Rate-of-change circuit — which block produces an output proportional to the time derivative (rate of change) of the input voltage?

Introduction / Context:Classic analog building blocks include differentiators and integrators. A differentiator accentuates rapid changes and edges, while an integrator accumulates the signal over time. This question checks recognition of the block that outputs a signal proportional to the input’s rate of change.

Given Data / Assumptions:

• Continuous-time, small-signal linear analysis.
• Ideal op-amp representations where relevant.
• Focus on functional behavior, not implementation details.

Concept / Approach:The ideal differentiator realizes v_out(t) ∝ dv_in(t)/dt. In frequency terms, it multiplies by jomega, boosting high-frequency components. In contrast, an integrator scales by 1/(jomega), emphasizing low-frequency content. Mixers and “averagers” are different functions altogether.

Step-by-Step Solution:

Verification / Alternative check:Standard inverting differentiator: a series capacitor at the input and a feedback resistor produce v_out = -RC * dv_in/dt (sign and scale depend on topology).

Why Other Options Are Wrong:

• Integrator: outputs an integral, not a derivative.
• Curve averager: not a standard calculus block; averaging is not differentiation.
• Mixer: multiplies/combines signals (e.g., RF), not a time derivative.

Common Pitfalls:Assuming “averaging fast changes” is differentiation; conflating noise amplification (a practical differentiator issue) with the ideal mathematical operation.

Final Answer:Differentiator.