Q: Antenna Diameter and Maximum Radar/Link Range If the antenna diameter D is increased fourfold (D → 4D) while other parameters remain unchanged, the maximum detectable/usable range increases by what factor?

Introduction / Context: This question examines how the physical aperture of an antenna influences maximum range in systems governed by the basic radar/link equation. Understanding how gain scales with diameter helps predict range improvements from larger dishes. Given Data / Assumptions: Dish efficiency and transmit power remain the same. Carrier frequency (hence wavelength λ) is constant. Receiver sensitivity/threshold does not change. Concept / Approach: For an aperture antenna, gain G ∝ (D/λ)^2 (assuming constant efficiency). In the basic radar equation, maximum range R_max ∝ (Pt * G^2 * σ / S_min)^(1/4) for monostatic radar, implying R_max ∝ G^(1/2) when other terms are fixed. Therefore, if G ∝ D^2, then R_max ∝ (D^2)^(1/2) = D. Step-by-Step Solution: 1) Start from aperture gain: G ∝ (D/λ)^2.2) Range scaling: R_max ∝ G^(1/2) for the monostatic case, holding other parameters constant.3) Substitute: R_max ∝ ((D/λ)^2)^(1/2) = D/λ, with λ constant, so R_max ∝ D.4) Therefore, if D becomes 4D, R_max becomes 4 * R_max (a factor of 4). Verification / Alternative check: Link-budget style reasoning for a receive-only long-haul link similarly shows SNR improving with G, which when mapped to threshold produces the same linear-in-D range scaling for fixed λ and thresholds. Why Other Options Are Wrong: 2 (either entry): Underestimates improvement; it would imply R_max ∝ sqrt(D), which is incorrect for this setup. 0.2: Nonsense direction (range would not decrease when increasing diameter). 1: Would suggest no improvement, contrary to gain scaling. Common Pitfalls: Confusing field strength (∝ 1/R) with power (∝ 1/R^2), or mixing aperture efficiency changes with pure diameter changes. Final Answer: 4

Q: Magnetron Interaction Forces: Assess the Assertion–Reason Pair Assertion (A): In the magnetron interaction space, an electron experiences three forces: electric force, magnetic force, and centrifugal force. Reason (R): Microwave generation in a mag…

Introduction / Context: Magnetrons operate with a crossed-field configuration: a static radial electric field and an axial magnetic field. Electrons emitted by the cathode interact with RF fields supported by anode cavities. This question distinguishes real (inertial-frame) forces from fictitious forces and checks the conceptual basis of crossed-field interaction. Given Data / Assumptions: Electrons are analyzed in an inertial (laboratory) frame. Fields include a dc electric field (radial) and a dc magnetic field (axial), with superposed RF fields from anode cavities. No rotating non-inertial frame is invoked. Concept / Approach: In an inertial frame, the electron is subject to the Lorentz force: F = q(E + v × B). There is no real centrifugal force unless one shifts to a rotating reference frame, in which that term is fictitious. Hence only electric and magnetic forces act physically. Meanwhile, the reason statement about microwave generation via interaction of EM fields with electrons in crossed dc fields is correct and captures the operating principle. Step-by-Step Solution: 1) Evaluate A: It lists electric, magnetic, and centrifugal forces. The centrifugal term is not a physical (inertial) force, so A is wrong.2) Evaluate R: Crossed E–B fields causing electron spoke formation and RF interaction is accurate; R is correct.3) Relationship: Since A is false but R is true, the correct choice is “A is wrong but R is correct”. Verification / Alternative check: Textbook treatments derive electron orbits solely from Lorentz force; any “centrifugal” description is a rotating-frame convenience, not an independent force. Why Other Options Are Wrong: Options A/B/C assume A is correct; it is not. Option E claims both are wrong; R is right. Common Pitfalls: Interpreting curvature of trajectories as evidence of a third physical force; conflating frame-dependent pseudo-forces with actual forces. Final Answer: A is wrong but R is correct

Q: VSWR → ∞: What Does It Say About the Load Termination? If the voltage standing-wave ratio (VSWR) on a line is infinite, which termination(s) are implied?

Introduction / Context: VSWR quantifies mismatch on a transmission line. It is related to the reflection coefficient magnitude |Γ| by VSWR = (1 + |Γ|) / (1 − |Γ|). Understanding extreme values of VSWR helps identify the nature of the terminating load. Given Data / Assumptions: Lossless (or low-loss) line near the termination. Standard definition of VSWR and reflection coefficient. Concept / Approach: VSWR tends to infinity when |Γ| → 1. An open circuit (ZL → ∞) gives Γ = +1; a short circuit (ZL = 0) gives Γ = −1; both have |Γ| = 1. Thus either open or short produces an infinite VSWR. Finite complex impedances produce |Γ| < 1 and hence finite VSWR. Step-by-Step Solution: 1) Start with VSWR relation: VSWR = (1 + |Γ|) / (1 − |Γ|).2) For open: Γ = +1 → VSWR → ∞.3) For short: Γ = −1 → VSWR → ∞.4) Any finite ZL ≠ 0, ∞ results in |Γ| < 1 (for passive lossy loads), giving finite VSWR. Verification / Alternative check: Smith chart positions at the extreme right (open) and left (short) lie on the |Γ| = 1 circle, which corresponds to infinite VSWR. Why Other Options Are Wrong: Short only (a) or open only (c) are incomplete since both cause VSWR → ∞. “Complex impedance” is vague and not necessarily |Γ| = 1. A purely reactive load of finite magnitude typically still yields |Γ| < 1 unless it is open/short. Common Pitfalls: Assuming all reactive loads give infinite VSWR; forgetting that only |Γ| = 1 causes infinite ratio. Final Answer: either (a) or (c)

Q: Impedance Inversion Techniques on Transmission Lines Which of the following realizations provides impedance inversion (Z_in ≈ Z0^2 / ZL) at a design frequency?

Introduction / Context: Impedance transformation is central to RF matching. A specific and powerful case is impedance inversion, where a load impedance is transformed roughly to Z0^2 / ZL using a transmission-line section. Given Data / Assumptions: Lossless line sections at a specific frequency. Characteristic impedance Z0 chosen for desired transformation. Single-frequency (narrowband) behavior is acceptable for design. Concept / Approach: A quarter-wave (λ/4) line produces the inversion: at electrical length 90 degrees, Z_in = (Z0^2) / ZL. This is the classic impedance inverter used in filters and matching networks. Half-wave lines repeat the load (Z_in ≈ ZL), while single stubs (open or short) mainly provide reactive compensation rather than full inversion. Step-by-Step Solution: 1) Recall λ/4 transformer relation: Z_in = Z0^2 / ZL.2) Compare: λ/2 line has Z_in ≈ ZL (apart from losses).3) Stubs supply susceptance to tune standing waves; they do not, by themselves, realize the inverter relation.4) Therefore, the quarter-wave line is the correct choice. Verification / Alternative check: Filter theory and matching texts describe quarter-wave inverters as fundamental elements in band-pass/band-stop networks. Why Other Options Are Wrong: Short/open stubs: supply reactive tuning, not inversion. Half-wave line: repeats the load. L-pad: resistive network causing loss, not ideal inversion. Common Pitfalls: Assuming any 90-degree stub acts as an inverter; only the through line of 90 degrees with Z0 set appropriately provides Z0^2 / ZL transformation. Final Answer: a quarter wave line

Q: Circular Waveguide Modes: Identify the Dominant (Lowest-Cutoff) Mode True or False: In a circular waveguide, the TE21 mode has the lowest cutoff frequency.

Introduction / Context: Circular waveguides support discrete TE and TM modes, each with a specific cutoff frequency determined by boundary conditions and Bessel function roots. Identifying the dominant mode is crucial in design. Given Data / Assumptions: Uniform, air-filled circular waveguide. Standard mode nomenclature TE_mn, TM_mn. Concept / Approach: The dominant (lowest-cutoff) mode in a circular waveguide is TE11, not TE21. The cutoff for each mode is proportional to the corresponding root divided by the diameter. TE21 has a higher-order root and therefore a higher cutoff frequency than TE11. Step-by-Step Solution: 1) List candidates: TE11, TM01, TE21, etc.2) Compare normalized cutoff constants: the smallest corresponds to TE11.3) Conclude: TE21 does not have the lowest cutoff; the statement is false. Verification / Alternative check: Mode charts show TE11 as dominant with λ_c ≈ 1.706 * D; TE21 has a larger constant and thus higher cutoff. Why Other Options Are Wrong: “True” variants contradict standard charts. “TM01 always dominant” is also incorrect; TM01 usually has higher cutoff than TE11 in air-filled circular guides. Common Pitfalls: Confusing circular with rectangular guides (dominant TE10 in rectangular) or mixing TE/TM ordering. Final Answer: False

Q: High-Frequency Line Parameters: Inductive Reactance vs AC Resistance Assertion (A): At high frequencies, the line's inductive reactance is very large compared to its ac resistance. Reason (R): Due to skin effect, the ac resistance of a conductor is…

Introduction / Context: Distributed-parameter lines exhibit frequency-dependent reactance and resistance. This item tests understanding of how inductive reactance and skin-effect resistance scale with frequency and whether one explains the other. Given Data / Assumptions: Inductance per unit length L is approximately constant with frequency over a wide range. Inductive reactance X_L = ω * L increases linearly with frequency. AC resistance R_ac rises with frequency due to skin effect. Concept / Approach: At high frequencies, X_L commonly exceeds R_ac, so the line is more inductive than resistive. Skin effect makes R_ac > R_dc, but that fact does not cause X_L to be large; rather, X_L is large because ω is large. Thus A is true, R is true, but R does not explain A. Step-by-Step Solution: 1) Evaluate A: With increasing frequency, X_L = ωL grows, and typically X_L ≫ R_ac for RF conductors.2) Evaluate R: Skin depth decreases with frequency, shrinking effective cross-section and raising R_ac versus R_dc, so R is true.3) Causality check: The largeness of X_L is due to ωL, not because R_ac increased; hence R does not explain A. Verification / Alternative check: Measurements on coax or twin-lead at RF show reactance dominating series resistance in many bands, despite rising R_ac. Why Other Options Are Wrong: Option A: Incorrect causal link. Option C/D/E conflict with correctness of A and R. Common Pitfalls: Attributing frequency behavior of reactance to resistive mechanisms; ignoring that both R_ac and X_L increase with frequency but for different physical reasons. Final Answer: Both A and R are correct but R is not correct explanation of A

Question 1

VSWR of a perfectly short-circuited line A transmission line with characteristic impedance Z0 = 300 ∠ 0 Ω is terminated in a perfect short circuit. What is the standing wave ratio (VSWR) on the line?

Accepted Answer

Introduction: VSWR quantifies the severity of standing waves caused by impedance mismatch. Extreme cases—open or short circuits—yield the largest possible mismatch magnitude. Recognizing these limits is essential for interpreting line measurements and avoiding damage due to high voltage/current antinodes. Given Data / Assumptions: Lossless/low-loss line with Z0 = 300 Ω (purely real). Load is an ideal short circuit: ZL = 0 Ω. Single-frequency steady-state excitation. Concept / Approach: The reflection coefficient is Γ = (ZL − Z0)/(ZL + Z0). For ZL = 0, Γ = −1, so |Γ| = 1. VSWR relates to |Γ| by VSWR = (1 + |Γ|)/(1 − |Γ|). When |Γ| = 1, the denominator becomes zero, implying VSWR → ∞. Physically, voltage nodes and antinodes repeat every λ/2 with zero voltage at the load and very large maxima along the line (limited by losses and source constraints). Step-by-Step Solution: 1) Compute Γ for a short: Γ = −1.2) Take magnitude: |Γ| = 1.3) Substitute into VSWR formula → VSWR = ∞. Verification / Alternative check: Smith Chart: a short is at the leftmost extreme; the SWR circle radius corresponds to |Γ| = 1, confirming infinite VSWR. Why Other Options Are Wrong: VSWR of 1 denotes a perfect match; finite values like 2 or 10 imply partial mismatch; 0 is nonphysical for SWR. Common Pitfalls: Confusing SWR with return loss; thinking that “very large” is still finite for an ideal short—mathematically it is infinite. Final Answer: ∞.

Question 2

Reflex klystron oscillator – Number of resonant cavities used In a reflex klystron microwave oscillator, how many resonant cavities are employed in the RF interaction structure?

Accepted Answer

Introduction: Klystron families include multi-cavity klystron amplifiers and reflex klystron oscillators. The reflex type achieves oscillation by reflecting the electron beam back through the same gap so that it interacts twice with a single resonator, establishing the correct phase for sustained oscillations. Given Data / Assumptions: Reflex klystron with cathode, input/output cavity (single gap), and negatively biased reflector (repeller). Beam traverses the cavity region, is turned around by the reflector, and re-traverses the gap. Proper reflector voltage sets the transit-time condition (n + 3/4 mode) for oscillation. Concept / Approach: Unlike multi-cavity amplifiers (separate buncher and catcher), the reflex klystron uses one resonant cavity. Energy exchange occurs twice with the same cavity as the bunched beam returns. Frequency is primarily determined by the cavity resonance and the reflector bias that fixes transit time. Step-by-Step Solution: 1) Identify device class: reflex (oscillator) rather than multi-cavity amplifier.2) Note presence of a single resonant cavity gap.3) Recognize role of reflector in providing feedback and phase synchronization. Verification / Alternative check: Standard reflex klystron schematics label a single re-entrant cavity and a repeller electrode; additional cavities are not used. Why Other Options Are Wrong: Two/three cavities describe other klystron types; “none” is impossible since a resonator is required to define frequency and store RF energy. Common Pitfalls: Confusing reflex klystrons with multi-cavity klystron amplifiers; misattributing multiple gaps to oscillators. Final Answer: one cavity resonator.

Question 3

Gunn devices and GaAs – Negative differential mobility statement Gallium arsenide (GaAs) is known for intervalley transfer under high electric field. State whether GaAs exhibits negative differential mobility (basis of the Gunn effect).

Accepted Answer

Introduction: Negative differential mobility (NDM) occurs when carrier drift velocity decreases with increasing electric field beyond a threshold. In certain III-V semiconductors, such as GaAs, intervalley transfer moves electrons from a low-mass, high-mobility valley to a higher-mass, lower-mobility satellite valley, causing NDM and enabling Gunn oscillations. Given Data / Assumptions: n-type GaAs under high electric fields (typically a few kV/cm). Room-temperature operation is of interest. Bulk transport dominated by intervalley transfer (Ridley–Watkins–Hilsum mechanism). Concept / Approach: As the electric field rises above threshold, electron population shifts to satellite valleys with larger effective mass. The average drift velocity peaks and then falls with further field increase, giving negative differential mobility (dv_d/dE < 0). Spatial domains form and drift across the sample, generating microwave oscillations without a traditional p-n junction. Step-by-Step Solution: 1) Increase field to exceed threshold → intervalley transfer begins.2) Mobility (and thus drift velocity for a given field) decreases with additional field → NDM region.3) Device structures exploit this via appropriate length and bias to sustain moving high-field domains → Gunn oscillation. Verification / Alternative check: Velocity-field characteristics of GaAs show a peak near the threshold field followed by a decline, confirming NDM at room temperature in n-type material. Why Other Options Are Wrong: “False” contradicts well-known transport curves; limiting to <1 GHz or p-type is incorrect; cryogenic temperatures are not required for Gunn operation in GaAs. Common Pitfalls: Confusing negative differential resistance in tunnel diodes with NDM in bulk GaAs; overlooking the need for n-type doping and device length criteria. Final Answer: True.

Question 4

Microwave Link Planning: Why Are Terrestrial Microwave Repeaters Typically Spaced About 50 km Apart?  Consider the main physical and system-level reasons that set practical hop length limits for point-to-point microwave links operating over the Eart…

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Introduction / Context: Terrestrial microwave links are designed as line-of-sight hops between towers. A common rule of thumb is that hops are roughly a few tens of kilometers (often near 30–60 km), with 50 km frequently cited. This question probes the dominant limiting factor for long, ground-based paths. Given Data / Assumptions: Frequency is in the microwave band where propagation is essentially line-of-sight with modest diffraction. Earth is curved, so the radio horizon is finite even with tall towers. Standard refractivity (k-factor around 4/3) is assumed for effective Earth radius in path design. Concept / Approach: The maximum path length is fundamentally constrained by geometry: the Earth's curvature hides distant terrain and reduces clearance through the Fresnel zone. Even if transmit power and receiver sensitivity are favorable, insufficient geometric clearance causes strong diffraction loss and fading. Atmospheric attenuation and equipment limits matter, but geometry dominates the first-order spacing choice. Step-by-Step Solution: 1) Radio horizon for a tower of height h (meters) is approximately 3.57 * (sqrt(h1) + sqrt(h2)) in km (using the 4/3 Earth assumption).2) For typical tower heights (e.g., tens of meters), the sum of horizons tends to a few tens of kilometers, often around 40–60 km.3) Adequate Fresnel zone clearance requires additional margin; this often fixes practical spacing near 50 km for many deployments.4) Therefore, Earth's curvature (and associated clearance) is the primary reason for typical 50 km spacing. Verification / Alternative check: Link budgets that ignore curvature predict longer ranges, but path profiles including terrain and bulge show clearance loss. Raising tower heights or adding passive/active repeaters is used to extend range, confirming the geometric limit. Why Other Options Are Wrong: Atmospheric attenuation: Significant at high rain rates or very high bands, but not the first-order limiter at ~50 km. Output tube power: Modern solid-state radios and dishes mitigate this; geometry still dominates. Excessive dc voltage: Not relevant to hop distance selection. Common Pitfalls: Confusing free-space path loss with geometric blocking; neglecting Fresnel zone clearance and k-factor variability. Final Answer: because of earth's curvature

Question 5

Antenna Diameter and Maximum Radar/Link Range  If the antenna diameter D is increased fourfold (D → 4D) while other parameters remain unchanged, the maximum detectable/usable range increases by what factor?

Accepted Answer

Introduction / Context: This question examines how the physical aperture of an antenna influences maximum range in systems governed by the basic radar/link equation. Understanding how gain scales with diameter helps predict range improvements from larger dishes. Given Data / Assumptions: Dish efficiency and transmit power remain the same. Carrier frequency (hence wavelength λ) is constant. Receiver sensitivity/threshold does not change. Concept / Approach: For an aperture antenna, gain G ∝ (D/λ)^2 (assuming constant efficiency). In the basic radar equation, maximum range R_max ∝ (Pt * G^2 * σ / S_min)^(1/4) for monostatic radar, implying R_max ∝ G^(1/2) when other terms are fixed. Therefore, if G ∝ D^2, then R_max ∝ (D^2)^(1/2) = D. Step-by-Step Solution: 1) Start from aperture gain: G ∝ (D/λ)^2.2) Range scaling: R_max ∝ G^(1/2) for the monostatic case, holding other parameters constant.3) Substitute: R_max ∝ ((D/λ)^2)^(1/2) = D/λ, with λ constant, so R_max ∝ D.4) Therefore, if D becomes 4D, R_max becomes 4 * R_max (a factor of 4). Verification / Alternative check: Link-budget style reasoning for a receive-only long-haul link similarly shows SNR improving with G, which when mapped to threshold produces the same linear-in-D range scaling for fixed λ and thresholds. Why Other Options Are Wrong: 2 (either entry): Underestimates improvement; it would imply R_max ∝ sqrt(D), which is incorrect for this setup. 0.2: Nonsense direction (range would not decrease when increasing diameter). 1: Would suggest no improvement, contrary to gain scaling. Common Pitfalls: Confusing field strength (∝ 1/R) with power (∝ 1/R^2), or mixing aperture efficiency changes with pure diameter changes. Final Answer: 4

Question 6

Magnetron Interaction Forces: Assess the Assertion–Reason Pair  Assertion (A): In the magnetron interaction space, an electron experiences three forces: electric force, magnetic force, and centrifugal force. Reason (R): Microwave generation in a mag…

Accepted Answer

Introduction / Context: Magnetrons operate with a crossed-field configuration: a static radial electric field and an axial magnetic field. Electrons emitted by the cathode interact with RF fields supported by anode cavities. This question distinguishes real (inertial-frame) forces from fictitious forces and checks the conceptual basis of crossed-field interaction. Given Data / Assumptions: Electrons are analyzed in an inertial (laboratory) frame. Fields include a dc electric field (radial) and a dc magnetic field (axial), with superposed RF fields from anode cavities. No rotating non-inertial frame is invoked. Concept / Approach: In an inertial frame, the electron is subject to the Lorentz force: F = q(E + v × B). There is no real centrifugal force unless one shifts to a rotating reference frame, in which that term is fictitious. Hence only electric and magnetic forces act physically. Meanwhile, the reason statement about microwave generation via interaction of EM fields with electrons in crossed dc fields is correct and captures the operating principle. Step-by-Step Solution: 1) Evaluate A: It lists electric, magnetic, and centrifugal forces. The centrifugal term is not a physical (inertial) force, so A is wrong.2) Evaluate R: Crossed E–B fields causing electron spoke formation and RF interaction is accurate; R is correct.3) Relationship: Since A is false but R is true, the correct choice is “A is wrong but R is correct”. Verification / Alternative check: Textbook treatments derive electron orbits solely from Lorentz force; any “centrifugal” description is a rotating-frame convenience, not an independent force. Why Other Options Are Wrong: Options A/B/C assume A is correct; it is not. Option E claims both are wrong; R is right. Common Pitfalls: Interpreting curvature of trajectories as evidence of a third physical force; conflating frame-dependent pseudo-forces with actual forces. Final Answer: A is wrong but R is correct

Question 7

VSWR → ∞: What Does It Say About the Load Termination?  If the voltage standing-wave ratio (VSWR) on a line is infinite, which termination(s) are implied?

Accepted Answer

Introduction / Context: VSWR quantifies mismatch on a transmission line. It is related to the reflection coefficient magnitude |Γ| by VSWR = (1 + |Γ|) / (1 − |Γ|). Understanding extreme values of VSWR helps identify the nature of the terminating load. Given Data / Assumptions: Lossless (or low-loss) line near the termination. Standard definition of VSWR and reflection coefficient. Concept / Approach: VSWR tends to infinity when |Γ| → 1. An open circuit (ZL → ∞) gives Γ = +1; a short circuit (ZL = 0) gives Γ = −1; both have |Γ| = 1. Thus either open or short produces an infinite VSWR. Finite complex impedances produce |Γ| < 1 and hence finite VSWR. Step-by-Step Solution: 1) Start with VSWR relation: VSWR = (1 + |Γ|) / (1 − |Γ|).2) For open: Γ = +1 → VSWR → ∞.3) For short: Γ = −1 → VSWR → ∞.4) Any finite ZL ≠ 0, ∞ results in |Γ| < 1 (for passive lossy loads), giving finite VSWR. Verification / Alternative check: Smith chart positions at the extreme right (open) and left (short) lie on the |Γ| = 1 circle, which corresponds to infinite VSWR. Why Other Options Are Wrong: Short only (a) or open only (c) are incomplete since both cause VSWR → ∞. “Complex impedance” is vague and not necessarily |Γ| = 1. A purely reactive load of finite magnitude typically still yields |Γ| < 1 unless it is open/short. Common Pitfalls: Assuming all reactive loads give infinite VSWR; forgetting that only |Γ| = 1 causes infinite ratio. Final Answer: either (a) or (c)

Question 8

Impedance Inversion Techniques on Transmission Lines  Which of the following realizations provides impedance inversion (Z_in ≈ Z0^2 / ZL) at a design frequency?

Accepted Answer

Introduction / Context: Impedance transformation is central to RF matching. A specific and powerful case is impedance inversion, where a load impedance is transformed roughly to Z0^2 / ZL using a transmission-line section. Given Data / Assumptions: Lossless line sections at a specific frequency. Characteristic impedance Z0 chosen for desired transformation. Single-frequency (narrowband) behavior is acceptable for design. Concept / Approach: A quarter-wave (λ/4) line produces the inversion: at electrical length 90 degrees, Z_in = (Z0^2) / ZL. This is the classic impedance inverter used in filters and matching networks. Half-wave lines repeat the load (Z_in ≈ ZL), while single stubs (open or short) mainly provide reactive compensation rather than full inversion. Step-by-Step Solution: 1) Recall λ/4 transformer relation: Z_in = Z0^2 / ZL.2) Compare: λ/2 line has Z_in ≈ ZL (apart from losses).3) Stubs supply susceptance to tune standing waves; they do not, by themselves, realize the inverter relation.4) Therefore, the quarter-wave line is the correct choice. Verification / Alternative check: Filter theory and matching texts describe quarter-wave inverters as fundamental elements in band-pass/band-stop networks. Why Other Options Are Wrong: Short/open stubs: supply reactive tuning, not inversion. Half-wave line: repeats the load. L-pad: resistive network causing loss, not ideal inversion. Common Pitfalls: Assuming any 90-degree stub acts as an inverter; only the through line of 90 degrees with Z0 set appropriately provides Z0^2 / ZL transformation. Final Answer: a quarter wave line

Question 9

Circular Waveguide Modes: Identify the Dominant (Lowest-Cutoff) Mode  True or False: In a circular waveguide, the TE21 mode has the lowest cutoff frequency.

Accepted Answer

Introduction / Context: Circular waveguides support discrete TE and TM modes, each with a specific cutoff frequency determined by boundary conditions and Bessel function roots. Identifying the dominant mode is crucial in design. Given Data / Assumptions: Uniform, air-filled circular waveguide. Standard mode nomenclature TE_mn, TM_mn. Concept / Approach: The dominant (lowest-cutoff) mode in a circular waveguide is TE11, not TE21. The cutoff for each mode is proportional to the corresponding root divided by the diameter. TE21 has a higher-order root and therefore a higher cutoff frequency than TE11. Step-by-Step Solution: 1) List candidates: TE11, TM01, TE21, etc.2) Compare normalized cutoff constants: the smallest corresponds to TE11.3) Conclude: TE21 does not have the lowest cutoff; the statement is false. Verification / Alternative check: Mode charts show TE11 as dominant with λ_c ≈ 1.706 * D; TE21 has a larger constant and thus higher cutoff. Why Other Options Are Wrong: “True” variants contradict standard charts. “TM01 always dominant” is also incorrect; TM01 usually has higher cutoff than TE11 in air-filled circular guides. Common Pitfalls: Confusing circular with rectangular guides (dominant TE10 in rectangular) or mixing TE/TM ordering. Final Answer: False

Question 10

High-Frequency Line Parameters: Inductive Reactance vs AC Resistance  Assertion (A): At high frequencies, the line's inductive reactance is very large compared to its ac resistance. Reason (R): Due to skin effect, the ac resistance of a conductor is…

Accepted Answer

Introduction / Context: Distributed-parameter lines exhibit frequency-dependent reactance and resistance. This item tests understanding of how inductive reactance and skin-effect resistance scale with frequency and whether one explains the other. Given Data / Assumptions: Inductance per unit length L is approximately constant with frequency over a wide range. Inductive reactance X_L = ω * L increases linearly with frequency. AC resistance R_ac rises with frequency due to skin effect. Concept / Approach: At high frequencies, X_L commonly exceeds R_ac, so the line is more inductive than resistive. Skin effect makes R_ac > R_dc, but that fact does not cause X_L to be large; rather, X_L is large because ω is large. Thus A is true, R is true, but R does not explain A. Step-by-Step Solution: 1) Evaluate A: With increasing frequency, X_L = ωL grows, and typically X_L ≫ R_ac for RF conductors.2) Evaluate R: Skin depth decreases with frequency, shrinking effective cross-section and raising R_ac versus R_dc, so R is true.3) Causality check: The largeness of X_L is due to ωL, not because R_ac increased; hence R does not explain A. Verification / Alternative check: Measurements on coax or twin-lead at RF show reactance dominating series resistance in many bands, despite rising R_ac. Why Other Options Are Wrong: Option A: Incorrect causal link. Option C/D/E conflict with correctness of A and R. Common Pitfalls: Attributing frequency behavior of reactance to resistive mechanisms; ignoring that both R_ac and X_L increase with frequency but for different physical reasons. Final Answer: Both A and R are correct but R is not correct explanation of A

Question 11

Broadband Matching on Transmission Lines: Best Practical Choice  For impedance matching over a range of frequencies (wider than a single spot frequency), which of the following is generally preferred?

Accepted Answer

Introduction / Context: Matching networks on transmission lines can be narrowband or relatively broadband. The choice of element affects how the match holds over frequency changes, which matters in wideband communications and instrumentation. Given Data / Assumptions: Goal: One network that provides acceptable match over a frequency range, not just at a single design frequency. Passive, transmission-line based components are used. Concept / Approach: A single shunt or series stub placed at a specific location precisely cancels reactance at one frequency; the match rapidly degrades away from that frequency. A double-stub tuner (two reactive elements separated along the line) offers extra degrees of freedom and can maintain a better match across a modest band, making it the practical choice among the listed options. Step-by-Step Solution: 1) Evaluate single-stub: Excellent for spot-frequency match; narrowband.2) Evaluate double-stub: Two-tuning elements allow improved broadband behavior and easier placement when load position is constrained.3) Balun: Balances/unbalances lines, not a broadband matching tool by itself.4) Directional coupler: A measurement/splitting device, not a primary impedance-matching network.5) Fixed quarter-wave transformer: Precise at one frequency; narrowband due to electrical length dispersion. Verification / Alternative check: Smith chart synthesis of double-stub networks shows wider acceptable SWR bandwidth compared to single-stub or fixed λ/4 transformers for many loads. Why Other Options Are Wrong: Balun and coupler solve different problems; single-stub and λ/4 are narrowband by nature. Common Pitfalls: Overestimating bandwidth of a single reactive tuner; ignoring that true broadbanding often needs multi-section transformers or distributed-element filters beyond two-stub capability. Final Answer: double stub

Question 12

Traveling-Wave Tube (TWT): Identify Correct Statements  Consider the following statements about a TWT: 1) It uses thermionic emission. 2) It uses an attenuator inside the slow-wave structure. 3) It is inherently a resonant device. 4) It has broad ba…

Accepted Answer

Introduction / Context: A TWT is a linear-beam vacuum device that amplifies RF by continuous interaction between an electron beam and an RF wave on a slow-wave structure (often a helix). It is commonly used where high gain and wide bandwidth are needed. Given Data / Assumptions: Statements concern emission mechanism, internal stabilization, bandwidth, and resonance. We consider a conventional helix TWT. Concept / Approach: 1) Thermionic emission is used to generate the electron beam. 2) An attenuator is inserted along the helix to suppress feedback and oscillation. 3) A TWT is not inherently resonant like a klystron cavity oscillator; it is a traveling-wave amplifier. 4) One of the main virtues of TWTs is their broad bandwidth. Step-by-Step Solution: Evaluate 1: True—electron guns rely on thermionic cathodes.Evaluate 2: True—attenuator prevents self-oscillation by absorbing backward waves.Evaluate 3: False—TWTs are non-resonant broad-band devices; gain is not set by discrete cavity resonance.Evaluate 4: True—slow-wave interaction yields wideband amplification. Verification / Alternative check: Standard references list internal attenuators and wideband response as defining features distinguishing TWTs from multi-cavity klystrons. Why Other Options Are Wrong: Any option including statement 3 is incorrect. Options excluding statement 2 or 4 omit correct facts about TWT design. Common Pitfalls: Confusing TWTs with cavity-based devices; assuming all vacuum devices are narrowband resonators. Final Answer: 1, 2 and 4

Question 13

Gunn Diode Fundamentals: Assertion–Reason  Assertion (A): A Gunn diode is a transferred-electron device. Reason (R): A Gunn oscillator exploits the transferred-electron effect to generate microwave oscillations.

Accepted Answer

Introduction / Context: Gunn diodes are widely used microwave sources in the centimeter/millimeter bands. Their physics is fundamentally different from ordinary pn diodes because the effect arises from bulk transport in certain semiconductors. Given Data / Assumptions: Material has multiple valleys in its conduction band (e.g., GaAs, InP). Under appropriate electric fields, electrons transfer to a higher-mass satellite valley. Negative differential mobility/negative differential resistance can occur. Concept / Approach: The transferred-electron effect shifts carriers from a low-mass, high-mobility valley to a high-mass, lower-mobility valley when the electric field exceeds a threshold. This reduces drift velocity with increasing field, creating negative differential resistance that supports oscillation. A Gunn “diode” has no pn junction; oscillation comes from domain formation in the bulk and interaction with an external resonant circuit or cavity. Step-by-Step Solution: 1) Assess A: Gunn devices operate by transferred-electron physics—true.2) Assess R: Gunn oscillators indeed use this effect to achieve microwave generation—true.3) Causality: R accurately explains why we classify Gunn devices as transferred-electron devices and how they oscillate. Verification / Alternative check: V–E (velocity–field) characteristics for GaAs show a peak then a decline, confirming negative differential mobility consistent with the transferred-electron effect. Why Other Options Are Wrong: Any option denying A or R conflicts with established device physics; saying R does not explain A ignores the direct link between the effect and the device category. Common Pitfalls: Assuming a pn junction exists; conflating tunnel diodes (quantum tunneling) with Gunn diodes (bulk transferred-electron effect). Final Answer: Both A and R are correct and R is correct explanation of A

Question 14

Reflex klystron oscillator — device classification and typical performance Identify which description best characterizes a reflex klystron used as a microwave source (single-cavity with repeller).

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Introduction / Context: Reflex klystrons are single-cavity vacuum microwave oscillators that use a negatively biased repeller to produce electron bunching and sustain RF oscillations. They are valued for simplicity and tunability, not for power or efficiency. This item checks whether you recognize their typical power and efficiency class. Given Data / Assumptions: Device: reflex klystron oscillator (single cavity + repeller). Application domain: laboratory sources, low-to-moderate output. Compare notions of power capability and efficiency against other microwave tubes. Concept / Approach: Reflex klystrons operate by velocity modulation in one cavity; electrons are turned back by the repeller and arrive at the gap in phase with the RF field. With only one interaction gap and no multi-cavity energy extraction, conversion efficiency is limited, and output power is modest compared to multi-cavity klystrons, traveling-wave tubes, or magnetrons. Step-by-Step Solution: 1) Identify the architecture: single cavity + repeller → simple, compact, tunable.2) Recognize energy extraction is limited → modest RF output.3) Typical efficiencies are comparatively low; power levels are in the low-watt to tens-of-watts range in many lab units.4) Therefore, it is accurately described as a low power device. Verification / Alternative check: Compare with two-cavity or multi-cavity klystron amplifiers and TWTs, which deliver higher power and efficiency due to multiple interaction regions and optimized extraction cavities. Why Other Options Are Wrong: High power device: contradicts the known limits of single-cavity extraction. High efficiency device: efficiency is limited; not a strong trait of reflex klystrons. Both (a) and (b): mutually inconsistent; cannot be low and high power simultaneously. Common Pitfalls: Confusing reflex klystrons with multi-cavity klystrons or magnetrons, which can achieve much higher power. Final Answer: low power device

Question 15

Magic-tee (hybrid-T) — equivalent circuit coefficients under matched conditions If all four ports of a magic-tee are perfectly matched and isolated as per its ideal equivalent circuit, what are the magnitude coefficients at the E- and H-plane arms?

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Introduction / Context: A magic-tee (hybrid-T) combines an E-plane tee and an H-plane tee, giving one sum port, one difference port, and two collinear ports. In its ideal, matched form, it behaves like a 3 dB power divider/combiner with specific voltage (or field) coupling coefficients. This question tests recognition of those ideal coefficients. Given Data / Assumptions: All ports are perfectly matched and isolated as per the ideal model. Power is split equally between appropriate arms in the ideal case. Lossless operation. Concept / Approach: In the ideal magic-tee, equal split implies each coupled arm receives half the power. Half-power corresponds to a field (or voltage) magnitude factor of 1/√2 relative to the incident wave. Numerically, 1/√2 ≈ 0.707, which serves as the normalized coefficient in the equivalent circuit or scattering matrix description. Step-by-Step Solution: 1) Equal split → each output gets 50% of power.2) Voltage/field scaling for half power is 1/√2.3) Thus nE = nH = 1/√2 ≈ 0.707 under ideal matched conditions. Verification / Alternative check: From the canonical S-matrix of an ideal 3 dB hybrid, the relevant scattering magnitudes are 1/√2 with appropriate phase relationships and port isolation. Why Other Options Are Wrong: 2: Would imply quadruple power, which violates conservation in a passive divider. Mixed 2 and 0.707: Inconsistent with symmetry and equal split. Common Pitfalls: Confusing power ratios with voltage ratios; forgetting that 3 dB corresponds to a voltage factor of 0.707, not 0.5. Final Answer: nE = nH = 0.707

Question 16

Rectangular waveguide TE10 — maximum power handling expression Given Z_TE as the wave impedance for TE waves, E_d as the maximum dielectric strength of the insulating medium, and a (broad wall width) and b (narrow wall height) as the guide dimension…

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Introduction / Context: Waveguide power handling is limited by the maximum electric field before breakdown. For a rectangular waveguide operating in the dominant TE10 mode, there is a known relationship between the peak electric field amplitude and the time-average power flow. This item asks you to identify the correct formula using Z_TE and breakdown strength E_d. Given Data / Assumptions: Mode: TE10 in a rectangular waveguide. Dimensions: a (broad wall), b (narrow wall). Z_TE: wave impedance for the TE mode at the operating frequency. E_d: maximum allowable electric field (dielectric strength) of the filling medium. Lossless, steady-state sinusoidal fields. Concept / Approach: For TE10, the transverse electric field varies as cos(pi x/a) across the broad wall, with a maximum field magnitude E_max. The average power flow P in a lossless guide can be written in terms of E_max as P = (a * b / 4) * (E_max^2 / Z_TE). Setting E_max = E_d at the threshold gives the maximum permissible power P_max. The 1/4 factor arises from spatial averaging of the field distribution across the guide cross-section. Step-by-Step Solution: 1) Start from the general TE10 average power expression: P = (a * b / 4) * (E_max^2 / Z_TE).2) At breakdown threshold, E_max → E_d.3) Substitute to obtain P_max = (a * b / 4) * (E_d^2 / Z_TE). Verification / Alternative check: Dimensional analysis: E^2 / Z_TE has dimensions of power density; multiplying by area ab yields power. The quarter factor is consistent with integrating cos^2(pi x/a) over the width. Why Other Options Are Wrong: (ab/2)(E_d^2Z_TE): incorrectly places Z_TE in numerator and has the wrong factor. (ab)(E_d/Z_TE): linear in E_d rather than quadratic; dimensionally inconsistent. (a/b)*(E_d^2/Z_TE): incorrect geometric dependence; power scales with cross-sectional area, not aspect ratio alone. Common Pitfalls: Forgetting the spatial averaging factor; mixing up Z_TE in numerator vs denominator; ignoring mode dependence of field distribution. Final Answer: P_max = (a * b / 4) * (E_d^2 / Z_TE)

Question 17

Assertion (A): An IMPATT diode can be used in both amplifiers and oscillators at microwave frequencies. Reason (R): An IMPATT diode works because it has a low resistance. Choose the correct relationship between A and R.

Accepted Answer

Introduction / Context: IMPATT (Impact Avalanche Transit-Time) diodes are widely used for microwave generation and amplification. The principle relies on avalanche multiplication and carrier transit-time phase delay that together yield negative resistance at RF, enabling gain and oscillation. This item probes whether you can distinguish the true operating reason from an incorrect simplification. Given Data / Assumptions: Device operates in avalanche mode with a drift region. Applications include oscillators (free-running or injection-locked) and amplifiers. Evaluation is about correctness and causality of A and R. Concept / Approach: IMPATT diodes exhibit negative differential resistance at microwave frequencies due to the combined phase shift from avalanche build-up and carrier transit delay. This negative resistance supports oscillations and provides gain in amplifier topologies. Merely having a “low resistance” does not explain these behaviors and is not the operative mechanism. Step-by-Step Solution: 1) A is true: IMPATTs are used in both oscillators and amplifiers.2) R claims “low resistance” as the reason; this is incorrect. The key is negative resistance from avalanche + transit time.3) Therefore, the correct evaluation is A true, R false. Verification / Alternative check: Small-signal models show a negative real part of impedance in the intended frequency range; practical circuits bias for avalanche and tune for the desired frequency. Why Other Options Are Wrong: Options implying R is correct contradict device physics. Common Pitfalls: Confusing negative resistance with merely small ohmic resistance; overlooking the role of impact ionization timing. Final Answer: A is correct but R is wrong

Question 18

Assertion (A): Microstrip is very commonly used in microwave integrated circuits (MICs) and RF printed-circuit designs. Reason (R): Microstrip provides easy top-surface access, allowing active and passive discrete components to be conveniently mount…

Accepted Answer

Introduction / Context: Microstrip is a planar transmission-line technology fabricated on dielectric substrates with a top signal trace and a bottom ground plane. It dominates MIC and RF PCB layouts for filters, couplers, dividers, matching networks, and component interconnects. This item asks whether the stated reason explains its widespread use. Given Data / Assumptions: Planar process on common substrates (FR-4, alumina, Rogers-type laminates). Components (chips, packaged devices) can be soldered or bonded to the top metal. Ground reference is the continuous bottom plane. Concept / Approach: Microstrip’s popularity arises from ease of fabrication, compatibility with PCB processes, straightforward component access on the top surface, and broad design tool support. The open top makes component mounting and tuning practical. While microstrip has dispersion and radiation compared to stripline, its assembly and prototyping advantages keep it prevalent in MICs. Step-by-Step Solution: 1) Confirm A: microstrip is widely used in MIC/RF-PCB practice.2) Evaluate R: easy top access directly supports mounting/connection → a primary practical reason for its dominance.3) Hence both are correct and R explains A. Verification / Alternative check: Standard MIC textbooks and application notes show abundant microstrip implementations with top-side component placement and tuning stubs. Why Other Options Are Wrong: B: R is indeed causally related. C/D: Deny true statements. Common Pitfalls: Assuming stripline is always superior; neglecting assembly and prototyping convenience in practical RF design choices. Final Answer: Both A and R are correct and R is correct explanation of A

Question 19

PIN diode structure identification — doping profile and intrinsic region Which option correctly describes the doping of a PIN diode's terminals relative to its intrinsic (i) region?

Accepted Answer

Introduction / Context: PIN diodes are essential RF/microwave control elements acting as current-controlled resistors at high frequency. Their behavior arises from a specific layer structure that enables charge storage in the intrinsic region, giving a nearly resistive RF impedance under forward bias and high impedance under reverse bias. Given Data / Assumptions: Device under test: PIN diode. Asked: which doping configuration is correct around the intrinsic layer. Concept / Approach: A PIN diode consists of a heavily doped p-type region (p+), an intrinsic (undoped or very lightly doped) region (i), and a heavily doped n-type region (n+). The heavy doping lowers series resistance and facilitates charge injection; the i-region stores charge, yielding a controllable RF resistance with bias current. Step-by-Step Solution: 1) Identify intrinsic layer at the center: symbol ‘‘i’’.2) Heavily dope both ends: p+ and n+ to reduce series resistance.3) Therefore the correct description is p+ – i – n+. Verification / Alternative check: Equivalent-circuit models and datasheets emphasize the low series resistance due to heavy terminal doping and the bias-dependent resistance from the stored charge in the i-region. Why Other Options Are Wrong: p+ and n (lightly doped n): not the standard PIN structure. p- and n-: incorrectly suggests lightly doped terminals; would raise series resistance. None of the above: incorrect because p+ – i – n+ is standard. Common Pitfalls: Confusing PIN with ordinary PN junction; overlooking the role of the intrinsic region in RF behavior. Final Answer: p+ and n+ layers separated by i layer

Question 20

Radio-navigation acronym expansion — LORAN What does the term ‘‘LORAN’’ stand for in radio navigation systems?

Accepted Answer

Introduction / Context: LORAN is a historic hyperbolic radio navigation system used by maritime and aviation users to determine position using time-difference measurements of pulsed transmissions from fixed stations. This question checks the expansion of the acronym. Given Data / Assumptions: Acronym: LORAN. Context: radio navigation, not broadcasting. Concept / Approach: LORAN expands to ‘‘Long Range Navigation.’’ It uses synchronized transmitters; by measuring time differences between received pulses, the receiver derives hyperbolic lines of position whose intersection yields location. Step-by-Step Solution: 1) Recognize category: radio navigation systems.2) Match acronym meaning: Long Range Navigation.3) Select the option stating exactly this phrase. Verification / Alternative check: Standard references and historical documentation of LORAN (e.g., LORAN-C) all expand to Long Range Navigation. Why Other Options Are Wrong: TV/cable transmission options are unrelated to navigation and the LORAN technology. Either (b) or (c) aggregates two incorrect broadcasting terms. Common Pitfalls: Assuming LORAN is tied to television due to the ‘‘long range’’ wording; misclassifying it as a general communications acronym. Final Answer: long range navigation

Microwave Communication Questions