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:
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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(B). 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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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
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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.
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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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(D). 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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(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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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 to1
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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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(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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(B). 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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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
KC3EWK
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The higher the transmission line loss, the more the SWR will read artificially low.
SWR is the product of an RF wave moving forward on a transmission line interacting with reflected power moving in the opposite direction. The interaction of these two waves is what forms the standing wave. For any fixed amount of input power an increase in reflected power will correlate with an increase in SWR.
Assume we have two transmission lines that have a terminating load that causes total reflection (Infinite SWR) at the far end. If the transmission line is lossless, the amount of forward and reflected power measured at any point along the line will be equal, and the apparent SWR measured at the input will be the actual SWR of the load.
Lossless transmission line
50W Fwd ------0 dB loss------> 50W
50W Rev <-----0 dB loss------- 100% Reflection of 50W
Apparent SWR measured at input = ∞:1
However, lines are lossy in the real world and power will be dissipated linearly along its length. In a lossy line we can measure forward and reflected power at the input. The forward power will not have been attenuated by traveling through the transmission line at all, but the reflected power will have been attenuated as it traveled the length of the transmission line twice.
If the attenuation is sufficient enough there would be no measurable reflected power whatsoever, despite that the transmission line in our thought experiment has total reflection (infinite SWR)
Lossy transmission line
50W Fwd ------10 dB loss------> 5W
0.5W Rev <-----10 dB loss------- 100% Reflection of 5W
Apparent SWR measured at input = 1.2:1
Although the latter case with 10 dB loss is somewhat exaggerated, it is typical for an amateur station to measure SWR at the input to a coax feedline which does have losses, and the operator should be aware of the truth of this measurement.
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