Output of a first-order low-pass at cutoff: If a low-pass filter’s maximum output is 15 V, what is the output (magnitude) at the critical frequency (cutoff)?

Introduction / Context:For first-order RC filters, the magnitude at the critical (cutoff) frequency fc equals 1/sqrt(2) of the passband (maximum) magnitude. This corresponds to the well-known -3 dB point.

Given Data / Assumptions:

• Max (passband) output magnitude = 15 V.
• Critical frequency fc defined for a first-order low-pass.
• Ideal components and standard -3 dB definition.

Concept / Approach:At fc, |H(jωc)| = 1/√2 ≈ 0.707 of the passband gain. Therefore V_out(fc) = 0.707 * V_max.

Step-by-Step Solution:V_out(fc) = 0.707 * 15 VV_out(fc) ≈ 10.606 VRounded to two decimals: 10.60 V

Verification / Alternative check:-3 dB corresponds to a power ratio of 0.5 and a voltage ratio of 0.707. Since filters are typically specified by voltage gain with equal source/load impedances, using 0.707 is appropriate.

Why Other Options Are Wrong:

• 0 V: The filter does not null the output at cutoff.
• 15 V: That is the passband maximum, not at fc.
• 21.21 V: Larger than the maximum; impossible.

Common Pitfalls:Mixing the -3 dB point with -6 dB or confusing voltage and power ratios. Always remember 0.707 for voltage at fc in first-order responses.

Final Answer:10.60 V