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Butterworth Filters

Imagine a filter whose frequency response in the passband is as flat as a perfectly calm lake — not a single ripple anywhere. That is the defining property of the Butterworth filter, described by British engineer Stephen Butterworth in his 1930 paper "On the Theory of Filter Amplifiers." Where other filter families sacrifice flatness to gain sharper rolloff, the Butterworth refuses to compromise: it achieves the most perfectly flat magnitude response of any realizable filter family. This makes it the natural choice for any application where signal amplitude must be preserved accurately across the entire passband — audio and sensor signal conditioning, measurement front ends, and harmonic filters where the wanted signal must emerge undistorted while the harmonics are rejected.

The Butterworth filter is also the simplest of the classical filter families to design. Its component values can be derived from straightforward formulas or read directly from normalized prototype tables, then scaled to any cutoff frequency and impedance level. You do not need to be a mathematician to build one; with this lesson, a clear design procedure, and a few components from the parts drawer, you can build a properly specified filter from first principles.

What you will learn in this lesson: the mathematical definition of the Butterworth response, the normalized low-pass prototype and why it matters, how to scale prototypes to any frequency and impedance, normalized component value tables for orders 1 through 7, practical LC low-pass filter design, and how to verify your design using the stopband attenuation formula.

The Butterworth Magnitude Response

The Butterworth filter is defined by its magnitude-squared frequency response. For a low-pass filter of order n with cutoff frequency fc:

|H(f)|² = 1 / (1 + (f/fc)²ⁿ)
where n is the filter order (number of poles) and fc is the −3 dB cutoff frequency.

This formula tells you everything about how the filter behaves. At f = 0 (DC), (f/fc)²ⁿ = 0, so |H|² = 1 — full output, no attenuation. At f = fc, (f/fc)²ⁿ = 1, so |H|² = 1/2 — output power is half input power, which is −3 dB. At frequencies far above fc, (f/fc)²ⁿ grows very large and |H|² approaches zero — strong attenuation.

The critical insight is the exponent 2n. A first-order filter (n = 1) gives |H|² = 1/(1 + (f/fc)²). A seventh-order filter (n = 7) gives |H|² = 1/(1 + (f/fc)¹⁴). At twice the cutoff frequency (f = 2fc), the attenuation in dB is:

Attenuation (dB) = 10·log₁₀(1 + (f/fc)²ⁿ)
At f = 2·fc: n=1 → 7 dB; n=3 → 19 dB; n=5 → 31 dB; n=7 → 43 dB

These numbers are critically important for filter design. If you need, say, 40 dB of attenuation at twice the cutoff frequency, you need at least a 7th-order Butterworth. In practice, you choose fc so that the frequencies you must reject are as far above it as possible while the wanted signal stays comfortably inside the passband — every extra bit of frequency ratio lets a lower-order filter meet the requirement.

Butterworth filter frequency response curves for orders 1 through 7, showing increasing sharpness of rolloff and the common -3dB point at the cutoff frequency

Butterworth low-pass filter magnitude response for orders 1 through 7. All curves are exactly −3 dB at the cutoff frequency. Higher orders give steeper rolloff but the passband remains perfectly flat for all orders.

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The "Maximally Flat" Property

The term "maximally flat" has a precise mathematical meaning: the Butterworth response has the maximum number of derivatives equal to zero at f = 0. Practically, this means the response is flat to the highest possible degree in the passband — it does not begin to roll off until it is close to the cutoff frequency. All other classical filter families (Chebyshev, elliptic, Bessel) sacrifice some degree of passband flatness to gain some other property. The Butterworth makes no such compromise.

The consequence is that a Butterworth low-pass filter introduces essentially zero amplitude variation well inside its passband. A filter with fc = 9 MHz has a response less than 0.01 dB below the DC response at 7.0 MHz, and less than 0.05 dB down at 7.3 MHz. A signal anywhere in that range passes through with negligible amplitude change, while frequencies above 14 MHz are progressively attenuated.

The Normalized Low-Pass Prototype

The genius of filter design tables is the concept of the normalized prototype. Instead of computing component values for every combination of cutoff frequency, source impedance, and load impedance, mathematicians solved the problem once for a specific "normalized" case:

  • Cutoff frequency: 1 radian per second (ωc = 1 rad/s, or fc = 1/(2π) Hz)
  • Source and load impedance: 1 ohm

For these normalized conditions, the component values — all in units of ohms, henrys, or farads that happen to equal small numbers — are tabulated for each order n. The normalized prototype values are called g1, g2, g3, …, gn, where each g represents either a series inductor or a shunt capacitor in the ladder network.

To design a filter for any real cutoff frequency and impedance, you simply scale the normalized values. This two-step process (look up normalized values, then scale) makes filter design accessible without needing to solve differential equations from scratch each time.

The Ladder Topology

A Butterworth low-pass filter is built as a ladder network: alternating series and shunt elements. For a filter connected between equal source and load impedances (50 ohms in and 50 ohms out), the standard topology for odd-order filters is:

Series L₁ → Shunt C₂ → Series L₃ → Shunt C₄ → … → Series Lₙ
(alternating series inductors and shunt capacitors, starting and ending with series inductors for odd orders)

For even-order filters, or when source and load impedances are not equal, the topology may start with a shunt element instead. For filters with equal 50-ohm source and load, the standard odd-order topology starting with a series inductor is the most common choice — it provides a low-resistance path for DC (the series inductors present almost zero DC resistance), which is convenient when the filter also has to pass a DC bias or supply current.

Normalized Component Value Tables

The following table gives the normalized component values gk for Butterworth low-pass prototype filters of orders 1 through 7. For a prototype with source resistance RS = 1 Ω and cutoff frequency ωc = 1 rad/s, even-numbered g values are shunt capacitors (in farads) and odd-numbered g values are series inductors (in henrys). The last value gn+1 is the normalized load resistance (always 1 for Butterworth with equal terminations).

Order (n) g₁ g₂ g₃ g₄ g₅ g₆ g₇
12.0000——————
21.41421.4142—————
31.00002.00001.0000————
40.76541.84781.84780.7654———
50.61801.61802.00001.61800.6180——
60.51761.41421.93191.93191.41420.5176—
70.44501.24701.80192.00001.80191.24700.4450

Notice the symmetry: each row reads the same left-to-right as right-to-left. This is because the Butterworth filter has equal source and load impedances, making the filter symmetric — you can turn it around and it performs identically. Also notice that the middle value of each odd-order filter equals 2.0000; this is a mathematical property of the Butterworth normalization.

Frequency and Impedance Scaling

The normalized prototype values (all referenced to 1 ohm and 1 rad/s) must be scaled to match your actual design frequency and impedance. The scaling formulas are straightforward:

Actual inductor value: L = gk × R / (2π × fc) [henrys]
Actual capacitor value: C = gk / (R × 2π × fc) [farads]
where R = system impedance (typically 50 Ω) and fc = cutoff frequency in Hz

These formulas come from two facts: impedance scaling multiplies inductors by R and divides capacitors by R; frequency scaling divides all component values by 2π·fc. The two scalings are applied simultaneously in the formulas above.

Calculating Stopband Attenuation

Before committing to a specific filter order, it is worth calculating exactly how much attenuation your filter will provide at the harmonic frequencies. The formula is:

Attenuation (dB) = 10 × log₁₀(1 + (f / fc)²ⁿ)
where f is the frequency at which you want to know the attenuation, fc is the −3 dB cutoff, and n is the filter order.

For filter design, work backwards: you know the required attenuation (say, 43 dB) and the frequency ratio (fharmonic / fc), and you want to find the minimum order n:

n ≥ log₁₀(10^(A/10) − 1) / (2 × log₁₀(f / fc))
where A is the required attenuation in dB.

Always round n up to the next integer (since you cannot build a non-integer order filter). A slightly higher order gives more margin.

Worked Design Example: Square Wave to Sine Wave

A very common bench requirement is a clean sine wave from a square-wave clock — for testing an amplifier, for a reference signal, or to drive a circuit that needs a sinusoidal input. A square wave contains its fundamental plus odd harmonics at 3×, 5×, 7× and so on, with amplitudes of 1/3, 1/5, 1/7 of the fundamental. A low-pass filter that passes the fundamental and removes the harmonics turns it into a sine wave. Let's design one first for a case where Butterworth struggles, then for one where it shines.

Part A — when the harmonic is close to the cutoff:

Suppose the signal could be anywhere from 7.0 to 7.3 MHz and you need 50 dB of attenuation at the second harmonic, 14.0 MHz (a sine-wave oscillator with distortion produces even harmonics too). With fc = 9.0 MHz:

Ratio = 14.0 / 9.0 = 1.556

n ≥ log₁₀(10^(50/10) − 1) / (2 × log₁₀(1.556)) = 5.000 / (2 × 0.1920) = 13.0

A 13th-order filter is impractical. When the frequency to be rejected is less than about twice the cutoff, a Butterworth needs too many components; the Chebyshev and elliptic filters in the next two lessons do the same job with far fewer.

Part B — a 10 MHz square-wave clock:
  • Wanted: the 10 MHz fundamental
  • First unwanted component: the 3rd harmonic at 30 MHz (already 9.5 dB below the fundamental, since its amplitude is 1/3)
  • System impedance: 50 Ω
  • Cutoff frequency fc: 12 MHz; order: 5

From the table: g₁ = 0.6180, g₂ = 1.6180, g₃ = 2.0000, g₄ = 1.6180, g₅ = 0.6180

Calculating inductors (odd elements: g₁, g₃, g₅):

L₁ = L₅ = g₁ × R / (2π × fc) = 0.6180 × 50 / (2π × 12×10⁶)

= 30.90 / (75.40×10⁶) = 410 nH

L₃ = g₃ × R / (2π × fc) = 2.0000 × 50 / (75.40×10⁶) = 1.326 µH

Calculating capacitors (even elements: g₂, g₄):

C₂ = C₄ = g₂ / (R × 2π × fc) = 1.6180 / (50 × 75.40×10⁶)

= 1.6180 / (3.770×10⁹) = 429 pF → use 430 pF (an E24 value)

Final component values:

  • L₁ = L₅ ≈ 410 nH (series inductors at input and output)
  • C₂ = C₄ ≈ 430 pF (shunt capacitors to ground)
  • L₃ ≈ 1.33 µH (series inductor at center)

Verifying attenuation:

At 30 MHz (3rd harmonic): ratio = 30/12 = 2.5; attenuation = 10 × log₁₀(1 + 2.5¹⁰) = 10 × log₁₀(9,538) = 39.8 dB

At 50 MHz (5th harmonic): ratio = 4.17; attenuation = 10 × log₁₀(1 + 4.17¹⁰) ≈ 62 dB

At 10 MHz (the fundamental): ratio = 0.833; loss = 10 × log₁₀(1 + 0.833¹⁰) = 0.65 dB

The 3rd harmonic ends up 9.5 + 39.8 = 49 dB below the fundamental and the 5th about 76 dB below — a clean sine wave. Because the ratio here is 2.5 rather than 1.5, a modest 5th-order Butterworth does the job comfortably.

The worked example shows an important real-world lesson: Butterworth filters provide clean, flat passbands but need a higher order than a Chebyshev to meet the same specification when the frequencies to be rejected are close to the cutoff. When the stopband frequencies are well above the cutoff — roughly 2.5 times or more — or when passband flatness is the primary concern, the Butterworth is the simplest and best choice.

Converting a Low-Pass Butterworth to a High-Pass Filter

The normalized low-pass prototype can be converted to a high-pass filter by a simple element substitution: every series inductor becomes a series capacitor with value 1/gk, and every shunt capacitor becomes a shunt inductor with value 1/gk. After substitution, apply the same frequency and impedance scaling formulas. The result is a high-pass filter with −3 dB cutoff at the same fc and with the same rolloff rate, but now passing high frequencies and blocking low ones.

A 3rd-order Butterworth high-pass filter with fc = 100 MHz and R = 50 Ω would protect a 433 MHz receiver module from strong broadcast interference: it would attenuate signals below 100 MHz (where the powerful AM, shortwave and FM broadcast transmitters live) while passing 433 MHz with virtually no loss.

Practical Construction Tips for LC Butterworth Filters

Building a working Butterworth filter requires attention to the practical aspects of component selection, winding technique, and layout.

Inductor Construction

For LC low-pass filters below about 30 MHz, inductors are typically wound on powdered-iron toroid cores. T68-2 (red/clear) cores are popular for 1–30 MHz; T68-6 (yellow/clear) cores are used above 10 MHz. The number of turns for a given inductance can be calculated from the core's AL value (inductance per 100 turns, in µH). For a T68-2 core, AL ≈ 57 µH per 100 turns. Use the formula: N = 100 × √(L / AL) where L is in µH and AL is in µH/100 turns. For the 1.33 µH inductor above: N = 100 × √(1.33/57) = 15 turns. Above about 50 MHz, small air-wound coils or surface-mount chip inductors are used instead.

Wind inductors with the tightest practical winding, covering 60–80% of the core circumference, and secure the windings with a drop of coil dope or RTV silicone. Use enameled copper wire sized for the current: at signal levels, almost any wire works; for 100 W into 50 Ω, peak current is approximately √(2 × 100/50) = 2 A, so 24 AWG is adequate.

Capacitor Selection

LC filters need high-Q, stable capacitors. C0G (NP0) ceramic capacitors are the standard choice: they have low series resistance, capacitance that barely changes with temperature or voltage, and good self-resonance characteristics. Silver-mica capacitors are a traditional alternative for values under 1,000 pF. Avoid X7R and Z5U ceramics, which have high losses at high frequencies and a capacitance that changes with temperature and applied voltage, shifting the filter response. For filters carrying significant power, check that the capacitor's voltage rating exceeds the peak signal voltage.

The capacitor's voltage rating must exceed the peak RF voltage: for 100 W into 50 Ω, peak voltage = √(2 × P × R) = √(2 × 100 × 50) = 100 V. Use capacitors rated at 200 V minimum (500 V is safer).

Layout and Shielding

A poor physical layout can ruin an otherwise well-designed filter. Keep the filter in a shielded metal enclosure. Orient the inductors so their magnetic fields are perpendicular to each other (alternating horizontal and vertical, or with 90° rotation between adjacent toroids) to minimize mutual coupling. Minimize lead lengths and keep ground connections short and direct. Run the signal path in a straight line if possible, with input and output connectors on opposite ends of the enclosure.

⚖ Experiment: Test a Simple 3-Pole Low-Pass Filter at Audio Frequencies

This experiment demonstrates the Butterworth low-pass response using audio-frequency components that are easy to measure with a multimeter and tone generator app. You will build a 3rd-order Butterworth LPF with fc ≈ 1 kHz and observe the rolloff.

You will need:
  • 3 inductors: 2× 16 mH and 1× 32 mH (can use inductor kits or from old junk electronics)
  • 2 capacitors: 2× 10 µF non-polarized (film type) or 2× 4.7 µF
  • Breadboard and jumper wires
  • Signal source: a tone generator app on a smartphone, connected to a 3.5mm cable
  • Digital multimeter with AC voltage measurement
  • 1 kΩ load resistor
  1. Calculate the expected cutoff frequency: for a 3rd-order Butterworth with R = 1 kΩ, use values for 1 kHz. From the table, g₁ = g₃ = 1.000 and g₂ = 2.000. L₁ = L₃ = 1.000 × 1000 / (2π × 1000) = 159 mH. C₂ = 2.000 / (1000 × 2π × 1000) = 318 nF. Use whatever values are closest in your parts box and recalculate fc = R·g / (2π·L) for the inductor scaling.
  2. Build the ladder: series L₁ — shunt C₂ to ground — series L₃ — load resistor R = 1 kΩ to ground.
  3. Apply 1 V AC from the tone generator. Measure the output voltage across the load resistor at 100 Hz, 200 Hz, 500 Hz, 1 kHz (cutoff), 2 kHz, 5 kHz, and 10 kHz.
  4. Calculate the attenuation at each frequency in dB: Attenuation = 20 × log₁₀(Vout/Vin).
  5. Compare your measured values to the Butterworth formula. At the cutoff frequency, you should measure −3 dB (Vout = 0.707 × Vin). Above cutoff, verify the rolloff is approximately −18 dB/octave (×8 frequency gives ×8 more attenuation).
What you should see:

The output voltage should be essentially unchanged (within a fraction of a dB) from DC up to about half the cutoff frequency. At fc, the output is 3 dB below the passband. One octave above fc, the attenuation is about 9 dB additional (18 dB total for 3rd order = 6 dB/octave × 3 poles). Two octaves above, attenuation is about 27 dB additional. The flat passband and clean rolloff confirm the Butterworth response. Any significant ripple in the passband would indicate component mismatch or coupling between inductors — adjust positioning to minimize ripple.

Frequently Asked Questions

Why is the Butterworth cutoff defined at −3 dB, not −6 dB or some other level?

The −3 dB point is used because it represents the half-power point: at −3 dB, the output power is exactly half the input power, and the output voltage is 0.707 (= 1/√2) of the input. This is a natural boundary in the Butterworth mathematics: for an nth-order filter at f = fc, the term (f/fc)²ⁿ = 1 exactly, giving |H|² = 1/(1+1) = 0.5, which is −3 dB. Different filter types define their passband edge differently — Chebyshev, for example, is sometimes specified at the edge of the ripple band, which may be at −0.5 dB. The −3 dB convention is used for Butterworth because it is the natural mathematical breakpoint.

Can I use standard component values instead of the exact calculated values?

Yes, with some caveats. Standard-value capacitors (from the E12 or E24 series) and standard-value inductors are close enough for most applications. The cutoff frequency will shift slightly, and the passband flatness may deviate slightly from the ideal Butterworth response, but the attenuation at frequencies well above fc will be barely affected. In practice, many designers adjust the cutoff frequency slightly to allow use of nearest-standard-value components. For harmonic and anti-aliasing filters, component tolerances of ±5% are usually acceptable. For filters where the passband edge must be precise, tighter tolerances are required.

The component values seem different in other references. Why?

Two different normalizations are in common use. Some references normalize to ωc = 1 rad/s (so fc = 0.159 Hz), while others normalize to fc = 1 Hz (so ωc = 2π rad/s). The component values will differ by a factor of 2π between the two conventions. When using a table from any reference, check which convention it uses and apply the correct scaling formula. The values in this lesson use the ωc = 1 rad/s convention, which is most common in RF filter design literature.

Test Your Knowledge

Answer the questions below to check your understanding of this lesson. Every answer can be found in the lesson above.

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