RC network identification — high-pass or low-pass?: Consider a simple RC with a series capacitor feeding a resistor to ground; the output is taken across the resistor. Is this a first-order high-pass filter?

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:

  • Series capacitor C from input node to output node.
  • Resistor R from output node to ground.
  • Output measured across R; linear, time-invariant operation.


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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