MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/termsaaacosφφφyr2rr1xI2I16.013 - Electromagnetics and Applications Fall 2005 Lecture 20 - Dipole Arrays Prof. Markus Zahn December 1, 2005 I. Two Element Array in θ = π 2 plane (x-y plane) Far field (kr � 1, r � a) ˆπ Eˆ1 e−jkr1 Eˆ2 e−jkr2 π Eθ(r, θ =2 , φ) = jkr1 + jkr2 = ηHˆφ(r, θ =2 , φ) Eˆ1 Iˆ1dlk2η = − 4π Eˆ2 = − Iˆ2dlk2η 4π r2 ≈ r + a cos(φ), r1 ≈ r − a cos(φ) Eˆθ(r, θ = π, φ) = ηHˆφ(r, θ = π, φ) ≈ − k�2ηdl e−jkr � Iˆ1e +jka cos(φ) + Iˆ2e−jka cos(φ) � 2 24πj�kr � �� � � �� � array factor element factor Iejχ ˆAssume: Iˆ1 = I, ˆIˆ2 =ˆEˆ1 = E0, Eˆ2 = Eˆ0ejχ ⇒ Eθ(r, θ = π, φ) = ηHˆφ(r, θ = π, φ) = Eˆ0 e−jkr � e +jka cos(φ) + ejχe−jka cos(φ) � 2 2 jkr Eˆ0 e−jkr jχ/2 � e−j( χ 2 −ka cos(φ)) + ej( χ 2 −ka cos(φ)) � = ejkr � �� � 2 cos(− χ 2 +ka cos(φ)) =2Eˆ0 e−jkrejχ/2 cos � χ + ka cos(φ) � jkr − 2 � π � 1 Eˆθ2 2|Eˆ02 � χ� 2Sr(t, θ =2 , φ) =2 | η |= η(kr )|2 cos ka cos(φ) − 2 1 Image by MIT OpenCourseWare.Maxima: ka cos(φ) − χ 2 = ±mπ, m = 0, 1, 2, . . . Minima: ka cos(φ) − χ = ±(2m + 1)π , m = 0, 1, 2, . . . 2 2 Case Studies: Broadside: λ π2a = , χ = 0, ka = 2 2 2|Eˆ0|2 cos2 �π cos(φ) � �Sr� = η(kr )2 2 Endfire: λ π2a = , χ = π, ka = 2 2 �Sr� =2|Eˆ0|2 cos2 �π (cos(φ) − 1) � η(kr )2 2 λ 2πa 2πa π2a = ka = = = 2 ⇒ λ 4a 2 2a = λ ka = π⇒ λ π χ χ2a =2 ⇒ 2 cos(φ) − 2= ±mπ (maxima) ⇒ cos(φ) = π ± 2m π χ π χ 2 cos(φ) − 2= ±(2m + 1)2 (minima) ⇒ cos(φ) = π ± (2m + 1) 2From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.λ = 4a χ cos(φmax) cos(φmin) φmax φmin ± 0, π Broadside ±75.5◦ ±138.6◦ ±60◦ ±120◦ ±41.4◦ ±104.5◦ 82.8◦, 151◦ 51◦, 112◦ π ± (2m + 1) χ cos(φmax) cos(φmin) φmax φmin 0, ±90◦, 180◦ ±60◦ 75.5◦, 138.6◦ 41.4◦, 104.5◦ 68.0◦, 128.7◦ 29.0◦, 97.2◦ 0, 1 60◦, 120◦ 90◦, 0◦ − − − λ = 2a cos(φmax) =⇒ 3−π 1 0 0, π ±90.◦ Endfire π ± m, cos(φmin) =13π 2π 1 2χ 20 0 1 −−3 1 1 2, − 1 242453 31 71 848, , , 8480 0, 1, −, −, −, −1 81 43 81 27 83 5481 44 1 22 44 χ 2π 1 23π 4 π π π4 π 2 34From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.� � � � � � �� � � � II. An N Dipole Array (θ = π 2 ) lim r�na rn ≈ r − na cos(φ) −N ≤ n ≤ N Eˆθ r, θ = π, φ = ηHˆφ r, θ = π, φ 2 2 � +N�kηdl � Iˆnejkna cos(φ) e−jkr = −4πjr � −N �� � Array factor = AF Example: Iˆn = I0e−jnχ0 − N ≤ n ≤ N +NAF = I0 ejn(ka cos(φ)−χ0) −N Let β ≡ ej(ka cos(φ)−χ0) +NAF � S = = βn = β−N + β−N+1 + . . . + β−2 + β−1 + 1 + β + β2 + . . . + βN−1 + βN I0 −N − βN+1 β−N − βN+1 β−N− 21 − βN+ 21 S(1 − β) = β−N ⇒ S =1 − β = β−1/2 − β1/2 multiply by β−1/2 β−1/2 sin (N + 1 )(ka cos(φ) − χ0)S = �2 � sin 1 (ka cos(φ) − χ0)2 Maxima: ka cos(φ) − χ0 = 2nπ, n = 0, 1, 2, . . . Principal maximum at n = 0 cos(φ) = χka 0 ⇒ Minima: (N + 1 )(ka cos(φ) − χ0) = nπ, n = 1, 2, 3, . . . 2 5From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.Demonstration, N = 2 (2 dipole array) 32a = λ, χ0 = 0 2� � � � I ∝ cos2(ka cos(φ)) = cos2 2�π 3 �λ cos(φ) = cos2 3π cos(φ)�λ 4�2 2 3�π π 1Minima: cos(φ) = �cos(φ) = φ = 70.5◦2�2�⇒ 3 ⇒ π π3�2��cos(φ) = 3�2��⇒ cos(φ) = 1 ⇒ φ = 0◦ 3πMaxima: cos(φ) = 0 φ = 90◦2 ⇒ 3�π 2cos(φ) = �π cos(φ) = φ = 48.2◦2 ⇒ 3 ⇒ фIntensity pattern 6 Image by MIT OpenCourseWare.7 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with
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