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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The velocity factor (VF) of a conductor is a measure of how much slower an electromagnetic wave travels through a medium compared to its speed in a vacuum. It's expressed as a fraction or percentage of the speed of light. For example, a velocity factor of 0.8 means the wave travels at 80% of the speed of light in that medium. This factor is important in designing antennas and transmission lines to ensure signals are accurately transmitted and received.

Since this is a measure of velocity, not resistance, it is not affected by the length of the wire, impedance, or resistance -- those would affect the characteristics of the wave, but not the velocity of the wave.

The insulating dielectric material affects the creation of both electric and magnetic fields in the material, which affects the velocity of an electromagnetic wave traveling through the material.

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)

Silly hint: if you forget insulation on your water line in the winter, the velocity of the water transmission will be zero

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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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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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The question gives you the electrical length of the line. For any conductor with a velocity factor < 1, the physical length will be shorter.

The velocity factor of solid polyethylene dielectric coax is 66%. ¹

A 14.1MHz wave has a wavelength of 21.3m, a quarter of that is 5.3m, and ⅔ of that is 3.5m.

\begin{align} 300 \over 14.1 & = 21.3 \text{ meters}\\ \\ 21.3 \over 4 & = 5.3 \text{ meters}\\ \\ 5.3 * .66 & = 3.5 \text{ meters} \end{align}

Silly hint: 14.1 MHz -> 14 + 1 = 15 = 3 * 5 <- 3.5 meters

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14.1 is in the 20 meter band. \(1 \over 2\) wavelength is approx. 10.63 meters. The velocity factor of air-insulated, parallel conductor transmission line is approx. 1.0. The physical length is nearly the same as the electrical length.

Mathematically:

\[\begin{align} 300 \over 14.1 & = 21.3 \text{ meters} \\ \\ 21.3 \over 2 & = 10.65 \text{ meters} \end{align}\]

Silly way to remember: 10 is in the question and answer.

• Q: 14.10
• A: 10.6

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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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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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7.2MHz is in the 40 meter band. \(1 \over 4\) of 40 meters is 10 meters. The velocity factor of FOAMED polyethylene dielectric coax is typically 80%.

\begin{align} 300 \over 7.2 & = 41.67 \text{ meters} \\ \\ 41.67 \over 4 & = 10.4 \text{ meters} \\ \\ 80 \% \text{ of } 10.4 & \approx 8.3 \text{ meters} \end{align}

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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.

Hint: We can build a sort of "truth table" to answer all of these questions.

• \(\frac{1}{4} \lambda\) Open = LOW

• \(\frac{1}{4} \lambda\) Closed = HIGH

• \(\frac{1}{2} \lambda\) Open = HIGH

• \(\frac{1}{2} \lambda\) Closed = LOW

To build this table, you just need to remember "QOL" (Quarter, Open, Low) - and you have the starting pair value from which you can build the table.

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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/

Hint: We can build a sort of "truth table" to answer all of these questions.

• \(\frac{1}{4} \lambda\) Open = LOW

• \(\frac{1}{4} \lambda\) Closed = HIGH

• \(\frac{1}{2} \lambda\) Open = HIGH

• \(\frac{1}{2} \lambda\) Closed = LOW

To build this table, you just need to remember "QOL" (Quarter, Open, Low) - and you have the starting pair value from which you can build the table.

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