Page 565 - Engineering Electromagnetics, 8th Edition

CHAPTER 14 ELECTROMAGNETIC RADIATION AND ANTENNAS 547

Using (100) and (101), the effective area of the Hertzian is
P L2 3 2 2 2
A e2 (θ 2 ) = = λ sin (θ 2 )[m ] (102)
S r1 8π
The directivity for the Hertzian dipole, derived in Example 14.1, is
3
2
D 2 (θ 2 ) = sin (θ 2 ) (103)
2
Comparing Eqs. (102) and (103), we find the relation that we are looking for: the
effective area and directivity for any antenna are related through

4π
D(θ, φ) = A e (θ, φ) (104)
λ 2

We can now return to Eq. (96) and use Eq. (104) to rewrite the ratio of the power
delivered to the receiving antenna load to the total power radiated by the transmitting
antenna: this yields an expression that involves the simple product of the effective
areas, known as the Friis transmission formula:

P L2 A e2 (θ 2 ,φ 2 ) D 1 (θ 1 ,φ 1 ) A e1 (θ 1 ,φ 1 )A e2 (θ 2 ,φ 2 )
= = (105)
2 2
P r1 4πr 2 λ r
The result can also be expressed in terms of the directivities:

P L2 λ 2
= D 1 (θ 1 ,φ 1 )D 2 (θ 2 ,φ 2 ) (106)
P r1 (4πr) 2
These results provide an effective summary of what was discussed in this section,
by way of giving us a very useful design tool for a free space communication link.
Again, Eq. (105) assumes lossless antennas in the far zones of each other and gives
the power dissipated by a load that is conjugate matched to the receiving antenna
impedance.

D14.10. Given: an antenna having a maximum directivity of 6 dB and op-
erating at wavelength, λ = 1m. What is the maximum effective area of the
antenna?

Ans. 1/π m 2

D14.11. The power of 1 mW is dissipated by the matched load of a receiving
2
antenna of a 1-m effective area. This antenna is positioned at the center of the
main beam of the transmitting antenna, located 1.0 km away. What total power
is radiated by the transmitter if its directivity is (a) 10dB, (b) 7dB?

Ans. 4π kW; 8π kW