Question 1

Transmission line propagation constant – Formula in terms of primary constants Given the per-unit-length parameters R (series resistance), L (series inductance), G (shunt conductance), and C (shunt capacitance), what is the correct expression for th…

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Introduction: The propagation constant γ characterizes how voltage and current waves attenuate and change phase as they travel along a transmission line. It depends on both the series and shunt per-unit-length parameters of the line and varies with frequency in a way captured by the telegrapher's equations. Engineers use γ to predict signal integrity, dispersion, and attenuation. Given Data / Assumptions: Uniform transmission line with parameters R, L, G, C (per meter). Sinusoidal steady state with angular frequency ω = 2πf. Linear, time-invariant behavior; standard reference impedance. Concept / Approach: From the telegrapher's equations, the dispersion relation is γ^2 = (R + jωL) * (G + jωC). Taking the principal square root gives γ = α + jβ, where α is the attenuation constant and β is the phase constant. Special cases include the low-loss line approximation and the lossless line where R = G = 0. These limits provide useful checks on the general expression. Step-by-Step Solution: 1) Write per-unit-length impedance Z′ = R + jωL and admittance Y′ = G + jωC.2) Use the relation γ = √(Z′Y′) → γ = √((R + jωL)(G + jωC)).3) Decompose into α and β if needed for design: γ = α + jβ. Verification / Alternative check: In the lossless limit (R = G = 0): γ = j * ω * √(LC). In low-loss lines, α ≈ (R/2) * √(C/L) + (G/2) * √(L/C) and β ≈ ω * √(LC), consistent with the exact formula, providing a sanity check for approximations used in hand calculations. Why Other Options Are Wrong: Ratios (options B, C) are dimensionally inconsistent for γ. Option D incorrectly separates α and β as independent square roots. Option E applies only to the lossless special case, not the general expression that includes R and G. Common Pitfalls: Forgetting frequency dependence; misusing the lossless formula when R and G are nonzero; confusing α and β with impedance/admittance magnitudes rather than with the complex square root of their product. Final Answer: γ = sqrt((R + jωL) * (G + jωC)).

Question 2

Broadside vs. end-fire linear arrays – Beamwidth and directive gain Let L be the overall element-center length of a uniform linear array and λ be the wavelength. Consider the statements:  The major-lobe width between first nulls (BWFN) of a broadsid…

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Introduction: Array radiation characteristics depend on the progressive phase and element spacing. Broadside arrays concentrate energy normal to the array axis; end-fire arrays point energy along the axis. This changes beamwidth and directivity relationships even when the physical aperture length L is fixed, which is vital when trading gain against beam steering or end-fire sensing requirements. Given Data / Assumptions: Uniform linear array with fixed overall length L. Free space, identical isotropic or identical element patterns. Broadside: progressive phase 0; End-fire: progressive phase chosen for end-fire maximum (e.g., Hansen-Woodyard adjustment). Concept / Approach: Broadside and end-fire arrays exhibit different array factors, yielding different first-null positions and thus different beamwidth between first nulls (BWFN). Their directivities are also generally different for the same L because end-fire requires destructive interference off-axis and constructive interference on-axis; classic approximations show broadside tends to have a narrower main lobe for a given L, while end-fire often exhibits a wider main lobe and somewhat lower directivity than broadside for the same aperture extent. Step-by-Step Solution: 1) Compare array factors: broadside AF has maximum at θ = 90°; end-fire AF has maximum at θ = 0° or 180°.2) Compute approximate BWFN: broadside BWFN ≈ 2λ/L; end-fire BWFN is typically larger (different dependence), hence statement 1 is true.3) Directivity D differs because beamwidth and side-lobe structure differ → statement 2 is true.4) Statement 3 claims a constant directivity of 4 for broadside, which is false; directivity scales with electrical aperture (roughly ∝ L/λ), not a fixed number. Verification / Alternative check: Textbook plots and closed-form approximations confirm distinct beamwidths for broadside vs. end-fire arrays of equal length and differing directivities as L/λ varies. Numerical array-factor simulations produce the same qualitative differences. Why Other Options Are Wrong: Any choice including statement 3 is incorrect because directivity is not a fixed constant of 4 for all broadside arrays. Choosing only statement 1 ignores the known difference in D for the two types of arrays under equal L. Common Pitfalls: Assuming directivity is independent of aperture; conflating BWFN with half-power beamwidth (HPBW); forgetting that element pattern and mutual coupling can modify results, though the fundamental trend holds across practical designs. Final Answer: 1 and 2 only.

Question 3

Microstrip in a metal enclosure – What happens to radiation loss? If a microstrip line is fully enclosed in a conductive housing (e.g., cavity or shielded box) while keeping the substrate and trace geometry otherwise unchanged, how are radiation los…

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Introduction: Open microstrip structures can radiate because part of the electromagnetic field is in air above the substrate. Enclosing the line within a conductive housing alters the boundary conditions and tends to confine fields, thus reducing radiation leakage and external coupling. This is a common technique for improving electromagnetic compatibility (EMC) and measurement repeatability. Given Data / Assumptions: Microstrip geometry (conductor over ground plane with dielectric substrate) remains the same. A conductive enclosure is added at some distance above and around the line. Operating frequencies at which the enclosure does not strongly resonate. Concept / Approach: Radiation from microstrip primarily comes from fringing fields and discontinuities (bends, steps). A metallic enclosure provides shielding that converts some of the open microstrip characteristics toward a shielded microstrip or stripline-like environment, better confining fields and reducing radiation. Care must be taken to avoid cavity resonances; nevertheless, in practical lab setups for detector heads, couplers, and filters, enclosures are used precisely to reduce radiation and improve isolation. Step-by-Step Solution: 1) Identify radiation mechanisms: open boundary fringing and discontinuities that excite free-space modes.2) Add conductive walls → fields terminate on the enclosure → less radiation to free space and reduced susceptibility.3) Result: lower radiation losses and improved EMC, provided apertures are minimized and the lid spacing avoids cavity resonances. Verification / Alternative check: EM simulations comparing open vs. enclosed microstrip show reduced radiated power with enclosure. Measurement practice often encloses sensitive sections to stabilize S-parameters against environmental coupling. Why Other Options Are Wrong: Capacitance/inductance do not go to zero; quasi-TEM propagation remains valid in shielded microstrip/stripline within practical dimensions. Claiming increased radiation contradicts shielding physics unless the enclosure is poorly designed with large slots. Common Pitfalls: Neglecting slot/aperture leaks or enclosure resonances which can re-introduce radiation if improperly implemented; forgetting to include via fences and RF gasketing in real products. Final Answer: Radiation losses are reduced compared with an open microstrip.

Question 4

Laboratory crystal detector (microwave power measurement) – Typical standing wave ratio In standard laboratory setups using a crystal (or Schottky) detector for relative microwave power measurements, what is a representative VSWR on the detector por…

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Introduction: Microwave crystal detectors are commonly used in labs for relative power measurements (for example, in slotted-line or coupler setups). Their input match affects measurement accuracy; thus the typical standing wave ratio (VSWR) at the detector port is specified and controlled to keep mismatch ripple and calibration errors small. Given Data / Assumptions: Detector used in its square-law region with appropriate impedance matching. Standard waveguide/coax fixtures with typical adapters and short leads. Goal is relative power measurement, not precision absolute power metrology. Concept / Approach: VSWR represents the degree of mismatch: VSWR = (1 + |Γ|) / (1 − |Γ|). Practical lab detectors are designed to keep reflections modest; a commonly quoted value around 1.25:1 corresponds to |Γ| ≈ 0.111, which is sufficiently low to reduce standing-wave errors while remaining achievable without overly complex matching networks over a useful bandwidth. Step-by-Step Solution: 1) Choose a typical, achievable match for detector heads → VSWR ≈ 1.25.2) Convert to reflection coefficient: |Γ| = (VSWR − 1) / (VSWR + 1) ≈ 0.25/2.25 ≈ 0.111.3) Recognize this keeps mismatch error small for relative readings and eases calibration using power meters or comparators. Verification / Alternative check: Datasheets for laboratory crystal detectors (waveguide or coax) commonly list VSWR values between 1.2 and 1.5 across band; 1.25 is a representative nominal figure balancing bandwidth and match complexity. Why Other Options Are Wrong: VSWR = 0.1 is impossible (minimum is 1). VSWR values of 5 or 10 are far too mismatched for lab measurement accuracy. 2.5 is higher than desirable and would produce large ripple in standing-wave systems and poor repeatability. Common Pitfalls: Confusing return loss (in dB) with VSWR; forgetting that good but not perfect matches are typical for broadband detector heads and that adapters and connectors also contribute to the overall VSWR seen by the system. Final Answer: 1.25.

Question 5

Loading of a Transmission Line: What Does “Loading” Mean?  In classical telephony and RF transmission engineering, “loading a line” refers to a deliberate design change to the line parameters. Which description best matches the technical meaning of …

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Introduction / Context: In wireline telephony and low-frequency RF practice, the term “loading” has a precise historical meaning: adding inductance to a line to improve its transmission over distance. Understanding what parameter is changed—and how it is changed—helps interpret classic loading-coil designs and modern distributed implementations. Given Data / Assumptions: The line is a two-wire or multi-conductor transmission line intended for voice/RF. Loading is implemented either with discrete loading coils at intervals or by making the series inductance per unit length larger (distributed loading). The design goal is improved amplitude response and reduced dispersion over a band of interest. Concept / Approach: A transmission line is characterized by per-unit-length R, L, G, and C. “Loading” specifically targets the series inductance L to shape the propagation constant and characteristic impedance. Increasing L (either by coils at spacing d, or continuously with magnetic materials or conductor geometry) flattens the attenuation over the voice band and pushes the line toward distortionless conditions (R/L ≈ G/C). Thus, the correct description is an increase of distributed inductance. Step-by-Step Solution: 1) Identify what “loading” historically meant: insertion of loading coils → increases effective L per unit length.2) Generalize to distributed loading: modify construction so L is higher everywhere, not only at coil locations.3) Conclude: loading means increasing distributed inductance of the line. Verification / Alternative check: Classical Heaviside condition for distortionless transmission is R/L = G/C. If R and G are fixed, increasing L can help meet the condition in the operating band, which corroborates the purpose of loading. Why Other Options Are Wrong: “Increasing inductance of line” (generic) omits the crucial notion of per-unit-length distribution and could be misconstrued as a single lump; “decreasing” variants contradict the observed practice; “adding shunt capacitance” increases C and is the opposite of classical inductive loading. Common Pitfalls: Confusing lumped one-time inductors with periodic/distributed loading; assuming loading improves high-frequency bandwidth (it typically narrows usable HF response while improving voice-band behavior). Final Answer: increasing distributed inductance of line

Question 6

Short-Circuited Stubs at the Source End of a Rectangular Waveguide  A rectangular waveguide has two short-circuited stubs (each of some length) connected at the source end for tuning. Which statement best describes the nature of the currents in the …

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Introduction / Context: Waveguide tuners commonly use short-circuited stubs to supply reactive compensation at the input, adjusting the apparent load seen by the source. Understanding where real power flows versus where reactive energy is stored is central to microwave matching theory. Given Data / Assumptions: Two short-circuited stubs are attached at the source end of a rectangular waveguide. The stubs are lossless and function as reactive elements (tunable susceptances). The main guide carries the forward traveling wave toward the load. Concept / Approach: Short-circuited stubs do not deliver average real power because the net time-average power at the short is zero (voltage is zero at the short, current is large and reactive). The stubs exchange energy with the field over an RF cycle (reactive current). The main longitudinal section transports real power toward the load and returns any reflected power toward the source. Thus, longitudinal current is associated with net power flow, whereas the stub currents are predominantly reactive. Step-by-Step Solution: 1) Short-circuited stub termination → instantaneous energy storage; average power at the short is zero.2) The equivalent input of a shorted stub is purely reactive (jB or −jX) at many lengths.3) The through waveguide section supports nonzero time-average Poynting vector → real power flow (power current). Verification / Alternative check: Smith chart analysis shows shorted stubs present susceptance or reactance; microwave tuners combine them to cancel load reactance without contributing power themselves. Why Other Options Are Wrong: Option B swaps roles erroneously; option C is vague and blurs power versus reactive flow; option D contradicts the existence of reactive currents; option E is false since ideal shorted stubs do not supply average power. Common Pitfalls: Assuming large currents imply large real power; forgetting that at a short, voltage is zero so average power is zero. Final Answer: the longitudinal current is power current and current in stubs is reactive

Question 7

Definition Check: Transmission-Line Discontinuity  A transmission-line discontinuity is defined as any interruption in the uniformity (geometry or material parameters) of the line. Is this statement correct?

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Introduction / Context: Accurate matching and modeling at RF/microwave frequencies requires recognizing what counts as a discontinuity. Discontinuities cause reflections and local field distortions, affecting return loss and bandwidth. Given Data / Assumptions: A “uniform” line has constant R, L, G, C per unit length or constant cross-section and material fill. Any geometric, dielectric, or conductive change breaks uniformity. We are considering general transmission structures: coax, microstrip, stripline, and waveguide. Concept / Approach: A discontinuity is any localized change in impedance or propagation constant. Examples include steps in width/height, connectors, bends, junctions, posts, irises, tapers, or dielectric inserts. These can be modeled by equivalent reactive/ resistive networks that create reflections (nonzero Γ) even if losses are low. Step-by-Step Solution: 1) Start from uniform line: Z0 and β constant → no reflection.2) Introduce change (e.g., width step): Z0 changes locally → Γ ≠ 0 → discontinuity exists.3) Therefore, the definition stating “any interruption in uniformity” is correct. Verification / Alternative check: Field solvers and Smith chart experiments verify that even small geometry changes introduce measurable S11, demonstrating discontinuity behavior. Why Other Options Are Wrong: Limitations to coax (C) or “resistive only” (D) are unfounded; discontinuities can be purely reactive. Threshold VSWR (E) is arbitrary; any nonzero reflection qualifies. Common Pitfalls: Ignoring very small features as “negligible”; at high frequency, even small steps can matter. Final Answer: True

Question 8

Polarization of Man-Made Noise Near the Earth  Interference generated by electrical equipment and automobile ignition systems, when observed close to the Earth's surface, is typically:

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Introduction / Context: Receiving antennas often encounter impulsive man-made noise from power lines, motors, and vehicle ignition. Knowing the usual polarization helps select antenna polarization to reduce interference and improve signal-to-noise ratio. Given Data / Assumptions: Observation is near the Earth where many emitters resemble vertical conductors (e.g., wires, leads, vehicle ignition harnesses). Local ground effects favor vertical radiation from such structures. Receiving systems can choose polarization. Concept / Approach: Short vertical wires and monopole-like structures predominantly radiate vertical polarization. Many sources of man-made interference near ground approximate vertical current elements; thus, their radiation near the surface is largely vertically polarized. Horizontal polarization is less common for these incidental emitters. Step-by-Step Solution: 1) Model interference sources as short monopoles (vertical current element).2) Recognize that the electric field is mainly vertical for such sources.3) Conclude that received interference close to Earth tends to be vertically polarized. Verification / Alternative check: EM measurements near power lines and vehicle ignition systems consistently show stronger coupling to vertically polarized antennas, validating the practical guidance. Why Other Options Are Wrong: Horizontal polarization (A) is not the dominant case for these emitters; “not polarized” (D) ignores the structured geometry; circular (E) is not typical of random switching/ignition noise; option C is incomplete/ambiguous. Common Pitfalls: Assuming noise is random in polarization; geometry of conductors sets a preferred polarization. Final Answer: vertically polarized

Question 9

Identify the Balanced Transmission Line  Which of the following transmission lines is inherently a balanced line (equal and opposite conductors with respect to ground)?

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Introduction / Context: Balanced versus unbalanced transmission determines how currents return and how a line couples to nearby structures. Balanced lines carry equal and opposite currents on two conductors with minimal common-mode radiation. Given Data / Assumptions: “Balanced” means symmetrical conductors with respect to ground. We compare common RF interconnect types. Concept / Approach: TV twin lead is the classic two-wire balanced line: both conductors are equivalent and neither is tied to ground. Coaxial line is unbalanced (outer conductor often at ground potential). Microstrip and single-ended stripline use a single signal conductor over/within a ground reference, also unbalanced. Step-by-Step Solution: 1) Examine geometry: twin-lead has two symmetric wires → balanced.2) Coax: central conductor plus outer shield → unbalanced.3) Microstrip/stripline (single-ended): one signal trace referenced to ground → unbalanced. Verification / Alternative check: Baluns (balanced-to-unbalanced transformers) are used to interface twin-lead to coax, highlighting the difference. Why Other Options Are Wrong: Coax and microstrip/stripline are unbalanced by default; waveguide supports TE/TM modes and is not a two-conductor balanced line. Common Pitfalls: Assuming “differential stripline” equals “strip line” generically; the default single-ended stripline is unbalanced unless specifically routed differentially as a pair. Final Answer: TV twin lead

Question 10

Ridge Waveguide vs. Rectangular Waveguide: Bandwidth, Size, and Loss  Consider the statements:  The usable bandwidth of a ridge waveguide is greater than that of a standard rectangular waveguide.  For a given dominant-mode (TE10) cutoff frequency, a…

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Introduction / Context: Ridge waveguides insert one or more ridges into a rectangular guide to lower cutoff and extend bandwidth. The trade-offs include fabrication complexity and increased loss due to higher surface currents near the ridges. Given Data / Assumptions: Comparison at the same target dominant-mode (TE10) cutoff frequency. Air-filled metallic guides with similar materials and finish. Focus on qualitative trends (bandwidth, physical size, conductor loss). Concept / Approach: Adding ridges increases the capacitance-like loading across the broad wall, lowering TE10 cutoff. For a given cutoff, the external dimensions can be reduced. The ridges also broaden the single-mode region, giving larger usable bandwidth. However, the narrowed field regions and enhanced surface currents around the ridge edges increase ohmic losses compared with a smooth rectangular guide. Step-by-Step Solution: 1) Bandwidth: ridge geometry supports wider single-mode bandwidth → statement 1 true.2) Size for same cutoff: lowered cutoff allows smaller outer dimensions → statement 2 true.3) Conductor loss: concentrated currents elevate loss compared with smooth walls → statement 3 true. Verification / Alternative check: Design charts and measured Q values show ridge guides have lower unloaded Q (higher conductor loss) and wider operational band than standard guides. Why Other Options Are Wrong: Any option omitting one of the correct statements is incomplete; “All” would imply further claims beyond 1–3 that are not presented here. Common Pitfalls: Confusing lowered cutoff with lower attenuation at all frequencies; while cutoff is lower, conductor losses may increase. Final Answer: 1, 2 and 3 only

Question 11

Frequencies Used in Radio Astronomy  To study electromagnetic radiation from astronomical objects, radio astronomers primarily use which part of the spectrum?

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Introduction / Context: Radio astronomy spans a broad range of frequencies, but practical observations from the ground cluster where the ionosphere and atmosphere are most transparent and antennas are feasible. Identifying the most commonly used region is helpful for system design and interference mitigation. Given Data / Assumptions: Ground-based observations (not space telescopes). Typical atmospheric windows and ionospheric transparency. Desire for reasonable angular resolution with practical dish sizes. Concept / Approach: The lower HF range is significantly affected by ionospheric absorption and refraction, limiting ground-based astronomical use. Much radio astronomy occurs in the centimeter to millimeter bands (microwaves), including key spectral lines such as the 1.42 GHz neutral hydrogen line (L band, at the edge of microwaves) and many molecular lines in the tens to hundreds of GHz. Microwaves offer atmospheric windows and allow high-resolution imaging with dishes and interferometers. Step-by-Step Solution: 1) Check atmospheric/ionospheric windows: microwaves contain strong windows where absorption is low.2) Telescope practicality: wavelengths from centimeters to millimeters allow manageable dish sizes with good resolution.3) Therefore, “microwaves” is the most accurate selection among the options provided. Verification / Alternative check: Major observatories (VLA, ALMA) operate predominantly in microwave and millimeter-wave bands, confirming practice. Why Other Options Are Wrong: Low/medium frequencies are constrained by the ionosphere and RFI; “either (a) or (b)” is too broad; infrared belongs to optical/IR astronomy, not radio astronomy. Common Pitfalls: Equating “radio” only with HF/VHF; modern radio astronomy spans up to hundreds of GHz. Final Answer: microwaves

Question 12

Reflex Klystron Lab Practice: Why Use Square-Wave Modulation and Square-Law Detection?  In a typical laboratory setup with a reflex klystron source and crystal detector, the output is examined while applying a square-wave modulation to the reflector…

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Introduction / Context: Reflex klystrons are often characterized in the lab with simple amplitude-detection receivers using crystal diodes. A small modulation of the reflector voltage creates an amplitude variation that the detector converts to DC for display or measurement. Given Data / Assumptions: Crystal (Schottky) detector used at low RF input power. Detector biased (or un-biased) so that it behaves in the square-law region. Square-wave modulation is applied at low frequency to sweep or dither the operating point. Concept / Approach: In the square-law region, detector output voltage V_out is approximately proportional to input RF power P_in (V_out ∝ P_in). Applying a low-rate square-wave modulation toggles the RF amplitude slightly, letting the detector deliver a corresponding square-wave DC variation that is easy to measure. This technique linearizes power readings and enables lock-in–style measurements without elaborate RF instrumentation. Step-by-Step Solution: 1) Operate the detector in square-law region → output is proportional to RF power.2) Modulate the klystron with a small square wave on the reflector → produce small-signal AM of the RF.3) Detect the AM with the square-law diode → obtain a low-frequency square wave proportional to power. Verification / Alternative check: Detector transfer curves show a quadratic region at low powers; demodulated outputs track input power changes linearly in that region. Why Other Options Are Wrong: Ease of generation (A) is incidental; (C) is incorrect—frequency may still vary with reflector voltage; (D) oversimplifies; (E) is false—matching is still important. Common Pitfalls: Driving outside square-law region (saturation) breaks proportionality; using excessive modulation that distorts measurement. Final Answer: crystal diode operates in the square-law region of its i–v characteristic

Question 13

Attenuator Loss vs. Forward Gain in a Traveling-Wave Tube (TWT)  A helix TWT includes an internal attenuator to suppress unwanted feedback. How does the loss of this attenuator compare with the forward gain of the tube?

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Introduction / Context: Helix TWTs often insert a resistive attenuator (lossy section) along the slow-wave structure. Its role is to damp backward waves and reflections that could drive oscillations, ensuring stable broadband amplification. Given Data / Assumptions: TWT designed as an amplifier (not oscillator). Attenuator is a localized lossy region or a distributed loss segment. Net small-signal gain of the TWT is typically tens of dB. Concept / Approach: The attenuator introduces only a few decibels of loss—just enough to kill unwanted feedback paths—while the electron-beam/helix interaction provides tens of decibels of forward gain. Therefore, the attenuator loss is much less than the forward gain so that the device still exhibits strong net amplification. Step-by-Step Solution: 1) Recognize purpose: suppress backward-wave build-up and cavity-like oscillations.2) Typical magnitudes: attenuator 3–10 dB; forward gain 20–50 dB (varies by design).3) Hence, loss ≪ gain, preserving net positive gain. Verification / Alternative check: Manufacturer datasheets show TWT gain budgets allocating small loss margins to internal attenuators while maintaining high overall gain. Why Other Options Are Wrong: “Almost equal” or “much higher” would negate amplifier utility; “almost infinite” is meaningless; “exactly the same” is unrealistic and unnecessary. Common Pitfalls: Confusing the attenuator with input/output couplers or incorrectly assuming high losses are required; excessive loss would simply reduce gain too much. Final Answer: much less than forward gain of tube

Question 14

Waveguide Construction: Identify the Incorrect Statement  For metallic waveguides used at microwave frequencies, which of the following statements is wrong?

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Introduction / Context: Waveguides are hollow metallic structures guiding TE/TM modes. Fabrication choices affect loss, bandwidth, and durability. Knowledge of correct construction principles avoids high-loss designs. Given Data / Assumptions: Normal microwave bands (e.g., X, Ku) with good conductors. Interest in qualitative correctness of statements about materials, geometry, and skin depth. Concept / Approach: At microwave frequencies, skin depth δ is very small; to minimize conductor loss and ensure mechanical strength, wall thickness is made many skin depths, not “very small compared to δ”. Rectangular and circular cross-sections are standard; brass, aluminium, and copper-plated surfaces are commonly used. Air is the usual dielectric fill to keep loss low. Step-by-Step Solution: 1) Evaluate materials: brass/aluminium are common → statement A correct.2) Shapes: rectangular and circular dominate → statement B correct.3) Dielectric: usually air → statement C correct.4) Skin depth: thickness must exceed several δ → D is incorrect as written.5) Copper plating lowers surface resistance → statement E correct. Verification / Alternative check: Skin-effect formulas show conductor loss varies with surface resistance; plating with high-conductivity copper/silver reduces loss and is common practice. Why Other Options Are Wrong: Only D contradicts standard engineering practice; the others reflect widely used constructions. Common Pitfalls: Misinterpreting “skin depth” and assuming any thickness suffices; in reality, multiple skin depths are required for low loss and structural integrity. Final Answer: The wall thickness is very small as compared to skin depth

Question 15

Magic-tee (hybrid-T) used as a continuous-wave (CW) duplexer Ports 1 and 2 are the collinear arms, port 3 is the E-plane (difference) arm, and port 4 is the H-plane (sum) arm. How should port 1, port 2, port 3 (E arm), and port 4 (H arm) be connecte…

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Introduction / Context: The magic-tee (hybrid-T) is a four-port waveguide junction formed by combining an H-plane tee (sum port) and an E-plane tee (difference port). It is often used as a simple continuous-wave (CW) duplexer so that a transmitter and a receiver can share a single antenna while maintaining isolation. Given Data / Assumptions: Port 4 is the H-arm (sum); port 3 is the E-arm (difference). Ports 1 and 2 are the collinear arms. Ideal, matched magic-tee behavior with correct phasing. Concept / Approach: Key properties: (1) Exciting the H-arm splits power equally and in phase to the collinear arms and is isolated from the E-arm. (2) Exciting the E-arm splits power equally but 180 degrees out of phase to the collinear arms and is isolated from the H-arm. To duplex CW, place the transmitter at the H-arm to feed the antenna (via one collinear arm) while isolating the receiver at the E-arm; terminate the unused collinear arm in a matched load to preserve the match and absorb the split power. Step-by-Step Solution: Assign port 4 (H-arm) → CW transmitter to exploit in-phase splitting.Assign port 3 (E-arm) → receiver for isolation from the transmitter.Use port 2 (one collinear arm) → antenna to radiate.Terminate port 1 (other collinear arm) → matched load to absorb the equal split (≈3 dB loss). Verification / Alternative check: The ideal S-matrix has S34 = S43 = 0 (E–H isolation). With the transmitter at the H-arm, the receiver at the E-arm sees minimal leakage in the ideal case; practical systems still add limiters and isolators to protect the receiver. Why Other Options Are Wrong: B/C/D misplace the transmitter/receiver or the matched termination, breaking isolation or match. Common Pitfalls: Expecting zero loss—one collinear arm must be terminated, giving a 3 dB split; ignoring receiver protection in real hardware. Final Answer: matched load, antenna, receiver and CW transmitter

Question 16

Rectangular waveguide cutoff condition — TE20 mode Let a be the broad-wall width and b be the narrow-wall height of a rectangular waveguide. What is the cutoff wavelength λ_c for the TE20 mode?

Accepted Answer

Introduction / Context: Each TEmn mode in a rectangular waveguide has a specific cutoff wavelength and frequency determined by the cross-sectional dimensions a (broad wall) and b (narrow wall). Recognizing these standard results is essential for band selection and single-mode operation. Given Data / Assumptions: Metal rectangular waveguide with dimensions a and b. Mode: TE20 (m = 2, n = 0). Lossless ideal walls for cutoff calculation. Concept / Approach: The general formula is λ_c = 2 / sqrt((m/a)^2 + (n/b)^2). For TE m0, this simplifies to λ_c = 2a/m. Substituting m = 2 gives λ_c = a. Step-by-Step Solution: Start: λ_c = 2 / sqrt((m/a)^2 + (n/b)^2).For TE20 → m=2, n=0 → λ_c = 2 / (2/a) = a.Therefore, the cutoff wavelength for TE20 equals the broad dimension a. Verification / Alternative check: Special cases: λ_c(TE10) = 2a and λ_c(TE01) = 2b; by the same pattern, TE20 → a. Handbooks and mode charts confirm this relationship. Why Other Options Are Wrong: a + b or a − b: no linear combination arises from the correct formula. a2 + b2: does not match the inverse-root dependence on dimensions. Common Pitfalls: Confusing TE10 dominance with TE20 cutoff; forgetting dependence on the m index across the broad wall. Final Answer: a

Question 17

Assertion–Reason: Relationship between TE10 and TE01 in rectangular waveguides Assertion (A): In a rectangular waveguide, the TE01 mode is simply the TE10 mode rotated by 90°. Reason (R): Rotating a rectangular waveguide by 90° interchanges the broa…

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Introduction / Context: Mode indices (m, n) in a rectangular waveguide count field variations across the broad and narrow dimensions. TE10 and TE01 are closely related by symmetry and physical rotation of the guide, which swaps the roles of a and b. Given Data / Assumptions: Dimensions: a (broad), b (narrow). Standard TE mode definitions: TE10 has one half-wave across a and none across b; TE01 has none across a and one across b. Concept / Approach: Physically rotating the guide by 90° interchanges a and b. The field pattern for TE10 in the rotated guide corresponds to TE01 in the original labeling. Hence the assertion is true, and the reason correctly explains it. Step-by-Step Solution: Identify TE10 → m=1 across a, n=0 across b.Rotate guide by 90° → a and b swap.Resulting pattern maps to TE01 (m=0, n=1) in the original coordinate system. Verification / Alternative check: Cutoff wavelengths confirm symmetry: λ_c(TE10) = 2a and λ_c(TE01) = 2b; rotation exchanges these values consistently. Why Other Options Are Wrong: B: The reason directly explains the assertion; it is not incidental. C/D: Incorrectly negate a true relationship derived from symmetry. Common Pitfalls: Mixing TE and TM families; assuming rotation changes the mode family instead of relabeling indices. Final Answer: Both A and R are correct and R is correct explanation of A

Question 18

Waveguide twists in microwave systems — primary function What is the principal purpose of a waveguide twist in a microwave system?

Accepted Answer

Introduction / Context: Waveguide plumbing includes components to route energy and control polarization. A waveguide twist is a gradual rotation of the guide about its axis, used to rotate the polarization of the dominant mode without introducing significant reflections or mode conversion. Given Data / Assumptions: Dominant TE10 operation in rectangular waveguide. Twist is sufficiently gradual for minimal mode conversion. Concept / Approach: In TE10, the electric field is oriented across the narrow dimension. Twisting the guide slowly about its axis rotates this electric field direction while keeping the propagation direction and the mode the same. Therefore, the primary role is polarization rotation, not steering or changing modes. Step-by-Step Solution: Start with TE10 polarized along a transverse axis.Apply an axial twist of angle θ → the field orientation rotates by θ.Maintain smooth geometry → reflections and higher-order mode excitation remain negligible. Verification / Alternative check: Measured return loss and insertion loss of properly designed twists show low reflection and negligible mode conversion, confirming polarization rotation as the function. Why Other Options Are Wrong: B: Changing direction is done by bends or elbows. C/D: Mode changes are not the intended function; abrupt twists can cause conversion but proper design avoids it. Common Pitfalls: Confusing polarization rotation with physical bending; using too short a twist that induces unwanted modes. Final Answer: change the plane of polarization of a propagating wave

Question 19

Field distribution in TE10 mode — location of maximum E-field For the dominant TE10 mode in a rectangular waveguide, where does the electric field have its maximum magnitude?

Accepted Answer

Introduction / Context: Knowing where fields peak inside a waveguide is critical for breakdown assessment, probe placement, and coupling design. In TE10, the dominant transverse electric field varies sinusoidally across the broad wall dimension. Given Data / Assumptions: Rectangular waveguide with width a (broad wall) and height b (narrow wall). Mode: TE10 (m = 1, n = 0). Concept / Approach: The principal field component varies as a cosine function across the a-dimension and is uniform across b. Boundary conditions enforce zero tangential electric field at the conducting walls (x = 0 and x = a), with a maximum at the midpoint x = a/2. Thus, the electric field maximum lies along the center line of the broad wall. Step-by-Step Solution: Model transverse field: E(x) ∝ cos(π x / a).At x = 0 and x = a → conductor boundary condition forces field components to zero.Maximum occurs at x = a/2 → center line across the broad wall. Verification / Alternative check: Standard field plots and coupling-probe placements show strongest coupling at x = a/2 for TE10, confirming the location of the E-field maximum. Why Other Options Are Wrong: B: The field is zero at the side walls, not maximum. C/D: Contradict the sinusoidal distribution and boundary conditions. Common Pitfalls: Confusing TE10 with higher-order modes; misapplying boundary conditions at conductor walls. Final Answer: along centre line of broad wall (i.e. x = a/2)

Question 20

Mode analogy between circular and square waveguides The TM01 mode in a circular waveguide is most closely analogous (in symmetry and nodal pattern) to which TM mode in a square waveguide?

Accepted Answer

Introduction / Context: Although circular and rectangular (square) waveguides have different eigenfunctions (Bessel versus sines/cosines), useful analogies can be drawn based on symmetry and the number of field variations across the aperture. This helps engineers transfer intuition between geometries. Given Data / Assumptions: Circular waveguide mode: TM01 (no azimuthal variation, single radial lobe with central maximum). Square waveguide modes: TMmn with m and n variations along the two transverse axes. Concept / Approach: TM01 in circular has an on-axis maximum and no angular variation. The closest square-guide analogue is a mode with one half-wave variation along one axis and none along the other, namely TM10 (or by symmetry TM01). From the provided options, TM10 best matches the symmetry and nodal structure of circular TM01. Step-by-Step Solution: Identify TM01 features: no angular dependence; single central lobe; first radial zero at the wall.Map to square: one transverse variation, the other uniform → TM10 in the option set.Select TM10 as the best analogy. Verification / Alternative check: Mode charts show the lowest TM in circular is TM01; in rectangular/square, the simplest TM with a single variation is TM10/TM01, supporting the analogy. Why Other Options Are Wrong: TM11: two transverse variations, not analogous in symmetry to TM01. TM20/TM02: higher-order patterns with two half-waves along one axis. Common Pitfalls: Assuming identical cutoff constants; analogies are qualitative, based on symmetry and nodal structure. Final Answer: TM10 mode in square waveguide

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