Active high-pass filter insight — effect of increasing capacitance on the critical (cutoff) frequency:

Difficulty: Easy

Correct Answer: Correct

Explanation:

Introduction / Context:In first-order active high-pass filters (e.g., op-amp based), the critical frequency establishes where passband behavior begins. Designers often adjust the capacitor to place the cutoff at the desired frequency; understanding the direction of change avoids needless trial and error on the bench.

Given Data / Assumptions:

  • Single-pole high-pass topology (active or passive behaves similarly for cutoff math).
  • Standard cutoff definition: f_c = 1 / (2π R C) for the relevant R and C in the high-pass network.
  • Linear component behavior.

Concept / Approach:The cutoff frequency is inversely proportional to the product R * C. Increasing C raises the time constant τ = R * C and therefore reduces f_c. This property lets you shift the corner down in frequency without altering R (useful where R sets bias or noise performance).

Step-by-Step Solution:

Start with f_c = 1 / (2π R C).Hold R fixed, increase C → denominator increases.Therefore, f_c decreases proportionally to 1/C.Conclusion: the statement is correct.

Verification / Alternative check:Example: double C → f_c halves; halve C → f_c doubles. Bode plots confirm the shift along the frequency axis without changing the slope magnitude (±20 dB/decade per pole).

Why Other Options Are Wrong:

“Incorrect” would require a topology where C does not enter the time constant inversely, which is not the case for standard first-order high-pass filters.

Common Pitfalls:Changing R instead of C and unintentionally altering input bias or source loading; confusing high-pass and low-pass forms.

Final Answer:Correct

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