Answer :
The power received by the helicopter is significantly higher than that received by the large aircraft despite the shorter distance, due to the inverse square relationship between the distance and the power received. The helicopter is about 13,000 times closer than the large aircraft, but it receives nearly 650 times more power.
The question is talking about an ATC (air traffic control) radar that has a frequency of 3 GHz, an average transmit power of 1 kW, and an antenna gain of 4000.
The goal is to compare the power received at the antenna of a large aircraft 20 nautical miles away and a helicopter at 3 miles distance sailing.
ATC radar uses a parabolic reflector with a diameter of 4 meters. So, the effective area of the antenna can be calculated as follows:
[tex]$$A_e = \frac{\pi D^2}{4}$$[/tex]
where Ae is the effective area, D is the diameter of the antenna, and pi is the mathematical constant 3.14.The effective area of the ATC antenna is given by:
[tex]$$A_e = \frac{\pi D^2}{4}$$ $$A_e = \frac{\pi (4)^2}{4}$$ $$A_e = 12.56 m^2$$[/tex]
The formula for calculating power received can be stated as follows:
[tex]$$P_r = P_t G_t G_r \frac{\lambda^2}{(4\pi R)^2}$$[/tex]
where Pt is the transmitted power, Gt and Gr are the gains of the transmitting and receiving antennas, respectively, lambda is the wavelength of the signal, and R is the distance between the two antennas.
To solve for the power received, we must calculate the other values first.
The wavelength of the signal can be calculated using the formula:
[tex]$$\lambda = \frac{c}{f}$$[/tex]
where c is the speed of light and f is the frequency of the signal. So, the wavelength of the signal can be determined as follows:
[tex]$$\lambda = \frac{c}{f} = \frac{3 \times 10^8 m/s}{3 \times 10^9 Hz} = 0.1 m$$[/tex]
The power received by a large aircraft at a distance of 20 nautical miles can be calculated as follows:
[tex]$$P_r = P_t G_t G_r \frac{\lambda^2}{(4\pi R)^2}$$$$P_r = (1000)(4000)(4000) \frac{(0.1)^2}{(4\pi (20 \times 1852))^2}$$$$P_r = 3.41 \times 10^{-12} W$$[/tex]
On the other hand, the power received by a helicopter at a distance of 3 nautical miles can be calculated as follows:
[tex]$$P_r = P_t G_t G_r \frac{\lambda^2}{(4\pi R)^2}$$$$P_r = (1000)(4000)(4000) \frac{(0.1)^2}{(4\pi (3 \times 1852))^2}$$$$P_r = 2.19 \times 10^{-9} W$$[/tex]
As a result, the power received by the helicopter is significantly higher than that received by the large aircraft despite the shorter distance, due to the inverse square relationship between the distance and the power received.
The helicopter is about 13,000 times closer than the large aircraft, but it receives nearly 650 times more power.
To know more about power visit;
brainly.com/question/29575208
#SPJ11
