RL integrator topology: In an RL integrating circuit (a first-order low-pass using RL), the output voltage is taken across the resistor, not the inductor. True or false?

Introduction / Context:Integrator and differentiator realizations with first-order networks depend on where the output is measured. Confusing the output node leads to the wrong frequency response (HP vs LP) and erroneous time-domain behavior.

Given Data / Assumptions:

• Series RL network driven by a source.
• Goal: integration-like (low-pass) behavior.
• Steady-state sinusoidal analysis is valid; time constants define transient response.

Concept / Approach:

An RL low-pass (integrator-like) response is obtained when the output is taken across the resistor: H(jω) = R / (R + jωL). This passes low frequencies (|H| → 1 as ω → 0) and attenuates high frequencies with the 20 dB/dec slope. Taking the output across the inductor yields a high-pass, not an integrator.

Step-by-Step Solution:

Verification / Alternative check:

Apply a step input: across R (low-pass), voltage rises exponentially toward the step (integration of input slope); across L (high-pass), you see a decaying spike (differentiation of the step).

Why Other Options Are Wrong:

• Choosing “True” swaps the node definitions: output across L produces a high-pass response, not an integrating low-pass.

Common Pitfalls:

Equating “integrator” strictly with ideal 1/(s) without acknowledging that first-order RL/RC circuits provide approximate integration only when time-constants are chosen appropriately relative to input waveforms.

Final Answer:

False