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

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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:
Background image of page 2
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 ).
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 4
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
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 6
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
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 =
Background image of page 8
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!
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon