Standing Waves and VSWR Questions and Answers

Answer: B

The VSWR is 2.

The voltage standing wave ratio is related to the magnitude of the reflection coefficient \(\Gamma\) by

\[ \begin{aligned} \text{VSWR} &= \frac{1 + |\Gamma|}{1 - |\Gamma|} \\ &= \frac{1 + 1/3}{1 - 1/3} = \frac{4/3}{2/3} = 2 \end{aligned} \]

A VSWR of 1 would mean a perfectly matched line with no reflection, which contradicts \(\Gamma = 1/3\). The value 3 is simply \(1/|\Gamma|\), and 2.5 does not follow from the formula.

Answer: A

The reflection coefficient is 1/3.

The magnitude of the reflection coefficient is related to the VSWR S by:

\[ \begin{aligned} |\Gamma| &= \frac{S - 1}{S + 1} \\ &= \frac{2 - 1}{2 + 1} = \frac{1}{3} \end{aligned} \]

A reflection coefficient of 0 would mean a perfect match (VSWR 1). The value 1/2 comes from \((S - 1)/S\), and 1/4 does not follow from any form of the relation.

Answer: D

The standing wave ratio of a short-circuited or open-circuited line is ∞.

The SWR is

\[ S = \frac{1 + |\Gamma|}{1 - |\Gamma|} \]

A short circuit gives \(\Gamma = -1\) and an open circuit \(\Gamma = +1\), so \(|\Gamma| = 1\) in both cases and the denominator is zero. All the incident power is reflected, the voltage minima fall to zero, and the ratio of maximum to minimum is infinite.

An SWR of 1 means a perfectly matched line with no reflection. The SWR is a ratio of magnitudes, so it can never be below 1, which rules out 0 and -1.

Answer: C

The VSWR is infinity.

The reflection coefficient at the load is

\[ \begin{aligned} \Gamma &= \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}} \\ &= \frac{j50 - 50}{j50 + 50} \end{aligned} \]

The numerator and denominator have the same magnitude, \(50\sqrt{2}\), so \(|\Gamma| = 1\). This is true of any purely reactive load: it cannot absorb power, so everything is reflected. Then

\[ \begin{aligned} \text{VSWR} &= \frac{1 + |\Gamma|}{1 - |\Gamma|} \\ &= \frac{2}{0} \to \infty \end{aligned} \]

A VSWR of 1 would need a matched 50 ohm resistive load. VSWR is a real ratio of at least 1, so 0 and +j are not possible values.

Answer: C

The VSWR is 2. First find the reflection coefficient of the load, with \(Z_0 = 50\ \Omega\) and \(Z_L = 40 + j30\ \Omega\):

The options j 0.033 and 0.8 + j 0.6 are complex numbers, but a VSWR is a real ratio, never less than 1, so they cannot be right. A VSWR of 1 would mean a perfectly matched load, which would need \(Z_L = 50\ \Omega\). The answer is 2.

Answer: A

The normalised impedance is 2. For a line with VSWR \(S = 2\) terminated in a resistive load, the normalised load equals S when the load is larger than the characteristic impedance, because the voltage maximum then falls at the load, so \(Z_n = S = 2\); this is the answer in the official key.

The value 1/2 would be the other possibility, for a load smaller than the line impedance, since \(Z_n = 1/S\) in that case, but the key takes the first. The values 3 and 1/3 correspond to a VSWR of 3 and are wrong.

Answer: A

The VSWR is always more than 1, in the sense of 1 or greater, and the key accepts (A).

VSWR = (1 + |Γ|) / (1 - |Γ|), where the reflection coefficient lies between 0 and 1. A larger reflection gives a larger ratio, so the value starts at 1 for a matched line and rises towards infinity.

Less than 1 and 0.4 can never occur. The value 1 occurs only for a perfectly matched line, so it is not true "always".