Allpass Examples

• The simplest allpass filter is a unit-modulus gain
  $$H(z) = e^{j\phi}$$
  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:
  $$H(z) = e^{j\phi} z^{-K}$$
  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
  $$H(z) = e^{j\phi} z^{-K} \frac{\tilde{A}(z)}{A(z)}$$
  where $K\geq 0$ ,
  $$A(z) \isdef 1 + a_1 z^{-1}+ a_2 z^{-2} + \cdots + a_Nz^{-N},$$
  and
  $$\begin{eqnarray*} \tilde{A}(z)&\isdef & z^{-N}\overline{A}(z^{-1})\\ &=& \overline{a_N} + \overline{a_{N-1}}z^{-1}+ \overline{a_{N-2}}z^{-2}+ \cdots + \overline{a_1} z^{-(N-1)} + z^{-N}. \end{eqnarray*}$$
  

  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
  $$H(s) = e^{j\phi} \frac{A(-s)}{A(s)}$$
  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
  $$e^{-s\tau} = 1 - \tau s + \frac{\tau^2}{2} s^2 + \cdots$$
  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
  $$H_i(s) = \frac{s+p_i}{s-p_i}.$$
  The frequency response of such a term is
  $$H_i(j\omega) = \frac{j\omega+p_i}{j\omega-p_i} = - \frac{p_i+j\omega}{\overline{p_i+j\omega}}$$
  which is obviously unit magnitude.