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Allpass Filters

This appendix addresses the general problem of characterizing all digital allpass filters, including multi-input, multi-output (MIMO) allpass filters. As a result of including the MIMO case, the mathematical level is a little higher than usual for this book. The reader in need of more background is referred to [84,37,98].Our first task is to show that losslessness implies allpass.

Definition: A linear, time-invariant filter $ H(z)$ is said to be lossless if it preserves signal energy for every input signal. That is, if the input signal is $ x(n)$, and the output signal is $ y(n) = (h\ast x)(n)$, then we have

$\displaystyle \sum_{n=-\infty}^{\infty} \left\vert y(n)\right\vert^2 =
\sum_{n=-\infty}^{\infty} \left\vert x(n)\right\vert^2.

In terms of the $ L2$ signal norm $ \left\Vert\,\,\cdot\,\,\right\Vert _2$, this can be expressed more succinctly as

$\displaystyle \left\Vert\,y\,\right\Vert _2^2 = \left\Vert\,x\,\right\Vert _2^2.

Notice that only stable filters can be lossless, since otherwise $ \left\Vert\,y\,\right\Vert$ can be infinite while $ \left\Vert\,x\,\right\Vert$ is finite. We further assume all filters are causalC.1 for simplicity. It is straightforward to show the following:

Theorem: A stable, linear, time-invariant (LTI) filter transfer function $ H(z)$ is lossless if and only if

$\displaystyle \left\vert H(e^{j\omega})\right\vert = 1, \quad \forall \omega.

That is, the frequency response must have magnitude 1 everywhere over the unit circle in the complex $ z$ plane.

Proof: We allow the signals $ x(n),y(n)$ and filter impulse response $ h(n)$ to be complex. By Parseval's theorem [84] for the DTFT, we have,C.2 for any signal $ x\leftrightarrow X$,

$\displaystyle \left\Vert\,x\,\right\Vert _2 = \left\Vert\,X\,\right\Vert _2


$\displaystyle \left\Vert\,x\,\right\Vert _2^2 \isdef \sum_{n=-\infty}^\infty \l...
...rt X(e^{j\omega})\right\vert^2 d\omega
\isdef \left\Vert\,X\,\right\Vert _2^2.

Thus, Parseval's theorem enables us to restate the definition of losslessness in the frequency domain:

$\displaystyle \left\Vert\,X\,\right\Vert _2^2 = \left\Vert\,Y\,\right\Vert _2^2 = \left\Vert\,H\cdot X\,\right\Vert _2^2

where $ Y=HX$ because the filter $ H$ is LTI. Thus, $ H$ is lossless by definition if and only if

$\displaystyle \frac{1}{2\pi}\int_{-\pi}^{\pi} \left\vert X(e^{j\omega})\right\v...
...{j\omega})\right\vert^2\left\vert X(e^{j\omega})\right\vert^2 d\omega. \protect$ (C.1)

Since this must hold for all $ X(e^{j\omega})$, we must have $ \left\vert H(e^{j\omega})\right\vert=1$ for all $ \omega$, except possibly for a set of measure zero (e.g., isolated points which do not contribute to the integral) [73]. If $ H(z)$ is finite order and stable, $ H(e^{j\omega})$ is continuous over the unit circle, and its modulus is therefore equal to 1 for all $ \omega\in[-\pi,\pi]$. $ \Box$

We have shown that every lossless filter is allpass. Conversely, every unity-gain allpass filter is lossless.

Previous: Elementary Filter Problems
Next: Allpass Examples

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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See for details.


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