RC high-pass topology and phase behavior For a standard RC high-pass filter (series capacitor, shunt resistor), is the output taken across the resistor and does the output lead the input in phase for much of the spectrum?

Difficulty: Easy

Correct Answer: True

Explanation:


Introduction / Context:
Knowing where to measure the output in first-order filters is the key to predicting both magnitude and phase behavior. The RC high-pass places the capacitor in series and the resistor to ground, with the output across the resistor.



Given Data / Assumptions:

  • Series capacitor C, shunt resistor R to ground.
  • Output node is the junction between C and R (across R).
  • Transfer function: H(jω) = jωRC / (1 + jωRC).


Concept / Approach:

The magnitude of H rises from near 0 at very low frequency to ~1 at high frequency (20 dB/decade slope). The phase is positive (lead) at low frequency, approaching +90°, and tends to 0° at high frequency, meaning the output generally leads the input except in the asymptotic high-frequency limit.


Step-by-Step Solution:

Write H(jω) = jωRC / (1 + jωRC).Magnitude: |H| = (ωRC) / sqrt(1 + (ωRC)^2).Phase: ∠H = 90° − arctan(ωRC), which is a positive angle (lead) for finite ω.Therefore, output is across R and exhibits phase lead relative to input over most of the band.


Verification / Alternative check:

At ω = 1 / RC, |H| = 1 / sqrt(2) ≈ 0.707 and phase = +45°, a clear lead. At ω → ∞, phase → 0°, consistent with diminishing lead as the passband flattens.


Why Other Options Are Wrong:

  • “False” contradicts the textbook topology and phase result.
  • “True only above cutoff” misses the fact that the phase lead is strongest below cutoff.
  • Component ideality and value ratios do not alter the qualitative behavior.


Common Pitfalls:

Swapping output node to the capacitor (which would create a low-pass) or assuming phase lag because many filters lag; here, the high-pass leads.


Final Answer:

True

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