Difficulty: Easy
Correct Answer: Yes, it is a first-order high-pass filter
Explanation:
Introduction / Context:Recognizing canonical RC topologies is essential for quick design judgements. Two complementary first-order forms exist: RC low-pass (R in series, C to ground, output across C) and RC high-pass (C in series, R to ground, output across R).
Given Data / Assumptions:
Concept / Approach:For the series-C / shunt-R network, the transfer magnitude |H(jω)| = |V_out/V_in| increases with frequency: |H| = ωRC / √(1 + (ωRC)^2). At low frequency (ω → 0), the capacitor blocks, |H| → 0. At high frequency (ω → ∞), the capacitor is a short, |H| → 1. This is precisely the behavior of a first-order high-pass filter, with cutoff f_c = 1/(2πRC).
Step-by-Step Solution:
Identify topology: series C, shunt R, output across R.Recall high-pass transfer |H| = ωRC / √(1 + (ωRC)^2).Low frequency → |H| ≈ 0; high frequency → |H| ≈ 1.Therefore classify as first-order high-pass.Verification / Alternative check:Bode plot shows +20 dB/decade slope below f_c and flat passband above f_c; phase approaches +90 degrees at high frequency.
Why Other Options Are Wrong:
Low-pass: opposite component placement (R series, C shunt, output across C).Band-pass and all-pass require different structures.“Cannot be determined” is incorrect; topology uniquely defines it.Common Pitfalls:Confusing where the output is taken; swapping where you probe flips high-pass vs low-pass.
Final Answer:Yes, it is a first-order high-pass filter
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