Question 1

Antenna reference for directive gain: Which reference antenna is used to define directive gain in antenna engineering?

Accepted Answer

Introduction / Context:Directive gain compares an antenna's maximum radiation intensity to that of a reference antenna with the same total radiated power. Understanding the reference standard is key to correctly interpreting gain specifications in datasheets and link budgets. Given Data / Assumptions: Directive gain (not power gain) is defined relative to an isotropic radiator. The isotropic radiator radiates uniformly in all directions (4π steradians). Concept / Approach:By definition, directive gain D = U_max / (P_rad / (4π)), where U_max is the maximum radiation intensity and P_rad is total radiated power. The denominator corresponds to an isotropic radiator's uniform intensity. Hence, the isotropic antenna is the reference for directive gain. Step-by-Step Solution: Recall that dBi (decibels relative to isotropic) is the common unit for gain.A half-wave dipole's directivity is about 1.64 (2.15 dBi), which illustrates that the dipole is not the reference; it has gain above isotropic. Verification / Alternative check:Standards and textbooks define directive gain with the isotropic reference, while power gain may also be reported relative to isotropic (dBi) or to a dipole (dBd = dBi − 2.15 dB). Why Other Options Are Wrong: Half-wave dipole / elementary doublet / infinitesimal dipole: real or canonical antennas, but not the reference for directive gain. Common Pitfalls: Confusing dBi (isotropic reference) with dBd (dipole reference). Final Answer: Isotropic

Question 2

Noise modeling in BJTs: How many fundamental noise sources are typically modeled for a bipolar junction transistor in small-signal analysis?

Accepted Answer

Introduction / Context:Accurate noise modeling is crucial for low-noise amplifiers and precision analog design. For small-signal BJTs, an equivalent input-referred noise model captures the essential random processes with a minimal set of sources. Given Data / Assumptions: Small-signal, midband operation. Equivalent noise model uses input-referred sources. Low-frequency 1/f noise may be considered separately. Concept / Approach:In the widely used two-source model, a BJT is represented by an input-referred noise voltage source (primarily due to base resistance thermal noise and correlated effects) in series with the input, and an input-referred noise current source (primarily due to base/collector shot noise) in parallel with the input. These two sources are sufficient to predict output noise with source resistance and circuit gain. Step-by-Step Solution: Represent base spreading resistance thermal noise as an equivalent voltage source v_n.Represent base/collector junction shot noise as an equivalent current source i_n at the input.Use standard noise analysis to compute total output noise given source resistance and transfer functions. Verification / Alternative check:Many texts present more detailed physical sources (thermal, shot, flicker), but the input-referred representation condenses them to two uncorrelated sources for most design calculations. Why Other Options Are Wrong: 1: inadequate to capture both voltage and current noise contributions.3 or 4: overcounts relative to the standard compact two-source representation used in design. Common Pitfalls: Confusing physical processes (shot, thermal, flicker) with the simplified input-referred model (two sources). Final Answer: 2

Question 3

In antenna grounding for communication towers, what does an effective earth mat consist of and at approximately what burial depth should the radials be installed?

Accepted Answer

Introduction / Context:An earth mat (also called a radial ground system) is critical in radio and communication towers to ensure personnel safety, provide a low-impedance return for lightning and fault currents, and stabilize the radio-frequency (RF) ground. The design impacts lightning protection, electromagnetic compatibility, and system efficiency, especially at HF/VHF where ground losses can dominate. Given Data / Assumptions: Tower base requires an earth mat for RF grounding and lightning dissipation. Options vary by the number of radials and their burial depth. Soil is assumed to be normal (not permafrost or extremely rocky). Concept / Approach: Ground systems lower the effective ground resistance/impedance. Many moderately shallow radials create wide-area contact with soil, reducing RF ground losses more effectively than a few deep conductors. At RF, current flows primarily near the surface; therefore, shallow, numerous radials are preferred to deep burial. Typical practice is 15–120 radials (application-dependent), laid 0.1–0.5 m below grade, with 0.3 m (30 cm) being a common, practical depth balancing protection and installation ease. Step-by-Step Solution: Identify best RF grounding practice → multiple shallow radials beat a few deep ones for RF impedance reduction.Compare depths: 5 cm is too shallow (exposed to damage, drying); 3 m is unnecessary and costly for RF performance.Typical recommended depth ≈ 0.3 m for robustness and performance → matches the option with many radials at ~30 cm depth. Verification / Alternative check: Broadcast standards and lightning protection guides consistently show that radial systems of many conductors at shallow depth minimize RF losses and provide effective surge paths, corroborating the 30 cm practice. Why Other Options Are Wrong: One or two shallow radials (5 cm): inadequate coverage and mechanically vulnerable. Deep 3 m burial: unnecessary for RF; does not maximize the near-surface current spread. Single deep rod only: good for DC/low-frequency lightning current, not optimal for RF ground. Common Pitfalls: Confusing DC grounding (deep rods) with RF grounding (wide-area shallow radials). Underestimating soil stratification and moisture variation. Final Answer: A large number of radials from the tower base buried about 30 cm deep

Question 4

In the CCIR-B (625-line, 50 Hz field rate) television system, what is the time interval between the start of one horizontal sync (H-sync) pulse and the start of the next?

Accepted Answer

Introduction / Context:Horizontal timing in analog TV standards like CCIR-B (PAL, 625 lines) determines line frequency, picture sampling along each line, and the placement of sync pulses. Knowing the line period is fundamental for scan timing, sync separation, and deflection system design. Given Data / Assumptions: CCIR-B standard: 625 lines per frame, interlaced, 50 fields per second. Horizontal line frequency ≈ 15.625 kHz. We need the period between successive H-sync pulse starts (i.e., one full line period). Concept / Approach: The horizontal line period T_H is the reciprocal of the line frequency f_H. In CCIR-B, f_H ≈ 15,625 Hz, so T_H ≈ 1 / 15,625 s. Step-by-Step Solution: f_H ≈ 15,625 Hz.T_H = 1 / f_H = 1 / 15,625 s ≈ 64 × 10^-6 s.Therefore, the time from one H-sync start to the next is approximately 64 μs. Verification / Alternative check: Standard PAL timing tables list a 64 μs line period with ~52 μs active video and ~12 μs blanking/sync, confirming the 64 μs figure. Why Other Options Are Wrong: 6.4 μs and 0.64 μs: off by factors of 10 and 100; much faster than CCIR-B line rates. 640 μs: off by factor of 10; would imply 1,562.5 Hz line rate, incorrect. 32 μs: half the correct value; not used in this standard. Common Pitfalls: Confusing field period (20 ms) and line period (64 μs), or mixing NTSC (525/60) and PAL (625/50) parameters. Final Answer: 64 μs

Question 5

A cordless telephone that uses separate radio frequencies for transmission from the base to the handset and from the handset back to the base is operating in which mode?

Accepted Answer

Introduction / Context:Telephony requires two-way communication. Two common approaches are duplex (simultaneous talk and listen) and half-duplex (talk or listen at a time). Cordless systems often implement frequency-division duplex (FDD) with distinct uplink and downlink frequencies. Given Data / Assumptions: Base and handset transmit on separate frequencies. No time alternation is described, implying simultaneous paths are possible. Concept / Approach: If two distinct frequencies are used concurrently—one for base-to-handset and another for handset-to-base—the system is frequency-division duplex (FDD), i.e., duplex operation. Half-duplex would share a single channel over time (time-division), not separate frequencies simultaneously. Step-by-Step Solution: Identify the presence of separate, simultaneous frequency paths → implies duplex.Rule out half-duplex → would not need two different concurrent frequencies; it alternates direction over one channel.Thus, the correct operating mode is duplex. Verification / Alternative check: Consumer cordless standards (e.g., early analog cordless phones) commonly used paired channels (uplink/downlink) to allow full-duplex conversation. Why Other Options Are Wrong: Half-duplex: not simultaneous; usually one channel shared in time. Either/neither: contradicts the clear FDD description. Simplex: one-way only, inapplicable to telephony. Common Pitfalls: Confusing duplex (simultaneous two-way) with half-duplex (alternating one-way) or assuming time-division when frequency-division is stated. Final Answer: Duplex arrangement (simultaneous two-way using different frequencies)

Question 6

In microphone specifications, the signal-to-noise ratio (SNR) is defined as the ratio of the output level produced by a standard reference sound pressure to the output in the absence of any sound. Which reference pressure (in pascals) is commonly us…

Accepted Answer

Introduction / Context:Signal-to-noise ratio (SNR) for microphones quantifies how much usable signal is obtained relative to internal self-noise. A consistent reference sound pressure is required so manufacturer ratings are comparable across devices and datasheets. Given Data / Assumptions: We need the standard reference sound pressure used to define SNR in many practical microphone specs. The options are 0.5, 0.2, 0.1, and 0.01 Pa (pascals). Concept / Approach: Common industry practice uses a fixed reference tone and level. While 1 Pa corresponds to 94 dB SPL and is widely used for sensitivity ratings, many SNR expressions in practice also cite 0.1 Pa (≈ 74 dB SPL) as a lower-level reference to highlight usable output above self-noise for typical conversational levels. Given the provided options, 0.1 Pa is the most standard among them for such SNR definitions in basic question banks. Step-by-Step Solution: Translate SPL to pressure: SPL (dB) = 20 * log10(p / 20e-6).For p = 0.1 Pa → SPL ≈ 20 * log10(0.1 / 20e-6) ≈ 74 dB SPL.This level is reasonable for SNR comparison relative to microphone self-noise and matches common didactic references. Verification / Alternative check: Manufacturers often provide both sensitivity at 1 Pa (94 dB SPL) and self-noise in dBA; SNR can be extrapolated at alternate reference levels such as 0.1 Pa. Of the choices offered, 0.1 Pa is the accepted value in many exam contexts. Why Other Options Are Wrong: 0.5 Pa and 0.2 Pa: Higher than typical reference used in such SNR questions. 0.01 Pa: Very low (≈ 54 dB SPL), not the standard teaching reference here. 1.0 Pa: Common for sensitivity, but not among the provided answer set in this question. Common Pitfalls: Confusing microphone sensitivity reference (often 1 Pa) with the SNR reference used in particular problem sets. Always check which reference the question expects. Final Answer: 0.1 Pa

Question 7

In the video intermediate-frequency (IF) strip of a color television receiver, which types of frequency traps are commonly used to shape the response and reject unwanted components?

Accepted Answer

Introduction / Context:Color TV IF strips must pass the desired video spectrum while rejecting the sound carrier, the color subcarrier image components, and adjacent-channel interference. Traps are frequency-selective networks added to notch out specific frequencies without unduly distorting the video passband. Given Data / Assumptions: Receiver is a color TV with a standard video IF around 38.9–45.75 MHz (standard-dependent). Different traps (series, shunt, absorption) can be used. Concept / Approach: Shunt and series traps implement notch responses with different insertion behaviors. Absorption traps (loosely coupled resonators) absorb energy at unwanted frequencies to create a deep notch while minimally disturbing adjacent passband regions. In practice, designers mix these techniques to achieve the required response template. Step-by-Step Solution: Identify unwanted lines (e.g., sound carrier spur, chroma-related components, adjacent channel).Choose traps: shunt for shunting unwanted frequency to ground; series to block the unwanted path; absorption for a broad, controlled notch.Combine as needed to meet the channel mask and preserve luminance/chrominance fidelity. Verification / Alternative check: Service manuals and classic IF design texts show multiple trap types in one IF strip to precisely shape the response curve. Why Other Options Are Wrong: Limiting to one trap type rarely meets stringent mask requirements. “None” contradicts standard IF design practice. Common Pitfalls: Believing a single trap type suffices; underestimating the role of coupling factor and Q in realizing deep, narrow notches without ringing. Final Answer: All of the above

Question 8

Consider the following statements about frequency modulation (FM): (1) The amplitude of an FM wave remains constant. (2) FM is generally more immune to noise than AM for the same received power. (3) FM broadcasts operate in the upper VHF and UHF ran…

Accepted Answer

Introduction / Context:FM is ubiquitous in broadcast audio and TV sound because of noise performance and capture effect. Understanding what is truly characteristic of FM helps avoid design and exam pitfalls. Given Data / Assumptions: (1) Amplitude of FM: does it stay constant? (2) Noise immunity: FM vs AM. (3) Broadcast bands: common FM services. (4) Relative equipment simplicity: FM vs AM. Concept / Approach: (1) In ideal FM, information is carried by instantaneous frequency deviation; amplitude is constant (limited by limiters). (2) FM exhibits better noise immunity due to amplitude limiting and pre-emphasis/de-emphasis, for the same received C/N. (3) Broadcast FM (radio) uses VHF; TV sound historically used FM in VHF and UHF channels—thus “upper VHF and UHF” is accepted in broadcast contexts. (4) FM RF/IF and demodulation chains (discriminators, PLL) are generally more complex than AM, so (4) is not correct. Step-by-Step Solution: Mark true: (1) constant amplitude; (2) better noise immunity; (3) broadcast usage in upper VHF and UHF contexts.Mark false: (4) FM gear is not simpler than AM.Therefore, the correct set is statements 1, 2, and 3. Verification / Alternative check: Block diagrams of FM receivers include limiters and discriminators; AM uses envelope detectors, typically simpler. Spectrum allocations place FM radio in VHF; TV sound (FM) occurred in both VHF and UHF bands. Why Other Options Are Wrong: Including (4) is incorrect; FM is not simpler than AM. Excluding (2) ignores FM's noise advantage; excluding (3) ignores TV FM sound allocations historically in UHF/VHF. Common Pitfalls: Equating only “FM radio” with VHF and forgetting TV sound; assuming FM equipment simplicity comparable to AM. Final Answer: 1, 2, 3

Question 9

In color television systems, how are the chrominance subcarrier and its sidebands fitted into the standard channel bandwidth without severe interference to the luminance spectrum?

Accepted Answer

Introduction / Context:Color TV adds chrominance information via a subcarrier and sidebands to an existing luminance signal within a fixed channel width. Clever spectral placement prevents visible cross-talk patterns while maintaining compatibility with monochrome receivers. Given Data / Assumptions: Chrominance is modulated onto a subcarrier (e.g., PAL ~4.43 MHz). Luminance spectrum contains strong line-rate harmonics. Goal: fit chroma within the same channel without objectionable interference. Concept / Approach: Frequency interleaving places the subcarrier at a frequency offset such that its sidebands fall between dominant luminance spectral lines (which are concentrated at multiples of half the line frequency), effectively interleaving spectra and reducing cross-luminance and cross-color artifacts. Step-by-Step Solution: Recognize the line-locked nature of luminance spectral lines.Choose subcarrier frequency related to half the line frequency to interleave chroma.Result: chroma sidebands occupy spectral valleys between luminance components. Verification / Alternative check: PAL/NTSC design uses subcarrier frequencies carefully tied to line frequency fractions (e.g., 283.75 * f_H in NTSC) to achieve interleaving. Why Other Options Are Wrong: “Adjustment/Changing/Amalgamation” are vague and do not capture the line-locked interleaving principle. Time-division techniques are not how analog color subcarrier coexists in the channel. Common Pitfalls: Assuming simple filtering suffices; in reality, precise frequency planning is essential for compatibility and artifact control. Final Answer: Frequency interleaving

Question 10

As radio frequency increases, how does the absorption (attenuation) of a ground wave by the Earth’s surface typically change?

Accepted Answer

Introduction / Context:Ground-wave propagation dominates at low and medium frequencies. Understanding frequency-dependent ground losses helps in selecting operating bands and predicting coverage for AM broadcast and LF/MF communication systems. Given Data / Assumptions: Ground wave hugs the Earth's surface and interacts with lossy ground. Soil has finite conductivity and permittivity, causing frequency-dependent attenuation. Concept / Approach: As frequency rises, the ground's effective loss tangent and surface impedance lead to greater attenuation of waves traveling along the surface. Hence, ground-wave range shortens with higher frequency, motivating MF/LF for long-range ground-wave services. Step-by-Step Solution: Consider the surface-wave attenuation factor which increases with frequency for typical earth conductivities.Higher f → smaller skin depth in soil → more loss → shorter ground-wave range.Therefore, absorption increases with frequency. Verification / Alternative check: Coverage charts for AM broadcast show reduced ground-wave range at higher MF compared to lower MF/LF bands for comparable power and soil conditions. Why Other Options Are Wrong: Decreases/Remains same: contradict empirical data and surface-wave theory. Independence above a threshold is not accurate for real soil parameters. Common Pitfalls: Confusing sky-wave behavior (ionospheric) with ground-wave; assuming free-space loss applies unchanged to surface waves. Final Answer: Increases with increasing frequency

Question 11

How does the velocity of sound waves in air vary with temperature, assuming normal atmospheric conditions?

Accepted Answer

Introduction / Context:The speed of sound in gases is a fundamental acoustic property that influences microphone calibration, room acoustics, sonar, and atmospheric acoustics. Temperature dependence is particularly important for precise measurement and system design. Given Data / Assumptions: Ideal-gas approximation for air. Neglect humidity and wind for the basic relation. Concept / Approach: For an ideal gas, speed of sound c = sqrt(γ * R * T / M), where γ is ratio of specific heats, R is the universal gas constant, M is molar mass, and T is absolute temperature. Thus, c ∝ sqrt(T). Step-by-Step Solution: Write relationship: c = sqrt(γ * R * T / M).Hold γ, R, and M constant for air → c ∝ sqrt(T).Therefore, as T increases, c increases with the square root of T. Verification / Alternative check: Empirical formula: c ≈ 331 m/s + 0.6 m/s per °C at moderate temperatures; this approximates the sqrt(T) dependence around standard conditions. Why Other Options Are Wrong: Constant: contradicts both theory and measurement. Directly proportional to T or inversely proportional to T: incorrect functional dependence. Independent of humidity/temperature: humidity has a small effect; temperature a major one. Common Pitfalls: Assuming a linear dependence across all temperatures; the linear relation is only a local approximation. Final Answer: Directly proportional to the square root of absolute temperature

Question 12

In antenna systems, a counterpoise (an elevated or radial ground network) is used under which site condition to provide an effective RF ground return?

Accepted Answer

Introduction / Context:A counterpoise is a network of wires or radials used to provide a low-impedance RF return when the Earth does not provide a good ground. It is essential for efficient radiation and matched operation of vertical antennas and coils connected to ground in high-resistivity locations. Given Data / Assumptions: Soil conductivity varies with location; deserts and rocky terrains are high resistivity. Question asks where counterpoise is used to compensate for poor natural ground. Concept / Approach: In high resistivity soil, ground losses are high; RF currents cannot return efficiently via the Earth, increasing system loss and detuning. A counterpoise creates an artificial ground plane close to the antenna feed, reducing loss and stabilizing impedance. Step-by-Step Solution: Assess ground quality: high resistivity → poor RF ground.Apply counterpoise: install elevated or buried radials to create large-area capacitive coupling to Earth, lowering effective ground impedance.Result: improved radiation efficiency and stable tuning. Verification / Alternative check: Field practice in AM broadcast and HF installations: counterpoises used where soil is poor; over seawater (low resistivity), natural ground is often excellent and counterpoise may be unnecessary. Why Other Options Are Wrong: Low/medium resistivity soils provide acceptable natural ground; counterpoise is not critical. Over sea water: natural ground is already excellent; a counterpoise is rarely required. Common Pitfalls: Confusing lightning safety grounding (which always benefits from rods) with RF ground return needs; the counterpoise addresses RF, not just DC. Final Answer: High resistivity soil (poor natural ground)

Question 13

Modem classification: A modem is generally classified as low-speed if the data rate it can handle is up to what value?

Accepted Answer

Introduction / Context:A modem (modulator-demodulator) is an essential communication device that converts digital signals into analog for transmission and back again at the receiving end. Historically, modems have been classified based on their data transmission speed in bits per second (bps). This classification helps identify applications for low, medium, and high-speed data transfer. Given Data / Assumptions: We are considering traditional wired telecommunication modems. Low-speed classification is being asked in terms of maximum data rate in bps. Choices range between 100 and 600 bps. Concept / Approach:Low-speed modems were widely used in early telecommunication systems, typically handling telegraphy, teletypewriters, and early data terminals. These were classified as handling speeds up to 600 bps. Beyond this, modems were considered medium-speed (1200 bps to a few kbps) and high-speed (tens of kbps and above). Step-by-Step Solution: Check classification boundaries: Early CCITT (now ITU-T) recommendations often marked 600 bps as the cutoff between low-speed and medium-speed modems.Options 100, 250, 400 bps are within the low-speed range but not standard cutoff points.600 bps is the recognized threshold value for classifying low-speed modems. Verification / Alternative check:Referencing communication system textbooks confirms that 'low-speed' was defined up to 600 bps, often used for telegraph services and simple data exchange applications. Why Other Options Are Wrong: 'upto 100 bps': too restrictive; excludes 300–600 bps modems widely used in early systems.'upto 250 bps' or '400 bps': arbitrary, not recognized classification thresholds.'upto 1200 bps': corresponds to early medium-speed modems, not low-speed. Common Pitfalls: Assuming that 'low-speed' means extremely slow like 100 bps only.Not realizing that classification has standard boundaries defined by telecommunication standards. Final Answer: upto 600 bps

Question 14

Carrier wave properties: Does a carrier wave itself carry information before modulation is applied?

Accepted Answer

Introduction / Context:A carrier wave is a high-frequency sinusoidal signal used in communication systems. It acts as a vehicle for information by being modulated in amplitude, frequency, or phase. This question checks whether learners understand the role of a carrier before modulation. Given Data / Assumptions: Carrier wave is defined as Acos(2πfct + φ), with fixed amplitude, frequency, and phase. No modulation is applied in the base case. Concept / Approach:A pure unmodulated carrier is deterministic and contains no message or information. Only after modulation (AM, FM, PM, etc.) does the carrier convey useful information. Therefore, the correct answer is 'False.' Step-by-Step Solution: Consider Acos(2πfct). Without modulation, it has constant amplitude and frequency.Information requires variation corresponding to a message signal.Thus, an unmodulated carrier contains no message information. Verification / Alternative check:Communication system references emphasize that the carrier is a 'blank canvas' for modulation. Spectrum analysis of a pure carrier shows a single frequency spike with no sidebands, confirming no information content. Why Other Options Are Wrong: 'True': incorrect, as an unmodulated carrier has no information.'Only in digital systems' / 'Only in wireless transmission': irrelevant distinctions; same concept applies everywhere.'Depends on amplitude': amplitude alone without modulation does not encode information. Common Pitfalls: Confusing a carrier's ability to carry information with the statement that it already contains information. Final Answer: False

Question 15

Audio systems: A woofer, which is designed to reproduce low-frequency sounds, should be connected to the input through which type of filter?

Accepted Answer

Introduction / Context:Woofers are specialized loudspeakers designed to handle bass and sub-bass frequencies, typically below 200 Hz. To ensure they receive only the appropriate frequency range, crossover filters are applied. Given Data / Assumptions: Woofer frequency range: ~20 Hz to 200 Hz. Input audio contains the full frequency spectrum. A filter is required to ensure only low frequencies reach the woofer. Concept / Approach:A low pass filter allows low-frequency signals to pass while attenuating higher-frequency components. This ensures the woofer does not attempt to reproduce mid or high frequencies, preventing distortion and ensuring fidelity. Step-by-Step Solution: Full audio range is split: woofer (lows), midrange speaker (mids), tweeter (highs).Crossover circuit assigns ranges using filters: low-pass for woofer, band-pass for midrange, high-pass for tweeter.Therefore, correct filter for woofer = low-pass filter. Verification / Alternative check:Commercial audio crossover designs confirm woofers are fed via low-pass sections tuned at crossover frequencies (100–200 Hz in home audio; 80–120 Hz in car systems). Why Other Options Are Wrong: High-pass: used for tweeters.Band-pass: used for midrange drivers.Band-stop / notch: used to eliminate narrow bands of interference, not for speaker crossovers. Common Pitfalls: Believing woofers can handle all audio input; while physically possible, distortion and inefficiency result without filtering. Final Answer: low pass filter

Question 16

In a colour television (TV) receiver front end, what is the primary function for which a varactor (varicap) diode is employed?

Accepted Answer

Introduction / Context:Varactor (also called varicap) diodes are junction diodes designed to exploit the voltage-dependent junction capacitance. In television receivers—especially colour TV front ends—precise, stable, and remote-controllable tuning over radio-frequency (RF) channels is essential. Varactors enable voltage-controlled tuning of oscillators and RF tuned circuits without mechanical variable capacitors. Given Data / Assumptions: The device in question is a varactor diode used within a colour TV receiver. Receiver requires RF and local oscillator tuning across assigned TV channels. Options include detection, rectification, and tuning functions. Concept / Approach: A reverse-biased varactor diode exhibits a capacitance Cj(Vr) that decreases as reverse voltage increases. By placing varactors in resonant circuits (RF tuned circuits, local oscillator tanks), the resonant frequency f0 = 1 / (2 * pi * sqrt(L * Ctotal)) can be swept by adjusting a DC tuning voltage. This provides electronic channel selection, fine tuning, and automatic frequency-control features. Detection and rectification are not the varactor’s intended roles in TV receivers; those are performed by detector diodes or rectifier diodes in the power supply, respectively. Step-by-Step Solution: Identify device characteristics: a varactor is a voltage-variable capacitor realized by a reverse-biased junction.Match device capability to TV receiver needs: electronic tuning of RF/LO circuits requires a variable capacitor controlled by a DC voltage.Conclude: the varactor diode in a colour TV receiver is used for tuning. Verification / Alternative check: Service manuals and RF textbooks list multiple varactors in front-end tuner modules for band selection and fine tuning controlled by a tuning voltage (often from a phase-locked loop or tuning potentiometer network). Why Other Options Are Wrong: Detection: performed by dedicated detector stages (envelope/product detectors), not by varactors. Rectification: a task for rectifier diodes in the power supply, not for varactors. Both detection and rectification: conflates unrelated roles. AGC action: controlled gain elements, not varactor capacitance. Common Pitfalls: Confusing the generic term “diode” with specific diode types; assuming any diode can detect or rectify. Varactor diodes are specialized for capacitance variation, not for power rectification or demodulation. Final Answer: Electronic tuning (voltage-controlled capacitance)

Question 17

Frequency-shift keying (FSK) is most commonly applied in which practical communication service?

Accepted Answer

Introduction / Context:Frequency-shift keying (FSK) is a digital modulation method that conveys binary symbols by shifting the frequency of a carrier between discrete values. Understanding which services predominantly use FSK clarifies the mapping between modulation families and application domains in communication systems. Given Data / Assumptions: FSK = digital keying by frequency shifting between symbol states. We compare with telephony, satellite analog TV, broadcast radio, etc. Concept / Approach: FSK was historically and is currently used in telegraphy and low-to-moderate data-rate links (e.g., RTTY, paging, low-speed modems, wireless sensor links). These applications require robust symbol recognition under noise and fading at modest bandwidths. Analog telephony and analog broadcast systems primarily use amplitude or frequency modulation of continuous waveforms—not symbol-by-symbol frequency hopping between tones as in FSK. Step-by-Step Solution: Map the modulation: FSK → digital symbol mapping using multiple tones.Map the service: telegraphy/data keying systems (e.g., RTTY) → classic FSK usage.Conclude FSK is mostly used in telegraphy / digital keying contexts. Verification / Alternative check: Standard textbooks list FSK, MSK, GFSK for telemetry, paging, and control channels; telegraphy is the archetypal example of FSK usage. Why Other Options Are Wrong: Telephony (analog PSTN) uses analog voice; not digital FSK over the entire call path. Analog satellite TV used FM video carriers, not symbol-by-symbol FSK. AM/FM radio broadcasting uses analog modulation, not FSK for program audio. FM stereo pilot is a single fixed tone (19 kHz), not FSK. Common Pitfalls: Assuming “frequency modulation” used in broadcast FM is the same as “frequency-shift keying.” FM is continuous deviation; FSK uses discrete frequency states for symbols. Final Answer: Telegraphy / data keying (digital keying systems)

Question 18

Among the three basic BJT configurations (common emitter, common base, common collector), which configuration exhibits the highest input impedance?

Accepted Answer

Introduction / Context:Small-signal amplifier design often starts by choosing a BJT configuration based on input/output impedance and gain needs. Recognizing which configuration gives the highest input impedance is vital for buffering and impedance matching, especially when interfacing high-impedance signal sources. Given Data / Assumptions: Compare CE, CB, and CC configurations qualitatively. Assume typical transistor parameters (moderate beta, room temperature), linear region. Concept / Approach: The common collector (emitter follower) presents a high input impedance because the input sees the base, and the emitter follows with near-unity gain, reducing the effective base current demand. CB has very low input impedance (input at emitter), and CE has moderate input impedance set by r_pi in parallel with biasing networks. Thus, CC > CE >> CB for input impedance, generally speaking. Step-by-Step Solution: Recall impedance ranking: Z_in(CC) is highest, Z_in(CE) moderate, Z_in(CB) lowest.Match to application: buffers use CC to avoid loading a source.Therefore, the configuration with the highest input impedance is CC (emitter follower). Verification / Alternative check: Hybrid-pi or T-model analyses show CC input resistance roughly beta times the emitter resistance seen, plus base bias network, giving large values compared to CE or CB under similar conditions. Why Other Options Are Wrong: CE: moderate input impedance; not the highest. CB: very low input impedance; the lowest of the three. Push–pull and cascode are multi-transistor topologies, not the basic single-transistor configurations asked. Common Pitfalls: Equating voltage gain with impedance; CC has near-unity voltage gain but excels as a buffer due to its high input and low output impedance. Final Answer: Common collector (CC, emitter follower)

Question 19

In a colour picture tube (CRT) used in television, what does “degaussing” refer to and why is it performed?

Accepted Answer

Introduction / Context:Colour CRTs employ a shadow mask and phosphor triads. Residual magnetization of metallic parts (mask, cabinet shielding) distorts electron beam landing, causing colour purity errors and blotches. Degaussing corrects this by demagnetizing ferromagnetic components near the display area. Given Data / Assumptions: Colour CRT with a shadow mask. Presence of residual magnetism from transport, nearby speakers, or Earth's field exposure. Degaussing coil present around the CRT face. Concept / Approach: Degaussing applies a decaying alternating magnetic field that randomizes magnetic domains, driving the net magnetization toward zero. This restores correct beam landing on red, green, and blue phosphors, improving colour purity. It does not increase or decrease coil resistance for time constants; such options confuse power/deflection circuit design with degaussing function. Step-by-Step Solution: Identify the symptom: colour blotches and purity shifts indicate magnetized metal near the CRT.Apply degaussing: an AC field that decays to zero demagnetizes the mask and shield.Result: improved colour purity and correct beam convergence/landing. Verification / Alternative check: TV service practice: degaussing thermistor circuits or manual degaussing wands are used after relocation or magnet exposure to restore picture quality. Why Other Options Are Wrong: Adding magnetism: opposite of the required action. Altering coil resistance/time constant: not the purpose of degaussing; that relates to different circuits. Mechanical alignment: relates to focus/convergence, not demagnetization. Common Pitfalls: Confusing convergence with purity; degaussing fixes magnetic contamination, while convergence adjusts beam geometry/electromagnetic alignment. Final Answer: Removing residual magnetism from the shadow mask, shield, and nearby metal parts

Question 20

In radio propagation terminology, a space wave can consist of which component paths?

Accepted Answer

Introduction / Context:Space-wave propagation dominates in VHF, UHF, and microwave links. Understanding what constitutes the space wave clarifies link budgets, fading mechanisms, and path profiles used in terrestrial and line-of-sight systems such as TV broadcast and point-to-point microwave links. Given Data / Assumptions: Space wave refers to near-line-of-sight propagation not dependent on ionospheric reflection. Possible components include direct and ground-reflected rays. Concept / Approach: The space wave typically includes the direct LOS component and a ground-reflected component (forming a two-ray model). At higher frequencies, diffraction around obstacles and tropospheric refraction/scatter may also contribute, but the classical definition emphasizes LOS and Earth-reflected rays, excluding ionospheric reflection (sky wave). Step-by-Step Solution: Define space wave: LOS plus often-present ground-reflected component.Exclude sky wave: ionospheric refraction is not part of the space-wave model.Therefore, a space wave may be direct, Earth-reflected, or a combination of both. Verification / Alternative check: Two-ray propagation models used in microwave engineering explicitly consider the direct and ground-reflected components of the space wave to predict fading/notching. Why Other Options Are Wrong: Only direct or only reflected: incomplete definitions. Ionospheric (sky) wave and troposcatter alone are different mechanisms, not the basic space wave. Common Pitfalls: Confusing sky-wave propagation (HF) with space-wave (VHF+). Space wave is primarily LOS and ground-reflection based. Final Answer: Direct wave or the wave reflected from the Earth's surface (or both)

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