All-pole IIR filters

October 2021: I originally wrote this for a different website. I have adapted the article to this blog.

Part 1 of this article describes generalized formulas for any 2-pole polynomial, no-zero, lowpass or highpass, infinite impulse response (IIR) filter. We then extend the 2-pole filter to a generalization for any even-order all-pole polynomial filter. For those who want answers without effort, Part 2 of this article shows recipes for constructing some traditional lowpass and highpass IIR filters, as well as tests of these filters.

I don't show the mathematics I went through in going from one step to the next. The intent here is to provide formulas for reference without having to know the details of how to derive them.

Part 1. Generalized calculations for any even-pole polynomial lowpass/highpass filter, having no zeros

The filters described in this article are all-pole filters with no zeros. That is, an all-pole filter has a frequency response function that goes infinite (poles) at specific frequencies, but there are no frequencies where the response function is zero. Basically the filter function (also called the transfer function) is a ratio with a constant in the numerator and a polynomial in the denominator. The poles (roots of the polynomial) are located in the negative frequency space, so the actual frequency response (the positive space) is well behaved.

Here we show how to calculate the IIR filter coefficients for the family of filters having two poles (i.e. a second order polynomial in the denominator of the transfer function). Next, we demonstrate how to break down a 4th-order polynomial filter into two different 2-pole stages. This is important if, for example, you want a 4-pole filter of a particular type; for example the properties of a 4-pole Bessel cannot be realized simply by stacking two 2-pole Bessels.

Generalized 2-pole lowpass filter

First, define the frequency response of the filter as an inverted 2nd-degree polynomial function $H(s)$. This is called the Transfer Function in the $s$ plane, where $s$ is the imaginary axis. The amplitude response versus frequency is a function along this axis, analogous to the frequency ranging from −∞ to +∞, with DC at the origin. A generalized 2-pole lowpass transfer function has the form:

$$H(s) = \frac{g}{s^2+ps+g}$$

where $g$ is a gain constant causing $H(0)=1$, and $p$ is a polynomial coefficient. If you have a coefficient on $s^2$, you can divide the numerator and denominator by that coefficient and absorb it into $p$ and $g$.

For example, for the specific filters shown in Part 2 below, the transfer functions are:

Butterworth: $H(s)=\frac{1}{s^2+\sqrt{2}s+1}$
Critically-damped: $H(s)=\frac{1}{s^2+2s+1}$
Bessel: $H(s)=\frac{3}{s^2+3s+3}$

If you know only the complex roots $\alpha \pm j\beta$ of your 2nd-degree polynomial in the denominator (which usually means the constants $g$ and $p$ aren't nice clean integers), then use

$$p= -2\alpha$$ $$g = \alpha^2+\beta^2$$

Next, we calculate the cutoff frequency correction factor $c$. To do this, we substitute $j\omega$ for $s$ in $H(s)$, where $j=\sqrt{-1}$. Then, multiply this by its complex conjugate (i.e. substituting $-j$ for $j$ to get the squared amplitude response:

$$|H(\omega)|^2 = H(j\omega)H(-j\omega) = \frac{g^2}{\omega^4 - (2g - p^2)\omega^2 + g^2}$$

Cascading the filters $n$ times is equivalent to raising $|H(\omega)|^2$ to the $n$-th power. Because we want the 3 dB cutoff frequency (where the squared amplitude is 0.5) to be the same for any number of cascades, simply set $|H(\omega)|^{2n}=\frac{1}{2}$ and solve for $\omega$, using the quadratic formula to solve for $\omega^2$. The correction factor $c$ is simply the inverse of the square root of the result. In other words:

$$c^2 = \frac{1}{\omega^2} = \frac{2}{2g - p^2 +\sqrt{ (2g-p^2)^2 - 4g^2 (1-\sqrt[n]{2}) }}$$

Take the square root of the above to get the correction factor $c$, which we apply to the cutoff frequency $f_0$ in the next step, to preserve the 3 dB cutoff position for any number of $n$ filter passes.

Because this is an analog filter, we need to make another adjustment to the cutoff frequency, to "warp" it from the analog world to the digital world. For the rest of the calculations, we use the adjusted digital cutoff frequency $\omega_0$:

$$\omega_0 = \tan\left(\pi c \frac{f_0}{f_s}\right)$$

where $f_0$ is the analog cutoff frequency and $f_s$ is the sampling frequency.

Now we need the coefficients of the filter. To do this, we perform a bilinear transformation from the $s$ domain to the $z$ domain. This isn't as hard as it sounds; it's just algebraic manipulation. Basically it involves substituting $s = \frac{z-1}{\omega_0(z+1)}$ in $H(s)$, collecting the powers of $\frac{1}{z}$ together in the numerator and denominator, normalizing things so that the coefficient for $z^0$ in the denominator is 1, and reading off the coefficients. The $a_i$ coefficients (which are multiplied by the most recent $x_i$ inputs) are in the numerator and the $b_i$ coefficients (which are multiplied by the most recent $y_i$ outputs) appear as their negative in the denominator. After simplifying, the coefficients are:

$$a_0 = \frac{g\omega_0^2}{1+p\omega_0+g\omega_0^2}$$ $$a_1 = 2a_0$$ $$a_2 = a_0$$ $$b_1 = 2a_0\left(\frac{1}{g\omega_0^2}-1\right)$$ $$b_2 = 1-(a_0+a_1+a_2+b_1)$$

Finally, calculate the filter:

$$y_k = a_0 x_k + a_1 x_{k-1} + a_2 x_{k-2} + b_1 y_{k-1} + b_2 y_{k-2}$$

Generalized 2-pole highpass filter

To get a highpass filter, use $\omega_0 = 1/\tan\left(\frac{\pi f_0}{c f_s}\right)$ as the adjusted digital cutoff frequency (that is, invert $c$ before applying to $f_0$ and invert the tangent to get $\omega_0$), and calculate the lowpass coefficients. Then, negate the coefficients $a_1$ and $b_1$ — but be sure you calculate $b_2$ before negating those coefficients! Applying these coefficients in the final filter formula above results in a highpass filter with a 3 dB cutoff at $f_0$.

Generalized 4-pole filter

A 4th-order polynomial filter has a transfer function of the form:

$$H(s) = \frac{g}{s^4+ps^3+qs^2+rs+g}$$

Although we can perform the same bilinear transformation here as with the 2nd-order filter, and obtain a one-pass fourth-order filter, we'll run into trouble if we do this. We'd find that the filter coefficients may have magnitudes different by several orders (say, one coefficient might be 0.013 and another might be 128.2), resulting in instability due to the finite precision of the computer. Using double precision helps, but the real solution is to decompose the 4th-order filter into a cascade of two 2nd-order filters.

Usually, a 4th-order polynomial cannot be factored readily into two clean 2nd-order polynomials. However, we know that the polynomal can be rewritten in terms of its roots; i.e. the four values $s= \{R_1, R_2, R_3, R_4\}$ that cause the polynomial to equal zero. Therefore the transfer function becomes:

$$H(s) = \frac{g}{(s-R_1) (s-R_2) (s-R_3) (s-R_4)}$$

...and we know that $g$ is the product of all the roots; i.e. $g = R_1 R_2 R_3 R_4$.

We can calculate the complex roots of a 4th-order polynomial using a number of online tools. For example, Wolfram Alpha is a free web-based math solver that can solve for the roots of a 4th-order polynomial.

Let's use the 4-pole Bessel transfer function as an example. The general form of an n-pole Bessel transfer function is:

$$H(s) = \frac{ \frac{(2n)!}{2^{n} n!}} {\sum_{i=0}^n \frac{(2n-i)!}{2^{n-i} i! (n-i)! } s^n}$$

In our case, $g$ is the numerator. The 4-pole Bessel transfer function is then:

$$H(s) = \frac{105}{s^4+10s^3+45s^2+105s+105}$$

Wolfram Alpha shows that the four roots of the denominator are:

$$R_1, R_2 = \alpha_1 \pm j\beta_1 = -2.10378939717963 \pm 2.65741804185675 j$$ $$R_3, R_4 = \alpha_2 \pm j\beta_2 = -2.89621060282037 \pm 0.86723412893450 j$$

Setting $g_1 = R_1 R_2 = \alpha_1^2 + \beta_1^2$ and $g_2 = R_3 R_4 = \alpha_2^2 + \beta_2^2$ and combining the corresponding root pairs in the denominator of $H(s)$, we get a product of two second-order transfer functions:

$$H(s) = \frac{g_1}{s^2 - 2\alpha_1 s + g_1} \cdot \frac{g_2}{s^2 - 2\alpha_2 s + g_2}$$

We set $p = -2\alpha$ and solve for the coefficients of each 2-pole filter stage separately, as described in the previous section.

Our remaining problem involves the frequency correction factor $c$, which must be the same for each stage. This must be calculated, as with the 2-pole filter, by solving for $\omega$ in $H{j\omega) H(-j\omega) = \frac{1}{2}. This is a quartic function of \(\omega^2$, and unlike quadratics, quartics are non-trivial to solve. Wolfram Alpha can solve them. For a 4-pole Bessel lowpass filter, the correction factor $c = \frac{1}{\omega} = \frac{1}{2.1139176749042}$.

After all this, the 4-pole filter is then realized by feeding the input data into the first stage, and feeding that output into the second stage.

Note that the two second-order filter stages making the whole 4th-order filter are different. That's why cascading two identical 2-pole filters doesn't result in a 4-pole filter of the same type. That is, if you want a fourth-order Bessel, you can't start out by cascading two second-order Bessels; you must instead decompose the fourth-order Bessel into two different second-order filters (which aren't themselves Bessels).

Any even-order filter can be decomposed into a cascade of 2-pole filters in the same manner described above, although for higher-order filters you likely must resort to numerical approximation techniques to solve for the polynomial coefficients and for the frequency correction factor.

Part 2. Recipes for 2-Pole IIR filters

You can download an Excel spreadsheet (43 kilobytes, right-click and "save as") that demonstrates the generalized 2-pole polynomial filter. You can play with the g and p polynomial coefficients to see how the filter reacts to an input signal consisting of a square wave impulse followed by Gaussian noise.

Calculation steps

Definitions:

$n$ = number of filter passes (cascaded filters)
$\lambda_0$ = 3 dB cutoff wavelength
$f_0\ = 1 / \lambda_0$ = 3 dB cutoff frequency
$f_s$ = sample frequency
$f_c$ = corrected cutoff frequency
$\omega_0$ = digital domain cutoff frequency
$x_k$ = input value at index $k$
$y_k$ = filter output value at index $k$

Filter type	Low Pass			High Pass
Filter type	Butter-worth	Critically damped	Bessel	Butter-worth	Critically damped	Bessel
3 dB cutoff correction $c$	$$\frac{1}{ \sqrt[4]{\sqrt[n]{2}-1} }$$	$$\frac{1}{\sqrt{\sqrt[2n]{2}-1}} $$	$$\frac{1}{\sqrt{3}\sqrt{ \sqrt{\sqrt[n]{2}-\frac{3}{4}}-\frac{1}{2}} } $$	$$\sqrt[4]{\sqrt[n]{2}-1}$$	$$\sqrt{\sqrt[2n]{2}-1}$$	$$\sqrt{3}\sqrt{ \sqrt{\sqrt[n]{2}-\frac{3}{4}}-\frac{1}{2}}$$
Polynomial coefficients	$$g=1$$ $$p=\sqrt{2}$$	$$g=1$$ $$p=2$$	$$g=3$$ $$p=3$$	$$g=1$$ $$p=\sqrt{2}$$	$$g=1$$ $$p=2$$	$$g=3$$ $$p=3$$
Corrected cutoff frequency $f_c$	$$f_c = c \frac{f_0}{f_s}$$			$$f_c = \frac{1}{2} - c \frac{f_0}{f_s}$$
Ensure $f_c$ OK for stability constraint	$$0 \lt f_c \lt \frac{1}{8}$$ (unstable when $f_c \ge \frac{1}{4}$)			$$\frac{3}{8} \lt f_c \lt \frac{1}{2}$$ (unstable when $f_c \le \frac{1}{4}$)
Warp cutoff freq from analog to digital domain	$$\omega_0 = \tan \pi f_c $$
Filter coefficient calculations	$$K_1 = p \omega_0$$ $$K_2 = g \omega_0^2$$ $$A_0 = \frac{K_2}{1+K_1+K_2}$$ $$A_1 = 2 A_0$$ $$A_2 = A_0$$ $$B_1 = 2 A_0 \left(\frac{1}{K_2}-1\right)$$ $$B_2 = 1 - (A_0+A_1+A_2+B_1)$$
Filter coefficients	$$a_0 = A_0$$ $$a_1 = A_1$$ $$a_2 = A_2$$ $$b_1 = B_1$$ $$b_2 = B_2$$			$$a_0 = A_0$$ $$a_1 = -A_1$$ $$a_2 = A_2$$ $$b_1 = -B_1$$ $$b_2 = B_2$$
Filter (1 pass)	$$y_k = a_0 x_k + a_1 x_{k-1} + a_2 x_{k-2} + b_1 y_{k-1} + b_2 y_{k-2}$$

Notes on the three filters

As shown in the table above, a highpass filter is constructed from a lowpass filter by inverting $c$ and $\omega_0$, and negating the lowpass coefficients $a_1$ and $b_1$. Note that the identity $\frac{1}{\tan \pi x} = \tan[\pi(\frac{1}{2}-x)]$ is used above to accomplish the inversion of $\omega_0$ for the highpass filter.
The Bessel has the most linear phase response, with a stopband frequency response midway between the Butterworth and critically-damped filters. See the figure below.
Cascading n 2-pole Butterworth filters does not yield a 2n-pole Butterworth filter. The same is true for Bessel. Cascading two Butterworths, for example, results in what is known as a "Linkwitz-Riley" filter.
On the other hand, cascading critically-damped filters results in another critically-damped filter.
You can cancel the overshoot of a Butterworth with the lag of the critically-damped filter to obtain another critically-damped filter that has the positive features of both. For example, a Butterworth of cutoff frequency $2f_0$ cascaded with a two critically-damped filters of cutoff $f_0$ results in a critically-damped filter that has less lag than the original critically-damped filter cascaded three times, yet greater stopband attenuation (steeper falloff). Similarly, an arithmetic average of a 1-pass $2f_0$ Butterworth and a 2-pass critically-damped filter of cutoff $f_0$ results in a critically-damped filter having a frequency response similar to the 1-pass Butterworth (i.e. better than a 1-pass critical), and with a lag in between the Butterworth and the 2-pass critically-damped.

Tests

Frequency Response

The following frequency response curves show the performance of each filter above, for different combinations of cutoff frequency and number of filters cascaded together (passes). These plots also demonstrate that the cutoff frequency correction $c$ above does indeed cause the 3 dB cutoff frequency to occur in the same place for each filter, regardless of the number of passes.

The frequency responses shown resulted from using a chirp as the input signal (i.e., a sinewave that increases in frequency over time). The chirp contained 32,000 data points and covered over 5 octaves, with the wavelength $1/f$ decreasing linearly to allow each frequency to have nearly the same number of cycles approximating that frequency. The plotted curves are maximum squared-amplitude output levels, updated each cycle. The curves look noisy at the short wavelengths (high frequencies) due to the chirp wavelength nearing the Nyquist limit of 2, resulting in inconsistent wave amplitudes in each cycle. The sampling frequency $f_s = 1$ in all cases.

Temporal response to a step function

The following time-domain responses to a step function show the performance of each filter, for the same cutoff frequency (1/80 in this case), plotted on the same time scale, for different number of passes. We make the following observations:

All high-pass filters oscillate, even when critically damped. For a step response one can generally expect a critically-damped filter to exhibit one less zero-crossing than the number of poles in the filter (and the more one cascades filters, the more poles one has). Underdamped filters like the Butterworth oscillate more.
The critically-damped filter has the fastest rise time, yet the Bessel appears to converge faster.
The Bessel actually has a microscopic amount of overshoot, so that it appears to settle faster than the critically-damped filter; however, the amplitude of the Bessel's oscillations in the highpass case are higher than the critically-damped filter.
Although the Butterworth filter has the flattest passband response, sharpest rolloff, and steepest falloff of all the filters shown, in the time domain it suffers from overshoot. The lowpass Butterworth also exhibits the most lag.

[Highpass 1-pass Temporal Step Response]

[Highpass 3-pass Temporal Step Response]

References

Young-Hoo Kwon, Butterworth Digital Filters
D. Gordon E. Robertson and James J. Dowling, "Design and responses of Butterworth and critically damped digital filters," Journal of Electromyography and Kinesiology 13 (2003) pp 569-573.
S.D. Murphy and D.G.E. Robertson, "Construction of a High-pass Digital Filter," presentation to the North American Conference on Biomechanics II (CSB VII), Chicago, USA, 1992.
Peter Nachtwey, Bessel High Pass
R.W. Erickson, Filter Circuits, 10 March 1999.
C.R. Bond, Filter Algorithms, Parameters and Software for Electronic Circuits
Usenet thread in comp.dsp, "Wanted: critically-damped high-pass IIR filter."

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