Choose low-pass or high-pass from the output node

The output node changes the filter

An RC series network can be a low-pass when output is across the capacitor, or a high-pass when output is across the resistor. The unloaded high-pass transfer is $H(s)=sRC/(1+sRC)$. At DC the capacitor blocks and output is zero; at high frequency the ideal capacitor impedance tends to zero and output approaches input. A low-pass has the complementary first-order form $1/(1+sRC)$.

For R = 100 kΩ and C = 1 μF, a high-pass corner is 1.591549 Hz. At 10 Hz, magnitude is approximately 0.987570 and phase lead approximately 9.043°. This can remove a DC offset while passing a slowly varying AC signal, but it also removes a genuine DC sensor value. Decide whether the desired information includes DC before adding a coupling capacitor.

Source and load alter the ideal response

Suppose the source has 10 kΩ resistance, the series capacitor is 1 μF, and the resistor to ground is 100 kΩ. At high frequency, gain is 100/(10+100) = 0.909091. The time constant is (10 kΩ + 100 kΩ) × 1 μF = 0.11 s, and corner is 1.446863 Hz. The simple 100 kΩ × 1 μF formula missed both the passband loss and the pole shift.

For a low-pass driven through source resistance, include that resistance in the series R. With 1 kΩ source resistance plus a 10 kΩ filter resistor and 100 nF capacitor, unloaded corner becomes 144.686312 Hz. A subsequent resistive load changes DC gain and the capacitor's effective resistance again.

Select what you want to reject

A first-order low-pass is not a brick wall. If fc = 100 Hz, a 1 kHz interference component is reduced only to 0.099504 of its input amplitude, approximately −20.043 dB. Decide whether that is enough before increasing complexity. Cascading stages requires considering loading or inserting a buffer. Keep passband gain, cutoff reference, signal phase, and interference attenuation in the design record; “use a filter” is not an acceptance criterion.