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: 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: 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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