In a vacuum the velocity of light is a well known constant and in a wire an electrical signal travels at a speed related to the speed of light and the location and makeup of the wire. Rather than quote for a wire the mile per hour or meter per second speed or something similar that can be confused with another representation, instead the speed at which the signal travels as a fraction of the speed of light in a vacuum is shown as a percentage value.

Answer: The velocity of the wave in the transmission line divided by the velocity of light in a vacuum

Hint: Only the correct answer has divided by the velocity of light in a vacuum.

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Velocity factor (VF) equals the inverse of the square root of the dielectric constant \(\kappa\) through which the signal passes.

\(\text{VF}=\frac{1}{\sqrt{\kappa}}\)

Some common VFs:

• 95-99% open wire ladder line
• 82% RG-8X (foamed polyethylene dielectric)
• 66% RG-213 (solid polyethylene dielectric)

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Electrical signals move more slowly in a coaxial cable than in air because of the velocity factor of the coaxial cable which is always significantly smaller than 1 due to the dielectric materials used.

A list of typical velocity factors for different transmission lines: Wikipedia.org/wiki/Velocity_factor

Hint: Electrical in the question and Electrical in the answer.

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Radio waves in space travel at the speed of light, but the speed of the radio wave traveling along a wire depends on what's wrapped around the wire.

Coaxial cable has a insulation wrapped around the center conductor - we call this tube of insulation the "dielectric." The dielectric separates the center conductor from the braid.

If you wrap anything around a wire, the speed of the radio wave traveling along that wire will be slower than the speed of light. If the dielectric is made of solid polyethylene, the speed of the radio wave is only 0.66 times the speed of light.

The factor we apply to the speed of light to get the actual speed of the radio wave is called the "velocity factor."

In this case, the velocity factor is 0.66.

Easy cross-out is 2.7 as nothing can be faster than the speed of light.

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Given parameters and known constants:

• the speed of light \(c = 3 \times 10^8 \text{ m/s}\)
• the velocity factor (VF) of this coax = \(0.66\)
• the frequency \(f = 14.1 \text{ MHz}\)

We're looking for the approximate physical length of a piece of this coax that, at the given \(f\), is electrically \(\frac{1}{4}λ\) long.

Determine the wavelength (\(λ\)) from the frequency: \begin{align} \lambda &= \frac{c}{f}=\frac{c\text{ in Mm/s}}{f\text{ in MHz}}=\frac{300}{14.1}\\ &\approx21.3\text{ m} \end{align}

We're interested in an electrical length of \(\frac{1}{4}λ\): \begin{align} \frac{1}{4}λ &= \frac{1}{4} \times (21.3 \text{ m}) \\ &= 5.3 \text{ m} \end{align}

Electricity moves more slowly in this coax than the speed of light; the slowdown is defined by the velocity factor (VF). \begin{align} \text{physical length}&=\text{electrical length}\times\text{VF}\\ &=5.3\times 0.66\\ &=3.5 \text{ m} \end{align}

Another way:
Estimate the speed of light in a polyethylene cable: \(c_r = (300\times0.66)\approx 200 \text{ m/s}\) ;
One-quarter wavelength long at 14.1 MHz is equivalent to one wavelength at 56.4 MHz (\(4\times14.1 = 56.4\)) (seconds)
length of coax = \(200\div56.4 = 3.5\) meters

To estimate without remembering the velocity factor of polyethylene (PE) simply calculate a quarter wave length of 14.1 MHz in vacuum which is about (300/14.1 MHz)/4 = 5.4 m. EM moves a great deal slower in PE than vacuum so the length should be shorter. => 3.5 meters is the best match.

Of course, you could just remember the 0.66 Velocity Factor and do the actual math!

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To estimate the answer simply remember that the speed of light is about 300 million meters per second. You can therefore divide 300 by the frequency in MHz to get an estimate for the wavelength \(\lambda\), in a vacuum:
\[\lambda=\frac{300\:\text{Mm/s}}{14.10\:\text{MHz}} = 21.277\:\text{m}\]

Then divide by 2 to get one-half wavelength: \[\frac{1}{2}\lambda=\frac{21.277\:\text{m}}{2} = 10.639\:\text{m}\approx10\:\text{meters}\]

The Velocity factor of air-insulated, parallel conductor transmission line is 0.95 through 0.99. (https://en.wikipedia.org/wiki/Velocity_factor) So the physical length should be between 10.639 x 0.95 = 10.107 meters, and 0.639 x 0.99 = 10.533 meters.

The closest answer is 10 meters.

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The insulating material surrounding a transmission line is called the dielectric. Every type of dielectric material has some loss associated with it, and this loss increases with frequency. Vacuum and air dielectrics have almost zero loss, while other materials have higher losses.

Since the electric fields surrounding ladder line are mostly in air there is little dielectric loss, compared to the electric fields in RG-58 coax which are mostly contained within the dielectric material of the coax.

This means that ladder line has lower loss at 50 MHz compared to RG-58 coax using polyethylene dielectric.

Memory Aid: (L)adder (L)ine = (L)ower (L)oss

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The velocity factor (VF), of a transmission medium is the speed at which a wavefront of a signal passes through the medium, relative to the speed of light.

The speed of radio signals in a vacuum, for example, is the speed of light, and so the velocity factor of a radio wave in a vacuum is unity, or 100%.

Source: Wikipedia - Velocity factor

Hint/memory aid: The correct choice has the word velocity in the answer; it is the closest in definition to speed.

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The general relationship between wavelength, speed and frequency is: \begin{align} \text{wavelength }\lambda\text{ [m]}&=\frac{\text{speed of light }c\:\text{[Mm/s]}}{\text{frequency }f\:\text{[MHz]}}\\ \lambda&=\frac{300}{7.2}\approx42\:\text{m} \end{align}

Because it question asks for a quarter wavelength, divide the wavelength by 4, or multiply by a quarter: \[\frac{1}{4}\times 42\text{ m}=10.5\text{ m}\]

Multiplying the wavelength with the velocity factor (VF) of a solid polyethylene dielectric coaxial transmission line will give you the electrical wavelength:

\[(\text{electrical wavelength})=\] \[(\text{wavelength }\lambda)\times (\text{velocity factor VF})\]

\[10.5 \:\text{m}\times0.66=6.93\approx6.9\:\text{m}\]

Quick math: \[\frac{300}{7.2}\cdot \frac{0.66}{4}=6.875 \]

\(300\) is the speed of light constant, \(0.66\) is the velocity factor, so \(6.875\) is the electrical wavelength

Note that the velocity factor on polyethylene coax (\(66\)%) is MUCH lower than in a single bare copper wire (\(95\)%), which you would use for dipole leg lengths. Don't accidentally forget it's coax and choose \(10\) m; that is the wrong answer!

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If the far end of a transmission line is a short circuit it appears as inductance (think of two pieces of wire bridged together at one end)

If the far end of a transmission line is open it appears as a capacitance (think two separate pieces if wire like a capacitor schematic symbol)

Full Explanation:

The input impedance of a lossless short circuited line is:

\[Z_{SC}=jZ_0 \tan (\beta l)\]

• \(j\) is the imaginary unit
• \(Z_0\) is the characteristic impedance of the line
• \(\beta = \frac{2 \pi}{\lambda}\) -- the phase constant of the line
• \(l\) is the physical length of the line.

Thus, depending on whether \({\tan(\beta l)}\) is positive or negative, the stub will be inductive or capacitive, respectively. In the case of a \(1 \over 8\) wavelength transmission line it is positive, thus the answer to the question.

Silly trick: I remember this and the other 1/8 answer by thinking of Open or Shorted (O or S) and the imaginary QSO: “Are you from California? SÍ, Orange County.” (S.I., O.C.)

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Section 20.3.1 of the 2014 edition of the ARRL Handbook simply states that a line less than 1/4 wavelength and open at the far end appears as a capacitance. It also states that if the far end is a short circuit it appears as inductance.

I remember this by picturing the open ended line as two parallel conductors (which is a capacitor).

Silly trick: The way I remember this is by looking to see if it says open, and if it says open, than the answer has "a", if not then it has "an".

Silly trick 2: I remember this by thinking of Open or Shorted (O or S) and the imaginary QSO: “Are you from California? SÍ, Orange County.” (S.I., O.C.)

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Caution: Do not confuse the answer with the answer to E9F13, which has the opposite answer. The only difference between the two items is that in this item, the question asks about the line being open at the far end, whereas the in the next question, it is shorted at the far end.

Lengths of transmission line can be used to transform an impedance. Thinking in terms of the Smith Chart, one complete revolution around the Smith Chart is 1/2-wavelength. That is, 1/2-wavelength (and multiples thereof) of transmission line leave the impedance seen at the generator unchanged. This also means that 1/4-wavelength of transmission line rotates the impedance 180 degrees on the Smith Chart. Since an open circuit is located on the edge of the Smith Chart at one end of the resistance axis, a 180 degree rotation moves it to the opposite end of the resistance axis - the short circuit. So the generator sees a very low impedance when an open is transformed by 1/4-wavelength of transmission line.

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Caution: Do not confuse the answer with the answer to E9F12, which has the opposite answer. The only difference between the two items is that in this item, the question asks about the line being shorted at the far end, whereas the in the next question, it is open at the far end.

A wave traveling down a \(\lambda/ 4\) transmission line lags the generator by 90° as it arrives at the far end of the line and is 180° out of phase as it returns (via the short) to the generator. Since the return signal is low when the source is high, the shorted transmission line appears to the generator like an open circuit or high impedance.

https://www.allaboutcircuits.com/textbook/alternating-current/chpt-14/impedance-transformation/

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Think about these questions in terms of the amplitude profile of a standing wave in the transmission line.

At a half wavelength, the end of the transmission line matches the amplitude of the input to the transmission line when a standing wave exists, as such shorting it will take energy out of the standing wave and be seen as a short at the generator. - which is seen as a low impedance to current flow when a voltage is applied.

Contrast this to a 1/4 wave long transmission line, where a peak amplitude of a standing wave at the generator will correspond to a node at the end of the transmission line. If we don't terminate the endpoint in this case the reflected wave will be 180deg out of phase and at all points the generator will have to work against itself to cancel its own reflections, which is seen as a low impedance to current flow when a voltage is applied. If we terminate the 1/4 wave, the node is held as a node and the reflected waves are dissipated, which will make a resonant element and a high impedance to current flow.

Hint: 1/2 (height) that is shorted is very low.

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When a line is open, all of the current reflects back to the source.

If the transmission line is 1/2 wavelength long, it takes an entire wavelength of time to reflect. This means the current arrives back delayed one cycle and in phase.

Because the current arrives back in phase, it looks like to the source no current is being accepted by the line. This is an infinite impedance. Of course, a real transmission line has losses and not all of the current will arrive back, so instead it is a very high value but not actually infinite.

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Air is added to a solid dielectric to make foam. Air has the lowest loss with the highest propagation velocity next to a vacuum. Therefore you will have lower loss and a higher velocity factor. But the solid dielectric provides more insulation so foam has a lower safe operating voltage limit, meaning it can handle less power.

Hint: "All other parameters are the same" = "All of these answers are correct" -kd2ocb

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