The RC low-pass filter takes its output across the capacitor and passes the bass. Swap the parts — take the output across the resistor instead — and you get its mirror image: the RC high-pass filter, which blocks the bass and passes the treble. It is the circuit behind DC-blocking coupling capacitors, tweeter crossovers, and every audio input that must keep DC out. This post explains why it works and works a complete example with a 1 kHz cutoff.

Figure 1: The RC high-pass filter (a), and its frequency response (b). The gain is near zero at low frequency, −3 dB at the cutoff, then flat at 0 dB — rising 20 dB for every ×10 of frequency below the cutoff. (Schematic, not to scale.)

Why it works — capacitive reactance again

The capacitor's reactance is X_C = 1 / (2πfC). At low frequency X_C is huge, so almost all of the input voltage drops across the capacitor and almost nothing is left across the resistor — the output. At high frequency X_C is tiny, so the full input appears across the resistor. The resistor and capacitor form a frequency-dependent voltage divider, this time with V_out taken across the resistor:

Gain = V_out / V_in = R / √(R² + X_C²)

The cutoff frequency

The cutoff (or −3 dB) frequency is the same formula as for the low-pass filter — it is where X_C equals R. At that point the gain is 1/√2 = 0.707, the output power is half the input power, and this time the output leads the input by 45°:

f_c = 1 / (2πRC)

The same R and C also set the time constant τ = R × C, which is how the filter responds to sudden changes — the step-response twin of the frequency response.

Worked example — a 1 kHz high-pass filter

Design a filter with R = 1.6 kΩ and C = 100 nF, driven by a 1 V sine wave — the same values as the low-pass example, so the two filters can be compared directly.

Step 1 — cutoff frequency: f_c = 1 / (2π × 1600 × 100 × 10⁻⁹) = 1 / (1.005 × 10⁻³) = 995 Hz — about 1 kHz. The time constant is τ = 1600 × 100 × 10⁻⁹ = 160 µs.

Step 2 — 100 Hz (well below cutoff): X_C = 1 / (2π × 100 × 100 × 10⁻⁹) = 15,915 Ω. Gain = 1600 / √(1600² + 15915²) = 0.100 (−20.0 dB), so V_out = 0.100 V. The 100 Hz signal is cut to a tenth; it leads the input by arctan(15915 / 1600) = 84.3°.

Step 3 — 1 kHz (at the cutoff): X_C = 1 / (2π × 1000 × 100 × 10⁻⁹) = 1,592 Ω, almost exactly R. Gain = 1600 / √(1600² + 1592²) = 0.709 (−2.99 dB), so V_out = 0.709 V, and the phase lead is 44.8°.

Step 4 — 10 kHz (ten times the cutoff): X_C = 159 Ω. Gain = 1600 / √(1600² + 159²) = 0.995 (−0.04 dB), so V_out = 0.995 V — the signal passes almost untouched. The phase lead is now only 5.7°. Every ×10 of frequency below the cutoff costs another 20 dB: at 10 Hz the output would be about 0.01 V.

Design notes

To design a filter, pick the cutoff you need and choose R and C from f_c = 1 / (2πRC). A practical starting point is C = 100 nF, then R = 1 / (2π × f_c × C). Keep R between about 1 kΩ and 100 kΩ, for the same reasons as the low-pass filter. The commonest use is as a coupling capacitor: put the capacitor in series with an audio signal and the resistor to ground, and DC is blocked while the audio passes — but remember the cutoff must sit below the lowest frequency you want to keep, or the bass will be thinned out. If you need a sharper cut-off than 20 dB per decade, cascade two RC stages with a buffer (an op-amp follower) between them.

Key points to remember

• An RC high-pass filter takes the output across the resistor: high frequencies pass, low frequencies are blocked, because X_C = 1 / (2πfC) is huge at low frequency.
• Cutoff frequency f_c = 1 / (2πRC); at f_c the gain is 0.709 (−3 dB) and the output leads the input by about 45°.
• Worked example: R = 1.6 kΩ, C = 100 nF → f_c = 995 Hz, τ = 160 µs; a 1 V input gives 0.100 V at 100 Hz, 0.709 V at 1 kHz and 0.995 V at 10 kHz.
• Below the cutoff the gain rises 20 dB per decade — ×10 frequency, ×10 output — and flattens at 0 dB above it.
• The classic use is the coupling capacitor: it blocks DC and passes audio, as long as the cutoff sits below the lowest frequency to keep.