A 1 kHz Filter Cut a 1 kHz Signal to 70.7%

A 2.00 V peak sensor tone entered a first-order RC low-pass at its stated cutoff. The output was about 1.414 V—not nearly 2.00 V.

1. Trace the RC Signal Path

A resistor and capacitor form a frequency-dependent divider. Low frequencies pass strongly; higher frequencies are increasingly attenuated.

2. Find the Response Knee

Cutoff is the bend in a first-order low-pass response, not a zero-loss boundary between frequencies that pass and frequencies that stop.

3. Use the Magnitude Equation

For a first-order RC low-pass, |H| = 1/√(1 + (f/fc)²). The frequency ratio determines the output-to-input amplitude.

4. Evaluate the Cutoff Point

At f = fc, the ratio becomes 1/√2 ≈ 0.707. The output voltage amplitude is already about 29.3% below the input.

5. Convert Gain Into Voltage

Multiply 2.00 V peak by 0.707. The expected output is about 1.414 V peak, so the measurement is evidence—not a mystery loss.

6. Design From the Allowed Loss

To retain at least 90% amplitude at 1 kHz, set fc to at least about 2.06 kHz. Add margin for real component tolerances.

7. Choose Parts, Then Verify

Use fc = 1/(2πRC), select available values, include tolerances and loading, then measure the actual response with a frequency sweep.

Specify Passband Performance, Not Just Cutoff

Define the amplitude you must preserve at the signal of interest. Then place cutoff, choose components and verify the built circuit against that goal.