Correct answer: terminating the line in its characteristic impedance

A transmission line appears infinitely long when there are no reflections from its far end. This condition occurs when the line is terminated in its characteristic impedance \(Z_0\).

When the load impedance equals \(Z_0\):

• all the incident power is absorbed by the load
• no reflected wave is generated
• the source cannot distinguish the line from an infinitely long one

Mathematically, the reflection coefficient becomes zero:

\[ \Gamma = \frac{Z_L - Z_0}{Z_L + Z_0} = 0 \quad \text{when } Z_L = Z_0 \]

• shorting the line at the end causes total reflection.
• leaving the line open at the end also causes total reflection.
• increasing the standing wave ratio above unity indicates reflections, the opposite of an infinite line condition.

Therefore, any length of transmission line can be made to appear infinitely long by terminating it in its characteristic impedance.

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Correct answer: physical dimensions and relative positions of the conductors

The characteristic impedance \(Z_0\) of a transmission line is determined by its distributed inductance \(L\) and capacitance \(C\) per unit length:

\[ Z_0 = \sqrt{\frac{L}{C}} \]

These values depend on the physical size, spacing, and arrangement of the conductors, as well as the dielectric material between them.

• The length of the line does not affect its characteristic impedance.
• The load connected to the line affects reflections and standing waves, not \(Z_0\).
• While losses may vary with frequency, the characteristic impedance is set by the line’s geometry and dielectric properties.

Therefore, the characteristic impedance is determined by the physical dimensions and relative positions of the conductors.

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Correct answer: 52 ohm

The characteristic impedance of a transmission line is determined by its physical construction:

• conductor spacing
• conductor size
• dielectric material

It does not depend on the length of the line.

Therefore, cutting the line from 20 m to 10 m does not change its characteristic impedance.

• 13, 26, and 39 ohm would imply dependence on length, which is incorrect.

Therefore, the impedance remains 52 ohm.

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Correct answer: 50 ohm coaxial cable

A quarter-wave ground plane antenna typically has a feedpoint impedance close to:

\[ \approx 50\ \Omega \]

Therefore, a 50 ohm coaxial cable provides a good impedance match, allowing efficient power transfer.

• 300 ohm balanced line is a poor match.
• 75 ohm line is closer but still mismatched.
• 300 ohm coaxial cable is not a standard feeder.

Therefore, the best match is 50 ohm coaxial cable.

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Correct answer: 50 ohm

Most modern transmitters are designed with an output impedance of approximately:

\[ 50\ \Omega \]

This standard is widely used for:

• coaxial feedlines
• antennas
• RF equipment

It provides a good compromise between:

• power handling

• signal loss

• 75 \(\Omega\) is common in TV systems.

• 25 \(\Omega\) and 100 \(\Omega\) are not typical standards.

Therefore, the nominal output impedance is 50 ohm.

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Correct answer: reduced antenna radiation

When there is an impedance mismatch between the transmitter (power amplifier) and the antenna:

• some of the RF power is reflected back toward the transmitter
• less power is delivered to the antenna

This results in:

• reduced radiation efficiency

• higher SWR on the feedline

• Key clicks are related to keying, not impedance.

• Modulation percentage is unrelated.

• DC current does not necessarily decrease.

Therefore, the result is reduced antenna radiation.

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Correct answer: less RF power being radiated

Losses in a transmission line (due to resistance, dielectric loss, etc.) cause some of the RF energy to be dissipated as heat before it reaches the antenna.

This results in:

• reduced power delivered to the antenna

• therefore reduced radiated signal strength

• Losses do not create a 1:1 SWR.

• Reflections are caused by impedance mismatch, not loss itself.

• Energy transfer is worsened, not improved.

Therefore, losses result in less RF power being radiated.

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Correct answer: standing waves are produced in the feedline

If the characteristic impedance of the feedline does not match the antenna input impedance, part of the transmitted power is reflected back toward the transmitter.

This reflection interferes with the forward wave, producing:

• voltage and current variations along the line
• standing waves

These are indicated by an SWR greater than:

\[ 1:1 \]

• Heat may result from losses, but this is not the primary effect.
• SWR increases, not drops, from 1:1.
• The antenna will still radiate some signal.

Therefore, standing waves are produced in the feedline.

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Correct answer: infinite

A transmission line of length \(\lambda/4\) has the property of impedance inversion.

The input impedance of a transmission line is:

\[ Z_{\text{in}} = \frac{Z_0^2}{Z_L} \]

where:

• \(Z_0\) is the characteristic impedance of the line
• \(Z_L\) is the load impedance at the far end

If the far end is short-circuited:

\[ Z_L = 0 \]

Substituting:

\[ Z_{\text{in}} = \frac{Z_0^2}{0} \rightarrow \infty \]

So a short circuit at the end of a quarter-wave line appears as an open circuit at the input.

• Zero impedance occurs at the shorted end itself.
• 5 \(\Omega\) and 150 \(\Omega\) are not valid results of impedance inversion.

Therefore, the impedance seen at the input is infinite.

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Correct answer: a standing wave ratio meter

A standing wave ratio (SWR) meter measures the ratio of forward power to reflected power in a transmission line.

If power is not being transferred efficiently to the antenna, some of it is reflected back toward the transmitter.

SWR is defined as:

\[ \text{SWR} = \frac{V_{\text{forward}} + V_{\text{reflected}}}{V_{\text{forward}} - V_{\text{reflected}}} \]

A high SWR indicates poor impedance matching and inefficient power transfer to the antenna.

• An antenna tuner matches impedance but does not measure power transfer.
• A dummy load replaces the antenna for testing.
• A keying monitor checks transmitter keying.

Therefore, the correct instrument is a standing wave ratio meter.

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Correct answer: open-wire feeder

Transmission line loss is mainly due to conductor resistance and dielectric losses in the insulating material between conductors.

Open-wire feeder has:

• Wide conductor spacing, reducing capacitance.
• Minimal dielectric material between the conductors (mostly air), which has very low loss.

Since air is a much better dielectric than plastic insulation, open-wire feeder typically has lower loss than coaxial cable, especially at higher frequencies.

• Twisted flex has closely spaced conductors and lossy insulation, giving higher loss.
• Coaxial cable contains dielectric material which introduces additional loss.
• Mains cable is not designed for RF transmission and has high loss.

Therefore, open-wire feeder exhibits the lowest transmission line loss.

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A radio wave in free space travels with the speed of light. When a wave travels on a transmission line, it travels slower, travelling through a dielectric/insulation. The speed at which it travels on a line compared to the free-space velocity is known as the "velocity factor". Typical figures are: Twin line 0.82, Coaxial cable 0.66, (free space 1.0). So a wave in a coaxial cable travels at about 66% of the speed of light (as an example). In practice this means that if you have to cut a length of coaxial transmission line to be a half-wavelength long (for, say, some antenna application), the length of line you cut off will have to be 0.66 of the free-space length that you calculated.

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Correct answer: coaxial cable

Coaxial cable is designed with:

• an inner conductor
• an insulating dielectric
• an outer shield

The outer shield protects the signal from external influences such as:

• moisture
• soil contact
• nearby objects

This makes it suitable for burial (especially when rated for direct burial).

• Twinlead and open-wire lines are exposed and easily affected by moisture and nearby materials.
• Their impedance would change if buried.

Therefore, the suitable feedline is coaxial cable.

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