ) ( x) ( y) k t G = Iv J0 ( ktρ) ρ π ωε 0 ob ωε rc 1 δ δ δ rc ( ) ( ) ( x) ( y) Note: The ntegral oe not converge when =, an t mt be nterprete n a lmtng ene. 9