Practice Problems (KEY)

Practice Problems (KEY)#

Communications#

  1. T/F: Line-of-sight communications are the primary way of communicating with satellites.

  2. T/F: The distance that ground waves travel increases as the frequency of the signal increases.

  3. T/F: Sky wave propagation requires “blobs” of air in the troposphere to reflect from.

  4. T/F: A waveguide acts as a high-pass filter.

  5. Eisenhower Tunnel is approximately 5 m wide and 4 m tall. Assuming the tunnel acts as a waveguide, which of the following frequencies could be used to communicate inside the tunnel?

    a. A CB radio operating at 27 MHz
    b. An AM radio station operating at 94.5 MHz
    c. An FM station operating at 101.5 MHz
    d. A Family Radio Service transmitter operating at 462 MHz

To determine the operating frequencies capable of passing through this tunnel, we use the waveguide cutoff frequency:

\[ f_{c/o} = \frac{c}{2a} = \frac{3 \times 10^{8}}{2 \cdot 5} = 30{,}000{,}000\ \text{Hz} = 30\ \text{MHz} \]

Because a waveguide acts as a high-pass filter, any frequency greater than 30 MHz will pass through the tunnel.

  1. Which form of wireless communication works best at night?

    a. Line-of-sight (LOS)
    b. Ground wave
    c. Sky wave
    d. Forward scatter

  1. Which frequency band attenuates rapidly in the atmosphere, but whose short wavelengths permit very precise measurements?

    a. Very low frequencies (VLF)
    b. High frequencies (HF)
    c. Ultra high frequencies (UHF)
    d. Extremely high frequencies (EHF)

  1. If your eyes are 6 ft above sea level, how far is the horizon?

\[ r = \sqrt{2h} = \sqrt{2 \cdot 6} = \sqrt{12} = 3.46\ \text{mi} = 5.58\ \text{km} \]

  1. An A-10 needs to establish LOS radio contact with a Special Forces team on the ground. If the A-10 is flying at 200 ft and the team is on a 32 ft ridgeline, how close must the A-10 get to establish LOS?

\[ r_{\text{LOS}} = \sqrt{2h_{\text{SOF}}} + \sqrt{2h_{\text{A-10}}} = \sqrt{2 \cdot 200} + \sqrt{2 \cdot 32} = 28\ \text{mi} = 45.08\ \text{km} \]

  1. A fighter aircraft orbiting at 5000 ft AGL is searching for a bomber at 100 ft AGL. If the aircraft are 93 miles apart, does the fighter have LOS with the bomber?

\[ r_{\text{LOS}} = \sqrt{2h_{\text{fighter}}} + \sqrt{2h_{\text{bomber}}} = \sqrt{2 \cdot 5000} + \sqrt{2 \cdot 100} = 114\ \text{mi} \]

Since \(114\ \text{mi} > 93\ \text{mi}\), the fighter does have LOS with the bomber.

Antennas#

  1. Calculate the required length of a half-wave dipole designed for a 144 MHz radio transceiver.

\[ \lambda = \frac{c}{f} = \frac{3 \times 10^{8}\ \text{m/s}}{144 \times 10^{6}\ \text{Hz}} = 2.083\ \text{m} \]
\[ \text{dipole antenna length} = \frac{\lambda}{2} = \frac{2.083\ \text{m}}{2} = 1.04\ \text{m} \]

  1. Calculate the required length of a quarter-wave monopole designed for an 800 kHz AM radio transceiver.

\[ \lambda = \frac{c}{f} = \frac{3 \times 10^{8}\ \text{m/s}}{800 \times 10^{3}\ \text{Hz}} = 375\ \text{m} \]
\[ \text{monopole antenna length} = \frac{\lambda}{4} = \frac{375\ \text{m}}{4} = 93.75\ \text{m} \]

  1. A circular parabolic dish has an efficiency of 91%, a radius of 1 m, and a gain of 300. What frequency is the antenna designed for?

\[ G = \left(\frac{2\pi r}{\lambda}\right)^2 \eta \]

So,

\[ \lambda = 2\pi r\sqrt{\frac{\eta}{G}} = 2\pi(1\ \text{m})\sqrt{\frac{0.91}{300}} = 0.346\ \text{m} = 346\ \text{mm} \]
\[ f = \frac{c}{\lambda} = \frac{3 \times 10^{8}\ \text{m/s}}{0.346\ \text{m}} = 867\ \text{MHz} \]

  1. A circular parabolic dish with a diameter of 600 mm is used to receive a satellite TV signal at 18 GHz. The dish has an efficiency of 90%. What is the dish’s gain?

\[ \lambda = \frac{c}{f} = \frac{3 \times 10^{8}\ \text{m/s}}{18 \times 10^{9}\ \text{Hz}} = 0.0167\ \text{m} = 16.7\ \text{mm} \]
\[ G = \left(\frac{2\pi(0.3\ \text{m})}{0.0167\ \text{m}}\right)^2(0.9) = 11512 \]

  1. T / F: Increasing the carrier frequency in a communication system increases the size of the necessary antenna.

The size of the antenna is inversely proportional to frequency. As frequency increases, antenna size decreases.

  1. A TACAMO aircraft is used to send messages to a submarine using a carrier frequency of 30 kHz. How long must the antenna be if a dipole antenna is used?

A dipole antenna must be half the wavelength of the carrier frequency.

\[ \lambda = \frac{c}{f} = \frac{3 \times 10^{8}}{30 \times 10^{3}} = 10{,}000\ \text{m} = 10\ \text{km} \]
\[ \text{dipole antenna length} = \frac{\lambda}{2} = \frac{10{,}000\ \text{m}}{2} = 5000\ \text{m} = 5\ \text{km} \]

The dipole antenna should be 5 km long.

  1. T / F: Antenna gain tells us how much the antenna can focus power.

Antenna gain tells us how much energy is focused in certain directions relative to antenna orientation.

  1. A circular dish antenna is used to communicate with a satellite system at 4.6 GHz. The dish has a diameter of 300 mm and an efficiency of 86%. What is the gain of the antenna?

The gain of a parabolic dish antenna is:

\[ G = \left(\frac{2\pi r}{\lambda}\right)^2 \eta \]

First, find wavelength:

\[ \lambda = \frac{c}{f} = \frac{3 \times 10^{8}\ \text{m/s}}{4.6 \times 10^{9}\ \text{Hz}} = 0.06522\ \text{m} = 65.22\ \text{mm} \]

Then compute gain:

\[ G = \left(\frac{2\pi(0.150\ \text{m})}{0.06522\ \text{m}}\right)^2(0.86) = 179.6 \]

Gain is unitless.

  1. T / F: An antenna can be used to transmit signals more easily than it can be used to receive signals.

There is no difference between an antenna’s ability to transmit and receive.

  1. Which of the following is an advantage of low-frequency carrier signals compared to high-frequency carrier signals?

a. They can be transmitted or received using smaller antennas.
b. They can support signals of greater bandwidth.
c. They can travel farther.

A is not true because lower-frequency signals have longer wavelengths, which require larger antennas.

B is not true because higher frequencies provide more available bandwidth. Bandwidth is approximately 1% of the carrier frequency.

C is true. Lower-frequency signals are not limited to line-of-sight. They can also follow the curvature of the Earth or reflect from the atmosphere, depending on frequency.

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