Your Answer : (Not Answered)

Correct Answer : A

Statements I and II are correct. For a loss-less line with air as the dielectric the phase velocity is the speed of light, so the wavelength at 15 MHz is

The 12 m length is therefore 12/20 = 0.6 wavelengths, so statement I is correct, and statement IV, which says the length is a quarter wavelength, is wrong. The phase constant is

so statement II is correct and statement III, which gives 0.2 pi, is wrong. Check: \(\beta l = 0.1\pi \times 12 = 1.2\pi\) rad, which is 0.6 of a full cycle of 2 pi. The answer is options I and II.

Your Answer : (Not Answered)

Correct Answer : A

Statements I and III are correct, so option (A) is right. In a rectangular waveguide the guided wavelength is \(\lambda_{g} = \lambda_{0}/\sqrt{1-(f_{c}/f)^{2}}\), which is always greater than the free space wavelength, so statement I is true. The phase velocity times the group velocity equals the square of the velocity of light, \(v_{p}v_{g} = c^{2}\), so statement III is true. Statement II is false, because a hollow metal waveguide cannot carry a TEM mode, as it has only one conductor. Statement IV is not accepted: the waves bounce between the walls in a zig-zag, but they do not travel along all four walls. Options (B), (C) and (D) include II or IV.

Your Answer : (Not Answered)

Correct Answer : A

For the dominant TE10 mode of a rectangular waveguide, the cut-off frequency depends only on the broader dimension a, which is 6 cm here, and the medium inside, which is air:

The value 5 GHz would be the cut-off of the TE20 mode, which is c/a, or alternatively of a guide of half the width; the narrower dimension of 4 cm gives \(c/2b = 3.75\) GHz for the TE01 mode. The options 25 MHz and 20.5 GHz are out by orders of magnitude. The answer is 2.5 GHz.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : B

The correct matching is a-ii, b-i, c-iv, d-iii.

• Klystron: electrons are velocity-modulated and bunched, and their kinetic energy is converted into RF energy, giving RF amplification (ii).
• Magnetron: a crossed-field oscillator that runs from a DC anode supply and converts that DC power into microwave power (i).
• Gunn diode: a transferred-electron device used as an oscillator in the range of about 10 GHz up towards 1 THz (iv).
• PIN diode: works by high-level injection of carriers into its intrinsic region, which makes it a current-controlled resistor at microwave frequencies (iii).

Only one code pairs the klystron with RF amplification by kinetic energy, its defining principle.

Your Answer : (Not Answered)

Correct Answer : A

A line is matched when \(Z_L = z_0\).

When the load impedance equals the characteristic impedance of the line, the reflection coefficient is zero:

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

All the incident power is absorbed by the load, there is no reflected wave, and the VSWR is 1.

A load of \(Z_0/2\) or \(2Z_0\) gives a reflection coefficient of magnitude 1/3 and a VSWR of 2. \(\sqrt{Z_0}\) is not even dimensionally an impedance, so it cannot be a matching condition.

Your Answer : (Not Answered)

Correct Answer : A

The true statement is that the TM\(_{10}\) mode does not exist.

For TM modes in a rectangular waveguide, the longitudinal electric field is \(E_{z} \propto \sin(m\pi x/a)\sin(n\pi y/b)\). If either m or n is zero, \(E_{z}\) vanishes everywhere and the whole field is zero, so TM\(_{10}\), TM\(_{01}\) and the like cannot exist; the lowest TM mode is TM\(_{11}\).

TE\(_{10}\) certainly exists: it is the dominant mode, with the lowest cut-off frequency. So TE\(_{10}\) and TM\(_{10}\) do not both exist. The cut-off frequency \(f_{c} = \frac{c}{2}\sqrt{(m/a)^{2} + (n/b)^{2}}\) increases, not decreases, as m and n increase.

Your Answer : (Not Answered)

Correct Answer : B

Single-mode operation occurs for 2 GHz \(\lt f \lt\) 3 GHz.

In a guide filled with a dielectric of index \(n\), waves travel at \(v = c/n = 2.4 \times 10^{8}\) m/s. The cut-off frequencies are \(f_{mn} = \frac{v}{2}\sqrt{(m/a)^{2} + (n/b)^{2}}\) with \(a = 6\) cm and \(b = 4\) cm:

\[ \begin{aligned} f_{10} &= \frac{2.4 \times 10^{8}}{2 \times 0.06} \\ &= 2\ \text{GHz} \\ f_{01} &= \frac{2.4 \times 10^{8}}{2 \times 0.04} \\ &= 3\ \text{GHz} \end{aligned} \]

Between these only TE₁₀ propagates. The other ranges do not match these cut-offs; without the dielectric the window would be 2.5 to 3.75 GHz, which is not offered either.

Your Answer : (Not Answered)

Correct Answer : C

A waveguide acts as a high pass filter.

Each mode in a waveguide has a cut-off frequency, set by its dimensions. Below the cut-off of the dominant mode, waves do not propagate but decay rapidly along the guide; above it they pass with little loss. For a rectangular guide of width a, the dominant \(TE_{10}\) mode has

\[ f_c = \frac{c}{2a} \]

so the guide passes frequencies above \(f_c\) and blocks those below, which is high-pass behaviour, not low-pass, band-pass or band-stop.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : C

The principal mode of a coaxial line is the TEM mode.

A coaxial line has two separate conductors, so it supports the transverse electromagnetic mode, in which both electric and magnetic fields lie wholly in the transverse plane. TEM has no cut-off frequency and is the dominant mode used in normal operation.

TE and TM modes, of which TE₁₁ is the lowest in a coaxial line, appear only above their cut-off frequencies and are avoided by keeping the line's dimensions small compared with the wavelength.

Your Answer : (Not Answered)

Correct Answer : A

The stub cancels the susceptance.

In single stub matching the stub is placed at the point on the line where the normalised admittance is 1 + jb, so that its conductance already matches the line. A short- or open-circuited stub of suitable length adds a susceptance of −jb, cancelling the unwanted susceptance and leaving a matched admittance of 1.

The conductance is matched by choosing the stub's position, not by the stub. Permeability is a property of the medium, and the stub works in admittance terms on the shunt susceptance, not on an inductance as such.

Your Answer : (Not Answered)

Correct Answer : C

The dominant mode is TE₁₀.

The dominant mode of a waveguide is the mode with the lowest cut-off frequency. For a rectangular guide of width a (greater than height b), the cut-off wavelength of the TE_mn mode is:

\[ \lambda_{c} = \frac{2}{\sqrt{(m/a)^{2} + (n/b)^{2}}} \]

It is longest, \(\lambda_{c} = 2a\), for m = 1, n = 0, so TE₁₀ propagates first. TE₁₁, TE₁₂ and TE₂₁ have higher cut-off frequencies. (TM modes need both m and n non-zero, so TM₁₁ is the lowest TM mode.)

Your Answer : (Not Answered)

Correct Answer : C

The transferred electron device is the Gunn diode.

In gallium arsenide, electrons in a strong field transfer from a low-energy, high-mobility valley of the conduction band to a higher, low-mobility valley. Above a threshold field the average velocity therefore falls as the field rises, a negative differential resistance, and the Gunn diode uses it to generate microwave oscillations.

The klystron, the travelling-wave tube and the magnetron are vacuum tubes. They work by velocity modulation and bunching of an electron beam, or by electrons moving in crossed electric and magnetic fields, not by transferred electrons in a semiconductor.

Your Answer : (Not Answered)

Correct Answer : B

The duplexer in a RADAR system is a microwave switch.

A radar using one antenna for both transmitting and receiving needs a duplexer to connect the antenna to the transmitter during the high-power pulse and to the receiver during the listening time between pulses, while protecting the sensitive receiver from the transmitted pulse. It does this with switching devices such as TR (transmit-receive) and ATR tubes, ferrite circulators or PIN diode switches.

It generates no signal, as an oscillator would, does not amplify, and is not an active filter.

Your Answer : (Not Answered)

Correct Answer : D

The klystron operates on velocity modulation.

In a klystron, the electron beam passes through the gap of an input (buncher) cavity, where the RF field speeds up some electrons and slows down others. Over the drift space the fast electrons catch up with the slow ones, so the beam forms bunches: velocity modulation is turned into current (density) modulation. The bunched beam then gives energy to the RF field in the output (catcher) cavity, producing amplification, or in a reflex klystron, oscillation.

Amplitude, frequency and pulse modulation are ways of impressing information on a carrier, not the mechanism by which a klystron works.

Your Answer : (Not Answered)

Correct Answer : D

The line should be 39 km long.

For exactly three pulses to be on the line, it must hold three pulse widths and the two gaps between them:

\[ \begin{aligned} t &= 3 \times 30 + 2 \times 20 = 130\ \mu\text{s} \\ L &= 3 \times 10^{8} \times 130 \times 10^{-6} \\ &= 39000\ \text{m} = 39\ \text{km} \end{aligned} \]

3.9 km is the same length with the power of ten slipped by one. 25 km is not a combination of three pulses, and 2.5 km would hold less than a single pulse, since one 30 μs pulse alone is 9 km long.

Your Answer : (Not Answered)

Correct 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.