Wave-shaping with RL networks: With a square-wave input applied to a series RL differentiator and the output taken across the inductor, what output waveform shape is observed?

Introduction / Context:First-order RL and RC networks are used for simple differentiation and integration of waveforms. Knowing where the output is taken (across R or across L) determines whether you see sharp spikes, exponential edges, or low-pass behavior. This is crucial in pulse circuits, timing, and EMI considerations.

Given Data / Assumptions:

• Series RL circuit driven by an ideal square wave.
• Output measured across the inductor L.
• Differentiator condition: time constant small compared to the square-wave period.

Concept / Approach:Voltage across an inductor is v_L = L * di/dt. When the input steps at each square-wave transition, current attempts to change from one steady value to another exponentially, so di/dt is largest at the transition and then decays toward zero. Consequently, v_L exhibits pulses that start at a high value and decay exponentially back to zero each half-cycle—an exponential-shaped pulse at each edge.

Step-by-Step Solution:

Verification / Alternative check:Time constant τ = L/R. If τ ≪ half-period, pulses are narrow compared to the flat portions. Oscilloscope traces clearly show exponentially decaying lobes of alternating polarity.

Why Other Options Are Wrong:

• dc: Only after a long time in a pure inductorless circuit; here the inductor voltage averages to zero between edges.
• Short time-duration pulses: True for output across R (current spikes), but across L the shape is explicitly exponential.
• Only a phase shift: RL differentiators produce transient wave-shapes, not mere sinusoidal phase shifts.

Common Pitfalls:Confusing where the output is taken; swapping R and L changes the observed waveform. Ensure the differentiator condition (small τ) to get distinct pulses.

Final Answer:exponential-shaped pulses