Equivalent Circuit of an Antenna – Input Radiation Resistance R

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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 radiatedθ Wrad is then φ θ θ W = P ⋅ dS dS = R 2 sin dφd rad ∫ avg S θ W = U ( , )dΩ dΩ =sin dθdφ rad ∫ S where U(θ,φ) is the power flow through a unit solid angle, and is called the radiation intensity (W/sr). θ 2 U ( ,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. θ U (θ,φ) U (θ,φ) GD ( ,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. θ U (θ,φ) G p ( ,φ) = Win / 4π 6 Example

Find the directive gain of a Hertzian dipole. 1 1 P = Re()E × H * = Eθ H avg 2 2 φ η (Idl )2 β P = 2 sin 2 θ avg 32π 2 r 2 o η (Idl )2 β U = P r 2 = 2 sin 2 θ avg 32π 2 o

I H φ

Eθ 7 Example and then

θ U (θ,φ) GD ( ,φ) = U avg sin 2 = 2ππ θ ()sin 2 sin d d / 4 ∫∫00 θ θ 3 θ φ = sin 2 θ π 2

8 Example

9 Example

Directive gain and the directivity of the Hertzian dipole θ φ 3 2 GD ( , ) = sin θ θ 2 π 3 ∴ D = G ( / 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 ππ Find the radiation resistance of a Hertzian dipole θ 2 θ φ Pr = Pavg sin d d ∫∫00 2η 2 π β 2ππ I ()dl 2 3 = o sin d d 32 2 ∫∫00θ θ φ Iπ2 ()dlη 2β = 2 12 2 o 2 I 2  π  dl   = 80 2    2   λ   2 π 2  dl  ∴ Rr = 80    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 )]

12 Linear dipole antenna

Knowing the current distribution I(z), we can sum up the η fieldsθ due to the infinitesimal segments on the antenna using the resultsπ of the Hertzian dipole. θ φ j I e − jkR E = ˆ o m F ( ) 2 R π θ θ jI e − jkR H = ˆ m β F ( )a I 2 R θ φ 2h cos()h cos − cos βh F ( ) = 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 π θ   cos cosθ  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

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: λ 2 A = πG (θ,φ) e 4 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

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 λ 2 = G P π4 D2 avg λ 2 GD1GD2 Pt ()4πr 2

P λ 2 L (Friis Equation) ∴ = 2 GD1GD2 Pt ()4πr