Question 1

Traveling-Wave Tube (TWT) – Does the RF wave amplitude stay constant along the helix? In a TWT amplifier, as the electron beam interacts with the slow-wave helix and transfers energy to the RF signal, does the amplitude of the resultant traveling wa…

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Introduction: A traveling-wave tube (TWT) is a linear-beam microwave amplifier in which an electron beam exchanges energy with an RF wave that propagates on a slow-wave structure (commonly a helix). The core idea is amplification by continuous interaction; therefore, the RF amplitude does not stay constant—it increases along the device until saturation mechanisms limit further growth. Given Data / Assumptions: Slow-wave helix provides phase velocity close to the electron beam velocity. Input RF is injected at the helix input; an electron beam travels parallel to the helix. Small-signal to large-signal operation considered, including saturation effects. Concept / Approach: Velocity-modulated electrons bunch as they travel, transferring kinetic energy to the RF wave. Power conservation dictates that a decrease in beam kinetic energy corresponds to an increase in RF power. Hence the RF amplitude grows along the helix in the gain region. Only in the absence of beam–wave coupling (for example, with beam off or detuned) would the wave remain roughly constant (subject to helix loss), which is not the amplifier regime. Step-by-Step Solution: 1) Inject small RF at the helix input.2) Beam enters; synchronism causes velocity modulation and bunching.3) Bunched electrons deliver energy to the wave, causing amplitude growth toward the output.4) Near saturation, space-charge, debunching, and circuit loss limit further gain. Verification / Alternative check: Measured gain per unit length (dB/cm) and output saturation power in TWT datasheets verify increasing RF amplitude along the helix, not a constant-amplitude condition. Why Other Options Are Wrong: Option A: Contradicts the principle of amplification. Option C: Trivial condition (no beam means no gain). Option D: Refers to a different device (BWO). Option E: Losslessness does not enforce constant amplitude in the presence of energy transfer from the beam. Common Pitfalls: Confusing a passive, matched transmission line (constant amplitude absent loss) with an active beam–wave device that supplies gain. Final Answer: False.

Question 2

Rectangular waveguide – Which mode cannot propagate? For a standard hollow rectangular waveguide (perfectly conducting walls), identify the mode designation that is not supported as a propagating TE or TM mode.

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Introduction: Rectangular waveguides support a family of discrete transverse electric (TE) and transverse magnetic (TM) modes. The allowed indices depend on boundary conditions. Recognizing which index combinations are valid is essential for mode charts, cutoff calculations, and avoiding spurious propagation in microwave hardware. Given Data / Assumptions: Lossless, air-filled rectangular waveguide. Mode notation TEmn or TMmn, where m and n are nonnegative integers describing half-wave variations along the wide (a) and narrow (b) dimensions. Dominant mode is TE10 for most rectangular guides. Concept / Approach: For TE modes, at least one index must be nonzero (m, n not both zero). That allows index combinations such as TE10, TE01, TE11, etc. For TM modes, both indices must be nonzero because a TM mode requires both transverse variations to satisfy boundary conditions with Ez ≠ 0 and Hz = 0. Therefore, TMm0 or TM0n do not exist as propagating modes. Step-by-Step Solution: 1) Evaluate TE candidates: TE15 and TE12 are valid TE modes (m or n may be zero, but here both are nonzero anyway).2) Evaluate TM candidates: TM11 is valid since both indices are nonzero.3) TM10 has n = 0 → violates TM existence condition.4) Conclude TM10 cannot propagate in a hollow rectangular guide. Verification / Alternative check: Standard tables list TEmn for m,n ≥ 0 with not both zero; TMmn for m,n ≥ 1—confirming that TM10 is disallowed. Why Other Options Are Wrong: TE15, TE12, TE01, and TM11 are all permissible modes (though their cutoffs differ). Common Pitfalls: Assuming TM modes are allowed with a zero index like TE; they are not. Final Answer: TM10.

Question 3

Microwave amplification – Which device is typically used as a microwave amplifier (not an oscillator)? Select the device that is widely used to provide gain for microwave signals over a broad bandwidth in linear-beam technology.

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Introduction: Microwave systems need components that either generate RF (oscillators) or amplify existing RF (amplifiers). Knowing which classical vacuum or solid-state device provides amplification rather than oscillation is fundamental in radar, satellite links, and EW systems. Given Data / Assumptions: Comparison among common microwave devices: TWT, magnetron, reflex klystron, Gunn, IMPATT. Focus on primary/typical use, not every exotic configuration. Linear-beam vs crossed-field categories. Concept / Approach: The TWT is a linear-beam device where an electron beam continuously interacts with a slow-wave structure to transfer energy to an RF wave, providing wideband gain. Magnetrons, reflex klystrons, Gunn, and IMPATT diodes are primarily oscillators—they self-generate RF via negative resistance or crossed-field interaction, not linear amplification of an input signal. Step-by-Step Solution: 1) Identify which device provides gain to an externally supplied input RF: TWT.2) Magnetron: produces RF autonomously (oscillator).3) Reflex klystron: single-cavity oscillator using transit-time feedback.4) Gunn/IMPATT: two-terminal devices with negative resistance used as oscillators. Verification / Alternative check: Block diagrams of satellite transponders and radar transmit chains commonly show TWTAs (TWT amplifiers) as the high-power amplifier stage. Why Other Options Are Wrong: They are oscillator-class parts by design and do not provide linear amplification of an applied RF over broad bandwidths. Common Pitfalls: Confusing a reflex klystron (oscillator) with multi-cavity klystron amplifiers; assuming any device with gain can be used as an amplifier—negative resistance oscillators are generally not linear amplifiers. Final Answer: TWT (Traveling-Wave Tube).

Question 4

Reflex klystron – Transit-time condition (in RF cycles) for sustained oscillations Let n be any integer. For a reflex klystron oscillator, the electron transit time through the repeller space must satisfy what condition (expressed in number of RF cy…

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Introduction: A reflex klystron is a single-cavity microwave oscillator that depends on electron bunches returning to the cavity gap in the proper phase to deliver energy to the RF field. The required timing is specified as a transit-time condition in terms of RF cycles. Given Data / Assumptions: Single re-entrant cavity (gap) and a negatively biased repeller electrode. Anode voltage sets initial beam velocity; repeller voltage sets turnaround distance/time. Oscillation occurs in discrete “n + 3/4” modes, with integer n ≥ 0. Concept / Approach: The phase condition for sustained oscillation is commonly written in radians as ωT ≈ (2n + 3/2) * π. Dividing by 2π converts radians to cycles: T / T_rf ≈ (2n + 3/2) * π / (2π) = n + 3/4. This ensures returning electrons encounter a decelerating RF field, transferring kinetic energy to the cavity and reinforcing the oscillation. Step-by-Step Solution: 1) Start from the klystron phase (Barkhausen) condition.2) Express the round-trip transit time T in RF cycles: T/T_rf = n + 3/4.3) Interpret: 3/4 cycle additional phase ensures electrons arrive when the gap field is opposite to their velocity.4) Conclude viable oscillation modes are labeled by n (e.g., 3/4, 7/4, 11/4, ... cycles). Verification / Alternative check: Tuning curves show distinct reflector-voltage bands corresponding to different n + 3/4 modes, confirming the timing condition. Why Other Options Are Wrong: Other linear expressions (2n − 1, 2(n − 1), n + 1/2, n + 1/4) do not meet the standard phase requirement and would not consistently yield positive power transfer. Common Pitfalls: Confusing radians and cycles; forgetting that increasing repeller magnitude changes the transit time and therefore the oscillation mode and frequency. Final Answer: n + 3/4.

Question 5

Circular waveguide – Cutoff wavelength for TE11 mode in terms of diameter If D is the inner diameter of a circular waveguide, what is the cutoff wavelength λc for the TE11 mode, expressed as a multiple of D?

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Introduction: Designing circular waveguides requires accurate cutoff relations for allowed TE/TM modes. TE11 is usually the dominant mode in circular guides (lowest cutoff), analogous to TE10 in rectangular guides. Expressing λc in terms of the physical diameter D helps in quick sizing and band allocation. Given Data / Assumptions: Air-filled, perfectly conducting circular waveguide. TE11 mode uses the first root of the derivative of the Bessel function: x′11 ≈ 1.841. Radius a = D/2. Concept / Approach: For circular guides, the cutoff frequency is f_c = (c / 2π) * (x′mn / a). The corresponding cutoff wavelength is λ_c = 2πa / x′mn. Substituting TE11 data yields λc relative to diameter. Step-by-Step Solution: 1) Use λc = 2πa / x′11.2) Substitute a = D/2 → λc = 2π(D/2) / 1.841 = (πD) / 1.841.3) Compute numeric factor: π / 1.841 ≈ 1.706.4) Therefore λc ≈ 1.706 D. Verification / Alternative check: Cutoff charts for circular WG list TE11 with λc/D ≈ 1.706; higher modes have smaller λc multiples and thus higher cutoff frequencies. Why Other Options Are Wrong: 2.11 D corresponds to other roots/modes; 0.82 D and 0.41 D are too small for dominant TE11; 3.54 D is roughly 2× the correct value. Common Pitfalls: Confusing TE versus TM roots (x′mn vs xmn) or using radius a where the formula expects diameter D. Final Answer: 1.706 D.

Question 6

Directional coupler performance terminology – What two parameters are standard? In microwave engineering, which pair of terms are universally used to characterize a directional coupler's performance?

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Introduction: Directional couplers are four-port passive devices that sample a defined fraction of power traveling in one direction on a transmission line or waveguide. The two most widely cited figures of merit are how much power is sampled and how well the device discriminates between forward and reverse waves. Given Data / Assumptions: Lossless/low-loss, reciprocal passive directional coupler. Ports: input, through, coupled, and isolated. Single-frequency or narrowband characterization. Concept / Approach: Coupling (in dB) quantifies the intended fraction of forward power appearing at the coupled port, typically defined as 10 * log10(P_in / P_coupled). Directivity (in dB) measures the device’s ability to distinguish direction, defined as the ratio of power coupled from the forward wave to that from the reverse wave at the same coupled port. High directivity means minimal contamination from reflections, enabling accurate VSWR and return-loss measurements. Step-by-Step Solution: 1) Identify the sampled fraction → coupling.2) Identify forward-versus-reverse discrimination → directivity.3) Other useful parameters (insertion loss, isolation) are derived but not the two primary names requested. Verification / Alternative check: Datasheets specify “Coupling: 10 dB, 20 dB, …” and “Directivity: ≥ 25 dB, ≥ 30 dB,” confirming the standard terminology. Why Other Options Are Wrong: “Gain” applies to active devices, not passive couplers. Isolation is related but is not paired as the defining duo in specifications; noise figure is irrelevant for a passive component in this context. Common Pitfalls: Confusing directivity with isolation; directivity compares forward vs reverse coupling at the same port, whereas isolation is port-to-port leakage. Final Answer: coupling and directivity.

Question 7

Logic with microwave devices – Gunn diode and GaAs MESFET usage Consider these statements:  A Gunn diode can be used in combinational logic circuits.  A Gunn diode can be used in sequential logic circuits.  Both a Gunn diode and a GaAs MESFET can be…

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Introduction: Logic implementation at microwave and very high speeds has explored various device physics beyond mainstream silicon CMOS, including transferred-electron devices (TEDs, e.g., Gunn diodes) and GaAs field-effect transistors (MESFETs). Understanding which devices have been used for practical logic is important in the history of high-speed digital design and specialized RF control logic. Given Data / Assumptions: Gunn diodes exhibit negative differential mobility and are commonly used as oscillators. GaAs MESFETs are three-terminal transistors widely used in GaAs logic (e.g., direct-coupled FET logic). “Logic circuits” means circuits with bistability or Boolean operations at high speed, not just simple RF switching. Concept / Approach: Although two-terminal Gunn diodes lack gain in the conventional sense, their negative differential resistance and domain dynamics have been used to realize sequential logic such as bistable elements, frequency dividers, and triggered circuits where hysteresis or oscillation-to-lock behavior is exploited. Meanwhile, GaAs MESFETs underpin entire logic families (combinational and sequential), so the statement that both Gunn and GaAs MESFET can be used in logic is broadly accurate when considering specialized, non-mainstream logic implementations. Step-by-Step Solution: 1) Identify Gunn capability: negative-resistance devices can implement latching/relaxation behavior → sequential logic feasible.2) Identify GaAs MESFET logic: DCFL, SCFL, etc., implement both combinational and sequential logic.3) Assess combinational use of Gunn: limited and uncommon due to lack of isolation and gain → generally not used for robust combinational gates.4) Conclude statements 2 and 3 are acceptable, while 1 is not generally supported. Verification / Alternative check: Publications on TED-based logic demonstrate flip-flops and dividers using Gunn-effect dynamics; commercial digital IC families are GaAs MESFET-based, not Gunn-based, corroborating the conclusions. Why Other Options Are Wrong: Options A, B, and D rely on statement 1, which overstates Gunn applicability to combinational logic. Option E ignores the well-established role of GaAs MESFETs in logic. Common Pitfalls: Assuming “oscillator-only” means “no logic”; sequential logic can be built from oscillatory or hysteretic elements. Conversely, mistaking this for mainstream digital logic—Gunn-based logic is niche. Final Answer: 2 and 3 only.

Question 8

Magnetron fields – Orientation of the applied magnetic field In a conventional cylindrical cavity magnetron, what is the orientation of the externally applied magnetic field relative to the cathode axis?

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Introduction: A cavity magnetron is a crossed-field microwave oscillator. “Crossed-field” means that the static electric field (radially outward from the central cathode to the anode cavities) and the static magnetic field are orthogonal. Recognizing the correct orientation of these fields is fundamental to understanding electron motion and spoke formation in the interaction space. Given Data / Assumptions: Cylindrical geometry with a central thermionic cathode and anode block containing resonant cavities. Static radial electric field from cathode to anode. Uniform static magnetic field applied by a magnet assembly. Concept / Approach: The magnetic field is applied axially, i.e., parallel to the cathode (and tube) axis. With E radial and B axial, the Lorentz force causes electrons to follow curved paths with azimuthal drift (E × B motion). Proper magnetic field magnitude establishes synchronous interaction with the π-mode fields of the cavities, enabling efficient RF generation. Step-by-Step Solution: 1) Identify geometry: cylinder → cathode at center, anode around periphery.2) Electric field is radial due to the anode–cathode voltage.3) Choose B to be axial (parallel to the cathode axis) to create crossed-field conditions (E ⟂ B).4) Resulting electron spokes interact with the RF cavity fields to sustain oscillations. Verification / Alternative check: Construction diagrams show pole pieces above and below the anode block, producing an axial B field through the interaction space. Why Other Options Are Wrong: Perpendicular to the axis (radial or tangential B) would not produce the standard crossed-field configuration; “inclined” is not how magnetrons are biased; claiming no magnetic field contradicts the device’s operating principle. Common Pitfalls: Confusing “perpendicular to E” with “perpendicular to the axis.” In a magnetron, E is radial, B is axial (parallel to axis) so E ⟂ B even though B ∥ axis. Final Answer: parallel to the axis of cathode.

Question 9

Single-Mode Operation in an Air-Filled Rectangular Waveguide (Dominant TE10 Only)  To ensure that only the dominant TE10 mode propagates, what practical lower and upper operating frequency limits are chosen relative to the cutoff of TE10 and the nex…

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Introduction / Context: Rectangular waveguides are designed to operate in single-mode—typically the dominant TE10 mode—to avoid dispersion jumps, mode conversion, and pattern distortion. This question asks for the practical frequency window that ensures only TE10 propagates in an air-filled guide. Given Data / Assumptions: Air-filled rectangular waveguide. Dominant mode is TE10 with cutoff frequency fc(TE10). The next higher mode is commonly TE20 with cutoff fc(TE20). We seek a practical, not purely theoretical, operating band. Concept / Approach: A safe single-mode band avoids the vicinity of fc(TE10) (where attenuation is high and fields are very dispersive) and also stays below fc(TE20) to prevent multimode propagation. A widely used engineering rule is to operate from about 1.25 * fc(TE10) up to about 0.95 * fc(TE20). This builds margin for manufacturing tolerances, temperature/refractive-index variations, and network transitions. Step-by-Step Solution: 1) Lower edge: choose f ≥ ~1.25 * fc(TE10) (≈ 25% above cutoff) to reduce near-cutoff loss/dispersion.2) Upper edge: choose f ≤ ~0.95 * fc(TE20) (≈ 5% below next cutoff) to avoid exciting TE20.3) The resulting band supports only TE10 with healthy margin. Verification / Alternative check: Standard waveguide band tables (e.g., WR-series) reflect this single-mode window; connectors and components are specified so normal operation falls within these limits. Why Other Options Are Wrong: 50% or 25% gaps around both cutoffs are unnecessarily wide or mis-centered. Operating at either cutoff is impractical due to high attenuation or multi-mode onset. Common Pitfalls: Confusing cutoff frequency margins with bandwidth of a component; forgetting that higher-order modes start at their own cutoffs, not at harmonics of TE10. Final Answer: about 25% above fc for TE10 and 5% below fc for TE20 mode respectively

Question 10

Pulse Radar Ambiguity Limit from Maximum Range  A pulse radar has a maximum unambiguous range of 120 km. What is the maximum allowable pulse repetition frequency (PRF) to maintain unambiguous reception?

Accepted Answer

Introduction / Context: In pulse radar, echoes from a given pulse must return before the next pulse is transmitted to avoid range ambiguity. The pulse repetition frequency (PRF) and the maximum unambiguous range are therefore inversely related. Given Data / Assumptions: Maximum unambiguous range R_max = 120 km = 120,000 m. Propagation speed c ≈ 3 * 10^8 m/s. Two-way travel time governs the timing between pulses. Concept / Approach: The maximum unambiguous range is R_max = c / (2 * PRF). Rearranging gives PRF_max = c / (2 * R_max). This ensures the echo from the farthest target returns before the next pulse is transmitted. Step-by-Step Solution: 1) Compute denominator: 2 * R_max = 2 * 120,000 = 240,000 m.2) Compute PRF_max: PRF_max = 3 * 10^8 / 240,000 = 1250 s^-1.3) Therefore, the maximum allowable PRF for unambiguous reception is 1250 Hz. Verification / Alternative check: Checking units: meters per second divided by meters gives s^-1, i.e., Hz; the magnitude 1250 Hz corresponds to a pulse repetition interval of 0.8 ms, which matches a round-trip time to 120 km. Why Other Options Are Wrong: 330 and 625 are too low (overly conservative), while 2500 and 8330 exceed the allowable limit and would cause range ambiguities. Common Pitfalls: Forgetting the factor of 2 for the round trip; mixing Hz with kHz; using one-way time instead of two-way. Final Answer: 1250

Question 11

Microstrip vs. Stripline: Identify a Key Disadvantage of Microstrip Interconnects  Compared to embedded stripline, which drawback most commonly affects microstrip implementations on printed circuit boards?

Accepted Answer

Introduction / Context: RF designers frequently choose between microstrip (a surface conductor over a dielectric with a ground plane below) and stripline (a conductor sandwiched between two ground planes). Understanding their electromagnetic behavior is essential for low-loss, low-noise layouts. Given Data / Assumptions: Standard PCB materials and thicknesses. Comparable characteristic impedance targets. No special shielding enclosures. Concept / Approach: Microstrip is an open structure: its fields are partly in air and partly in dielectric. This makes it more susceptible to radiation and coupling to nearby structures. Stripline, being fully enclosed by ground planes, confines fields better, reducing radiation and crosstalk and often offering more consistent impedance and velocity. Step-by-Step Solution: 1) Identify open vs enclosed geometry: microstrip is open; stripline is enclosed.2) Open fields in microstrip increase fringing and radiation, especially at bends and discontinuities.3) Hence, among the listed choices, increased likelihood of radiation is the characteristic disadvantage for microstrip. Verification / Alternative check: EM simulation and near-field scanning measurements typically show higher radiation from microstrip bends, stubs, and transitions than equivalent stripline designs. Why Other Options Are Wrong: Microstrip is perfectly suited to PCB fabrication (so A is wrong), is not bulkier (C), and is generally simpler/cheaper (D). Vacuum operation (E) is irrelevant. Common Pitfalls: Assuming microstrip always has higher loss; actual loss depends on dielectric, copper finish, and geometry. Final Answer: are more likely to radiate

Question 12

Transferred-Electron (Gunn) Devices: Relevant Time Constants  Consider these time parameters in Gunn devices: (1) domain growth time constant, (2) transit time, (3) dielectric relaxation time in the positive mobility regime. Which of these are used …

Accepted Answer

Introduction / Context: Gunn diodes (transferred-electron devices) rely on formation and motion of high-field space-charge domains in materials like GaAs. Several characteristic times control whether domains form, how they evolve, and how they interact with external circuits. Given Data / Assumptions: (1) Domain growth time constant: time for a nascent perturbation to grow into a mature domain. (2) Transit time: time for the domain to traverse the active region. (3) Dielectric relaxation time (positive mobility regime): time for charge perturbations to relax via conduction and permittivity. Concept / Approach: All three times are fundamental to TED physics. Dielectric relaxation competes with negative differential mobility; when relaxation is fast, perturbations dissipate instead of growing. The domain growth time sets how quickly a domain can form once the field exceeds threshold. Transit time links device length and oscillation frequency via domain formation and collapse at the anode. Step-by-Step Solution: 1) If dielectric relaxation is too short, domains are suppressed; operation shifts to LSA-like modes.2) If growth time is long relative to transit, stable domains may not form; oscillations weaken.3) Proper device length sets transit time to target frequency ranges (GHz). Verification / Alternative check: Device design charts relate length and doping to domain velocity and the transit-time-limited frequency, while relaxation and growth times determine operating regimes (domain vs LSA). Why Other Options Are Wrong: Omitting any one of these neglects a required piece of Gunn device dynamics and leads to incorrect regime prediction. Common Pitfalls: Confusing dielectric relaxation with recombination lifetime; equating transit time with period without considering domain dynamics. Final Answer: 1, 2, and 3

Question 13

Circular Waveguide Dominant Mode Identification  In a uniform, air-filled circular waveguide of radius r, which mode is dominant (i.e., has the lowest cutoff frequency)?

Accepted Answer

Introduction / Context: Each circular waveguide mode has a specific cutoff determined by Bessel-function roots. The dominant mode is the one with the lowest cutoff frequency and is preferred for low-loss, single-mode operation. Given Data / Assumptions: Air-filled circular waveguide. Standard TE/TM mode nomenclature. Concept / Approach: The dominant mode in circular waveguide is TE11. Its cutoff depends on the first root of the derivative of J1 (often denoted x'11 or x′11 ≈ 1.841), giving a cutoff wavelength λc ≈ (2π * r) / 1.841 and hence λc ≈ 3.41 * r (equivalently λc ≈ 1.706 * D, where D = 2r). TE11 has a lower cutoff than TM01 or TE01 for the same radius. Step-by-Step Solution: 1) List candidate dominant modes: TE11, TM01, TE01, TM11.2) Compare normalized cutoff constants: the smallest corresponds to TE11.3) Conclude TE11 is dominant. Verification / Alternative check: Mode charts and design tables for circular waveguides universally show TE11 as dominant for air-filled uniform guides. Why Other Options Are Wrong: TM01 and TE01 have higher cutoff constants; TM11 is not the lowest either. Common Pitfalls: Confusing circular with rectangular guides (dominant TE10 in rectangular) or misreading TE/TM orderings. Final Answer: TE11

Question 14

Circular Polarizer Function  A circular polarizer converts an incident linearly polarized wave into a circularly polarized wave (of appropriate handedness) when aligned and tuned correctly. True or False?

Accepted Answer

Introduction / Context: Polarization control is crucial in satellite links, radar, and instrumentation. A circular polarizer is a network (waveplate, septum, or multilayer structure) that changes the polarization state of an electromagnetic wave. Given Data / Assumptions: Incident field is linearly polarized at the required 45° orientation to the polarizer axes. Device is designed for a frequency band of interest. Concept / Approach: A circular polarizer imposes a 90° phase shift between two orthogonal linear components of equal amplitude. When the input is linearly polarized at 45° to those axes, the outputs in the two eigen-axes are equal in magnitude; the quarter-wave phase delay converts the field into circular polarization (left- or right-hand depending on orientation and sign of the phase shift). Step-by-Step Solution: 1) Resolve the linear input into two orthogonal components (Ex, Ey) of equal magnitude.2) Impose a ±90° phase shift via the polarizer (quarter-wave behavior).3) The combined field has equal magnitudes and a quadrature phase, which is the definition of circular polarization. Verification / Alternative check: Stokes parameter analysis shows the output has normalized parameters consistent with circular polarization (Q ≈ 0, U ≈ 0, V ≈ ±1). Why Other Options Are Wrong: “False” contradicts the operating principle. Real devices have finite bandwidth, not zero bandwidth, and exist for waveguide and free-space implementations. Common Pitfalls: Feeding at 0° or 90° instead of 45°, leading to ellipticity or no conversion; ignoring bandwidth and tolerance limits. Final Answer: True

Question 15

Rectangular Waveguide Cutoff Wavelength for TM11  For a rectangular waveguide of width a and height b, what is the cutoff wavelength λc for the TM11 mode (air-filled, uniform guide)?

Accepted Answer

Introduction / Context: Cutoff characteristics of rectangular waveguides depend on geometry and mode indices. Correct formulas prevent design errors in frequency planning and component selection. Given Data / Assumptions: Mode TMmn with m = 1, n = 1. Uniform, air-filled rectangular guide with dimensions a (width) and b (height). Standard boundary conditions for perfect electric conductors. Concept / Approach: The general cutoff relation for TE/TM modes in a rectangular guide is λc = 2 / sqrt((m/a)^2 + (n/b)^2). Substituting m = 1 and n = 1 yields λc = 2 / sqrt((1/a)^2 + (1/b)^2). This expression assumes a and b are in meters so that λc is in meters. Step-by-Step Solution: 1) Start with λc(m,n) = 2 / sqrt((m/a)^2 + (n/b)^2).2) Plug in m = 1, n = 1 → λc = 2 / sqrt((1/a)^2 + (1/b)^2).3) Confirm dimensional consistency: a and b in meters give λc in meters. Verification / Alternative check: Equivalent forms can be derived from the cutoff frequency fc = (c/2) * sqrt((m/a)^2 + (n/b)^2); using λc = c/fc yields the same expression. Why Other Options Are Wrong: Algebraic variants including ab or π terms here are incorrect for the rectangular-guide cutoff formula; the correct relation depends on the quadratic sum of inverse dimensions. Common Pitfalls: Confusing cutoff wavelength with guide wavelength; mixing TE and TM presence/absence conditions on m and n (TM requires both m and n nonzero). Final Answer: λc = 2 / sqrt((1/a)^2 + (1/b)^2)

Question 16

VSWR Pattern with Resistive Load after Short-Termination Check  A 75 Ω line is first short-circuited and minima locations are noted. The short is then replaced by an unknown purely resistive load, and the minima locations do not change. If the measu…

Accepted Answer

Introduction / Context: Standing-wave patterns on transmission lines depend on the magnitude and phase of the reflection coefficient Γ. Comparing patterns for different terminations helps diagnose the load impedance. Given Data / Assumptions: Characteristic impedance Z0 = 75 Ω. Initial termination is a short (Γ = −1, phase = π), minima at the load. Second termination is a purely resistive load with VSWR = 3 and identical minima locations. Concept / Approach: For a real load R, Γ = (R − Z0) / (R + Z0) is real (phase 0 for R > Z0, phase π for R < Z0). The position of minima near the load is set by the phase of Γ. If minima locations are unchanged from the short, Γ must retain phase π, implying R < Z0. The VSWR relation gives |Γ| = (S − 1) / (S + 1). Step-by-Step Solution: 1) Compute |Γ| from S = 3 → |Γ| = (3 − 1) / (3 + 1) = 0.5.2) Since minima positions match the short, Γ is negative (phase π): Γ = −0.5.3) Solve (R − 75) / (R + 75) = −0.5 → R − 75 = −0.5R − 37.5 → 1.5R = 37.5 → R = 25 Ω. Verification / Alternative check: Substitute R = 25 Ω: Γ = (25 − 75) / (25 + 75) = −50 / 100 = −0.5, S = (1 + 0.5) / (1 − 0.5) = 3, consistent with observations. Why Other Options Are Wrong: 50 Ω gives Γ = −0.2 → S = 1.5; 225 Ω and 250 Ω yield positive Γ (phase 0), shifting minima; 75 Ω gives Γ = 0, no standing waves. Common Pitfalls: Ignoring the phase of Γ when comparing patterns; forgetting the VSWR–Γ magnitude relation. Final Answer: 25 Ω

Question 17

Parametric amplifier — what nonlinearity does it exploit for gain? A parametric amplifier achieves amplification by periodically varying a circuit parameter using a high-frequency pump. Which kind of nonlinearity is fundamentally required?

Accepted Answer

Introduction / Context: Parametric amplifiers are low-noise RF/microwave/optical devices that achieve gain by periodically varying a circuit parameter with a pump signal. The core idea is energy transfer from the pump to the signal via reactive, time-varying elements rather than by dissipative active devices. This question asks which kind of nonlinearity is essential for true parametric amplification. Given Data / Assumptions: Amplifier is driven by an external high-frequency pump. Objective is signal gain through a time-varying circuit parameter. Options distinguish resistive vs reactive nonlinearity. Concept / Approach: Paramps rely on a nonlinear reactance (capacitance or inductance) whose value depends on instantaneous voltage/current. Classic implementations use varactor diodes (voltage-dependent capacitance) or Josephson junction inductance. When a pump modulates the reactance, mixing occurs between pump, signal, and idler frequencies, enabling power flow from the pump to the signal without ohmic dissipation as the primary mechanism. Purely resistive nonlinearity produces conversion loss and distortion rather than low-noise gain, and it does not implement the parametric energy-exchange principle. Step-by-Step Solution: 1) Define parametric action: gain from time-varying parameter → requires reactive element whose value can be modulated.2) Identify suitable devices: varactor (C(V)) or nonlinear inductors (L(I)) supply the needed nonlinearity.3) Energy transfer pathway: pump power is converted into signal and idler via mixing relations set by the reactive nonlinearity.4) Resistive nonlinearity mainly dissipates energy and raises noise, conflicting with the low-noise aim of paramps. Verification / Alternative check: Standard small-signal models show gain equations proportional to the pump-induced variation of reactance; dissipation must be minimized to maintain the characteristic low noise temperature associated with paramps. Why Other Options Are Wrong: Non linear resistance: does not realize true parametric energy exchange and adds loss. Either (a) or (b): only reactive nonlinearity fits parametric operation. Neither: contradicts the definition of a parametric device. Common Pitfalls: Confusing mixers based on resistive elements with parametric amplifiers; assuming any nonlinearity suffices for parametric gain. Final Answer: non linear reactance

Question 18

Effect of loading a transmission line (e.g., with loading coils in telephony): what happens to Z0, the attenuation constant α, and the phase constant β? Select the correct combined qualitative effect.

Accepted Answer

Introduction / Context: In classical telephony and low-RF practice, long cables were sometimes ‘‘loaded’’ by inserting series inductance (loading coils) to improve voice-band transmission. Understanding how loading alters the characteristic impedance (Z0), the attenuation constant (α), and the phase constant (β) helps explain why loading extends useful bandwidth and reduces distortion in the intended band. Given Data / Assumptions: Loading means adding series inductance per unit length (effective L increases). Capacitance per unit length C is essentially unchanged. We consider qualitative, band-limited effects near the passband created by loading. Concept / Approach: For a low-loss line: Z0 ≈ sqrt(L/C). Adding inductive loading raises L, so Z0 increases. The attenuation constant α depends on R and G relative to the reactive terms jωL and jωC; by increasing L, the reactive impedance dominates more strongly over R in the voice band, which typically reduces α in the passband. The phase constant β ≈ ω * sqrt(L * C), so increasing L raises β for a given frequency. Hence, the combined effect is higher Z0, lower α in the designed band, and higher β. Step-by-Step Solution: 1) Z0 trend: Z0 ≈ sqrt(L/C) → increases with L.2) α trend: for low-loss conditions, stronger reactive terms reduce the relative impact of R, so α decreases over the loaded passband.3) β trend: β ≈ ω * sqrt(L*C) → increases with L.4) Therefore, the correct qualitative set is: Z0 ↑, α ↓, β ↑. Verification / Alternative check: Frequency-response plots of loaded lines show flatter amplitude and phase characteristics across the design band with reduced attenuation compared to the same cable without loading coils. Why Other Options Are Wrong: A: Claims α increases, contradicting the typical passband reduction. C: Predicts all three decrease; inconsistent with Z0 ≈ sqrt(L/C). D: Predicts Z0 and β decrease; both should increase when L increases. Common Pitfalls: Assuming loading always improves loss across all frequencies; in reality, loading shapes a passband and introduces cutoff behavior outside it. Final Answer: increase in Z0 and β but decrease in α

Question 19

Reflex klystron timing: if the magnitude of the repeller voltage is increased (i.e., made more negative), how does the electron transit time to return to the cavity gap change?

Accepted Answer

Introduction / Context: A reflex klystron is a single-cavity microwave oscillator in which electrons are emitted, pass through the cavity gap, are turned back by a negatively biased repeller electrode, and return to the gap. The transit time determines the bunching phase and thus oscillation frequency. This item asks how making the repeller voltage more negative affects that transit time. Given Data / Assumptions: Repeller electrode is biased negative relative to the cavity. Increasing repeller voltage means increasing its negative magnitude. Electron motion is ballistic between interactions, neglecting space-charge complications for first-order reasoning. Concept / Approach: A more negative repeller increases the retarding electric field. Electrons penetrate less deeply into the repeller space before being turned around, so their path length shortens. With a stronger retarding field, their turnaround occurs sooner, reducing the round-trip time to the gap. Therefore, the transit time decreases as the magnitude of the negative repeller bias is increased. Step-by-Step Solution: 1) Start with the velocity-modulated beam leaving the cavity.2) Increase |V_repeller| → stronger deceleration.3) Electrons reverse sooner → shorter trajectory.4) Shorter path length → reduced transit time. Verification / Alternative check: Reflex klystron tuning curves show frequency increasing with more negative repeller voltage over particular regions, consistent with decreased transit time setting a different bunching condition. Why Other Options Are Wrong: Increased: contradicts the physical effect of stronger repulsion. Same: ignores electric-field strength dependence. Either: not ambiguous for a fixed geometry; direction is deterministic. Common Pitfalls: Confusing sign conventions; forgetting that the relevant change is in the magnitude of negative bias, not toward zero. Final Answer: will be decreased

Question 20

Magnetron operation — order of magnetic flux density used for electron cycloidal motion Identify the correct order of the magnetic flux density used in a typical microwave magnetron.

Accepted Answer

Introduction / Context: A crossed-field magnetron uses a static electric field between cathode and anode cavities and a perpendicular magnetic field to force electrons into complex orbits that interact with RF fields. The magnetic field strength must be on the order of teslas to sustain proper interaction at microwave frequencies. The question asks for the correct order and unit identification. Given Data / Assumptions: Magnetic flux density B is measured in Wb/m^2 (tesla). Typical magnetron B fields are fractions of a tesla up to about a tesla, depending on design. Other options list incorrect units or irrelevant quantities. Concept / Approach: The physically meaningful unit for magnetic flux density is Wb/m^2 (T). Orders near 0.1–1 T are common in high-power continuous or pulsed magnetrons. Values written as m/s are velocities, not flux densities, and Wb/s is a rate of change of flux, also not magnetic flux density. Therefore, 1 Wb/m^2 correctly represents the order of B used in a magnetron. Step-by-Step Solution: 1) Identify correct unit for B → Wb/m^2 = tesla.2) Magnetron operation requires strong magnetic fields around tenths of a tesla.3) Among the options, only 1 Wb/m^2 matches both the dimension and realistic magnitude. Verification / Alternative check: Engineering references for S-band magnetrons show design B values roughly 0.1–0.3 T, consistent with the order-of-unity tesla scale used in multiple magnetron families. Why Other Options Are Wrong: 1 mWb/s: time derivative of flux, not flux density. 10^6 m/s and 10^12 m/s: linear velocities, not magnetic flux density. Common Pitfalls: Confusing B (tesla) with changing flux (dΦ/dt) or with electron velocity units. Final Answer: 1 Wb/m2

Microwave Communication Questions