First-order passive low-pass and high-pass filters, solved in any direction: the cutoff frequency from the parts you have, or the part you need for a cutoff you want — with gain and phase at any frequency, a Bode plot, and the nearest values you can actually buy.

How this is calculated

A first-order passive filter is one resistor and one reactive part. The reactance of the capacitor or inductor changes with frequency while the resistor's does not, so the two split the input signal in a ratio that slides with frequency. The corner of that slide is the cutoff frequency:

RC: f_c = 1 / (2π·R·C) RL: f_c = R / (2π·L)

Both reduce to the same thing, f_c = 1/(2π·τ), where the time constant is τ = R·C for RC and τ = L/R for RL. That is why swapping the circuit type above, with values that give the same τ, changes nothing about the response — the two are the same filter built from different parts.

With x = f / f_c, the amplitude and phase are:

low-pass: |H| = 1/√(1+x²), φ = −arctan(x)
high-pass: |H| = x/√(1+x²), φ = 90° − arctan(x)

Gain in decibels is 20·log₁₀|H| — the voltage form, because a filter divides voltage. Our decibel converter moves between ratios and dB in both conventions if you need the power form.

Reading the numbers

The cutoff frequency is not where the filter starts working — it is where the output has already fallen to 70.7 % of the input, which is −3.01 dB, exactly half the power. The filter is already doing something an octave before that, and it is still passing a tenth of the signal a decade after it.

Past the corner, a first-order filter rolls off at 6 dB per octave, or 20 dB per decade, forever. That is gentle. If you need 40 dB of rejection you are two full decades past the cutoff, which is usually too far to be useful — that is the moment to reach for a higher-order filter rather than a smaller capacitor.

The phase shift is the part people forget. A first-order filter turns the phase by 45° at the cutoff and approaches 90° far beyond it, and that rotation starts two decades early: at a tenth of the cutoff frequency a low-pass has already shifted the phase by 5.7°, while costing only 0.04 dB of level. If you are filtering something that will later be summed with an unfiltered copy of itself — a crossover, a feedback loop, a parallel signal path — the phase, not the amplitude, is what will bite you.

Choosing real components

The maths gives one exact number; the shop sells a dozen. Passive parts come in E-series decades — E12 is twelve values per decade (10 % steps, the usual capacitor grid), E24 is twenty-four (5 %), and resistors go further to E96 at 1 %. The calculator gives you the exact value and then the nearest real one in both series, with the cutoff you will actually get.

Then there is tolerance. A ±10 % resistor with a ±20 % ceramic capacitor gives a cutoff anywhere in a ±30 % band, and a Class-2 ceramic (X7R, Y5V) drifts further with temperature and applied voltage — a Y5V part can lose most of its capacitance at rated voltage. If the corner frequency matters, put the precision in the capacitor: buy a film or C0G/NP0 capacitor at 5 % and trim with the resistor, which is the cheaper and more stable part to specify tightly.

Finally, loading. Every formula on this page assumes the source driving the filter has zero output impedance and nothing draws current from the output. A real source resistance adds to R in a low-pass and lowers the cutoff; a real load resistance sits in parallel with the output and does the same. The rule of thumb that keeps you out of trouble: make the filter's resistance at least ten times the source impedance and at most a tenth of the load impedance. When that is impossible, buffer the filter with an op-amp follower and the ideal assumptions become true again.

FAQ

My built filter's cutoff is 20 % off — what did I do wrong?

Probably nothing. Stack the tolerances first: a 10 % resistor and a 20 % capacitor already allow ±30 %, before temperature. Then check loading — if the next stage's input impedance is not far above your resistor, or the source cannot drive it stiffly, the corner moves. Measure the parts you actually soldered before suspecting the circuit; two multimeter readings usually explain the whole discrepancy.

Does the resistor value matter if only the product R·C sets the cutoff?

Very much. 1 kΩ with 160 nF and 1 MΩ with 160 pF give the same cutoff and behave nothing alike. The low-resistance version loads its source harder and burns more current but is immune to stray capacitance and picks up little noise. The high-resistance version is quiet on the supply but generates more thermal noise, is vulnerable to a few picofarads of board and cable capacitance, and can be thrown off by the input bias current of whatever follows it. For audio and general signal work, keeping the resistor between about 1 kΩ and 100 kΩ avoids both ends.

Can I cascade two RC sections for a steeper slope?

You can, but not by simply repeating the same section — the second stage loads the first, so the result is not two independent poles. Both slopes reach 12 dB per octave far out, but the corner moves and it moves further than people expect. Two identical sections separated by a buffer put the combined −3 dB point at 0.644 of the single-section cutoff; wire them directly together and the loading drags it down to 0.374. Either buffer the sections, make the second section's impedance about ten times the first, or design a real second-order filter — and in every case set the corner from the combined response, not from one section's 1/(2πRC).

RC or RL — which should I use?

RC, almost always. To get an audio-band cutoff from an inductor you need henries: a 10 kΩ RL low-pass at 1.6 kHz calls for a 1 H inductor, a large, expensive, lossy part with its own series resistance and a magnetic field that both radiates and picks up hum. Capacitors of the equivalent value are cheap and inert. RL filters earn their place where inductors are small and capacitors misbehave — RF, switching-supply output filtering, and anywhere the low-frequency path has to stay DC-coupled through a low resistance.

What is the time constant actually good for?

It is the same filter viewed in the time domain rather than the frequency domain. After a step input, a low-pass output covers 63 % of the distance in one τ and 99 % in five; the 10–90 % rise time is 2.2·τ. That is the number you want when the RC is a switch debounce, a power-on delay, an envelope follower or a sample-and-hold droop budget — and it is why a fast edge and a low cutoff cannot coexist. For musical time constants in milliseconds, the delay & BPM calculator converts them to note values.

Is a high-pass the same thing as a DC blocker or coupling capacitor?

Yes — a series capacitor into the next stage's input resistance is an RC high-pass, whether or not anyone drew it that way. Put the following stage's input impedance in the resistance box above to find where your coupling network actually rolls off. This is the single most common accidental filter in analogue design, and the reason a circuit that measured flat on the bench loses bass into a different load.

How is this different from the speaker crossover calculator?

Different job, different assumptions. This page is for line-level and signal-level filtering into a high-impedance, essentially resistive load. A speaker crossover drives a driver whose impedance swings with frequency and rises with the voice coil's inductance, so the design starts from a nominal impedance and uses aligned second- and fourth-order topologies — Butterworth, Linkwitz-Riley — chosen for how the two drivers sum acoustically, not just for a corner frequency.

Why is the gain slightly negative even far below a low-pass cutoff?

Because a passive filter can only ever remove signal. At a hundredth of the cutoff a low-pass still costs 0.0004 dB, which is nothing, but it is never exactly zero — the capacitor always draws some current. Any filter that shows gain above unity is active, and has an amplifier in it somewhere.

RC & RL filter calculator