Butterworth Filter: What is it? (Design & Applications)

Contents

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Key learnings:

Butterworth Filter Definition: A Butterworth filter is defined as a signal processing filter designed for a maximally flat frequency response in the passband.
Design Complexity: Designing a Butterworth filter involves selecting the appropriate filter order and components, usually capacitors, to achieve the desired flatness in frequency response.
First-order Filters: First-order Butterworth filters attenuate frequencies above a certain cutoff, rolling off at -20 dB per decade.
Applications: Butterworth filters are crucial in audio processing and communications to ensure clear signal transmission without distortion.
Comparison with Chebyshev: Unlike Chebyshev filters, Butterworth filters provide a smoother frequency response without ripples, making them ideal for audio applications.

What is a Butterworth Filter?

A Butterworth filter has a magnitude response that is maximally flat at the centre of its passband. Engineers therefore call it a maximally flat magnitude filter. British physicist Stephen Butterworth described the response in his 1930 paper On the Theory of Filter Amplifiers.

The low-pass Butterworth magnitude decreases monotonically from the passband into the stopband, without ripple. “Passband” does not mean bandpass filter; low-pass, high-pass, band-pass and band-stop filters all have passbands. An nth-order low-pass response approaches a stopband slope of -20n dB per decade.

Butterworth describes a response, not one circuit. A passive implementation may use an inductor and capacitor ; an active RC design uses op-amps, resistors and capacitors; a digital filter uses coefficients. Filter order is the degree of the transfer-function denominator, not simply the capacitor count.

This article focuses on a low-pass active RC implementation. A first-order high pass filter can swap the series and shunt roles of resistance and capacitance, but higher-order transformations require the correct topology and component equations.

Butterworth Low Pass Filter Design

An ideal brick-wall low-pass passes all frequencies below its edge and rejects all frequencies above it. No finite-order causal circuit has that discontinuous response. A Butterworth approximation trades transition width for a smooth, ripple-free magnitude.

Higher order gives a steeper transition but needs more poles and greater control of component tolerances, noise, op-amp bandwidth and dynamic range. A correctly designed Butterworth response has no passband ripple; excessive ripple indicates implementation error, component variation or a different approximation.

For the standard low-pass normalisation, magnitude is down 3.0103 dB at cutoff frequency. The response is maximally flat near DC, not exactly constant all the way to cutoff. A first-order response approaches -20 dB per decade above the transition.

Each additional pole adds 20 dB per decade to the asymptotic attenuation, so a second-order low-pass approaches -40 dB per decade. A second-order Butterworth section has quality factor Q = 1/√2, about 0.707. Higher-order filters require different Q values for their individual second-order sections.

The below figure shows the frequency response of the Butterworth filter for various orders of the filter.

Frequency Response of Butterworth Filter

The following magnitude form describes an nth-order Butterworth low-pass:

$H(j\omega) = \frac{1}{\sqrt{1+\varepsilon^2(\frac{\omega}{\omega_C})^{2n}}}$

Where:
n = filter order,
ω = evaluation angular frequency
ω_C = reference cutoff angular frequency
ε = a scale or edge-loss parameter, not passband gain A_max

The next relation links ε to the selected edge magnitude. In the usual -3.0103 dB Butterworth normalisation, ε equals 1.

$H_1 = \frac{H_0}{\sqrt{1+\varepsilon^2}}$

Where:
H₁ = magnitude at the specified passband edge
H₀ = DC or passband reference magnitude

First-order Lowpass Butterworth Filter

A low-pass filter passes lower-frequency components with limited attenuation and increasingly attenuates components above its transition region.

A first-order transfer function has one pole. The active circuit below uses one RC pole followed by a non-inverting op-amp stage.

This example is an active Low pass filter because it contains an op-amp. The non-inverting gain is set by feedback resistor R₁ and R_F, while R and C set the ideal pole frequency. The op-amp must have adequate gain-bandwidth, slew rate, input range and output drive.

Apply the complex-impedance voltage divider rule at Va to find the capacitor voltage:

$V_a = \frac{-jX_C}{R-jX_C} V_{in}$

$V_a = \frac{-j(\frac{1}{2\pi f C})}{R-j (\frac{1}{2\pi f C})} V_{in}$

$V_a = \frac{-j}{2\pi fRC - j} V_{in}$

$V_a = \frac{V_{in}}{1-\frac{2 \pi fRC}{j}}$

$V_a = \frac{V_{in}}{1+j2 \pi fRC}$

For an ideal non-inverting op-amp stage:

$V_0 = \left( 1+ \frac{R_f}{R_1} \right) V_a$

$V_0 = \left( 1+ \frac{R_f}{R_1} \right) \frac{V_{in}}{1+j2 \pi fRC}$

$\frac{V_0}{V_a} = \frac{A_f}{1+j\frac{f}{f_c}}$

Where,

$A_f = 1 + \frac{R_F}{R_1}$

A_f = Gain of filter in Passband

$f_c = \frac{1}{2 \pi RC}$

f_c = Cutoff Frequency
f = Operating Frequency

$\frac{V_0}{V_a} = \left|\frac{V_0}{V_a} \right|\angle \phi$

$\left|\frac{V_0}{V_a} \right| = \frac{A_f}{\sqrt{1+j \left( \frac{f}{f_c} \right) ^2}}$

$\phi = - \tan^{-1} \left( \frac{f}{f_c} \right)$

At very low frequency, f<<f_c
$\left|\frac{V_0}{V_a} \right| \approx A_f (Constant)$
At cutoff frequency, f= f_c
$\left|\frac{V_0}{V_a} \right| = \frac{A_f}{\sqrt{2}} = 0.707A_f$
At high frequency, f> f_c

$\left|\frac{V_0}{V_a} \right| < A_f$

The figure shows the first-order low-pass magnitude response. Two protected equations above contain notation errors. The transfer-function left side should reference input rather than Va. The magnitude denominator should be a real square root without j.

Frequency Response of First-order Low Pass Butterworth Filter

Second-order Butterworth Filter

A second-order response has two poles. The figure shows one Sallen-Key-style active low-pass implementation with two resistors and two capacitors.

Second-order Low Pass Butterworth Filter

In this circuit, resistor R and R_F set non-inverting op-amp gain. Components R₂, R₃, C₂ and C₃ determine pole frequency and Q together, so tolerance and loading affect both.

The two RC networks interact through the feedback path; they are not independent back-to-back first-order filters. R_L is the load resistance, which should be high enough not to disturb the designed response.

Higher-order Butterworth filters can be built by cascading first- and second-order sections. Each section must use the pole frequency and Q assigned by the nth-order Butterworth pole set.

The following derivation analyses the illustrated second-order circuit.

Apply Kirchhoff’s current law at node V₁:

$I_1 = I_2 + I_3$

(1) $\begin{equation*} \frac{V_{in}-V_1}{R_2} = \frac{V_1-V_0}{\frac{1}{sC_2}}+\frac{V_1-V_a}{R_3} \end{equation*}$

Use the potential-divider relation at node V_a:

$V_a = V_1\left[ \frac{\frac{1}{sC_3}}{R_3+\frac{1}{sC_3}} \right]$

$V_a = V_1\left[ \frac{\frac{1}{sC_3}}{\frac{R_3{sC_3} +{1}}{sC_3}} \right]$

$V_a = \frac{V_1}{1+sR_3C_3}$

$V_1 = V_a (1+sR_3C_3)$

Substitute V₁ into equation 1. The protected first expanded line has 1/R3 in a denominator where the preceding KCL relation uses R3; verify this derivation independently before using component values.

$\frac{V_{in}-V_a (1+sR_3C_3)}{R_2} = \frac{V_a (1+sR_3C_3)-V_0}{\frac{1}{sC_2}} + \frac{V_a (1+sR_3C_3)-V_a}{\frac{1}{R_3}}$

$\frac{V_{in}}{R_2} -\frac{V_a (1+sR_3C_3)}{R_2} = \frac{V_a (1+sR_3C_3)}{\frac{1}{sC_2}} -\frac{V_0}{\frac{1}{sC_2}} + \frac{V_a (1+sR_3C_3)}{R_3}-\frac{V_a}{R_3}$

$\frac{V_{in}}{R_2} + \frac{V_0}{\frac{1}{sC_2}} = \frac{V_a (1+sR_3C_3)}{\frac{1}{sC_2}} + \frac{V_a (1+sR_3C_3)}{R_2} + \frac{V_a (1+sR_3C_3)}{R_3} -\frac{V_a}{R_3}$

$\frac{V_{in}}{R_2} +V_0 sC_2 = V_a\left[sC_2(1+sR_3C_3) + \frac{(1+sR_3C_3)}{R_2} + \frac{(1+sR_3C_3)}{R_3} - \frac{1}{R_3} \right]$

$\frac{V_{in} +V_0 sC_2 R_2}{R_2} = V_a\left[ \frac{ R_3 R_2 sC_2(1+sR_3C_3) + R_3(1+sR_3C_3) + R_2(1+sR_3C_3) - R_2}{R_2R_3} \right]$

$R_3(V_{in} +V_0 sC_2 R_2) = V_a\left[ R_3 R_2 sC_2(1+sR_3C_3) + R_3(1+sR_3C_3) + R_2(1+sR_3C_3) - R_2\right]$

$R_3 V_{in} +V_0 sC_2 R_2 R_3 = V_a\left[ (1+sR_3C_3) \left( R_3 R_2 sC_2 + R_3 + R_2 \right) - R_2\right]$

$V_a = \frac{ R_3 V_{in} +V_0 sC_2 R_2 R_3 }{ (1+sR_3C_3) \left( R_3 R_2 sC_2 + R_3 + R_2 \right) - R_2 }$

For the ideal non-inverting op-amp:

$V_0 = A_f V_a$

Where,

$A_f = 1+\frac{R_f}{R_1} = Gain \, of \, filter \, in \, passband$