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Explain the significance of the term 'effective area of an antenna'. Derive the relationship between effective area and directivity of any antenna.

Mumbai University > Electronics Engineering > Sem 5 > Electromagnetic Engineering

Marks: 10 Marks

Year: May 2015, 2016

1 Answer
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A useful parameter of calculating the received power of an antenna is the effective area or effective aperture(Ae).

The gain is related to effective area by following relationship:

$G=\dfrac{4πA_e}{λ^2}$

And $A_e=k_a.A$

Where,

λ → wave length

$A_e →$ Effective antenna aperture

A → Physical area of antenna

$k_a →$ Antenna aperture efficiency.

Consider, a two antenna communication system as shown below-

enter image description here

The antennas are displayed at a distance R and are assumed to be lined up with respect to polarization and directivity.

The directivity of antenna 1 is DT and its effective area is AeT. Similarly, directivity of antenna 2 is DR and it’s effective area is AeR.

The totally radiated power by antenna 1 is PT. If antenna 1 were isotropic radiator, the power density from distance R from antenna 1 would be,

$S_o=\dfrac{P_T}{4πR^2}$

Power density ST at distance R is

$S_T=S_o D_T=\dfrac{P_T.D_T}{4πR^2}$

Power received by antenna 2 PR is

$P_R=S_T A_e R=\dfrac{P_T D_T A_e R}{4πR^2}$

AeR is effective aperture of antenna 2

$D_T A_e R=\dfrac{P_R}{P_T}4πR^2$

Now if antenna 2 is transmitted we get,

$D_R.A_e T=\dfrac{P_R}{P_T} 4πR^2 \\ \dfrac{D_T}{A_e T}=\dfrac{D_R}{A_e R}$

For short dipole: $A_e=\dfrac{3}{8π} λ^2 \\ D=\dfrac{3}{2}$

Thus, for any antenna $\dfrac{λ^2}{4π}=\dfrac{A_e}{D}$

This gives us the required relation between effective apperture Ae and directivity D of an antenna

$$D=\dfrac{4πR^2}{λ^2}$$

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