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 \(6\text{ dB}\)".

Hint: The ideal value is the highest value.

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.

Mnemonic hint: High Ratio = Maximum Efficiency.

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Essentially, there is a point of diminishing returns when you're trying to maximize forward gain, after which the front lobe spreads out more and front to back ratio starts to fall off again.

There's a great explanation on the ARRL forum: http://www.arrl.org/forum/topics/view/118

Now: https://web.archive.org/web/20200805195036/http://www.arrl.org/forum/topics/view/118

Hint: Just remember Maximum gain, Minimum ratio

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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.

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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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Radiation resistance represents the portion of an antenna’s impedance that corresponds to power successfully radiated as radio waves, as if the power were dissipated by a resistor. A higher radiation resistance generally means the antenna radiates more efficiently.

At the resonant frequency, radiation resistance is maximized, resulting in the most efficient radiation. Below resonance, the impedance becomes more reactive (usually capacitive), reducing the radiation resistance and making the antenna less efficient at radiating energy.

Thus, radiation resistance decreases below resonance.

Memory aid:
Below resonance = Resistance decreases. "Below" = Decrease.

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A wide, flat strap (preferably NOT braided, same reason as NOT stranded) has more surface area than an equivalent gauge wire. RF energy travels on the surface of the conductor. Greater surface area = greater conductor. Commercial broadcast stations have solid 00 ga. RF ground cables.

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The best ground is a low-resistance connection to earth. While a water pipe might accomplish that, ground rods is the answer they want on the test.

The reason why the water pipe is not the best answer is that usually the condition of the pipe and its connection to the ground is unknown. For example, it is likely to be corroded. ground rods are clad in copper and designed such that they don't have the corrosion problems of the average water pipe.

Another reason is that the metal section of a water pipe may be very short because someone at some point replaced the water main with PVC pipe but you're unable to see it since it is buried.

The other answers are wrong because: a resistor doesn't improve your ground, it only makes it worse, and a series RF choke either does nothing or makes things worse.

Hint: Ground=Earth.

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