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I have been reading on conjugate quadrature filters (CQF). The linked article explains that, if we have a lowpassfilter $h_0$, we can create a set of filters for a perfect reconstruction filterbank. The analysis highpass filter $h_1$ and synthesis low pass/highpass filters $f_0$ and $f_1$ can be generated as follows:

\begin{align} h_1(n)&=-\left(-1\right)^n h_0\left(L-1-n\right)\\ f_0(n)&=h_0\left(L-1-n\right)\\ f_1(n)&=-\left(-1\right)^n h_0(n)=-h_1\left(L-1-n\right). \end{align}

This works excellent for a set of Smith-Barnwell coefficients. Yet it does not work for every lowpass filter that could be designed. For instance a Kaiser-Bessel windowed sinc-function with a cutoff at $\pi/2$ (as described in the article as well) does not lead to perfect reconstruction.

Are there any particular requirements on $h_0$/$H_0$ before this will work ?

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So the central question here is what the restrictions for the filters of this class of filterbanks (i.e. $N=2$-filter bank with the HPF generated from a lowpass/highpass transform of the LPF) is.

Let us for that first introduce a couple of helping definitions:

$$\begin{align} \mathbf{h}(z) &= \begin{bmatrix}h_0(z)\\h_1(z)\end{bmatrix}&\text{analysis filterbank matrix}\\ \mathbf{r}(z) &= \begin{bmatrix}r_0(z)\\r_1(z)\end{bmatrix}&\text{reconstruction filterbank matrix}\\ \end{align}$$

With these in place: Let's consider the polyphase TypeI representation $A(z)$ (analysis) and TypeII representation $R(z)$ (reconstruction) of these filter banks (by the way, this is why we have these different types of polyphase deconstructions); I hope my notation is clear, $h_{1,0}$ is the zeroth polyphase component of the high pass filter.

$$\begin{align} \mathbf{h}(z) &= \begin{bmatrix} h_{0,0}^{\mathrm I}(z^2) & h_{0,1}^{\mathrm I}(z^2)\\ h_{1,0}^{\mathrm I}(z^2) & h_{1,1}^{\mathrm I}(z^2) \end{bmatrix} \begin{bmatrix} z^{-0}\\ z^{-1} \end{bmatrix}\\ &= \mathbf A(z^2) \mathbf e_z(z)\\ \mathbf{r}^T(z) &= \begin{bmatrix} z^{-1} & z^{-0} \end{bmatrix} \begin{bmatrix} r_{0,0}^{\mathrm {II}}(z^2) & r_{0,1}^{\mathrm {II}}(z^2)\\ r_{1,0}^{\mathrm {II}}(z^2) & r_{1,1}^{\mathrm {II}}(z^2) \end{bmatrix}\\ &= \mathbf{ \tilde e}_z(z) \mathbf R(z^2) \\ \end{align}$$

Now, looking at this, we simply define perfect reconstruction "intuitively" as the case where applying $\mathbf R$ to the output of $\mathbf A$ leads to the input of $\mathbf A$, of course delayed, and maybe scaled with a constant $c$:

$$\begin{align} h,\,r &\text{ form a perfect reconstruction fb}\\ &\iff\\ \mathbf R(z)\mathbf A(z) &= c z^{-k} \mathbf I_N & c \text{ const.},\, k\in\mathbb N \end{align}$$

That gives us a clear method to move forward:

$$\begin{align} \mathbf R(z)\mathbf A(z) &= \begin{bmatrix} h_{0,0}^{\mathrm {II}}(z) & h_{0,1}^{\mathrm {II}}(z)\\ h_{1,0}^{\mathrm {II}}(z) & h_{1,1}^{\mathrm {II}}(z) \end{bmatrix} \begin{bmatrix} r_{0,0}^{\mathrm I}(z) & r_{0,1}^{\mathrm I}(z)\\ r_{1,0}^{\mathrm I}(z) & r_{1,1}^{\mathrm I}(z) \end{bmatrix}\\ &= \begin{bmatrix} h_{0,0}^{\mathrm {II}}(z)r_{0,0}^{\mathrm I}(z) + h_{0,1}^{\mathrm {II}}(z)r_{1,0}^{\mathrm I}(z) & h_{0,0}^{\mathrm {II}}(z)r_{0,1}^{\mathrm I}(z) + h_{0,1}^{\mathrm {II}}(z)r_{1,1}^{\mathrm I}(z)\\ h_{1,0}^{\mathrm {II}}(z)r_{0,0}^{\mathrm I}(z) + h_{1,1}^{\mathrm {II}}(z)r_{1,0}^{\mathrm I}(z) & h_{1,0}^{\mathrm {II}}(z)r_{0,1}^{\mathrm I}(z) + h_{1,1}^{\mathrm {II}}(z)r_{1,1}^{\mathrm I}(z) \end{bmatrix}\\ &\overset!= \begin{bmatrix}c z^{-k}&0\\0&c z^{-k}\end{bmatrix} \end{align}$$

EDIT now, let's do our jobs of inserting our conditions:

$$r_0 = h_0 (L-1-n)$$

notice that the substitution of $n$ by $L-1-n$ implies that the typeI polyphase decomposition of $r_0$ is the same as the typeII decomposition of $h_0$!

Hence, the diagonal entries of above matrix get a lot easier to calculate:

$$\begin{align} \mathbf R(z)\mathbf A(z) &=\begin{bmatrix} h_{0,0}^{\mathrm {II}}(z)r_{0,0}^{\mathrm I}(z) + h_{0,1}^{\mathrm {II}}(z)r_{1,0}^{\mathrm I}(z) & h_{0,0}^{\mathrm {II}}(z)r_{0,1}^{\mathrm I}(z) + h_{0,1}^{\mathrm {II}}(z)r_{1,1}^{\mathrm I}(z)\\ h_{1,0}^{\mathrm {II}}(z)r_{0,0}^{\mathrm I}(z) + h_{1,1}^{\mathrm {II}}(z)r_{1,0}^{\mathrm I}(z) & h_{1,0}^{\mathrm {II}}(z)r_{0,1}^{\mathrm I}(z) + h_{1,1}^{\mathrm {II}}(z)r_{1,1}^{\mathrm I}(z) \end{bmatrix}\\ &=\begin{bmatrix} h_{0,0}^{\mathrm {II}}(z)h_{0,0}^{\mathrm {II}}(z) + h_{0,1}^{\mathrm {II}}(z)-h_{1,0}^{\mathrm{II}}(z) & h_{0,0}^{\mathrm {II}}(z)h_{0,1}^{\mathrm {II}} + h_{0,1}^{\mathrm {II}}(z)-h_{1,1}^{\mathrm{II}}(z)\\ h_{1,0}^{\mathrm {II}}(z)h_{0,0}^{\mathrm{II}}(z) + h_{1,1}^{\mathrm {II}}(z)-h_{1,0}^{\mathrm{II}}(z) & h_{1,0}^{\mathrm {II}}(z)h_{0,1}^{\mathrm{II}}(z) + h_{1,1}^{\mathrm {II}}(z)-h_{1,1}^{\mathrm{II}}(z) \end{bmatrix}\\ &=\begin{bmatrix} h_{0,0}^{\mathrm {II}}(z)h_{0,0}^{\mathrm {II}}(z) - h_{0,1}^{\mathrm {II}}(z)h_{1,0}^{\mathrm{II}}(z) & h_{0,0}^{\mathrm {II}}(z)h_{0,1}^{\mathrm {II}} - h_{0,1}^{\mathrm {II}}(z)h_{1,1}^{\mathrm{II}}(z)\\ h_{1,0}^{\mathrm {II}}(z)h_{0,0}^{\mathrm{II}}(z) - h_{1,1}^{\mathrm {II}}(z)h_{1,0}^{\mathrm{II}}(z) & h_{1,0}^{\mathrm {II}}(z)h_{0,1}^{\mathrm{II}}(z) h_{1,1}^{\mathrm {II}}(z)h_{1,1}^{\mathrm{II}}(z) \end{bmatrix}\\ \end{align}$$

To achieve the zero entries in the identity matrix, it is necessary that

$$\begin{align} h_{0,1}^{\mathrm {II}} h_{0,0}^{\mathrm {II}}(z)&= h_{0,1}^{\mathrm {II}}(z)h_{1,1}^{\mathrm{II}}(z)\\ h_{1,0}^{\mathrm {II}}(z)h_{0,0}^{\mathrm{II}}(z) &= h_{1,0}^{\mathrm{II}}(z)h_{1,1}^{\mathrm {II}}(z) \end{align}$$

This breaks down to filters that need to have zeros at every $N$th position, since the 2-polyphased decomposition of a HPF/LPF transformed HPF is either the same as the original's decomposition, or the original's decomposition with a factor of $-1$.

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  • $\begingroup$ How do you obtain a cutoff frequency of pi/2 from this + does it imply any form restrictions on $H_0$ ? $\endgroup$ – user7488 Aug 14 '16 at 9:55
  • $\begingroup$ You don't – you just insert your $h_0$, $h_1$, $r_0$ and $r_1$ and check whether they fulfill the last equation :) $\endgroup$ – Marcus Müller Aug 14 '16 at 10:16
  • $\begingroup$ Although an extended mathematical explanation, it merely states that to satisfy the perfect reconstruction requirement the combination of lowpass and highpass filters should lead to perfect reconstruction. As far as I see it you do not go back to setting out requirements on $h_0$. $\endgroup$ – user7488 Aug 14 '16 at 17:56
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    $\begingroup$ well, $h_{0,0}=h_{1,1}$ is a strong condition, isn't it, given that $h_1 = -(-1)^n h_0$! $\endgroup$ – Marcus Müller Aug 14 '16 at 17:59
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    $\begingroup$ @Werner Van Belle The so-called Shannon wavelet could be of interest en.wikipedia.org/wiki/Shannon_wavelet $\endgroup$ – Laurent Duval Aug 14 '16 at 19:40
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So, let's take this in two steps. I'm too tired and can't answer both aspects, but let's explain the lowpass/highpass transform in an isolated answer:

Your $h_1$ is $h_0(L-1-n)$ multiplied with a complex sinusoid – becasue really, $-(-1)^n=-e^{j2\pi \frac n2}$, and that means you're multiplying with a sinusoid of frequency $\frac{f_\text{nyquist}}2$ in time domain, which is equivalent to a shift of half the Nyquist bandwidth in frequency domain!

LPF HPF transform

This is really just the easiest low-pass/high-pass transform: you "push" the passband from 0 on the frequency axis so far that the maximum amplitude is just at the corner of the Nyquist band.

Now, depending on how you define "cut-off" frequency, you might get into a lot of troubles when getting your perfect reconstruction.

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  • $\begingroup$ That (-1)^n performs a frequency shifting was clear to me. The question was what properties should the LPF ($h_0$) satisfy in order to have perfect reconstruction. $\endgroup$ – user7488 Aug 14 '16 at 3:43

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