Q: Microwave applications across communication services Microwaves are used in which of the following systems?

Introduction / Context: Microwave frequencies (roughly 1–30 GHz, extending into millimetre wave) underpin many communication infrastructures. Recognizing their broad applicability helps learners connect spectrum, antennas, and link design with real-world services. Given Data / Assumptions: Telephony includes point-to-point microwave backhaul and cellular microwave links. Radio broadcast includes studio-transmitter links and satellite program feeds. TV systems include terrestrial microwave relays, satellite distribution, and wireless video links. Concept / Approach: Although legacy consumer broadcast bands (AM/FM) sit below microwaves, the distribution networks and contribution links behind both radio and TV often use microwave relays or satellites. Telephony likewise uses microwave backhaul between towers, and modern cellular systems use microwave and millimetre-wave for small cells and fronthaul/backhaul. Step-by-Step Solution: Identify microwave roles: high-capacity line-of-sight links and satellite downlinks.Map to services: telephony backhaul; radio/TV studio–transmitter links and satellite distribution.Conclude that all listed systems employ microwaves in some part of their chain. Verification / Alternative check: Industry frequency allocation charts show numerous microwave bands designated for fixed service (FS), broadcast auxiliary services (BAS), and satellite services supporting telephony and TV/radio distribution. Why Other Options Are Wrong: Choosing only one system ignores the ubiquity of microwave backbone links across multiple services. Common Pitfalls: Confusing consumer receive frequencies with network distribution layers; overlooking that much content delivery relies on microwave hops and satellite paths even if the last-mile broadcast is at VHF/UHF. Final Answer: all of the above

Q: Radar antenna beamwidth — preferred for detection and angular resolution For most radar applications, how should the antenna beamwidth be chosen?

Introduction / Context: Radar systems determine target direction by the pointing of a directive antenna. Angular resolution and clutter rejection depend strongly on beamwidth. Selecting an appropriate beamwidth is a fundamental radar design decision, closely tied to antenna aperture size and operating frequency. Given Data / Assumptions: Narrower beamwidth improves angular resolution and reduces clutter/side illumination. Practical limits set by antenna size, frequency, and scanning method. Concept / Approach: Beamwidth (approximately proportional to wavelength/aperture) dictates how finely the radar can discriminate two close targets and how much ground/sea clutter enters the receiver. A very narrow beam concentrates energy and sensitivity in a small angular sector, improving detection range (antenna gain) and angle accuracy. Hence, radars strive for the narrowest feasible beam within mechanical and cost constraints. Step-by-Step Solution: Increase aperture or frequency to narrow the beam.Observe that narrower beams raise gain → longer detection ranges for a given transmit power.Conclude that “very narrow” is preferred to maximize resolution and minimize clutter. Verification / Alternative check: Phased-array radars achieve sub-degree beams via large apertures, illustrating the performance benefits of narrow beams in operational systems. Why Other Options Are Wrong: Very wide: poor angle resolution, high clutter; used only for surveillance with different trade-offs. Neither too narrow nor very wide / either (b) or (c): vague and not optimal for most tracking/search radars aiming for high resolution. Common Pitfalls: Ignoring sidelobe control; a narrow mainlobe with high sidelobes still invites interference—hence tapering and array weighting are also important. Final Answer: very narrow

Q: Rectangular resonant cavity — mode existence rules Which of the following modes cannot exist in a rectangular resonant cavity?

Introduction / Context: Electromagnetic modes in a rectangular cavity are classified as TEmnp or TMmnp, where m, n, p count half-wave variations along the a, b, and d dimensions. Existence rules differ for TE and TM families because of boundary conditions on tangential fields at conducting walls. Given Data / Assumptions: Perfectly conducting rectangular cavity with dimensions a, b, d. Mode indices m, n, p are nonnegative integers. Concept / Approach: For TM modes in a cavity, the axial electric field component must be nonzero and satisfy boundary conditions on all six walls. This enforces that all three indices m, n, and p must be nonzero (≥ 1). If any of m, n, or p is zero, the required field configuration cannot satisfy the boundary conditions for TM in a closed cavity. In contrast, TE modes may have one index equal to zero, provided not all are zero simultaneously. Step-by-Step Solution: Check TM110: p = 0 → violates TM requirement that m, n, p ≥ 1 → cannot exist.Check TE110: allowed; TE permits a zero or more in indices as long as fields satisfy boundary conditions.Check TE011: allowed for the same reason (TE with m = 0 is permissible).Check TM111: allowed; all indices nonzero. Verification / Alternative check: Standard cavity-mode tables list TMmnp only with m, n, p ≥ 1, while TEmnp may include zeros. This corroborates the exclusion of TM110. Why Other Options Are Wrong: TE110 and TE011: valid TE modes with appropriate boundary conditions. TM111: all indices nonzero → valid TM mode. Common Pitfalls: Assuming waveguide rules carry over directly; in a closed cavity, TM requires three standing-wave variations. Final Answer: TM110

Q: Klystron amplifier — gap transit-time requirement In a klystron amplifier cavity, the electron transit time across the gap should be:

Introduction / Context: Klystron amplifiers rely on velocity modulation at the input cavity gap, followed by bunching and energy extraction at the output cavity. If electrons spend too long traversing the gap, the RF field changes phase significantly during transit, reducing effective modulation and gain. This is the classic “transit-time effect.” Given Data / Assumptions: Single gap within an RF cavity. RF frequency f with period T = 1/f. Uniform field approximation across a small gap length compared to the electron path. Concept / Approach: Define the gap transit angle ψ = ω * τ, where τ is electron time in the gap. For efficient modulation, ψ should be sufficiently small, commonly taken as ≤ π/2 (i.e., τ ≤ T/2). This ensures electrons experience substantially uniform field polarity while crossing the gap. Tighter criteria (e.g., τ ≤ T/4) further reduce distortion but are not strictly necessary in many practical designs; the widely cited limit is “less than half a cycle.” Step-by-Step Solution: Compute τ from electron velocity and gap length.Form transit angle ψ = ω * τ.Impose ψ ≤ π/2 → τ ≤ T/2 → “less than half cycle.” Verification / Alternative check: Klystron theory shows gain decreasing as ψ grows; design texts specify small transit angles, often targeting ψ around 1 radian or less, which is below π/2. Why Other Options Are Wrong: Less than one cycle: too loose; significant phase reversal can occur within one period. Less than quarter cycle: overly strict for general requirement; not mandatory in all designs. More than quarter cycle: can be acceptable in some cases, but beyond half cycle clearly degrades modulation; the accepted rule is “less than half cycle.” Common Pitfalls: Confusing gap transit time with drift region bunching time; ignoring relativistic effects at very high voltages. Final Answer: less than half cycle

Q: Hybrid ring (rat-race) coupler — basic properties Consider a hybrid ring. Evaluate the following statements: 1) Ports 1 and 2 are decoupled (isolated). 2) Ports 3 and 4 are coupled. 3) A signal entering either port 1 or 2 splits equally between port…

Introduction / Context: The hybrid ring (rat-race) is a 3 dB four-port coupler built from a ring one-and-a-half wavelengths in circumference with ports spaced 90 degrees. It produces sum and difference combinations with port-to-port isolation depending on which port is used as input. Given Data / Assumptions: Ideal, lossless hybrid ring with ports equally spaced. Standard port labeling where opposite ports act as isolated pairs for specific excitations. Concept / Approach: When excited at one port, the signal splits into two paths around the ring that recombine with appropriate phase at the output ports, giving equal magnitudes at two ports and isolation at the fourth. For the common labeling, ports 1 and 2 are isolated from each other when driven at 3 or 4, and a drive at 1 (or 2) divides equally between 3 and 4. Symmetry implies reciprocal statements when driving the opposite pair. Step-by-Step Solution: Drive port 1 → equal outputs at ports 3 and 4; port 2 isolated.Drive port 2 → equal outputs at ports 3 and 4; port 1 isolated.Drive port 3 or 4 → equal outputs at ports 1 and 2; the opposite ring port is isolated accordingly. Verification / Alternative check: Ideal S-matrix of a rat-race confirms 3 dB splits with port pairs isolated depending on excitation. Practical units show finite directivity but follow the same pattern. Why Other Options Are Wrong: Any subset (B, C, D) omits at least one true statement about equal splitting or isolation inherent to the hybrid ring. Common Pitfalls: Confusing port numbering schemes; mixing hybrid ring behavior with that of a magic-tee or directional coupler. Final Answer: 1, 2, 3 and 4

Question 1

Waveguide operation exactly at cutoff In a rectangular waveguide, what happens to the flow of electromagnetic energy at the cutoff frequency?

Accepted Answer

Introduction / Context: Propagation in waveguides occurs only above the cutoff frequency for a given mode. At and below cutoff, fields become evanescent and do not transport real power along the guide. Understanding this boundary clarifies why systems specify operating bands well above cutoff to ensure low loss and stable group velocity. Given Data / Assumptions: Rectangular metallic waveguide supporting a specific TE/TM mode. Operating at frequency f = f_c (mode cutoff). Lossless idealization for basic reasoning. Concept / Approach: At cutoff, the longitudinal propagation constant β equals 0, so the phase advance per unit length is zero and group velocity collapses to 0. Poynting vector averaged over a cross-section is proportional to β (through power–velocity relations), so average power flow goes to zero. For f < f_c, β becomes imaginary; fields decay exponentially and still do not carry real power. Step-by-Step Solution: For a given mode, compute β = sqrt(k^2 − k_c^2).At f = f_c, k = k_c → β = 0.Power flow P ∝ v_g * stored energy; with group velocity v_g → 0 at cutoff, P → 0. Verification / Alternative check: Measured transmission near cutoff shows rapidly increasing attenuation as frequency approaches f_c from above, and below f_c only near-field (non-propagating) coupling occurs. Why Other Options Are Wrong: Infinite/50%/10%: none aligns with the physics of β = 0 and zero group velocity at cutoff. Common Pitfalls: Assuming nonzero power because fields exist—standing fields do not guarantee net forward energy transport; confusing stored reactive energy with transmitted power. Final Answer: the flow of electromagnetic energy is zero

Question 2

Microwave applications across communication services Microwaves are used in which of the following systems?

Accepted Answer

Introduction / Context: Microwave frequencies (roughly 1–30 GHz, extending into millimetre wave) underpin many communication infrastructures. Recognizing their broad applicability helps learners connect spectrum, antennas, and link design with real-world services. Given Data / Assumptions: Telephony includes point-to-point microwave backhaul and cellular microwave links. Radio broadcast includes studio-transmitter links and satellite program feeds. TV systems include terrestrial microwave relays, satellite distribution, and wireless video links. Concept / Approach: Although legacy consumer broadcast bands (AM/FM) sit below microwaves, the distribution networks and contribution links behind both radio and TV often use microwave relays or satellites. Telephony likewise uses microwave backhaul between towers, and modern cellular systems use microwave and millimetre-wave for small cells and fronthaul/backhaul. Step-by-Step Solution: Identify microwave roles: high-capacity line-of-sight links and satellite downlinks.Map to services: telephony backhaul; radio/TV studio–transmitter links and satellite distribution.Conclude that all listed systems employ microwaves in some part of their chain. Verification / Alternative check: Industry frequency allocation charts show numerous microwave bands designated for fixed service (FS), broadcast auxiliary services (BAS), and satellite services supporting telephony and TV/radio distribution. Why Other Options Are Wrong: Choosing only one system ignores the ubiquity of microwave backbone links across multiple services. Common Pitfalls: Confusing consumer receive frequencies with network distribution layers; overlooking that much content delivery relies on microwave hops and satellite paths even if the last-mile broadcast is at VHF/UHF. Final Answer: all of the above

Question 3

Microwave resistors — why very small wall (film) thickness? In practical microwave resistors (e.g., thin-film or resistive cards), the conductive film or wall thickness is made as small as possible primarily to:

Accepted Answer

Introduction / Context: At microwave frequencies, any conductor behaves not only as a resistor but also as an inductor and capacitor due to geometry and current distribution (skin effect). For resistive elements used as terminations or absorbers, parasitic reactances degrade broadband matching. Minimizing these parasitics is central to good RF performance. Given Data / Assumptions: Resistive element implemented as a thin film or resistive card in a waveguide or microstrip. Goal: achieve broadband, nonreactive resistance close to the target value. Operating in the microwave regime where skin depth is small. Concept / Approach: Parasitic inductance arises from loop area and current path thickness/length. A thinner conductive layer shortens current path height and lowers stored magnetic energy, thus reducing series inductance. While sheet resistance can be set by material resistivity and film thickness, the primary microwave rationale for thin walls/films is to reduce inductive reactance so the part behaves resistively over a wider band. Excess thickness can also aggravate discontinuities and excite higher-order fields. Step-by-Step Solution: Identify microwave error sources: series L and shunt C.Reduce L by minimizing film thickness and loop area.Select sheet resistance (ohms/square) via material and geometry to hit the desired R value without resorting to thick, inductive conductors. Verification / Alternative check: Thin-film attenuators and terminations specify low VSWR across band when implemented with very thin resistive layers and tapered geometries, confirming inductance reduction benefits. Why Other Options Are Wrong: High/low R alone miss the RF reason; R can be achieved with geometry. The thinness target is primarily to curb inductance. Maximize capacitive effect is counterproductive; designers minimize C as well via layout. Common Pitfalls: Assuming DC resistance selection forces thick films; at RF, parasitics dominate and must be minimized for flat response. Final Answer: reduce inductance

Question 4

Radar antenna beamwidth — preferred for detection and angular resolution For most radar applications, how should the antenna beamwidth be chosen?

Accepted Answer

Introduction / Context: Radar systems determine target direction by the pointing of a directive antenna. Angular resolution and clutter rejection depend strongly on beamwidth. Selecting an appropriate beamwidth is a fundamental radar design decision, closely tied to antenna aperture size and operating frequency. Given Data / Assumptions: Narrower beamwidth improves angular resolution and reduces clutter/side illumination. Practical limits set by antenna size, frequency, and scanning method. Concept / Approach: Beamwidth (approximately proportional to wavelength/aperture) dictates how finely the radar can discriminate two close targets and how much ground/sea clutter enters the receiver. A very narrow beam concentrates energy and sensitivity in a small angular sector, improving detection range (antenna gain) and angle accuracy. Hence, radars strive for the narrowest feasible beam within mechanical and cost constraints. Step-by-Step Solution: Increase aperture or frequency to narrow the beam.Observe that narrower beams raise gain → longer detection ranges for a given transmit power.Conclude that “very narrow” is preferred to maximize resolution and minimize clutter. Verification / Alternative check: Phased-array radars achieve sub-degree beams via large apertures, illustrating the performance benefits of narrow beams in operational systems. Why Other Options Are Wrong: Very wide: poor angle resolution, high clutter; used only for surveillance with different trade-offs. Neither too narrow nor very wide / either (b) or (c): vague and not optimal for most tracking/search radars aiming for high resolution. Common Pitfalls: Ignoring sidelobe control; a narrow mainlobe with high sidelobes still invites interference—hence tapering and array weighting are also important. Final Answer: very narrow

Question 5

Rectangular resonant cavity — mode existence rules Which of the following modes cannot exist in a rectangular resonant cavity?

Accepted Answer

Introduction / Context: Electromagnetic modes in a rectangular cavity are classified as TEmnp or TMmnp, where m, n, p count half-wave variations along the a, b, and d dimensions. Existence rules differ for TE and TM families because of boundary conditions on tangential fields at conducting walls. Given Data / Assumptions: Perfectly conducting rectangular cavity with dimensions a, b, d. Mode indices m, n, p are nonnegative integers. Concept / Approach: For TM modes in a cavity, the axial electric field component must be nonzero and satisfy boundary conditions on all six walls. This enforces that all three indices m, n, and p must be nonzero (≥ 1). If any of m, n, or p is zero, the required field configuration cannot satisfy the boundary conditions for TM in a closed cavity. In contrast, TE modes may have one index equal to zero, provided not all are zero simultaneously. Step-by-Step Solution: Check TM110: p = 0 → violates TM requirement that m, n, p ≥ 1 → cannot exist.Check TE110: allowed; TE permits a zero or more in indices as long as fields satisfy boundary conditions.Check TE011: allowed for the same reason (TE with m = 0 is permissible).Check TM111: allowed; all indices nonzero. Verification / Alternative check: Standard cavity-mode tables list TMmnp only with m, n, p ≥ 1, while TEmnp may include zeros. This corroborates the exclusion of TM110. Why Other Options Are Wrong: TE110 and TE011: valid TE modes with appropriate boundary conditions. TM111: all indices nonzero → valid TM mode. Common Pitfalls: Assuming waveguide rules carry over directly; in a closed cavity, TM requires three standing-wave variations. Final Answer: TM110

Question 6

Klystron amplifier — gap transit-time requirement In a klystron amplifier cavity, the electron transit time across the gap should be:

Accepted Answer

Introduction / Context: Klystron amplifiers rely on velocity modulation at the input cavity gap, followed by bunching and energy extraction at the output cavity. If electrons spend too long traversing the gap, the RF field changes phase significantly during transit, reducing effective modulation and gain. This is the classic “transit-time effect.” Given Data / Assumptions: Single gap within an RF cavity. RF frequency f with period T = 1/f. Uniform field approximation across a small gap length compared to the electron path. Concept / Approach: Define the gap transit angle ψ = ω * τ, where τ is electron time in the gap. For efficient modulation, ψ should be sufficiently small, commonly taken as ≤ π/2 (i.e., τ ≤ T/2). This ensures electrons experience substantially uniform field polarity while crossing the gap. Tighter criteria (e.g., τ ≤ T/4) further reduce distortion but are not strictly necessary in many practical designs; the widely cited limit is “less than half a cycle.” Step-by-Step Solution: Compute τ from electron velocity and gap length.Form transit angle ψ = ω * τ.Impose ψ ≤ π/2 → τ ≤ T/2 → “less than half cycle.” Verification / Alternative check: Klystron theory shows gain decreasing as ψ grows; design texts specify small transit angles, often targeting ψ around 1 radian or less, which is below π/2. Why Other Options Are Wrong: Less than one cycle: too loose; significant phase reversal can occur within one period. Less than quarter cycle: overly strict for general requirement; not mandatory in all designs. More than quarter cycle: can be acceptable in some cases, but beyond half cycle clearly degrades modulation; the accepted rule is “less than half cycle.” Common Pitfalls: Confusing gap transit time with drift region bunching time; ignoring relativistic effects at very high voltages. Final Answer: less than half cycle

Question 7

Hybrid ring (rat-race) coupler — basic properties Consider a hybrid ring. Evaluate the following statements: 1) Ports 1 and 2 are decoupled (isolated). 2) Ports 3 and 4 are coupled. 3) A signal entering either port 1 or 2 splits equally between port…

Accepted Answer

Introduction / Context: The hybrid ring (rat-race) is a 3 dB four-port coupler built from a ring one-and-a-half wavelengths in circumference with ports spaced 90 degrees. It produces sum and difference combinations with port-to-port isolation depending on which port is used as input. Given Data / Assumptions: Ideal, lossless hybrid ring with ports equally spaced. Standard port labeling where opposite ports act as isolated pairs for specific excitations. Concept / Approach: When excited at one port, the signal splits into two paths around the ring that recombine with appropriate phase at the output ports, giving equal magnitudes at two ports and isolation at the fourth. For the common labeling, ports 1 and 2 are isolated from each other when driven at 3 or 4, and a drive at 1 (or 2) divides equally between 3 and 4. Symmetry implies reciprocal statements when driving the opposite pair. Step-by-Step Solution: Drive port 1 → equal outputs at ports 3 and 4; port 2 isolated.Drive port 2 → equal outputs at ports 3 and 4; port 1 isolated.Drive port 3 or 4 → equal outputs at ports 1 and 2; the opposite ring port is isolated accordingly. Verification / Alternative check: Ideal S-matrix of a rat-race confirms 3 dB splits with port pairs isolated depending on excitation. Practical units show finite directivity but follow the same pattern. Why Other Options Are Wrong: Any subset (B, C, D) omits at least one true statement about equal splitting or isolation inherent to the hybrid ring. Common Pitfalls: Confusing port numbering schemes; mixing hybrid ring behavior with that of a magic-tee or directional coupler. Final Answer: 1, 2, 3 and 4

Microwave Communication Questions