Question

A transmitter with a 50 W output power is using 6dBi antenna to transmit 3GHz signal. The receiving antenna has a 3dBi gain and minimum noise floor is −80dBm.

What is the maximum distance that the signal can be transmitted.

Answer 1

The maximum distance that the signal can be transmitted is approximately 2.08 kilometers.

Step-by-step explanation:

To determine the maximum distance that the signal can be transmitted, we need to consider the power of the transmitter, the gain of the antennas, and the minimum noise floor. The formula used to calculate the maximum distance is:

Maximum Distance = √ ((Transmitter Power * Transmitter Gain) / (4 * π * Minimum Noise Floor * Receiver Gain * Frequency))

Plugging in the given values:

Transmitter Power = 50 W, Transmitter Gain = 6 dBi, Receiver Gain = 3 dBi, Minimum Noise Floor = -80 dBm, Frequency = 3 GHz (3000 MHz)

Converting dBm to Watts: -80 dBm = 10^((-80-30)/10) = 10^(-11) Watts = 0.00000000001 Watts

Plugging in the values:

Maximum Distance = √ ((50 * 10^(-3)) / (4 * π * 0.00000000001 * 3000 * 10^6)) = √ (0.00000005 / (4 * 3.14159 * 0.00000000001 * 3000 * 1000000000)) = 2.08 * 10^6 meters

Therefore, the maximum distance that the signal can be transmitted is approximately 2.08 kilometers.