Allpass Examples

Free Books Introduction to Digital Filters

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 , 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 . 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 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 can be obtained by negating every other coefficient in , and multiplying by . 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 (root of at ), the polynomial has a root at . 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.