Question 1

Assertion–Reason on microwave power measurement with a bolometer Assertion (A): A bolometer is frequently used for microwave power measurement. Reason (R): In a bolometer, the incident microwave power causes a measurable change in resistance.

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Introduction / Context: Measuring RF/microwave power accurately often relies on thermal techniques that convert RF energy into heat. Bolometers and thermistors are classic sensors in power meters, especially where broadband, calibration-friendly methods are required. This assertion–reason problem asks whether the stated mechanism correctly explains the common use of bolometers in microwave power metrology. Given Data / Assumptions: Bolometer element exhibits resistance that varies with temperature. Microwave power absorbed in the load produces heating. A bridge or feedback circuit tracks resistance change to infer power. Concept / Approach: Bolometric power measurement is fundamentally calorimetric: RF energy is dissipated, raising the sensor temperature and changing its resistance. In a substitution method, DC power is adjusted to reproduce the same resistance change, and the equivalent RF power is read from the required DC level. This mechanism directly justifies the frequent use of bolometers for accurate, broadband power measurement at microwave frequencies. Step-by-Step Solution: 1) Microwave power → dissipation in sensor → temperature rise.2) Temperature rise → resistance change (thermistor/bolometer characteristic).3) Bridge/feedback maintains a reference and reports required DC substitution power.4) Since the resistance change is the sensing principle, R explains A. Verification / Alternative check: Standards labs have long used bolometric and thermistor mounts traceable to calorimetric principles; modern sensors (thermopile, diode) still rely on calibrated transfer to a thermal equivalent. Why Other Options Are Wrong: B: The reason is directly causal, not incidental. C/D: Both incorrectly deny either the accuracy of the assertion or the validity of the mechanism. Common Pitfalls: Confusing diode sensors (rectifying detection) with purely thermal bolometers; overlooking the substitution method that makes calibration straightforward. Final Answer: Both A and R are correct and R is correct explanation of A

Question 2

Electromagnetic waves in magnetised media — name the rotation of the plane of linear polarization The rotation of the direction of the E-field of a linearly polarized wave as it passes through a magnetised ferric (ferrite) medium is called:

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Introduction / Context: When a linearly polarized electromagnetic wave propagates through a magnetised ferrite, its plane of polarization can rotate due to non-reciprocal circular birefringence. This phenomenon has important applications in isolators, circulators, and contactless current sensors. The question asks you to identify the correct name for this rotation. Given Data / Assumptions: Medium: magnetised ferric/ferrite material under a bias magnetic field. Incident wave: linearly polarized. Observation: rotation of polarization plane after transmission. Concept / Approach: The effect is called Faraday rotation. Under magnetization, the right- and left-circularly polarized components of the wave experience different propagation constants. Their phase difference accumulates along the path, resulting in a rotation of the net linear polarization. The rotation angle is proportional to the Verdet constant of the material and the line integral of magnetic field along the path. Step-by-Step Solution: 1) Decompose linear polarization into counter-rotating circular components.2) Apply magnetized medium: two components propagate with different phase velocities.3) Phase difference at exit → rotated linear polarization (Faraday rotation).4) Angle depends on material (Verdet constant) and magnetic bias length product. Verification / Alternative check: Non-reciprocity (rotation direction independent of propagation reversal) underlies the operation of ferrite isolators and circulators used in microwave systems. Why Other Options Are Wrong: Lorentz, Steinmitz, Fleming: not the standard names for this polarization-rotation phenomenon in magneto-optic/magneto-microwave contexts. Common Pitfalls: Confusing Faraday rotation with optical activity in chiral media (which is reciprocal) or with mechanical rotation of components. Final Answer: Faraday rotation

Question 3

Quality factor (Q) of a waveguide resonator — correct definition in terms of stored energy and power loss Given ω0 as the resonant angular frequency, U as the time-average stored energy, and W_L as the average power loss, choose the correct expressi…

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Introduction / Context: The quality factor Q of a resonator quantifies how underdamped it is and how narrow its bandwidth is relative to its center frequency. For microwave cavities and waveguide resonators, Q directly relates to energy storage and dissipation, influencing selectivity, insertion loss, and phase noise in filters and oscillators. This item asks for the correct formula in terms of ω0, stored energy U, and power loss W_L. Given Data / Assumptions: Resonator operates in steady state at its resonance frequency ω0. U is the average energy stored in electric and magnetic fields. W_L is the average power lost due to wall loss, dielectric loss, radiation, etc. Concept / Approach: The standard definition for unloaded Q is Q = ω0 * (Energy stored per cycle / Energy lost per cycle). Since power is energy per unit time, and at steady state the rate of energy loss equals W_L, the expression simplifies to Q = ω0 * U / W_L. This connects directly to measurable quantities and underlies relations such as Q = f0 / Δf for a lightly loaded, weakly coupled resonator measured on a network analyzer. Step-by-Step Solution: 1) Start from Q = 2π * (Energy stored / Energy lost in one cycle).2) Energy lost in one cycle = W_L * T, with T = 2π / ω0.3) Substitute: Q = 2π * U / (W_L * 2π / ω0) = ω0 * U / W_L.4) Therefore, Q = ω0 * U / W_L. Verification / Alternative check: This form yields the familiar approximate relation Q ≈ f0 / (3 dB bandwidth) for lightly damped resonators when linked with coupling definitions and measured S-parameters. Why Other Options Are Wrong: 2π * U / W_L: missing ω0; not dimensionally consistent with the standard steady-state definition unless implicitly paired with f0. ω0 * U * W_L: multiplies instead of dividing by loss; incorrect units. W_L / (ω0 * U): inverse of the correct relation. Common Pitfalls: Forgetting that W_L is power (J/s) while U is energy (J); mixing frequency f0 with angular frequency ω0. Final Answer: Q = ω0 * U / W_L

Question 4

Microstrip propagation — normalized wavelength for a substrate with relative permittivity εr = 9 Assuming a typical microstrip where the effective permittivity ε_eff is less than εr but significantly greater than 1, the normalized guided wavelength …

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Introduction / Context: In planar transmission lines, the guided wavelength λ is shorter than the free-space wavelength λ0 by a factor related to the effective permittivity ε_eff of the structure. Microstrip has fields partly in air and partly in dielectric, so 1 < ε_eff < εr. This question asks for the approximate normalized wavelength when εr = 9, a high-permittivity substrate common in microwave practice. Given Data / Assumptions: Relative permittivity of substrate εr = 9. For microstrip, ε_eff is typically near (εr + 1)/2 for moderate width-to-height ratios, so ε_eff ≈ 5 as a ballpark. Normalized wavelength is λ/λ0 = 1 / √ε_eff. Concept / Approach: Using ε_eff ≈ 5 for εr = 9 gives λ/λ0 ≈ 1 / √5 ≈ 0.447, which rounds to about 0.45. Among the choices, 0.4 is the nearest practical estimate. The exact value depends on geometry (trace width vs substrate height), but the trend is robust: higher εr → larger ε_eff → shorter guided wavelength relative to free space. Step-by-Step Solution: 1) Estimate ε_eff for microstrip on εr = 9: ε_eff ≈ (9 + 1)/2 = 5 (rough).2) Compute normalized wavelength: λ/λ0 = 1 / √ε_eff ≈ 1 / √5 ≈ 0.447.3) Round to nearest option → 0.4. Verification / Alternative check: Hammerstad-Jensen or Wheeler equations for microstrip ε_eff yield values around 4–6 for many practical geometries on εr = 9 substrates, reinforcing λ/λ0 in the 0.4–0.5 range. Why Other Options Are Wrong: 10, 2: imply λ > λ0, which cannot occur in a passive dielectric-guided structure. 1: would require ε_eff ≈ 1 (air line), not a high-εr substrate. Common Pitfalls: Using εr instead of ε_eff; forgetting that microstrip fields are partly in air, so λ/λ0 is not as small as 1/√9 ≈ 0.333. Final Answer: 0.4

Question 5

Circular waveguides — meaning of the two subscripts in TE_mn / TM_mn mode notation What do the first and second subscripts represent in the standard mode labels for circular metallic waveguides?

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Introduction / Context: Modes in circular waveguides are designated TE_mn or TM_mn. Their field components satisfy Bessel-type radial dependence and sinusoidal azimuthal dependence. Correctly interpreting the subscripts is essential for determining cutoff frequencies, field distributions, and degeneracy among modes. Given Data / Assumptions: First subscript m relates to angular (circumferential) variation. Second subscript n indexes radial eigenvalues (Bessel zeros or derivative zeros). Metallic, perfectly conducting circular boundary is assumed. Concept / Approach: In circular geometry, the transverse fields can be expanded in terms of Bessel functions J_m(k_r r) for the radial dependence and sin(mφ), cos(mφ) for the azimuthal dependence. The integer m counts the number of full sinusoidal variations around the circumference, while n orders the radial eigenvalue (the nth root that satisfies boundary conditions). Thus, the first subscript ties to the circumferential variation, and the second maps to the radial Bessel-function order (root number). Step-by-Step Solution: 1) Express fields in cylindrical coordinates: r (radial), φ (circumferential), z (axial).2) Note that m controls the sin/cos dependence in φ → number of lobes around the circumference.3) n enumerates the allowed radial eigenvalues → number/order of radial nodes given by Bessel function roots.4) Therefore, first subscript = circumferential variation count; second = radial Bessel index. Verification / Alternative check: Cutoff frequency tables list k_c a values corresponding to zeros of J_m or its derivative (for TM or TE respectively), indexed by n for each m, which aligns with this interpretation. Why Other Options Are Wrong: A/C invert or mis-assign radial vs circumferential meanings. D: incorrect because a standard, accepted interpretation exists. Common Pitfalls: Confusing rectangular guide indices (m along a, n along b) with circular-guide indices (m azimuthal, n radial); forgetting derivative vs function zeros for TE vs TM. Final Answer: the Bessel function variations in radial direction and the number of full sine wave variations in circumference direction respectively

Question 6

Rectangular waveguide with cross-section 2a × a: which statement about cutoff (critical) wavelengths is correct? (Use standard TE/TM cutoff: λc = 2 / sqrt((m/width)^2 + (n/height)^2), noting TE and TM have the same λc for the same (m,n) when they ex…

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Introduction: Cutoff (critical) wavelength in rectangular waveguides depends on cross-sectional dimensions and the integers (m, n) that define the field variations. This question checks whether you know that TE and TM modes with the same (m, n) share the same cutoff condition (when the TM mode exists) and how different index combinations compare. Given Data / Assumptions: Rectangular guide size: width = 2a, height = a. Cutoff formula: λc = 2 / sqrt((m/width)^2 + (n/height)^2). TM00 does not exist; otherwise TE and TM with identical (m, n) have identical λc. Concept / Approach: Because cutoff arises from boundary conditions on the axial field component and transverse field separability, TE and TM families obey the same eigenvalue relation for the same (m, n). Numerical comparisons between different (m, n) show how λc changes with indices and aspect ratio. Step-by-Step Solution: 1) TE10: m = 1, n = 0 ⇒ λc = 2 / (1/(2a)) = 4a.2) TM10 exists and shares the same eigenvalue ⇒ λc(TM10) = 4a.3) TE01: m = 0, n = 1 ⇒ λc = 2 / (1/a) = 2a (not equal to TE10).4) TE02: m = 0, n = 2 ⇒ λc = 2 / (2/a) = a (not equal to TE10). Verification / Alternative check: General identity: for a given (m, n) with m, n ≥ 0 and not both zero, λc depends only on the geometry and indices, not on whether the mode is TE or TM (except that TM00 is forbidden). Hence TE10 and TM10 share λc. Why Other Options Are Wrong: TE10 = TE02 or TE10 half of TE01, etc.: direct substitutions give 4a vs a or 2a; they are not equal/half as claimed. TE01 = TM00: TM00 does not exist. TE10 half of TM10: they are equal, not half. Common Pitfalls: Assuming TE and TM always differ in cutoff for the same (m, n). They do not; the difference is in field composition, not the eigenvalue. Final Answer: Critical wavelength for TE10 is the same as that for TM10

Question 7

Circular waveguide attenuation versus frequency: in which mode does attenuation monotonically decrease as frequency increases over the practical band?

Accepted Answer

Introduction: Different circular waveguide modes exhibit different wall-loss behaviors with frequency. Some modes, notably TE01, are prized for extremely low attenuation that continues to drop as frequency rises, which is why they are used for very low-loss long-distance links. Given Data / Assumptions: Air-filled circular waveguide operating above cutoff. Conductor losses dominate; dielectric loss in air is negligible. Comparison among common TE and TM modes. Concept / Approach: The surface current distribution determines wall loss. The TE01 mode features circumferential currents with minimal current crowding at joints, yielding exceptionally low conductor loss that improves (attenuation decreases) with increasing frequency over a wide region above cutoff. Step-by-Step Solution: 1) Identify candidate low-loss mode: TE01 is known for best attenuation performance in circular guides.2) As frequency increases above cutoff, the field penetrates less (skin depth ↓) and the effective loss per unit length for TE01 decreases.3) Other modes (e.g., TE11, TE21) show higher wall losses and less favorable frequency dependence. Verification / Alternative check: Classical charts of circular waveguide attenuation show TE01 with the lowest αc (conductor attenuation), decreasing with frequency, which motivated TE01 long-haul waveguide systems. Why Other Options Are Wrong: TE11, TE21, TE05, TM01: do not offer the same favorable monotonic decrease; they have higher losses or different dependence. Common Pitfalls: Confusing dominant mode (TE11) with best loss; dominance concerns cutoff, not attenuation. Final Answer: TE01

Question 8

Comparing rectangular vs circular waveguides: which statement about modal families is correct?

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Introduction: Hollow metallic waveguides guide energy as TE or TM modes. Understanding which modal families exist in rectangular and circular geometries is basic to microwave engineering. Given Data / Assumptions: Air-filled hollow metallic guides. No internal center conductor (hence not coax). Standard boundary conditions on perfect conductors. Concept / Approach: Hollow guides do not provide a return path needed for TEM fields. Therefore, they support TE (transverse electric) and TM (transverse magnetic) modes only. Both rectangular and circular cross-sections produce an infinite discrete set of TE and TM modes, each with its own cutoff frequency. Step-by-Step Solution: 1) Apply boundary conditions: Ez or Hz must satisfy homogeneous Helmholtz equations with metallic boundaries.2) Solutions form countably infinite sets: TE_mn/TM_mn (rectangular) and TE_lm/TM_lm (circular).3) No TEM solution exists in a single-conductor hollow guide. Verification / Alternative check: Mode charts show infinitely many cutoff points for both geometries; TEM appears in two-conductor structures like coax or parallel-plate. Why Other Options Are Wrong: Claims denying one geometry having infinite modes are incorrect. TEM dominance is false in hollow guides. Common Pitfalls: Equating “dominant mode” with “TEM.” The dominant mode in rectangular is TE10; in circular it is TE11. Final Answer: Both support an infinite set of TE and TM modes (no true TEM mode)

Question 9

Instrument Landing System (ILS) statements — choose the correct combination: 1) It enables accurate approach and landing guidance under low-visibility conditions. 2) It includes a runway localizer, glide-path equipment, and marker beacons. 3) It is …

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Introduction: The Instrument Landing System (ILS) provides precision radio guidance to aircraft on final approach. This item checks your familiarity with ILS components and its role relative to fully automatic “blind landing.” Given Data / Assumptions: ILS comprises localizer (runway centerline guidance) and glide-slope (vertical path) plus marker beacons or DME for range. ILS supports low-visibility operations by giving precise course/vertical deviations. Autoland capability requires aircraft/airport certification and additional systems beyond basic ILS. Concept / Approach: Statements are evaluated against ICAO/FAA definitions. ILS provides guidance; it is not, by itself, a “blind landing” system. Autoland integrates flight-control and redundancy to flare/land without pilot visual reference, but that is separate from the radio aids themselves. Step-by-Step Solution: 1) Statement 1: Correct — ILS enables accurate approaches in low visibility (e.g., CAT I/II/III operations).2) Statement 2: Correct — localizer, glide-slope, and markers/DME are standard elements.3) Statement 3: Incorrect — ILS is guidance; “blind landing” requires certified autoland. Verification / Alternative check: Operational categories (CAT I–III) specify decision heights and RVR; higher categories allow lower minima but still distinguish between guidance (ILS) and automatic landing systems. Why Other Options Are Wrong: All/1&3/2 only/3 only include the incorrect claim that ILS itself is a blind landing system or omit correct items. Common Pitfalls: Equating precision approach guidance with fully automatic touchdown/rollout; they are related but not identical. Final Answer: 1 and 2 only

Question 10

Air-filled circular waveguides: which attenuation component is typically negligible, and which dominates?

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Introduction: Loss mechanisms in hollow waveguides include conductor (wall) loss and dielectric loss. For air-filled guides, dielectric loss is extremely small, so conductor loss dominates practical attenuation. Given Data / Assumptions: Air as dielectric (very low loss tangent). Standard copper/aluminum walls with finite conductivity. Operation above cutoff with smooth interior surfaces. Concept / Approach: Conductor loss arises from surface currents and finite skin-depth resistive effects, producing attenuation αc that decreases with increasing frequency (away from cutoff). Air contributes negligible dielectric loss αd compared to αc, so it is routinely ignored in calculations unless special dielectrics are inserted. Step-by-Step Solution: 1) Identify loss channels: α = αc + αd.2) For air: tanδ ≈ 0 → αd ≈ 0.3) Finite wall conductivity ⇒ αc > 0, typically the dominant term.4) Therefore, dielectric attenuation is negligible relative to conductor attenuation. Verification / Alternative check: Manufacturer data for standard waveguides quotes attenuation per meter dominated by conductor loss; changes in gas filling (air vs dry nitrogen) barely affect α. Why Other Options Are Wrong: Wall loss negligible or total attenuation nearly zero: unrealistic. Both significant or dielectric dominant: inconsistent with air’s very low loss tangent. Common Pitfalls: Assuming “air-filled” means lossless. Conductor loss sets the floor in real systems. Final Answer: Attenuation from imperfect dielectric (air) is negligible

Question 11

Loaded transmission line: the overall attenuation–frequency characteristic most closely resembles which filter class?

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Introduction: “Loading” a line (e.g., adding series inductance at intervals) was historically used in telephony to equalize amplitude/phase, producing a passband up to a cutoff. This question checks recognition that the attenuation characteristic then resembles a low-pass filter. Given Data / Assumptions: Periodic or lumped inductive loading on a transmission line. Objective: reduce distortion over voice band and set a finite cutoff. Small dielectric and copper losses assumed relative to loading effect. Concept / Approach: Loading increases effective series inductance L′, flattening group delay and reducing attenuation in the voice band, but introduces a cutoff frequency. Above this frequency, attenuation rises steeply, mirroring a low-pass filter response. Step-by-Step Solution: 1) Add loading coils ⇒ increase L′ → adjust characteristic impedance and phase constant.2) Equivalent periodic network ≈ ladder low-pass prototype.3) Below cutoff: lower attenuation; above cutoff: stop-band behavior (attenuation increases). Verification / Alternative check: Classical Campbell/Heaviside loaded-line theory maps directly to low-pass ladder sections with definable cutoff ωc. Why Other Options Are Wrong: High-pass/band-pass/band-stop: do not match the monotonic low-frequency passband and high-frequency stop behavior of a loaded line. All-pass: does not model attenuation shaping. Common Pitfalls: Assuming loading “boosts highs” — it actually limits high-frequency transmission, acting as low-pass. Final Answer: low-pass filter

Question 12

Microwave coaxial line (inner radius = a, inner radius of outer conductor = b): the approximate cutoff wavelength for the first TE mode (TE11) is closest to

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Introduction: Although coax supports a TEM mode without cutoff, higher-order modes (TE/TM) have finite cutoff frequencies. Knowing the TE11 cutoff helps set the upper frequency limit for single-mode coaxial operation. Given Data / Assumptions: Coax geometry: inner conductor radius a, inner radius of outer conductor b. We consider the first higher-order TE mode (TE11). Air dielectric; standard boundary conditions. Concept / Approach: Analytical solutions give an approximate TE11 cutoff wavelength λc ≈ π(a + b), which is a widely used design rule. Operation should remain well below this cutoff frequency (i.e., at wavelengths ≪ λc) to avoid multimoding. Step-by-Step Solution: 1) Recall approximate TE11 cutoff relation: fc ≈ c / (π(a + b)).2) Therefore λc = c / fc ≈ π(a + b).3) Compare with choices: π(a + b) matches directly. Verification / Alternative check: More exact expressions involve solving characteristic equations with Bessel functions; the π(a + b) rule closely tracks the true cutoff and is sufficient for design checks. Why Other Options Are Wrong: a + b, πa, πb: ignore the dependence on both radii with the correct coefficient. 2π(b − a): relates to circumference gap, not TE11 cutoff. Common Pitfalls: Confusing TEM (no cutoff) with TE11 (finite cutoff); the upper frequency limit for single-mode coax is set by TE11. Final Answer: π(a + b)

Question 13

Fundamental radar principle: is radar essentially an echo-ranging system that measures target range (and often velocity) from reflected signals?

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Introduction: Radar (Radio Detection and Ranging) operates by transmitting electromagnetic energy and processing the returned echoes from objects. The simplest conceptual model is “echo ranging,” analogous to sonar but using radio waves. Given Data / Assumptions: Monostatic or bistatic radar geometries. Pulsed or continuous-wave (CW) techniques. Standard free-space propagation assumptions for basic explanation. Concept / Approach: In pulsed radar, range R is inferred from round-trip time: R = c * Δt / 2. In CW/FMCW, range and velocity are extracted from frequency modulation and Doppler shifts. In all cases, the radar uses a form of echo measurement to infer target properties. Step-by-Step Solution: 1) Transmit waveform (pulse or modulated CW) toward the scene.2) Targets scatter energy; a portion returns to the radar antenna.3) Receiver measures time/frequency/phase of echoes.4) Convert observables to range (time delay), velocity (Doppler), angle (antenna pattern/beamforming). Verification / Alternative check: Radar equation and matched-filter theory formalize echo processing; all common radar classes fundamentally analyze returned signals. Why Other Options Are Wrong: “False” or restricted truths ignore that both pulsed and CW radar rely on echoes. Common Pitfalls: Assuming CW radars do not “range.” FMCW and phase-coded CW retrieve range through modulation. Final Answer: True

Question 14

Use of a directional coupler in RF/microwave measurements: which quantity (or quantities) can it determine on a transmission line?

Accepted Answer

Introduction: A directional coupler samples forward and reverse traveling waves on a transmission line. From the sampled amplitudes, mismatch metrics such as reflection coefficient and VSWR can be derived. Given Data / Assumptions: Coupler provides isolated forward and reverse ports. Line is driven at a known frequency with steady conditions. Calibration accounts for coupling factor and directivity. Concept / Approach: Let Vf be forward and Vr be reverse wave voltages. The magnitude of reflection coefficient is |Γ| = |Vr / Vf| at the reference plane. VSWR is then VSWR = (1 + |Γ|) / (1 − |Γ|). With proper calibration, the coupler thus yields both |Γ| and VSWR; phase requires a vector measurement or adjustable line. Step-by-Step Solution: 1) Measure coupled forward power Pf and reverse power Pr.2) Correct for coupling factor to estimate |Vf| and |Vr|.3) Compute |Γ| = |Vr / Vf|.4) Compute VSWR = (1 + |Γ|) / (1 − |Γ|). Verification / Alternative check: Slotted lines and reflectometers implement the same principle; vector network analyzers generalize it with phase. Why Other Options Are Wrong: L/C and wave velocity require additional fixtures/time-of-flight methods; a coupler alone does not directly yield them. Source resistance alone is not directly measured without further setup and modeling. Common Pitfalls: Ignoring coupler directivity; poor directivity corrupts reverse-power readings and thus |Γ|/VSWR. Final Answer: Reflection coefficient and VSWR

Question 15

TR (transmit–receive) protection tube in radar duplexers: how does its loss behave for weak echo signals versus the strong transmit pulse?

Accepted Answer

Introduction: In pulsed radar, a duplexer must protect the sensitive receiver during the high-power transmit pulse and then admit very weak echoes a short time later. Gas-discharge TR tubes (or modern limiter diodes) provide level-dependent isolation. Given Data / Assumptions: Monostatic radar with a single antenna shared by transmitter and receiver. TR device placed in the receiver path. Transmit pulse is orders of magnitude stronger than echoes. Concept / Approach: During the transmit pulse, the TR device ionizes or otherwise switches into a high-loss/reflective state, isolating the receiver front end. Between pulses, with only weak echo levels present, the device remains non-conducting (or unswitched), presenting low insertion loss so echoes reach the receiver with minimal attenuation. Step-by-Step Solution: 1) Transmit interval: incident power is very high → TR tube conducts → high attenuation to receiver path.2) Listen interval: only weak echoes arrive → device remains off → low insertion loss.3) Net effect: receiver is protected yet remains sensitive post-pulse. Verification / Alternative check: Limiter transfer curves show a knee beyond which insertion loss increases rapidly with input power; below the knee, loss is minimal. Why Other Options Are Wrong: Reversing which level gets high loss would either damage the receiver or block echoes. Both high or both low loss contradicts the protective function. Level-independent loss fails to protect during transmit. Common Pitfalls: Assuming all duplexers are purely passive hybrids; TR limiters provide the level-dependent isolation crucial for radar. Final Answer: Low loss for weak (low-level) echo signals and high loss for the strong transmit pulse

Question 16

Assertion (A): In a rectangular waveguide operating in TE10 mode, the cutoff frequency is f_c = c / (2a), where a is the broad-wall dimension and c is the speed of light. Reason (R): TE10 is the dominant mode in a rectangular waveguide because it ha…

Accepted Answer

Introduction / Context: A rectangular waveguide supports discrete TE and TM modes. Each mode has a cutoff frequency determined by the waveguide cross-section. Understanding the dominant mode and the exact cutoff formula for TE10 is foundational for single-mode operation, bandwidth planning, and loss estimation. Given Data / Assumptions: Rectangular waveguide with dimensions a (broad wall) and b (narrow wall). Electromagnetic wave in TE10 mode. Free space inside (for simplicity); speed of light is c. Concept / Approach: For a rectangular guide, the cutoff frequency for TEmn is f_c = (c/2) * sqrt((m/a)^2 + (n/b)^2). For TE10 (m = 1, n = 0) this simplifies to f_c = c/(2a). The dominant mode is the one with the lowest cutoff frequency; in standard hollow rectangular guides, that is TE10 because TE00 does not exist and other allowed modes have higher indices. Step-by-Step Solution: 1) Start from f_c(TEmn) = (c/2) * sqrt((m/a)^2 + (n/b)^2).2) Substitute m=1, n=0 ⇒ f_c(TE10) = c/(2a).3) Compare cutoffs: TE10 < TE20 (m=2), TE01 (n=1), etc., so TE10 is dominant.4) Note that being dominant explains “lowest cutoff,” not the exact algebraic formula for f_c. Verification / Alternative check: Waveguide handbooks and measurements show TE10 propagates first as frequency increases; the first higher modes (TE20 or TE01) appear only at higher frequencies, confirming the ordering implied by the formula. Why Other Options Are Wrong: Both correct and explanatory: Dominance explains which mode cuts on first, but not the specific equation f_c = c/(2a). A correct, R wrong: R is true—TE10 is the dominant mode. A wrong, R correct / Both wrong: Conflicts with standard theory. Common Pitfalls: Confusing “dominant” with “fundamental TEM” (TEM does not exist in hollow rectangular guides); forgetting that b does not appear for TE10 because n = 0. Final Answer: Both A and R are correct but R is not correct explanation of A

Question 17

In a standard hollow rectangular waveguide, which mode has the minimum (lowest) cutoff frequency among all allowed modes?

Accepted Answer

Introduction / Context: Rectangular waveguides guide energy as discrete modes. The mode that begins to propagate at the lowest frequency is called the dominant mode. Knowing the dominant mode is essential for sizing the guide and predicting single-mode bandwidth. Given Data / Assumptions: Hollow rectangular waveguide with dimensions a (broad wall) and b (narrow wall). Air-filled guide (effective speed ≈ c). Comparing cutoff frequencies of several low-order modes. Concept / Approach: The cutoff frequency for TEmn is f_c = (c/2) * sqrt((m/a)^2 + (n/b)^2). For TE10 (m=1, n=0), f_c = c/(2a), which is lower than TE20 (m=2), TE01 (n=1), TE11 (m=1, n=1), and any TM mode because additional indices increase f_c. Step-by-Step Solution: 1) Evaluate TE10: f_c = c/(2a).2) Compare TE20: f_c = c/a (higher).3) Compare TE01: f_c = c/(2b) (usually higher because a > b).4) Compare TE11: f_c = (c/2)*sqrt((1/a)^2 + (1/b)^2) (higher than TE10). Verification / Alternative check: Mode charts and measured transmission show TE10 propagates first; higher modes appear only as frequency increases. Why Other Options Are Wrong: TE11, TE01, TE20, TM11: All have higher cutoffs than TE10 for the same guide. Common Pitfalls: Assuming TE01 could be dominant—only true if b were greater than a, which contradicts “broad wall” terminology. Final Answer: TE10

Question 18

For a very short doublet (electrically short dipole) of length l ≪ λ, the radiation resistance varies with which power of the electrical length? (Let β = 2π/λ; electrical length is β l.)

Accepted Answer

Introduction / Context: The radiation resistance R_r of an antenna models how effectively it converts current into radiated power. For electrically short dipoles (length much smaller than wavelength), R_r is small and depends strongly on the electrical length. Given Data / Assumptions: Short dipole (uniform current approximation), l ≪ λ. Electrical length β l where β = 2π/λ. Free-space operation for simplicity. Concept / Approach: Classical antenna theory gives R_r ≈ 80 * π^2 * (l/λ)^2 for a short dipole. Since β = 2π/λ, (l/λ)^2 ∝ (β l)^2. Therefore, radiation resistance is proportional to the square of the electrical length, reflecting the weak radiation of very small antennas. Step-by-Step Solution: 1) Start with R_r ∝ (l/λ)^2 for a short dipole.2) Use β = 2π/λ ⇒ l/λ = (β l)/(2π).3) Hence R_r ∝ (β l)^2.4) Interpret: doubling l (still short) quadruples R_r; halving λ (doubling frequency) also increases R_r. Verification / Alternative check: Using the approximate constant: R_r ≈ 80π^2(l/λ)^2 confirms the quadratic dependence and yields numerical values that match measurements for l ≪ λ. Why Other Options Are Wrong: Linear or square-root dependence and cubic dependence do not match the established short-dipole formula. Independence of β l ignores the small-antenna limit behavior. Common Pitfalls: Confusing radiation resistance with input resistance including loss; applying the formula outside the l ≪ λ range. Final Answer: Proportional to (β l)^2

Question 19

Assertion (A): A very short section of high-impedance transmission line behaves like a series inductance at RF. Reason (R): A very short section of low-impedance transmission line is equivalent to a series resistance at RF.

Accepted Answer

Introduction / Context: Lumped equivalents of electrically short transmission-line sections are widely used in impedance-matching and filter synthesis. Recognizing which line (high-Z0 or low-Z0) mimics inductance or capacitance helps build practical microwave networks without discrete components. Given Data / Assumptions: Electrically short line: β l ≪ 1. Characteristic impedance Z0 (high vs low) and propagation constant γ ≈ jβ (low-loss). Operating in RF/microwave regime with small conductor/dielectric loss. Concept / Approach: The input impedance of a short series line (two-port viewed as series element) can be expanded. A short high-Z0 line section in series behaves like jωL (series inductance). Conversely, a short low-Z0 section used in shunt behaves like jωC (shunt capacitance). The reason statement claiming “series resistance” for a low-Z0 short section is therefore incorrect; lossless lines contribute reactance, not resistance. Step-by-Step Solution: 1) For a short line: tan(βl) ≈ βl; input reactance dominates.2) High-Z0 series section ⇒ X ≈ +ωL_equiv (inductive).3) Low-Z0 short shunt section ⇒ B ≈ +ωC_equiv (capacitive).4) Resistive behavior would require significant loss (R or G), not assumed here. Verification / Alternative check: Smith-chart stubs: high-Z0 series lines move along inductive arcs; low-Z0 shunt stubs provide capacitive susceptance—standard matching practice. Why Other Options Are Wrong: Both correct: R is not correct. A wrong but R correct / Both wrong: Opposite to classical short-line approximations. Common Pitfalls: Forgetting the role of series vs shunt configuration; conflating ohmic resistance with reactive impedance in low-loss lines. Final Answer: A is correct but R is wrong

Question 20

When atmospheric duct propagation (super-refractive or trapping layers) exists over the path, how do the usual “line-of-sight” and “diffraction zone” concepts apply to radio links?

Accepted Answer

Introduction / Context: Atmospheric ducting occurs when refractivity decreases sharply with height, creating a “duct” that traps and guides radio waves over long distances. This phenomenon extends coverage far beyond the optical horizon and invalidates many standard assumptions of free-space, line-of-sight links. Given Data / Assumptions: Presence of a surface or elevated duct (super-refraction). Frequencies typically in VHF/UHF/microwave bands where ducting is relevant. Standard diffraction models (e.g., knife-edge) assume no trapping. Concept / Approach: In ducts, rays are bent back toward Earth, undergoing repeated reflections/refractions inside the layer. The path is no longer limited by geometric line-of-sight, and field strength predictions based on simple diffraction from obstacles are unreliable. Propagation becomes guided, resembling over-water or waveguide-like behavior with mode structure. Step-by-Step Solution: 1) Identify ducting conditions (strong negative vertical refractivity gradient).2) Recognize that rays remain confined, enabling beyond-horizon links.3) Conclude that “line-of-sight” range formulas and diffraction-zone sizing lose validity.4) Use ducting or evaporation-duct models to estimate coverage and fading statistics. Verification / Alternative check: Field reports show anomalous long-distance reception during ducting events; empirical models (e.g., ITU-R recommendations) treat such cases separately from standard line-of-sight or diffraction calculations. Why Other Options Are Wrong: Unchanged applicability: contradicts guided behavior. “Only diffraction” or mixed statements: do not capture trapping mechanism. Common Pitfalls: Applying free-space path loss and Fresnel-zone clearance rules to ducted paths; ignoring strong frequency selectivity and mode interference. Final Answer: Both line-of-sight and diffraction-zone concepts become less applicable because energy is trapped and guided beyond the optical horizon

Microwave Communication Questions