Q: Pulse spectra insight: The flat (constant) portions of a pulse waveform correspond primarily to low-frequency or near-DC spectral components. True or false?

Introduction / Context: Time-domain features map to frequency-domain content. Recognizing which parts of a pulse contribute low versus high frequencies helps in filter selection, EMI control, and edge-rate specifications for digital systems. Given Data / Assumptions: A rectangular-like pulse with finite rise and fall times. Piecewise description: flat top/bottom and sharp edges. Fourier decomposition applies under linear, time-invariant assumptions. Concept / Approach: Slowly varying or constant segments correspond to low time derivatives and therefore weaker high-frequency content. In contrast, rapid transitions (edges) have large time derivatives and inject higher-frequency harmonics. A perfect DC level (flat, infinite duration) is purely zero frequency; a finite flat segment concentrates energy at lower harmonics compared to edges. Step-by-Step Solution: Model the pulse as a sum of a DC component plus harmonics.Note that constant sections imply dV/dt ≈ 0 ⇒ minimal high-frequency content from those regions.Edges behave like differentiated impulses, supplying wideband high-frequency energy.Therefore, flat regions are dominated by low-frequency components. Verification / Alternative check: Spectrum analyzer of a pulse shows harmonic amplitudes decreasing with frequency; sharpening edges (reducing rise time) spreads energy to higher frequencies, while lengthening the flat portion largely affects low-order harmonics and DC content. Why Other Options Are Wrong: “False” would misattribute the high-frequency energy to flat parts, contrary to Fourier analysis and practical EMI observations. Common Pitfalls: Assuming the pulse top alone defines bandwidth; in reality, edge rates dominate required bandwidth. Confusing repetition rate (sets line spacing) with edge rate (sets envelope extent). Final Answer: True

Q: RL integrator topology: In an RL integrating circuit (a first-order low-pass using RL), the output voltage is taken across the resistor, not the inductor. True or false?

Introduction / Context: Integrator and differentiator realizations with first-order networks depend on where the output is measured. Confusing the output node leads to the wrong frequency response (HP vs LP) and erroneous time-domain behavior. Given Data / Assumptions: Series RL network driven by a source. Goal: integration-like (low-pass) behavior. Steady-state sinusoidal analysis is valid; time constants define transient response. Concept / Approach: An RL low-pass (integrator-like) response is obtained when the output is taken across the resistor: H(jω) = R / (R + jωL). This passes low frequencies (|H| → 1 as ω → 0) and attenuates high frequencies with the 20 dB/dec slope. Taking the output across the inductor yields a high-pass, not an integrator. Step-by-Step Solution: Define ZR = R and ZL = jωL in series.For output across R: HLP(jω) = R / (R + jωL) ⇒ low-pass (integrator-like in time domain).For output across L: HHP(jω) = jωL / (R + jωL) ⇒ high-pass (differentiator-like).Therefore, stating that an RL integrator takes output across L is incorrect. Verification / Alternative check: Apply a step input: across R (low-pass), voltage rises exponentially toward the step (integration of input slope); across L (high-pass), you see a decaying spike (differentiation of the step). Why Other Options Are Wrong: Choosing “True” swaps the node definitions: output across L produces a high-pass response, not an integrating low-pass. Common Pitfalls: Equating “integrator” strictly with ideal 1/(s) without acknowledging that first-order RL/RC circuits provide approximate integration only when time-constants are chosen appropriately relative to input waveforms. Final Answer: False

Q: RC integrator fault case: If the capacitor in an RC integrator goes open-circuit, the output will simply equal the input voltage. True or false?

Introduction / Context: Troubleshooting analog signal-shaping circuits requires understanding failure modes. An open capacitor radically changes an RC integrator's behavior and output node impedance. Given Data / Assumptions: Standard RC integrator: input → resistor → node → capacitor to ground; output is taken at the capacitor node. Capacitor fails open (infinite impedance). Measurement uses a high-impedance instrument (typical DMM/oscilloscope). Concept / Approach: With the capacitor open, no current flows through the branch, so the node becomes floating except for leakage and measurement loading. The network no longer forms a divider; there is no guaranteed equality between input and output—often the output is undefined or follows stray capacitances, not the intended signal. Step-by-Step Solution: Original: Vout = voltage across C; with C present, the circuit integrates when RC ≫ signal period.Fault: C → open ⇒ branch impedance → ∞; current through R is ~0.Node after R floats; Vout is not clamped and depends on leakage paths, probe capacitance, and parasitics.Therefore, Vout ≠ Vin in any predictable way; the statement is false. Verification / Alternative check: Replace the opened capacitor with a very large value (simulate open); observe Vout becoming noisy, drifting, or holding a random charge. In contrast, shorting the capacitor (not typical failure) would force Vout ≈ 0, not Vin. Why Other Options Are Wrong: “True” assumes the node magically mirrors input; without a conductive path, the node cannot reproduce Vin and becomes high-impedance and indeterminate. Common Pitfalls: Confusing an open capacitor with removing the resistor; conflating probe-through effects (a scope probe can inadvertently provide a small capacitance that momentarily couples input to output). Final Answer: False

Q: RC differentiator topology: In a basic RC differentiating circuit, the output is taken across the resistor (R), producing a high-pass, edge-emphasizing response. True or false?

Introduction / Context: RC networks can approximate time differentiation when the time constant is chosen properly relative to the input wave shape. Correct node selection determines whether the circuit differentiates or integrates. Given Data / Assumptions: Series RC circuit driven by a source. Output measured across the resistor. Assume τ = R*C is small compared with the significant time scales of the input pulse width. Concept / Approach: Taking the output across R yields H(jω) = jωRC / (1 + jωRC), a first-order high-pass response. In the time domain, for τ much smaller than the input waveform period, Vout ≈ τ * dVin/dt, which emphasizes rapid changes (edges) and suppresses slow variations (DC). Step-by-Step Solution: Write impedances: ZC = 1/(jωC), ZR = R.Voltage divider: Vout = Vin * ZR / (ZR + ZC) = Vin * (jωRC)/(1 + jωRC).For ω ≫ 1/RC (edge content), |H| → 1; for ω ≪ 1/RC (slow content), |H| → 0.With τ small, approximate Vout(t) ≈ τ * dVin/dt (differentiator). Verification / Alternative check: Apply a step input. The output is a narrow positive spike on the rising edge and a negative spike on the falling edge, characteristic of differentiation. Scope traces confirm high-pass behavior and edge emphasis. Why Other Options Are Wrong: Answering “False” would imply the resistor node does not provide differentiation; however, moving the output to the capacitor node produces the opposite (integration/low-pass). Common Pitfalls: Using too large τ so that the circuit no longer approximates d/dt; forgetting that real differentiators amplify noise and may require limiting networks to maintain stability. Final Answer: True

Q: DC component definition: The DC component of a pulse waveform equals its average (mean) value over time, not the peak value. True or false?

Introduction / Context: Separating a signal into DC and AC components is fundamental in electronics and signal processing. Misidentifying the DC component leads to errors in biasing, power calculations, and coupling decisions. Given Data / Assumptions: A periodic pulse waveform with defined amplitude and duty cycle. Stationary statistics (periodic steady-state). Standard Fourier series interpretation: DC component is the zero-frequency term. Concept / Approach: The DC component is the average value over one full period: Vdc = (1/T) * ∫0^T v(t) dt. Peak value is the maximum instantaneous amplitude and generally exceeds the average unless duty cycle is 100% with no variation. Therefore, equating DC component with peak value is incorrect. Step-by-Step Solution: For a rectangular pulse: v(t) = Vp during Ton, 0 otherwise.Compute Vdc = (Vp * Ton) / T = Vp * duty_cycle.Compare with Vpeak = Vp. Unless duty_cycle = 1, Vdc < Vpeak.Therefore, DC component ≠ peak value in general. Verification / Alternative check: Measure with a DMM in DC mode (averaging) and an oscilloscope for peak. For a 25% duty 5 V pulse, Vdc = 1.25 V while Vpeak = 5 V, demonstrating the distinction. Why Other Options Are Wrong: “True” would only hold for a constant (non-pulsed) signal; the prompt claims it for pulses, which is generally false. Common Pitfalls: Confusing RMS with average; RMS for pulses depends on duty and equals √(duty)*Vpeak for a 0/Vpeak pulse, which is also not the DC component. Final Answer: False

Q: Integrator time-scale rule: In a practical integrator, when the input pulse width is much shorter than the circuit time constant, the output does not reproduce the input shape; it tends toward a small triangular/linear ramp proportional to the area.…

Introduction / Context: First-order integrators (RC or RL low-pass forms) approximate mathematical integration only under specific time-scale relationships. Getting this relationship backward is a common source of lab and design errors. Given Data / Assumptions: RC integrator with τ = R*C (the same reasoning applies to RL low-pass). Input is a pulse of width Tw. “Transient time” refers to the circuit time constant τ. Concept / Approach: For good integration, τ should be much larger than Tw (τ ≫ Tw). In this regime, the output cannot follow the rapid input changes; instead, it responds slowly, producing a small-amplitude ramp whose height is proportional to the input area, not the input shape. Therefore, the statement that the output “approaches the shape of the input” is false under τ ≫ Tw. Step-by-Step Solution: Assume τ ≫ Tw (integrator condition).During the pulse, capacitor voltage increases approximately linearly: ΔVout ≈ (1/RC) * ∫ Vin dt = (Vin * Tw)/RC.After the pulse, the output decays slowly toward its prior value with time constant τ.The waveform is a small ramp, not a rectangular replica of the input. Verification / Alternative check: Oscilloscope observation with adjustable τ: as τ increases relative to Tw, the output top rounds into a shallow ramp; as τ decreases (τ ≪ Tw), the circuit behaves more like a follower, not an integrator, and the output resembles the input—opposite of the integrator condition. Why Other Options Are Wrong: “True” confuses integrator and follower regimes; reproducing the input shape requires bandwidth comparable to or greater than the input spectrum, not the limited bandwidth of an integrator. Common Pitfalls: Mixing “differentiator” and “integrator” criteria; forgetting that component tolerances and source/load impedances affect effective τ and integration quality. Final Answer: False

Q: RL differentiator topology: In an RL differentiating circuit (first-order high-pass using RL), the output voltage is taken across the resistor. True or false?

Introduction / Context: For RL networks, which node yields differentiation (high-pass, edge emphasis) and which yields integration (low-pass) depends on whether the output is across the inductor or the resistor. This affects timing circuits, snubbers, and pulse-shaping. Given Data / Assumptions: Series RL network driven by a voltage source. Seeking differentiator behavior (high-pass). Assume small τ = L/R relative to input edge times for strong differentiation. Concept / Approach: Taking the output across the inductor gives a high-pass transfer H(jω) = jωL / (R + jωL). Output across the resistor yields the complementary low-pass H(jω) = R / (R + jωL). Therefore, an RL differentiator uses the inductor node, not the resistor node. Step-by-Step Solution: Write impedances: ZL = jωL, ZR = R.High-pass output across L: magnitude rises with frequency, approaching 1 as ω → ∞.Low-pass output across R: magnitude falls with frequency beyond the break ω = R/L.Hence, “output across R” for an RL differentiator is incorrect. Verification / Alternative check: Apply a step input. The inductor node shows a sharp spike (dV/dt response) that decays with τ = L/R, consistent with differentiation. The resistor node shows a conventional exponential toward the step value (integration/low-pass). Why Other Options Are Wrong: Choosing “True” reverses the RL roles compared with RC networks and leads to an unintended low-pass output. Common Pitfalls: Assuming duality with RC without swapping which element the output is taken across; the dual network places the output across the reactive component for RL differentiators. Final Answer: False

Q: Pulse spectrum rule of thumb: The rising and falling edges (transitions) of a pulse waveform contain the higher-frequency components, while the flat portions are low-frequency. True or false?

Introduction / Context: Bandwidth in digital systems is set largely by edge rates, not by repetition rate alone. Recognizing that transitions carry the high-frequency energy is central to signal integrity and EMI control. Given Data / Assumptions: Pulse waveform with finite rise and fall times. Fourier decomposition applies. Edges are localized rapid changes; flats are near-constant segments. Concept / Approach: High-frequency content scales with how quickly a signal changes. Mathematically, the derivative of a step contains an impulse-like component, which has a broad spectrum. Therefore, the transitions (where dV/dt is large) inject higher-frequency harmonics; the constant regions (dV/dt ≈ 0) contribute primarily DC and low-order components. Step-by-Step Solution: Approximate a pulse as the superposition of two step edges and a flat interval.Each step contributes a wideband spectrum with slowly decaying harmonic amplitudes.Shorter rise/fall times (steeper edges) spread energy farther into high frequencies.Lengthening the flat interval mainly alters the DC and low-order harmonic content, not the high-frequency envelope. Verification / Alternative check: Spectrum analyzer measurements show that tightening edge times increases high-frequency amplitudes, even if the repetition rate is unchanged. Time-domain simulators reveal overshoot/ringing sensitivity driven by edge spectra, not the steady level. Why Other Options Are Wrong: “False” contradicts both Fourier theory and practical EMI experience, where edges dominate radiated and conducted emissions. Common Pitfalls: Equating “higher frequency” with “higher repetition rate” only; overlooking that a slow clock with very fast edges still demands high-bandwidth interconnects and proper termination. Final Answer: True

Question 1

Which statements about an ideal capacitor's behavior are correct in circuit theory (choose the best overall statement)?

Accepted Answer

Introduction: Capacitors are energy-storage elements whose current-voltage relationship is i = C * dv/dt. This relationship yields several widely used rules of thumb about transient and steady-state behavior. The question asks you to identify the consolidated truth among the listed statements. Given Data / Assumptions: Ideal capacitor with capacitance C (no leakage, ESR, or ESL). Standard circuit-theory interpretations of "instantaneous" and "dc steady state". Small-signal, linear operating region. Concept / Approach: Because i = C * dv/dt, an instantaneous voltage step (large dv/dt) implies a very large current impulse; conversely, changing the capacitor's voltage requires time (non-instantaneous) unless infinite current is available. In dc steady state (dv/dt = 0), the current is zero, so the ideal capacitor behaves as an open circuit. Step-by-Step Solution: 1) Voltage cannot jump instantly: that would require infinite current since dv/dt would be infinite.2) For fast transients, the capacitor presents very low impedance, effectively shunting rapid changes and acting "like a short" for the highest-frequency components.3) At dc steady state, dv/dt = 0, so i = 0 and the capacitor behaves as an open circuit.4) Therefore, each individual statement is correct; the combined best choice is "All of the above". Verification / Alternative check: Frequency-domain view: Xc = 1 / (2 * pi * f * C). As f → ∞, Xc → 0 (short-like). As f → 0 (dc), Xc → ∞ (open-like). This confirms the time-domain reasoning. Why Other Options Are Wrong: Any single statement alone (a, b, or c) is incomplete; the question asks for the best comprehensive truth. None of the above: contradicts standard capacitor behavior taught in basic circuit theory. Common Pitfalls: Confusing "short to instantaneous changes" with "short to all AC". Real capacitors have finite impedance and parasitics; only very high-frequency components see near-short behavior. Final Answer: All of the above

Question 2

In a simple RC integrator, if the capacitor accidentally becomes shorted (zero impedance), what happens to the output measured across the capacitor node?

Accepted Answer

Introduction: An RC integrator typically consists of a series resistor feeding a capacitor to ground, with the output taken across the capacitor. This question explores the failure mode when the capacitor is shorted, a common troubleshooting scenario that produces a distinctive symptom at the output node. Given Data / Assumptions: Passive RC integrator: input → series R → node → capacitor to ground, output at the node. Steady operation with a bounded input signal. "Shorted capacitor" means the capacitor's impedance has collapsed to ~0 Ω. Concept / Approach: If the shunt element (capacitor) is a short, the output node is directly clamped to ground potential. Regardless of the input waveform, the node voltage cannot rise above approximately 0 V because any attempted change is immediately diverted to ground through the short. Step-by-Step Solution: 1) In a healthy RC integrator, the capacitor charges/discharges, creating an output proportional to the time-integral of the input.2) With a shorted capacitor, the output node is effectively a direct connection to ground.3) Therefore, the measured output voltage is ≈ 0 V for all practical purposes.4) The circuit no longer integrates; the input sees a series resistor into a ground short, which may stress the source. Verification / Alternative check: Measure with a multimeter or oscilloscope: the node reads near 0 V despite input changes. Upstream, current is limited only by the series resistor, confirming the fault signature. Why Other Options Are Wrong: Same as input: impossible; the node is clamped to ground. None of the above: incorrect because 0 V is the expected reading. Is at ground vs would measure zero volts: both descriptions point to 0 V, but the most precise measurement-oriented choice is "would measure zero volts". Common Pitfalls: Assuming the integrator will simply "stop integrating" but still pass some signal. A hard short forces the output to 0 V; any remaining signal appears only as current through the series resistor. Final Answer: would measure zero volts

Question 3

For an RC differentiating circuit driven by a periodic pulse waveform, which statement correctly captures the two possible regimes relating the pulse width (tw) to the circuit time constant (tau)?

Accepted Answer

Introduction: In RC differentiators, performance depends on comparing the input pulse width tw to the circuit time constant tau = R * C. Designers commonly distinguish between "long" pulses (tw much greater than tau) and "short" pulses (tw much less than tau) to anticipate output spike shapes and amplitudes. Given Data / Assumptions: First-order RC differentiator with time constant tau. Periodic pulse input with width tw and period Tperiod (not used directly here). Rule-of-thumb boundary at ~5 * tau separates long and short regimes. Concept / Approach: When tw ≥ 5 * tau, the capacitor has time to charge and discharge substantially within each pulse edge, yielding narrow, well-defined spikes primarily at transitions. When tw < 5 * tau, the circuit operates in a strongly differentiating regime where the output resembles brief spikes with limited decay during the pulse, emphasizing edges over the flat top. Step-by-Step Reasoning: 1) Define tau = R * C as the characteristic time scale.2) For tw ≥ 5 * tau, treat the pulse as "long"; the capacitor's exponential response largely settles between edges.3) For tw < 5 * tau, treat the pulse as "short"; output is dominated by edge spikes, with minimal settling between edges.4) Hence the two practical conditions are tw ≥ 5 * tau or tw < 5 * tau. Verification / Alternative check: Simulate with typical values (e.g., R = 1 kΩ, C = 1 nF → tau = 1 µs). Compare outputs for tw = 10 µs (≥ 5 * tau) versus tw = 0.2 µs (< 5 * tau). The first shows distinct charge/discharge, the second shows sharp, narrow spikes. Why Other Options Are Wrong: Options a and b omit the complementary short-pulse regime or duplicate the same inequality; option c covers only short cases; option e is a single value and not a regime. Common Pitfalls: Treating 5 * tau as an exact boundary; it is a heuristic. Real component tolerances and loading shift the practical split, but the two regimes remain conceptually valid. Final Answer: tw ≥ 5Τ or tw < 5Τ

Question 4

Filter behavior mapping: An RL integrator behaves as which kind of first-order filter, and an RC differentiator behaves as which kind, respectively?

Accepted Answer

Introduction: First-order RL and RC networks can be arranged to integrate or differentiate signals. Each arrangement corresponds to a classic filter response, useful for quick intuition about frequency-domain behavior and for selecting components in signal conditioning. Given Data / Assumptions: RL integrator topology: output taken so that low-frequency components are passed and fast changes are attenuated. RC differentiator topology: output taken so that high-frequency transitions are emphasized. Linear, time-invariant approximation with ideal components. Concept / Approach: Integrators pass low-frequency components (they "accumulate" slowly varying signals) and attenuate high-frequency content, which is the hallmark of a low-pass filter. Differentiators emphasize rapid changes—high-frequency components—while attenuating slow variations, the hallmark of a high-pass filter. Step-by-Step Solution: 1) RL integrator: with appropriate node selection, the inductor resists rapid current changes; the output follows the averaged (integrated) behavior, passing lows → low-pass.2) RC differentiator: the capacitor's impedance decreases with frequency, creating spikes at edges and passing highs → high-pass.3) Therefore, the correct mapping is low-pass for the RL integrator and high-pass for the RC differentiator.4) This mapping aligns directly with Bode magnitude slopes: ±20 dB/dec around the cutoff. Verification / Alternative check: Frequency-domain impedances: Xl = 2 * pi * f * L increases with f (blocking highs in the integrator path); Xc = 1 / (2 * pi * f * C) decreases with f (passing highs in the differentiator path). The resulting transfer functions match low-pass and high-pass behaviors, respectively. Why Other Options Are Wrong: Low-pass, low-pass / high-pass, high-pass: both entries cannot simultaneously describe integrator and differentiator behavior. High-pass, low-pass: reverses the correct mapping. Band-pass, band-stop: require at least second-order behavior or combined networks, not simple first-order integrator/differentiator forms. Common Pitfalls: Confusing where the output is taken in the network; swapping nodes can convert an apparent integrator into a different response. Always confirm topology and node definitions. Final Answer: low-pass, high-pass

Question 5

Pulse spectra insight: The flat (constant) portions of a pulse waveform correspond primarily to low-frequency or near-DC spectral components. True or false?

Accepted Answer

Introduction / Context: Time-domain features map to frequency-domain content. Recognizing which parts of a pulse contribute low versus high frequencies helps in filter selection, EMI control, and edge-rate specifications for digital systems. Given Data / Assumptions: A rectangular-like pulse with finite rise and fall times. Piecewise description: flat top/bottom and sharp edges. Fourier decomposition applies under linear, time-invariant assumptions. Concept / Approach: Slowly varying or constant segments correspond to low time derivatives and therefore weaker high-frequency content. In contrast, rapid transitions (edges) have large time derivatives and inject higher-frequency harmonics. A perfect DC level (flat, infinite duration) is purely zero frequency; a finite flat segment concentrates energy at lower harmonics compared to edges. Step-by-Step Solution: Model the pulse as a sum of a DC component plus harmonics.Note that constant sections imply dV/dt ≈ 0 ⇒ minimal high-frequency content from those regions.Edges behave like differentiated impulses, supplying wideband high-frequency energy.Therefore, flat regions are dominated by low-frequency components. Verification / Alternative check: Spectrum analyzer of a pulse shows harmonic amplitudes decreasing with frequency; sharpening edges (reducing rise time) spreads energy to higher frequencies, while lengthening the flat portion largely affects low-order harmonics and DC content. Why Other Options Are Wrong: “False” would misattribute the high-frequency energy to flat parts, contrary to Fourier analysis and practical EMI observations. Common Pitfalls: Assuming the pulse top alone defines bandwidth; in reality, edge rates dominate required bandwidth. Confusing repetition rate (sets line spacing) with edge rate (sets envelope extent). Final Answer: True

Question 6

RL integrator topology: In an RL integrating circuit (a first-order low-pass using RL), the output voltage is taken across the resistor, not the inductor. True or false?

Accepted Answer

Introduction / Context: Integrator and differentiator realizations with first-order networks depend on where the output is measured. Confusing the output node leads to the wrong frequency response (HP vs LP) and erroneous time-domain behavior. Given Data / Assumptions: Series RL network driven by a source. Goal: integration-like (low-pass) behavior. Steady-state sinusoidal analysis is valid; time constants define transient response. Concept / Approach: An RL low-pass (integrator-like) response is obtained when the output is taken across the resistor: H(jω) = R / (R + jωL). This passes low frequencies (|H| → 1 as ω → 0) and attenuates high frequencies with the 20 dB/dec slope. Taking the output across the inductor yields a high-pass, not an integrator. Step-by-Step Solution: Define ZR = R and ZL = jωL in series.For output across R: HLP(jω) = R / (R + jωL) ⇒ low-pass (integrator-like in time domain).For output across L: HHP(jω) = jωL / (R + jωL) ⇒ high-pass (differentiator-like).Therefore, stating that an RL integrator takes output across L is incorrect. Verification / Alternative check: Apply a step input: across R (low-pass), voltage rises exponentially toward the step (integration of input slope); across L (high-pass), you see a decaying spike (differentiation of the step). Why Other Options Are Wrong: Choosing “True” swaps the node definitions: output across L produces a high-pass response, not an integrating low-pass. Common Pitfalls: Equating “integrator” strictly with ideal 1/(s) without acknowledging that first-order RL/RC circuits provide approximate integration only when time-constants are chosen appropriately relative to input waveforms. Final Answer: False

Question 7

RC integrator fault case: If the capacitor in an RC integrator goes open-circuit, the output will simply equal the input voltage. True or false?

Accepted Answer

Introduction / Context: Troubleshooting analog signal-shaping circuits requires understanding failure modes. An open capacitor radically changes an RC integrator's behavior and output node impedance. Given Data / Assumptions: Standard RC integrator: input → resistor → node → capacitor to ground; output is taken at the capacitor node. Capacitor fails open (infinite impedance). Measurement uses a high-impedance instrument (typical DMM/oscilloscope). Concept / Approach: With the capacitor open, no current flows through the branch, so the node becomes floating except for leakage and measurement loading. The network no longer forms a divider; there is no guaranteed equality between input and output—often the output is undefined or follows stray capacitances, not the intended signal. Step-by-Step Solution: Original: Vout = voltage across C; with C present, the circuit integrates when RC ≫ signal period.Fault: C → open ⇒ branch impedance → ∞; current through R is ~0.Node after R floats; Vout is not clamped and depends on leakage paths, probe capacitance, and parasitics.Therefore, Vout ≠ Vin in any predictable way; the statement is false. Verification / Alternative check: Replace the opened capacitor with a very large value (simulate open); observe Vout becoming noisy, drifting, or holding a random charge. In contrast, shorting the capacitor (not typical failure) would force Vout ≈ 0, not Vin. Why Other Options Are Wrong: “True” assumes the node magically mirrors input; without a conductive path, the node cannot reproduce Vin and becomes high-impedance and indeterminate. Common Pitfalls: Confusing an open capacitor with removing the resistor; conflating probe-through effects (a scope probe can inadvertently provide a small capacitance that momentarily couples input to output). Final Answer: False

Question 8

RC differentiator topology: In a basic RC differentiating circuit, the output is taken across the resistor (R), producing a high-pass, edge-emphasizing response. True or false?

Accepted Answer

Introduction / Context: RC networks can approximate time differentiation when the time constant is chosen properly relative to the input wave shape. Correct node selection determines whether the circuit differentiates or integrates. Given Data / Assumptions: Series RC circuit driven by a source. Output measured across the resistor. Assume τ = R*C is small compared with the significant time scales of the input pulse width. Concept / Approach: Taking the output across R yields H(jω) = jωRC / (1 + jωRC), a first-order high-pass response. In the time domain, for τ much smaller than the input waveform period, Vout ≈ τ * dVin/dt, which emphasizes rapid changes (edges) and suppresses slow variations (DC). Step-by-Step Solution: Write impedances: ZC = 1/(jωC), ZR = R.Voltage divider: Vout = Vin * ZR / (ZR + ZC) = Vin * (jωRC)/(1 + jωRC).For ω ≫ 1/RC (edge content), |H| → 1; for ω ≪ 1/RC (slow content), |H| → 0.With τ small, approximate Vout(t) ≈ τ * dVin/dt (differentiator). Verification / Alternative check: Apply a step input. The output is a narrow positive spike on the rising edge and a negative spike on the falling edge, characteristic of differentiation. Scope traces confirm high-pass behavior and edge emphasis. Why Other Options Are Wrong: Answering “False” would imply the resistor node does not provide differentiation; however, moving the output to the capacitor node produces the opposite (integration/low-pass). Common Pitfalls: Using too large τ so that the circuit no longer approximates d/dt; forgetting that real differentiators amplify noise and may require limiting networks to maintain stability. Final Answer: True

Question 9

DC component definition: The DC component of a pulse waveform equals its average (mean) value over time, not the peak value. True or false?

Accepted Answer

Introduction / Context: Separating a signal into DC and AC components is fundamental in electronics and signal processing. Misidentifying the DC component leads to errors in biasing, power calculations, and coupling decisions. Given Data / Assumptions: A periodic pulse waveform with defined amplitude and duty cycle. Stationary statistics (periodic steady-state). Standard Fourier series interpretation: DC component is the zero-frequency term. Concept / Approach: The DC component is the average value over one full period: Vdc = (1/T) * ∫0^T v(t) dt. Peak value is the maximum instantaneous amplitude and generally exceeds the average unless duty cycle is 100% with no variation. Therefore, equating DC component with peak value is incorrect. Step-by-Step Solution: For a rectangular pulse: v(t) = Vp during Ton, 0 otherwise.Compute Vdc = (Vp * Ton) / T = Vp * duty_cycle.Compare with Vpeak = Vp. Unless duty_cycle = 1, Vdc < Vpeak.Therefore, DC component ≠ peak value in general. Verification / Alternative check: Measure with a DMM in DC mode (averaging) and an oscilloscope for peak. For a 25% duty 5 V pulse, Vdc = 1.25 V while Vpeak = 5 V, demonstrating the distinction. Why Other Options Are Wrong: “True” would only hold for a constant (non-pulsed) signal; the prompt claims it for pulses, which is generally false. Common Pitfalls: Confusing RMS with average; RMS for pulses depends on duty and equals √(duty)*Vpeak for a 0/Vpeak pulse, which is also not the DC component. Final Answer: False

Question 10

Integrator time-scale rule: In a practical integrator, when the input pulse width is much shorter than the circuit time constant, the output does not reproduce the input shape; it tends toward a small triangular/linear ramp proportional to the area.…

Accepted Answer

Introduction / Context: First-order integrators (RC or RL low-pass forms) approximate mathematical integration only under specific time-scale relationships. Getting this relationship backward is a common source of lab and design errors. Given Data / Assumptions: RC integrator with τ = R*C (the same reasoning applies to RL low-pass). Input is a pulse of width Tw. “Transient time” refers to the circuit time constant τ. Concept / Approach: For good integration, τ should be much larger than Tw (τ ≫ Tw). In this regime, the output cannot follow the rapid input changes; instead, it responds slowly, producing a small-amplitude ramp whose height is proportional to the input area, not the input shape. Therefore, the statement that the output “approaches the shape of the input” is false under τ ≫ Tw. Step-by-Step Solution: Assume τ ≫ Tw (integrator condition).During the pulse, capacitor voltage increases approximately linearly: ΔVout ≈ (1/RC) * ∫ Vin dt = (Vin * Tw)/RC.After the pulse, the output decays slowly toward its prior value with time constant τ.The waveform is a small ramp, not a rectangular replica of the input. Verification / Alternative check: Oscilloscope observation with adjustable τ: as τ increases relative to Tw, the output top rounds into a shallow ramp; as τ decreases (τ ≪ Tw), the circuit behaves more like a follower, not an integrator, and the output resembles the input—opposite of the integrator condition. Why Other Options Are Wrong: “True” confuses integrator and follower regimes; reproducing the input shape requires bandwidth comparable to or greater than the input spectrum, not the limited bandwidth of an integrator. Common Pitfalls: Mixing “differentiator” and “integrator” criteria; forgetting that component tolerances and source/load impedances affect effective τ and integration quality. Final Answer: False

Question 11

RL differentiator topology: In an RL differentiating circuit (first-order high-pass using RL), the output voltage is taken across the resistor. True or false?

Accepted Answer

Introduction / Context: For RL networks, which node yields differentiation (high-pass, edge emphasis) and which yields integration (low-pass) depends on whether the output is across the inductor or the resistor. This affects timing circuits, snubbers, and pulse-shaping. Given Data / Assumptions: Series RL network driven by a voltage source. Seeking differentiator behavior (high-pass). Assume small τ = L/R relative to input edge times for strong differentiation. Concept / Approach: Taking the output across the inductor gives a high-pass transfer H(jω) = jωL / (R + jωL). Output across the resistor yields the complementary low-pass H(jω) = R / (R + jωL). Therefore, an RL differentiator uses the inductor node, not the resistor node. Step-by-Step Solution: Write impedances: ZL = jωL, ZR = R.High-pass output across L: magnitude rises with frequency, approaching 1 as ω → ∞.Low-pass output across R: magnitude falls with frequency beyond the break ω = R/L.Hence, “output across R” for an RL differentiator is incorrect. Verification / Alternative check: Apply a step input. The inductor node shows a sharp spike (dV/dt response) that decays with τ = L/R, consistent with differentiation. The resistor node shows a conventional exponential toward the step value (integration/low-pass). Why Other Options Are Wrong: Choosing “True” reverses the RL roles compared with RC networks and leads to an unintended low-pass output. Common Pitfalls: Assuming duality with RC without swapping which element the output is taken across; the dual network places the output across the reactive component for RL differentiators. Final Answer: False

Question 12

Pulse spectrum rule of thumb: The rising and falling edges (transitions) of a pulse waveform contain the higher-frequency components, while the flat portions are low-frequency. True or false?

Accepted Answer

Introduction / Context: Bandwidth in digital systems is set largely by edge rates, not by repetition rate alone. Recognizing that transitions carry the high-frequency energy is central to signal integrity and EMI control. Given Data / Assumptions: Pulse waveform with finite rise and fall times. Fourier decomposition applies. Edges are localized rapid changes; flats are near-constant segments. Concept / Approach: High-frequency content scales with how quickly a signal changes. Mathematically, the derivative of a step contains an impulse-like component, which has a broad spectrum. Therefore, the transitions (where dV/dt is large) inject higher-frequency harmonics; the constant regions (dV/dt ≈ 0) contribute primarily DC and low-order components. Step-by-Step Solution: Approximate a pulse as the superposition of two step edges and a flat interval.Each step contributes a wideband spectrum with slowly decaying harmonic amplitudes.Shorter rise/fall times (steeper edges) spread energy farther into high frequencies.Lengthening the flat interval mainly alters the DC and low-order harmonic content, not the high-frequency envelope. Verification / Alternative check: Spectrum analyzer measurements show that tightening edge times increases high-frequency amplitudes, even if the repetition rate is unchanged. Time-domain simulators reveal overshoot/ringing sensitivity driven by edge spectra, not the steady level. Why Other Options Are Wrong: “False” contradicts both Fourier theory and practical EMI experience, where edges dominate radiated and conducted emissions. Common Pitfalls: Equating “higher frequency” with “higher repetition rate” only; overlooking that a slow clock with very fast edges still demands high-bandwidth interconnects and proper termination. Final Answer: True

Question 13

RC integrating circuit – Where is the output taken? In a classic single-pole RC integrator used for wave-shaping at low frequencies (input across series R, C to ground), the output voltage is measured across which component?

Accepted Answer

Introduction / Context: RC integrating circuits are foundational in analog electronics for converting fast changing inputs into slowly varying outputs (for example, turning a square wave into an approximate triangular waveform). Knowing where to measure the output is essential to obtaining the intended integration behavior. Given Data / Assumptions: Single-pole RC network: input applied to a series resistor R, capacitor C connected from node to ground. We operate in the integrating region where RC is large compared with the signal period (i.e., RC >> T of the highest significant frequency component). Linear, time-invariant behavior; ideal R and C (no parasitics). Concept / Approach: An RC integrator outputs a voltage proportional to the time integral of the input. Placing the capacitor to ground and taking the output at the capacitor node results in a low-pass response whose magnitude falls with frequency and whose phase tends to −90 degrees at high frequency—characteristics that approximate mathematical integration for suitable RC and input spectrum. Step-by-Step Solution: Arrange components: input → R → node Vout; from node Vout → C → ground.Impedance of C: Zc = 1/(jωC).Voltage division: Vout = Vin * (Zc / (R + Zc)).For ωRC ≫ 1 (integrating region), Vout ≈ Vin / (jωRC) → proportional to 1/ω, indicating integration.Therefore, the output is the voltage across the capacitor. Verification / Alternative check: Differentiate the capacitor current–voltage relation: iC = C * dVout/dt. Because the input current flows through R into C (neglecting load), the capacitor voltage is proportional to the integral of the current and thus the integral of the input (scaled by R). Measuring across R would instead yield a differentiated waveform at high frequency, not the integrator's intended output. Why Other Options Are Wrong: Across R: that tends toward differentiation at high frequency. Across the series combination or input source: that simply reproduces Vin. A separate series load node is not part of the standard single-pole integrator topology. Common Pitfalls: Choosing R too small or C too small such that ωRC is not ≫ 1; loading the output heavily and shifting the pole; confusing integrator (output at C) with differentiator (output across R). Final Answer: Across the capacitor (C).

Question 14

Signals and systems terminology – Definition of pulse response Is the reaction (output) of a circuit to a specified pulse input called the “pulse response”?

Accepted Answer

Introduction / Context: Characterizing how circuits react to input waveforms is central to analysis and design. While “impulse response” is ubiquitous in linear systems, engineers also use “pulse response” when the test input is a finite-width pulse. This question checks correct terminology. Given Data / Assumptions: Pulse input means a finite duration, finite amplitude rectangular or shaped pulse. Circuits may be linear or nonlinear; the term “pulse response” is descriptive of the experiment. Time-invariant behavior is not strictly required for using the term, though it simplifies analysis. Concept / Approach: The “pulse response” is simply the output that results when a circuit is driven by a pulse. In LTI systems, it can be predicted by convolving the input pulse with the system’s impulse response. The term does not mean the same as “impulse response,” which corresponds to an idealized zero-width, unit-area input. Step-by-Step Solution: Define impulse response h(t): output for δ(t) input.Define pulse p(t): finite width τ and amplitude A.For LTI systems: y(t) = p(t) * h(t) (convolution).Therefore, the circuit’s reaction to p(t) is indeed its “pulse response.” Verification / Alternative check: Construct a unit-step response s(t) and express a rectangular pulse as the difference of two steps. The resulting output is the difference of two step responses, corroborating that “pulse response” is a valid, conventional term. Why Other Options Are Wrong: Calling it “impulse response” is incorrect; an impulse is an idealized limiting case. The response exists for nonlinear circuits as well; the name is still descriptive even though superposition does not apply. Common Pitfalls: Confusing impulse with finite pulses; assuming only digital circuits use pulses. Analog filters and amplifiers are routinely characterized by pulse testing to check slew rate, overshoot, and settling. Final Answer: Yes, this is the standard definition (True).

Question 15

RC integrator — single narrow pulse behavior: “The output voltage of an RC integrator will eventually equal the amplitude of a single input pulse if the pulse width is less than 5 time constants.” Decide whether this statement is valid.

Accepted Answer

Introduction / Context: An RC integrator is designed so that the capacitor voltage represents the time integral (area) of the input waveform, particularly when the input changes much faster than the circuit’s time constant τ = R * C. The prompt asks whether a single narrow pulse (width < 5τ) will make the output eventually equal the pulse amplitude. This clarifies a common misconception between an integrator and a follower. Given Data / Assumptions: Series RC integrator with output taken across the capacitor. Single rectangular input pulse of width T_p where T_p < 5τ. Initial capacitor voltage is zero; ideal components, linear operation. Concept / Approach: For a finite pulse width T_p, the capacitor charges only for time T_p, reaching V_out(T_p) = V_in * (1 − exp(−T_p / τ)). If T_p is small compared to τ, the exponential term exp(−T_p / τ) ≈ 1 − (T_p / τ), giving V_out ≈ V_in * (T_p / τ) which is much less than V_in. Thus, the output does not “equal the pulse amplitude.” Instead, it is a smaller value proportional to the pulse area V_in * T_p divided by τ. Step-by-Step Solution: Define τ = R * C and pulse width T_p < 5τ.Compute capacitor rise over T_p: V_out(T_p) = V_in * (1 − exp(−T_p / τ)).For T_p ≪ τ, approximate V_out ≈ V_in * (T_p / τ), clearly less than V_in.After the pulse ends, the capacitor discharges back toward zero with time constant τ; it never “catches up” to the input amplitude of that already-ended pulse. Verification / Alternative check: Simulate with τ = 10 ms and T_p = 2 ms: V_out(T_p) = V_in * (1 − e^(−0.2)) ≈ 0.181 * V_in, far below V_in. As T_p approaches several τ, V_out approaches V_in, but the statement restricts T_p to be less than 5τ and claims “equal,” which is not guaranteed nor typical. Why Other Options Are Wrong: Correct: Not correct; integrators respond to pulse area, not necessarily to amplitude. Valid only if the capacitor is ideal / pulse amplitude small: Ideal parts or small amplitude do not make V_out equal to V_in for short pulses. Common Pitfalls: Confusing RC charging to source (a follower with long pulses) with integrating behavior for short pulses; assuming the 5τ “near steady-state” rule for steps applies to narrow pulses. Final Answer: Incorrect.

Question 16

Differentiator function — DC output claim: “A differentiator circuit can convert a pulse input into a nearly constant DC output.” Assess the validity of this statement for a standard RC differentiator.

Accepted Answer

Introduction / Context: An RC differentiator emphasizes rapid changes in the input signal and suppresses slowly varying components. Designers often use it to detect edges of pulses and to create narrow trigger spikes. The claim that such a circuit produces a nearly constant DC output from a pulse input reverses its intended function and needs to be evaluated critically. Given Data / Assumptions: Simple RC differentiator: series capacitor followed by a resistor to ground with output taken across the resistor. Input: a rectangular pulse or a train of pulses. Small-signal, linear, time-invariant behavior; no active elements. Concept / Approach: Differentiation is mathematically v_out ∝ dv_in/dt. For a rectangular pulse, dv/dt is large at the rising and falling edges and nearly zero during the flat top. Consequently, the differentiator outputs a positive spike at the rising edge and a negative spike at the falling edge, with almost zero output between edges (apart from small exponential tails). There is no mechanism to produce a steady DC level from a constant input segment in a passive RC differentiator. Step-by-Step Solution: Model the input as step up, constant interval, step down.At step up: dv/dt is large positive ⇒ positive spike across the resistor.During the constant interval: dv/dt ≈ 0 ⇒ output decays toward 0 V.At step down: dv/dt is large negative ⇒ negative spike. Verification / Alternative check: Scope captures of an RC differentiator fed by a pulse train show symmetric positive/negative spikes bracketing each pulse, with the mean output near zero. This is standard edge-detection behavior used in timing and trigger circuits. Why Other Options Are Wrong: Correct: Contradicts the fundamental purpose of a differentiator. Valid only with a very large capacitor / very low frequency: Changing RC alters spike width, not the fact that steady DC is rejected. Common Pitfalls: Confusing differentiators with integrators or low-pass filters that can produce average/DC components; overlooking that the differentiator’s transfer magnitude increases with frequency and tends to zero at DC. Final Answer: Incorrect.

Question 17

Differentiator in steady state — average output: During steady-state operation with a symmetrical pulse train input, is the average output voltage of an RC differentiator approximately zero volts?

Accepted Answer

Introduction / Context: RC differentiators respond to changes, not steady levels. For periodic pulse inputs, the output consists of alternating spikes at each transition. Understanding the average (DC) component at the output is important for coupling considerations, measurement offsets, and avoiding unintended biasing of downstream stages. Given Data / Assumptions: Standard RC differentiator with output across the resistor. Input is a repetitive pulse train without DC offset. Linear, time-invariant operation; steady state reached. Concept / Approach: A differentiator ideally has zero gain at DC. For a symmetrical pulse train that rises and falls to the same baseline, the positive spike at each rising edge and the negative spike at each falling edge have equal and opposite areas (charge transfer through the capacitor). Over a complete period, the time-average (integral over one cycle divided by the period) is therefore approximately zero, ignoring minor leakage and parasitic effects. Step-by-Step Solution: Represent the pulse as a sum of a DC component and AC transitions.The differentiator blocks DC; the AC transitions produce spikes whose average over a full period cancels.Compute average over one period: ∫ v_out dt over T is ≈ 0 for equal positive/negative spike areas.Conclude the average (DC) output is approximately 0 V. Verification / Alternative check: On an oscilloscope with AC coupling disabled, the baseline of the differentiated waveform hovers near 0 V for symmetric pulses. Any small average offset typically points to leakage, unequal edges, or measurement bias, not to inherent DC generation by the ideal differentiator. Why Other Options Are Wrong: Incorrect: Conflicts with the differentiator’s DC blocking property. Applies only if duty cycle is 50%: Even with other duty cycles, equal and opposite edge contributions can keep the average near zero as long as the waveform returns to the same baseline each cycle. Applies only if there is DC offset: A DC offset at the input would shift the baseline but a pure differentiator does not pass steady DC. Common Pitfalls: Assuming unequal spikes due to saturation or slew limits; using a non-symmetric waveform with different high/low levels, which can introduce a small average. Final Answer: Correct.

Question 18

RC integrator capacitor voltage — how it changes: “The voltage across the capacitor in an RC integrator cannot change exponentially; it can change only instantaneously.” Judge this statement.

Accepted Answer

Introduction / Context: Capacitor behavior in first-order RC networks is a cornerstone of electronics. A common misunderstanding is that capacitor voltage can jump instantly. In reality, the defining property of a capacitor is that current is proportional to the rate of change of voltage, which leads to an exponential change in voltage for step inputs—not an instantaneous jump—when driven through a resistor. Given Data / Assumptions: RC integrator topology with capacitor connected to ground and driven through a resistor. Ideal components for the conceptual model. Input waveforms include steps and pulses. Concept / Approach: In an RC network, i_C = C * dv_C/dt. For a finite series resistance, dv_C/dt is finite for finite current; therefore, v_C cannot change instantaneously. For a step input, the solution is v_C(t) = V_final * (1 − exp(−t/τ)) + V_initial * exp(−t/τ), which is exponential in time. Only an ideal zero-impedance source directly across a capacitor (or infinite current) could enforce an instantaneous change—neither occurs in a standard RC integrator. Step-by-Step Solution: Identify that the current into the capacitor is limited by the series resistor.Relate current and voltage rate: i_C = C * dv_C/dt ⇒ finite i_C gives finite dv_C/dt.For a step input, write the first-order solution showing exponential charging/discharging with τ = R * C.Conclude that capacitor voltage changes exponentially, not instantaneously, in an RC integrator. Verification / Alternative check: Oscilloscope measurements of an RC charge curve show the well-known exponential approach to a new level, reaching ~63% in 1τ, ~86% in 2τ, ~95% in 3τ, ~98% in 4τ, and ~99% in 5τ. None of these represent an instantaneous jump. Why Other Options Are Wrong: Correct: Opposite of the physical behavior in an RC network. Valid only for ideal zero-ESR capacitors / very high frequency: ESR or frequency does not permit instantaneous voltage steps without infinite current. Common Pitfalls: Confusing an observed fast but finite rise with an instantaneous jump; ignoring series resistance or source limitations that necessarily enforce exponential transitions. Final Answer: Incorrect.

Question 19

Differentiator output polarity — falling edge behavior: In a standard RC differentiator (output taken across the resistor), does the output go negative on the falling edge of a positive input pulse?

Accepted Answer

Introduction / Context: Polarity of the spikes produced by differentiators matters when designing edge-triggered circuits and pulse shapers. A standard passive RC differentiator places a capacitor in series with the input and a resistor to ground, taking the output across the resistor. Understanding the sign of the output on rising and falling edges is essential for trigger polarity selection. Given Data / Assumptions: Series capacitor, shunt resistor, output across the resistor. Input: a positive rectangular pulse returning to zero baseline. Ideal, linear behavior; no saturation. Concept / Approach: At a rising edge (positive step), current flows through the capacitor into the resistor, dropping voltage across the resistor that is positive at the top node, so the output shows a positive spike. At a falling edge (negative step), current reverses direction; the resistor voltage reverses its polarity, producing a negative spike. Between edges, current decays toward zero and the output returns toward zero as the capacitor charges/discharges. Step-by-Step Solution: Model the falling edge as a negative step: ΔV = −V_pulse.The capacitor current is i_C = C * dV/dt, which is negative at the instant of the falling edge.Negative current through the resistor produces a negative voltage across it (top node goes negative relative to ground).Therefore, the output spike at the falling edge is negative. Verification / Alternative check: Oscilloscope traces of a differentiator driven by a pulse train show a positive spike at each rising edge and a corresponding negative spike at each falling edge, matching the sign convention presented here. Why Other Options Are Wrong: Incorrect: Ignores the direction reversal of current at the falling edge. Conditional options: Duty cycle or exact R/Xc ratio shape spike width/amplitude but not the sign at an ideal instant. Common Pitfalls: Taking the output across the capacitor instead of the resistor (which changes interpretation); losing the sign convention on current direction at edges. Final Answer: Correct.

Question 20

Differentiator with very long pulses — spike approximation: When the input pulse width is much longer than 5 time constants (T_p ≫ 5τ), can the output of an RC differentiator be considered a pair of narrow spikes at the transitions?

Accepted Answer

Introduction / Context: Edge detection is a primary use case for RC differentiators. When a rectangular pulse is very wide compared with the circuit time constant, the differentiator s output between edges is negligible. Recognizing when the spike approximation is valid helps simplify timing analysis in digital interfacing and trigger circuitry. Given Data / Assumptions: RC differentiator: series C, shunt R, output across R. Pulse width T_p much larger than 5τ = 5 * R * C. Symmetric high and low levels with a return to baseline. Concept / Approach: At each transition, dv/dt is large, producing a current impulse through the capacitor and a correspondingly sharp voltage across the resistor. After a few time constants, the capacitor current decays toward zero and the output settles very close to 0 V for the remainder of the pulse plateau. With T_p ≫ 5τ, most of the period is flat at zero output, and the response looks like two spikes per pulse—one positive at the rising edge and one negative at the falling edge. Step-by-Step Solution: Identify condition: T_p ≫ 5τ ensures complete decay between edges.At edges: i_C = C * dv/dt is large → narrow output spikes.Between edges: i_C ≈ 0 → v_out ≈ 0.Thus the waveform is well-approximated by impulse-like spikes at transitions. Verification / Alternative check: Simulations with τ = 1 ms and pulses of width 50 ms show spikes of a few milliseconds width and nearly zero output elsewhere. Scope captures in labs confirm this classical behavior used for trigger generation. Why Other Options Are Wrong: Incorrect: Contradicts standard differentiator operation for T_p ≫ 5τ. True only if the capacitor is leaky / only for certain duty cycles: Leakage and duty cycle adjust amplitudes and baselines but do not negate the spike approximation under the stated condition. Common Pitfalls: Using T_p only slightly greater than τ and expecting perfect spikes; forgetting that finite source and load impedances shape spike width and amplitude. Final Answer: Correct.

Time Response of Reactive Circuits Questions