Login or Register for FREE!
Subelement G9

Antennas and Feed Lines

Section G9B

Basic antennas

What is one disadvantage of a directly fed random-wire HF antenna?

  • It must be longer than 1 wavelength
  • Correct Answer
    You may experience RF burns when touching metal objects in your station
  • It produces only vertically polarized radiation
  • It is more effective on the lower HF bands than on the higher bands

(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

Last edited by N8GCU. Register to edit

Tags: arrl chapter 7 arrl module 31

Which of the following is a common way to adjust the feed-point impedance of a quarter wave ground-plane vertical antenna to be approximately 50 ohms?

  • Slope the radials upward
  • Correct Answer
    Slope the radials downward
  • Lengthen the radials
  • Shorten the radials

With a minimal impact to antenna performance, angling the radials downward instead of placing them horizontally can increase the feed point impedance. W5ALT states that sloping downward at approximately 45 degrees below horizontal is sufficient to obtain feed point impedance of about 50 ohms:

Silly Hint: Ground -> Down

https://web.archive.org/web/20190228062257/http://www.comportco.com:80/~w5alt/antennas/notes/ant-notes.php?pg=21

Last edited by kd7bbc. Register to edit

Tags: arrl chapter 7 arrl module 28

Which of the following best describes the radiation pattern of a quarter-wave, ground-plane vertical antenna?

  • Bi-directional in azimuth
  • Isotropic
  • Hemispherical
  • Correct Answer
    Omnidirectional in azimuth

A quarter-wave ground plane is sort of like a vertical dipole, it radiates horizontally in all directions. So, that leaves out bi-directional.

Isotropic is a theoretical antenna, a point source that radiates in all directions. A quarter-wave ground-plane radiates very little "up" and "down" - so it's certainly not an isotropic radiator.

Hemisperical also theoretical and would be like isotropic/spherical, but with no transmission on one side of a bisecting plane. If the bisection were horizontal, it would have equal horizontal and vertical transmission above (or below) that midline. A quarter-wave ground-plane is again disqualified because it radiates very little "up" (or "down").

Last edited by serif. Register to edit

Tags: arrl chapter 7 arrl module 28

What is the radiation pattern of a dipole antenna in free space in a plane containing the conductor?

  • Correct Answer
    It is a figure-eight at right angles to the antenna
  • It is a figure-eight off both ends of the antenna
  • It is a circle (equal radiation in all directions)
  • It has a pair of lobes on one side of the antenna and a single lobe on the other side

A standard dipole antenna radiates in a doughnut pattern in 3-D, but when examined in a polar diagram will resemble a figure-8 at a right angle to the antenna axis.

"The radiation pattern of a half-wave dipole antenna [is] that the direction of maximum sensitivity or radiation is at right angles to the axis of the RF antenna."

Dipole Radiation Pattern & Polar Diagram

Silly Hint: The correct answer is the only option containing the word "right".

Last edited by greg.johnston. Register to edit

Tags: arrl chapter 7 arrl module 28

How does antenna height affect the horizontal (azimuthal) radiation pattern of a horizontal dipole HF antenna?

  • If the antenna is too high, the pattern becomes unpredictable
  • Antenna height has no effect on the pattern
  • Correct Answer
    If the antenna is less than 1/2 wavelength high, the azimuthal pattern is almost omnidirectional
  • If the antenna is less than 1/2 wavelength high, radiation off the ends of the wire is eliminated

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

Last edited by dogshed. Register to edit

Tags: arrl chapter 7 arrl module 28

Where should the radial wires of a ground-mounted vertical antenna system be placed?

  • As high as possible above the ground
  • Parallel to the antenna element
  • Correct Answer
    On the surface of the Earth or buried a few inches below the ground
  • At the center of the antenna

(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".

Last edited by N8GCU. Register to edit

Tags: arrl chapter 7 arrl module 28

How does the feed-point impedance of a 1/2 wave dipole antenna change as the antenna is lowered below 1/4 wave above ground?

  • It steadily increases
  • Correct Answer
    It steadily decreases
  • It peaks at about 1/8 wavelength above ground
  • It is unaffected by the height above ground

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

Last edited by ironcal67. Register to edit

Tags: arrl chapter 7 arrl module 28

How does the feed point impedance of a 1/2 wave dipole change as the feed point is moved from the center toward the ends?

  • Correct Answer
    It steadily increases
  • It steadily decreases
  • It peaks at about 1/8 wavelength from the end
  • It is unaffected by the location of the feed point

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

Last edited by zaslager1997. Register to edit

Tags: arrl chapter 7 arrl module 28

Which of the following is an advantage of a horizontally polarized as compared to a vertically polarized HF antenna?

  • Correct Answer
    Lower ground reflection losses
  • Lower feed-point impedance
  • Shorter radials
  • Lower radiation resistance

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

Last edited by kd7bbc. Register to edit

Tags: arrl chapter 7 arrl module 28

What is the approximate length for a 1/2 wave dipole antenna cut for 14.250 MHz?

  • 8 feet
  • 16 feet
  • 24 feet
  • Correct Answer
    33 feet

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


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}


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}


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.

Last edited by kd7bbc. Register to edit

Tags: arrl chapter 7 arrl module 28

What is the approximate length for a 1/2 wave dipole antenna cut for 3.550 MHz?

  • 42 feet
  • 84 feet
  • Correct Answer
    132 feet
  • 263 feet

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


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!


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.

Last edited by rwil46. Register to edit

Tags: arrl chapter 7 arrl module 28

What is the approximate length for a 1/4 wave vertical antenna cut for 28.5 MHz?

  • Correct Answer
    8 feet
  • 11 feet
  • 16 feet
  • 21 feet

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


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.


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}

Last edited by jeff00seattle. Register to edit

Tags: arrl chapter 7 arrl module 28

Go to G9A Go to G9C