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Analog Allpass Filters

It turns out that analog allpass filters are considerably simpler mathematically than digital allpass filters (discussed in §B.2). In fact, when working with digital allpass filters, it can be fruitful to convert to the analog case using the bilinear transform (§I.3.1), so that the filter may be manipulated in the analog $ s$ plane rather than the digital $ z$ plane. The analog case is simpler because analog allpass filters may be described as having a zero at $ s=-\overline{p}$ for every pole at $ s=p$ , while digital allpass filters must have a zero at $ z=1/\overline{p}$ for every pole at $ z=p$ . In particular, the transfer function of every first-order analog allpass filter can be written as

$\displaystyle H(s) = e^{j\phi}\frac{s+\overline{p}}{s-p} $

where $ \phi\in[-\pi,\pi)$ is any constant phase offset. To see why $ H(s)$ must be allpass, note that its frequency response is given by

$\displaystyle H(j\omega) = e^{j\phi}\frac{j\omega+\overline{p}}{j\omega-p} = - e^{j\phi}\frac{\overline{j\omega-p}}{j\omega-p}, $

which clearly has modulus 1 for all $ \omega$ (since $ \vert\overline{z}/z\vert=1,\,\forall z\neq 0$ ). For real allpass filters, complex poles must occur in conjugate pairs, so that the ``allpass rule'' for poles and zeros may be simplified to state that a zero is required at minus the location of every pole, i.e., every real first-order allpass filter is of the form

$\displaystyle H(s) = \pm\frac{s+p}{s-p}, $

and, more generally, every real allpass transfer function can be factored as

$\displaystyle H(s) = \pm\frac{(s+p_1)(s+p_2)\cdots(s+p_N)}{(s-p_1)(s-p_2)\cdots(s-p_N)}. \protect$ (E.12)

This simplified rule works because every complex pole $ p_i$ is accompanied by its conjugate $ p_k=\overline{p_i}$ for some $ k\in[1:N]$ .

Multiplying out the terms in Eq.(E.12), we find that the numerator polynomial $ B(s)$ is simply related to the denominator polynomial $ A(s)$ :

$\displaystyle H(s) = \pm(-1)^N\frac{A(-s)}{A(s)} = \pm(-1)^N\frac{s^N - a_{N-1}s^{N-1} + \cdots - a_1 s + a_0}{s^N + a_{N-1}s^{N-1} + \cdots + a_1 s + a_0} $

Since the roots of $ A(s)$ must be in the left-half $ s$ -plane for stability, $ A(s)$ must be a Hurwitz polynomial, which implies that all of its coefficients are nonnegative. The polynomial

$\displaystyle A(-s)=A\left(e^{j\pi}s\right) $

can be seen as a $ \pi $ -rotation of $ A(s)$ in the $ s$ plane; therefore, its roots must have non-positive real parts, and its coefficients form an alternating sequence.

As an example of the greater simplicity of analog allpass filters relative to the discrete-time case, the graphical method for computing phase response from poles and zeros (§8.3) gives immediately that the phase response of every real analog allpass filter is equal to twice the phase response of its numerator (plus $ \pi $ when the frequency response is negative at dc). This is because the angle of a vector from a pole at $ s=p$ to the point $ s=j\omega$ along the frequency axis is $ \pi $ minus the angle of the vector from a zero at $ s=-p$ to the point $ j\omega$ .


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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition)
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Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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