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
Last edited by qubit. Register to edit
Tags: arrl chapter 7 arrl module 33
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
Last edited by ironcal67. Register to edit
Tags: arrl chapter 7 arrl module 33
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))
Last edited by ko4ssd. Register to edit
Tags: arrl chapter 7 arrl module 33
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
Last edited by koamp. Register to edit
Tags: arrl chapter 7 arrl module 33
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
Last edited by qubit. Register to edit
Tags: arrl chapter 7 arrl module 33
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
Last edited by gconklin. Register to edit
Tags: arrl chapter 7 arrl module 33
(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)
Last edited by N8GCU. Register to edit
Tags: arrl chapter 7 arrl module 33
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
Last edited by breathlessblizzard. Register to edit
Tags: arrl chapter 7 arrl module 33
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)
Last edited by mk2019. Register to edit
Tags: arrl chapter 7 arrl module 33
(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)
Last edited by qubit. Register to edit
Tags: arrl chapter 7 arrl module 33
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
Last edited by kf0jfq. Register to edit
Tags: arrl chapter 7 arrl module 33
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
Last edited by kermitjr. Register to edit
Tags: arrl chapter 7 arrl module 33
Higher transmission line losses mean a larger portion of the reflected power will be absorbed, thus leading to an artificially lower SWR reading.
Last edited by jahman. Register to edit
Tags: arrl chapter 7 arrl module 33