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: C — correct impedance match between transmitter and antenna

Efficient power transfer between a transmitter and its antenna requires that the impedance of the load (antenna system) matches the output impedance of the transmitter. When impedances are matched, maximum power flows from the source to the load with minimum reflection. If there is a mismatch, some power is reflected back toward the transmitter rather than being radiated, reducing efficiency and potentially stressing the transmitter's output stage.

The principle of maximum power transfer states that maximum power is delivered when the source impedance equals the load impedance. In amateur radio practice, transmitters and feed lines are commonly designed around a standard impedance of 50 Ω, and antennas or antenna tuners are adjusted to present this same impedance to the transmitter.

• A. High load impedance — A high load impedance relative to the source causes a severe mismatch and significant power reflection; it is not a general requirement for efficient transfer.
• B. Low load impedance — Similarly, a low load impedance causes a mismatch and reflected power; neither extreme on its own is correct.
• D. High standing wave ratio — A high SWR is a direct indicator of impedance mismatch and increased reflected power, which is the opposite of what is wanted for efficient transfer.

Therefore, efficient power transfer from a transmitter to an antenna requires a correct impedance match between the two.

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Correct answer: C — braid and insulation around a central conductor

Coaxial cable ("coax") consists of a central inner conductor surrounded by a dielectric (insulating) material, which is then wrapped by a tubular outer conductor (the braid), and finally covered by a protective outer jacket. The braid serves as both the return conductor and an electromagnetic shield, keeping RF energy contained within the cable and preventing external interference from entering.

• A — a single conductor: A single conductor alone cannot form a transmission line; it has no return path and provides no shielding.
• B — two parallel conductors separated by spacers: This describes open-wire (ladder line or twin-lead) feedline, not coaxial cable.
• D — braid and insulation twisted together: Twisting the braid and insulation together does not describe any standard feedline construction; the braid must surround the central conductor concentrically to function correctly.

Therefore, coaxial feedline is defined by its concentric construction — a central conductor, a dielectric spacer, and an outer braided shield — which gives it its characteristic impedance and shielding properties.

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Correct answer: D — transfer maximum power to the antenna

Matching the transmission line at the transmitter end ensures that maximum power is transferred from the transmitter into the feedline and ultimately to the antenna. This is the principle of impedance matching: when the source impedance equals the load impedance (conjugate match), reflected power is minimised and forward power is maximised. A mismatch causes standing waves (high SWR) on the line, meaning some power is reflected back toward the transmitter rather than being radiated.

• A — prevent frequency drift: Frequency stability is a function of the oscillator and transmitter design, not the feedline match. A mismatched line does not cause frequency drift in a well-designed transmitter (though it can stress some older transmitter designs).
• B — overcome fading of the transmitted signal: Fading (QSB) is a propagation phenomenon caused by multipath interference and ionospheric variation. Matching the feedline has no effect on signal fading at the receiver.
• C — ensure the radiated signal has the intended polarisation: Polarisation is determined by the physical orientation of the antenna elements, not by matching at the transmitter end.

Therefore, the primary reason to match the transmission line at the transmitter end is to transfer maximum power to the antenna by minimising reflected power and standing waves on the feedline.

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Correct answer: C — excessive heat being produced in the transmitter output stage

A transmitter's output stage is designed to work into a specific load impedance (typically 50 Ω). When the antenna or feedline is damaged, the impedance presented to the final amplifier deviates from this design value, causing a high SWR (Standing Wave Ratio). The output transistors or valves can no longer transfer power efficiently into the load; instead, much of the energy is dissipated as heat within the output stage components. In severe cases this can destroy the final amplifier transistors or valves.

• A — the driver stage not delivering power to the final: The driver stage is largely unaffected by a mismatched load at the antenna; it sees the input impedance of the final stage, not the antenna.
• B — the output tuned circuit breaking down: While high SWR can stress components, the primary and most immediate effect is thermal dissipation in the active devices, not breakdown of the tuned circuit itself.
• D — loss of modulation in the transmitted signal: Modulation is generated earlier in the transmitter chain and is not directly disrupted by an impedance mismatch at the output.

Therefore, a damaged antenna or feedline creates an impedance mismatch that causes the transmitter's output stage to dissipate excessive heat rather than radiate the power as intended.

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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: C — reduced transfer of RF energy to the antenna

Standing waves occur on a transmission line when the load impedance (the antenna) does not match the characteristic impedance of the feedline. Energy travelling toward the antenna is partially reflected back toward the transmitter, creating a pattern of voltage and current maxima and minima along the line — the standing wave pattern. This reflected energy does not reach the antenna and instead bounces back and forth, increasing resistive losses in the line and potentially stressing the transmitter's output stage. The Standing Wave Ratio (SWR) is the standard measure of this mismatch; a high SWR means significant reflection and reduced efficiency.

• A is incorrect — maximum energy transfer only occurs when the line is matched (SWR = 1:1); standing waves indicate a mismatch, not maximum transfer.
• B is incorrect — standing waves are a direct consequence of imperfect impedance matching, not perfect matching.
• D is incorrect — a well-constructed, balanced transmission line is designed not to radiate regardless of SWR; radiation from the feedline is a separate (and undesirable) phenomenon usually caused by common-mode currents, not standing waves per se.

Therefore, standing waves on a non-resonant transmission line indicate an impedance mismatch that results in reflected energy and reduced transfer of RF power to the antenna.

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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 — disable the unit not being used

When a transmit/receive (T/R) switching system connects a single antenna alternately to a transmitter and a receiver, the unit that is not in use must also be disabled (or isolated). This is because the transmitter, even when nominally "switched off" the antenna, can still generate RF that reaches the receiver through stray coupling — damaging the sensitive front-end components. Equally, a live transmitter path connected even briefly to a receiver input can destroy low-noise amplifiers or mixer stages. Disabling the idle unit ensures both protection and clean signal switching.

• B — disconnect the antenna tuner: The antenna tuner can remain in circuit for both transmit and receive; disconnecting it is unnecessary and would upset the impedance match.
• C — ground the antenna on receive: Grounding the antenna during receive would short out the incoming signal entirely, making reception impossible.
• D — switch between power supplies: The transmitter and receiver typically share a common power supply or have independent supplies; switching supplies is not a function of a T/R antenna-switching system.

Therefore, a proper antenna-sharing T/R switch must also disable whichever unit is not currently connected to the antenna, protecting the receiver from transmitter RF and ensuring clean operation.

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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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Correct answer: D — coaxial cable

When a feedline must run close to grounded metal objects (metal pipes, guttering, building frames, etc.), coaxial cable is the correct choice. Coaxial cable has a self-shielding construction: the outer conductor (braid or foil) surrounds the inner conductor completely, confining the RF field entirely within the cable. This means nearby metal objects have virtually no effect on the impedance or the signal being carried.

Open-wire and twinlead feedlines are balanced transmission lines whose fields extend outside the conductors. If these external fields interact with grounded metal, the line's characteristic impedance is disturbed, standing waves increase, and losses rise significantly.

• A — 75 ohm twinlead: A parallel-conductor balanced line; its external field is disrupted by nearby metal, causing impedance changes and increased losses.
• B — 300 ohm twinlead: The most common TV-ribbon type; also a balanced line with an external field, making it highly susceptible to detuning by adjacent metalwork.
• C — 600 ohm open-wire: An open balanced line with the largest conductor spacing and the greatest external field exposure; the worst choice near grounded metal objects.

Therefore, coaxial cable is the only feedline type suitable for routing near grounded metal objects, because its shielded construction isolates the signal from external conductors.

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