Solution to Exam in Antenna Theory Time: 18 March 2010, at 8.00–13.00. Location: Polacksbacken, Skrivsal You may bring: Laboratory reports, pocket calculator, English dictionary, RådeWestergren: “Beta”, Nordling-Österman: “Physics Handbook”, or comparable handbooks. Six problems, maximum five points each, for a total maximum of 30 points. 1. a) If the complex electric field is denoted E(r), find the corresponding instantaneous (timedependent) electric field E(r,t). (1p) b) The array factor of a N-element uniform array can be written sin N2 ψ , AF = sin 21 ψ where ψ = kd cos θ + β is the progressive (total) phase shift. Specify the condition for β for a i) broadside array; ii) end-fire-array; iii) phased (or scanning) array. (2p) c) A half-wavelength dipole has the input impedance (73 + j42.5) Ω. What is the input impedance of a quarter-wavelength monopole placed directly above an infinite perfect electric conductor? (1p) d) A folded half-wavelength dipole has an input resistance of approximately i) 50 Ω; ii) 75 Ω; iii) 150 Ω; iv) 300 Ω; v) 600 Ω Solution a) E(r,t) = Re{E(r)e jω t } b) i) Broadside array: ψ = kd cos 90◦ + β = 0 ⇒ β = 0 ψ = kd cos 0◦ + β = 0 ⇒ β = −kd ii) End-fire array: ψ = kd cos 180◦ + β = 0 ⇒ β = kd iii) Phased (or scanning) array: ψ = kd cos θ0 + β = 0 ⇒ β = −kd cos θ0 c) Zmonopole = 21 Zdipole = (36.5 + j21.25) Ω d) Impedance of a folded half-wavelength dipole: iv) 300Ω 1 (1p) 2. Consider a very thin finite length dipole of length l which is symmetrically positioned about the origin with its length directed along the z axis according to the figure. In the far-field region the condition that the maximum phase error should be less than π /8 defines the inner boundary of that region to be r = 2l 2 /λ . For r ≤ 2l 2 /λ , we are in the radiating near-field region and the far-field approximation is not valid. By allowing a maximum phase error of less than π /8, p show that the inner boundary of this region is at r = 0.62 l 3 /λ . Hint: The vector potential is given by e− jkR ′ µ Ie (x′ , y′ , z′ ) dl . A= 4π R Expand R, where the higher order terms become more important as the distance to the antenna decreases. Note that r′ = z′ ẑ. Z z R l/2 r r θ y φ x −l/2 Solution Apply the theorem of cosine to the triangle in the figure: s ′ 2 z′ z 2 2 ′2 ′ R = r + z − 2rz cos θ or R = r 1 − 2 cos θ + r r (1) Expand the square root using ′ 2 √ z′ z x x2 x3 4 (2) 1 + x = 1 + − + + O(x ) where x = −2 cos θ + 2 8 16 r r " " ′ 2 # ′ 2 #2 z z 1 z′ z′ 1 R = r 1+ −2 cos θ + −2 cos θ + − 2 r r 8 r r " # " ′ 2 #3 z z′ 4 1 z′ (3) + +O −2 cos θ + 16 r r r " ( ′ 3 ′ 4 # z′ 2 2 z z z′ 1 z′ 2 1 − cos θ − 4 cos θ + = r 1 − cos θ + r 2 r 8 r r r " #) ′ 4 ′ 3 z z 1 3 cos θ + O (4) − 8 16 r r " #) ( ′ 3 1 z′ 2 1 z′ 4 z′ z = r 1 − cos θ + (1 − cos2 θ ) + cos θ (1 − cos2 θ ) + O r 2 r 2 r r (5) = r − z′ cos θ + 1 r z′2 2 sin2 θ + 2 1 r2 z′3 2 cos θ sin2 θ + . . . (6) In the far-field region, the two first terms are used as the approximation for R and the third term is the error. In this case, we consider the radiating near-field region and we have to approximate R with the three first terms 1 z′2 2 ′ sin θ (7) R ≈ r − z cos θ + r 2 and the error is given by the fourth term 1 z′3 2 cos θ sin θ r2 2 (8) The maximum error is found when ∂ 1 z′3 z′3 2 θ sin θ cos sin θ (− sin2 θ + 2 cos2 θ ) = 0 = ∂ θ r2 2 2r2 (9) We note that θ = 0 or 180◦ give no maximum because they make the fourth term equal to zero. So, we must have the maximum error for √ θ = arctan(± 2) (10) The maximum phase error ≤ π8 , which gives k 1 z′3 cos θ sin2 θ r2 2 √ z′ =l/2, θ =arctan(± 2) = π l3 π 2π 1 1 2 √ = √ ≤ 2 λ r 3 3 12 3λ r2 8 We solve for r and obtain r l3 r ≥ 0.62 λ (11) (12) which shows that the inner boundary of the radiating near-field region is r l3 r = 0.62 λ (13) QED 3 3. An infinitesimal horizontal electric dipole of length l and constant electric current I0 is placed parallel to the y axis a height h = λ /2 above an infinite electric ground plane. a) Find the spherical E- and H-field components radiated by the dipole in the far-zone. b) Find the angles of all the nulls of the total field. Solution ψ̂ z r̂ a) Introduce a new set of spherical coordinates (r, ψ , χ ), where ψ̂ is given in the figure and χ̂ = r̂ × ψ̂ . θ̂ r1 r θ kI0 l − jkr1 e sin ψ ψ̂ 4π r1 kI0 l − jkr2 Eψ2 ψ̂ = jη e sin ψ ψ̂ 4π r2 p p sin ψ = 1 − cos2 ψ = 1 − |r̂ · ŷ| q = 1 − sin2 θ sin2 φ Eψ1 ψ̂ = jη r2 h ψ y π −θ x h φ (1) (2) (3) Theorem of cosine: 2 + h2 − 2rh cos θ r12 = rq r1 = r h 1 − 2 hr cos θ + r1 = r 1 − hr cos θ + 21 r1 ≈ r − h cos θ h 2 r i h 2 r 2 + h2 − 2rh cos(π − θ ) r22 = rq 2 √ r2 = r 1 + 2 hr cos θ + hr , but 1 + x ≈ 2x h i 2 2 r2 = r 1 + hr cos θ + 21 hr , but hr ≪ hr (4) r2 ≈ r + h cos θ Far-field approximation: r1 ≈ r − h cos θ r1 ≈ r + h cos θ r1 ≈ r2 ≈ r for phases for amplitudes (5) (6) q h i kI0 l − jkr = jη 1 − sin2 θ sin2 φ e jkh cos θ − e− jkh cos θ ψ̂ = [kh = π ] E= e 4π r q kI0 l − jkr = jη 1 − sin2 θ sin2 φ [2 j sin(π cos θ )] ψ̂ , 0 ≤ θ ≤ π /2 (7) e {z } 4 π r | {z }| AF EF q kI0 l − jkr 1 1 − sin2 θ sin2 φ [2 j sin(π cos θ )] χ̂ , 0 ≤ θ ≤ π /2 H = r̂ × E = j e η 4π r (8) Eψ1 ψ̂ + Eψ2 ψ̂ b) Nulls of the AF: AF = 2 j sin(π cos θ ) = 0 ⇒ π cos θ = nπ , ⇒ cos θ = n, (9) n = 0, ±1, ±2, . . . n = −1, 0, 1 4 (10) (11) n = −1 : cos θ = −1 ⇒ θ = 180◦ > 90◦ n=0: cos θ = 0 ⇒ θ = 90◦ ⇒ θ = 0◦ n=1: cos θ = 1 ⇒ not a null (12) Nulls of the EF: ⇒ ⇒ 1 − sin2 θ sin2 φ = 0 (13) sin θ sin φ = 1 2 2 sin θ = ±1 ⇒ θ = 90 ◦ (14) simultaneous with simultaneous with sin φ = ±1 φ = 90 ◦ θ = 90◦ is already a null for the AF, so the EF does not introduce any new nulls. Nulls of the total field for θ = 0◦ and 90◦ , regardsless of value of φ . 5 (15) (16) 4. A four-element uniform array has its elements placed along the z axis with distance d = λ /2 between them according to the figure below. a) Derive the array factor and show that it can be written as sin(2ψ ) , where ψ is the sin(ψ /2) progressive phase shift between the elements. b) In order to obtain maximum radiation along the direction θ = 0◦ , where θ is measured from the positive z axis, determine the progressive phase shift ψ . c) Find all the nulls of the array factor. z d y d x d Solution z θ E∼ a) r3 d y d d ∑ Far-field approximation for phases: 3 r1 = r + d cos θ 2 1 r2 = r + d cos θ 2 1 r3 = r − d cos θ 2 3 r4 = r − d cos θ 2 and for amplitudes: r r2 r1 1 2 d cos θ e− j(krn +βn ) e− jkr AF ≈ rn r n=1 4 r4 3 2 d cos θ r1 ≈ r2 ≈ r3 ≈ r4 ≈ r AF = e− j 2 (kd cos θ +β ) + e− j 2 (kd cos θ +β ) + e j 2 (kd cos θ +β ) + e j 2 (kd cos θ +β ) 3 1 1 − j 23 ψ = [ψ = kp cos θ + β ] = e − j 32 ψ +e +e j 21 ψ +e j 23 ψ (2) (3) (4) (5) (6) (7) (8) j3ψ (9) 3 AF e jψ − 1 = e− j 2 ψ e j4ψ − 1 (11) =e 1 + e jψ + e j2ψ + e − j 21 ψ 3 (1) Use the expression for a geometric series or form AF e jψ − AF, where AF e jψ = 3 ψ ψ ψ ψ ψ −j j2 j3 j4 j e 2 e +e +e +e . 3 (10) AF e jψ − AF = e− j 2 ψ e j4ψ − 1 or 6 Solving for the AF gives AF = e− j 2 ψ 3 = j2ψ (e j2ψ − e− j2ψ ) 3 3 e j4ψ − 1 sin ψ − j 23 ψ e = e = e− j 2 ψ e j 2 ψ 1 1 1 ψ j e −1 sin 12 ψ e j 2 ψ (e j 2 ψ − e− j 2 ψ ) sin ψ sin 12 ψ Q.E.D. (12) (13) b) For maximum along θ = 0◦ all sources must be in phase, i.e., ψ (θ = 0◦ ) = 0: ψ = (kd cos θ + β )|θ =0◦ , d=λ /2 = 0 2π λ cos 0◦ + β = 0 ⇒ β = −π ⇒ λ 2 ⇒ ψ = π (cos θ − 1) 2ψ =0 c) Null of the AF ⇒ AF = sin sin ψ2 First we investigate the nominator: sin[2π (cos θ − 1)] = 0 ⇒ ⇒ ⇒ ⇒ sin 2ψ = 0 and (14) (15) (16) sin ψ2 6= 0 (17) n = 0, ±1, . . . 2π (cos θ − 1) = nπ , n cos θ − 1 = 2 n 2=n , n = −4, −3, −2, −1, 0 cos θ = 1 + = 2 2 (18) (19) (20) Then we insert the values of n and obtain the angle θ and simultaneously check the denominator: ⇒ θ = 180◦ ⇒ sin π2 (cos θ − 1) = 0 ⇒ Not a null! n = −4 ⇒ cos θ = −1 n = −3 ⇒ cos θ = −1/2 ⇒ θ = 120◦ ⇒ sin π2 (cos θ − 1) 6= 0 OK! π ◦ ⇒ θ = 90 ⇒ sin 2 (cos θ − 1) 6= 0 OK! n = −2 ⇒ cos θ = 0 n = −1 ⇒ cos θ = 1/2 ⇒ θ = 60◦ ⇒ sin π2 (cos θ − 1) 6= 0 OK! ⇒ θ = 0◦ ⇒ sin π2 (cos θ − 1) = 0 ⇒ Not a null! n = −0 ⇒ cos θ = 1 (21) Nulls for θ = 60◦ , 90◦ , 120◦ 7 5. Two identical constant current loops with radius a are placed a distance d apart according to the figure below. Determine the smallest radius a and the smallest separation d so that nulls are formed in the directions θ = 0◦ , 60◦ , 90◦ , 120◦ , and 180◦ , where θ is the angle measured from the positive z axis. z a y d a x Solution Study the element factor(EF) and the array factor (AF) separately. EF: The electric field from a constant current loop is Eφ = η kaI0 − jkr e J1 (ka sin θ ) 2r (1) The Bessel function J1 (ka sin θ ) has zeros for ka sin θ = 0, 3.8317, ... ka sin θ = 0 ⇒ ka sin θ = 3.8317 θ = 0◦ or θ = 180◦ 3.8317 3.8317λ ⇒ a= = k sin θ 2π sin θ (2) (3) Since the elements are fed with equal phases (and amplitudes), the AF must have a maximum in the broadside direction θ = 90◦ and not a null. Therefore, the null for θ = 90◦ must come from the EF, so a= 3.8317 = 0.61λ 2π sin 90◦ (4) AF We must choose d so that AF(θ = 60◦ = AF(θ = 120◦ ) = 0 z r1 r Far-field approximation: r2 x d/2 θ _d cos θ 2 d/2 _d cos θ 2 AF = e r1 = r − d2 cos θ r2 = r + d2 cos θ y r1 ≈ r2 ≈ r jk d2 cos θ +e AF(θ = 60 ) = 0 ◦ − jk d2 cos θ ⇒ ⇒ ⇒ for phases (5) for amplitudes (6) kd = 2 cos cos θ 2 πd πd ◦ cos cos 60 = cos =0 λ 2λ πd π = (2n + 1) , n = 0, ±1, ±2, . . . 2λ 2 n = 0, ±1, ±2, . . . d = (2n + 1)λ , 8 (7) (8) (9) (10) Choose d = λ as the smallest distance. 2π λ cos θ = 2 cos(π cos θ ) AF = 2 cos λ 2 (11) We have to check the zero at θ = 120◦ : π AF(θ = 120◦ ) = 2 cos(π cos 120◦ ) = 2 cos − =0 2 OK! (12) For zeros along θ = 0◦ , θ = 60◦ , θ = 90◦ , θ = 120◦ , and θ = 180◦ , we choose the radius of the loops to be a = 0.61λ and the spacing between the loops as d = λ . 9 6. Design a linear array of isotropic elements placed along the z axis such that the nulls of the array factor occur at θ = 60◦ , θ = 90◦ , and θ = 120◦ . Assume that the elements are spaced a distance d = λ /4 apart and that β = 45◦ . a) Sketch and label the visible region on the unit circle. b) Find the required number of elements. c) Determine the excitation coefficients. N Hint: The array factor of an N-element linear array is given by AF = ∑ an ej(n−1)ψ , where n=1 ψ = kd cos θ + β . Use the representation z = ejψ . Solution a) The visible region: AF = N N n=1 n=1 ∑ an e j(n−1)ψ = ∑ anzn−1 = a1 + a2 z + · · · + aN zN−1 = (z − z1 )(z − z2 ) · · · · · (z − zN−1 ) ψ = kd cos θ + β = θ = 0◦ ⇒ θ = 90◦ ⇒ θ = 180◦ ⇒ z = e jψ ⇒ (1) π π 2π 5λ cos θ − = (5 cos θ − 1) λ 8 4 4 (2) π (5 − 1) = π 4 π π ψ = (0 − 1) = − 4 4 3π π ψ = (−5 − 1) = − 4 2 ψ= |z| = 1 ⇒ (3) (4) (5) unit circle (6) When θ varies from 0◦ to 180◦ , ψ varies from π to −3π /2 ψ = − 32π , θ = 180◦ Visible region (the whole unit circle and and additional quarter) ψ ψ = π , θ = 0◦ ψ = − π4 , θ = 90◦ b) Nulls: θ1 = 0◦ ⇒ ψ1 = π θ2 = 90◦ ⇒ ψ2 = − θ3 = 180◦ ⇒ π 4 3π ψ3 = − 2 ⇒ ⇒ ⇒ 10 z1 = e jψ1 = e jπ = −1 1 z2 = e jψ2 = e− jπ /4 = √ (1 − j) 2 z3 = e jψ3 = e− j 3π 2 =j (7) (8) (9) AF = (z − z1 )(z − z2 )(z − z3 ) = a1 + a2 z + a3 z2 + a4 z3 ⇒ (10) 4 elements (a1 , a2 , a3 , a4 ) are needed (11) c) Excitation coefficients: (z − z1 )(z − z2 )(z − z3 ) = . . . = z3 − z2 (z1 + z2 + z3 ) +z (z1 z2 + z2 z3 + z3 z1 ) −z1 z2 z3 {z } | {z } | {z } | −a3 a2 (12) a1 1 1 a1 = −z1 z2 z3 = −(−1) √ (1 − j) j = √ (1 + j) (13) 2 2 √ 1 1 a2 = z1 z2 + z2 z3 + z3 z1 = (−1) √ (1 − j) + √ (1 − j) j + j(−1) = j 2−1 2 2 (14) 1 1 1 (15) a3 = −(z1 + z2 + z3 ) = − −1 + √ − j √ + j = 1 − √ (1 − j) 2 2 2 a4 = 1 (16) 11