b) The array factor of a N-element uniform array can be written

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Benedict Parker
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1 to Eam in Antenna Theo Time: 18 Mach 010, at Location: Polacksbacken, Skivsal You ma bing: Laboato epots, pocket calculato, English ictiona, Råe- Westegen: Beta, Noling-Östeman: Phsics Hanbook, o compaable hanbooks. Si poblems, maimum five points each, fo a total maimum of 30 points. 1. a) If the comple electic fiel is enote E(), fin the coesponing instantaneous (timeepenent) electic fiel E(, t). (1p) b) The aa facto of a N-element unifom aa can be witten AF = sin( N ψ) sin ( 1 ψ), whee ψ = k cos θ + β is the pogessive (total) phase shift. Specif the conition fo β fo a i) boasie aa; ii) en-fie-aa; iii) phase (o scanning) aa. (p) c) A half-wavelength ipole has the input impeance (73 + j4.5) Ω. What is the input impeance of a quate-wavelength monopole place iectl above an infinite pefect electic conucto? (1p) ) A fole half-wavelength ipole has an input esistance of appoimatel i) 50 Ω; ii) 75 Ω; iii) 150 Ω; iv) 300 Ω; v) 600 Ω (1p) a) E(,t) = Re{E()e jωt } b) i) Boasie aa: ψ = k cos90 + β = 0 β = 0 { ψ = k cos0 ii) En-fie aa: + β = 0 β = k ψ = k cos180 + β = 0 β = k iii) Phase (o scanning) aa: ψ = k cosθ 0 + β = 0 β = k cos θ 0 c) Z monopole = 1 Z ipole = (36.5+ j1.5) Ω ) Impeance of a fole half-wavelength ipole: iv) 300Ω 1

2 . Consie a ve thin finite length ipole of length l which is smmeticall positione about the oigin with its length iecte along the ais accoing to the figue. In the fa-fiel egion the conition that the maimum phase eo shoul be less than π/8 efines the inne bouna of that egion to be = l /λ. Fo l /λ, we ae in the aiating nea-fiel egion an the fa-fiel appoimation is not vali. B allowing a maimum phase eo of less than π/8, show that the inne bouna of this egion is at = 0.6 l 3 /λ. Hint: The vecto potential is given b A = µ I e (,, e jkr ) 4π R l. Epan R, whee the highe oe tems become moe impotant as the istance to the antenna eceases. Note that = ẑ. l/ θ R φ l/ Appl the theoem of cosine to the tiangle in the figue: R = + cosθ o R = 1 cosθ + ( ) (1) Epan the squae oot using ( ) 1+ = O(4 ) whee = cosθ + () [ R = 1+ 1 ( ) ] [ cos θ + 1 ( ) ] 8 cosθ + [ + 1 ( ) ] 3 ( ) ] +O[ 4 16 cosθ + (3) { = 1 cos θ + 1 ( ) [ ( 1 ) ) cos 3 ) ] 4 θ 4( cosθ +( 8 1 ( ) 3 ( 16 8 cos 3 ) ]} 4 θ +O[ (4) { = 1 cos θ + 1 ( ) (1 cos θ)+ 1 ( ) 3 ( cos θ(1 cos ) ]} 4 θ)+o[ (5) = cosθ + 1 ( sin θ )+ 1 ( 3 ) cosθ sin θ +... (6)

3 In the fa-fiel egion, the two fist tems ae use as the appoimation fo R an the thi tem is the eo. In this case, we consie the aiating nea-fiel egion an we have to appoimate R with the thee fist tems R cos θ + 1 ( ) sin θ (7) an the eo is given b the fouth tem ( 1 3 ) cos θ sin θ (8) The maimum eo is foun when [ ( 1 3 )] θ cos θ sin θ = 3 sinθ( sin θ + cos θ) = 0 (9) We note that θ = 0 o 180 give no maimum because the make the fouth tem equal to eo. So, we must have the maimum eo fo θ = actan(± ) (10) The maimum phase eo π 8, which gives k 1 3 cosθ sin θ =l/, θ=actan(± = π ) λ We solve fo an obtain l λ = πl3 1 3λ π 8 which shows that the inne bouna of the aiating nea-fiel egion is l 3 = 0.6 λ (11) (1) (13) QED 3

4 3. An infinitesimal hoiontal electic ipole of length l an constant electic cuent I 0 is place paallel to the ais a height h = λ/ above an infinite electic goun plane. a) Fin the spheical E- an H-fiel components aiate b the ipole in the fa-one. b) Fin the angles of all the nulls of the total fiel. h h 1 θ ψ π θ φ ˆψ ˆ ˆθ a) Intouce a new set of spheical cooinates (,ψ, χ), whee ˆψ is given in the figue an ˆχ = ˆ ˆψ. E 1 ψ ˆψ = jη ki 0 l e jk 1 sin ψ ˆψ 4π 1 (1) E ψ ˆψ = jη ki 0 l e jk sin ψ ˆψ 4π () sinψ = 1 cos ψ = 1 ˆ ŷ = 1 sin θ sin φ (3) Theoem of cosine: 1 = + h hcos θ = + h hcos(π θ) 1 = 1 h cosθ +( ) h = 1+ h cosθ +( ) h, but 1+ [ 1 = 1 h cosθ + 1 ( h ) ] [ = 1+ h cos θ + 1 ( h ) ], but ( ) h h 1 hcosθ +hcosθ (4) Fa-fiel appoimation: } 1 hcosθ 1 +hcosθ fo phases (5) 1 fo amplitues (6) E = E 1 ψ ˆψ + E ψ ˆψ = jη ki 0 l [ 4π e jk 1 sin θ sin φ e jkhcos θ e jkhcos θ] ˆψ = [kh = π] = jη ki 0 l 4π e jk 1 sin θ sin φ [ j sin(π cosθ)] ˆψ, 0 θ π/ (7) } {{ } EF } {{ } AF H = 1 η ˆ E = j ki 0 l 4π e jk 1 sin θ sin φ [ j sin(π cosθ)] ˆχ, 0 θ π/ b) Nulls of the AF: AF = j sin(π cosθ) = 0 (9) π cosθ = nπ, n = 0,±1,±,... (10) cos θ = n, n = 1,0,1 (11) (8) 4

5 (1) n = 1 : cosθ = 1 θ = 180 > 90 not a null n = 0 : cosθ = 0 θ = 90 n = 1 : cosθ = 1 θ = 0 Nulls of the EF: 1 sin θ sin φ = 0 (13) sin θ sin φ = 1 (14) sin θ = ±1 simultaneous with sinφ = ±1 (15) θ = 90 simultaneous with φ = 90 (16) θ = 90 is alea a null fo the AF, so the EF oes not intouce an new nulls. Nulls of the total fiel fo θ = 0 an 90, egasless of value of φ. 5

6 4. A fou-element unifom aa has its elements place along the ais with istance = λ/ between them accoing to the figue below. a) Deive the aa facto an show that it can be witten as sin(ψ), whee ψ is the sin(ψ/) pogessive phase shift between the elements. b) In oe to obtain maimum aiation along the iection θ = 0, whee θ is measue fom the positive ais, etemine the pogessive phase shift ψ. c) Fin all the nulls of the aa facto. θ 4 3 a) E 4 e j(k n+β n ) n=1 n e jk AF (1) Fa-fiel appoimation fo phases: 1 = + 3 cos θ () = + 1 cos θ (3) 1 cosθ 1 3 = 1 cos θ (4) 4 = 3 cos θ (5) 3 cosθ an fo amplitues: (6) AF = e j 3 (k cos θ+β) + e j 1 (k cosθ+β) + e j 1 (k cosθ+β) + e j 3 (k cosθ+β) (7) = [ψ = kpcos θ + β] = e j 3 ψ + e j 1 ψ + e j 1 ψ + e j 3 ψ (8) =e j 3 ψ ( 1+e jψ + e jψ + e j3ψ) (9) Use the epession fo a geometic seies o fom AF e jψ AF, whee AF e jψ = e j 3 ψ ( e jψ + e jψ + e j3ψ + e j4ψ). o AF e jψ AF = e j 3 ψ ( e j4ψ 1 ) (10) AF ( e jψ 1 ) = e j 3 ψ ( e j4ψ 1 ) (11) 6

7 Solving fo the AF gives AF = e j 3 ψ e j4ψ 1 e jψ 1 = e j 3 ψ e jψ e j 1 ψ (e jψ e jψ ) (e j 1 ψ e j 1 ψ ) = e j 3 ψ e j 3 ψ sin ψ sin 1 ψ (1) = sinψ sin 1 ψ Q.E.D. (13) b) Fo maimum along θ = 0 all souces must be in phase, i.e., ψ(θ = 0 ) = 0: ψ = (k cos θ + β) θ=0, =λ/ = 0 (14) π λ λ cos0 + β = 0 β = π (15) ψ = π(cos θ 1) (16) c) Null of the AF AF = sinψ sin ψ Fist we investigate the nominato: = 0 sinψ = 0 an sin ψ 0 sin[π(cos θ 1)] = 0 (17) π(cos θ 1) = nπ, n = 0,±1,... (18) cosθ 1 = n (19) cosθ = 1+ n = = n, n = 4, 3,, 1, 0 (0) Then we inset the values of n an obtain the angle θ an simultaneousl check the enominato: n = 4 cosθ = 1 θ = 180 sin [ π (cos θ 1)] = 0 Not a null! n = 3 cosθ = 1/ θ = 10 sin [ π (cos θ 1)] 0 OK! n = cosθ = 0 θ = 90 sin [ π (cos θ 1)] 0 OK! n = 1 cosθ = 1/ θ = 60 sin [ π (cos θ 1)] 0 OK! n = 0 cosθ = 1 θ = 0 sin [ π (cos θ 1)] = 0 Not a null! (1) Nulls fo θ = 60, 90, 10 7

8 5. Two ientical constant cuent loops with aius a ae place a istance apat accoing to the figue below. Detemine the smallest aius a an the smallest sepaation so that nulls ae fome in the iections θ = 0, 60, 90, 10, an 180, whee θ is the angle measue fom the positive ais. a a Stu the element facto(ef) an the aa facto (AF) sepaatel. EF: The electic fiel fom a constant cuent loop is E φ = η kai 0 e jk J 1 (kasin θ) (1) The Bessel function J 1 (kasin θ) has eos fo kasinθ = 0, ,... kasinθ = 0 θ = 0 o θ = 180 () kasinθ = a = k sin θ = λ π sinθ Since the elements ae fe with equal phases (an amplitues), the AF must have a maimum in the boasie iection θ = 90 an not a null. Theefoe, the null fo θ = 90 must come fom the EF, so a = = 0.61λ (4) π sin90 AF We must choose so that AF(θ = 60 = AF(θ = 10 ) = 0 (3) / / 1 θ _ cos θ _ cos θ Fa-fiel appoimation: 1 = cosθ } = + cosθ fo phases (5) 1 fo amplitues (6) ( ) AF = e jk cosθ + e jk k cosθ = cos cosθ (7) AF(θ = 60 ) = 0 ( π cos λ π cos 60 ) ( ) π = cos = 0 (8) λ λ = (n+1)π, n = 0,±1,±,... (9) = (n+1)λ, n = 0,±1,±,... (10) 8

9 Choose = λ as the smallest istance. ( ) π λ AF = cos λ cosθ = cos(π cos θ) (11) We have to check the eo at θ = 10 : AF(θ = 10 ) = cos(π cos10 ) = cos ( π ) = 0 OK! (1) Fo eos along θ = 0, θ = 60, θ = 90, θ = 10, an θ = 180, we choose the aius of the loops to be a = 0.61λ an the spacing between the loops as = λ. 9

10 6. Design a linea aa of isotopic elements place along the ais such that the nulls of the aa facto occu at θ = 60, θ = 90, an θ = 10. Assume that the elements ae space a istance = λ/4 apat an that β = 45. a) Sketch an label the visible egion on the unit cicle. b) Fin the equie numbe of elements. c) Detemine the ecitation coefficients. Hint: The aa facto of an N-element linea aa is given b AF = ψ = k cos θ + β. Use the epesentation = e jψ. N a n e j(n 1)ψ, whee n=1 a) The visible egion: AF = N n=1 a n e j(n 1)ψ = N n=1 a n n 1 = a 1 + a + + a N N 1 = ( 1 )( ) ( N 1 ) (1) ψ = k cos θ + β = π λ 5λ 8 cosθ π 4 = π (5cos θ 1) () 4 θ = 0 ψ = π (5 1) = π (3) 4 θ = 90 ψ = π 4 (0 1) = π 4 θ = 180 ψ = π 4 ( 5 1) = 3π (4) (5) = e jψ = 1 unit cicle (6) When θ vaies fom 0 to 180, ψ vaies fom π to 3π/ ψ = 3π, θ = 180 ψ = π, θ = 0 ψ Visible egion (the whole unit cicle an an aitional quate) ψ = π 4, θ = 90 b) Nulls: θ 1 = 0 ψ 1 = π 1 = e jψ 1 = e jπ = 1 (7) θ = 90 ψ = π 4 = e jψ = e jπ/4 = 1 (1 j) (8) θ 3 = 180 ψ 3 = 3π 3 = e jψ 3 = e j 3π = j (9) 10

11 AF = ( 1 )( )( 3 ) = a 1 + a +a 3 + a 4 3 (10) 4 elements (a 1,a,a 3,a 4 ) ae neee (11) c) Ecitation coefficients: ( 1 )( )( 3 ) =... = 3 ( ) +( }{{} ) }{{} 1 3 (1) }{{} a 3 a a 1 a 1 = 1 3 = ( 1) 1 (1 j) j = 1 (1+ j) (13) a = = ( 1) 1 (1 j)+ 1 ( ) (1 j) j+ j( 1) = j 1 (14) ( a 3 = ( ) = 1+ 1 j 1 ) ( + j = 1 1 )(1 j) (15) a 4 = 1 (16) 11

Physics Courseware Physics II Electric Field and Force

Physics Courseware Physics II Electric Field and Force Physics Cousewae Physics II lectic iel an oce Coulomb s law, whee k Nm /C test Definition of electic fiel. This is a vecto. test Q lectic fiel fo a point chage. This is a vecto. Poblem.- chage of µc is