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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 [83,37,98].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 causalD.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 [83] for the DTFT, we have,D.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

$\displaystyle \frac{1}{2\pi}\int_{-\pi}^{\pi} \left\vert X(e^{j\omega})\right\v...
...{j\omega})\right\vert^2\left\vert X(e^{j\omega})\right\vert^2 d\omega. \protect$ (D.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) [70]. 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.

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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (August 2006 Edition).
Copyright © 2007-02-02 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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