Butterworth Filters — Maximally Flat LC Filter Design
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 — including the transmitter harmonic low-pass filter, where the fundamental 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. A ham radio operator building a transmitter output filter does not need to be a mathematician; with this lesson, a clear design procedure, and a few components from the junk box, you can build a properly specified harmonic filter from first principles.
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:
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:
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 transmitter filter design. The FCC requires at least 43 dB of harmonic suppression. If your filter's cutoff is at fc and the second harmonic falls at 2·fc, you need at minimum a 7th-order Butterworth to achieve 43 dB. In practice, the second harmonic is usually well above 2·fc (you set fc above the operating frequency but below the second harmonic), giving you more than an octave of headroom and allowing a lower-order filter to meet the requirement.
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.
View LargerThe "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 for transmitter filtering is that a Butterworth low-pass filter introduces essentially zero amplitude variation across the ham band. A 40m filter with fc = 9 MHz will have a frequency response that is less than 0.01 dB below the DC response at 7.0 MHz, and less than 0.05 dB down at 7.3 MHz. This is perfect — the entire 40m band passes through with negligible distortion, while the harmonics 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:
(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 ham radio transmitter 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 smooth, low-attenuation path for DC current (the series inductors present zero DC resistance) which simplifies biasing in some transmitter circuits.
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 |
|---|---|---|---|---|---|---|---|
| 1 | 2.0000 | — | — | — | — | — | — |
| 2 | 1.4142 | 1.4142 | — | — | — | — | — |
| 3 | 1.0000 | 2.0000 | 1.0000 | — | — | — | — |
| 4 | 0.7654 | 1.8478 | 1.8478 | 0.7654 | — | — | — |
| 5 | 0.6180 | 1.6180 | 2.0000 | 1.6180 | 0.6180 | — | — |
| 6 | 0.5176 | 1.4142 | 1.9319 | 1.9319 | 1.4142 | 0.5176 | — |
| 7 | 0.4450 | 1.2470 | 1.8019 | 2.0000 | 1.8019 | 1.2470 | 0.4450 |
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 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:
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:
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: 40m Transmitter LPF
Let's design a low-pass filter for a 100 W 40m transmitter that must attenuate the second harmonic by at least 50 dB. This matches a typical home-built or commercial HF transmitter requirement.
- Operating frequency: 7.000–7.300 MHz (40m)
- System impedance: 50 Ω
- Cutoff frequency fc: 10.5 MHz (safely above 7.3 MHz, well below 14.0 MHz)
- Required attenuation at second harmonic (14.0 MHz): ≥50 dB
- Power handling: 100 W continuous
Step 1: Calculate the frequency ratio at the second harmonic.
Ratio = fharmonic / fc = 14.0 MHz / 10.5 MHz = 1.333
Step 2: Determine the minimum filter order.
n ≥ log₁₀(10^(50/10) − 1) / (2 × log₁₀(1.333))
= log₁₀(10⁵ − 1) / (2 × log₁₀(1.333))
= log₁₀(99999) / (2 × 0.1249)
= 4.9999 / 0.2498
= 20.0 — this seems very high!
Wait — let's reconsider. A ratio of 1.333:1 is very close to the passband. Let's raise fc to reduce the ratio more. Raise fc to 9.0 MHz:
Ratio = 14.0 / 9.0 = 1.556
n ≥ log₁₀(99999) / (2 × log₁₀(1.556)) = 5.000 / 0.3849 = 13.0
Still high. Let's try fc = 7.5 MHz and check only at the 3rd harmonic (21.0 MHz), which the FCC also regulates:
At 14.0 MHz, ratio = 14.0/7.5 = 1.867; needed n = 5.000 / (2 × 0.2711) = 9.2 → n=10
The difficulty with a Butterworth at ratio <2 is clear. The solution in practice is to set fc to get ratio ≥ 2 between the second harmonic and the cutoff. For a 40m filter (operating to 7.3 MHz), set fc = 9.0 MHz so the second harmonic at 14.6 MHz falls at ratio 14.6/9.0 = 1.62 — still only moderate improvement.
For genuinely challenging cases like 40m (second harmonic close to 20m band), a Chebyshev filter is recommended over Butterworth. The Butterworth shines when the ratio fc to fstopband is larger — for example, a VHF harmonic filter where you need to suppress 288 MHz (second harmonic of 144 MHz) while passing 144 MHz with fc = 180 MHz gives ratio 288/180 = 1.6, still moderate, but at UHF it's impractical anyway.
Better application: 160m LPF — fc = 2.5 MHz, second harmonic at 3.5 MHz, ratio = 3.5/2.5 = 1.4. Or more practically, a 10m LPF with fc = 35 MHz and harmonics at 70 MHz (ratio = 2.0).
Practical design: 10m transmitter LPF, 5th-order Butterworth
Specification: fc = 35 MHz, R = 50 Ω, 5th order, Butterworth
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 × 35×10⁶)
= 30.90 / (219.9×10⁶) = 140.5 nH
L₃ = g₃ × R / (2π × fc) = 2.0000 × 50 / (2π × 35×10⁶)
= 100.0 / (219.9×10⁶) = 454.7 nH ≈ 455 nH
Calculating capacitors (even elements: g₂, g₄):
C₂ = C₄ = g₂ / (R × 2π × fc) = 1.6180 / (50 × 2π × 35×10⁶)
= 1.6180 / (10.996×10⁹) = 147.1 pF
Final component values:
- L₁ = L₅ ≈ 140 nH (series inductors at input and output)
- C₂ = C₄ ≈ 147 pF (shunt capacitors to ground)
- L₃ ≈ 455 nH (series inductor at center)
Verifying attenuation at 70 MHz (second harmonic of 10m):
Ratio = 70/35 = 2.0; Attenuation = 10 × log₁₀(1 + 2¹⁰) = 10 × log₁₀(1025) = 30.1 dB
Hmm — only 30 dB at the second harmonic. For 43 dB from the FCC, we need to increase the order or lower fc. A 7th-order Butterworth with same fc = 35 MHz:
Attenuation at 70 MHz = 10 × log₁₀(1 + 2¹⁴) = 10 × log₁₀(16385) = 42.1 dB — just barely sufficient.
Or lower fc to 28 MHz (just above the 10m band): second harmonic at 60 MHz, ratio = 60/28 = 2.14:
5th-order: Attenuation = 10 × log₁₀(1 + 2.14¹⁰) = 10 × log₁₀(1+1667) = 32.2 dB — still insufficient for 5th order.
This demonstrates why Chebyshev is generally preferred for transmitter harmonic filters.
The worked example shows an important real-world lesson: Butterworth filters provide clean, flat passbands but may require higher order than a Chebyshev to meet the same harmonic specification. For transmitter harmonic suppression where the second harmonic is less than 3:1 away from the cutoff frequency (as is almost always the case in HF), Chebyshev is more component-efficient. Butterworth shines when stopband frequencies are well above the cutoff or when passband flatness is the primary concern.
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 VHF receiver from HF interference: it would attenuate signals below 100 MHz (where the powerful AM broadcast and HF shortwave transmitters live) while passing 144 MHz and above with only 3 dB of insertion loss at the cutoff and virtually none at 145 MHz and above.
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 HF transmitter low-pass filters (below 30 MHz), inductors are typically wound on powdered-iron toroid cores. T68-2 (red/clear) cores are popular for 1–30 MHz with moderate power; 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 (or 0.57 µH for 10 turns, 57 nH for 10 turns — each turn contributes about 0.57 nH for this core size). Use the formula: N = 100 × √(L / AL) where L is in µH and AL is in µH/100-turns.
Wind inductors with the tightest practical winding, covering 60–80% of the core circumference, and secure the windings with a drop of Q-dope or RTV silicone. Use enameled copper wire sized for the transmitter power level: for 100 W into 50 Ω, peak current is approximately √(2 × 100/50) = 2 A, so 24 AWG is adequate; for 500 W use 20 AWG or heavier.
Capacitor Selection
Transmitter harmonic filters must use high-Q capacitors that can handle the voltages involved. Silver-mica capacitors are the traditional choice for values under 1000 pF at HF: they have low series resistance, stable capacitance over temperature, and excellent self-resonance characteristics. For values above 1000 pF or for very high power (above 500 W), NP0 ceramic capacitors (labeled C0G) offer similar performance and higher voltage ratings at lower cost. Avoid X7R and Z5U ceramics, which have high dissipation factors at RF and temperature-dependent capacitance that shifts the filter response with operating temperature.
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.
- 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
- 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.
- Build the ladder: series L₁ — shunt C₂ to ground — series L₃ — load resistor R = 1 kΩ to ground.
- 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.
- Calculate the attenuation at each frequency in dB: Attenuation = 20 × log₁₀(Vout/Vin).
- 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).
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 harmonic-filter applications. The cutoff frequency will shift slightly, and the passband flatness may deviate slightly from the ideal Butterworth response, but the harmonic attenuation at a frequency 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 transmitter harmonic filters, component tolerances of ±5% are acceptable. For precision IF filtering, 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. Every answer can be found in the lesson above.