ECE 6340 Intermediate EM Waves. Fall Prof. David R. Jackson Dept. of ECE. Notes 22

Transcription

1 ECE 634 Intemediate EM Waves Fall 6 Pof. David R. Jackson Dept. of ECE Notes

2 Radiation z Infinitesimal dipole: I l y kl << x z Cuent model: J i z I Il = = x y x y z z = l y x y x

3 Radiation (cont.) Define Then x x x, + Px ( ) = x othewise J( xyz,, ) = ( Il) PxPyPz ( ) ( ) ( ) i z As x, Px ( ) δ ( x) Letting x, y, z o J ( x, y, z) = ( Il) δ( x) δ( y) δ( z) i z J ( x, y, z) = ( Il) δ ( ) i z 3

4 Radiation (cont.) Maxwell s Equations: E = jωµ H i H = J + jωε E v c () () H = (3) E = ρ / ε (4) Note: ε c accounts fo conductivity. i J( xyz,, ) = zil ˆ( ) δ ( ) Fom (3): H = A (5) µ H Note: The Haington book uses = Most efeences use A B = A 4

5 Radiation (cont.) Fom () and (5): E = jωµ ( A) µ o ( E+ jω A) = o Hence E+ jω A= Φ E = jω A Φ (6) The function Φ is called the potential function. It is uniquely defined, even though voltage is not. This is the mixed potential fom fo E. 5

6 Radiation (cont.) Substitute (5) and (6) into (): i H = J + jωε E c ( A) = J i + jωε c[ jω A Φ] µ i A k A J j ( ) = µ ωµε Φ c Use the vecto Laplacian identity: A ( A) ( A) The vecto Laplacian has this nice popety in ectangula coodinates: A= xˆ A + yˆ A + zˆ A x y z 6

7 Radiation (cont.) Hence i A A k A J j ( ) = µ ωµε Φ c o i A+ k A= µ J + A + jωµεc Φ [ ( ) ] Loenz Gauge: Note: Loenz, not Loentz (as in Loentz foce law)! Choose A= jωµε c Φ Then we have i A+ k A= µ J k = ω µε c 7

8 Radiation (cont.) Take the z component: A + k A = µ J i z z z Hence we have z A + k A = µ ( Il) δ( ) z Note: A + k A = x A + k A = y x y so A x = A = y 8

9 Radiation (cont.) A + k A = µ ( Il) δ( ) z Spheical shell model of 3D delta function: z z S a a The A z field should be symmetic (a function of only) since the souce is. x Shell y In spheical coodinates: δ() = δ a 4π a ( ) Note : V a ( ) ( ) δ dv = δ 4π d = a + 9

10 Radiation (cont.) Assume A() = R () z Fo we have that A + k A = z z so that d dr kr + = d d Next, let R () = h () R( ) dr() () () d = h () h() = h h

11 Radiation (cont.) We then have: d h ( h h) + k = h ( ) h () + kh () = d h + h h + k = Solution: jk h() = Ae + Be + jk We choose exp (-jk) fo > a to satisfy the adiation condition at infinity. We choose sin(k) fo < a to ensue that the potential is finite at the oigin. ( Recall that R( ) = h ( )/. )

12 Radiation (cont.) Hence ( ) h jk Ae, > a = B sin ( k ), < a The R function satisfies d dr ( ) ( ) ( ) + k R = µ Il δ = µ Il δ a 4 d d π a h () k h() µ ( Il) δ a 4 + = π a ( ) h () + k h() = µ ( Il) δ a 4 π a ( ) ( )

13 Radiation (cont.) We equie ( + ) h( = a ) h a (BC #) Also, fom the diffeential equation we have a a a ( ) h () d + k h() d = µ ( Il) δ a d 4π a a a a ( + ) ( ) h a h a = µ ( Il) 4 π a (BC #) 3

14 Radiation (cont.) Recall ( ) h jk Ae, > a = B sin ( k ), < a Hence ( ) jka Ae = B sin ka (BC #) jka A( jk ) e Bk cos ( ka) = µ ( Il) 4 π a (BC #) 4

15 Radiation (cont.) Substituting Ae -jka fom the fist equation into the second, we have jk Bsin ka Bk cos ka = µ ( Il) 4 π a ( ) ( ) ( ) Hence, we have Fom BC # we then have B = µ ( Il) 4 π a k ( j sin ( ka) + cos ( ka) ) A jka sin ( ka) ( ) + ( ) e = µ ( Il) 4 π a k ( j sin ka cos ka ) 5

16 Radiation (cont.) Letting a, we have jka sin ( ka) ( ) + ( ) e A = µ ( Il) 4 π a k ( j sin ka cos ka ) µ ( Il) 4 π Hence so A = µ ( Il) 4 π Az R h Ae ( Il) e a 4 π jk jk ( ) = ( ) = ( ) = = µ ( > ) 6

17 Radiation (cont.) Final esult: z I l y x J ( x, y, z) = ( Il) δ ( ) i z jk µ e A = zˆ ( Il) 4π ( ˆ cos ˆsin ) µ = θ θ θ ( Il) 4π e jk Note: It is the moment Il that is impotant. 7

18 Extensions (a) Dipole not at the oigin z y x µ A = zˆ ( Il) 4π e jk 8

19 Extensions (cont.) z (b) Dipole not in the z diection ˆp y x µ A = pˆ ( Il) 4π e jk 9

20 Extensions (cont.) (c) Volume cuent z i J ( ) y Conside the z component of the cuent: i J dv ds dl = = i zˆ J ds dl z zi ˆ dl x Hence jk jk µ e i µ e d A = zˆ ( Idl) = J dv 4 π 4 π jk µ i e A = J ( ) dv 4π V The same esult is obtained fo the cuent components in the x and y diections, so this is a geneal esult.

21 Extensions (cont.) (d) Volume, suface, and filamental cuents i J dv i J dv i s J ds ( volume cuent density) ( suface cuent density) ˆ I d ( filamentay (wie) cuent) jk µ i e J ( ) dv ( ) 4π volume cuent density V jk µ i e A( ) = J s( ) ds ( ) 4π suface cuent density S jk µ ˆ( ) i e I ( ) d ( ) 4π filament of cuent C

22 Extensions (cont.) Filament of cuent (useful fo wie antennas): jk µ ( ) ˆ( ) i e A = I ( ) d 4π C I ˆ B A C Note: The cuent I i is the cuent that flows in the diection of the unit tangent vecto along the contou.

23 Fields To find the fields: H = A µ E = H ( valid fo ). jωε c Altenative way to find the electic field: E = jω A Φ = jω A+ A jωµε c ( ) Recall : A= jωµε c Φ 3

24 Fields (cont.) Results fo infinitesimal, z-diected dipole at the oigin: E Il jk = ηe + cosθ π jk Il jk Eθ = ( jωµ ) e sinθ 4 π + + jk ( jk) H φ Il jk = ( jk) e + sinθ 4π jk with η = µ ε c 4

25 Note on Dissipated Powe Note: If the medium is lossy, thee will be an infinite amount of powe dissipated by the infinitesimal dipole. P d = ωε V ε ε π π ε π π ( ) ωε Eθ + E sinθ dφ dθ d E dv = ωε E sinθ dφ dθ d 4 = ωε (powe dissipation inside of a small spheical egion V ε of adius ε) ε ( ) ( ) N N π θ E E d ε ( ) 3 4 Il ωε ( π ) + + ( ) 3 3 4πωε c = d The N supescipt denotes that the θ dependence is suppessed in the fields. π π sin cos θsinθ dθ = Note: θsinθ dθ =

26 Fa Field Fa-field of dipole: ( k ) E O jωµ jk Eθ ~ ( Il) e sinθ 4π jk jk Hφ ~ ( Il) e sinθ 4π behaves as Note that E H θ φ ωµ = = η k The fa-zone field acts like a homogeneous plane wave. k = kˆ, = ˆ, k = k 6

27 Radiated Powe (lossless case) P Re ( ) ˆ ad = E H ds S The adius of the sphee is chosen as infinite in ode to simplify the fields. = = = = Re Re η π η ππ π S S * EθH φ ds E θ ds E sinθ dθdφ θ E sinθ dθ θ η We assume lossless media hee (η is eal). 7

28 Radiated Powe (cont.) o π π ωµ Pad = ( Il) (sin θ) sinθ dθ η 4π π 3 π ωµ = ( Il) sin η 4π θ dθ Integal esult: π sin 3 θ dθ = 4 3 Hence Pad π ωµ 4 = ( Il) η 4π 3 8

29 Radiated Powe (cont.) Simplify using ωµ = kη P ad π kη 4 = ( Il) η 4π 3 o P ad = k ( Il) η π Use k = π λ Then Pad = 4πη ( Il) πλ 9

30 Radiated Powe (cont.) We then have P ad l πη = I λ 3 [ W] Note: The dipole adiation becomes significant when the dipole length is significant elative to a wavelength. 3

31 Potential Function We do not need the potential function in ode to calculate the fields, but we can obtain the potential function if we wish. E = jω A Φ E = jω A+ A jωµε c Needs Φ ( ) Recall : c Does not need Φ A= jωµε Φ i Anothe fom: J = ( away fom the dipole) E = H J jωε jωε c c i = ( A) J Does not need Φ jωµε jωε c i (fom Ampee s law) c 3

32 Potential Function (cont.) The potential function is given by: ( ) i ρv jk Φ ( ) = e dv 4πε V c whee i i J j v = ωρ ( ) (This is left as a HW poblem.) Note: The potential function is uniquely defined, even at high fequency, but voltage is not. In statics, potential = voltage. 3

33 Fa-Field Radiation Fom Abitay Cuent i J ( ) z y x jk µ i e A = J ( x, y, z ) dv 4π V 33

34 Fa- Field (cont.) ( ) ( ) ( ) = x x + y y + z z ( ) ( x y z x y z ) ( xx yy zz ) = / = + = + ˆ = + / x Use Taylo seies expansion: x 8 + x ~ + x

35 Fa- Field (cont.) ˆ ˆ = + + O 3 8 = ˆ+ ( ) ( ˆ) O + As ˆ 35

36 Fa- Field (cont.) ˆ Physical intepetation: z Fa-field point (at infinity) ˆ y x ( sinθcosφ) ( sinθsinφ) ( cosθ) ˆ = xˆ + yˆ + zˆ 36

37 Fa- Field (cont.) Using this appoximation, we have: ˆ jk jk i e e i jk ˆ µ A() = J ( x, y, z ) dv ~ µ J ( ) e dv 4π 4π V V Define: ψ () e jk ( sinθcosφ) ( sinθsinφ) ( cosθ) k kˆ = xˆ k + yˆ k + zˆ k (This is the k vecto fo a plane wave popagating towads the obsevation point.) and i + jk a( θφ, ) J ( ) e dv V 37

38 Fa- Field (cont.) The we have whee µ A ( )~ ψ a θφ, 4π ( ) ( ) ψ () = e jk i + jk a( θφ, ) = J ( ) e dv V ( sinθcosφ) ( sinθsinφ) ( cosθ) k = x k + y k + z k 38

39 Fouie Tansfom Intepetation i + jk a( θφ, ) = J ( ) e dv V vecto aay facto Denote k k k x y z = = = ksinθ cosφ ksinθsinφ kcosθ Then i + jkx ( x + ky y + kz z ) a( θφ, ) = J ( x, y, z ) e dv V The vecto aay facto is the 3D Fouie tansfom of the cuent. 39

40 Magnetic Field We next calculate the fields fa away fom the souce, stating with the magnetic field. H = A µ In the fa field the spheical wave acts as a plane wave, and hence jk = jkˆ We then have H = jk A µ ( )( ˆ ) Recall µ A~ ψ ( ) a( θφ, ) 4π Hence H jk µ = ψ µ 4π a ( ) ( )( ˆ ) 4

41 Electic Field We stat with jk H ~ ( ˆ a(, θφ) ) ψ() 4π The fa-zone electic field is then given by E = H jωε jωε c c ( jkˆ ) jk = jωε 4π c H ( jkˆ) ( ˆ a) µ = jω ψ a 4π ( ˆ ( ˆ )) ψ 4

42 Electic Field (cont.) We next simplify the expession fo the fa-zone electic field. µ E jω ψ a θφ 4π µ = jω ψat ( θφ, ) 4π ( ) ( ˆ ( ˆ, )) Recall that µ A ( )~ ψ θφ, 4π ( ) a( ) Hence E jω A t 4

43 Electic Field (cont.) Hence, we have ( θφ) E ~ jωa,, t Note: The exact field is E = jω A Φ The exact electic field has all thee components (, θ, φ) in geneal, but in the fa field thee ae only (θ, φ) components. 43

44 Relation Between E and H Stat with jk H a 4π jk = a 4π Hence ( ˆ ) ( ˆ ) t ψ ψ jk E = ˆ ψ 4 π jωµψ / (4 π ) µ E jω ψa t θφ 4π (, ) Simplify using k ω µε c ε c = = = ωµ ωµ µ η H ˆ E η ( ) o E H θ φ Eφ = η, = η H θ 44

45 Summay of Fa Field Recipe E ~ jωa,, t ( θφ) (keep only θ and φ components of A) ψ () µ A (, θφ, )~ ψ aθφ, 4π e jk i + jk = a( θφ, ) = J ( ) e dv x y z H ˆ E η ( ) ( ) V ( ) ( sinθcosφ) ( sinθsinφ) ( cosθ) k = kx + ky + kz = k x + k y + k z 45

46 Summay of Fa Field Recipe To specialize to any dimensionality, we have: i + jk a( θφ, ) = J ( ) e dv V i + jk a( θφ, ) = J ( ) e ds S s (, ) ˆ + jk a θφ = ( ) I( ) e d C 46

47 Poynting Vecto S = E H ~ Et H t = ˆ + ˆ ˆ + ˆ ˆ ( * * = EH EH ) θ φ φ θ * EE EE ˆ θ θ φ φ = + * * η η ( ) ( ) * * θeθ φeφ θhθ φhφ 47

48 Poynting Vecto (cont.) E E E θ φ ˆ * * * η η η S = ˆ + = Assuming a lossless media, E E θ φ E S = ˆ + = ˆ η η η In a lossless medium, the Poynting vecto is puely eal (no imaginay powe flow in the fa field). 48

49 Fa-Field Citeion Antenna (in fee-space) D D : Maximum diamete How lage does have to be to ensue an accuate fa-field appoximation? z The antenna is enclosed by a cicumscibing sphee of diamete D. D / y x D/ 49

50 Fa-Field Citeion (cont.) ( ˆ ) = ˆ + + O Let φ = maximum phase eo in the exponential tem inside the integand. Eo φ = k ( ˆ ) max Neglect Hence k k( D/ ) φ < = max 5

51 Fa-Field Citeion (cont.) φ = max kd 8 Set π o φmax = [ ad ] (.5 ) 8 Then kd π = 8 8 5

52 Fa-Field Citeion (cont.) kd π D D π λπ λ = = = Hence, we have the estiction D λ Faunhofe citeion fo φ = max π 8 [ ad] 5

53 Fa-Field Citeion (cont.) Note on choice of oigin: The diamete D can be minimized by a judicious choice of the oigin. This coesponds to selecting the best possible mounting point when mounting an antenna on a otating measuement platfom. D D = D a D = D a λ Da D = diamete of cicumscibing sphee D a = diamete of antenna Da Mounted at cente Mounted at edge 53

54 Fa-Field Citeion (cont.) It is best to mount the antenna at the cente of the otating platfom. Platfom (gound plane) Antenna Rotating pedestal 54

55 Wie Antenna z Wie antenna in fee space +h I (z') y x -h Assume ( ) I( z ) = Isin k h z -h +h 55

56 Wie Antenna (cont.) (, ) i + jk a θφ = J ( ) e dv V ( ) i J dv zˆ I( z ) dz Hence + h + jk a( θφ, ) = zˆ I( z ) e dz = = h + h ( sinθcosφ sinθsinφ cosθ) ˆ ( ) j xk yk zk z I z e dz h + h ( cosθ ) ˆ ( ) + jz k z I z e dz h 56

57 The esult is Recall that Wie Antenna (cont.) + h + jz ( k ) ( ) cos θ a( θφ, ) = zˆ Isin k h z e dz h cos( kh cos θ ) cos( kh ) a( θφ, ) = zˆ ( I) k sin θ µ E jωat jω ψ() at(, θφ) 4π Fo the wie antenna we have ( ) ( ( ˆ )) ˆ ˆ ( ˆ ) a ( θ, φ) = ˆ θ ˆ θ z a + φ φ za t z z = ˆ θ ( sinθ) a z 57

58 Wie Antenna (cont.) Hence jk µ e E ~ jω θ( sin θ) az( θφ, ) 4π ˆ o jk ˆ e E ~ θ ( jωµ sin θ ) az( θ, φ) 4π jk ˆ e cos( kh cos θ ) cos( kh ) = θ ( jωµ sin θ ) ( I) 4π k sin θ Next, simplify using ωµ k = η 58

59 Wie Antenna (cont.) We then have E ~ ˆ θ jη I ( ) jk e cos( kh cos θ ) cos( kh ) π sinθ Note: The patten goes to zeo at θ =, π. (You can veify this by using L Hôpital s ule.) 59

60 Wie Antenna: Radiated Powe ππ Pad = Eθ sinθdθdφ η = π ( π ) θ sin η π = η E θdθ π Eθ sin d θ θ π kh θ η sin π cos( cos ) cos( kh) = I sinθdθ η π θ 6

61 Radiated Powe (cont.) o [ cos( khcos θ ) cos( kh) ] π η Pad = I dθ 4π sinθ 6

62 Input Resistance [ cos( khcos θ ) cos( kh) ] π η 4π sinθ Pad = I dθ Cicuit model of antenna: Z in I () R in jx in Pad = Rin I() Pad Rin = I() 6

63 Input Resistance (cont.) ( ) I() = Isin k h z = I sin( kh) R in π η [ ] cos( kh cos θ ) cos( kh ) = π sin( kh ) sinθ dθ λ/ Dipole: (esonant) h λ π 4 =, kh = R 73 [ ] in Ω 63

64 Example z A small loop antenna (The fa field does not vay with φ.) x a y I a << λ (The cuent is appoximately unifom). I ( φ ) I (, ) ˆ + jk a θφ = ( ) I( ) e d so C (, ) ˆ + jk a θφ = I φ e adφ C k = kx x + kyy + kz z = ka sinθcosφ Assume φ = : E φ = k sin x = k θ k = y E x = acosφ ˆ φ yˆ = cosφ y 64

65 Example (cont.) z a Hence π φ = y = φ = φ jkasinθcosφ a a I cosφ e ad x I y Assume φ = : By symmety, we have a = The x component of the aay facto cancels fo points φ and -φ. x At φ = : ( ) ( ) ( ) x y x y a ˆ ˆ ˆ ˆ ˆ ˆ ˆ θ = θ xa + ya = a θ x + a θ y = 65

66 Example (cont.) Identity: Hence π jx cosφ cos ( ) φ e dφ = j πj x ( ) ( sin ) a j ai J k a x ka sinθ φ = π θ We then have jk µ e = ω 4 π Eφ j aφ jk µ e ( ) ( sin ) = jω j π aij ka θ 4π 66

67 Example (cont.) Hence jk ωµ ai e = ˆ E φ J ka ( sinθ) Appoximation of Bessel function: x J ( x), x<< The field can be thought of as coming fom a small vetical magnetic dipole, as we will discuss late (fom duality). The esult is then E jk ˆ ωµ kai e sinθ 4 φ ( ) Kl = jωµ π a I 67