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Chapters

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Embedded SystemsFPGAElectronics

Introduction to Digital Filters
    Allpass Filters
       Allpass Examples

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Allpass Examples

  • The simplest allpass filter is a unit-modulus gain

    $\displaystyle H(z) = e^{j\phi} $

    where $ \phi$ can be any phase value. In the real case $ \phi$ can only be 0 or $ \pi $, in which case $ H(z)=\pm 1$.

  • A lossless FIR filter can consist only of a single nonzero tap:

    $\displaystyle H(z) = e^{j\phi} z^{-K} $

    for some fixed integer $ K$, where $ \phi$ is again some constant phase, constrained to be 0 or $ \pi $ in the real-filter case. Since we are considering only causal filters here, $ K\geq 0$. As a special case of this example, a unit delay $ H(z)=z^{-1}$ is a simple FIR allpass filter.

  • The transfer function of every finite-order, causal, lossless IIR digital filter (recursive allpass filter) can be written as

    $\displaystyle H(z) = e^{j\phi} z^{-K} \frac{\tilde{A}(z)}{A(z)} $

    where $ K\geq 0$,

    $\displaystyle A(z) \isdef 1 + a_1 z^{-1}+ a_2 z^{-2} + \cdots + a_Nz^{-N}, $

    and

    \begin{eqnarray*} \tilde{A}(z)&\isdef & z^{-N}\overline{A}(z^{-1})\\ &=& \overl... ...ne{a_{N-2}}z^{-2}+ \cdots + \overline{a_1} z^{-(N-1)} + z^{-N}. \end{eqnarray*}

    We may think of $ \tilde{A}(z)$ as the flip of $ A(z)$. For example, if $ A(z)=1+1.4z^{-1}+0.49z^{-2}$, we have $ \tilde{A}(z)=0.49+1.4z^{-1}+z^{-2}$. Thus, $ \tilde{A}(z)$ is obtained from $ A(z)$ by simply reversing the order of the coefficients and conjugating them when they are complex.

  • For analog filters, the general finite-order allpass transfer function is

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

    where $ K\geq 0$, $ A(s) = s^N + a_1 s^{N-1} + \cdots + a_{N-1}s + a_N$. The polynomial $ A(-s)$ can be obtained by negating every other coefficient in $ A(z)$, and multiplying by $ z^N$. In analog, a pure delay of $ \tau$ seconds corresponds to the transfer function

    $\displaystyle e^{-s\tau} = 1 - \tau s + \frac{\tau^2}{2} s^2 + \cdots $

    which is infinite order. Given a pole $ p_i$ (root of $ A(s)$ at $ s=p_i$), the polynomial $ A(-s)$ has a root at $ s = -p_i$. Thus, the poles and zeros can be paired off as a cascade of terms such as

    $\displaystyle H_i(s) = \frac{s+p_i}{s-p_i}. $

    The frequency response of such a term is

    $\displaystyle H_i(j\omega) = \frac{j\omega+p_i}{j\omega-p_i} = - \frac{p_i+j\omega}{\overline{p_i+j\omega}} $

    which is obviously unit magnitude.

Previous: Allpass Filters
Next: Paraunitary FiltersC.4

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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 http://ccrma.stanford.edu/~jos/ for details.


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