The Butterworth filter is the gold standard when you need a maximally flat passband with zero ripple. Whether you are designing an active Sallen-Key audio crossover or a passive LC anti-aliasing stage for an ADC, the math dictates your component tolerances and op-amp bandwidth requirements. Below is the complete mathematical framework, normalized hardware tables, and step-by-step derivations to take a Butterworth design from theory to the bench.

The Core Butterworth Filter Formula & Symbol Definitions

The defining characteristic of a Butterworth filter is that its magnitude response is as flat as mathematically possible in the passband. The squared magnitude response for a low-pass Butterworth filter is given by:

|H(jω)|2 = 1 / [1 + (ω / ωc)2n]

In practical bench work, we rarely use linear voltage ratios. We use decibels (dB) to express attenuation. The attenuation formula (expressed as a positive dB loss) is:

AdB = 10 · log10[1 + (f / fc)2n]

Table 1: Butterworth Filter Formula Symbol Definitions
Symbol Parameter Unit Practical Context
|H(jω)| Magnitude Response V/V (Linear) Ratio of output voltage to input voltage at frequency ω.
AdB Attenuation dB Signal loss relative to the DC passband gain. Always positive in this formula.
ω Angular Frequency rad/s ω = 2πf. Used in transfer function derivations.
f Target Frequency Hz The specific frequency you are evaluating for attenuation.
ωc Cutoff Angular Frequency rad/s The -3 dB point in radians per second (ωc = 2πfc).
fc Cutoff Frequency Hz The -3.01 dB corner frequency. Defines the passband edge.
n Filter Order Dimensionless Determines the roll-off rate (n × 20 dB/decade). Must be an integer.

Normalized Prototype Values for Hardware Implementation

Knowing the attenuation formula is only half the battle. To actually build a passive LC Butterworth filter, you need the Butterworth polynomials. Rather than solving complex roots on the bench, hardware engineers use normalized prototype values (g-values). These assume a source resistance of 1 Ω and a cutoff frequency of 1 rad/s.

According to standard filter design theory detailed by All About Circuits, you scale these normalized values to your actual system impedance (R0) and desired cutoff frequency (fc) using the following hardware scaling laws:

  • Actual Inductor: Lk = (R0 · gk) / (2π · fc)
  • Actual Capacitor: Ck = gk / (2π · fc · R0)
Table 2: Normalized Low-Pass Butterworth Prototype Elements (gk)
Order (n) g1 g2 g3 g4 g5
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

Bench Tip: For a 5th-order filter with a 50 Ω source and a 10 MHz cutoff, g1 (0.6180) scales to an inductor of L1 = (50 · 0.6180) / (2π · 107) = 491.8 nH. Always use air-core or high-Q powdered iron cores at RF; standard ferrites will introduce parasitic resistance that destroys the Butterworth flatness.

Rearranged Forms for Design Variables

When designing active filters (like cascaded Sallen-Key stages), you usually start with frequency and attenuation requirements and work backward to find the required order and cutoff. Here are the algebraically rearranged forms of the core attenuation formula.

1. Solving for Filter Order (n)

Given a passband frequency (fp) with maximum allowable loss (Ap), and a stopband frequency (fs) requiring minimum attenuation (As):

n ≥ log10[ (10A_s/10 - 1) / (10A_p/10 - 1) ] / [ 2 · log10(fs / fp) ]

Note: Because physical filter stages must be integers, always round the calculated n UP to the next whole number.

2. Solving for Cutoff Frequency (fc)

If you know the desired attenuation (AdB) at a specific frequency (f) and your chosen order (n):

fc = f / (10A_dB/10 - 1)1 / (2n)

3. Solving for Target Frequency (f)

To find the exact frequency where a specific attenuation occurs, given fc and n:

f = fc · (10A_dB/10 - 1)1 / (2n)

Worked Examples with Unit Tracking

Abstract formulas cause mistakes on the bench. Let us track the units and intermediate steps through two common design scenarios.

Problem 1: Determining Required Filter Order for an ADC Anti-Aliasing Stage

Scenario: You are designing an anti-aliasing filter for a 16-bit ADC sampling at 20 kSPS. Your passband extends to fp = 2 kHz, where you can tolerate a maximum droop of Ap = 0.5 dB. At the Nyquist frequency (fs = 10 kHz), you need at least As = 50 dB of attenuation to prevent aliasing. What order Butterworth filter is required?

  1. Identify Variables: fp = 2000 Hz, Ap = 0.5 dB, fs = 10000 Hz, As = 50 dB.
  2. Calculate Stopband Linear Power Ratio: 1050/10 - 1 = 105 - 1 = 99,999.
  3. Calculate Passband Linear Power Ratio: 100.5/10 - 1 = 100.05 - 1 ≈ 1.1220 - 1 = 0.1220.
  4. Calculate the Ratio of Ratios: 99,999 / 0.1220 ≈ 819,663.9.
  5. Apply the Logarithm (Numerator): log10(819,663.9) ≈ 5.9136.
  6. Calculate Frequency Ratio (Denominator base): fs / fp = 10000 / 2000 = 5.
  7. Apply the Logarithm (Denominator): 2 · log10(5) ≈ 2 · 0.69897 = 1.3979.
  8. Divide to find n: n ≥ 5.9136 / 1.3979 ≈ 4.23.

Result: Since n must be an integer, you must round up to n = 5. A 5th-order Butterworth filter (implemented as two cascaded Sallen-Key stages and one first-order RC stage) is required.

Problem 2: Calculating Actual Attenuation at a Specific Interference Frequency

Scenario: You have built a 4th-order (n = 4) active Butterworth low-pass filter with a -3 dB cutoff frequency of fc = 1.5 kHz. A known switching noise spur exists on your board at f = 8 kHz. How much attenuation (in dB) will the filter provide at this exact spur frequency?

  1. Identify Variables: n = 4, fc = 1500 Hz, f = 8000 Hz.
  2. Calculate Frequency Ratio: f / fc = 8000 / 1500 = 5.3333.
  3. Raise to the Power of 2n: (5.3333)2·4 = (5.3333)8.
  4. Compute the Exponent: 5.33338 ≈ 643,392.5.
  5. Add 1 (from the formula denominator): 1 + 643,392.5 = 643,393.5.
  6. Apply the Logarithm and Scale: AdB = 10 · log10(643,393.5).
  7. Final Calculation: 10 · 5.80847 ≈ 58.08 dB.

Result: The filter will attenuate the 8 kHz switching spur by 58.08 dB. (Note: A 4th-order filter rolls off at 80 dB/decade. The distance from 1.5 kHz to 8 kHz is roughly 0.72 decades, yielding an approximation of 0.72 × 80 = 57.6 dB, which closely matches our exact calculation).

Assumptions, Unit Traps, and Realistic Magnitudes

Formulas assume ideal components. The real world introduces parasitics that will break your Butterworth response if you are not careful. As noted in Electronics Tutorials, understanding the boundaries of the math is just as critical as the math itself.

Core Assumptions of the Formula

  • Maximally Flat Passband: The formula assumes zero ripple in the passband. If your application can tolerate 0.5 dB of ripple, a Chebyshev Type I filter will give you a much steeper transition band for the same order n.
  • Monotonic Roll-off: The attenuation increases smoothly and continuously. There are no transmission zeros (notches) in the stopband, unlike elliptic (Cauer) filters.
  • Ideal Source and Load: The prototype g-values assume perfect impedance matching. In active filters, the op-amp output impedance must be near zero, and the input impedance of the next stage must be infinitely high.

Unit Mistakes That Break the Math

⚠️ The Radians vs. Hertz Trap:
The transfer function H(jω) strictly uses radians per second (ω). The attenuation formula AdB uses Hertz (f). If you plug ω = 10,000 rad/s directly into the Hertz formula without dividing by 2π, your calculated attenuation will be wildly incorrect. Always verify whether your variable is f (Hz) or ω (rad/s) before hitting 'equals' on the calculator.
⚠️ The Linear vs. Decibel Trap:
The order calculation formula requires Ap and As to be entered as positive dB values representing loss. If your passband gain is specified as -0.5 dB, you must use Ap = 0.5 in the formula. Inserting negative numbers into the 10A/10 term will result in a fractional power ratio less than zero, breaking the logarithm.

Realistic Magnitudes and Bench Limits

What does a 'normal' answer look like? In theoretical textbooks, you might see n = 12 or n = 15. On a physical PCB, an order greater than n = 8 is exceptionally rare for passive LC filters and n = 6 is the practical ceiling for active op-amp filters.

Why? Component sensitivity. In a 6th-order Sallen-Key Butterworth filter, the required Q-factor for the final stage approaches 5. This means your capacitor ratio must be exact. If you use standard X7R ceramic capacitors, their capacitance drops by 20% to 50% under DC bias voltage, and their dielectric absorption introduces phase errors. This shifts the pole locations, destroying the maximally flat Butterworth response and turning it into a lumpy, peaking mess.

The Hardware Fix: For any active Butterworth filter above n = 2, you must use C0G/NP0 dielectric capacitors and 1% tolerance metal film resistors. Furthermore, ensure your op-amp's Gain-Bandwidth Product (GBW) is at least 100× higher than the filter's cutoff frequency (fc). If you are building a 50 kHz Butterworth filter, a standard LM358 (GBW ≈ 1 MHz) will introduce severe phase shift and amplitude droop near the corner frequency. Upgrade to an OPA2134 or AD8605 to maintain the mathematical ideal.