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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},


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

    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.

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