BMEN90033 · Week 8
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BMEN90033 · WEEK 8 · ACTIVE FILTER FAMILIES

From filter shape to pole and zero placement

A buffered cascade of first-order stages can only place poles on the negative real axis. Active topologies, by contrast, allow each pole pair to sit anywhere in the left half plane, and zeros to be introduced on the imaginary axis. The four classical lowpass families differ only in where they place these singularities. This page builds each placement up from its geometric reason: the pole circle of the Butterworth, the pole ellipse of the Chebyshev Type I, the imaginary-axis zeros of the Chebyshev Type II, and the combined pole-zero pattern of the elliptic.

butterworth chebyshev elliptic s-plane geometry
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01from cascade to family

Why pole placement defines a filter family

A cascade of $N$ identical buffered first-order stages has $N$ repeated real poles at $s = -1/RC$. The magnitude response is

$$|H(j\omega)| = \frac{1}{(1 + (\omega/\omega_c)^2)^{N/2}},$$

which has slope $-20N\,\text{dB/decade}$ at high frequency but a soft transition near $\omega_c$. The response only reaches its asymptotic slope well into the stopband. The passband edge is gradual because every pole sits on the same point on the negative real axis: there is no resonance to keep the magnitude up near the edge.

Once an op-amp synthesises complex pole pairs, each pair has two degrees of freedom: the distance from the origin sets the natural frequency, and the angle from the negative real axis sets the damping. A pole close to the imaginary axis produces a nearly resonant response that lifts the magnitude near that frequency. Choosing how to spend these degrees of freedom across $N$ poles, and whether to add zeros on the imaginary axis, defines a filter family.

The four classical lowpass families

The prototypes are shown on a normalised frequency scale. Butterworth, Chebyshev Type I, and elliptic responses use $\omega_p = 1\,\text{rad/s}$ as the passband edge; Chebyshev Type II is referenced to a stopband edge $\omega_s = 1\,\text{rad/s}$. In each case, the magnitude is at most unity. They differ in the optimisation criterion:

The plot to the right shows the magnitude response of each family at $N = 4$, all with the same normalised edge marker. Read the geometry from right to left: more constraint on the magnitude shape pushes the poles into more specific locations. Butterworth is the most relaxed, elliptic the most constrained.
magnitude response · N = 4 lowpass, four families
02butterworth

Butterworth: poles on a circle

The Butterworth response is defined by a single condition on the magnitude:

$$|H(j\omega)|^2 = \frac{1}{1 + (\omega/\omega_c)^{2N}}.$$

This makes the first $2N - 1$ derivatives of $|H(j\omega)|^2$ vanish at $\omega = 0$. The passband is as flat as a rational function of degree $N$ allows, hence the name "maximally flat".

Where the poles come from

The defining magnitude condition implies, on the imaginary axis,

$$H(s)\,H(-s) = \frac{1}{1 + (s/j\omega_c)^{2N}}.$$

The $2N$ roots of $1 + (s/j\omega_c)^{2N} = 0$ are evenly spaced on a circle of radius $\omega_c$ centred at the origin. Exactly $N$ roots lie in the left half plane and $N$ lie in the right half plane. A causal stable filter takes the $N$ left-half-plane roots as the poles of $H(s)$:

$$s_k = \omega_c \, \exp\!\left(j \, \frac{\pi}{2}\,\frac{2k - 1 + N}{N}\right), \quad k = 1, 2, \ldots, N.$$

The geometric reason for the circle

The condition $|H(j\omega)|^2 = 1/(1 + (\omega/\omega_c)^{2N})$ depends only on the magnitude $|s| = \omega$ along the imaginary axis. Any rotation of the pole pattern about the origin leaves the product $|s_k - j\omega|$ unchanged when summed in pairs across the conjugate symmetric set, so the only constraint is on the radius. Equal spacing distributes the $N$ poles to keep the magnitude response symmetric about $\omega_c$, and the unit-radius circle is what places every pole at the same distance from the origin. That equidistance is what makes every pole contribute equally to the rolloff, giving the smoothest possible passband edge.

A pole closer to the imaginary axis produces a sharper resonance at its imaginary frequency. With every Butterworth pole on the same circle, the closest pole to $j\omega$ for any given $\omega \le \omega_c$ is always nearby, but never too nearby. No single pole dominates, so the passband is flat rather than rippled.
order N 4 probe ω off |·| ∠· 3D view
s-plane · Butterworth poles, normalised to $\omega_c = 1$
Bode response · maximally flat magnitude
03chebyshev type i

Chebyshev Type I: poles on an ellipse

The Chebyshev Type I response trades the maximally flat passband for a sharper transition. The magnitude obeys

$$|H(j\omega)|^2 = \frac{1}{1 + \varepsilon^2 \, T_N^2(\omega/\omega_c)},$$

where $T_N$ is the Chebyshev polynomial of the first kind. Across the passband, $T_N$ remains bounded between $-1$ and $+1$, so $|H(j\omega)|^2$ oscillates between $1$ and $1/(1 + \varepsilon^2)$. The ripple amplitude is set by $\varepsilon$.

From ripple to ellipse

Solving $H(s)H(-s) = $ constant for the poles on the locus where $T_N$ oscillates gives a closed form. Let $\mu = (1/N)\,\sinh^{-1}(1/\varepsilon)$. The $N$ left-half-plane poles are

$$\small \begin{aligned} s_k &= -\sinh\mu \, \sin\theta_k \;+\; j \cosh\mu \, \cos\theta_k, \\ \theta_k &= (2k - 1)\,\frac{\pi}{2N}, \qquad k = 1, \ldots, N. \end{aligned}$$

These coordinates are the parametric equation of an ellipse with semi-minor axis $\sinh\mu$ along the real axis and semi-major axis $\cosh\mu$ along the imaginary axis. As the ripple shrinks, $\varepsilon \to 0$ sends $\mu \to \infty$ and $\sinh\mu / \cosh\mu \to 1$, so the ellipse becomes asymptotically circular after the usual frequency renormalisation.

The geometric reason for the ellipse

The same Chebyshev polynomials that minimise the maximum deviation on $[-1, 1]$ over polynomials of degree $N$ have evenly spaced extrema in the angle variable. Mapping those extrema onto the imaginary axis and lifting them off through a real-axis offset $\sinh\mu$ gives the ellipse. The horizontal compression by the factor $\sinh\mu / \cosh\mu < 1$ pulls every pole closer to the imaginary axis than its Butterworth counterpart, which sharpens each pole's resonance and accelerates the rolloff. The price is paid in the passband: the closer poles produce multiple peaks where the response touches unity, separated by dips of $1/\sqrt{1 + \varepsilon^2}$.

Larger ripple means smaller $\mu$, a flatter ellipse, and poles even closer to the imaginary axis. For the same order, a Chebyshev Type I rolls off faster than a Butterworth in direct proportion to how much passband ripple is tolerated. The trade is geometric: ellipse aspect ratio sets passband ripple; Butterworth is the zero-ripple limit after renormalisation.
order N 4 ripple ε 0.35 probe ω off |·| ∠· 3D view
s-plane · Chebyshev I poles on the ellipse
Bode response · equiripple in passband
04chebyshev type ii

Chebyshev Type II: zeros on the imaginary axis

The Type II response (also called inverse Chebyshev) inverts the Type I trade. The passband is monotone, and the equiripple behaviour is moved into the stopband:

$$|H(j\omega)|^2 = \frac{1}{1 + \dfrac{1}{\varepsilon^2 \, T_N^2(\omega_s/\omega)}}.$$

Here $\omega_s$ is the stopband edge. The argument $\omega_s/\omega$ inverts the frequency variable, so the Chebyshev oscillation that was in the passband for Type I is mapped into the stopband.

Where the zeros come from

Whenever $T_N(\omega_s/\omega) = 0$, the reciprocal term diverges and the magnitude goes to zero. The zeros occur at $\omega_s/\omega = \pm \cos\theta_k$, that is

$$\omega_{z,k} = \frac{\omega_s}{\cos\theta_k}, \qquad \theta_k = (2k - 1)\,\frac{\pi}{2N}, \qquad k = 1, \ldots, \lfloor N/2 \rfloor.$$

These zeros sit on the imaginary axis above the stopband edge. Each zero punches a notch into the response: since $|j\omega - j\omega_{z,k}| = 0$ exactly at $\omega = \omega_{z,k}$, the magnitude is forced to zero there. The notches alternate with rebounds because, between adjacent zeros, the ratio of zero distances to pole distances determines the stopband ripple envelope.

Where the poles come from

The poles of a Chebyshev Type II filter are the reciprocals of a Chebyshev Type I prototype with the same $N$ and ripple parameter:

$$s_{k,\,\text{II}} = \frac{1}{s_{k,\,\text{I}}^*}\bigg|_{\text{LHP}}.$$

Geometrically, the Chebyshev I ellipse is reflected through the unit circle. Poles that were close to the imaginary axis in the Type I prototype move outward toward $|s| \to \infty$, and poles that were close to the real axis move inward toward the imaginary axis. The Type II pole pattern is the "inversion" of the Type I ellipse, which is why this filter is sometimes called the inverse Chebyshev.

The geometric reason for the placement

Notches on the imaginary axis are the only way a rational $H(s)$ can produce stopband ripple at finite frequencies. The pole reciprocation is dictated by the requirement that the passband be monotone: the product of pole distances along the passband must vary smoothly, with no peaks. Inverting the Chebyshev I ellipse achieves exactly this. The result is a flat passband, a sharper transition than Butterworth (because of the imaginary-axis zeros pulling the magnitude down), and equiripple stopband attenuation set by $\varepsilon$ (smaller $\varepsilon$ means a lower stopband ripple envelope between the notches).

Chebyshev II is the right choice when the stopband must meet a specified attenuation with controlled equiripple peaks between transmission zeros. The penalty is that the magnitude does not roll to zero monotonically: it drops sharply, rebounds, drops again, and asymptotes to a non-zero stopband floor between the highest zero and infinity for $N$ even.
order N 4 stopband ε 0.10 probe ω off |·| ∠· 3D view
s-plane · Chebyshev II poles and imaginary-axis zeros
Bode response · equiripple in stopband
05elliptic (cauer)

Elliptic: ripple in both bands

The elliptic, or Cauer, response is the magnitude that achieves the steepest possible transition for a given order, ripple, and stopband attenuation. It generalises the Chebyshev family by allowing equiripple behaviour in both bands at once:

$$|H(j\omega)|^2 = \frac{1}{1 + \varepsilon^2 \, R_N^2(\omega, \xi)},$$

where $R_N(\omega, \xi)$ is the elliptic rational function with selectivity parameter $\xi = \omega_s/\omega_p$. $R_N$ is bounded between $\pm 1$ on the passband and has an equiripple envelope in the stopband, with poles that become transmission zeros of $H(s)$. This makes both bands equiripple after the $1/(1 + \varepsilon^2 R_N^2)$ transformation.

Pole and zero placement

Two sets of singularities appear:

The closed form involves Jacobi elliptic functions:

$$\omega_{z,r} = \frac{1}{\kappa \, \mathrm{sn}\!\left(\dfrac{(2r - 1) K(\kappa^2)}{N},\, \kappa^2\right)}, \qquad r = 1, \ldots, \lfloor N/2 \rfloor,$$

where $\mathrm{sn}$ is the Jacobi sine and $K$ the complete elliptic integral, both evaluated at modulus $\kappa = 1/\xi$. The pole locations come from a related elliptic transcendental equation involving a second modulus tied to $\varepsilon$.

The geometric reason for the placement

An elliptic filter spends every degree of freedom on the transition. Each pole-zero pair locally steepens the transition: the zero on the imaginary axis at $\omega_{z,r}$ pulls the magnitude down sharply, and the matched pole just inside the passband edge maintains the passband edge against that zero. The pairs are placed so that the local minima in the passband are equal and the local maxima in the stopband are equal, by the equioscillation theorem. No other rational placement of the same number of poles and zeros achieves a steeper transition for the same ripple and floor.

Setting the stopband attenuation to infinity collapses the elliptic into a Chebyshev Type I; setting the passband ripple to zero collapses it into a Chebyshev Type II; setting both to their respective limits recovers the Butterworth. The elliptic is the parent; the other three families are special cases obtained by giving up either the passband ripple, the stopband ripple, or both.
order N 4 probe ω off |·| ∠· 3D view
s-plane · elliptic poles and zeros
Bode response · equiripple in both bands
06side by side

The four families together

Plotted at the same order on a normalised frequency scale, the four families show what each trade buys:

The s-plane plot below overlays the pole and zero patterns. Read the geometry: the Butterworth poles sit on a circle, the Chebyshev I poles on a horizontally compressed ellipse, the Chebyshev II poles on the reciprocal ellipse with zeros on the imaginary axis, and the elliptic poles inside a Chebyshev I ellipse with zeros also on the imaginary axis. Each step toward sharper transition pulls the low-frequency poles closer to the imaginary axis and, for Type II and elliptic, adds zeros that punch notches into the stopband.

The choice of family is a choice about where ripple is acceptable. Biopotential amplifiers use Bessel filters when flat group delay is the priority and Butterworth filters when a flat magnitude passband is the priority; antialiasing stages favour elliptic or Chebyshev II for sharp transitions; pulse-detection front ends sometimes favour Chebyshev I for its fast rolloff with the simplest topology. The geometry on the s-plane is a direct picture of the underlying compromise.
order N 4
s-plane · poles and zeros, four families overlaid
Bode response · magnitude and phase, four families