Antenna Fundamentals

Antenna Fundamentals - NUS/ECE EE4101 Antenna Fundamentals...

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Hon Tat Hui Antennas NUS/ECE EE4101 1 Antenna Fundamentals 1 Introduction Antennas are device designed to radiate electromagnetic energy efficiently in a prescribed manner. It is the current distributions on the antennas that produce the radiation. Usually these current distributions are excited by transmission lines or waveguides. Transmission line Current distributions Antenna

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Hon Tat Hui Antennas NUS/ECE EE4101 2 2 Antenna Parameters 2.1 Poynting Vector and Power Density Instantaneous Poynting vector: () ( ) {} 2 ,, , , Re , , Re , , (W/m ) jt xyzt xyze ωω pE H EH Average Poynting vector: { } * 2 av 1 R e ( W / m ) 2 xyz PE H Time expressions: E ( x , y , z , t ) H ( x , y , z , t ) Phasor expressions: E ( x , y , z ) H ( x , y , z ) Note that Poynting vector is a real vector. Its magnitude gives the instantaneous or average power density of the electromagnetic wave. Its direction gives the direction of the power flow at that particular point. Note:
Hon Tat Hui Antennas NUS/ECE EE4101 3 2.2 Power Intensity ( ) 2 av W/sr Ur = P sr = steradian, unit for measuring the solid angle. Solid angle Ω is the ratio of that part of a spherical surface area S subtended at the centre of a sphere to the square of the radius of the sphere. r S Ω () 2 sr S r Ω= Spherical surface The solid angle subtended by a whole spherical surface is therefore: (sr) 4 4 2 2 π = = Ω r r o Note that U is a function of direction ( θ , φ ) only and not distance ( r ).

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Hon Tat Hui Antennas NUS/ECE EE4101 4 2.3 Radiated Power * rad av 1 Re[ ] (W) 2 ss P =⋅ = × ∫∫ Pd s E H d s ww P av ϕ Antenna n ds ˆ sin 2 φ θ d d r = Note that the integration is over a closed surface with the antenna inside and the surface is sufficiently far from the antenna (far field conditions). r θ
Hon Tat Hui Antennas NUS/ECE EE4101 5 Example 1 Find the total average radiated power of a Hertzian dipole. Solution {} av 2 2 2 2 2 11 Re Re 22 1 Re sin (W/m ) 24 EH E EE kId r θφ θ θθ ηη ηθ π ∗∗ = × ⎧⎫ == ⎨⎬ ⎩⎭ = r rr r PE H a aa a A

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Hon Tat Hui Antennas NUS/ECE EE4101 6 rad av 2 2 2 2 2 00 2 sin sin 24 (W) 3 s P kId rd d r Id ππ ηθ θ φθ π ηπ λ =⋅ = ∫∫ rr Pd s aa A A w
Hon Tat Hui Antennas NUS/ECE EE4101 7 Example 2 Find the total average radiated power of a half-wave dipole. For a half-wave dipole: Solution ( ) [ ] cos 2 cos 60 , sin jkr θ θ m eE Ej I H r φ π θ η ⎛⎞ == ⎜⎟ ⎝⎠ {} 2 av 2 2 2 2 2 15 cos[( / 2)cos ] (W/m ) sin m E I r πθ = = r r Pa a

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Hon Tat Hui Antennas NUS/ECE EE4101 8 {} rad av 2 2 2 2 2 00 2 2 0 15 cos[( / 2)cos ] sin sin cos [( / 2)cos ] 30 (W) sin s m m P I rd d r Id ππ π πθ θ φθ =⋅ = ∫∫ rr Pd s aa w The above remaining integral can be evaluated numerically to give: 2 rad 36.54 (W) m PI =
Hon Tat Hui Antennas NUS/ECE EE4101 9 Hence for a λ /4 monopole over a ground plane with a maximum current at its base = I m , the radiated power is half that of a /2 dipole, i.e., 2 rad 18.27 (W) m PI = Why?? Think about it!

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