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/terms6.013 - Electromagnetics and Applications Fall 2005 Lecture 21 - Receiving Antennas Prof. Markus Zahn December 6, 2005 I. Review of Transmitting Antennas (Short Dipoles ) =zdldzeff∫(0)(0)(z)1dl 2IvIvIvdl2+� A. Far fields (r � λ) Eˆ0 Id leffk2η Eˆθ = ηHˆφ = jkr sin(θ)e−jkr, Eˆ0 = − ˆ4π , η = µ� B. Intensity �Sr� �Sr� = 12 Re � E ¯ˆ× H¯ˆ∗ � = 21 η |Eˆ¯ |2 = 21 η |kEˆ20r|2 sin2(θ)2 ˆ42η�2 =1 |Idleff|2k�sin2(θ)ηk�222�16π2�rˆ2k2η = |Idleff|sin2(θ)32π2r2 C. Total time average power �P � � π � 2π �P � = dθ dφ �Sr� r 2 sin(θ) 0 0 Id leff= | ˆ|2ηk2 12π =1 ˆ2R R =2πη � dleff �2 radiation resistance2|I| ⇒ 3 λ D. Gain G(θ, φ) = �Sr� �P � /(4πr2) = �|Idlˆ�eff�� |2��η�sin2(θ) · 12�π(�πr��2)k24�2�r2�Id l�eff�� �8�32��π2��| ˆ| η��k2 3 = sin2(θ)2 1 Image by MIT OpenCourseWare.� � � � � � � � � � � � � EoS1V1d1I1abce+yx___For d1 << λθϕZII. Receiving Antennas In absence of receiving antenna: E¯ inc = E¯ 0, H¯ inc = H¯ 0. With d1 � λ, over size scale of antenna, E¯ 0 and H¯ 0 are approximately spatially uniform. In presence of receiving antenna, electric and magnetic fields are perturbed so that tangential E ¯ and normal H ¯ are zero along the perfectly conducting length of the antenna. E ¯= E¯ 0 + E¯ 1 (1) H ¯= H¯ 0 + H¯ 1 (2) Surface S1 above intimately hugs the antenna so that E ¯ × da¯ = da E¯ 0 + E¯ 1 × n¯ = 0 (tangential E ¯ = 0) (3) S1 S1 +d1/2 da n¯ × (H¯ 0 + H¯ 1) = da K ¯= dz I¯ 1(z) = I1deff ¯iz (4) S1 S1 −d1/2 Another useful relationship: da (E¯ 0 × H¯ 0) n¯ = ( E¯ 0 × H¯ 0) da n¯ = 0 (5)· · S1 S1 Integral of normal over closed surface is zero: dV �f = n, Take f = 1, �f = 0 = da ¯ (6) da f ¯ n = 0 V S S Scalar Triple Product Identity: (¯a × ¯ b) c¯ = ¯a (¯ b × c¯) (7)· · (Interchange of cross and dot) Complex power supplied by receiving antenna ¯ P = da Sˆn¯ (8)· S1 Sˆ¯= 12(E ¯ˆ× Hˆ¯ ∗) = 12 � (Eˆ¯ 0 + Eˆ¯ 1) × (Hˆ¯ 0 ∗ + Hˆ¯ 1 ∗) � ¯ ¯ ¯ ¯ ¯ ¯ =1 � Eˆ0 × (Hˆ0 ∗ + Hˆ∗) + Eˆ1 × (Hˆ0 ∗ + Hˆ∗) � (9) �2 �1 �� 1 ��� �� P = da Sˆ¯ n¯ =1 da E¯ˆ0 × (H¯ˆ0 ∗ + H¯ˆ1 ∗) n¯ + da E¯ˆ1 × (H¯ˆ0 ∗ + H¯ˆ1 ∗) n¯ (10) · 2 · · S1 S1 S1 2 Image by MIT OpenCourseWare.� � � � �� � � � � �� ��� � ¯ ¯ da Eˆ0 × (H¯ˆ0 ∗ + Hˆ1 ∗) n¯ = da E¯ˆ0 · (H¯ˆ0 ∗ + H¯ˆ1 ∗) × n¯· S1 S1 = E¯ˆ0 · da (Hˆ¯ 0 ∗ + Hˆ¯ 1 ∗) × n¯S1 = −E¯ˆ0 · Iˆ1deff ¯iz (from (4)) (11)�� ��� � �� � da Eˆ¯ 1 × (Hˆ¯ 0 ∗ + Hˆ¯ 1 ∗) n¯ = da E¯ˆ1 × H¯ˆ0 ∗ n¯ + da E¯ˆ1 × H¯ˆ1 ∗ n¯ (12)· · · S1 S1 S1 |Iˆ1|2(R+jX) where R is the radiation resistance and X is the antenna reactance �� ��� � da Eˆ¯ 1 × H¯ˆ0 ∗ · n¯ = − da H¯ˆ0 ∗ × Eˆ¯ 1 · n¯S1 S1 ¯ ¯ ¯ = − da Hˆ0 ∗ × (Eˆ0 + Eˆ1) n¯· �S1 � � �� � = − da Hˆ¯ 0 ∗ (Eˆ¯ 0 + E¯ˆ1) × n¯ = −H¯ 0 ∗ da Eˆ¯ 0 + E¯ˆ1 × n¯· · S1 S1 = 0 (from (3)) (13) P = 12 � −Eˆ¯ 0 · Iˆ1deff ¯iz + |Iˆ¯ 1|2(R + jX) � = 12Vˆ1Iˆ1 ∗ Thevenin Equivalent Circuit VˆT H = Vˆoc = −E¯ˆ0 · d¯ eff (d¯ eff = deff ¯iz) 3 Image by MIT OpenCourseWare.III. Transmitting and Receiving Antennas A. Circuit Description I1I2V1V2Two port network^^^^ˆV1 = ˆI1Z11 + ˆI2Z12 ˆV2 = ˆI1Z21 + ˆI2Z22 Z12 = ˆV1 ˆI2 ��� ˆI1=0 Z21 = ˆV2 ˆI1 ��� ˆI2=0 Reciprocity Theorem: Z12 = Z21 B. Antenna Thevenin Equivalent Circuits Z11 = R1 + jX1Z22 = R2 + jX2I1I2Vth1Vth2V1V2Z22TransmitterReceiver (Balanced load Z*22 to cancel reactance X2) *ˆVth1 = ˆI2Z12 = − � ˆ¯ E1 · dl = − ˆ¯ E1 · dleff ˆVth2 = ˆI1Z21 = − ˆ¯ E2 · dleff 4 Image by MIT OpenCourseWare.Image by MIT OpenCourseWare.1 |Vˆth2/2|2 =1 |Vˆth2|2 =1 |Eˆ2dleff � sin(�θ2 )|2 �P2� =2 R2 8 R2 8 2πη dleff 3 λ 1 �Eˆ�2�2 1 �ˆ��2 sin2(θ) 3λ2 �P2� = Arec(θ, φ) �Sr� = Arec(θ, φ) · 2 | η�|=8 |E2|2πη�· 3 λ2 λ2 Arec(θ, φ) = sin2(θ) = Grec(θ, φ)2 4π 4π C. Representative Parameters 1. Minimum received power ≈ 10−20 watts For total transmitted power of 1 watt, how far away can the receive r be at f = 1 GHz? �Ptrans� λ2 �Prec� =4πr2 Gtrans Grec 4π� �� �� �� � �Sr � Arec(θ,φ) fλ = c λ = c =3 × 108 = .3 m ⇒ f 109 3 Gtrans = Grec = sin2(θ) (for short dipoles) (identical transmitting and receiving antennas) 2 π 3 Take θ = Gtrans = Grec = 2 ⇒ 2 � �2 2 Ptrans λ r = GtransGrecPrec 4π � �� �21 9 .3 = 10−20 4 4π = 1.28 × 1017 m2 r = 3.58 × 108 m = 3.58 × 105 km ≈ 200, 000 miles 2. For data transmission, receivers need Eb > 4 × 10−20 Joules/bit Power received = M Eb where M is the data rate, bits/s 10−9 watts received power allows M = 10E−b 9 = 4×1010−−920 = .25 × 1011 bits/s 1 CD = 700 × 106 bytes = 5600 × 106 bits (1 byte = 8 bits) M = .25 × 1011 bits/sec ≈ 4.5 CD/sec 3. Distance is not a barrier to wireless communications r = 1 lightyear = 3 × 108 m/s 3 × 107 s/yr = 9 × 1015 m/yr ·Ptrans =? c f = 3GHz λ = = .1 m ⇒ f M = 1 bit/s, Eb = 4 × 10−20 Joules/bit Prec = MEb = 4 × 10−20 Watts Gtrans = Grec = 107 5� �24πr Prec λPtrans = GtransGrec� �2 4 × 10−20 4π(9×1015) .1 = 1014 = 512 Watts For M = 2.4 kb/s ⇒ Ptrans ≈ 1.2 MW (with a 1 year delay each way) 4. Optical Communications: E = hf, h = 6.625 × 10−34 Joule-sec (Planck’s Constant) a. Radio Photons …
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