To understand how a differentiator shapes its output, which parts of the pulse response must be analyzed?
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AOnly the response to the rising pulse edge
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BOnly the interval between rising and falling edges
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COnly the response to the falling pulse edge
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DAll of the above
Answer
Correct Answer: All of the above
Explanation
Introduction / Context:Differentiators emphasize changes in the input signal. For pulse waveforms, the output is dominated by edge behavior, but the inter-edge interval also affects baseline and droop. This question probes comprehensive reasoning about the entire pulse cycle.
Given Data / Assumptions:
- First-order RC or RL differentiator operating on rectangular pulses.
- Finite time constants and practical component values.
Concept / Approach:The output to a pulse includes a positive spike at the rising edge (for standard polarity), a transient settling toward baseline between edges (with exponential decay governed by time constant), and a negative spike at the falling edge. Ignoring any portion yields an incomplete understanding of amplitude, symmetry, and baseline shift.
Step-by-Step Solution:
Rising edge: Large dv/dt produces a spike whose polarity follows the edge direction and whose initial magnitude reflects the abrupt step.Between edges: The output decays exponentially toward baseline; the decay rate depends on RC or L/R, affecting pulse-to-pulse overlap and droop.Falling edge: Opposite-polarity spike appears as the input steps back; its shape mirrors the rising-edge response in magnitude (for symmetric pulses).Verification / Alternative check:Time-domain convolution (input with differentiator impulse response) or Laplace-domain multiplication (H(s) with input step pairs) shows contributions at each interval, confirming all segments matter to the net waveform.
Why Other Options Are Wrong:
- Considering only one edge or only the middle interval misses essential dynamics; all three regions determine output symmetry, amplitude, and baseline.
Common Pitfalls:
- Assuming ideal impulses with zero-width edges, which hides practical finite-time constant effects.
- Neglecting recovery time between edges, leading to baseline shift and distortion.
Final Answer:All of the above