(B). One disadvantage of a directly fed random-wire antenna is that you may experience RF burns when touching metal objects in your station.

The simple single wire length of the random-wire antenna acts as both feed line and antenna and is directly connected to the transmitter. Because of the proximity of the antenna to your equipment, RF energy can "feed back" to your equipment and create RF hot spots. Proper grounding of the antenna and equipment cases is important to reduce burn risk.

For more info see Wikipedia: Random-wire antenna

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(D). An advantage of downward sloping radials on a quarter wave ground-plane antenna is that they bring the feed-point impedance closer to 50 ohms.

This type of antenna consists of a vertical 1/4 wavelength element and four horizontal 1/4 wavelength radials. These radial wires make up the ground plane. The advantage of sloping these radials from the horizontal plane downwards towards a 45 degree angle, is that the feed point impedance increases from about 35 ohms closer to 50 ohms. This impedance matches with the standard 50 ohm impedance of standard coaxial cable. Matching impedances is important at this point as it reduces the standing wave ratio (SWR) of the antenna system, reducing reflected signal and increasing signal power throughput.

For more info see Wikipedia: Monopole antenna, Coaxial cable

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(B). The feed point impedance of a ground-plane antenna increases when its radials are changed from horizontal to downward sloping.

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This type of antenna consists of a vertical 1/4 wavelength element and four horizontal 1/4 wavelength horizontal radials. These radials make up the ground plane. By sloping the horizontal elements downward, the feed-point impedance increases from about 35 ohms to 50 ohms (at a 45 degree angle).

This is important because matching the feed-point impedance with the 50 ohm impedance of standard coaxial cable will decrease the standing wave ratio (SWR), reducing reflected power and increasing signal strength throughput.

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For more info see Wikipedia: Monopole antenna, Coaxial cable

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(A). The low angle azimuthal radiation pattern of an ideal half-wavelength dipole antenna installed 1/2 wavelength high and parallel to the Earth is a figure-eight at right angles to the antenna.

The maximum radiatiated signal from a 1/2-wavelength dipole at 1/2-wavelength above the ground is at right angles or 90 degrees from the dipole. The pattern drops off to zero on the antenna's axis. So this pattern forms the opposite loops of the figure 8.

For more info see Wikipedia: Dipole antenna

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When a horizontal dipole antenna is close to the ground, the signals reflected from the ground cause more of the signal to be directed back at high vertical angles. This changes the "figure 8" pattern to a more omnidirectional "doughnut".

For more info see Wikipedia: Dipole antenna

Hint: Look for answer with Azimuthal

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(C). The radial wires of a ground-mounted vertical antenna system should be placed on the surface or buried a few inches below the ground.

By placing the radial wires at or just below ground level, you can create an artificial ground screen to act as a more effective ground plane. Depending on your type of installation and the ground conductivity you may need to install 8, 16, 32 (multiples of 2) or more 1/4 wavelength or longer radial wires to form this "ground screen".

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As the antenna is lowered from 1/4 wave above ground, the feed-point impedance of a 1/2 wave dipole antenna steadily decreases.

Placing the antenna at least 1/4 wave above ground is optimal for this type of antenna. In addition to decreasing the feed-point impedance of the antenna, lowering the antenna below 1/4 wavelength above the ground will also greatly alter the radiation pattern of the antenna.

Memory tip: Decreasing the height, decreases the impedance.

For more info see Wikipedia: Dipole antenna

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As the feed-point location is moved from the center toward the ends of a 1/2 wave dipole, the feed point impedance steadily increases.

The center of a 1/2 wave dipole antenna is usually the best place to mount the feed-point. The impedance is lowest at this point at about 72 ohms, which is close to matching the feed line impedance of 75-ohm coaxial cable. By moving the feed point toward the ends of the antenna, the impedance increases steadily and can reach a level of several thousand ohms!

Silly hint: As you move from Center to Ends you move up in the alphabet.

For more info see Wikipedia: Dipole antenna

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Lower ground reflection losses is an advantage of a horizontally polarized as compared to vertically polarized HF antenna.

By polarizing the antenna horizontally, currents are induced along the surface of the ground, which has lower reflection losses. This horizontal polarization also has the advantage in that waves reflected from the ground will recombine with the non-reflected signal wave pattern of the antenna and form a stronger signal.

For more info see Wikipedia: Antenna_ Polarization

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The approximate length for a 1/2-wave dipole antenna cut for 14.250 MHz is 33 feet.

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There are many factors that will affect the amount of length needed for the 1/2 wave dipole antenna, such as the physical characteristics of the wire or nearby conductive sources. But the easiest way to solve this problem is to remember that for a 1/2 wave dipole "4 - 6 - 8, Who do we appreciate?!".

Divide the value of 468 by the frequency in MHz and this will give you the approximate length needed in feet.

\begin{align} \text{Length} &= \frac{468}{14.250\ \text{MHz}}\\ &= 32.842\ \text{feet}\\ &\approx 33\ \text{feet} \end{align}

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For a different method:
I'm not good at keeping extra constants in my head, so I start from the basics: \[\lambda \times f = c\]

Everything's in basic SI units here:

• \(\lambda\) is in meters
• \(f\) (frequency) is in Hertz
• \(c\) (speed of light) is in meters per second

A full wavelength would be:

\[\lambda = \frac{c}{f}\]

But only want half of that to make our half-wave dipole:

\begin{align} \text{Half-wave} &= \frac{1}{2} \times \frac{c}{f}\\ &= \frac{ 3.00 \times 10^8\text{ m/s}}{2 \times 14.250\text{ MHz}}\\ &= \frac{ 3.00 \times 10^8\text{ m/s}}{28.500 \times 10^6\text{ Hz}}\\ &= \frac{ 3.00 \times 10^2\text{ m}}{28.500}\\ &= 10.53\text{ m} \end{align}

At approximately \(3.28\text{ feet/meter}\): \begin{align} 10.53\text{ m} \times 3.28\text{ feet/meter} &= 34.5\text{ feet}\\ &\approx 33\text{ feet} \end{align}

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Alternative explanation without exponents:

\begin{align} \frac{1}{2} \times \frac{300}{14.250} &= 10.5 \text{ meters}\\ &\approx 34.4\text{ feet} \end{align}

\(1 \over 2\) = the half wavelength in the question and \(300 \over \text{frequency}\) is the formula we learned studying for the Technician Class license. Peasy.

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The approximate length for a 1/2-wave dipole antenna cut for 3.550 MHz is 131 feet.

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There are several factors that will in actuality affect the length needed for this antenna. However, the simplest way to get the approximate length needed is to remember " 4 - 6 - 8 Who do we appreciate?!". Dividing the value of 468 by the frequency in MHz will give you the approximate length needed in feet.

For this question:

\[Length = \frac{468}{3.550\text{ MHz}} = 131.8\text{ feet}\]

So choose the answer closest to this value which is: 131 feet. Yikes! I hope you have lots of space for mounting such an antenna!

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Alternate Method #1:

The fewer tricks I have to remember, the better, so I start from the basics:

\[\lambda \times f = c\]

Everything's in base SI units here:

• \(\lambda\) (wavelength) is in meters
• \(f\) (frequency) is in Hertz
• \(c\) (speed of light) is \(3.00 \times 10^8\text{ m/s}\)

Convert our frequency to be in Hertz:

\begin{align} 3.550\text{ MHz} &= 3.550 \times 10^6\text{ Hz} \end{align}

Let's calculate a full wavelength:

\begin{align} \lambda &= \frac{c}{f}\\ &= \frac{3.00 \times 10^8\text{ m/s}}{3.550 \times 10^6\text{ Hz}}\\ &= \frac{3.00 \times 10^2}{3.550}\text{ m}\\ &= 84.5\text{ m} \end{align}

This question is asking for a \(\frac{1}{2}\) wavelength antenna, so:

\[\frac{1}{2} \times 84.5\text{ m} = 42.3\text{ m}\]

Convert to feet:

\begin{align} 42.3\text{ m} \times \frac{3\frac{1}{4}\text{ ft}}{\text{m}} &= 137\text{ ft}\\ &\approx 131\text{ ft} \end{align}

Alternative Method #2:

We can solve this by using the formula for finding the frequency of a signal given 2/3 variables (where one is fixed to always be 300):

\[Frequency (MHz) = \frac{300}{\text{Wavelength (meters)}}\]

You can determine the wavelength in meters using:

\[Wavelength = \frac{300}{\text{Frequency (MHz)}}\]

Hence:

\[\frac{300}{3.550\text{ MHz}} = 84.50\text{ m}\]

Now, we multiply this by 3 (remember, 1 meter is approximately 1 yard or 3 feet) and get \(253.52\text{ ft}\). Now, since the question asks for the length of a 1/2 wave antenna, we divide this number by 2 to get \(126.76\text{ ft}\). The closest number to this is 131ft, so you have your answer.

A good way to memorize the frequency formula is to use the Ohm's law triangle, but replace V (voltage, on top) with 300, I (current) with "wl" for wavelength (meters), and R with "f" for frequency (megahertz).

A simple addition tip for the answer to this specific question: Add the digits in the provided MHz, that is, add 3.550 (3+5+5+0 = 13) which becomes the first two digits of the correct answer to the question, 1 and 3 (13) of 132 ft. The last digit of the correct answer which is "2," is the amount of digits of the sum you got by adding the MHz digits together. Therefore the answer is 132 ft.

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The approximate length of wire to cut for a 1/4 wavelength vertical antenna for 28.5 MHz is 8 feet.

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The easy rule we have learned for the 1/2 wave dipole " 4 - 6 - 8 Who do we appreciate?!" can be modified to suit the 1/4 wave situation.

Simply divide the value of 468 and you get 234. Then divide this new value by the frequency in MHZ to solve for the approximate length in feet. So the rule for the 1/4 wave dipole is " 2 - 3 - 4 Will get your 1/4 wave out the door!"

For this question:

\[Length = \frac{234}{28.5\text{ MHz}} = 8.21\text{ feet}\]

You could also use the "4-6-8" rule and then divide the result by 2:

\[Length = \frac{\frac{468}{28.5\text{ MHz}}}{2} = \frac{16.42}{ 2} = 8.21\text{ feet}\]

Then choose the answer closest to this value which is (A): 8 feet.

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Alternate Method:

The fewer tricks I have to remember, the better, so I start from the basics:

\[\lambda \times f = c\]

Everything's in base SI units here:

• \(\lambda\) (wavelength) is in meters
• \(f\) (frequency) is in Hertz
• \(c\) (speed of light) is \(3.00 \times 10^8\text{ m/s}\)

Convert our frequency to be in Hertz: \begin{align} 28.5\text{ MHz} &= 28.5 \times 10^6\text{ Hz}\\ &= 2.85 \times 10^7\text{ Hz} \end{align}

Let's calculate a full wavelength:

\begin{align} \lambda &= \frac{c}{f}\\ &= \frac{3.00 \times 10^8\text{ m/s}}{2.85 \times 10^7\text{ Hz}}\\ &= \frac{3.00 \times 10}{2.85}\text{ m}\\ &= 10.5\text{ m} \end{align}

This question is just asking for a 1/4 wavelength antenna, so:

\[\frac{1}{4} \times 10.5\text{ m} = 2.63\text{ m}\]

Convert to feet: \begin{align} 2.63\text{ m} \times \frac{3\frac{1}{4}\text{ ft}}{\text{m}} &= 8.55\text{ ft}\\ &\approx 8\text{ ft} \end{align}

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