Question

The available power required at a receiving antenna is 10^-6 Watts. Transmitting and receiving antennas have gain of 40 dB each. The carrier frequency is 4 GHz and the distance between them is 30 miles. The required transmitter power is
1. 0.64 W
2. 0.71 W
3. 0.78 W
4. 0.32 W

Answer 1

Correct Answer - Option 1 : 0.64 W

Concept:

The power received by the receiving antenna is given by:

\({P_r} = \;{P_t}{\rm{\;}}{G_t}{G_r}{\left( {\;\frac{{\rm{λ\;}}}{{4\pi r}}\;} \right)^2}\)

Where:

P_t is the power transmitted from the antenna,

G_t is the Gain of the transmitting antenna,

G_r is the Gain of the receiving antenna,

λ is the wavelength of the carrier signal,

r is the distance between the transmitting and receiving antenna.

And the relation between wavelength and frequency is given by:

λ = c/f

Where:

c is the velocity of light, which is a constant (c = 3 × 10⁸)

f is the frequency of the given signal.

Given:

P_r = 10^-6

G_L = 40 dB

f_c = 4 GHz

r = 30 miles

Analysis:

\({P_r} = {P_t}{G_t}{G_r}{\left( {\frac{λ }{{4\pi r}}} \right)^2}\)

\({G_t} = {G_r} = 40\;dB = {10^4}\)

\(λ = \frac{{3\; × \;{{10}^8}}}{{4\; × \;{{10}^9}}} = 0.075\;m\)

1 mile = 1.6 km = 1600 m

30 miles = 1600 × 30 = 48000 m 

\({10^{ - 6}} = {P_t} × (\frac{{{{10}^4}\; × \;{{10}^4}\; ×\; 0.075\; × \;0.075}}{{4 \;×\; 3.14\; × \;48000 \;×\; 4 \;× \;3.14\; ×\; 48000}})\)

\({P_t} = \frac{{{{10}^{ - 6}}\; × \;4\; × \;3.14\; ×\; 48\;× \;4\; × \;3.14\; × \;48\; × \;100}}{{{{10}^4}\; ×\; 0.075\; ×\; 0.075}}\)

= 0.64 W