Q: Ridge waveguide vs. standard rectangular waveguide – Usable bandwidth comparison How does the usable bandwidth of a properly designed ridge waveguide compare with that of a conventional rectangular waveguide (same outer cross-section)?

Introduction: Ridge waveguides introduce one or two metallic ridges inside a rectangular waveguide to modify field distribution and dispersion. The goal is to lower cutoff, flatten impedance variation, and extend single-mode or low-loss bandwidth for practical components such as couplers, filters, and transitions. Given Data / Assumptions: Comparison at the same external size envelope. Properly designed ridge geometry (gap and ridge height selected for the band). Focus on usable (single-mode/low-VSWR) bandwidth. Concept / Approach: By concentrating the electric field in the ridge gap, the effective capacitance increases and the TE10 cutoff frequency is reduced. This widens the frequency region between the new (lower) cutoff and higher-order mode onsets, enabling a broader useful passband with better match and power handling for many applications. Step-by-Step Solution: 1) Add ridge → lower fc(TE10).2) Adjust gap/height → manage impedance and higher-order mode spacing.3) Result → broader single-mode/low-loss operational band than standard rectangular guide. Verification / Alternative check: Design curves and EM simulations of ridge guides show extended bandwidths used in broadband transitions and couplers. Why Other Options Are Wrong: Options A, C, D contradict the purpose of the ridge. Option E is incorrect; ridge guides still support TE modes. Common Pitfalls: Assuming ridges always increase loss; properly designed ridges maintain acceptable attenuation over a wider band. Final Answer: is higher than that of rectangular waveguide.

Q: Why short-circuited stubs are preferred to open-circuited stubs in RF/microwave matching? Choose the main reason an open-circuited stub is often avoided compared to a short-circuited stub of the same electrical length.

Introduction: Single-stub tuning is a classic transmission-line method for impedance matching. Designers often prefer short-circuited stubs over open-circuited ones, particularly at high frequencies where unintended radiation and coupling become significant concerns. Given Data / Assumptions: Lossless/low-loss line; identical Z0 for main line and stub unless otherwise designed. Same electrical length for fair comparison. Practical implementation with finite grounding and fringing fields. Concept / Approach: An open end supports strong fringing electric fields, which can radiate and pick up environmental coupling. A short-circuited end presents a well-defined boundary with minimal radiation and provides a robust RF ground at the end. Both open and short stubs can synthesize the same reactance values, but the shorted stub generally offers better repeatability and lower radiation loss. Step-by-Step Solution: 1) Compare boundary conditions: open end → high E-field fringe; short end → high current, confined magnetic field.2) Radiation risk is higher for open stubs due to strong external E-fields.3) Hence, for similar matching range, the short-circuited stub is preferred. Verification / Alternative check: Near-field scans and EM simulation show higher external fields around open-ended stubs, confirming higher radiation susceptibility. Why Other Options Are Wrong: Option A: Fabrication difficulty is not the fundamental reason. Option B: Z0 choice is design-dependent, not an intrinsic limitation of open stubs. Option D: Both open and short stubs can realize a full reactance range. Option E: Irrelevant claim. Common Pitfalls: Assuming open stubs cannot realize inductive susceptance; they can, but with higher radiation risk at high frequencies. Final Answer: is liable to radiate.

Q: Rectangular waveguide – Effect of increasing internal air pressure on maximum power handling If the air pressure inside a rectangular waveguide is increased (with all other factors unchanged), how does its maximum RF power-handling capacity change?

Introduction: Waveguide power handling is limited by RF breakdown of the gas (usually air) at the wall/field maxima. The breakdown threshold depends on pressure and gap according to gas discharge physics (Paschen-like behavior). Pressurizing waveguides (or using SF6) is a standard technique to extend power capacity. Given Data / Assumptions: Same geometry, frequency, and surface finish. Gas is air; only internal pressure is increased within safe mechanical limits. No condensation or contamination introduced. Concept / Approach: Increasing pressure raises the breakdown electric field strength, delaying ionization and arcing. Since maximum power for a given guide mode scales with allowable peak field before breakdown, higher breakdown field translates directly to higher permissible power. Step-by-Step Solution: 1) Breakdown field E_bd increases with pressure in the practical operating region.2) For a given mode, P_max ∝ E_bd^2 times cross-sectional factors.3) Therefore, raising pressure increases P_max. Verification / Alternative check: High-power radar and broadcast systems routinely use pressurized guides to achieve higher P_max and reduce multipaction/rf arcing. Why Other Options Are Wrong: Option A/C contradict observed practice; Option D is too vague; Option E is incorrect because breakdown field is not constant—it varies with pressure and gap. Common Pitfalls: Ignoring moisture/contaminants that can lower breakdown in practice; assuming metal loss, not breakdown, always limits power (often breakdown is the primary constraint). Final Answer: is increased.

Q: Klystron amplifier – Typical noise figure What is a representative noise figure (in dB) for a conventional multi-cavity klystron amplifier used as a high-power RF chain stage (not a low-noise front end)?

Introduction: Noise figure (NF) characterizes how much a device degrades the signal-to-noise ratio. Klystron amplifiers are efficient high-power transmit amplifiers but are not designed as low-noise receivers. Understanding typical NF values helps in cascade noise budgeting and in distinguishing power amplifiers from low-noise front-end devices. Given Data / Assumptions: Conventional multi-cavity klystron operated as a power amplifier. Not cryogenic or specially optimized for low noise. Room-temperature operation and standard beam voltages. Concept / Approach: High-power beam devices prioritize efficiency, gain, and linearity over ultra-low noise. Typical NF values are tens of dB, far higher than the 1–3 dB range associated with LNAs. Therefore, a representative figure around 25 dB is appropriate for an ordinary klystron amplifier stage. Step-by-Step Solution: 1) Identify device class: power klystron (not LNA).2) Compare with LNA benchmarks: 1–3 dB typical for LNAs; much higher for klystrons.3) Choose representative NF: ≈ 25 dB. Verification / Alternative check: System design texts place klystron NF in the tens of dB. Receiver front ends use low-noise solid-state devices instead. Why Other Options Are Wrong: 1–3 dB or below are LNA-class, not power klystrons. “Zero” is physically impossible. Common Pitfalls: Assuming all amplifiers used in RF chains have LNA-like noise; conflating gain with noise performance. Final Answer: 25 dB.

Q: Basic radar equation scaling – Effect of peak transmitted power on maximum range If the peak transmitted power P_t in a monostatic radar is increased by a factor of 81 (all other parameters unchanged), by what factor does the maximum detection range…

Introduction: The basic radar range equation shows that received echo power falls with R^4 (two-way propagation and spreading). Consequently, the maximum range R_max scales with the fourth root of peak transmitted power P_t. This question checks your ability to apply the 4th-root dependence quickly. Given Data / Assumptions: Monostatic radar; same antenna gains, target RCS, wavelength, receiver sensitivity, and SNR threshold. Only P_t is scaled by 81. Free-space propagation and the same processing gain. Concept / Approach: From the radar equation, keeping everything except P_t fixed, R_max ∝ (P_t)^(1/4). Therefore, the range factor equals the fourth root of the transmit-power factor. Step-by-Step Solution: 1) Power factor k = 81.2) Range factor = k^(1/4) = 81^(1/4).3) Note 81 = 3^4 → 81^(1/4) = 3. Verification / Alternative check: Doubling R requires 16× power (2^4). Here, 81× power gives 3× range, consistent with the 4th-power relationship. Why Other Options Are Wrong: 9 or 27 imply square-root or cube-root scaling; 81 implies linear scaling; √81 = 9 is a misapplied square-root relation. Common Pitfalls: Forgetting the two-way path loss gives R^4; using square-root scaling appropriate to one-way links, not radar. Final Answer: 3.

Q: Waveguide/cavity modes – Identify the incorrect statement about TE/TM transmission Which of the following statements is NOT correct regarding modes used to carry electromagnetic energy in hollow guides and resonant structures?

Introduction: Electromagnetic energy in hollow metal guides is carried in discrete modes, commonly labeled TE_mn or TM_mn for rectangular guides. The dominant mode is the lowest-cutoff mode that can propagate and is central to practical design. Given Data / Assumptions: Standard empty (air-filled) rectangular waveguide. Mode labels follow conventional definitions. Focus on conceptual correctness rather than numerical cutoff values. Concept / Approach: In TE modes, E_z = 0, so the electric field is everywhere transverse to the propagation direction. In TM modes, H_z = 0, so the magnetic field is everywhere transverse. The subscripts m, n indicate the number of half-wave field variations along the guide's two transverse dimensions. The dominant mode of a rectangular waveguide is TE10, not TM11. Step-by-Step Solution: 1) Recall TE definition: E has no longitudinal component → transverse everywhere.2) Recall TM definition: H has no longitudinal component → transverse everywhere.3) Identify dominant mode in rectangular WG: TE10 (lowest cutoff frequency).4) Therefore, the statement naming TM11 as dominant is incorrect. Verification / Alternative check: Cutoff tables list TE10 with the smallest fc for rectangular guides of given a × b cross-section. Why Other Options Are Wrong: Options A, B, C, and E correctly describe TE/TM definitions and the dominant mode for rectangular waveguide. Common Pitfalls: Confusing TE vs TM definitions; assuming symmetry implies the same dominant mode across shapes (it does not). Final Answer: TM11 is called the dominant mode.

Q: Eight-cavity magnetron – Total phase shift around the ring in π-mode For a magnetron with 8 identical cavity resonators, what is the total phase shift around the periphery in the π-mode (adjacent cavities 180° out of phase)?

Introduction: Multi-cavity magnetrons support discrete phase modes around their circular cavity arrays. The π-mode is preferred for stable, efficient operation and places adjacent cavities 180° out of phase. Given Data / Assumptions: Eight cavities equally spaced. π-mode: phase step of π between neighbors. Steady periodic boundary condition around the ring. Concept / Approach: With N cavities and a phase advance Δφ between adjacent cavities, the total phase shift around the periphery is N * Δφ. In π-mode, Δφ = π. Step-by-Step Solution: 1) N = 8 cavities.2) Δφ = π radians (adjacent cavities out of phase).3) Total phase shift = N * Δφ = 8 * π = 8π radians (± indicating direction of traversal). Verification / Alternative check: Mode charts for magnetrons label the π-mode with alternating polarity across adjacent cavities and total 8π phase around an 8-cavity ring. Why Other Options Are Wrong: 4π or 2π correspond to other fractional modes; 16π doubles the correct value; zero would mean all in phase (0-mode), not π-mode. Common Pitfalls: Confusing total phase around the ring with the phase relative to a fixed cavity; forgetting to multiply by the number of cavities. Final Answer: ± 8π radians around the periphery.

Question 1

At microwave frequencies, how does the skin depth of good conductors compare with the typical physical dimensions of most feed structures and conductors?

Accepted Answer

Introduction / Context: Skin depth δ determines how deeply RF currents penetrate into conductors. At microwave frequencies, δ becomes very small, concentrating current near the surface and impacting conductor loss, plating choices, and fabrication. Given Data / Assumptions: Good conductors (e.g., copper, silver) at GHz frequencies. Feed structures include coaxial inner/outer conductors, microstrip, waveguide flanges. δ ∝ 1/√f for a given material. Concept / Approach: Because δ decreases with √f, at microwave frequencies δ is typically a few micrometers or less—far smaller than mechanical dimensions (mil-scale or mm-scale). Hence current flows in a thin surface layer; thick plating beyond several skin depths yields diminishing returns for RF resistance. Step-by-Step Solution: 1) Use rule of thumb for copper: δ(μm) ≈ 66/√f(MHz).2) At 1 GHz ⇒ δ ≈ 2 μm; typical conductor thicknesses are tens of μm to mm.3) Since δ ≪ thickness, current is confined near the surface.4) Practical implication: surface roughness and plating quality dominate loss. Verification / Alternative check: Measured insertion loss of lines strongly correlates with surface finish at GHz even when metal is thick, confirming δ ≪ dimension scale. Why Other Options Are Wrong: Same order / much larger: contradicts δ ∝ 1/√f at microwave. Polarization-dependent only: δ is material/frequency dependent. Independent of frequency: incorrect. Common Pitfalls: Over-plating well beyond several δ; ignoring roughness (Huray/hammer effect) which raises effective resistance. Final Answer: Much smaller than feed dimensions

Question 2

In a rectangular waveguide, which listed mode corresponds to the trivial (non-propagating) solution where all field components are zero?

Accepted Answer

Introduction / Context: Waveguide modes are labeled TEmn or TMmn. Not every index pair represents a physical propagating solution. Some index choices lead to a trivial zero-field solution, which is excluded from the set of allowed modes. Given Data / Assumptions: Hollow rectangular waveguide with perfectly conducting walls (idealized). Mode families: TE and TM with integer indices m, n ≥ 0. Boundary conditions require tangential electric field to vanish at walls. Concept / Approach: For TM modes, both tangential electric field components must satisfy standing-wave conditions along the cross-section. If m = n = 0 (TM00), the field solution reduces to zero everywhere to satisfy boundary conditions—hence it is a non-existent (trivial) mode. By contrast, TE10 is the dominant physical mode. Step-by-Step Solution: 1) Consider TMmn with m or n equal to zero.2) Apply boundary conditions for the scalar potential forms; for m=n=0, fields cannot satisfy non-zero boundary conditions without contradiction.3) Conclude that TM00 implies E = H = 0 (trivial solution), not a propagating mode.4) Therefore, TM00 is excluded from the physical mode set. Verification / Alternative check: Standard mode tables omit TM00 for rectangular waveguides; laboratory measurements never observe TM00 propagation. Why Other Options Are Wrong: TM11, TM01, TM10, TE10: All describe legitimate non-trivial modes with finite cutoff (except TM00). Common Pitfalls: Assuming every index pair yields a mode; confusing the TE/TM naming with TEM, which does not propagate in hollow rectangular guides. Final Answer: TM00

Question 3

For a microstrip line printed on a substrate with relative permittivity ε_r, the effective dielectric constant ε_eff of the quasi-TEM mode (accounting for fringing fields in air) is typically

Accepted Answer

Introduction / Context: Microstrip supports a quasi-TEM mode with fields partly in the substrate and partly in air. Designers use an effective dielectric constant ε_eff to model phase velocity, impedance, and dispersion of this mixed-media line. Given Data / Assumptions: Microstrip conductor on a dielectric substrate (ε_r > 1) with air above. Operating in microwave region where quasi-TEM approximation is valid. Fringing fields extend significantly into air. Concept / Approach: Because a portion of the field lines pass through air (ε ≈ 1) and the rest through the substrate (ε_r), the modal effective permittivity satisfies 1 < ε_eff < ε_r. The exact value depends on geometry (conductor width, substrate thickness) and ε_r, and it approaches ε_r only for very wide traces or very thick/high-ε_r substrates that confine more field in the dielectric. Step-by-Step Solution: 1) Recognize mixed field distribution in air and dielectric.2) Model with ε_eff to capture average polarization effect.3) Conclude that ε_eff must lie between 1 and ε_r.4) Use empirical formulas (e.g., Hammerstad–Jensen) for numerical design. Verification / Alternative check: EM solvers and closed-form models consistently produce ε_eff values between 1 and ε_r for practical microstrip geometries. Why Other Options Are Wrong: ε_eff > ε_r or =1: Physically inconsistent for microstrip with finite substrate. Independent of ε_r: Contradicts the mixed-media nature. Common Pitfalls: Assuming ε_eff equals ε_r (that would be closer for stripline with full dielectric confinement). Final Answer: less than ε_r (but greater than 1)

Question 4

Consider the statements about a radar beacon (RACON / transponder):  It is also called a transponder.  It includes a pulse transmitter, a receiver, and an antenna.  It replies with its own fixed (preset) transmitter power once it detects an interrog…

Accepted Answer

Introduction / Context: Radar beacons (RACONs) and transponders are cooperative targets that detect interrogations and transmit replies at their own power level. They are used in maritime navigation aids, IFF (Identification Friend or Foe), and secondary surveillance radar systems to enhance detection and provide identification or bearing information. Given Data / Assumptions: The device is cooperative and responds to an external interrogation. It contains RF front-end receive circuitry, a timing/logic section, and a transmitter with antenna. Reply power is generated by the beacon’s transmitter, not by amplifying the incoming pulse. Concept / Approach: A transponder architecture includes a receiver to detect incoming pulses, a processing block to validate and time the response, and a transmitter to send a coded reply via an antenna. As long as the interrogation exceeds the detection threshold, the reply is transmitted at a preset power, largely independent of the received signal strength. This distinguishes cooperative beacons from passive reflectors. Step-by-Step Solution: 1) Statement 1: “Also called transponder” — correct; RACONs and SSR/IFF units are transponder types.2) Statement 2: “Includes pulse transmitter, receiver, and antenna” — correct and necessary for operation.3) Statement 3: “Replies with full/preset power even for weak triggers” — correct within limits; once detection threshold is crossed, reply power is determined by the beacon’s transmitter rating. Verification / Alternative check: Specifications for marine RACONs and aviation transponders list fixed output powers (e.g., watts or kilowatts peak) and minimum trigger levels, confirming independence of reply power from input amplitude once detected. Why Other Options Are Wrong: Omitting any of 1–3 leaves out an essential property of transponders or misstates their power behavior. Common Pitfalls: Confusing active transponders with passive retro-reflectors; assuming reply power scales with received power like a linear amplifier. Final Answer: All

Question 5

Microwave passive components – Equivalent circuit of a physical resistor when its physical length l is much smaller than the wavelength λ (l << λ) In the lumped-element region (electrically short body and leads), what is the most appropriate small-s…

Accepted Answer

Introduction: Real resistors at microwave frequencies are physical structures with finite lead lengths and electrode areas. However, when the physical length l is much smaller than the operating wavelength λ, the component behaves as an electrically short element. This question tests whether you recognize that, under the l << λ condition, the resistor is well approximated by a lumped resistance R, with any parasitic inductance (from leads) and capacitance (from pads/body) being comparatively small. Given Data / Assumptions: Operating frequency such that l << λ (electrically short). Linear, small-signal operation about the bias point. Physical resistor realized as a film or chip with short interconnects. Concept / Approach: The lumped-element model applies when phase variation across the device is negligible. Under l << λ, distributed effects (standing waves along the body) do not develop, and the element can be modeled by its lowest-order parameters. The dominant parameter of a resistor is its resistance R. Parasitic series inductance Ls and shunt capacitance Cp exist but, for sufficiently small l/λ, they have second-order influence and can often be ignored in first-pass analysis, especially for narrowband designs or moderate Q networks. Step-by-Step Solution: 1) Check electrical length: βl ≈ 2πl/λ → if βl << 1, the device is lumped.2) Identify dominant parameter: for a resistor, R dominates; Ls and Cp are parasitic.3) Form the equivalent circuit: an ideal R, optionally annotated with very small Ls and Cp for higher-accuracy models.4) Decide model order: for conceptual questions, the ideal R is the correct minimal equivalent under l << λ. Verification / Alternative check: EM or circuit simulations show that as frequency lowers (or size shrinks), the impedance approaches a constant real value R; reactance magnitudes diminish toward zero, validating the ideal-R approximation. Why Other Options Are Wrong: Option B/C: A real resistor cannot be purely inductive or capacitive. Option D: A λ/4 line is a distributed element, contradicting l << λ. Option E: Negative resistance describes active or special tunneling devices, not passive resistors. Common Pitfalls: Over-fitting parasitics in the lumped region; confusing the high-frequency distributed regime (where parasitics matter) with the explicitly stated l << λ case. Final Answer: An ideal lumped resistance R (dominant), with negligible parasitic inductance and capacitance.

Question 6

Ridge waveguide vs. standard rectangular waveguide – Usable bandwidth comparison How does the usable bandwidth of a properly designed ridge waveguide compare with that of a conventional rectangular waveguide (same outer cross-section)?

Accepted Answer

Introduction: Ridge waveguides introduce one or two metallic ridges inside a rectangular waveguide to modify field distribution and dispersion. The goal is to lower cutoff, flatten impedance variation, and extend single-mode or low-loss bandwidth for practical components such as couplers, filters, and transitions. Given Data / Assumptions: Comparison at the same external size envelope. Properly designed ridge geometry (gap and ridge height selected for the band). Focus on usable (single-mode/low-VSWR) bandwidth. Concept / Approach: By concentrating the electric field in the ridge gap, the effective capacitance increases and the TE10 cutoff frequency is reduced. This widens the frequency region between the new (lower) cutoff and higher-order mode onsets, enabling a broader useful passband with better match and power handling for many applications. Step-by-Step Solution: 1) Add ridge → lower fc(TE10).2) Adjust gap/height → manage impedance and higher-order mode spacing.3) Result → broader single-mode/low-loss operational band than standard rectangular guide. Verification / Alternative check: Design curves and EM simulations of ridge guides show extended bandwidths used in broadband transitions and couplers. Why Other Options Are Wrong: Options A, C, D contradict the purpose of the ridge. Option E is incorrect; ridge guides still support TE modes. Common Pitfalls: Assuming ridges always increase loss; properly designed ridges maintain acceptable attenuation over a wider band. Final Answer: is higher than that of rectangular waveguide.

Question 7

Why short-circuited stubs are preferred to open-circuited stubs in RF/microwave matching? Choose the main reason an open-circuited stub is often avoided compared to a short-circuited stub of the same electrical length.

Accepted Answer

Introduction: Single-stub tuning is a classic transmission-line method for impedance matching. Designers often prefer short-circuited stubs over open-circuited ones, particularly at high frequencies where unintended radiation and coupling become significant concerns. Given Data / Assumptions: Lossless/low-loss line; identical Z0 for main line and stub unless otherwise designed. Same electrical length for fair comparison. Practical implementation with finite grounding and fringing fields. Concept / Approach: An open end supports strong fringing electric fields, which can radiate and pick up environmental coupling. A short-circuited end presents a well-defined boundary with minimal radiation and provides a robust RF ground at the end. Both open and short stubs can synthesize the same reactance values, but the shorted stub generally offers better repeatability and lower radiation loss. Step-by-Step Solution: 1) Compare boundary conditions: open end → high E-field fringe; short end → high current, confined magnetic field.2) Radiation risk is higher for open stubs due to strong external E-fields.3) Hence, for similar matching range, the short-circuited stub is preferred. Verification / Alternative check: Near-field scans and EM simulation show higher external fields around open-ended stubs, confirming higher radiation susceptibility. Why Other Options Are Wrong: Option A: Fabrication difficulty is not the fundamental reason. Option B: Z0 choice is design-dependent, not an intrinsic limitation of open stubs. Option D: Both open and short stubs can realize a full reactance range. Option E: Irrelevant claim. Common Pitfalls: Assuming open stubs cannot realize inductive susceptance; they can, but with higher radiation risk at high frequencies. Final Answer: is liable to radiate.

Question 8

Rectangular waveguide – Effect of increasing internal air pressure on maximum power handling If the air pressure inside a rectangular waveguide is increased (with all other factors unchanged), how does its maximum RF power-handling capacity change?

Accepted Answer

Introduction: Waveguide power handling is limited by RF breakdown of the gas (usually air) at the wall/field maxima. The breakdown threshold depends on pressure and gap according to gas discharge physics (Paschen-like behavior). Pressurizing waveguides (or using SF6) is a standard technique to extend power capacity. Given Data / Assumptions: Same geometry, frequency, and surface finish. Gas is air; only internal pressure is increased within safe mechanical limits. No condensation or contamination introduced. Concept / Approach: Increasing pressure raises the breakdown electric field strength, delaying ionization and arcing. Since maximum power for a given guide mode scales with allowable peak field before breakdown, higher breakdown field translates directly to higher permissible power. Step-by-Step Solution: 1) Breakdown field E_bd increases with pressure in the practical operating region.2) For a given mode, P_max ∝ E_bd^2 times cross-sectional factors.3) Therefore, raising pressure increases P_max. Verification / Alternative check: High-power radar and broadcast systems routinely use pressurized guides to achieve higher P_max and reduce multipaction/rf arcing. Why Other Options Are Wrong: Option A/C contradict observed practice; Option D is too vague; Option E is incorrect because breakdown field is not constant—it varies with pressure and gap. Common Pitfalls: Ignoring moisture/contaminants that can lower breakdown in practice; assuming metal loss, not breakdown, always limits power (often breakdown is the primary constraint). Final Answer: is increased.

Question 9

Klystron amplifier – Typical noise figure What is a representative noise figure (in dB) for a conventional multi-cavity klystron amplifier used as a high-power RF chain stage (not a low-noise front end)?

Accepted Answer

Introduction: Noise figure (NF) characterizes how much a device degrades the signal-to-noise ratio. Klystron amplifiers are efficient high-power transmit amplifiers but are not designed as low-noise receivers. Understanding typical NF values helps in cascade noise budgeting and in distinguishing power amplifiers from low-noise front-end devices. Given Data / Assumptions: Conventional multi-cavity klystron operated as a power amplifier. Not cryogenic or specially optimized for low noise. Room-temperature operation and standard beam voltages. Concept / Approach: High-power beam devices prioritize efficiency, gain, and linearity over ultra-low noise. Typical NF values are tens of dB, far higher than the 1–3 dB range associated with LNAs. Therefore, a representative figure around 25 dB is appropriate for an ordinary klystron amplifier stage. Step-by-Step Solution: 1) Identify device class: power klystron (not LNA).2) Compare with LNA benchmarks: 1–3 dB typical for LNAs; much higher for klystrons.3) Choose representative NF: ≈ 25 dB. Verification / Alternative check: System design texts place klystron NF in the tens of dB. Receiver front ends use low-noise solid-state devices instead. Why Other Options Are Wrong: 1–3 dB or below are LNA-class, not power klystrons. “Zero” is physically impossible. Common Pitfalls: Assuming all amplifiers used in RF chains have LNA-like noise; conflating gain with noise performance. Final Answer: 25 dB.

Question 10

Basic radar equation scaling – Effect of peak transmitted power on maximum range If the peak transmitted power P_t in a monostatic radar is increased by a factor of 81 (all other parameters unchanged), by what factor does the maximum detection range…

Accepted Answer

Introduction: The basic radar range equation shows that received echo power falls with R^4 (two-way propagation and spreading). Consequently, the maximum range R_max scales with the fourth root of peak transmitted power P_t. This question checks your ability to apply the 4th-root dependence quickly. Given Data / Assumptions: Monostatic radar; same antenna gains, target RCS, wavelength, receiver sensitivity, and SNR threshold. Only P_t is scaled by 81. Free-space propagation and the same processing gain. Concept / Approach: From the radar equation, keeping everything except P_t fixed, R_max ∝ (P_t)^(1/4). Therefore, the range factor equals the fourth root of the transmit-power factor. Step-by-Step Solution: 1) Power factor k = 81.2) Range factor = k^(1/4) = 81^(1/4).3) Note 81 = 3^4 → 81^(1/4) = 3. Verification / Alternative check: Doubling R requires 16× power (2^4). Here, 81× power gives 3× range, consistent with the 4th-power relationship. Why Other Options Are Wrong: 9 or 27 imply square-root or cube-root scaling; 81 implies linear scaling; √81 = 9 is a misapplied square-root relation. Common Pitfalls: Forgetting the two-way path loss gives R^4; using square-root scaling appropriate to one-way links, not radar. Final Answer: 3.

Question 11

Waveguide/cavity modes – Identify the incorrect statement about TE/TM transmission Which of the following statements is NOT correct regarding modes used to carry electromagnetic energy in hollow guides and resonant structures?

Accepted Answer

Introduction: Electromagnetic energy in hollow metal guides is carried in discrete modes, commonly labeled TE_mn or TM_mn for rectangular guides. The dominant mode is the lowest-cutoff mode that can propagate and is central to practical design. Given Data / Assumptions: Standard empty (air-filled) rectangular waveguide. Mode labels follow conventional definitions. Focus on conceptual correctness rather than numerical cutoff values. Concept / Approach: In TE modes, E_z = 0, so the electric field is everywhere transverse to the propagation direction. In TM modes, H_z = 0, so the magnetic field is everywhere transverse. The subscripts m, n indicate the number of half-wave field variations along the guide's two transverse dimensions. The dominant mode of a rectangular waveguide is TE10, not TM11. Step-by-Step Solution: 1) Recall TE definition: E has no longitudinal component → transverse everywhere.2) Recall TM definition: H has no longitudinal component → transverse everywhere.3) Identify dominant mode in rectangular WG: TE10 (lowest cutoff frequency).4) Therefore, the statement naming TM11 as dominant is incorrect. Verification / Alternative check: Cutoff tables list TE10 with the smallest fc for rectangular guides of given a × b cross-section. Why Other Options Are Wrong: Options A, B, C, and E correctly describe TE/TM definitions and the dominant mode for rectangular waveguide. Common Pitfalls: Confusing TE vs TM definitions; assuming symmetry implies the same dominant mode across shapes (it does not). Final Answer: TM11 is called the dominant mode.

Question 12

Eight-cavity magnetron – Total phase shift around the ring in π-mode For a magnetron with 8 identical cavity resonators, what is the total phase shift around the periphery in the π-mode (adjacent cavities 180° out of phase)?

Accepted Answer

Introduction: Multi-cavity magnetrons support discrete phase modes around their circular cavity arrays. The π-mode is preferred for stable, efficient operation and places adjacent cavities 180° out of phase. Given Data / Assumptions: Eight cavities equally spaced. π-mode: phase step of π between neighbors. Steady periodic boundary condition around the ring. Concept / Approach: With N cavities and a phase advance Δφ between adjacent cavities, the total phase shift around the periphery is N * Δφ. In π-mode, Δφ = π. Step-by-Step Solution: 1) N = 8 cavities.2) Δφ = π radians (adjacent cavities out of phase).3) Total phase shift = N * Δφ = 8 * π = 8π radians (± indicating direction of traversal). Verification / Alternative check: Mode charts for magnetrons label the π-mode with alternating polarity across adjacent cavities and total 8π phase around an 8-cavity ring. Why Other Options Are Wrong: 4π or 2π correspond to other fractional modes; 16π doubles the correct value; zero would mean all in phase (0-mode), not π-mode. Common Pitfalls: Confusing total phase around the ring with the phase relative to a fixed cavity; forgetting to multiply by the number of cavities. Final Answer: ± 8π radians around the periphery.

Question 13

Reflex klystron oscillation condition – Role of anode and repeller voltages State whether a reflex klystron oscillates only for those combinations of anode voltage and repeller (reflector) voltage that yield a favorable electron transit time through…

Accepted Answer

Introduction: A reflex klystron is a single-cavity oscillator that relies on velocity modulation at the cavity gap and subsequent reflection of the electron beam by a negatively biased repeller electrode. Sustained oscillation requires that bunched electrons return to the gap in the correct phase to deliver energy to the RF field. Given Data / Assumptions: Single resonator (re-entrant cavity) and a repeller electrode with variable negative voltage. Anode voltage sets beam energy; repeller voltage sets turn-around distance/time. Small-signal oscillation requires loop phase and loop gain conditions (Barkhausen) to be met. Concept / Approach: For oscillation, the electron transit time T from the first pass through the gap, to reflection, and back to the gap must satisfy a phase condition commonly written as ωT ≈ (2πn + 3π/2) for integer n (modes often labeled n + 3/4). Only certain combinations of anode and repeller voltages produce a suitable T that yields positive power transfer to the cavity. Step-by-Step Solution: 1) Choose anode voltage → sets initial beam velocity v0.2) Adjust repeller voltage → sets reflection point and transit time T.3) When the phase condition is satisfied, returning bunches arrive when the gap field decelerates them → energy is delivered to the RF → oscillation builds.4) Outside these voltage combinations, the device does not meet loop conditions and oscillation ceases. Verification / Alternative check: Practical tuning curves show discrete voltage regions (modes) where output power peaks; between them, oscillation stops or weakens. Why Other Options Are Wrong: Claims about cryogenic temperatures, multiple resonators, or space-charge prerequisites are not fundamental to the oscillation condition in reflex klystrons. Common Pitfalls: Confusing the single-cavity reflex oscillator with multi-cavity klystron amplifiers; overlooking that repeller tuning controls frequency and output power via transit time. Final Answer: True.

Question 14

Microwave Link Engineering: Mitigating Rain Fade in Terrestrial and Satcom Paths  In microwave communication links, significant fading and attenuation can occur during rain events (rain fade). Which mitigation methods are most appropriate to solve o…

Accepted Answer

Introduction / Context: Rain fade is a dominant impairment for microwave and satellite links, especially at frequencies above about 10 GHz. Water droplets absorb and scatter radio waves, causing additional path loss and fast fading. This question asks which techniques best mitigate rain-induced degradation in real deployments. Given Data / Assumptions: Microwave links operate where precipitation can be intense. System design may include multiple frequencies, sites, and adaptive strategies. Goal is to maintain availability (uptime) and acceptable link margin during rain. Concept / Approach: Two robust strategies are path diversity and frequency diversity. Path diversity uses geographically separated routes or space diversity on the same path (two antennas spaced vertically or horizontally) so that deep fades are rarely simultaneous. Frequency diversity uses a second channel at a different frequency, since rain attenuation is strongly frequency dependent; the lower-frequency path often remains serviceable when the higher-frequency path is in deep fade. Step-by-Step Solution: 1) Identify main impairment: excess specific attenuation due to rain, leading to reduced link margin.2) Counter with diversity: add independent paths (space or route) so not all suffer peak fade simultaneously.3) Add frequency diversity: carry traffic on a second frequency band less sensitive to rain to ensure continuity.4) Combine with other measures (adaptive coding/modulation, higher antenna gain) for overall availability targets. Verification / Alternative check: ITU-R rain models show higher attenuation at higher frequencies; operational networks use dual-frequency and space-diverse links to achieve 99.99%+ availability. Why Other Options Are Wrong: Antenna replacement/feed tweaks (A) are maintenance, not targeted mitigation. Amplitude/phase trimming (B) is for calibration, not rain-specific. Polarization shifting and code diversity (C) can help robustness but are not primary cures for the propagation loss. Increasing cable DC current (E) is unrelated and unsafe. Common Pitfalls: Relying only on extra transmit power; during heavy rain, dynamic range may be insufficient. Ignoring site diversity spacing guidelines and frequency planning. Final Answer: path diversity and frequency diversity

Question 15

Microwave Connectors and Return Loss (SWR)  True or False: A good microwave connector should have a high standing-wave ratio (SWR).

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Introduction / Context: SWR (standing-wave ratio) is a measure of mismatch between a connector (or any component) and the system impedance. In microwave engineering, mismatch leads to reflections, ripple in passbands, and wasted power. Given Data / Assumptions: System reference impedance (commonly 50 Ω or 75 Ω). Connector is used at microwave frequencies. Objective is minimal reflection and maximum power transfer. Concept / Approach: Lower SWR corresponds to lower reflection coefficient |Γ| and higher return loss. A well-designed connector aims for SWR close to 1.0 over its specified band. High SWR indicates substantial mismatch, causing ripple, distortion, and heating. Step-by-Step Solution: 1) Recall: SWR = (1 + |Γ|) / (1 − |Γ|); SWR = 1 means perfect match.2) Good connectors advertise low VSWR (e.g., ≤ 1.15 or ≤ 1.2) across the band.3) Therefore, the statement “a microwave connector should have high SWR” is false. Verification / Alternative check: Datasheets list return loss and VSWR targets; better-grade connectors have higher return loss (e.g., > 20 dB) meaning lower SWR. Why Other Options Are Wrong: Any conditional “True” variants contradict basic impedance-matching practice; performance expectations do not reverse at short lengths or high frequencies. Common Pitfalls: Confusing “high return loss” (good) with “high SWR” (bad). Return loss high = reflections small; SWR high = reflections large. Final Answer: False

Question 16

Directional Coupler with 10 dB Coupling: Compute Coupled-Port Power  A 4-port directional coupler has a specified coupling factor of 10 dB. If 100 mW is applied at port 1 (matched, ideal lossless device), what power appears at the coupled port (port…

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Introduction / Context: Directional couplers sample a defined fraction of the input power to a secondary port while maintaining directionality. The coupling factor specifies how much lower the coupled power is relative to the input. Given Data / Assumptions: Coupling C = 10 dB. Input power Pin = 100 mW at port 1. Matched, ideal, and lossless coupler. Concept / Approach: By definition, coupling in dB is C = 10 * log10(Pin / Pcoupled). Solving for Pcoupled gives Pcoupled = Pin / 10^(C/10). Step-by-Step Solution: 1) Compute power ratio: 10^(C/10) = 10^(10/10) = 10.2) Pcoupled = 100 mW / 10 = 10 mW.3) Therefore, the coupled port (port 3) delivers 10 mW. Verification / Alternative check: Expressed as percentage, a 10 dB coupler samples 10% of the input power. 10% of 100 mW is 10 mW, consistent with the calculation. Why Other Options Are Wrong: 15 mW, 20 mW, and 25 mW correspond to 8.2 dB, 7 dB, and 6 dB coupling respectively. 1 mW corresponds to 20 dB coupling, not 10 dB. Common Pitfalls: Confusing coupling with insertion loss; mixing amplitude ratios (in dB for voltage) with power ratios. Final Answer: 10 mW

Question 17

Transferred-Electron (Gunn) Effect: Direction of Electron Transfer  In a Gunn diode operating above threshold field, electrons transfer between energy valleys in the conduction band. From which type of valley to which type do they predominantly tran…

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Introduction / Context: Gunn diodes are transferred-electron devices. Their negative differential mobility arises from field-induced transfer of carriers among multiple conduction-band valleys in certain semiconductors (e.g., GaAs). Given Data / Assumptions: Material with at least one low-energy central valley and higher-energy satellite valleys. Electric field exceeds a threshold so intervalley transfer occurs. Mobility in satellite valleys is lower due to higher effective mass. Concept / Approach: At low fields, electrons occupy the central valley and exhibit high mobility. At high fields, a significant fraction transfers to satellite valleys with lower mobility. Net drift velocity can then decrease with increasing field, producing negative differential resistance and enabling oscillation. Step-by-Step Solution: 1) Identify initial valley: central (high mobility).2) Apply high field: carriers scatter to satellite valleys.3) Mobility drops; drift velocity vs field shows a peak then a decline.4) Resulting negative differential resistance supports microwave oscillation in appropriate circuits. Verification / Alternative check: Velocity–field characteristics in GaAs show transfer at fields of the order of a few kV/cm, aligning with Gunn effect theory. Why Other Options Are Wrong: Option B reverses the physics; options C and D confuse band transport with device-level domain terminology; option E describes tunneling (tunnel diode), not the Gunn effect. Common Pitfalls: Assuming a pn junction is required; Gunn diodes are bulk devices without a junction. Final Answer: from high-mobility central valley to low-mobility satellite valley

Question 18

Satellite Communications Bands (Uplink/Downlink Ranges)  Which of the following options corresponds to a commonly used satellite communication frequency band?

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Introduction / Context: Satellite communications use several standardized microwave bands. Classic commercial links include C band, Ku band, and Ka band, with center frequencies chosen to balance atmospheric attenuation and bandwidth needs. Given Data / Assumptions: We are selecting from broad ranges, not precise allocations. Atmospheric losses are modest in C band and increase with frequency. Concept / Approach: The 3 to 6 GHz range corresponds to the C band, commonly used for satellite downlinks near 3.7–4.2 GHz and uplinks near 5.9–6.4 GHz in many regions. Lower ranges (kHz or MHz) are unsuitable for high-throughput satcom, while extremely high bands (100+ GHz) are specialized and face high atmospheric loss. Step-by-Step Solution: 1) Identify satcom-standard bands: C (≈ 4/6 GHz), Ku (≈ 12/14 GHz), Ka (≈ 20/30 GHz).2) Compare with options: only 3–6 GHz maps to a widely used band.3) Conclude the correct choice is 3 to 6 GHz. Verification / Alternative check: ITU and operator band plans document C-band allocations in this range. Why Other Options Are Wrong: 10–20 MHz and 50–70 MHz are HF/VHF, not satcom carriers. 100–120 GHz is experimental with high atmospheric attenuation. 200–300 kHz is LF. Common Pitfalls: Confusing radar bands with satcom bands; overlooking regional allocation differences. Final Answer: 3 to 6 GHz

Question 19

Phase Velocity in a Rectangular Waveguide (General Medium)  Let μr be the relative permeability and εr the relative permittivity of the filling medium. Let fc be the cutoff frequency of the chosen mode. The phase velocity vp (speed of a constant-pha…

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Introduction / Context: Unlike TEM transmission lines, waveguides support TE/TM modes with dispersive propagation. The phase velocity exceeds the medium wave speed as frequency approaches cutoff, a key concept in microwave engineering. Given Data / Assumptions: Uniform rectangular waveguide supporting a specific mode with cutoff fc. Filling medium characterized by μr and εr (lossless for this analysis). Operating frequency f > fc. Concept / Approach: The phase velocity in a waveguide is vp = ω / β. For TE/TM modes, β = k0 * sqrt(μr * εr) * sqrt(1 - (fc / f)^2), where k0 = ω / c. Therefore vp equals the medium wave speed divided by sqrt(1 - (fc/f)^2). Since the square root in the denominator is less than 1 above cutoff, vp exceeds the medium speed. Step-by-Step Solution: 1) Medium wave speed: v_med = c / sqrt(μr * εr).2) Propagation constant: β = (ω / v_med) * sqrt(1 - (fc / f)^2).3) Phase velocity: vp = ω / β = v_med / sqrt(1 - (fc / f)^2).4) Substitute v_med: vp = (c / sqrt(μr * εr)) / sqrt(1 - (fc / f)^2). Verification / Alternative check: Near cutoff (f → fc+), the denominator → 0+, so vp → ∞, consistent with dispersion theory. At high f ≫ fc, vp approaches the medium speed. Why Other Options Are Wrong: Option B inverts the dispersion; C and D misuse μrεr factors; E is dimensionally incorrect and ignores μr, εr. Common Pitfalls: Confusing phase velocity with group velocity; assuming vp cannot exceed c (it can; energy is carried at group velocity, not phase velocity). Final Answer: vp = (c / sqrt(μr * εr)) / sqrt(1 - (fc / f)^2)

Question 20

Characteristic Impedance of High-Frequency Coaxial Line  In a high-frequency coaxial transmission line, the characteristic impedance Z0 depends on which physical properties?

Accepted Answer

Introduction / Context: Coaxial lines are ubiquitous in RF systems. Their characteristic impedance Z0 determines matching, loss, and bandwidth performance. Given Data / Assumptions: Lossless approximation to reveal first-order dependencies. Coax defined by inner radius a and outer radius b, filled with dielectric of relative permittivity εr and permeability close to μr ≈ 1. Concept / Approach: For a lossless coax, Z0 = (60 / sqrt(εr)) * ln(b / a) (with μr ≈ 1). Thus Z0 depends on geometry (ratio b/a) and material permittivity εr. Both the insulating material and conductor radii matter. Step-by-Step Solution: 1) Recognize formula: Z0 ∝ ln(b/a) / sqrt(εr).2) Changing dielectric (εr) changes Z0 via 1/sqrt(εr).3) Changing geometry (b/a) alters the logarithmic factor and Z0. Verification / Alternative check: Design calculators for coax dimensions ask for εr and the two radii because both determine Z0. Why Other Options Are Wrong: Only-dielectric or only-geometry cannot fully determine Z0; braid angle affects shielding but not the first-order Z0 relation. Common Pitfalls: Assuming copper thickness or plating dominates Z0; those affect loss, not the primary impedance. Final Answer: both (a) and (b)

Microwave Communication Questions