Question 1

Vacuum tube indicator in classic radios: The EM84 tube in a radio receiver is used as a _____ device.

Accepted Answer

Introduction / Context: Many vintage radios used a special display tube to show tuning quality. The EM84 is a classic example: a bar-graph style tuning indicator that glows more or less depending on signal strength or AGC voltage. Given Data / Assumptions: EM84 (6E3P) is a vacuum indicator tube, not a power or small-signal amplifier. It responds to the receiver's control/AGC voltage. Concept / Approach: The EM84 produces a variable illuminated display proportional to the control voltage, guiding the user to the point of maximum signal during tuning—hence the term 'magic eye.' Step-by-Step Solution: Receive RF → detect → AGC/control voltage develops.AGC drives EM84 control grid.Displayed bar widens or narrows indicating better/worse tuning. Verification / Alternative check: Service manuals identify EM84 explicitly as a tuning indicator tube, distinct from audio or RF amplification stages. Why Other Options Are Wrong: Audio/RF amplifier and rectifier roles are performed by other tubes/semiconductors. EM84 is not a frequency converter. Common Pitfalls: Assuming every tube is an amplifier or rectifier; some are indicators. Final Answer: ‘‘Magic eye’’ tuning indicator (visual signal-strength display)

Question 2

Popular intermediate frequencies (IF) for microwave and radar receivers operating in the 1–10 GHz RF range are typically which of the following?

Accepted Answer

Introduction / Context: Microwave and radar receivers use down-conversion from gigahertz RF to a convenient IF where filtering and amplification are practical. Standardized IFs simplify filter design and system interoperability. Given Data / Assumptions: RF range: 1–10 GHz (L-, S-, C-, X-bands). Need IF high enough to separate images yet low enough for high-performance filters. Concept / Approach: Common IFs for microwave superhets include ~30 MHz, ~60 MHz, and ~70 MHz (and their near variants). These strike a balance between image rejection, filter availability (crystal/SAW), and achievable IF gains. Step-by-Step Solution: High RF (GHz) → mixer with LO → fixed IF (tens of MHz).Select IF values with mature filter tech (e.g., 70 MHz SAW/crystal filters).Achieve narrow, stable passbands and good dynamic range. Verification / Alternative check: Radar/receiver literature and module catalogs list 30/60/70 MHz IF chains as industry staples. Why Other Options Are Wrong: kHz IFs are far too low for modern microwave selectivity demands. GHz IFs defeat the purpose of down-conversion. 1.6/4.5/10.7 MHz are common in other contexts (e.g., 10.7 MHz for FM), not typical for microwave radar IF back-ends. Common Pitfalls: Assuming FM broadcast IF (10.7 MHz) is universal; microwave systems prefer higher IFs. Final Answer: 30, 60 and 70 MHz

Question 3

FM demodulation comparison: Which statement is NOT a true advantage of a phase discriminator over a slope detector?

Accepted Answer

Introduction / Context: FM demodulators convert frequency variations to voltage. The slope detector is a simple, single-tuned AM detector biased to operate on the slope of a resonance curve, whereas a phase discriminator (e.g., Foster–Seeley/ratio-type architectures) uses phase relationships with typically two resonant circuits to obtain a more linear and stable demodulation characteristic. Given Data / Assumptions: Slope detector employs one tuned circuit and is sensitive to amplitude changes. Phase discriminator commonly employs double-tuned circuits and is paired with limiter stages for AM rejection. We compare typical advantages in broadcast/communication receivers. Concept / Approach: Slope detectors are simple but have poor linearity, high amplitude sensitivity, and narrow effective deviation range. Phase discriminators improve linearity and, with limiters, greatly reduce amplitude noise. However, they do not use fewer tuned circuits; in fact, they generally use more (a key trade-off for performance). Step-by-Step Solution: List true advantages of phase discriminators: better linearity, better AM rejection with limiter, wider useful deviation range.Check tuned-circuit count: slope detector → one; phase discriminator → usually two.Therefore, “fewer tuned circuits” is false as an advantage. Verification / Alternative check: Receiver design texts show Foster–Seeley and ratio detectors using a double-tuned transformer or equivalent, confirming increased circuit complexity compared with a single tuned slope detector. Why Other Options Are Wrong (i.e., they are true advantages): Better linearity: phase detection around quadrature is inherently more linear than operating on a single slope. Greater immunity to AM (with limiter): limiters suppress amplitude noise before the discriminator. Alignment more robust than a slope point: not confined to a tiny slope region; discriminator balance can be centered. Wider dynamic range of deviation: less distortion over practical FM deviations than slope detectors. Common Pitfalls: Confusing phase discriminators with ratio detectors (a specific AM-insensitive variant) or assuming simplicity equals better performance. Final Answer: Fewer tuned circuits are required

Question 4

Effect of choosing a higher intermediate frequency (IF) in a superheterodyne receiver: which statement is correct?

Accepted Answer

Introduction / Context: In a superheterodyne receiver, the choice of intermediate frequency (IF) strongly affects image rejection, selectivity, and tuning (tracking) between the RF front end and the local oscillator (LO). Designers balance these factors when selecting a practical IF for a given band and application. Given Data / Assumptions: Higher IF increases frequency separation between RF and LO. IF selectivity depends on achievable Q and filter technology (bandwidth roughly scales with center frequency for a given Q). Adjacent-channel rejection depends on overall selectivity at IF. Concept / Approach: Raising IF improves image rejection because the image frequency moves farther from the desired RF, easing RF preselection. However, for a given practical Q, bandwidth BW ≈ f0/Q increases with center frequency, so making IF higher can make narrow selectivity harder, potentially worsening adjacent-channel rejection unless higher-Q filters are used. Also, a larger separation between LO and RF increases tracking difficulty across the tuning range, since the LO and RF tuned circuits must maintain a precise frequency relationship across the band. Step-by-Step Solution: Identify trade-off: Higher IF → better image rejection but harder selectivity and tracking.Among choices, only “increases tracking problems” is uniformly correct without extra caveats.Thus, select option stating tracking problems increase. Verification / Alternative check: Classic receiver design references recommend moderate IF choices (e.g., ~455 kHz AM broadcast) balancing image rejection and IF selectivity; higher IFs are used with crystal/ceramic/SAW filters to restore selectivity but otherwise complicate tracking. Why Other Options Are Wrong: Improves selectivity: not generally true for given Q; selectivity typically worsens unless better filters are used. Decreases tracking problems: opposite of reality. Improves adjacent-channel rejection automatically: depends on selectivity; higher IF alone does not ensure it. Improves both selectivity and image rejection without trade-offs: unrealistic; there is always a trade-off. Common Pitfalls: Assuming one parameter (IF) can optimize everything simultaneously; in practice, designers use multiple conversion stages or high-Q filtering (e.g., crystal/SAW) to manage compromises. Final Answer: Increases tracking problems in the RF and local oscillator tuning

Question 5

High-frequency noise mechanisms in electronic devices Which one of the following noise types becomes particularly significant at very high operating frequencies (for example, in microwave devices and fast transistors)?

Accepted Answer

Introduction / Context: Different noise mechanisms dominate in different frequency ranges. Understanding which mechanism rises in importance at high frequencies is essential for RF and microwave design, low-noise amplifiers, and high-speed semiconductor devices. Given Data / Assumptions: We are comparing common noise types: flicker (1/f), shot, thermal (Johnson), transit-time, and burst noise. Focus is on frequency dependence and physics in active devices (e.g., BJTs, FETs, diodes). Devices are biased in their normal operating regions without oscillation. Concept / Approach: Transit-time noise appears when the carrier transit time across the active region is comparable to the signal period. As frequency increases, the signal period decreases, causing phase lag and random fluctuations in current due to carrier transit statistics. This effect is negligible at low frequency but becomes prominent in microwave regimes. By contrast, flicker noise dominates at very low frequencies, and thermal and shot noise are broadband but do not specifically increase in relative importance solely because of very high frequency in the same way transit-time effects do. Step-by-Step Solution: Identify operating regime: very high frequency (short period).Compare carrier transit time with signal period.When transit time is not negligible, random arrival of carriers adds excess fluctuations → transit-time noise dominates. Verification / Alternative check: Microwave diodes (e.g., IMPATT, Gunn) and high-fT transistors exhibit behavior where carrier transit phenomena set performance limits and noise rises at high f. Why Other Options Are Wrong: Shot noise (b) is frequency-independent within the device bandwidth; flicker noise (c) dominates at low frequencies; thermal noise (d) is white and does not preferentially increase at high f due to the mechanism itself; burst noise (e) is low-frequency and process-related. Common Pitfalls: Assuming shot or thermal noise automatically dominates at all frequencies; forgetting that transit-time is tied to device geometry and carrier velocity. Final Answer: Transit-time noise

Question 6

Image frequency in a superheterodyne receiver Which statement correctly describes the image frequency in a superheterodyne architecture with an intermediate frequency (IF)?

Accepted Answer

Introduction / Context: In superheterodyne receivers, any RF signal that mixes with the local oscillator (LO) to produce the same IF as the desired signal is called the image. Image responses can degrade selectivity and cause interference if not handled properly in the RF front end. Given Data / Assumptions: Receiver uses one RF preselector, a mixer, and an IF strip centered at a fixed IF (e.g., 455 kHz). LO may be above or below the desired signal (high-side or low-side injection). IF filters pass the IF band around the fixed IF frequency. Concept / Approach: The image frequency is separated from the desired RF by 2IF. After mixing, both the desired signal and the image convert to the same IF. Therefore, once inside the IF chain, the IF filters cannot distinguish between them. Image rejection must be achieved ahead of the mixer by the RF tuned circuits or preselector filters with adequate quality factor and appropriate tracking. Step-by-Step Solution: Let desired RF be fS and LO be fLO.IF = |fLO − fS|.An image at fIM = fLO + IF or fLO − IF also mixes to IF → indistinguishable at IF.Hence, IF filters cannot reject the image; RF front-end must do it. Verification / Alternative check: Receiver textbooks show image-rejection ratio is largely set by the RF tuned circuits bandwidth relative to separation 2IF. Why Other Options Are Wrong: (a) Image depends on tuned frequency and IF; (b) it can fall inside the front-end tuning range; (c) image exists in AM, SSB, FM, etc.; (e) detectors operate at IF or baseband and do not perform image rejection. Common Pitfalls: Confusing image rejection with adjacent-channel selectivity; assuming IF filters alone solve image problems. Final Answer: It is not rejected by the IF tuned circuits

Question 7

Local oscillator setting in radio receivers In a standard superheterodyne receiver using high-side injection, how is the local oscillator (LO) frequency chosen relative to the incoming RF signal frequency?

Accepted Answer

Introduction / Context: Superheterodyne receivers translate the desired RF signal to a fixed intermediate frequency (IF) for consistent amplification and filtering. The local oscillator sets the difference frequency and thus determines the IF outcome and potential image frequency location. Given Data / Assumptions: Conventional high-side injection is assumed (common in broadcast receivers). IF is fixed (e.g., 455 kHz for AM broadcast). Mixer produces sum and difference frequencies. Concept / Approach: With high-side injection, the LO is set above the RF such that IF = fLO − fS. This places the image at fIM = fLO + IF, separated from fS by 2*IF, which can simplify front-end rejection. Although low-side injection is possible, most broadcast designs use high-side for practical tracking and image considerations. Step-by-Step Solution: Choose LO so that IF = fLO − fS.Therefore fLO > fS.Mixer outputs include fLO − fS (wanted IF) and fLO + fS (image-related sum), filtered by IF strip. Verification / Alternative check: Service manuals and receiver schematics show oscillator tracking curves where LO runs above the RF across the tuning range. Why Other Options Are Wrong: (b) describes low-side injection, which is not the standard assumption here; (c) would give IF = 0; (d) is incorrect because LO is precisely chosen; (e) is unrelated to superhet translation. Common Pitfalls: Confusing image location and injection side; forgetting that either injection side can be used depending on design. Final Answer: LO frequency is higher than the incoming signal frequency

Question 8

Receiver behavior at higher RF frequencies As the tuned RF frequency increases in a superheterodyne receiver, which of the following effects generally becomes worse and needs more careful design?

Accepted Answer

Introduction / Context: At higher RF frequencies, maintaining front-end performance becomes challenging due to component Q limits, stray capacitances, and fixed IF separation. Image rejection, in particular, depends on how well the RF tuned circuits attenuate signals spaced 2IF away from the desired frequency. Given Data / Assumptions: Fixed IF (e.g., 455 kHz in AM or 10.7 MHz in FM receivers). Single or dual RF tuned circuits ahead of the mixer. Component Q decreases relatively as frequency increases and parasitics rise. Concept / Approach: Image frequency sits at a separation of 2IF from the desired RF. As RF increases, the fractional separation (2IF / fS) shrinks, making it harder for finite-Q RF circuits to reject the image because their skirts are not infinitely steep. Therefore, image rejection typically degrades unless more selective preselection or higher IF (or multiple conversions) is used. Step-by-Step Solution: For a fixed IF, image offset = 2IF.As fS ↑, relative offset (2*IF / fS) ↓.Finite-Q RF filters provide less attenuation at smaller fractional offsets → poorer image rejection. Verification / Alternative check: Shortwave and VHF receivers use higher IFs or dual conversion to improve image rejection at high frequencies, confirming the trend. Why Other Options Are Wrong: (a) tracking typically becomes harder, not easier; (b) tracking can degrade, but image rejection deterioration is the core limitation; (c) IF selectivity does not inherently worsen with RF frequency since IF is fixed; (e) audio fidelity is not the primary issue tied to RF tuning frequency. Common Pitfalls: Confusing adjacent-channel selectivity (IF issue) with image rejection (RF issue); assuming raising LO alone fixes the image without improving RF filtering. Final Answer: Image frequency rejection will be poor

Question 9

Coupling used in IF stages of superheterodyne receivers Which type of coupling is typically employed between IF amplifier stages to obtain the desired bandwidth and selectivity response?

Accepted Answer

Introduction / Context: Intermediate frequency (IF) strips shape the receiver selectivity and gain. The coupling choice between IF stages strongly influences bandwidth, passband flatness, and skirt selectivity, which determine adjacent-channel rejection and audio fidelity for AM/FM/SSB receivers. Given Data / Assumptions: Superheterodyne topology with a fixed IF. Goals include controlled bandwidth and relatively steep skirts. Standard IF frequencies and practical transformer Q values. Concept / Approach: Double-tuned transformer coupling uses two resonant windings with adjustable coupling, producing a band-pass response that can be shaped for flatter tops and steeper skirts than a single-tuned stage. By adjusting mutual coupling, designers tailor bandwidth and ripple, achieving better selectivity than simple RC or single-tuned coupling. Step-by-Step Solution: Recognize IF strip needs a band-pass characteristic.Double-tuned transformers provide controllable bandwidth via coupling coefficient.This yields improved selectivity and flatter passband compared with single-tuned or RC coupling. Verification / Alternative check: Classic AM and FM receiver IF designs employ cascaded double-tuned transformers or ceramic filters achieving the desired response. Why Other Options Are Wrong: Capacitive or RC coupling (a, b) is not selective; single-tuned (d) has narrower, more peaked response and less control; direct coupling (e) provides no band-pass shaping. Common Pitfalls: Assuming more stages alone guarantee selectivity; bandwidth and shape factor depend on coupling and filter design. Final Answer: Double-tuned transformer coupling

Question 10

Power in frequency modulation (FM) In an FM signal with a given carrier amplitude, how does the total transmitted power change as the modulation index is varied?

Accepted Answer

Introduction / Context: FM is a constant-envelope modulation. Unlike AM, where carrier plus sidebands alter the total power with modulation depth, FM redistributes power among the carrier and multiple sidebands while keeping the overall power nearly constant for a constant carrier amplitude. Given Data / Assumptions: Carrier amplitude is fixed by the transmitter PA and limiter stages. Modulation index β = frequency deviation / modulating frequency. Linear, ideal FM without overdrive or limiter compression changes in amplitude. Concept / Approach: In FM, instantaneous frequency varies while amplitude stays constant. Bessel distributions determine how power shifts from the carrier to sidebands as β changes, but the sum of carrier and all sideband powers remains equal to the original carrier power (ignoring practical losses). Thus total radiated power is essentially independent of β. Step-by-Step Solution: Hold carrier amplitude constant.Vary modulation index β → alters relative sideband amplitudes.Sum of powers across carrier and sidebands remains constant → total power unchanged. Verification / Alternative check: Spectrum analyzer measurements show carrier amplitude may decrease at certain β zeros while sidebands grow, but integrated power stays constant. Why Other Options Are Wrong: (a), (b), (c), and (e) contradict constant-envelope FM theory; total power does not trend up or down with β under ideal conditions. Common Pitfalls: Confusing bandwidth expansion (Carson rule) with power change; equating FM with AM power behavior. Final Answer: Power remains essentially constant as the modulation index changes

Question 11

RF trimmer component identification In RF tuning networks and oscillators, a trimmer is primarily used as a small, adjustable component. It is basically which of the following?

Accepted Answer

Introduction / Context: RF circuits require precise tuning to maintain resonance and tracking across bands. Small adjustable components called trimmers provide fine calibration after manufacturing and during servicing. Recognizing their function aids in correct alignment procedures for receivers and transmitters. Given Data / Assumptions: The trimmer is a physically small adjustable device on RF coils or tuning capacitors. Used for final alignment, not large range tuning. Operates with low loss and good stability. Concept / Approach: Trimmers are typically miniature variable capacitors (or sometimes small adjustable inductances in different context), but in common usage the term refers to a variable capacitor used to tweak the resonant frequency of LC tanks, oscillator feedback networks, and tracking in multi-gang tuning assemblies. Step-by-Step Solution: Identify purpose: fine frequency alignment of an LC network.Most effective small change in resonance is via capacitance adjustment.Therefore, a trimmer is essentially a small variable capacitor. Verification / Alternative check: Receiver service manuals reference trimmer capacitors (CT) adjusted with a non-metallic screwdriver during alignment steps. Why Other Options Are Wrong: (a) Not a functional tuning element; (b) inductors used are called slug-tuned coils, not trimmers; (d) variable resistors are trimmers in DC bias circuits but not RF trimmers; (e) ferrite slugs adjust inductance in IF transformers, distinct from capacitor trimmers. Common Pitfalls: Confusing RF trimmer capacitors with bias trimmer potentiometers; misadjusting inductors when capacitance tweak is intended. Final Answer: A capacitor for fine capacitance adjustment

Question 12

Dominant internal noise source in a superheterodyne receiver In a typical superheterodyne radio receiver, which stage most often contributes the greatest amount to the overall noise figure (especially when there is no RF preamplifier)?

Accepted Answer

Introduction / Context: The overall noise performance of a receiver is primarily set by the earliest active stages because their noise is amplified by subsequent stages. When no RF amplifier is used, the mixer (converter) becomes the first active device and thus tends to dominate the noise figure. Given Data / Assumptions: No dedicated low-noise RF amplifier ahead of the mixer. Standard superheterodyne chain: antenna → tuned RF network → mixer/LO → IF → detector → audio. Power supply and audio stages do not determine RF noise figure. Concept / Approach: By Friis formula for cascaded noise, the noise contribution of later stages is divided by the gain of preceding stages. If the first active stage is the mixer and its conversion loss and noise figure are modest, its noise sets the floor. Adding an RF LNA with gain and low noise can dramatically reduce overall noise figure by suppressing the relative impact of mixer noise. Step-by-Step Solution: Identify first active stage → mixer.Apply Friis formula: overall NF ≈ NF1 + (NF2 − 1)/G1 + ...With G1 absent (no RF gain), mixer NF dominates. Verification / Alternative check: Practical receivers show improved sensitivity when a low-noise RF preamp precedes the mixer, confirming the mixer's dominant role in its absence. Why Other Options Are Wrong: (b) Power supply ripple or noise is filtered and does not define RF noise figure; (c) audio amplifier noise is after detection; (d) contributions are not equal; (e) detector noise is after significant IF gain. Common Pitfalls: Confusing hum or audio hiss with RF front-end noise; overlooking the importance of front-end gain before the mixer. Final Answer: Mixer (frequency converter) stage

Question 13

Common local oscillator choice up to shortwave bands A broadcast superheterodyne receiver (without an RF amplifier) uses IF = 455 kHz. For operation up to the upper limit of shortwave broadcasting, which oscillator topology is often used as the loca…

Accepted Answer

Introduction / Context: Local oscillator stability and ease of tuning over a wide frequency range are crucial in superheterodyne receivers. The choice of oscillator affects frequency drift, phase noise, and tuning linearity, all important up to shortwave bands (roughly 3–30 MHz). Given Data / Assumptions: Broadcast/shortwave coverage with IF around 455 kHz. Discrete LC oscillators are common in classic designs. Good frequency stability and practical component values are desired. Concept / Approach: The Colpitts oscillator (with capacitive feedback divider) is widely favored in RF oscillators due to good stability, ease of implementation, and suitable operating range for HF/shortwave. Clapp is a variant with improved stability but is less common in low-cost broadcast sets. Hartley (Armstrong) is also used but Colpitts is a frequent choice for compact, stable LO designs in this range. Step-by-Step Solution: Identify required frequency range (HF/shortwave).Select an LC oscillator with stable feedback network.Colpitts topology meets these needs in many broadcast receivers. Verification / Alternative check: Service literature for consumer radios commonly shows Colpitts LOs in AM shortwave sections. Why Other Options Are Wrong: (a) Hartley is used but less common in many broadcast LO sections; (c) Clapp offers high stability but is less typical in low-cost sets; (d) and (e) are not suitable RF LOs for these bands. Common Pitfalls: Assuming any LC topology is equivalent; ignoring mechanical layout and temperature compensation which also impact drift. Final Answer: Colpitts oscillator

Question 14

Shannon capacity of an AWGN channel A communication channel with additive white Gaussian noise has bandwidth 4 kHz and signal-to-noise ratio SNR = 15 (linear). What is the channel capacity?

Accepted Answer

Introduction / Context: The Shannon–Hartley theorem provides the maximum theoretical data rate (capacity) over a band-limited AWGN channel for a given SNR. It is central to link budgeting, modem design, and assessing the benefit of coding and bandwidth expansion. Given Data / Assumptions: Bandwidth B = 4 kHz. SNR (linear) = 15 → 1 + SNR = 16. A baseband real channel interpretation. Concept / Approach: Shannon capacity formula: C = B * log2(1 + SNR). Insert the given values and evaluate the logarithm in base 2. Recognize that 16 is a power of 2, simplifying computation. Step-by-Step Solution: Compute 1 + SNR = 1 + 15 = 16.Evaluate log2(16) = 4.Apply C = B * log2(1 + SNR) = 4000 * 4 = 16000 bits/s.Convert to kbps: 16000 / 1000 = 16 kbps (approx.). Verification / Alternative check: Using natural logs: C = 4000 * ln(16)/ln(2) = 4000 * 2.7726/0.6931 ≈ 16000 bps, matching the simple power-of-two result. Why Other Options Are Wrong: (a) 1.6 kbps is off by a factor of 10; (c) 32 kbps assumes log2(1 + SNR) = 8, incorrect; (d) 256 kbps is far too high for 4 kHz bandwidth; (e) 64 kbps exceeds the Shannon limit for the given SNR. Common Pitfalls: Confusing dB with linear SNR; forgetting to use base-2 logarithms; mixing up bandwidth units (Hz vs. kHz). Final Answer: 16 kbps

Question 15

Sampling and bandpass selection in AM systems An arbitrary signal m(t) has zero average value and is strictly band-limited to 3.2 kHz. It is uniformly sampled at a rate of 8 k samples/s (fs = 8 kHz). The sampled signal is then passed through an idea…

Accepted Answer

Introduction / Context: Sampling by an impulse train replicates a baseband spectrum around integer multiples of the sampling frequency. Selecting one of these replicated bands with a band-pass filter is a classic way to implement modulation by spectral translation. This question tests recognition of how an ideally sampled, band-limited signal appears after a narrow band-pass centered on a sampling harmonic. Given Data / Assumptions: Baseband is m(t), zero mean, band-limited to 3.2 kHz. Sampling frequency fs = 8 kHz. Ideal band-pass filter centered at 32 kHz with bandwidth 6.4 kHz (passes 32 ± 3.2 kHz). Ideal impulse-train sampling (no aperture effects). Concept / Approach: Impulse sampling multiplies m(t) by a Dirac comb in time, which in frequency creates repeated spectra M(f) centered at integer multiples kfs. Around f = 4fs = 32 kHz, the band 32 ± 3.2 kHz is a faithful copy of the baseband spectrum. Because the replica contains upper and lower sidebands around 32 kHz but no discrete carrier at exactly 32 kHz (mean of m(t) is zero), the selected output is a double-sideband suppressed-carrier (DSB-SC) signal. Step-by-Step Solution: Replicas appear at k*fs = …, 24, 32, 40 kHz, etc.The filter passes only the 32 ± 3.2 kHz band (28.8–35.2 kHz).This band contains symmetric upper and lower sidebands of m(t) shifted to 32 kHz.No discrete carrier exists at 32 kHz because m(t) has zero average (no impulse at DC to translate).Therefore the output is AM-DSB suppressed-carrier. Verification / Alternative check: If the message had a DC component, impulse sampling would create a spectral line at multiples of fs; filtering at 32 kHz could pass a tone at 32 kHz (a residual “carrier”). Here, zero mean removes that tone, confirming suppressed carrier. Why Other Options Are Wrong: AM-DSB with carrier: Requires a discrete line at the center frequency; not present for zero-mean m(t). AM-SSB with carrier: Single sideband would need asymmetrical filtering to remove one sideband, which is not the case. Exponentially decaying sines: That description matches damped sinusoids, unrelated to ideal filtering of sampled spectra. FM constant envelope: No frequency modulation mechanism is involved. Common Pitfalls: Confusing DSB-SC created by spectral replication with conventional product modulation; both yield identical spectra but the sampling origin often misleads students into expecting a residual carrier. Final Answer: AM-DSB signal with suppressed carrier

Question 16

Purpose of a noise limiter in AM broadcast receivers In an AM (amplitude modulation) receiver, what is the primary function of the built-in noise limiter circuit?

Accepted Answer

Introduction / Context: AM receivers are particularly vulnerable to impulsive noise sources because such impulses appear as brief, large-amplitude disturbances that the detector will demodulate as loud clicks and pops. A noise limiter is a non-linear circuit that clips or otherwise suppresses these transients before audio detection. Given Data / Assumptions: Standard broadcast AM receiver. Presence of impulsive noise sources like automotive ignition, lightning, and switching machinery. Limiter operates ahead of or around the detector stage. Concept / Approach: Impulse noise produces high-amplitude spikes over a wide bandwidth. A noise limiter (clipper, slicer, or gated device) reduces the amplitude of such spikes to protect the demodulator from large excursions, thereby improving the listener’s experience without substantially affecting normal program peaks. Step-by-Step Solution: Identify common impulsive noise sources: ignition systems, lightning, switching motors.Recognize that AM envelope detection converts spikes directly into audible clicks.A limiter clamps or clips these peaks before detection.Hence it mitigates noise from all listed sources. Verification / Alternative check: Service manuals show limiter diodes around the detector/IF stages labeled as “noise limiter” or “ANL” (automatic noise limiter), evidencing broad effectiveness against impulses irrespective of origin. Why Other Options Are Wrong: Each single source is only part of the picture; the limiter targets any impulsive noise, not just one category. The added option about audio dynamic range expansion is unrelated to the limiter’s clipping function. Common Pitfalls: Expecting the limiter to reduce continuous hiss; it is most effective on brief transients, not on wideband thermal noise. Final Answer: All of the above

Question 17

Cause of double spotting in superheterodyne receivers In a superheterodyne receiver, “double spotting” (hearing the same station at two nearby dial positions) most commonly results from:

Accepted Answer

Introduction / Context: Double spotting is a classic symptom in superheterodyne radios where a strong station appears at two different tuning points. Recognizing the root cause helps diagnose front-end selectivity and alignment issues. Given Data / Assumptions: Standard superhet architecture with RF stage(s), mixer/LO, and fixed IF. Observed double appearance of the same station. No special tracking filters beyond usual broadcast-band front end. Concept / Approach: The superhet converts the incoming signal by mixing with a local oscillator to a fixed intermediate frequency (IF). Signals at the image frequency (f_image = f_signal ± 2IF, depending on LO side) also mix to the same IF. If the RF front end does not sufficiently attenuate the image, both the desired and image signals pass to the mixer, causing “double spotting.” Step-by-Step Solution: Recall image relationship: f_image = f_signal ± 2IF.Insufficient preselector/RF filter rejection at f_image allows the image to reach the mixer.Mixer produces the same IF from both, so the same station is heard at two dial settings.Hence the principal cause is poor image rejection (inadequate RF selectivity). Verification / Alternative check: Improving or adding a tuned RF stage or a tracking preselector typically eliminates double spotting, confirming the diagnosis. Why Other Options Are Wrong: Poor sensitivity: Reduces signals, not duplications. High IF gain: Can aggravate audibility of images but is not the core cause without poor RF selectivity. Miscalibrated dial or LO drift: These cause frequency errors, not two valid conversion paths to the same IF. Common Pitfalls: Attempting to cure double spotting by only lowering IF gain; the correct fix is enhancing front-end filtering and alignment. Final Answer: Inadequate image-frequency rejection (insufficient RF selectivity)

Question 18

Effect of receiving frequency on selectivity in tuned-radio stages How does receiver selectivity typically change as the tuned receiving frequency is raised (with fixed-Q tuned circuits and the same IF)?

Accepted Answer

Introduction / Context: Selectivity characterizes how well a receiver rejects adjacent channels. With tuned circuits of roughly constant quality factor Q, the absolute bandwidth increases with frequency, so selectivity can degrade at higher tuned frequencies unless additional measures are taken. Given Data / Assumptions: Front-end tuned circuits have approximately constant Q. We compare behavior as RF tuning is increased within a band (e.g., medium wave). IF selectivity remains fixed. Concept / Approach: For a resonant circuit, fractional bandwidth ≈ 1/Q. Thus absolute bandwidth BW ≈ f0 / Q. As the center frequency f0 increases and Q remains the same, BW increases proportionally, making adjacent-channel rejection worse (poorer selectivity) at the high-frequency end of the band. Step-by-Step Solution: Assume Q is roughly constant with tuning.Compute BW: BW ≈ f0 / Q; as f0 rises, BW rises.Wider BW means more adjacent energy passes → worse selectivity.Therefore, selectivity is somewhat worse at higher receiving frequencies. Verification / Alternative check: Many broadcast receivers employ additional tracking circuits or stagger-tuned IF stages to keep effective selectivity more uniform across the band; without these, selectivity degrades at the high end. Why Other Options Are Wrong: “Better when raised”: Contradicts BW ∝ f0. “Better at IF only”: IF selectivity may be fixed, but the question concerns RF tuning. “Unchanged”: Only true if designs compensate Q or employ filters whose fractional BW changes appropriately. Common Pitfalls: Confusing Q (dimensionless) with absolute bandwidth; higher frequency with same Q means wider absolute BW and thus poorer selectivity. Final Answer: Somewhat worse when the receiving frequency is raised

Question 19

Passband requirement of tuned circuits in an AM broadcast receiver For standard AM broadcast reception, the composite RF/IF passband of the tuned circuits should be approximately equal to which value to pass the transmitted audio without excessive d…

Accepted Answer

Introduction / Context: AM broadcast signals carry two sidebands around the carrier. The required receiver passband is tied to the highest transmitted audio frequency to avoid cutting off significant information while still rejecting adjacent-channel interference. Given Data / Assumptions: Standard AM program audio bandwidth is up to about 10 kHz in many textbook examples (varies by region and practice; often less in crowded bands). Both upper and lower sidebands must be passed around the carrier. IF (e.g., 455 kHz) is the center of selectivity shaping, not the passband width itself. Concept / Approach: If the highest audio frequency is f_a (≈ 10 kHz in many illustrative cases), the required RF/IF passband to pass both sidebands is approximately 2 * f_a (about 20 kHz). This ensures minimal amplitude distortion of the modulating audio while maintaining reasonable adjacent-channel rejection. Step-by-Step Solution: Let f_a ≈ 10 kHz → sidebands at carrier ± 0…10 kHz.To pass both sidebands, the IF passband ≈ 2 * f_a ≈ 20 kHz.Selectivity can be tailored, but the fundamental width is set by the double-sideband content. Verification / Alternative check: Service data often cites AM IF bandwidths on the order of 6–12 kHz for narrow/communications and up to ~20 kHz for high-fidelity AM, aligning with the 2 * f_a guideline. Why Other Options Are Wrong: 455 kHz / >455 kHz / 1455 kHz: These are center frequencies or unrelated large values, not passband widths. 2 kHz: Much too narrow; would severely cut audio frequencies. Common Pitfalls: Confusing the IF center frequency (e.g., 455 kHz) with the passband width required to pass both sidebands. Final Answer: 20 kHz

Question 20

Demodulation of single-sideband (SSB) signals Which one of the following circuits cannot demodulate an SSB signal into intelligible audio?

Accepted Answer

Introduction / Context: SSB signals have no carrier and just one sideband. Demodulation requires a locally generated carrier (or a product detector) to translate the SSB spectrum back to audio. FM detectors are designed for frequency variations, not for amplitude/phase-coherent detection needed by SSB. Given Data / Assumptions: SSB reception requires carrier reinsertion or coherent detection. Receiver has access to a beat-frequency oscillator (BFO) or product detector. Phase discriminator is an FM detector type. Concept / Approach: A product detector multiplies the incoming SSB with a locally generated sinusoid near the original carrier frequency to shift the spectrum to audio. An FM phase discriminator responds to instantaneous frequency/phase deviations and does not recover amplitude-encoded SSB audio. Step-by-Step Solution: Identify detectors viable for SSB: product detector; mixer plus BFO; synchronous detectors.Recognize that an FM phase discriminator requires a frequency-modulated input to produce audio.Since SSB is not FM, a phase discriminator cannot demodulate it properly. Verification / Alternative check: Communications receivers implement SSB modes via BFO/product detectors; FM detectors are disabled in SSB/AM modes, confirming the incompatibility. Why Other Options Are Wrong: Product detector / BFO + mixer: Standard SSB demodulation methods. Balanced modulator as synchronous detector: When driven by a reinserted carrier, it functions as a coherent detector. Synchronous envelope detector: With a locked carrier, it will demodulate SSB/AM. Common Pitfalls: Assuming any “detector” works for all modulation types; detectors are modulation-specific. Final Answer: Phase discriminator (FM detector)

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