Antenna for EMC Prof. Tzong-Lin Wu EMC Lab. Department of Electrical Engineering National Taiwan University Introduction 1. Overview 2. Steps in evaluation of radiated fields 3. Ideal Dipole (Hertzian Dipole) 4. Antenna parameters 5. Dipole and Monopole 6. Small Loop Antenna 7. Antenna Arrays 8. Examples of Antenna 9. Antennas in communication systems 1 How antenna radiate: a single accelerated charged particle • The static electric field originates at charge and is directed radially away from charge. • At point A, the charge begins to be accelerated until reaching point B. • The distance between the circles is that distance light would travel in time △t, and △r=rb – ra = △t / c • Charge moves slowly compared to the speed of the light, ∴△r >> △z and two circles are concentric. • The electric field lines in the △r region are joined together because of required continuity of electrical lines in the absence of charges. • This disturbance expands outward and has a transverse component Et , which is the radiated field. • If charges are accelerated back and forth (i.e., oscillate), a regular disturbance is created and radiation is continuous. • This disturbance is directly analogous to a transient wave created by a stone dropped into a calm lake. How antenna radiate : Evolution of a dipole antenna from an open- circuited transmission line • Open-circuited transmission line 1. The currents are in opposite directions on the two wires and behaves as a standing wave pattern with a zero current magnitude at the ends and every half wavelength from the end. 2. The conductors guide the waves and the power resides in the region surrounding the conductors as manifested by the electric and magnetic fields. 3. Electric fields originate from or terminate on charges and perpendicular to the wires. 4. Magnetic fields encircle the wires. • Bending outward to form a dipole 1. The currents are no longer opposite but are both upwardly directed. 2. The bounded fields are exposed to the space. 3. The currents on the dipole are approximately sinusoidal. 4. The situation on the Fig. is the peak current condition. As time proceeds and current oscillation occur, the disturbed fields are radiated. 2 How antenna radiate : Time dynamics of the fields for a dipole antenna Closed loop • At t = 0, peak charges buildup occurs (positive on the upper half and negative on the lower half). current I = 0. • At t = T/4, +/- charges are neutralized and I is maximum. • Because there are no longer charges for the termination of electric fields, they form closed loop near the dipole. • At t = T/2, peak charges buildup again, but upper half is negative and lower half is positive. I = 0. • Notice the definition of λ/2 at t = T/2. Overview of antenna 3 Overview of antenna Antenna Concept (I) 4 Antenna Concept (II) Steps in evaluation of radiation fields : Solution of Maxwell equations for radiation problems) Magnetic vector potential 1 ∇⋅H =0 HA=∇× Electric scalar potential μ ∇×EjH =− ωμ ∇×()EjA +ω =0 EjA+=−∇ωφ ∇×HjEJ =ωε + ∇×∇×AAAjjAJ =∇() ∇⋅ −∇2 =ωμε ( − ω −∇ φ ) + μ ρ ∇⋅E = If we set ∇⋅Aj = − ωμεφ ε (Lorenz condition) Vector wave equation ∇+22AAJωμε =− μ Solution is e − jRβ AJ= zzz μ dv ' v' 4πR 5 Steps in evaluation of radiation fields (Far Fields) 1. Find A − jrβ e ' Far field A = μ ∫∫∫ Jejrrβ ˆ⋅ dvn 4π r v' − jrβ e ' Z-directed sources A = zJedvˆμ jrrβ ˆ⋅ ' ∫∫∫ z 4π r v' − jrβ e ' Z-directed line sources Az= ˆμ ∫∫∫ Ize()'cos'jzβθ⋅ dv 4π r v' Steps in evaluation of radiation fields (Far Fields) 2. Find E Far field E =−jAω Transverse to the propagation ˆ ˆ direction r EjAA=−ω(θφ θ + φ) Z-directed sources ˆ E = jAωsinθ θz 3. Find H 1 Far field (Plane wave) H = rEˆ× η 1 Z-directed source HE= φ η θ 6 Far-Field Conditions Parallel ray approximation for far-field calculations R = rrr−⋅ˆ ' 2D2 r > λ Far Field conditions D is the length of rD>> line source r >> λ Far-Field and Near-field Conditions 7 The ideal dipole (Hertzian dipole) : definition Ideal dipole: a uniform amplitude current that is electrically small with △z <<λ Δz /2 e− jRβ A = zIˆμ ∫ dz' −Δz /2 4π R ∵R～r for small dipole e− jrβ A =ΔμIz zˆ 4π r The ideal dipole (Hertzian dipole): E and H field 1 IzΔ 1 e− jrβ HA=∇× Hj=+β (1 ) sin θφˆ μ 4πβjr r IzΔ 11 e− jrβ 1 Ej= ωμθθ[1+− ] sin ˆ + E =∇×H 2 jωε 4()πββjrrr IzΔ 11 e− jrβ ηθ[− jr ] cos ˆ 2πβrrr2 Far field condition : βr >> 1 IzΔ e− jrβ Hj= ωμθφsin ˆ 4π r E ωμ μ IzΔ e− jrβ θ == ===ηπ120 377 Ej= β sinθθˆ Hφ βε 4π r 8 The ideal dipole (Hertziandipole): radiated power Power flowing density : (Unit : w/m*m) 11 IzΔ sin2 θ SEH=×=∗ ()2ωμβ rˆ 224π r 2 Total radiated power (Unit : W) ωμβ Δz P =⋅=Sds()40() Iz Δ=2222π I ∫∫ 12π λ Antenna Parameters: radiation pattern E The field pattern with its maximum value is 1 F(,)θφ = θ Eθ ,max For Hertzian dipole F(θ)=sinθ 9 Antenna Parameters: power pattern PF(,)θφ = (,)θ φ 2 It is worth noting that the field patter and power pattern are the same in decibles. PF(,)θφ= (,) θφ dB dB Power pattern parameters Antenna Parameters: Directivity Directivity: The ratio of the radiation intensity in a certain direction to the average radiation intensity. 1 ∗ 2 Re(E × Hr )⋅ ˆ UUr(,)θφ (,)/ θφ 2 D(,)θφ ==22 UUrPrave(,)θφ ave (,)/ θφ /4 π Radiation intensity: the power radiated in a given direction per unit solid angle 1 UEHrr(,)θφ =×⋅ Re(∗ ) 2 ˆ 2 Average radiation intensity: 1 UUdP(,)θ φθφπ=Ω= (,) /4 ave 4π ∫∫ 10 Antenna Parameters: Directivity When directivity is quoted as a single number without reference to a direction, maximum directivity is usually intended. U 4π D ==m U 2 ave ∫∫ Fd(,)θφ Ω Directivity of Hertzian dipole 1 IzΔ U (,)θ φωμβθ= ( )22 sin 24π 1 IzΔ ∴U = ()2ωμβ m 24π 1 IzΔ UP(,)θ φπ== /4 ( )2 βωμ ave 34π 3 ∴DdB==10log1.5 = 1.75 2 Antenna Parameters: Gain Gain : 4π times the ratio of radiation intensity in a given direction to The net power accepted by the antenna from the connected transmitter 4(,)πU θφ G(,)θφ = Pin • If all input power appeared as radiated power (Pin=P), Directivity = Gain. • In reality, some of the power is lost in the antenna absorbed by the antenna and nearby structures). • Radiation efficiency: P eerr= ,(0≤≤ 1) Pin ∴GeD= r 11 Antenna Parameters: antenna impedance, radiation efficiency • Input impedance Z AA= RjX+ A RA: Dissipation = Radiation + Ohmic loss XA: Power stored near the antenna 1122 PPP=+ = RI + R I in ohmic22 r A ohmic A • Efficiency PPRr er == = PPPin+ ohmic R A • For Ideal dipole Pz2 ωμβ Δ RIz==()80() Δ=222π r 2 I 2 12π λ I A Rr is very small since △z << λ Antenna Parameters: ideal dipole is an ineffective radiator Example For radiator △z = 1cm, f0 = 300MHz (λ= 1m) → Rr = 79mΩ ∴ It needs 3.6A for 1W of radiated power. For radiator △z = 1cm, f0 = 1GHz (λ= 30cm) → Rr = 0.87Ω ∴ It needs 1.5A for 1W of radiated power. 12 Antenna Parameters: How to increase the radiation resistance (efficiency) of the short dipole Practical short dipole with triangular-like current distribution 1. Capacitor-plate antenna: the top-plate supply the charge such that the current on the wire is constant Δz R = 80π 22 ( ) r λ Δz R = 20π 22 ( ) r λ Antenna Parameters: How to increase the radiation resistance (efficiency) of the short dipole 2. Transmission line loaded antenna Looks like uniform distribution if △z<<L Image theory 3. Inverted-L antenna or Inverted-F antenna 13 Antenna Parameters: effective aperture a. The effective aperture of an antenna, Ae ,is the ratio of power received in its load impedance, PRav , to the power density of the incident wave, S ,when the polarization of the incident wave and the polarization of receiving antenna ard matched: PR 2 Ae = ()in m Sav b. The maximum effective aperture Aem is the A e when the maximum power transferring to the load takes place, which means load impedance is the conjugate to the antenna impedance. c. Example: compute the Aem (maximum effective aperture) of a Hertzian dipole antenna. Ans: To solve Aem, two conditions should be kept in mind: (1)matched polarization for the incident θ Zˆ =−RjX wave and the antenna. L rad * (2)matched load. (ie. Z L = Z ant ) Eˆ Suppose the incident wave is arriving at an θ ˆ Z in =+RjXrad angle θ Zˆ =+R jX (1)the open-circuit voltage produced at the in rad Zˆ =−R jX terminals of the antenna is L rad Voc ˆ Vdloc = Eθ sinθ 14 (2)power density in the incident wave 2 1 Eθ Sav = 2 η0 (3)∵ the load is matched 2222 VEdloc θ sin θ ∴==the received power PR 88RRrad rad 2 112 ⎛⎞V ( ∵ I R=⎜⎟oc R ) 22⎜⎟rad ⎝⎠2Rrad (4)radiation resistance for Hertzian dipole 2 2 ⎛⎞dl R =80π ⎜⎟ rad ⎜⎟ ⎝⎠λ0 2 2 E λ 2 ∴=P θ 0 sin θ R 640π 2 P ⎛⎞2 λ2 2 λ 0 (5) ∴==AD()θ,φθθφR 1.5 sin⎜⎟0 =() , em S ⎜⎟4π 4π av ⎝⎠ d. For general antennas, the above relation hold. 4π ⇒=DA()θφ,2 em () θφ , for lossless antenna. λ0 4π ⇒=GA()θφ,2 em () θφ , for lossy antenna. λ0 The maximum effective aperture of an antenna used for reception is related to the directive gain in the direction of the incoming wave of that antenna when it is used for transmission.