RC filter cutoff frequency from given R and C A first-order RC filter (single pole) is built with R = 1.00 kΩ and C = 11.0 nF. Assuming an ideal op-amp buffer or negligible loading, what is the cutoff frequency (defined as the –3 dB frequency)?

Introduction / Context:First-order RC filters (high-pass or low-pass) have a single time constant τ = R * C and a cutoff frequency where the magnitude response is 1/√2 of the passband, commonly called the –3 dB point. This frequency sets bandwidth, rise time, and phase behavior for countless analog front ends.

Given Data / Assumptions:

• R = 1.00 kΩ.
• C = 11.0 nF.
• Single-pole RC, negligible loading (e.g., buffered output).
• Cutoff definition: fc = 1 / (2πRC).

Concept / Approach:The pole frequency for a single-pole RC is fc = 1 / (2πRC). With R and C provided, plug in the values directly. Ensure consistent SI units to avoid order-of-magnitude mistakes.

Step-by-Step Solution:

Verification / Alternative check:Use the rule-of-thumb fc ≈ 0.159 / (R[kΩ] * C[µF]). Here R = 1 kΩ and C = 0.011 µF, so fc ≈ 0.159 / (1 * 0.011) ≈ 14.45 kHz, which agrees with the precise calculation.

Why Other Options Are Wrong:

Common Pitfalls:Confusing rad/s with Hz; forgetting to convert nF to F; neglecting load resistance which can shift fc if not buffered.

Final Answer:14.47 kHz