Butterworth Filter Calculator
Use this calculator to plan ideal Butterworth filter order, cutoff frequency, attenuation, normalized poles, and stage Q values before moving into a circuit topology such as Sallen-Key.
FIL-009 deliberately calculates the Butterworth approximation itself. It does not synthesize component values for each active filter stage, which remains the scope of topology-specific calculators.
Engineering tool
Butterworth Filter Calculator
Calculate Butterworth order, low-pass and high-pass response, normalized poles, stage Q values, cutoff frequency, and sweep rows without performing topology-specific Sallen-Key synthesis.
Parameter panel
fp must be below fs.
fs must be above fp.
Result console
- Required Order
- 8
- Raw Calculated Order
- 7.61848
- Selected Cutoff
- 1.088119 kHz
- Cutoff Lower Bound
- 1.088119 kHz
- Cutoff Upper Bound
- 1.12469 kHz
- Second-Order Stages
- 4
- First-Order Stage
- No
- Specification
- Passes
Required order was rounded upward with ceil, not rounded to the nearest integer.
The next lower order does not satisfy the selected specification.
| Parameter | Value | Engineering Check |
|---|---|---|
| Passband attenuation | 1 dB | Limit <= 1 dB |
| Stopband attenuation | 42.2968 dB | Requirement >= 40 dB |
| Allowed cutoff range | 1.088119 kHz to 1.12469 kHz | Derived from passband and stopband constraints |
| N-1 check | Fails | The lower order should fail for a non-trivial rounded-up design |
Formula reference
Butterworth Formulas
Butterworth filters are maximally flat magnitude approximations with no ideal passband ripple. Low-pass and high-pass order calculations require different frequency-ratio direction.
Low-pass: |H(jf)| = 1 / sqrt(1 + (f/fc)^(2N))High-pass: |H(jf)| = 1 / sqrt(1 + (fc/f)^(2N))Attenuation = 10log10(1 + ratio^(2N))Gain dB = 20log10(|H|)At cutoff: |H| = 1/sqrt(2) = -3.0103 dBLow-pass order ratio = fs/fpHigh-pass order ratio = fp/fsN = ceil(log10[(10^(As/10)-1)/(10^(Ap/10)-1)] / [2log10(ratio)])Low-pass cutoff from attenuation: fc = f / (10^(A/10)-1)^(1/(2N))High-pass cutoff from attenuation: fc = f x (10^(A/10)-1)^(1/(2N))Stage Q from pole p: Q = 1/(-2 Re(p))Variable definitions
- N
- Butterworth filter order
- fc
- complete Butterworth cutoff frequency
- fp
- passband frequency
- fs
- stopband frequency
- Ap
- maximum passband attenuation in dB
- As
- minimum stopband attenuation in dB
- p
- normalized Butterworth pole
- Q
- second-order stage quality factor
Butterworth Formula Audit
| Low-Pass Magnitude Formula | |H| = 1 / sqrt(1 + (f/fc)^(2N)) |
|---|---|
| High-Pass Magnitude Formula | |H| = 1 / sqrt(1 + (fc/f)^(2N)) |
| Cutoff Definition | At f = fc, |H| = 1/sqrt(2), gain = -3.0103 dB. |
| Required Order Formula | N >= log10[(10^(As/10)-1)/(10^(Ap/10)-1)] / [2log10(ratio)] |
| Low-Pass Frequency Ratio | ratio = fs/fp, with fs > fp. |
| High-Pass Frequency Ratio | ratio = fp/fs, with fp > fs. |
| Passband Cutoff Constraint | Derived independently from Ap at fp. |
| Stopband Cutoff Constraint | Derived independently from As at fs. |
| Pole Formula | Stable normalized poles lie on the unit circle in the left-half s-plane. |
| Pole Count | Exactly N poles are generated. |
| Pole Stability | All generated poles have negative real part. |
| Stage Pairing | Complex conjugate pole pairs form second-order stages. |
| Stage Q Formula | Q = 1/(-2 Re(p)) for each complex pole pair. |
| Odd-Order Handling | Odd order includes one first-order stage at s = -1. |
| Asymptotic Roll-Off | 20N dB/decade or 6.0206N dB/octave far from cutoff. |
| Numerical Overflow Handling | Attenuation is calculated in log-domain and near-zero gain is displayed safely. |
Worked Examples
N=1 at fc
Known: f = fc
Gain = 0.70710678 and attenuation = 3.0103 dB.
N=2 at fc
Known: f = fc
The complete response is still -3.0103 dB.
N=8 at fc
Known: f = fc
All Butterworth orders share the same cutoff definition.
2nd-order LP stopband
Known: f = 10fc
Attenuation is approximately 40.0004 dB.
4th-order LP stopband
Known: f = 10fc
Attenuation is approximately 80 dB.
2nd-order HP stopband
Known: f = fc/10
Attenuation is approximately 40.0004 dB.
Low-pass passband limit
Known: f << fc
Gain approaches 1 V/V.
High-pass passband limit
Known: f >> fc
Gain approaches 1 V/V.
N=2 stage Q
Known: Pole-derived pair
Q = 0.70710678.
N=3 stage structure
Known: Odd order
One first-order stage plus one second-order stage with Q = 1.
N=4 Q set
Known: Two complex pairs
Q values are approximately 0.5411961 and 1.3065630.
N=5 Q set
Known: Two complex pairs plus real pole
Q values are approximately 0.618034 and 1.618034.
Pole magnitude
Known: Normalized poles
Every ideal Butterworth pole has magnitude 1.
Pole stability
Known: Left-half s-plane
Every pole has negative real part.
Odd-order real pole
Known: N odd
One real pole appears at s = -1.
Cutoff solver round-trip
Known: N + f + A
Solved fc reproduces the requested attenuation.
Order solver
Known: fp = 1 kHz, fs = 2 kHz, Ap = 1 dB, As = 40 dB
The raw order is rounded upward and verified.
N-1 verification
Known: Same specification
The next lower order fails unless the raw order is already at a boundary.
Unit conversion
Known: 1000 Hz and 1 kHz
Both represent the same frequency and produce identical response.
Log sweep
Known: 50 rows
Rows are finite, ordered, and bounded to avoid raw Infinity values.
Butterworth Filter
Ideal Butterworth filters have maximally flat magnitude response and no passband ripple.
Filter Order
Higher order produces a steeper transition but requires more stages and tighter implementation control.
Cutoff Frequency
The complete Butterworth response is -3.0103 dB at fc for every order.
Passband
The passband is the region intended to remain close to unity gain or selected passband gain.
Stopband
The stopband is the frequency region where a specified minimum attenuation is required.
Roll-Off
The asymptotic slope is 20N dB/decade, but points near cutoff require the exact formula.
Butterworth Poles
Poles lie on the normalized unit circle in the left-half s-plane.
Stage Q
Different pole pairs usually produce different second-order Q values.
Cascaded Filters
Individual stages are not all -3 dB at the overall cutoff; their product is the Butterworth response.
Sallen-Key Implementation
Use Sallen-Key pages for topology-specific component realization after stage Q is known.
Op-Amp Limitations
High-order active filters require op-amp bandwidth, slew rate, noise, and output swing checks.
Tolerance
Component tolerance changes realized pole locations, cutoff frequency, and Q.
Common Mistakes
Support reference
FAQ
What is a Butterworth filter?
A Butterworth filter is an ideal filter approximation with a maximally flat magnitude response and no passband ripple in the mathematical model.
Why is a Butterworth filter called maximally flat?
Its passband magnitude has as many zero derivatives at zero frequency as possible for the selected order, which creates a smooth response without ripple.
How do I calculate Butterworth filter order?
Use passband frequency, passband attenuation, stopband frequency, and stopband attenuation. The raw order is rounded upward with ceil because a lower integer order may fail the requirement.
What is the cutoff frequency of a Butterworth filter?
The cutoff frequency is the complete filter response frequency where ideal Butterworth gain is 1/sqrt(2), or approximately -3.0103 dB.
Why is the response -3 dB at cutoff?
At f = fc the normalized term equals one, so |H| = 1/sqrt(1 + 1), which is 1/sqrt(2) or -3.0103 dB.
How much roll-off does each filter order provide?
Each order contributes approximately 20 dB per decade, or 6.0206 dB per octave, far from the cutoff region.
How do I calculate Butterworth attenuation?
For low-pass filters, attenuation is 10log10(1 + (f/fc)^(2N)). For high-pass filters, attenuation is 10log10(1 + (fc/f)^(2N)).
What are Butterworth poles?
Butterworth poles are normalized stable transfer-function poles on the unit circle in the left-half s-plane. They determine the cascade stages and Q values.
How do I calculate the Q of each Butterworth stage?
For a complex pole pair with denominator s^2 + a s + 1, Q = 1/a. This calculator derives a from the pole real part using Q = 1/(-2 Re(p)).
Why do high-order Butterworth stages have different Q values?
Different complex pole pairs sit at different angles on the unit circle, so their damping terms and Q values are not identical.
How do I cascade Butterworth filter stages?
Cascade the first-order section when the order is odd and use second-order sections for complex pole pairs. Stage Q values should be implemented by an appropriate active filter topology.
Is every stage -3 dB at the cutoff frequency?
No. The complete cascaded Butterworth response is -3.0103 dB at fc. Individual second-order stages generally are not all -3 dB at the overall cutoff.
What is the difference between Butterworth and Sallen-Key?
Butterworth is a filter approximation and pole pattern. Sallen-Key is one circuit topology that can implement suitable second-order low-pass or high-pass stages.
How do component tolerances affect a Butterworth filter?
Component tolerances shift realized pole locations, cutoff frequency, Q, and attenuation. High-Q stages are especially sensitive and should be verified with SPICE and measurement.
Planned Engineering Guides
Planned guide
Butterworth Filters Explained
Planned guide
How to Choose Filter Order
Planned guide
Active Filter Stage Q and Component Tolerance
Planned guide
Butterworth vs Chebyshev vs Bessel Filters
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Engineering Disclaimer
This calculator uses ideal Butterworth magnitude, pole, order, and stage-Q relationships. It does not model real op-amp gain-bandwidth, slew rate, noise, component tolerance, loading, PCB parasitics, or topology-specific component synthesis. Verify critical filter designs using the selected circuit topology, SPICE, tolerance analysis, and bench measurement.