In an RC integrator used for pulse shaping, what happens if the capacitor becomes leaky (i.e., exhibits significant leakage resistance)?

Introduction / Context:
An RC integrator ideally uses a high-quality capacitor and a resistor to produce an output proportional to the time integral of the input (for appropriate RC/pulse-width ratios). Real capacitors, however, have leakage (modeled as a large but finite resistance in parallel with the capacitor). This question tests the practical impact of leakage on an RC integrator's performance.

Given Data / Assumptions:

• An RC integrator with series R and shunt C.
• The capacitor exhibits leakage, represented by a finite parallel resistance R_leak across C.
• Pulse or varying input signal; linear, small-signal analysis assumptions.

Concept / Approach:
The ideal time constant is tau_ideal = R * C. With leakage, the effective impedance across the capacitor is Z_C || R_leak, so the effective time constant becomes tau_eff = R * C_eff, where C_eff is reduced at low frequencies because the parallel path R_leak bleeds charge. This both reduces the effective time constant and alters the transfer function magnitude/phase, distorting amplitude and waveshape.

Step-by-Step Solution:

Verification / Alternative check:
Consider the Thevenin equivalent looking into the capacitor node. With R_leak present, the DC gain is reduced because the node cannot hold charge; time-domain step responses show faster decay toward ground, confirming a reduced effective time constant and lower peaks.

Why Other Options Are Wrong:

• 'The time constant will be effectively reduced': True.
• 'The waveshape of the output voltage across the capacitor is altered': True, due to droop and distortion.
• 'The amplitude of the output is reduced': True, peaks fall because charge leaks away.

Common Pitfalls:

• Assuming leakage only affects DC and not transient behavior.
• Ignoring that even large R_leak values can matter when pulse repetition intervals are long relative to tau.

Final Answer:
All of the above