RL pulse shaping — in a basic RL differentiator (step/pulse shaping network), is the output conventionally taken across the inductor to emphasize rapid changes (spikes at edges)?

Introduction / Context:
Differentiator networks generate outputs that are large during rapid input changes and small during steady levels. An RL differentiator uses the inductor’s property v_L = L * di/dt, producing voltage spikes at rising and falling edges of a pulse. Knowing where the output is taken (across which element) is essential for correct classification and design.

Given Data / Assumptions:

• Series RL excited by steps/pulses.
• Small-signal, linear behavior; ideal inductor.
• Load is high-impedance measurement so as not to disturb the network.

Concept / Approach:
For a series RL with an applied step, the current cannot change instantaneously, so the inductor initially takes nearly the entire input voltage and then its voltage decays exponentially as current builds. This “edge-peaked” v_L is proportional to di/dt and therefore implements differentiation. Hence, the conventional RL differentiator output is measured across the inductor.

Step-by-Step Solution:

Verification / Alternative check:
Transient plots show v_L peaks with exponential decay after each transition, while the resistor’s voltage tends to follow the slowly changing current, not the derivative highlight.

Why Other Options Are Wrong:

• Incorrect or “only for square waves”: differentiation property holds for any changing waveform; square waves just make spikes obvious.
• Constraints like R ≫ X_L or core material: affect shape and amplitude but not the fundamental placement of the output for differentiation.

Common Pitfalls:
Confusing RL differentiator with RC differentiator (where output is taken across the resistor).

Final Answer:
Correct — the output of an RL differentiator is taken across the inductor.