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:
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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