RL high-pass behavior: In a series RL high-pass filter, the output is taken across the inductor (L), and the output voltage leads the input phase. True or false?

Introduction / Context:RL and RC first-order networks can form either low-pass or high-pass filters depending on where the output is taken. Knowing which node provides high-pass behavior and the phase relationship is crucial for signal-conditioning and phase-lead networks.

Given Data / Assumptions:

• Series RL network with input applied across the series combination.
• High-pass output is required.
• Linear, time-invariant components; sinusoidal steady-state (phasor) analysis applies.

Concept / Approach:

In a series RL divider, taking output across L yields a high-pass response because |ZL| = ωL grows with frequency. The transfer function becomes H(jω) = jωL / (R + jωL), which approaches 1 at high frequency and 0 at low frequency. The phase of H is 90° − arctan(ωL/R), a positive angle for any ω > 0, meaning the output leads the input.

Step-by-Step Solution:

Verification / Alternative check:

At ω ≪ R/L, |H| ≈ ωL/R → 0 (blocked at low frequency). At ω ≫ R/L, |H| → 1 and phase → 0° to 90° lead limit; scope measurements confirm the output zero-crossing precedes the input for midband HP operation.

Why Other Options Are Wrong:

• “False” would imply either output is not across L for RL high-pass or that phase does not lead, contradicting both the impedance rise of L with frequency and the phasor angle of jωL/(R + jωL).

Common Pitfalls:

Swapping node choice: output across R is a low-pass. Also, forgetting that phase lead approaches +90° only ideally; with finite R, lead is between 0° and 90°.

Final Answer:

True