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Suppose that we have four filters $H_0(z), H_1(z), F_0(z),$ and $F_1(z)$ forming a classic perfect-reconstruction 2-channel filter bank: enter image description here Will the perfect reconstruction still be achieved if we interchange $F_i(z)$ with $H_i(z)$?
My thinking is to check if the condition that the aliasing transfer function is equal to zero is met. But I am confused that I do not know if the aliasing transfer function of the interchanged system: $F_0(-z)H_0(z)+F_1(-z)H_1(z)$ is equal to zero. Who could answer my question?

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I am not very familiarized with this structure, but applying the reasoning in [1] to your problem we can represent the analysis bank as $h(z) = \mathbf{E}(z^2) \mathbf{e}(z)$ and $f^T(z) = \mathbf{e}^T(z) R(z^2)$, with $e(z)^T = \begin{bmatrix}1 & z & z^2 &\ldots & z^n \end{bmatrix}$.

Denoting downsampling by $\downarrow(k)$, and upsampling by $\uparrow(k)$ we have $E(z^2)\downarrow(2)~\equiv~\downarrow(2)~E(z)$, and similarly $\uparrow(2) R(z^2) \equiv R(z)\uparrow(2)$, so that your entire system can be represented by

$$e(z)^T R(z^2)\downarrow(2) \uparrow(2) E(z^2) e(z) \equiv e(z) \downarrow(2)~R(z)E(z)\uparrow(2) e(z)^T $$

However, order to get perfect reconstruction $R(z) E(z)=I$, so $E(z)R(z)=I$ as well, and the swapping $E(z)$ and $R(z)$, gives a perfect reconstruction as well.

Acknowledgment: John Bofarull is whom actually pointed to fundamental condition for the proof.

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  • $\begingroup$ Since 'perfect reconstruction' is mentioned in the same question, such constraint implies F1=H0 and H1=F0 otherwise reconstruction is not possible. Even if F1=H0;H1=F0 decimation may render signal with alias, but let's say there's no alias. Then since branches are identical the questioned exchange is probably going to work. $\endgroup$ Commented Oct 24, 2022 at 0:53
  • $\begingroup$ Hi @JohnBofarull, I worked out the math for $F_1 = H_0$ and $F_0=H_1$, seems to work only for $z=0$, would you like to review if I made some mistake? $\endgroup$
    – Bob
    Commented Oct 24, 2022 at 6:57
  • $\begingroup$ Hi Bob, link [1] in your answer already contains the necessary and sufficient conditions to exchange Hi(z) Fi(z) and achieve perfect reconstruction. It seems Hi(z)*Fi(z)=I is a broad condition to satisfy reconstruction after swapping. $\endgroup$ Commented Nov 21, 2022 at 20:03
  • $\begingroup$ Shame on me, I missed the necessity for that condition. $\endgroup$
    – Bob
    Commented Nov 22, 2022 at 15:19

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