Communication Systems
Second-Order (RLC) Filter Calculator
Resonant frequency, Q, bandwidth, and damping for second-order low-pass, high-pass, and band-pass responses, with the Bode plot.
About this calculator
Two energy-storage elements make a second-order filter, and with them arrives everything first-order circuits cannot do: resonance, peaking, and a 40 dB/decade slope. The canonical responses are fully described by two numbers — the resonant (undamped natural) frequency f₀ and the quality factor Q — regardless of whether they are realized as a series RLC branch, an active biquad, or a mechanical resonator.
For the series RLC realization this calculator computes f₀ = 1/(2π√(LC)) and Q = (1/R)·√(L/C); alternatively, enter f₀ and Q directly for any realization. From them follow the −3 dB bandwidth of the band-pass response, BW = f₀/Q, and the damping ratio ζ = 1/(2Q), which classifies the transient behavior: underdamped and ringing below 1, critically damped at 1, sluggish above it.
Q also governs the frequency response's shape. The band-pass response peaks at f₀ with unity gain and falls off symmetrically; the low-pass and high-pass responses develop a resonant peak of approximately Q times the passband gain when Q exceeds 1/√2 ≈ 0.707 — the Butterworth (maximally flat) boundary. The Bode chart shows magnitude and phase together: the phase swings through 90° (band-pass) or 180° (low/high-pass) around resonance, and the higher the Q, the more abrupt the transition.
Magnitude and phase are evaluated exactly at your frequency of interest, with every formula worked in sequence. For plain resonance parameters without the response evaluation, the RC/RL/RLC time-constant calculator covers the same series/parallel Q arithmetic.
Assumptions
- Ideal L and C with all loss lumped into the single series resistance R; real inductor core loss and capacitor ESR lower the achieved Q.
- Canonical unity-gain second-order transfer functions — an active or coupled realization may scale the passband gain.
- The series branch is driven by an ideal source and the output is unloaded; finite source or load impedance re-damps the circuit.
When to use this calculator
Appropriate for
- Finding resonant frequency, Q, bandwidth, damping, and the Bode response of a canonical second-order filter
- Comparing low-pass, high-pass, and band-pass second-order responses, including peaking above Q = 0.707
- Teaching the f₀/Q/ζ relationships
Not suitable for
- Real circuits where inductor core loss, capacitor ESR, and source/load impedance lower the achieved Q
- Band-stop (notch) responses or higher-order/staggered-tuned filters outside the canonical second-order forms
- Precise filter realization without accounting for loading, tolerance, and the specific topology
What this calculator does not cover
- Second-order canonical forms only — band-stop (notch) responses and higher-order or staggered-tuned filters are outside this model.
- The Q entered or computed is the unloaded Q; source and load resistances in a real circuit lower the working (loaded) Q and widen the bandwidth.
- Frequency response only: step response overshoot and settling follow from ζ but are not computed here.
- As with every calculator on this site: results are preliminary and educational, are not verified for any specific installation, and must be reviewed against the applicable code edition and stamped by a licensed Professional Engineer before real-world use.
Frequently asked questions
How are Q, bandwidth, and damping ratio related?
They are three views of one property: BW = f₀/Q and ζ = 1/(2Q). A high-Q circuit is narrow and lightly damped — selective in frequency but ringing in time. Filter designers think in Q, control engineers in ζ; the circuit doesn't care.
Why does my low-pass filter peak above 0 dB?
Because Q exceeds 1/√2 ≈ 0.707. Beyond that, the resonance is underdamped enough that the magnitude rises near f₀ before rolling off — by roughly a factor of Q for high Q. At exactly Q = 0.707 the response is maximally flat (Butterworth).
Does this apply to filters that aren't literally RLC circuits?
Yes — the normalized second-order responses depend only on f₀ and Q, however they are realized: active Sallen-Key or biquad stages, crystal and ceramic resonators, even mechanical systems. Use the 'f₀ and Q directly' mode for those.
References
- Alexander, C. & Sadiku, M., Fundamentals of Electric Circuits, 5th ed., Ch. 14 (resonance and second-order filters)
- Sedra, A. & Smith, K., Microelectronic Circuits, 7th ed. (second-order filter functions and Q)
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