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Critically Sampled Perfect Reconstruction Filter Banks

A Perfect Reconstruction (PR) filter bank is any filter bank whose reconstruction is the original signal, possibly delayed, and possibly scaled by a constant [287]. In this context, critical sampling (also called ``maximal downsampling'') means that the downsampling factor is the same as the number of filter channels. For the STFT, this implies $ R=M=N$ (with $ M>N$ allowed for Portnoff windows).

As derived in Chapter 8, the Short-Time Fourier Transform (STFT) is a PR filter bank whenever the Constant-OverLap-Add (COLA) condition is met by the analysis window $ w$ and the hop size $ R$ . However, only the rectangular window case with no zero-padding is critically sampled (OLA hop size = FBS downsampling factor = $ N$ ). Perceptual audio compression algorithms such as MPEG audio coding are based on critically sampled filter banks, for obvious reasons. It is important to remember that we normally do not require critical sampling for audio analysis, digital audio effects, and music applications; instead, we normally need critical sampling only when compression is a requirement. Thus, when compression is not a requirement, we are normally interested in oversampled filter banks. The polyphase representation is useful in that case as well. In particular, we will obtain some excellent insights into the aliasing cancellation that goes on in such downsampled filter banks (including STFTs with hop sizes $ R>1$ ), as the next section makes clear.

Two-Channel Critically Sampled Filter Banks

Figure 11.15 shows a simple two-channel band-splitting filter bank, followed by the corresponding synthesis filter bank which reconstructs the original signal (we hope) from the two channels. The analysis filter $ H_0(z)$ is a half-band lowpass filter, and $ H_1(z)$ is a complementary half-band highpass filter. The synthesis filters $ F_0(z)$ and $ F_1(z)$ are to be derived. Intuitively, we expect $ F_0(z)$ to be a lowpass that rejects the upper half-band due to the upsampler by 2, and $ F_1(z)$ should do the same but then also reposition its output band as the upper half-band, which can be accomplished by selecting the upper of the two spectral images in the upsampler output.

\begin{psfrags} % latex2html id marker 29839\psfrag{x(n)}{\normalsize $x(n)$}\psfrag{x}{\normalsize $x$}\psfrag{(n)}{\normalsize $(n)$}\psfrag{y}{\normalsize $y$}\psfrag{v}{\normalsize $v$}\psfrag{H}{\normalsize $H$}\psfrag{F}{\normalsize $F$}\psfrag{z}{\normalsize $z$}\begin{figure}[htbp] \includegraphics[width=\twidth]{eps/cqf} \caption{Two-channel critically sampled filter bank.} \end{figure} \end{psfrags}

The outputs of the two analysis filters in Fig.11.15 are

$\displaystyle X_k(z) \eqsp H_k(z)X(z), \quad k=0,1.$ (12.16)

Using the results of §11.1, the signals become, after downsampling,

$\displaystyle V_k(z) \eqsp \frac{1}{2}\left[X_k(z^{1/2}) + X_k(-z^{1/2})\right], \; k=0,1.$ (12.17)

After upsampling, the signals become
$\displaystyle Y_k(z) \eqsp V_k(z^2)$ $\displaystyle =$ $\displaystyle \frac{1}{2}[X_k(z) + X_k(-z)]$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}[H_k(z)X(z) + H_k(-z)X(-z)],\; k=0,1.$  

After substitutions and rearranging, we find that the output $ \hat{x}$ is a filtered replica of the input signal plus an aliasing term:
$\displaystyle \hat{X}(z)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[H_0(z)F_0(z) + H_1(z)F_1(z)\right]X(z)$  
  $\displaystyle +$ $\displaystyle \frac{1}{2}\left[H_0(-z)F_0(z) + H_1(-z)F_1(z)\right]X(-z) \protect$ (12.18)

For perfect reconstruction, we require the aliasing term to be zero. For ideal half-band filters cutting off at $ \omega=\pi/2$ , we can choose $ F_0=H_0$ and $ F_1=H_1$ and the aliasing term is zero because there is no spectral overlap between the channels, i.e., $ H_0(-z)F_0(z)=H_1(z)H_0(z)=0$ , and $ H_1(-z)F_1(z)=H_0(z)H_1(z)=0$ . However, more generally (and more practically), we can force the aliasing to zero by choosing synthesis filters
$\displaystyle F_0(z)$ $\displaystyle =$ $\displaystyle \quad\! H_1(-z),$  
$\displaystyle F_1(z)$ $\displaystyle =$ $\displaystyle -H_0(-z). \protect$ (12.19)

In this case, synthesis filter $ F_0(z)$ is still a lowpass, but the particular one obtained by $ \pi$ -rotating the highpass analysis filter around the unit circle in the $ z$ plane. Similarly, synthesis filter $ F_1(z)$ is the $ \pi$ -rotation (and negation) of the analysis lowpass filter $ H_0(z)$ on the unit circle. For this choice of synthesis filters $ F_0$ and $ F_1$ , aliasing is completely canceled for any choice of analysis filters $ H_0$ and $ H_1$ .

Referring again to (11.18), we see that we also need the non-aliased term to be of the form

$\displaystyle A(z)$ $\displaystyle =$ $\displaystyle H_0(z)F_0(z) + H_1(z)F_1(z) \protect$ (12.20)

where $ A(z)$ is of the form

$\displaystyle A(z)\eqsp g\,z^{-d}.$ (12.21)

That is, for perfect reconstruction, we need, in addition to aliasing cancellation, that the non-aliasing term reduce to a constant gain $ g$ and/or delay $ d$ . We will call this the filtering cancellation constraint on the channel filters. Thus perfect reconstruction requires both aliasing cancellation and filtering cancellation.

Let $ {\tilde H}$ denote $ H(-z)$ . Then both constraints can be expressed in matrix form as follows:

$\displaystyle \left[\begin{array}{cc} H_0 & H_1 \\ [2pt] {\tilde H}_0 & {\tilde H}_1 \end{array}\right]\left[\begin{array}{c} F_0 \\ [2pt] F_1 \end{array}\right]\eqsp \left[\begin{array}{c} c \\ [2pt] 0 \end{array}\right]$ (12.22)

Substituting the aliasing-canceling choices for $ F_0$ and $ F_1$ from (11.19) into the filtering-cancellation constraint (11.20), we obtain

$\displaystyle g\,z^{-d}$ $\displaystyle =$ $\displaystyle H_0(z)H_1(-z) - H_1(z)H_0(-z). \protect$ (12.23)

The filtering-cancellation constraint is almost satisfied by ideal zero-phase half-band filters cutting off at $ \pi/2$ , since in that case we have $ H_1(-z)=H_0(z)$ and $ H_0(-z)=H_1(z)$ . However, the minus sign in (11.23) means there is a discontinuous sign flip as frequency crosses $ \omega=\pi/2$ , which is not equivalent to a linear phase term. Therefore the filtering cancellation constraint fails for the ideal half-band filter bank! Recall from above, however, that ideal half-band filters did work using a different choice of synthesis filters, relying instead on their lack of spectral overlap. The presently studied case from (11.19) arose from so-called Quadrature Mirror Filters (QMF), which are discussed further below. First, however, we'll look at some simple special cases.

Amplitude-Complementary 2-Channel Filter Bank

A natural choice of analysis filters for our two-channel critically sampled filter bank is an amplitude-complementary lowpass/highpass pair, i.e.,

$\displaystyle H_1(z) \eqsp 1-H_0(z)$ (12.24)

where we impose the unity dc gain constraint $ H_0(1)=1$ . Note that amplitude-complementary means constant overlap-add (COLA) on the unit circle in the $ z$ plane.

Substituting the COLA constraint into the filtering and aliasing cancellation constraint (11.23) gives

\begin{eqnarray*} g\,z^{-d} &=& H_0(z)\left[1-H_0(-z)\right] - \left[1-H_0(z)\right]H_0(-z) \\ [5pt] &=& H_0(z) - H_0(-z)\\ [5pt] \;\longleftrightarrow\;\quad a(n) &=& h_0(n) - (-1)^n h_0(n) \\ [5pt] &=& \left\{\begin{array}{ll} 0, & \hbox{$n$\ even} \\ [5pt] 2h_0(n), & \hbox{$n$\ odd} \\ \end{array} \right. \end{eqnarray*}

Thus, we find that even-indexed terms of the impulse response are unconstrained, since they subtract out in the constraint, while, for perfect reconstruction, exactly one odd-indexed term must be nonzero in the lowpass impulse response $ h_0(n)$ . The simplest choice is $ h_0(1)\neq 0$ .

Thus, we have derived that the lowpass-filter impulse-response for channel 0 can be anything of the form

$\displaystyle h_0 \eqsp [h_0(0), \bold{h_0(1)}, h_0(2), 0, h_0(4), 0, h_0(6), 0, \ldots] \protect$ (12.25)

or

$\displaystyle h_0 \eqsp [h_0(0), 0, h_0(2), \bold{h_0(3)}, h_0(4), 0, h_0(6), 0, \ldots]$ (12.26)

etc. The corresponding highpass-filter impulse response is then

$\displaystyle h_1(n) \eqsp \delta(n) - h_0(n).$ (12.27)

The first example (11.25) above goes with the highpass filter

$\displaystyle h_1 \eqsp [1-h_0(0), -h_0(1), -h_0(2), 0, -h_0(4), 0, -h_0(6), 0, \ldots]$ (12.28)

and similarly for the other example.

The above class of amplitude-complementary filters can be characterized in general as follows:

\begin{eqnarray*} H_0(z) &=& E_0(z^2) + h_0(o) z^{-o}, \quad E_0(1)+h_0(o)\eqsp 1, \, \hbox{$o$\ odd}\\ [5pt] H_1(z) &=& 1-H_0(z) \eqsp 1 - E_0(z^2) - h_0(o) z^{-o} \end{eqnarray*}

In summary, we see that an amplitude-complementary lowpass/highpass analysis filter pair yields perfect reconstruction (aliasing and filtering cancellation) when there is exactly one odd-indexed term in the impulse response of $ h_0(n)$ .

Unfortunately, the channel filters are so constrained in form that it is impossible to make a high quality lowpass/highpass pair. This happens because $ E_0(z^2)$ repeats twice around the unit circle. Since we assume real coefficients, the frequency response, $ E_0(e^{j2\omega})$ is magnitude-symmetric about $ \omega=\pi/2$ as well as $ \pi$ . This is not good since we only have one degree of freedom, $ h_0(o) z^{-o}$ , with which we can break the $ \pi/2$ symmetry to reduce the high-frequency gain and/or boost the low-frequency gain. This class of filters cannot be expected to give high quality lowpass or highpass behavior.

To achieve higher quality lowpass and highpass channel filters, we will need to relax the amplitude-complementary constraint (and/or filtering cancellation and/or aliasing cancellation) and find another approach.

Haar Example

Before we leave the case of amplitude-complementary, two-channel, critically sampled, perfect reconstruction filter banks, let's see what happens when $ H_0(z)$ is the simplest possible lowpass filter having unity dc gain, i.e.,

$\displaystyle H_0(z) \eqsp \frac{1}{2} + \frac{1}{2}z^{-1}.$ (12.29)

This case is obtained above by setting $ E_0(z^2)=1/2$ , $ o=1$ , and $ h_0(1)=1/2$ . The polyphase components of $ H_0(z)$ are clearly

$\displaystyle E_0(z^2)\eqsp E_1(z^2)\eqsp 1/2.$ (12.30)

Choosing $ H_1(z)=1-H_0(z)$ , and choosing $ F_0(z)$ and $ F_1(z)$ for aliasing cancellation, the four filters become

\begin{eqnarray*} H_0(z) &=& \frac{1}{2} + \frac{1}{2}z^{-1} \eqsp E_0(z^2)+z^{-1}E_1(z^2)\\ [5pt] H_1(z) &=& 1-H_0(z) \eqsp \frac{1}{2} - \frac{1}{2}z^{-1} \eqsp E_0(z^2)-z^{-1}E_1(z^2)\\ [5pt] F_0(z) &=& \;\;\, H_1(-z) \eqsp \frac{1}{2} + \frac{1}{2}z^{-1} \eqsp \;\;\,H_0(z)\\ [5pt] F_1(z) &=& -H_0(-z) \eqsp -\frac{1}{2} + \frac{1}{2}z^{-1} \eqsp -H_1(z). \end{eqnarray*}

Thus, both the analysis and reconstruction filter banks are scalings of the familiar Haar filters (``sum and difference'' filters $ (1\pm z^{-1})/\sqrt{2}$ ). The frequency responses are

\begin{eqnarray*} H_0(e^{j\omega}) &=&\;\;\,F_0(e^{j\omega}) \eqsp \frac{1}{2} + \frac{1}{2}e^{-j\omega}\eqsp e^{-j\frac{\omega}{2}} \cos\left(\frac{\omega}{2}\right)\\ [5pt] H_1(e^{j\omega}) &=& -F_0(e^{j\omega}) \eqsp \frac{1}{2} - \frac{1}{2}e^{-j\omega}\eqsp j e^{-j\frac{\omega}{2}} \sin\left(\frac{\omega}{2}\right) \end{eqnarray*}

which are plotted in Fig.11.16.

Figure 11.16: Amplitude responses of the two channel filters in the Haar filter bank.
\includegraphics[width=\twidth]{eps/haar}

Polyphase Decomposition of Haar Example


\begin{psfrags} % latex2html id marker 29979\psfrag{x(n)}{\normalsize $x(n)$}\psfrag{x}{\normalsize $x$}\psfrag{(n)}{\normalsize $(n)$}\psfrag{y}{\normalsize $y$}\psfrag{v}{\normalsize $v$}\psfrag{H}{\normalsize $H$}\psfrag{F}{\normalsize $F$}\psfrag{z}{\normalsize $z$}\begin{figure}[htbp] \includegraphics[width=\twidth]{eps/cqfcopy} \caption{Two-channel polyphase filter bank and inverse.} \end{figure} \end{psfrags}

Let's look at the polyphase representation for this example. Starting with the filter bank and its reconstruction (see Fig.11.17), the polyphase decomposition of $ H_0(z)$ is

$\displaystyle H_0(z) \eqsp E_0(z^2) + z^{-1}E_1(z^2) \eqsp \frac{1}{2}+\frac{1}{2}z^{-1}.$ (12.31)

Thus, $ E_0(z^2)=E_1(z^2)=1/2$ , and therefore

$\displaystyle H_1(z) \eqsp 1-H_0(z) \eqsp E_0(z^2)-z^{-1}E_1(z^2).$ (12.32)

We may derive polyphase synthesis filters as follows:

\begin{eqnarray*} \hat{X}(z) &=& \left[F_0(z)H_0(z) + F_1(z)H_1(z)\right] X(z)\\ &=& \left[\left(\frac{1}{2} + \frac{1}{2}z^{-1}\right)H_0(z) + \left(-\frac{1}{2}+\frac{1}{2}z^{-1}\right)H_1(z)\right]X(z)\\ &=& \frac{1}{2}\left\{\left[H_0(z)-H_1(z)\right] + z^{-1}\left[H_0(z) + H_1(z)\right]\right\}X(z) \end{eqnarray*}

The polyphase representation of the filter bank and its reconstruction can now be drawn as in Fig.11.18. Notice that the reconstruction filter bank is formally the transpose of the analysis filter bank [263]. A filter bank that is inverted by its own transpose is said to be an orthogonal filter bank, a subject to which we will return §11.3.8.

Figure 11.18: Polyphase representation of the general two-channel, critically sampled filter bank and its inverse.
\includegraphics[width=\twidth]{eps/poly2chan}

Figure 11.19: Figure 11.18 with downsamplers commuted inside the filter branches.
\includegraphics[width=\twidth]{eps/poly2chanfast}

Commuting the downsamplers (using the noble identities from §11.2.5), we obtain Figure 11.19. Since $ E_0(z)=E_1(z)=1/2$ , this is simply the OLA form of an STFT filter bank for $ N=2$ , with $ N=M=R=2$ , and rectangular window $ w=[1/2,1/2]$ . That is, the DFT size, window length, and hop size are all 2, and both the DFT and its inverse are simply sum-and-difference operations.


Quadrature Mirror Filters (QMF)

The well studied subject of Quadrature Mirror Filters (QMF) is entered by imposing the following symmetry constraint on the analysis filters:

$\displaystyle H_1(z) \eqsp H_0(-z)\quad \hbox{(QMF Symmetry Constraint)} \protect$ (12.33)

That is, the filter for channel 1 is constrained to be a $ \pi$ -rotation of filter 0 along the unit circle. This of course makes perfect sense for a two-channel band-splitting filter bank, and can form the basis of a dyadic tree band splitting, as we'll look at in §11.9.1 below.

In the time domain, the QMF constraint (11.33) becomes $ h_1(n) = (-1)^n h_0(n)$ , i.e., all odd-index coefficients are negated. If $ H_0$ is a lowpass filter cutting off near $ \omega=\pi/2$ (as is typical), then $ H_1$ is a complementary highpass filter. The exact cut-off frequency can be adjusted along with the roll-off rate to provide a maximally constant frequency-response sum.

Two-channel QMFs have been around since at least 1976 [51], and appear to be the first critically sampled perfect reconstruction filter banks. Moreover, the Princen-Bradley filter bank, the initial foundation of MPEG audio as we now know it, was conceived as the Fourier dual of QMFs [214]. Historically, the term QMF applied only to two-channel filter banks having the QMF symmetry constraint (11.33). Today, the term ``QMF filter bank'' may refer to more general PR filter banks with any number of channels and not obeying (11.33) [287].

Combining the QMF symmetry constraint with the aliasing-cancellation constraints, given by

\begin{eqnarray*} F_0(z) &=& \quad\! H_1(-z) \eqsp \quad\! H_0(z)\\ [5pt] F_1(z) &=& -H_0(-z) \eqsp -H_1(z), \end{eqnarray*}

the perfect reconstruction requirement reduces to

$\displaystyle g\,z^{-d} \eqsp H_0(z)F_0(z) + H_1(z)F_1(z) \eqsp H_0^2(z) - H_0^2(-z). \protect$ (12.34)

Now, all four filters are determined by $ H_0(z)$ .

It is easy to show using the polyphase representation of $ H_0(z)$ (see [287]) that the only causal FIR QMF analysis filters yielding exact perfect reconstruction are two-tap FIR filters of the form

\begin{eqnarray*} H_0(z) &=& c_0 z^{-2n_0} + c_1 z^{-(2n_1+1)}\\ H_1(z) &=& c_0 z^{-2n_0} - c_1 z^{-(2n_1+1)} \end{eqnarray*}

where $ c_0$ and $ c_1$ are constants, and $ n_0$ and $ n_1$ are integers. Therefore, only weak channel filters are available in the QMF case [ $ H_1(z)=H_0(-z)$ ], as we saw in the amplitude-complementary case above. On the other hand, very high quality IIR solutions are possible. See [287, pp. 201-204] for details. In practice, approximate ``pseudo QMF'' filters are more practical, which only give approximate perfect reconstruction. We'll return to this topic in §11.7.1.

The scaled Haar filters, which we saw gave perfect reconstruction in the amplitude-complementary case, are also examples of a QMF filter bank:

\begin{eqnarray*} H_0(z) &=& 1 + z^{-1}\\ [5pt] H_1(z) &=& 1 - z^{-1} \end{eqnarray*}

In this example, $ c_0=c_1=1$ , and $ n_0=n_1=0$ .

Linear Phase Quadrature Mirror Filter Banks

Linear phase filters delay all frequencies by equal amounts, and this is often a desirable property in audio and other applications. A filter phase response is linear in $ \omega$ whenever its impulse response $ h_0(n)$ is symmetric, i.e.,

$\displaystyle h_0(-n) \eqsp h_0(n)$ (12.35)

in which case the frequency response can be expressed as

$\displaystyle H_0(e^{j\omega}) \eqsp e^{-j\omega N/2}\left\vert H_0(e^{j\omega})\right\vert.$ (12.36)

Substituting this into the QMF perfect reconstruction constraint (11.34) gives

$\displaystyle g\,e^{-j\omega d} \eqsp e^{-j\omega N}\left[ \left\vert H_0(e^{j\omega})\right\vert^2 - (-1)^N\left\vert H_0(e^{j(\pi-\omega)})\right\vert^2\right].$ (12.37)

When $ N$ is even, the right hand side of the above equation is forced to zero at $ \omega=\pi/2$ . Therefore, we will only consider odd $ N$ , for which the perfect reconstruction constraint reduces to

$\displaystyle g\,z^{-j\omega d} \eqsp e^{-j\omega N}\left[ \left\vert H_0(e^{j\omega})\right\vert^2 + \left\vert H_0(e^{j(\pi-\omega)}\right\vert^2\right].$ (12.38)

We see that perfect reconstruction is obtained in the linear-phase case whenever the analysis filters are power complementary. See [287] for further details.

Conjugate Quadrature Filters (CQF)

A class of causal, FIR, two-channel, critically sampled, exact perfect-reconstruction filter-banks is the set of so-called Conjugate Quadrature Filters (CQF). In the z-domain, the CQF relationships are

$\displaystyle H_1(z) \eqsp z^{-(L-1)}H_0(-z^{-1}).$ (12.39)

In the time domain, the analysis and synthesis filters are given by

\begin{eqnarray*} h_1(n) &=& -(-1)^n h_0(L-1-n) \\ [5pt] f_0(n) &=& h_0(L-1-n) \\ [5pt] f_1(n) &=& -(-1)^n h_0(n) \eqsp - h_1(L-1-n). \end{eqnarray*}

That is, $ f_0=\hbox{\sc Flip}(h_0)$ for the lowpass channel, and each highpass channel filter is a modulation of its lowpass counterpart by $ (-1)^n$ . Again, all four analysis and synthesis filters are determined by the lowpass analysis filter $ H_0(z)$ . It can be shown that this is an orthogonal filter bank. The analysis filters $ H_0(z)$ and $ H_1(z)$ are power complementary, i.e.,

$\displaystyle \left\vert H_0{e^{j\omega}}\right\vert^2 + \left\vert H_1{e^{j\omega}}\right\vert^2 \eqsp 1$ (12.40)

or

$\displaystyle {\tilde H}_0(z) H_0(z) + {\tilde H}_1(z) H_1(z) \eqsp 1$ (12.41)

where $ {\tilde H}_0(z)\isdef \overline{H}_0(z^{-1})$ denotes the paraconjugate of $ H_0(z)$ (for real filters $ H_0$ ). The paraconjugate is the analytic continuation of $ \overline{H_0(e^{j\omega})}$ from the unit circle to the $ z$ plane. Moreover, the analysis filters $ H_0(z)$ are power symmetric, e.g.,

$\displaystyle {\tilde H}_0(z) H_0(z) + {\tilde H}_0(-z) H_0(-z) \eqsp 1 .$ (12.42)

The power symmetric case was introduced by Smith and Barnwell in 1984 [272]. With the CQF constraints, (11.18) reduces to

$\displaystyle \hat{X}(z) \eqsp \frac{1}{2}\left[H_0(z)H_0(z^{-1}) + H_0(-z)H_0(-z^{-1})\right]X(z) \protect$ (12.43)

Let $ P(z) = H_0(z)H_0(-z)$ , such that $ H_0(z)$ is a spectral factor of the half-band filter $ P(z)$ (i.e., $ P(e^{j\omega})$ is a nonnegative power response which is lowpass, cutting off near $ \omega=\pi/4$ ). Then, (11.43) reduces to

$\displaystyle \hat{X}(z) \eqsp \frac{1}{2}\left[P(z) + P(-z)\right]X(z) \eqsp -z^{-(L-1)}X(z)$ (12.44)

The problem of PR filter design has thus been reduced to designing one half-band filter $ P(z)$ . It can be shown that any half-band filter can be written in the form $ p(2n) = \delta(n)$ . That is, all non-zero even-indexed values of $ p(n)$ are set to zero.

A simple design of an FIR half-band filter would be to window a sinc function:

$\displaystyle p(n) \eqsp \frac{\hbox{sin}[\pi n/2]}{\pi n/2}w(n)$ (12.45)

where $ w(n)$ is any suitable window, such as the Kaiser window.

Note that as a result of (11.43), the CQF filters are power complementary. That is, they satisfy

$\displaystyle \left\vert H_0(e^{j \omega})\right\vert^2 + \left\vert H_1(e^{j \omega})\right\vert^2 \eqsp 2.$ (12.46)

Also note that the filters $ H_0$ and $ H_1$ are not linear phase. It can be shown that there are no two-channel perfect reconstruction filter banks that have all three of the following characteristics (except for the Haar filters):
  • FIR
  • orthogonal
  • linear phase
In this design procedure, we have chosen to satisfy the first two and give up the third.

By relaxing ``orthogonality'' to ``biorthogonality'', it becomes possible to obtain FIR linear phase filters in a critically sampled, perfect reconstruction filter bank. (See §11.9.)

Orthogonal Two-Channel Filter Banks

Recall the reconstruction equation for the two-channel, critically sampled, perfect-reconstruction filter-bank:

\begin{eqnarray*} \hat{X}(z) &=& \frac{1}{2}[H_0(z)F_0(z) + H_1(z)F_1(z)]X(z) \nonumber\\ [5pt] &+& \frac{1}{2}[H_0(-z)F_0(z) + H_1(-z)F_1(z)]X(-z) \end{eqnarray*}

This can be written in matrix form as

$\displaystyle \hat{X}(z) \eqsp \frac{1}{2} \left[\begin{array}{c} F_0(z) \\ [2pt] F_1(z) \end{array}\right]^{T} \left[\begin{array}{cc} H_0(z) & H_0(-z) \\ [2pt] H_1(z) & H_1(-z) \end{array}\right] \left[\begin{array}{c} X(z) \\ [2pt] X(-z) \end{array}\right]$ (12.47)

where the above $ 2 \times 2$ matrix, $ \bold{H}_m(z)$ , is called the alias component matrix (or analysis modulation matrix). If

$\displaystyle {\tilde {\bold{H}}}_m(z)\bold{H}_m(z) \eqsp 2\bold{I}$ (12.48)

where $ {\tilde {\bold{H}}}_m(z)\isdef \bold{H}_m^T(z^{-1})$ denotes the paraconjugate of $ \bold{H}_m(z)$ , then the alias component (AC) matrix is lossless, and the (real) filter bank is orthogonal.

It turns out orthogonal filter banks give perfect reconstruction filter banks for any number of channels. Orthogonal filter banks are also called paraunitary filter banks, which we'll study in polyphase form in §11.5 below. The AC matrix is paraunitary if and only if the polyphase matrix (defined in the next section) is paraunitary [287].

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Perfect Reconstruction Filter Banks
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Polyphase Decomposition