Module 6: Antennas 6.0 Introduction

Module 6: Antennas 6.0 Introduction In chapter 2, the fundamental concepts associated with electromagnetic radiation were examined. In this chapter, basic antenna concepts will be reviewed, and several types of antennas will be examined. In particular, the antennas commonly used in making EMC-related measurements will be emphasized. 6.1 The Radiation Mechanism Antennas produce fields which add in phase at certain points of space. Consider a loop of wire that carries a current. R1 d I 1 D R 2 d I 2 ¡ ¡ Here two elements of current d I 1 and d I 2 are separated by a distance D. The current elements are located at distances R1 and R2, respectively from a distant observation point. If ¢ λ R2-R1 0.1 D ¢ 0.1λ Then the fields produced by the current elements add out of phase, and the amount of radiation is small. However, if £ λ R2-R1 0.1 D £ 0.1λ Then the fields produced by the current elements add in phase, and the amount of radiation is large. -Reception Mechanism Electromagnetic fields which are incident upon an antenna induce currents on the surface of the antenna which deliver power to the antenna load. 6-1 induced current incident field load Z L impedance transmission line antenna 6.2 Radiated Power The power radiated by a distribution of sources is that power which passes through a sphere of infinite radius. This, therefore, is the power which leaves the vicinity of the source system, and never returns. In chapter 2 the time-average Poynting vector was presented ¤ ¤ ¤ 1 P = R e{E × H ∗} 2 At points far from the antenna (the radiation zone) − jkr ¦ ¦ ¥ µ ¥ e E (r ) ≈ − jω [θ N + φ N ] 4π r θ φ § ¨ § § × ( ) § r E r ( ) ≈ H r η where © © (θ φ) = ( © ′) jk (r ⋅r ′) ′ N , ∫ J S r e d v v is known as the “radiation vector.” The radiation vector is related to the vector potential by − jkr © © µ e A (r ,θ , φ) = N (θ , φ) 4π r with 6-2 − jkr © © µ © © e ( ) A (r ) ≈ J (r ′)e jk r ⋅r ′ d v ′ π ∫ S 4 r v Now in the radiation zone © ∗ © © 1 r × E P ≈ R e E × 2 η Using the vector identity A × ( B × C ) = B ( A ⋅ C ) − C ( A ⋅ B ) © ©© © © 1 P ≈ R e{(E ⋅ E ∗ )r − (E ⋅ r )E ∗} 2η © © ∗ ⋅ E E ≈ r ......because E ⋅ r = 0 2η 2 1 ωµ ∗ ∗ ≈ r [N θ N θ + N φ N φ ] 2η 4πr ∗ ∗ ∗ ∗ ......because [θ N θ + φ N φ ]⋅[θN θ + φ N φ ] = [N θ N θ + N φ N φ ] Finally 1 η 2 2 P ≈ r N + N r 2 8 λ2 θ φ This represents the average power flow density and lies in the direction of wave propagation. The power radiated through a sphere of infinite radius is given by W = lim ∫ n ⋅ P ds r →∞ s Applying the expression for the time average Poynting vector leads to π 2π η 1 2 2 = ⋅ + 2 θ θ φ W lim ∫ ∫ r r 2 2 N θ N φ r sin d d r →∞ λ 0 0 r 8 6-3 π 2 π η 2 2 = N + N sin θd θd φ ∫ ∫ λ2 θ φ 0 0 8 d W η 2 2 = = + Let K 2 N θ N φ ......”radiation intensity” d Ω 8 λ = power radiated per unit solid angle. ......where d Ω = sin θd θd φ . The total radiated power is then π 2 π W = ∫ ∫ K (θ , φ )d Ω. 0 0 6.3 Antenna Terminology Antenna Patterns Radiation pattern - A plot of the radiation characteristics of an antenna. There are two types of radiation patterns: 1. Power pattern - A plot of the radiated power at a constant radius. 2. Field pattern - A plot of the electric or magnetic field magnitude at a constant radius. An antenna pattern consists of a number of lobes. The largest lobe is usually called the main lobe, while the other smaller lobes are called side lobes. The minima between lobes are called nulls. main lobe side lobe null Radiation patterns are three-dimensional, but are usually measured and displayed as two- 6-4 dimensional patterns, which are sometimes called cuts. For most antennas, two cuts give a good representation of the three-dimensional pattern. The radiation patterns of linearly polarized antennas are often specified in terms of E- plane and H-plane patterns. The E-plane contains the direction of maximum radiation and the electric field vector. The H-plane contains the direction of maximum radiation and the magnetic field vector. E-plane E H-plane H No antenna has a truly isotropic pattern (one which is the same in all directions). Rather antennas (real ones anyway) tend to radiate more effectively in some directions rather than others. Directive gain - The ratio of the radiation intensity K(θ,φ) to the uniform radiation intensity for an isotropic radiator with the same total radiation power W. K (θ , φ ) 4π g (θ , φ ) = = K (θ , φ ) d W W 4π 4π .....where is the total power radiated by an isotropic radiator per unit solid angle. W Directivity - The maximum value of directive gain. Gain - Directivity expressed in dB. G = 10 log10 (directivity) = gain in dB Beamwidth - The beamwidth of a radiation pattern is the angle between the half-power points of the pattern. 6-5 3 dB (half power) φ points side lobe Radiation efficiency - The radiation efficiency of an antenna is the ratio of the power radiated by the antenna to the total power supplied to the antenna. The total power supplied to the antenna consists of the power radiated and the power given up to resistive losses. W = E + W W L ......where E = radiation efficiency W = power radiated WL = power lost Radiation resistance - The radiation resistance of an antenna is the equivalent resistance through which its input current must flow in order that the power dissipated in the resistance is equal to the total radiated power. 1 I I ∗R = W 2 0 0 r 2W = or R r ∗ ......radiation resistance I 0 I 0 ......where I0 is the input current to the antenna. From the stand point of the source that drives an antenna, radiation resistance is indistinguishable from Ohmic resistance. In both cases, the source must continuously supply energy to the antenna in order to keep the current amplitude constant with time. In the case of Ohmic resistance, this resistance converts energy into propagating electromagnetic waves. Input impedance - An arbitrary antenna with a pair of input terminals ‘a’ and ‘b’ is shown below. 6-6 Z in I in a V b Antenna When the antenna is not receiving power from waves generated by other sources the Thevenin equivalent circuit looking into the terminals of the antenna consists only of an impedance V = = + Z in R in jX in I in where Rin is the input resistance and Xin is the input reactance. The input resistance is the sum of two components = + Rin R ri R L where Rri is the input radiation resistance and RL is the input loss resistance. RL accounts for that portion of the input power that is dissipated as heat, while the input radiation resistance Rri accounts for power that is radiated by the antenna. Rri is related to Rr by 2 I = m ax R ri R r . I 0 Radiation efficiency can be expressed P R η = ra d = ri . r + Pin R ri R L 6.4 Hertzian Dipole The simplest radiation source consists of a short segment of current 6-7 z d z A , B , E 2 I r d z y − d z 2 x A Hertzian dipole consists of a uniform current I flowing in a short wire dz terminated by point charges. Here −d z d z δ ( )δ ( ) Iz x y ...... fo r ≤ z ≤ J (r ) = 2 2 0 ...... e lse w h ere It is seen that I = ∫ J ⋅ d s = ∫ ∫ Iδ ( x )δ (y )d xd y = I x y The charge associated with the current is found using the continuity equation 1 d J ∇ ⋅ J = − jωρ ⇒ ρ = − z jω d z The current density may be expressed d z d z J = Iz δ (x )δ (y )u z + − u z − 2 2 where u(t) represents the unit step function. 6-8 J z d z d z − 2 2 1 d d z d z ρ = Iδ ( x )δ (y ) u z + − u z − jω d y 2 2 I d z d z du(t) = δ ( x )δ (y )−δ z + + δ z − ......because = δ(t) jω 2 2 dt d z d z = (+) point charge at z = , (-) point charge at z = − 2 2 Vector potential − jkR µ e ( ) = 0 ( ′) ′ A r π ∫ J r dv 4 v R where R = r − r ′ ≈ r = r for r >> d z so, − jkr µ e ( ) ≈ 0 ( ′) ′ A r π ∫ J r d v 4 r v dz − jkr 2 µ e ≈ 0 δ ( ′)δ ( ′) ′ ′ ′ π ∫ ∫ x y d x d y ∫ Iz d z 4 r x y dz z ′=− 2 or − jkr µ e A(r ) ≈ z 0 Idz 4π r 6-9 now use z = r cosθ − θ sin θ to get − jkr µ e A (r ) ≈ (r c osθ − θ sin θ ) 0 Id z 4π r E-M fields φ ∂ ∂A = ∇ × = − r B A (rAθ ) r ∂r ∂θ µ φ ∂ ∂ e − jkr = 0 Idz (− sin θe − jkr )− co sθ 4π r ∂r ∂θ r µ φ θ sin − = 0 Id z jk sin θ + e jkr 4π r r so µ 1 + jkr B = φ 0 Id z sin θe − jkr 4π r 2 j = − ∇ × = E ωµ ε B at all points where J 0 0 0 ! j r! ∂ θ ∂ = − ( θ ) − ( ) ωµ ε θ ∂θ Bφ sin ∂ rB φ 0 0 r sin r r ! j Id z r! ∂ 1 + jkr θ ∂ 1 + jkr = − sin 2 θe − jkr − sin θe − jkr ωε π θ ∂θ 2 ∂ 0 4 r sin r r r r # " j Idz r 1 + jkr E = − 2 sin θ cosθe − jkr ωε π θ 2 0 4 r sin r # θ k jkr − 1 − jkr − sin θ − j [1 + jkr ]+ e − jkr r r r 2 6-10 2 2 ! j Id z ! 1 + jkr 1 + jkr − k r = − r 2 co sθ + θ sin θ e − jkr ωε π 3 3 0 4 r r now use 1 1 µ η µ = 0 = 0 .....where η = 0 ωε ω µ ε ε 0 ε 0 0 0 0 k 0 Then − jkr − jkr ! $ η η ! Id z e 2 2 Id z e 1 E = r 0 + co sθ + θ jωµ + 0 + sin θ .

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