Lossless Filters

To motivate the idea of paraunitary filters, let's first review some properties of lossless filters, progressing from the simplest cases up to paraunitary filter banks:

• A linear, time-invariant filter $H(z)$ is said to be lossless (or allpass) if it preserves signal energy. That is, if the input signal is $x(n)$ , and the output signal is $y(n) = (h\ast x)(n)$ , then we have
  $$\sum_{n=-\infty}^{\infty} \left\vert y(n)\right\vert^2 = \sum_{n=-\infty}^{\infty} \left\vert x(n)\right\vert^2$$
  In terms of the $L_2$ signal norm $\left\Vert\,\,\cdot\,\,\right\Vert _2$ , this can be expressed more succinctly as
  $$\left\Vert\,y\,\right\Vert _2^2 = \left\Vert\,x\,\right\Vert _2^2$$

• Notice that only stable filters can be lossless since, otherwise, $\left\Vert\,y\,\right\Vert=\infty$ . We further assume all filters are causal for simplicity.

• It is straightforward to show that losslessness implies
  $$\left\vert H(e^{j\omega})\right\vert = 1, \quad \forall \omega.$$
  That is, the frequency response must have magnitude 1 everywhere on the unit circle in the $z$ plane. Another way to express this is to write
  $$\overline{H(e^{j\omega})} H(e^{j\omega}) = 1, \quad\forall\omega,$$
  and this form generalizes to ${\tilde H}(z)H(z)$ over the entire the $z$ plane.

• The paraconjugate of a transfer function may be defined as the analytic continuation of the complex conjugate from the unit circle to the whole $z$ plane:
  $${\tilde H}(z) \mathrel{\stackrel{\mathrm{\Delta}}{=}}\overline{H}(z^{-1})$$
  where $\overline{H}(z)$ denotes complex conjugation of the coefficients only of $H(z)$ and not the powers of $z$ . For example, if $H(z)=1+jz^{-1}$ , then $\overline{H}(z) = 1-jz^{-1}$ . We can write, for example,
  $$\overline{H}(z) \mathrel{\stackrel{\mathrm{\Delta}}{=}}\overline{H\left(\overline{z}\right)}$$
  in which the conjugation of $z$ serves to cancel the outer conjugation.

  We refrain from conjugating $z$ in the definition of the paraconjugate becase $\overline{z}$ is not analytic in the complex-variables sense. Instead, we invert $z$ , which is analytic, and which reduces to complex conjugation on the unit circle.

  The paraconjugate may be used to characterize allpass filters as follows:

• A causal, stable, filter $H(z)$ is allpass if and only if
  $${\tilde H}(z) H(z) = 1$$
  Note that this is equivalent to the previous result on the unit circle since
  $${\tilde H}(e^{j\omega}) H(e^{j\omega}) = \overline{H}(1/e^{j\omega})H(e^{j\omega}) = \overline{H(e^{j\omega})}H(e^{j\omega})$$

• To generalize lossless filters to the multi-input, multi-output (MIMO) case, we must generalize conjugation to MIMO transfer function matrices:
  • A $p\times q$ transfer function matrix $\bold{H}(z)$ is lossless if it is stable and its frequency-response matrix $\bold{H}(e^{j\omega})$ is unitary. That is,
    $$\bold{H}^*(e^{j\omega})\bold{H}(e^{j\omega}) = \bold{I}_q$$
    for all $\omega$ , where $\bold{I}_q$ denotes the $q\times q$ identity matrix, and $\bold{H}^\ast(e^{j\omega})$ denotes the Hermitian transpose (complex-conjugate transpose) of $\bold{H}(e^{j\omega})$ :
    $$\bold{H}^*(e^{j\omega}) \mathrel{\stackrel{\mathrm{\Delta}}{=}}\overline{\bold{H}^T(e^{j\omega})}$$

  • Note that $\bold{H}^*(e^{j\omega})\bold{H}(e^{j\omega})$ is a $q\times q$ matrix product of a $q\times p$ times a $p\times q$ matrix. If $q>p$ , then the rank must be deficient. Therefore, we must have $p\geq q$ . (There must be at least as many outputs as there are inputs, but it's ok to have extra outputs.)

  • A lossless $p\times q$ transfer function matrix $\bold{H}(z)$ is paraunitary, i.e.,
    $${\tilde {\bold{H}}}(z) \bold{H}(z) = \bold{I}_q$$
    Thus, every paraunitary matrix transfer function is unitary on the unit circle for all $\omega$ . Away from the unit circle, paraunitary $\bold{H}(z)$ is the unique analytic continuation of unitary $\bold{H}(e^{j\omega})$ .