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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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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(D). The characteristic impedance of flat ribbon TV type twinlead is 300 ohms. This type of cable is made of up two parallel wires which are kept at a precise distance apart by encasing in a flat plastic ribbon. This type of ribbon is commonly used for TV radio antenna applications.

Commonly referred to as ladder line

For more info see Wikipedia: Twin lead

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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.

easy remember point to point in answer

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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

Memory hint: Attenuation increases when frequency increases

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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 to 1

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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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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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A standing wave ratio of 2:1 will result when connecting a 50 ohm feed line to a non-reactive load having 25 ohm impedance.

We can do this calculation using only the two absolute impedance values because we know the load impedance is not reactive.

Non-reactive means the load impedance is entirely resistive; there is no inductive or capacitive reactance component. This simplifies the calculation of the reflection coefficient:

\begin{align} \Gamma &= \frac{ \sqrt{ (50 - 25)^2}}{\sqrt{ (50 + 25)^2}}\\ &= \frac{25}{75}\\ &= \frac{1}{3} \end{align}

Now we can calculate the VSWR: \begin{align} \text{VSWR} &= \frac{ 1 + \Gamma }{ 1 - \Gamma}\\ &= \frac{ 1 + \frac{1}{3} }{ 1 - \frac{1}{3}}\\ &= 2 \end{align}

So the VSWR = 2 : 1

Somewhat conveniently for the student, the system of equations can be solved for the purely resistive situation and simplified to give the SWR as the simple ratio of the larger impedance to the smaller impedance, so in this case:

\begin{align} \text{VSWR} &= \frac{50}{25}\\ &= 2 : 1 \end{align}

No question on the Element 3 (General) exam requires a more complex VSWR calculation with a complex impedance. The simple ratio shown above is all you need to concern yourself with.

Always remember that the SWR cannot be smaller than 1:1. If your result is a ratio smaller than 1 : 1, calculate its reciprocal.

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

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We can do this calculation using only the two absolute impedance values because we know the load impedance is not reactive.

Non-reactive means the load impedance is entirely resistive; there is no inductive or capacitive reactance component. This simplifies the calculation of the reflection coefficient:

\begin{align} \Gamma &= \frac{ \sqrt { (50 - 300)^2 } }{ \sqrt { (50 + 300)^2 } }\\ &= \frac{ 250 }{ 350 }\\ &= \frac{ 5 }{ 7 } \end{align}

Next:

\begin{align} \text{VSWR} &= \frac{ 1 + \Gamma }{ 1 - \Gamma }\\ &= \frac{ 1 + \frac{ 5 }{ 7 } }{ 1 - \frac{ 5 }{ 7 } }\\ &= \frac{ \frac{ 12 }{ 7 } }{ \frac{ 2 }{ 7 } }\\ &= 6\ \ (i.e.\ 6 : 1)\\ \end{align}

Somewhat conveniently for the student, the system of equations can be solved for the purely resistive situation and simplified to give the SWR as the simple ratio of the larger impedance to the smaller impedance, so in this case:

\begin{align} \text{VSWR} &= \frac{ 300\ \Omega }{ 50\ \Omega }\\ &= 6\ \ (i.e.\ 6 : 1)\\ \end{align}

No question on the Element 3 (General) exam requires a more complex VSWR calculation with a complex impedance. The simple ratio shown above is all you need to concern yourself with. Always remember that the SWR cannot be smaller than 1:1. If your result is a ratio smaller than 1:1, calculate its reciprocal.

http://en.wikipedia.org/wiki/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

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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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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