Note that the gain of a parabolic antenna is governed by the following:

\[G = \frac{ 4\pi{A} }{ \lambda^2 }e_A\]

Where:

• \(A\) is the area of the antenna aperture (the mouth of the parabolic reflector)
• \(\lambda\) (lambda) is the wavelength of the radio waves
• \(e_A\) is a dimensionless "aperture efficiency" parameter between \(0\) and \(1\)

It is clear that by doubling the frequency, the wavelength is halved. Using proportional reasoning, we see that substituting \(\frac{\lambda}{2}\) for \(\lambda\) results in a change in \(G\) by a factor of \(4\).

In decibels, \(10\log_{10}(4)\) is equal to \(6.02\text{ dB}\). Hence, the correct answer is "Gain is increased by \(6\text{ dB}\)".

Silly memory aid: "para" means fo(u)r in Spanish, and you'd need 6 dB to quadruple power

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Hint: Only one answer has the word perpendicular in it.

Hint: only one answer has 'BOOM' in it :)

The key here is that the two Yagis are overlaid on the same boom. The result is two sets of elements, both pointing the same direction, with one set rotated 90° along the axis of the boom at right angles to the other.

The distractors talk about arranging the antennas in parallel or linearly. Neither of those things make for good circles – you have to have perpendicular angles out of phase to make the spiral wave.

Check out this great video from Khan Academy that explains linear and circular polarization of electromagnetic waves.

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Due to the fact that short verticals have a low radiation resistance, they are naturally ineffective so you will need to do whatever you can to make them as efficient as possible.

An HF mobile antenna loading coil should have a high ratio of reactance to resistance to minimize losses.

A high-Q loading coil should be placed near the center of the vertical radiator to minimize losses in a shortened vertical antenna.

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A small loading coil simply inserts a series inductive reactance that cancels capacitive antenna reactance. By using a mobile antenna loading coil you will minimize ground related losses. - KM4ARR

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Bandwidth is inversely proportional to quality factor Q, and \[Q = \frac{\text{reactance}}{\text{resistance}}\] Thus, adding reactance reduces (decreases) the bandwidth.

-robotoloco

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Eliminate Distractors:

Lower Q - Actually raises Q by tuning antenna to be resonant

Higher Losses - Not an advantage

Greater structural strength - Nonsense

Only real answer is Improved radiation efficiency

Top loading is a methodology which increases radiation resistance, hence efficiency, even if the ground plane is substandard; seemingly a ubiquitous vertical antenna shortcoming. A top loaded vertical antenna has several advantages over the conventional vertical, but the biggest advantage is that it's shorter in length.

Source: Antenna 013: 20 Meter Top Loaded Vertical

Maximizing Efficiency in HF Mobile Antennas

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The Q or Q-Factor, when it relates to antennas, is simply an inverse measure of the bandwidth in which that antenna is useable.

Q is defined as the center frequency divided by the bandwidth. So something with a higher Q would have a smaller bandwidth around its designed center frequency.

Example: You have a dipole which is made for 14.2 MHz and has a bandwidth (acceptable VSWR) of \(\pm 250\) kHz (0.5 MHz bandwidth)

The Q factor would be: \[Q=\frac{14.2}{0.5}=28.4\]

If the bandwidth were to be cut in half, the Q factor would double accordingly.

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The coil (inductor) is added to cancel out the capacitance already present in the circuit to try to achieve resonance.

It also facilitates a method to electrically shorten an antenna to "tune" to lower frequencies than the intended "designed" antenna length.

Silly hint: coil cancels capacitive reactance

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This question has two parts, concerning radiation resistance and capacitive reactance respectively.

For the first part: remember that radiation resistance is the amount of power which has been successfully radiated away by the antenna, described as though it was power dissipated by a resistor — so a higher radiation resistance implies a more efficient antenna, and maximum efficiency occurs at the resonant frequency. At other frequencies, efficiency decreases, which means radiation resistance decreases.

For the second part: the definition of capacitive reactance \(X_C\) in ohms is the reciprocal of the product of \(2π\), the frequency \(f\) in hertz, and the capacitance \(C\) in farads:

\[X_C=\frac{1}{2\pi fC}\]

This means the capacitive reactance \(X_C\) varies inversely with the frequency \(f\) — so if the frequency decreases, capacitive reactance increases.

Putting the two parts together, we get:

The radiation resistance decreases and the capacitive reactance increases.

For more details, see https://en.wikipedia.org/wiki/Antenna_(radio)#Current_and_voltage_distribution

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