Question 1

What is the principal advantage of a parametric amplifier in microwave receivers? (Consider noise performance, gain, and bandwidth.)

Accepted Answer

Introduction: Parametric amplifiers (paramps) exploit time-varying reactance (pumped capacitance) to transfer energy from a pump to the signal, enabling low-noise amplification at microwave frequencies. The key test here is your recall of what makes paramps attractive in front-end receivers. Given Data / Assumptions: Microwave receiver front-end application. Emphasis on noise figure (NF), gain, and bandwidth trade-offs. Externally pumped varactor or similar nonlinear element. Concept / Approach: Paramps are historically prized for extremely low noise figures—often competitive with cryogenic LNAs. They can provide useful gain, but the standout advantage is low NF, not maximum gain or widest bandwidth. Step-by-Step Solution: 1) Identify performance drivers in weak-signal reception: NF is paramount.2) Paramp operation shifts energy from a pump to the signal with minimal added noise.3) Result: excellent sensitivity improvement at the system level.4) Gain and bandwidth are adequate but typically not the dominant selling points compared to NF. Verification / Alternative check: Comparing cascaded noise figures (Friis formula) shows that a very low-NF first stage reduces overall system NF drastically, even with moderate gain. Why Other Options Are Wrong: Very high gain/broad bandwidth: paramps can have constraints; these are not their defining benefits. No pump source: incorrect—paramps require an external pump. Common Pitfalls: Assuming “amplifier” always implies bandwidth first. In receivers, the first stage NF often dominates overall performance. Final Answer: low noise figure

Question 2

A transmission line with Z0 = 300 ∠ 0 Ω is open-circuited at the far end. What is the resulting VSWR on the line?

Accepted Answer

Introduction: Voltage standing wave ratio (VSWR) quantifies mismatch on a line. An open circuit is a complete mismatch and produces maximum standing waves. This question checks your understanding of reflection magnitude and its effect on VSWR. Given Data / Assumptions: Characteristic impedance Z0 = 300 Ω (purely real). Load is an open circuit ⇒ ZL → ∞. Lossless line assumption for clarity. Concept / Approach: Reflection coefficient at the load: Gamma_L = (ZL − Z0) / (ZL + Z0). For an open circuit, |Gamma_L| = 1. VSWR is (1 + |Gamma|) / (1 − |Gamma|). Step-by-Step Solution: 1) Open circuit ⇒ |Gamma_L| = 1.2) VSWR = (1 + 1) / (1 − 1) = 2 / 0 ⇒ unbounded.3) Therefore, VSWR tends to infinity; the line exhibits maximum standing waves. Verification / Alternative check: At current nodes/voltage antinodes, the measured voltage can be much larger than the incident amplitude; the infinite VSWR idealizes this behavior in the lossless limit. Why Other Options Are Wrong: 0 or 0.5: VSWR cannot be less than 1. 1: indicates perfect match, not an open circuit. 2: corresponds to |Gamma| = 1/3, not a complete mismatch. Common Pitfalls: Mistaking infinite VSWR for infinite power; in reality, losses and source limitations cap voltages/currents, but the ratio idealizes to ∞. Final Answer: ∞

Question 3

Which of the following is a transferred-electron device (TED) used for microwave generation and amplification?

Accepted Answer

Introduction: Transferred-electron devices (TEDs) exploit negative differential resistance arising from electron transfer between energy valleys in certain semiconductors. This question checks whether you can identify which microwave diode operates on the Gunn effect. Given Data / Assumptions: Common microwave diodes: Gunn, IMPATT, BARITT, step-recovery, tunnel. TED behavior involves intervalley electron transfer under high electric fields. Concept / Approach: The Gunn diode is the archetypal TED, typically fabricated in GaAs or InP. When biased above a threshold field, it exhibits negative differential resistance and can oscillate at microwave frequencies without a pn junction. Step-by-Step Solution: 1) Recall that TED = device based on the Gunn effect.2) Identify which option corresponds to the Gunn effect.3) Gunn diode matches the description; others use different mechanisms. Verification / Alternative check: Frequency–field domain formation in Gunn devices (transit-time domains) is the hallmark of TED operation, absent in the other listed diodes. Why Other Options Are Wrong: IMPATT/BARITT: rely on impact ionization or barrier injection and transit-time effects, not TED behavior. Step-recovery diode: used for pulse sharpening/harmonics via charge storage. Tunnel diode: quantum tunneling device with its own negative resistance mechanism, not TED. Common Pitfalls: Confusing all negative-resistance devices as TEDs. Only Gunn devices specifically use transferred-electron physics. Final Answer: Gunn diode

Question 4

For a transmitting aperture antenna, the directive gain (directivity) is proportional to which geometric quantity of the antenna, all else equal?

Accepted Answer

Introduction: Directivity quantifies how concentrated an antenna’s radiation is in a particular direction. For aperture antennas (dishes, horns), directivity is tied to effective area and wavelength. This question checks awareness of that proportionality. Given Data / Assumptions: Far-field pattern consideration. Comparable aperture efficiencies across compared designs. Same operating wavelength. Concept / Approach: For an aperture antenna, D ≈ 4π * Ae / lambda^2, where Ae is the effective area. If efficiency is held roughly constant, Ae scales with the physical cross-sectional area A, so D ∝ A. Step-by-Step Solution: 1) Start from D ≈ 4π * Ae / lambda^2.2) With constant efficiency, Ae ∝ A (aperture area).3) Therefore, D ∝ A (linear proportionality to area). Verification / Alternative check: Doubling the diameter of a circular aperture quadruples the area and increases directivity by roughly 4×, consistent with D ∝ A. Why Other Options Are Wrong: Square, square-root, cube-root dependencies do not follow from the standard aperture relation. Perimeter alone does not determine directivity. Common Pitfalls: Confusing gain with directivity. Gain also includes radiation efficiency, but with fixed efficiency the proportionality remains to area. Final Answer: its cross-sectional (aperture) area

Question 5

A magnetron requires an externally applied magnetic field with flux lines oriented parallel to the cathode axis. Is this statement correct?

Accepted Answer

Introduction: Magnetrons are crossed-field microwave devices. They use an electric field (radial) and a magnetic field (axial) to control electron motion and generate microwave oscillations. This item checks basic device-physics orientation. Given Data / Assumptions: Cylindrical magnetron with central cathode and surrounding anode cavities. Static magnetic field applied externally. Vacuum conditions typical for microwave tubes. Concept / Approach: In a magnetron, the DC magnetic field is applied along the axis of the cathode (axial direction), while the DC electric field is primarily radial. The crossed E and B fields cause electrons to follow complex cycloidal paths, enabling interaction with RF cavity fields. Step-by-Step Solution: 1) Identify field geometry: E field radial; B field axial (parallel to cathode axis).2) Recognize that this orientation is essential for crossed-field operation.3) Therefore, the statement regarding flux lines parallel to the cathode axis is correct. Verification / Alternative check: Standard magnetron diagrams show solenoids or permanent magnets arranged to produce axial magnetic flux through the interaction space. Why Other Options Are Wrong: False/conditional options contradict the fundamental crossed-field design. Reflex klystrons are different devices; the statement pertains to magnetrons specifically. Common Pitfalls: Confusing magnetron geometry with klystron or TWT structures, where fields and interaction mechanisms differ. Final Answer: True

Question 6

High-speed digital logic systems prefer which kind of pulse widths for reliable high-data-rate operation (assuming proper synchronization and bandwidth)?

Accepted Answer

Introduction: Pulse width is tied to bit period and bandwidth. In high-speed logic, narrower bit periods (and thus narrower valid pulses) are inherent to higher data rates. The question probes conceptual links between timing, bandwidth, and pulse shaping. Given Data / Assumptions: Digital system where clock rate increases for higher throughput. Bandwidth and rise/fall times are adequate for the chosen technology. Jitter and metastability are managed by proper design. Concept / Approach: Bit period Tbit shrinks as data rate increases. Logic pulses representing bits must fit within Tbit, so their valid-high (or valid-low) time also becomes small. Hence, high-speed circuits “use” small-width pulses by necessity, while ensuring integrity via signal conditioning. Step-by-Step Solution: 1) Data rate R ↑ ⇒ Tbit = 1/R ↓.2) Minimum viable pulse width scales with Tbit and setup/hold requirements.3) With adequate bandwidth (fast edges), narrow pulses can propagate and be sampled reliably.4) Therefore, very small pulse widths are characteristic of high-speed logic. Verification / Alternative check: Observe technology nodes: from TTL to ECL to modern CMOS/SerDes, as line rates increase, allowable pulse widths and rise/fall times decrease accordingly. Why Other Options Are Wrong: Very large width: limits data rate, increases intersymbol interference due to overlapping symbols. Neither small nor large / larger than a clock period: contradicts the definition of high-speed operation. Common Pitfalls: Equating “narrow pulses” with “unreliable.” Reliability depends on bandwidth, jitter control, and timing margins, not merely width. Final Answer: pulses with very small width

Question 7

In antenna arrays, what is the correct way to obtain the total far-field at a distant observation point from the contributions of all individual elements? (Assume the point is in the far field and mutual coupling is already accounted for in the elem…

Accepted Answer

Introduction / Context: In antenna array theory, the radiation observed in the far field results from the superposition of the contributions of each element. Because fields are sinusoidal and have phase, the combination must preserve both amplitude and phase information, which is naturally handled by phasors. Given Data / Assumptions: Observation point is in the far field (Fraunhofer region). Element excitations and relative positions are known. Mutual coupling effects, if any, are already reflected in the element patterns/phases. Concept / Approach: The electromagnetic field is a vector quantity with time-varying phase. Superposition therefore requires vector (phasor) addition, not scalar addition. The array factor arises from summing complex exponentials representing element phases and spacings. Step-by-Step Solution: 1) Represent each element's far-field as a complex phasor: E_k = |E_k| * e^{j(phi_k)} with its angular dependence.2) Apply superposition: E_total(θ, φ) = Σ_k E_k(θ, φ).3) Separate common element pattern and geometry-dependent phase to obtain the array factor AF(θ, φ) = Σ_k a_k * e^{jψ_k}.4) Compute observable power only after the phasor sum: P ∝ |E_total|^2. Verification / Alternative check: If you incorrectly add magnitudes first, you will lose interference effects (lobes/nulls). Correct phasor addition predicts constructive and destructive interference that match measured patterns. Why Other Options Are Wrong: Algebraic sum of magnitudes: Ignores phase; cannot predict nulls/lobes. Arithmetic average of powers: Power is a squared quantity; averaging powers cannot reproduce interference. Sum of directivities: Directivity is a system property, not an element-level additive quantity. RMS sum of magnitudes: Still discards relative phase information. Common Pitfalls: Adding gains or powers instead of fields; forgetting that phase delays from spacing/feeding control the pattern. Final Answer: Phasor (vector) sum of the individual element fields

Question 8

A complex load Z_L = 75 − j50 Ω is connected to a lossless 75 Ω coaxial line at 10 GHz. What is the most practical single-stub matching method at microwave frequencies to achieve a perfect match?

Accepted Answer

Introduction / Context: At microwave frequencies, distributed matching techniques are favored because lumped components become lossy, parasitic-dominated, and hard to realize. A single-stub tuner on a transmission line is a standard and flexible solution. Given Data / Assumptions: Line is lossless with Z0 = 75 Ω. Load is Z_L = 75 − j50 Ω. Operating frequency is 10 GHz (wavelength is short; distributed elements are convenient). Concept / Approach: The single-stub shunt tuner works by first moving along the line from the load to a point where the normalized admittance has a real part of 1 (or a convenient value). A short-circuited shunt stub at that point cancels the residual susceptance to achieve a match. Step-by-Step Solution: 1) Normalize the load: z_L = Z_L/Z0 = (75 − j50)/75 = 1 − j(50/75) = 1 − j0.667.2) Convert to admittance at the load: y_L = 1/z_L (complex inversion).3) Move a distance d toward the generator until Re{y(d)} = 1 (conductance equal to 1/Z0).4) At that point, add a short-circuited shunt stub with susceptance b_stub = −Im{y(d)} to cancel the reactive part.5) With the susceptance canceled, the input admittance equals 1/Z0, giving a perfect match. Verification / Alternative check: This procedure is exactly what the Smith chart single-stub design accomplishes: rotate from the load toward the generator to a unity-conductance circle, then pick the stub length to supply equal and opposite susceptance. Why Other Options Are Wrong: Add a series inductance at the load: Practical but less flexible at 10 GHz; hard to realize accurately and does not exploit the line transformation. Short-circuited shunt stub exactly at the load: May not provide the precise susceptance needed because the transformed admittance at the load is not at a convenient point. Series capacitance at some distance: Series elements are less convenient for coax implementations and often require tight tolerances. Open-circuited series stub: Series stubs are harder to implement; open stubs risk radiation and dispersion. Common Pitfalls: Forgetting to move to a unity-conductance point before adding the stub; choosing an open stub in exposed environments; ignoring line losses and connector repeatability. Final Answer: Use a short-circuited shunt stub placed at a specific distance from the load

Question 9

Assertion (A): The condition for a distortionless transmission line is L = (R * C) / G. Reason (R): Line loading refers to adding series inductance to help meet the distortionless-line condition.

Accepted Answer

Introduction / Context: The distortionless condition in transmission-line theory ensures that signals propagate without waveform distortion. It imposes a specific relationship among primary constants R, L, G, and C so that attenuation is independent of frequency and phase velocity is constant. Given Data / Assumptions: Primary line constants: R (series resistance per unit length), L (series inductance per unit length), G (shunt conductance per unit length), C (shunt capacitance per unit length). Goal: Identify the correctness of the assertion and the reason. Concept / Approach: The correct Heaviside distortionless condition is R/L = G/C (equivalently L/C = R/G when all quantities are positive and finite). This makes the propagation constant's real part independent of frequency and the phase constant proportional to frequency. Step-by-Step Solution: 1) State the correct condition: R/L = G/C.2) Rearranging gives L/C = R/G, but not L = (R * C) / G.3) Therefore, the assertion L = (R * C) / G is dimensionally and conceptually incorrect.4) Line loading (historically adding series inductance, e.g., loading coils) increases L so that R/L approaches G/C, thereby reducing distortion on low-loss lines where G is small. Verification / Alternative check: In telephony, adding loading coils increased L to approximate R/L ≈ G/C, which improved voice-band fidelity over long pairs. Practical designs show flatter group delay and reduced dispersion when the Heaviside condition is approximated. Why Other Options Are Wrong: Both A and R correct: A is false because the stated equality is wrong. Both correct but not explanatory: Invalid since A is false. A correct, R wrong: Opposite of the truth; A is false and R (as a concept) is true. Both wrong: R is essentially correct about line loading adding inductance to meet the condition. Common Pitfalls: Memorizing an incorrect formula; confusing R/L = G/C with spurious rearrangements; forgetting that matching the ratio, not absolute values, yields distortionless propagation. Final Answer: A is wrong but R is correct

Question 10

To couple a coaxial (unbalanced) line to a parallel two-wire (balanced) line with minimal common-mode currents and good impedance transfer, which device should be used?

Accepted Answer

Introduction / Context: Coaxial lines are unbalanced, whereas parallel two-wire lines are balanced. Directly interconnecting them causes common-mode currents and radiation. A balun solves this by converting between balanced and unbalanced interfaces. Given Data / Assumptions: One side is a coaxial cable (unbalanced). Other side is a two-wire line (balanced). Goal: Proper coupling and impedance transfer with minimal radiation. Concept / Approach: A balun provides both a current/voltage transformation (if designed as an impedance transformer) and the balanced-to-unbalanced conversion to suppress common-mode currents. Step-by-Step Solution: 1) Identify line types: coax (unbalanced) and twin-lead (balanced).2) Choose a device that enforces equal and opposite currents on the balanced ports while isolating the shield from radiating.3) A balun satisfies these conditions and can also provide impedance transformation if needed. Verification / Alternative check: Practical RF systems (e.g., TV twin-lead to coax) employ 1:1 or 4:1 baluns to ensure proper mode conversion and matching. Why Other Options Are Wrong: Slotted line: Measurement tool, not a mode converter. Directional coupler: Samples power; not a balanced/unbalanced converter. Quarter-wave transformer: Matches impedances but does not perform mode conversion. Wilkinson divider: Power split/combining; not a balun. Common Pitfalls: Attempting to match impedance only without addressing mode conversion; ignoring common-mode radiation on the feedline. Final Answer: Balun (balanced-to-unbalanced transformer)

Question 11

Assertion (A): A magnetron is generally operated in the π-mode. Reason (R): In π-mode, its operating frequency is well separated from adjacent competing modes, enabling stable, efficient operation.

Accepted Answer

Introduction / Context: A magnetron is a crossed-field microwave oscillator whose resonant cavities can support multiple modes. Selecting a specific mode with good separation from others ensures stable frequency and high efficiency. Given Data / Assumptions: Resonant cavity chain supports several modes (e.g., 0, π/2, π, etc.). Design goal is to suppress mode jumping and enhance efficiency. Concept / Approach: The π-mode sets adjacent cavity RF fields 180 degrees apart, creating strong azimuthal RF fields that interact efficiently with the rotating electron spokes. This mode typically has better separation from neighboring modes, reducing the risk of mode competition. Step-by-Step Solution: 1) Identify that practical magnetrons include strapping to enforce π-mode operation.2) The π-mode provides stronger synchronous interaction and higher output.3) Frequency separation from adjacent modes in π-mode helps maintain frequency stability and efficiency. Verification / Alternative check: Magnetron design literature shows strapping techniques specifically to favor π-mode by shorting alternate cavities, raising unwanted mode frequencies, and ensuring stable π-mode oscillation. Why Other Options Are Wrong: Both correct but not explanatory: R directly explains why π-mode is chosen. A correct, R wrong: R is consistent with design practice. A wrong but R correct / Both wrong: Contrary to standard magnetron operation where π-mode is preferred. Common Pitfalls: Confusing π-mode with other modes used in different devices (e.g., klystrons or BWOs); assuming adjacent modes are harmless to stability. Final Answer: Both A and R are correct and R is the correct explanation of A

Question 12

In a backward-wave oscillator (BWO) using a slow-wave structure, which statement best describes the direction of propagation of the RF wave with respect to the electron beam and power flow?

Accepted Answer

Introduction / Context: A backward-wave oscillator is a vacuum-electron device where a slow-wave circuit supports a dispersion relation with negative slope, allowing phase to travel opposite to the direction of energy transport. Given Data / Assumptions: Slow-wave structure with a dispersion curve ω(β) of negative slope in the operating region. Electron beam provides energy for oscillation. Concept / Approach: In a BWO, group velocity (power flow) and phase velocity have opposite signs. The device exploits synchronism between the backward wave and the electron beam to sustain oscillation and enable electronic frequency tuning via beam voltage. Step-by-Step Solution: 1) Identify dispersion where dω/dβ < 0 ⇒ group velocity opposite to phase velocity.2) Power (group velocity) travels toward the RF output (forward).3) Phase fronts move toward the electron gun (backward), hence the name backward wave.4) Beam–wave interaction is maintained by choosing an operating point where the beam line intersects the backward portion of the dispersion curve. Verification / Alternative check: Experimental field probes in slow-wave structures show phase advance toward the gun while power is extracted at the output end. Why Other Options Are Wrong: Meanders back and forth: Not a correct physical description of propagating modes. Phase progresses only forward: Describes a forward-wave device, not a BWO. Either (a) or (c): Not contingent solely on beam voltage; it is set by dispersion. Both phase and power backward: Violates energy extraction at the output. Common Pitfalls: Confusing phase direction with power flow; assuming backward behavior depends purely on beam voltage rather than circuit dispersion. Final Answer: The RF phase progresses in the backward direction while power flows forward

Question 13

In a rectangular waveguide with cutoff frequency f_c for a given mode, how does attenuation behave relative to the operating frequency f?

Accepted Answer

Introduction / Context: Waveguides support propagation only above a cutoff frequency f_c that depends on the mode and cross-section. Below cutoff, fields are evanescent and decay rapidly with distance. Given Data / Assumptions: Uniform rectangular waveguide. Single operating mode considered with cutoff f_c. Low-loss metallic walls (typical microwave practice). Concept / Approach: For f < f_c, the propagation constant becomes imaginary and fields attenuate exponentially. For f > f_c, the mode propagates with finite attenuation primarily due to conductor and dielectric losses, which are relatively small. Step-by-Step Solution: 1) Determine if f is above or below f_c for the considered mode.2) If f < f_c ⇒ evanescent: E(z) ∝ e^{−αz} with large α (no true propagation).3) If f > f_c ⇒ propagating: β is real, α is small (losses only).4) Therefore attenuation is low above cutoff and very high below cutoff. Verification / Alternative check: Measurement with a network analyzer shows negligible transmission below cutoff, while above cutoff S21 rises sharply and only modestly degrades with wall losses. Why Other Options Are Wrong: Lowest at f = f_c: At cutoff, the guide is at the edge of propagation; attenuation tends to be large. Lowest when f < f_c: Below cutoff there is no propagation. Independent of f: Contradicts the defining property of cutoff. Always the same: Not true in dispersive, lossy structures. Common Pitfalls: Assuming modest leakage exists below cutoff; confusing conductor loss trends with cutoff behavior. Final Answer: Attenuation is low when f > f_c and very high when f < f_c

Question 14

Radar systems – Branched duplexer components required In radar transmit–receive switching, what gas-discharge protection tubes are required for a branched duplexer so that a single antenna can be shared safely by the transmitter and receiver?

Accepted Answer

Introduction: Radar front ends must use the same antenna for high-power transmission pulses and for receiving extremely weak echoes. A duplexer performs this switching and isolation. In classic pulse radars, the branched duplexer architecture relies on gas-discharge devices that become conductive only during transmit, thus protecting the sensitive receiver while minimally degrading receive performance. Given Data / Assumptions: Single radar antenna shared between transmitter and receiver. Pulse radar with high peak power during transmit and very low-level signals during receive. Need fast, reliable isolation and low insertion loss on receive. Concept / Approach: The branched duplexer uses two special tubes: the TR (Transmit-Receive) tube placed in the receiver branch and the ATR (Anti-Transmit-Receive) tube placed in the transmitter or antenna branch as required by the topology. During the transmit pulse, strong RF triggers ionization within these gas tubes, forcing them into a low-impedance state that reflects or shunts energy away from the receiver. Between pulses, the gas de-ionizes, returning the path to high impedance and permitting low-loss reception. Step-by-Step Solution: 1) Identify protection goal: prevent high-power transmit energy from reaching receiver.2) Use TR tube in receiver leg: it conducts during transmit, isolating and protecting the RF front end.3) Use ATR tube to improve isolation toward other branches and shape the energy flow toward the antenna.4) After the pulse, both tubes return to non-conducting state, enabling sensitive receive operation. Verification / Alternative check: Historical radar designs and training manuals specify the combination of TR and ATR tubes in branched duplexers for adequate isolation and low receive loss. Why Other Options Are Wrong: TR only or ATR only provide insufficient isolation in a branched network. No special tubes would expose the receiver to damage. A ferrite isolator alone is not adequate for the extremely high peak powers and the switching function required. Common Pitfalls: Confusing circulators/isolators with gas-tube duplexers; assuming one tube can cover all isolation needs; overlooking ionization recovery time in pulse design. Final Answer: Both TR and ATR tubes.

Question 15

Short-circuited transmission line – Where is the voltage minimum? A lossless transmission line is shorted at the far end. Identify the location of the first voltage minimum on the line relative to the short-circuited termination.

Accepted Answer

Introduction: Standing waves arise on transmission lines when the termination is not matched to the characteristic impedance. A perfect short circuit forces the voltage at the termination to zero, creating a fixed voltage minimum at that point and establishing a predictable sequence of minima and maxima along the line. Given Data / Assumptions: Lossless or low-loss line so standing-wave patterns are clear. Short-circuited load at the far end. Single-frequency steady-state sinusoidal excitation. Concept / Approach: For a shorted line, boundary condition is V = 0 at z = 0 (the load). The standing-wave solution is V(z) = V+ (e^{-jβz} − e^{jβz}) which equals −2j V+ sin(βz). This is zero at z = 0 and at every half-wavelength increment. Current is maximum at the short, because I(z) ∝ cos(βz). Thus the first and strongest voltage minimum is exactly at the shorted end. Step-by-Step Solution: 1) Apply boundary condition at short: V(0) = 0 implies a voltage node at the load.2) Standing-wave pattern gives nodes every λ/2 moving away from the load.3) Therefore, the minimum voltage occurs at the shorted end and repeats every λ/2 toward the source. Verification / Alternative check: Smith Chart view: a short corresponds to a reflection coefficient of −1. The normalized voltage standing-wave pattern shows a node at the load. Why Other Options Are Wrong: Option B: A node is not guaranteed at the source unless the length happens to place one there. Option C: Midway is not universal; it depends on electrical length. Option D: The position is definite due to boundary conditions. Option E: Voltage nodes recur every λ/2, not only at λ/4. Common Pitfalls: Mixing up positions of voltage and current extrema; assuming λ/4 behavior for all shorted lines; ignoring electrical length dependence of additional minima. Final Answer: At the far end (at the short itself).

Question 16

Klystron amplifiers – Applegate diagram and electron bunching State whether the bunching process in a multi-cavity klystron amplifier is correctly represented by an Applegate diagram.

Accepted Answer

Introduction: Multi-cavity klystron amplifiers use velocity modulation followed by drift-space bunching to convert beam kinetic energy into RF power at the output cavity. Engineers visualize the evolution of electron phases and velocities using the Applegate diagram, a time–position plot of representative electrons that shows how an initially uniform beam forms density bunches. Given Data / Assumptions: Electron beam passes through an input cavity that imposes velocity modulation. Electrons drift and form bunches due to velocity differences. Output (catcher) cavity extracts RF power from the bunched beam. Concept / Approach: The Applegate diagram tracks electron trajectories as lines whose slopes correspond to velocities. After modulation, faster electrons overtake slower ones, creating bunching lines that converge. This diagram is not an exact solution but an insightful representation to understand phase focusing, optimum drift length, and cavity tuning for maximum gain. Step-by-Step Solution: 1) Apply small-signal velocity modulation at the input cavity: v = v0 + Δv cos(ωt).2) In the drift region, faster electrons close the gap to slower electrons, producing density peaks.3) At the output cavity, the bunched beam induces strong RF currents that deliver power to the load. Verification / Alternative check: Simulation or beam-wave theory reproduces the bunching predicted qualitatively by the Applegate construction, validating its pedagogical use. Why Other Options Are Wrong: Limiting the diagram to low voltages, oscillators only, or zero space-charge is incorrect; it is a general visualization tool. Space-charge modifies but does not invalidate the representation. Common Pitfalls: Confusing Applegate diagrams with exact particle-in-cell results; ignoring space-charge effects when estimating drift length. Final Answer: True.

Question 17

Vacuum tubes – When does electron transit time matter most? Considering motion from cathode to anode, at what frequency range does the finite electron transit time significantly affect tube behavior and limit gain or phase response?

Accepted Answer

Introduction: In vacuum tube devices, electrons require a finite time to travel from the cathode to the anode or through interaction structures. When the RF period becomes comparable to this transit time, device performance degrades due to phase lag, reduced transconductance, and parasitic resonances. Understanding this limitation explains why classical triodes and pentodes struggle at microwave frequencies, while specialized tubes (klystrons, TWTs) use alternative interaction mechanisms. Given Data / Assumptions: Electrons accelerate through static fields between electrodes. RF excitation occurs at angular frequency ω = 2πf. Transit time τ across the gap is finite. Concept / Approach: Transit-time effects are small when ωτ << 1 and large when ωτ ≈ 1. As frequency increases, the phase of current relative to voltage shifts, effectively reducing the device's ability to deliver power or gain. Grid-to-cathode and anode-to-grid capacitances further worsen high-frequency response. Step-by-Step Solution: 1) Estimate τ from electron acceleration and gap distance.2) Compare τ to RF period T = 1/f.3) If τ is a significant fraction of T, expect gain roll-off and phase distortion. Verification / Alternative check: Empirical frequency response of conventional tubes shows rapid gain decrease into the VHF/UHF ranges, aligning with ωτ limits. Why Other Options Are Wrong: Low or midband frequencies do not challenge transit time. Saying both ranges or never ignores the physical time-of-flight constraint. Common Pitfalls: Assuming capacitance alone limits HF behavior; overlooking that electron inertia and finite velocity create fundamental phase delay. Final Answer: High frequencies.

Question 18

Radar duplexer – Primary purpose In a radar front end, what is the main function of a duplexer with respect to sharing one antenna between the transmitter and receiver without mutual interference?

Accepted Answer

Introduction: Radar systems typically use one antenna to transmit high-power pulses and receive extremely weak echoes. A duplexer is the device that switches and isolates the paths so a single antenna can serve both roles safely and efficiently. This concept is foundational in classic pulsed radars and remains important even where ferrite circulators or solid-state switches are used. Given Data / Assumptions: One antenna shared between transmitter and receiver. High peak power during transmit, very low-level signals during receive. Need for rapid switching and strong isolation. Concept / Approach: A duplexer provides high isolation from the transmitter port to the receiver port during the transmit pulse, protecting sensitive receiver stages. Between pulses, it presents a low-loss path from the antenna to the receiver while isolating the transmitter. Classic implementations include branched duplexers with TR/ATR tubes and ferrite circulator-based duplexers. Step-by-Step Solution: 1) Define the switching requirement and isolation levels for receiver protection.2) Provide a low-loss receive path during the listening interval.3) Ensure minimal detuning or distortion of the antenna pattern. Verification / Alternative check: Receiver survival ratings and measured isolation confirm proper duplexer operation in deployed radars. Why Other Options Are Wrong: Options A, B, and C misstate the function. Option E confuses duplexing with power combining; a duplexer does not increase antenna gain. Common Pitfalls: Assuming an isolator alone suffices; underestimating the necessary peak power handling and recovery time. Final Answer: Allow one antenna to be used for transmission and reception without interference.

Question 19

Free-space path loss comparison – 0.5 GHz versus 5 GHz Assuming the same link distance and antennas, how does free-space transmission loss at 5 GHz compare with that at 0.5 GHz?

Accepted Answer

Introduction: Free-space path loss (FSPL) quantifies spreading loss as a radio wave propagates without obstacles. It increases with frequency for a fixed distance and antenna aperture. Comparing 0.5 GHz and 5 GHz (a 10× frequency increase) illustrates the logarithmic dependence of FSPL on frequency. Given Data / Assumptions: Same transmitter power, range, and antenna gains (in dBi). Line-of-sight free-space propagation. Frequencies f1 = 0.5 GHz and f2 = 5 GHz. Concept / Approach: The standard formula in decibels is FSPL(dB) = 20 * log10(4πd / λ) = 32.44 + 20 * log10(d_km) + 20 * log10(f_MHz). At the same distance, the difference depends only on frequency ratio. When frequency increases by a factor of 10, FSPL increases by 20 * log10(10) = 20 dB. Step-by-Step Solution: 1) Compute Δ = 20 * log10(f2 / f1).2) Substitute f2/f1 = 5 / 0.5 = 10.3) Get Δ = 20 * log10(10) = 20 dB. Verification / Alternative check: Using wavelength form: FSPL ∝ 1/λ^2, and λ shrinks by 10× at 5 GHz, yielding 100× power density loss, which is 20 dB. Why Other Options Are Wrong: Less loss contradicts the formula. 700 dB is unrealistic for a 10× change. No change ignores the frequency term in FSPL. Common Pitfalls: Confusing antenna aperture scaling; if physical antenna size is constant, gain may rise with frequency, partially offsetting FSPL, but FSPL itself still increases by 20 dB. Final Answer: 20 dB more at 5 GHz.

Question 20

Reflex klystron oscillator – Potential of the focusing electrode Consider the focusing electrode used to shape and confine the electron beam near the cathode in a reflex klystron. Is this electrode normally held at a high potential compared to the c…

Accepted Answer

Introduction: A reflex klystron is a microwave oscillator in which an electron beam emitted by the cathode is velocity-modulated at a cavity gap, then turned back by a negatively biased reflector (repeller) electrode to pass the gap again in phase with the RF. Several electrodes shape the beam near the cathode region, including a focusing electrode (or focusing anode). Given Data / Assumptions: Standard reflex klystron construction with cathode, focusing electrode, resonant cavity, and negatively biased reflector. Goal of focusing electrode is beam shaping and confinement close to the cathode. Concept / Approach: The focusing electrode is typically at or near cathode potential, sometimes slightly negative relative to the cathode to compress the beam. It is not held at a high positive potential. In contrast, the reflector electrode is at a high negative potential relative to the cathode to reverse the electron beam and set the oscillation frequency through transit-time conditions. Step-by-Step Solution: 1) Identify electrode roles: focusing electrode shapes the beam; reflector returns the beam.2) Assign polarities: reflector is strongly negative; focusing electrode is near cathode potential, not high.3) Conclude the statement claiming a high potential on the focusing electrode is false. Verification / Alternative check: Textbook biasing diagrams show the focusing electrode tied to cathode or slightly negative, while the reflector has an adjustable high negative bias controlling mode and frequency. Why Other Options Are Wrong: Options C, D, and E suggest special cases that do not reflect normal design practice. The focusing electrode is not the element providing high potential; that function belongs to the reflector. Common Pitfalls: Confusing the focusing electrode with the reflector; assuming all control electrodes must be at high potential. Final Answer: False.

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