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Antenna gain measures how well an antenna directs radio frequency energy in a specific direction. It is typically expressed in dBi, which means decibels over isotropic, a theoretical point source antenna that sends signals equally in all directions.
To convert antenna gain from dBi to a linear scale, we use the formula: \( G = 10^{\frac{dB}{10}} \)
For our exercise, each antenna has a 30 dBi gain. In linear scale, this is: \( G = 10^{\frac{30}{10}} = 10^3 = 1000 \).
Higher gain antennas focus energy more tightly in a specific direction, allowing for greater distances and reduced interference from other directions.

The effective aperture area of an antenna is a measure of its ability to receive power from a radio wave. The formula for the effective aperture (\( A_e \)) is: \( A_e = \frac{G \times \frac{c^2}{4 \times \pi^2 \times f^2}} \)
Here, \( G \) is the gain (linear scale), \( c \) is the speed of light (\( 3 \times 10^8 \) m/s), and \( f \) is the frequency (\( 14 \times 10^9 \) Hz).
Substituting these values, we get: \( A_e = \frac{1000 \times (3 \times 10^8)^2}{4 \times \pi^2 \times (14 \times 10^9)^2} \).
Simplifying further, the effective aperture area comes out to be approximately 0.64 square meters.

Path loss represents the signal attenuation, or loss, as it travels from the transmitter to the receiver. It can be calculated using the Friis transmission equation.
The general form for free-space path loss (L) is: \( L = 20 \times \log_{10}(d) + 20 \times \log_{10}(f) - 147.55 \)
For a distance \( d = 10 \, km \) and frequency \( f = 14 \, GHz \): \( L = 20 \times \log_{10}(10^4) + 20 \times \log_{10}(14 \times 10^9) - 147.55 = 80 + 202.92 - 147.55 = 135.37 \, dB \).
This equation factors in the distance and frequency to give us the total signal loss from the transmitter to the receiver.

The Friis transmission equation estimates the power received by an antenna under ideal conditions. It states that the power received (\( P_r \)) is directly proportional to the transmitted power (\( P_t \)) and antenna gains, and inversely proportional to the square of the distance between them. The equation is: \[ P_r = P_t \times G_t \times G_r \times \left( \frac{\lambda}{4 \pi d} \right)^2 \]
Where \( \lambda \) is the wavelength, \( G_t \) and \( G_r \) are the gains of the transmitting and receiving antennas, and \( d \) is the distance. Given that \( G_t = G_r = 1000 \) and \( \lambda = \frac{c}{f} = \frac{3 \times 10^8}{14 \times 10^9} \), we can compute the received power under line of sight (LOS) conditions.

Power density roll-off describes how the power per unit area decreases as we move away from the source. This follows the inverse square law, given by: \( S(r) = \frac{P_t}{4 \times \pi \times r^2} \)
Here, \( S(r) \) is the power density at distance \( r \), and \( P_t \) is the transmitted power.
As \( r \) increases, the power density decreases by the square of the distance. Thus, if you double the distance \( r \), the power density becomes one-fourth. This principle is crucial in understanding how signal strength diminishes with distance, impacting the quality and reliability of communication links.

Line of Sight (LOS) propagation occurs when the transmitting and receiving antennas are in direct view of each other without any obstacles. This type of propagation is ideal for point-to-point communication systems as it minimizes path loss.
In our exercise, the antennas are mounted on masts 100m high, thus ensuring a clear LOS over the 10 km distance.
By avoiding obstacles and reflections, LOS propagation ensures the transmitted signal maintains its integrity, allowing for optimal communication performance.