Radiation resistance Antennas are designed for effective radiation of electromagnetic energy. I – Equivalent circuit of an antenna in R r – input radiation resistance Rr • Represents radiated energy – input loss resistance RL • Represents conduction and dielectric losses of the antenna – input reactance XA • represents the energy stored in the field near the antenna 1 Radiation resistance I 2 R The power radiated is equal to:W = in r rad 2 The power losses is I 2 R W = in L loss 2 If Win is the input power, the radiation efficiency is: Wrad Rr η r = = Win Rr + RL I in R r 2 Directive gain, directivity and gain Stronger in some Same intensity for all directions directions Isotropic Antenna (the reference antenna) 3 Directive gain, directivity and gain Let Pavg be the average Poynting vector which is the power flow density per unit area, 1 P = Re()E× H * avg 2 The total power radiated Wrad is then W = P ⋅ dS dS = R 2 sinθdθdφ rad ∫ avg S W = U (θ,φ)dΩ dΩ =sinθdθdφ rad ∫ S where U(θ,φ) is the power flow through a unit solid angle, and is called the radiation intensity (W/sr). 2 U (θ ,φ) = r Pavg 4 Directive gain, directivity and gain 5 Directive gain, directivity and gain Directive gain GD(θ,φ) Ratio of the radiation intensity in a particular direction(θ,φ) to the average radiation intensity. U (θ,φ) U (θ,φ) GD (θ,φ) = = U avg Wrad / 4π Directivity Maximum value of the directive gain in a certain direction. D = Max{Gd (θ,φ)} Power Gain Ratio of the radiation intensity in a given direction to the radiation intensity of a lossless isotropic radiator that has the same input power. U (θ,φ) G p (θ,φ) = Win / 4π 6 Example Find the directive gain of a Hertzian dipole. 1 1 P = Re()E × H * = E H avg 2 2 θ φ (Idl )2 P = η β 2 sin 2 θ avg 32π 2 r 2 o (Idl )2 U = P r 2 = η β 2 sin 2 θ avg 32π 2 o θ I H φ Eθ 7 Example and then U (θ,φ) GD (θ,φ) = U avg sin 2 θ = 2ππ ()sin 2 θ sinθdθdφ / 4π ∫∫00 3 = sin 2 θ 2 8 Example 9 Example Directive gain and the directivity of the Hertzian dipole 3 2 GD (θ,φ) = sin θ θ 2 3 ∴ D = G (π / 2,φ) = = 1.76 dB I d 2 Suppose the radiation efficiency is 46%, θ ∴G p = 0.46D = 0.69 = −0.16 dB (Wrad /Win = 0.46) 10 Example Find the radiation resistance of a Hertzian dipole 2ππ Pr = Pavg sinθdθdφ ∫∫00 2 2 2ππ I ()dl 2 3 = ηo β sin θdθdφ 32π 2 ∫∫00 I 2 ()dl 2 = η β 2 12π 2 o 2 I 2 dl = 80π 2 2 λ 2 2 dl ∴ Rr = 80π λ dl Suppose, = 0.01 ⇒ Rr = 0.08Ω Poor radiator !! λ 11 Linear dipole antenna In dipole antennas, the current magnitude along the dipole can be represented like in a transmission line, where 2h is the length of the dipole and z=0 at the center feed-point of the dipole. I ( z) = I m sin [β (h − z )] I 2h 12 Linear dipole antenna Knowing the current distribution I(z), we can sum up the fields due to the infinitesimal segments on the antenna using the results of the Hertzian dipole. − jkR ˆ jη o I m e E = θ F (θ ) 2πR jI e − jkR H = φˆ m F (θ )a I 2πR φ 2h cos()βh cos θ − cos βh F (θ ) = sin θ F(θ) is called the pattern function 13 Linear dipole antenna Antenna pattern – E-plane pattern (pattern function versus θ for a constant φ) – H-plane pattern (pattern function versus φ for a constant θ=π/2) 14 Linear dipole antenna At certain dipole lengths (≅ λ/2, λ…) called resonant lengths, the input impedance is purely resistive. For half- wavelength dipole, Z in ≈ Rr = 73 Ω The pattern pattern for a half-wavelength dipole is cos(βh cos θ ) − cos βh F (θ ) = sin θ π cos cos θ 2 = sin θ D = 1.64 15 Example -- Monopole A thin quarter-wavelength vertical antenna over a conducting ground is excited by a sinusoidal source at its base. Find the (a) radiation pattern, (b) resistance, and (c) directivity. 16 Example -- Monopole (a) The electromagnetic field in the upper half-space due to the quarter-wave vertical antenna is the same as that of the half-wave antenna. (b) The magnitude of the time-average Poynting vector holds0 ≤forθ ≤ π / 2 but the quarter-wave antenna radiates only into the half-space, its total radiated power is only half of a half-wave dipole. Therefore, the radiation resistance is Rr = Rr (half − wave dipole)/2 = 73/ 2 = 36.5 Ω (c) SameD as= 1 half-wave.64 dipole 17 Effective Area and Friis Equation Effective Area The effective area Ae of a receiving antenna is the ratio of the time-average power received to the time-average power density of the incident wave at the antenna. Ae = PL / Pavg It may be shown that is Ae related to the directive gain as: λ2 A = G (θ,φ) e 4π D 18 Effective Area and Friis Equation Friis Equation Consider two antennae separated by a distance r. The transmitting antenna transmits a total power Pt. A , G , P Ae1, GD1, Pt e2 D2 L r The time-average power density at the receiving antenna is P P = t G avg 4πr 2 D1 19 Effective Area and Friis Equation The power received to the load is PL = Pavg Ae2 λ2 = G P 4π D2 avg λ2 GD1GD2 Pt ()4πr 2 P λ2 L (Friis Equation) ∴ = 2 GD1GD2 Pt ()4πr 20.