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Given the following system function $$ H(z) =\frac{1}{1+\frac{2}{3}z^{-1}+\frac{5}{8}z^{-2}+\frac{2}{3}z^{-3}+z^{-4}} \tag{1} $$ We have to draw lattice structure for above system function

The denominator polynomial of the transfer function at lattice stage number $m=4$ is assumed as $A_m$ and now $A_{m-1}$ can be found with

$$\begin{align} A_{m-1}(z)&=\frac{A_{m}(z)-k_mB_{m}(z)}{1-k_m^{2}}\\ \text{with}\quad B_m(z)&=z^{-m}A_m(1/z)\\ \text{and}\quad A_0(z)&=B_0(z)=1\end{align}\tag{2}$$

where $k_m$ is the $m^{th}$ reflection coefficient. But since the $k_m$ value is 1 equation (2) becomes zero in denominator and hence cannot be solved.

I couldn't find the solution to this particular problem in some of my book or other site. How can I get the lattice coefficient for above system function?

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Note that all the filter's poles lie on the unit circle, i.e., the filter is unstable. It cannot be implemented by a lattice structure, for the very reason that you've found out yourself.

The given filter is the inverse of a linear phase FIR filter. Note that also linear phase FIR filters cannot be implemented using lattice structures.

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  • $\begingroup$ why can't it be implemented as an unstable lattice? $\endgroup$ – robert bristow-johnson Mar 18 '19 at 2:59
  • $\begingroup$ @robertbristow-johnson: Because the recursion for computing the reflection coefficients from the direct form coefficients breaks down, since you get $|k_m|=1$ and the denominator in the recursion becomes zero. You can implement an unstable lattice filter by simply choosing some $k_m$ as $k_m=\pm 1$ but I wouldn't know how to control the locations of the poles (other than that they are on the unit circle). $\endgroup$ – Matt L. Mar 18 '19 at 9:27
  • $\begingroup$ well, i know i can do it for a 2nd-order lattice. $\endgroup$ – robert bristow-johnson Mar 18 '19 at 19:36
  • $\begingroup$ @robertbristow-johnson: Give it a try then, I would be curious to see if it can be done for the given transfer function. $\endgroup$ – Matt L. Mar 18 '19 at 21:48
  • $\begingroup$ well, that recursion that you refer to is, i presume, what O&S discuss and it's not for cascaded second-order-sections (SOS). i haven't really done that (i always have busted my big filters into SOSes, usually don't use lattice, but once in a while i do). all's know is that if $$ H(z) = \frac{b_0 + b_1 z^{-1} + b_2 z^{-2}}{1 + a_1 z^{-1} + a_2 z^{-2}} $$ then $k_2=-a_2$, $k_1=\frac{-a_1}{a_2+1}$ and the feedforward taps are $\hat{b}_2=b_2$, $\hat{b}_1=b_1 - k_1(k_2-1)b_2$, and $\hat{b}_0=b_0 + k_1 b_1 - \big(k_1^2(k_2-1)-k_2 \big)b_2$ so the five coefficients i need are well defined. $\endgroup$ – robert bristow-johnson Mar 18 '19 at 21:59
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Watch after 30mins... above question is addressed.

Consider: $$ A_{4} (z) = 1 + h_1 z^{-1} + h_2 z^{-2} + h_1 z^{-3} + z^ {-4}$$ $$ A_4 (z) = A_3 (z) + k_4 * B_3 (z)$$ $$ A_4 (z) = A_3 (z) + k_4 * z^ {-4} (A_3 (z^ {-1})) $$


Try to break $A_4(z)$ into two equal halves:


$$ A_4 (z) = (1 + h_1 z^ {-1} + \frac{h_2}{2} z^ {-2} ) + z^ {-4} (1 + h_1 z^{1} + \frac{h_2}{2} z^{2} )$$

$$ A_3 (z) = A_2 (z) : k_3 = 0 , k_2 = \frac{h_2}{2} $$

$$A_1 (z ) = \frac{A_2 (z) - k_2 B_2 (z)}{1- k_2 ^ {2}} $$


Solving: $$ k_1 = \frac{h_1}{ 1 + \frac{h_2}{2}}$$


For above question : $$k_4 = 1$$ $$k_3 = 0$$ $$k_2 = \frac{\frac{5}{8}}{2} = \frac{5}{16}$$ $$k_1 = \frac{\frac{2}{3}}{1+ \frac{5}{16}} = \frac{32}{63} $$

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  • $\begingroup$ not to heap any work on @PeterK. but i think that $\LaTeX$ should be used in expressing this answer. if it were a smaller job i might fix it myself. $\endgroup$ – robert bristow-johnson Mar 13 at 20:02
  • $\begingroup$ @robertbristow-johnson OK! After I’ve finished my tea. About 30 minutes. $\endgroup$ – Peter K. Mar 13 at 20:24
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    $\begingroup$ i'll tweak it a little, @PeterK. just obvious stuff. i just didn't want to take time to deal with the content. $\endgroup$ – robert bristow-johnson Mar 13 at 21:29

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