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Subelement G9
Antennas and Feed Lines
Section G9A
Antenna feed lines: characteristic impedance and attenuation; SWR calculation, measurement, and effects; matching networks
Which of the following factors determine the characteristic impedance of a parallel conductor antenna feed line?
  • Correct Answer
    The distance between the centers of the conductors and the radius of the conductors
  • The distance between the centers of the conductors and the length of the line
  • The radius of the conductors and the frequency of the signal
  • The frequency of the signal and the length of the line

The characteristic impedance of a parallel conductor antenna feed line is determined by the distance between the centers of the conductors and the radius of the conductors.


Neither of the following have anything to do with the characteristic impedance of a parallel conductor antenna feed line:

  • The length of the line
  • The frequency of the signal

By eliminating all the answers containing these two things, the only answer remaining is the correct one.

For more info see Wikipedia: Characteristic impedance, Feed line

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What are the typical characteristic impedances of coaxial cables used for antenna feed lines at amateur stations?
  • 25 and 30 ohms
  • Correct Answer
    50 and 75 ohms
  • 80 and 100 ohms
  • 500 and 750 ohms

50 and 75 ohms are the typical characteristic impedances of coaxial cable used for antenna feed lines at amateur stations. The standardization of impedances for coaxial cables makes them very useful for antenna applications where impedances must be matched for optimum signal power output.

For more info see Wikipedia: Coaxial cables, Feed lines

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What is the typical characteristic impedance of "window line" parallel transmission line?
  • 50 ohms
  • 75 ohms
  • 100 ohms
  • Correct Answer
    450 ohms

Window line (also known as ladder line) generally has a 450 ohm impedance, however there are 300 ohm (twin lead, from older analog TV installations) and 600 ohm (open wire ladder line) variants. https://en.wikipedia.org/wiki/Twin-lead#Ladder_line

(A ladder is how you get to the highest place (pick the highest number))

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What might cause reflected power at the point where a feed line connects to an antenna?
  • Operating an antenna at its resonant frequency
  • Using more transmitter power than the antenna can handle
  • Correct Answer
    A difference between feed-line impedance and antenna feed-point impedance
  • Feeding the antenna with unbalanced feed line

The reason for the occurrence of reflected power at the point where a feed line connects to an antenna is because of a difference between feed-line impedance and antenna feed-point impedance.


Impedances of the feed-line and antenna feed-point should be matched to prevent power reflection as a standing wave (the greater the difference in impedance, the greater the reflected energy). Matching impedances optimizes the system for more complete signal power.

easy remember point to point in answer

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How does the attenuation of coaxial cable change as the frequency of the signal it is carrying increases?
  • Attenuation is independent of frequency
  • Correct Answer
    Attenuation increases
  • Attenuation decreases
  • Attenuation reaches a maximum at approximately 18 MHz

Attenuation is another word for loss, where some of the energy is converted to heat. All cables have some amount of loss, generally measured in loss in dB per unit length (e.g. feet or meters).

The higher the frequency, the higher the attenuation and loss.


Think about how heat works: it's molecules that are jiggling. The faster they jiggle, the higher the temperature. The higher the frequency, the more often electrons are bumping into molecules and setting them in motion.

A useful analogy is to think about people walking in a corridor. When there are few people they're much less likely to bump into each other. The more crowded it is, the more likely people are to bump into each other.

For more info see Wikipedia: Coaxial cable

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In what units is RF feed line loss usually expressed?
  • Ohms per 1000 feet
  • Decibels per 1000 feet
  • Ohms per 100 feet
  • Correct Answer
    Decibels per 100 feet

The values for RF feed line losses are usually expressed in dB per 100 ft.

The amount of signal loss through a substance is referred to as its attenuation. The attenuation of various feed lines, such as coaxial cables standardized in units of dB per 100 ft (or metric dB/meter). It is important to note that attenuation is also frequency dependant, and so the dB per 100 feet will often be expressed along with a standard frequency, such as for coax at 750 MHz.

For more info see Wikipedia: Coaxial cable, Attenuation

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What must be done to prevent standing waves on an antenna feed line?
  • The antenna feed point must be at DC ground potential
  • The feed line must be cut to a length equal to an odd number of electrical quarter wavelengths
  • The feed line must be cut to a length equal to an even number of physical half wavelengths
  • Correct Answer
    The antenna feed point impedance must be matched to the characteristic impedance of the feed line

(D). The antenna feed-point impedance must be matched to the characteristic impedance of the feed line to prevent standing waves on an antenna feed line.

When the impedances are not matched, the standing waves may be reflected back which will raise the feed line standing wave ratio (SWR). These reflections should be eliminated by matching impedances to maximize power output and reduce the SWR.

For more info see Wikipedia: Impedance matching, SWR (standing wave ratio)

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If the SWR on an antenna feed line is 5 to 1, and a matching network at the transmitter end of the feed line is adjusted to 1 to 1 SWR, what is the resulting SWR on the feed line?
  • 1 to 1
  • Correct Answer
    5 to 1
  • Between 1 to 1 and 5 to 1 depending on the characteristic impedance of the line
  • Between 1 to 1 and 5 to 1 depending on the reflected power at the transmitter

It won't help to adjust the transmitter end of the feed line, you need to adjust the antenna end of the line.

Sticking a matching network adjusted to 1:1 at the transmitter end of things will leave you with an unchanged standing wave ratio of 5:1.

For more info see Wikipedia: standing wave ratio (SWR)

Hint: The answer is in the question: the Standing Wave Ratio (SWR) of the feed line is 5 to 1

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What standing wave ratio will result when connecting a 50 ohm feed line to a non-reactive load having 200 ohm impedance?
  • Correct Answer
    4:1
  • 1:4
  • 2:1
  • 1:2

Remember, SWR is always 1:1 or greater. Thus, you can eliminate the distractors 1:2 and 1:4, which are both less than 1:1.

The standing wave ratio that will result from the connection of a 50-ohm feed line to a non-reactive load having a 200-ohm impedance is 4:1.

To calculate the standing wave ratio (SWR) in a case where the load is non-reactive, simply divide the greater impedance by the lesser impedance, thereby giving a value greater than one.

For this problem: \begin{align} \text{SWR} &= \frac{ 200\ \Omega }{ \ \ 50\ \Omega }\\ &= 4 \end{align}

We describe this as a "4:1 SWR".

For more info see Wikipedia: Standing wave ratio (SWR)

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What standing wave ratio will result when connecting a 50 ohm feed line to a non-reactive load having 10 ohm impedance?
  • 2:1
  • 50:1
  • 1:5
  • Correct Answer
    5:1

(D). The standing wave ratio that will result from the connection of a 50-ohm feed line to a non-reactive load having a 10-ohm impedance is 5:1.

In cases where the load is non-reactive, the SWR may be calculated by simply dividing the greater impedance value divided by the lesser impedance value (whichever fraction will give a result greater than 1).

In this problem:

\begin{align} \text{SWR} &= \frac{ 50\ \Omega }{ 10\ \Omega}\\ &= 5 \end{align}

...which we can express as a 5:1 SWR.

For more info see Wikipedia: Standing wave ratio (SWR)

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What standing wave ratio will result when connecting a 50 ohm feed line to a non-reactive load having 50 ohm impedance?
  • 2:1
  • Correct Answer
    1:1
  • 50:50
  • 0:0

The standing wave ratio that will result from the connection of a 50-ohm feed line to a non-reactive load having a 50-ohm impedance is 1:1.

Where the load is non-reactive, you can calculate the SWR by simply dividing the greater impedance value by the lesser value (giving a value or fraction greater than 1).

For this problem:

SWR = 50 Ω / 50 Ω = 1 or expressed as a 1:1 SWR

For more info see Wikipedia: Standing wave ratio

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What is the interaction between high standing wave ratio (SWR) and transmission line loss?
  • There is no interaction between transmission line loss and SWR
  • Correct Answer
    If a transmission line is lossy, high SWR will increase the loss
  • High SWR makes it difficult to measure transmission line loss
  • High SWR reduces the relative effect of transmission line loss

If transmission line is lossy, high SWR will increase the loss.

Simply put, an SWR reading measures match between transmitter, feedline, and antenna. If it is high ( an inefficient impedance match ) power is attenuated by reflecting it back, and it is lost in the feedline and amp finals.

https://en.wikipedia.org/wiki/Standing_wave_ratio

https://en.wikipedia.org/wiki/Feed_line

Note that lossy is a key word in the answer. As in "line loss is the lossy-est loss to lose"

KC3EWK

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What is the effect of transmission line loss on SWR measured at the input to the line?
  • Correct Answer
    The higher the transmission line loss, the more the SWR will read artificially low
  • The higher the transmission line loss, the more the SWR will read artificially high
  • The higher the transmission line loss, the more accurate the SWR measurement will be
  • Transmission line loss does not affect the SWR measurement

The higher the transmission line loss, the more the SWR will read artificially low.

SWR is defined as the ratio of forward power to reflected power.

Higher transmission line losses mean a larger portion of the reflected power will be absorbed, thus leading to an artificially lower SWR reading.

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