Engineering · Electrical Engineering · Signal Processing
Low-Pass Filter Calculator
Calculates the cutoff frequency, time constant, impedance, and phase angle for first-order RC and RL low-pass filters.
Calculator
Formula
f_c is the -3 dB cutoff frequency in Hz, R is resistance in ohms, C is capacitance in farads, and L is inductance in henries. The time constant is \tau = RC (or \tau = L/R). The phase angle at a given frequency f is \phi = -\arctan(f / f_c). The voltage gain (magnitude) is |H| = 1 / \sqrt{1 + (f/f_c)^2}.
Source: Sedra & Smith, Microelectronic Circuits, 7th Ed.; Horowitz & Hill, The Art of Electronics, 3rd Ed.
How it works
A first-order passive low-pass filter is the simplest frequency-selective network, constructed from just two components: either a resistor and capacitor (RC) or a resistor and inductor (RL). As frequency increases, the reactive component (capacitor or inductor) increasingly impedes or bypasses the signal path, causing the output amplitude to roll off at –20 dB per decade (–6 dB per octave) above the cutoff frequency. This makes the LPF indispensable for removing high-frequency noise, anti-aliasing before analog-to-digital conversion, smoothing PWM signals, and limiting bandwidth in audio and RF applications.
The –3 dB cutoff frequency f_c defines the boundary where the output power falls to half of its passband value (voltage drops to 1/√2 ≈ 0.707 of input). For an RC filter, f_c = 1 / (2πRC). For an RL filter, f_c = R / (2πL). The time constant τ = RC or τ = L/R represents the characteristic time for the filter's transient response to reach 63.2% of its final value. At any arbitrary signal frequency f, the voltage gain magnitude is |H(f)| = 1 / √[1 + (f/f_c)²] and the phase shift is φ = –arctan(f/f_c), indicating output lags input by up to –90° as frequency increases.
Practical applications span virtually every area of electronics: RC filters are widely used in audio equalizers, ADC input conditioning, and power supply decoupling due to their simplicity and low cost; RL filters appear in power electronics, EMI suppression, and motor drive outputs where inductors are already present. Understanding the frequency response, gain, and phase behaviour at the signal frequency of interest allows engineers to verify that a filter meets attenuation requirements and assess any phase distortion introduced into the signal path.
Worked example
Suppose you need to design an RC low-pass filter to remove high-frequency switching noise above 1.59 kHz from a sensor signal, using a standard 10 nF capacitor. What resistor value is needed, and what is the gain and phase shift at 5 kHz?
Step 1 — Find R from the cutoff frequency:
Rearranging f_c = 1/(2πRC): R = 1/(2π × f_c × C) = 1/(2π × 1590 × 10×10⁻⁹) ≈ 10,000 Ω (10 kΩ).
Step 2 — Verify cutoff frequency:
f_c = 1/(2π × 10,000 × 10×10⁻⁹) = 1/(628.3 × 10⁻⁶) ≈ 1,591.5 Hz. ✓
Step 3 — Time constant:
τ = RC = 10,000 × 10×10⁻⁹ = 100 µs.
Step 4 — Gain at 5 kHz:
f/f_c = 5000/1591.5 ≈ 3.142.
|H| = 1/√(1 + 3.142²) = 1/√(1 + 9.87) = 1/√10.87 ≈ 0.3033, or in dB: 20×log₁₀(0.3033) ≈ –10.36 dB. The filter reduces the 5 kHz component to about 30% of its original amplitude.
Step 5 — Phase shift at 5 kHz:
φ = –arctan(5000/1591.5) = –arctan(3.142) ≈ –72.3°. The output signal lags the input by 72.3° at 5 kHz.
Limitations & notes
This calculator models ideal first-order passive RC and RL filters only. Real-world components introduce non-ideal behaviour: capacitors have equivalent series resistance (ESR) and parasitic inductance; inductors have winding resistance and core losses; both degrade the idealized frequency response. The formulas assume a purely resistive source impedance and infinite load impedance — in practice, source and load impedances interact with the filter network, shifting the effective cutoff frequency. For cascaded multi-stage or active filters (Butterworth, Chebyshev, Sallen-Key), different design equations apply and this calculator should not be used directly. At very high frequencies (RF range), transmission line effects, parasitic coupling, and component self-resonance dominate and lumped-element models break down. Always validate with SPICE simulation and physical measurement for production designs, especially in noise-sensitive or safety-critical applications.
Frequently asked questions
What is the -3 dB cutoff frequency and why does it matter?
The –3 dB cutoff frequency is the point at which the filter's output power is reduced to exactly half (–3 dB) of its passband value, corresponding to a voltage ratio of 1/√2 ≈ 0.707. It is the standard definition of where a filter transitions from passing to attenuating signals, and it is the primary design specification for bandwidth and noise rejection.
What is the difference between an RC and an RL low-pass filter?
Both are first-order filters with identical frequency response shapes, but they use different components. An RC filter uses a series resistor and shunt capacitor; an RL filter uses a series inductor and shunt resistor. RC filters are more common in low-frequency signal processing due to smaller component size and cost, while RL filters are preferred in power electronics where inductors are present for energy storage or current smoothing.
How do I choose between an RC and RL low-pass filter for my design?
For audio, instrumentation, and general signal conditioning at frequencies below 1 MHz, RC filters are almost always preferred — capacitors are smaller, cheaper, and easier to tune than inductors. RL filters are the natural choice when you already have an inductor in the circuit (e.g., a switching power supply choke) or when DC resistance must be minimized in high-current paths where a series resistor would cause unacceptable power loss.
Can I cascade two first-order low-pass filters to get a steeper rolloff?
Yes — cascading two identical first-order filters gives a second-order response with a –40 dB/decade rolloff. However, the effective –3 dB frequency of the combined cascade shifts lower than each individual stage's f_c. For a two-stage RC cascade with the same R and C values and a high-impedance buffer between stages, the combined cutoff is approximately f_c_combined = f_c / √(√2 – 1) ≈ 0.644 × f_c. For a true maximally-flat Butterworth second-order response, a Sallen-Key active topology is recommended.
Why is my filter not attenuating as much as the calculator predicts?
The most common cause is load impedance: if the circuit connected after the filter has an impedance comparable to or lower than the filter's series element (R for an RC filter), it loads the filter and shifts the effective cutoff frequency upward, reducing attenuation. Other causes include component tolerances (±5–20% for standard capacitors), parasitic inductance in capacitors at high frequencies, and ground plane or layout issues in PCB implementations. Using a buffer amplifier (unity-gain op-amp) between the filter and load eliminates loading effects.
Last updated: 2025-01-15 · Formula verified against primary sources.