RC integrating circuit – Where is the output taken? In a classic single-pole RC integrator used for wave-shaping at low frequencies (input across series R, C to ground), the output voltage is measured across which component?

Introduction / Context:
RC integrating circuits are foundational in analog electronics for converting fast changing inputs into slowly varying outputs (for example, turning a square wave into an approximate triangular waveform). Knowing where to measure the output is essential to obtaining the intended integration behavior.

Given Data / Assumptions:

• Single-pole RC network: input applied to a series resistor R, capacitor C connected from node to ground.
• We operate in the integrating region where RC is large compared with the signal period (i.e., RC >> T of the highest significant frequency component).
• Linear, time-invariant behavior; ideal R and C (no parasitics).

Concept / Approach:

An RC integrator outputs a voltage proportional to the time integral of the input. Placing the capacitor to ground and taking the output at the capacitor node results in a low-pass response whose magnitude falls with frequency and whose phase tends to −90 degrees at high frequency—characteristics that approximate mathematical integration for suitable RC and input spectrum.

Step-by-Step Solution:

Verification / Alternative check:

Differentiate the capacitor current–voltage relation: iC = C * dVout/dt. Because the input current flows through R into C (neglecting load), the capacitor voltage is proportional to the integral of the current and thus the integral of the input (scaled by R). Measuring across R would instead yield a differentiated waveform at high frequency, not the integrator's intended output.

Why Other Options Are Wrong:

Across R: that tends toward differentiation at high frequency. Across the series combination or input source: that simply reproduces Vin. A separate series load node is not part of the standard single-pole integrator topology.

Common Pitfalls:

Choosing R too small or C too small such that ωRC is not ≫ 1; loading the output heavily and shifting the pole; confusing integrator (output at C) with differentiator (output across R).

Final Answer:

Across the capacitor (C).