Question

A VHF communications link operates over a large lake at a carrier frequency of 52MHz. The path length between transmitter and receiver is 18.6 km and the heights of the sites chosen for the location of the transmitter and receiver towers are 72.6 m and 95.2 m respectively.

Assuming that the cost of building a tower increases at a rate greater than linearly with tower height find the most economic tower heights which will take full advantage of the 'ground' reflected ray.

Answer 1

Using the wave speed equation and the respective antenna lengths, the calculated frequencies for the channels are approximately 77.3 MHz for the 1.94 m antenna and 199.2 MHz for the 0.753 m antenna, both of which fall within the VHF TV channels frequency ranges.

Step-by-step explanation:

The student's question involves calculating the frequencies for television channels based on the lengths of antennas designed for VHF (very high frequency) reception. The formula used to calculate frequency (f) is derived from the wave speed equation, where the speed of electromagnetic waves (c) in a vacuum is approximately equal to 3×108 m/s, and the wavelength (λ) is twice the length of the antenna, as antennas are typically designed to be half the wavelength they are receiving. So, the frequency is calculated as f = c / λ.

To find the frequencies for the given antennas, we calculate:

• For the 1.94 m antenna, the wavelength is 2 × 1.94 m = 3.88 m. Thus, the frequency is f = 3×108 m/s / 3.88 m ≈ 77.3 MHz.
• For the 0.753 m antenna, the wavelength is 2 × 0.753 m = 1.506 m. Thus, the frequency is f = 3×108 m/s / 1.506 m ≈ 199.2 MHz.

These frequencies fall within the typical VHF TV channels ranges of 54 to 88 MHz and 174 to 222 MHz.