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?


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.

  • $\begingroup$ why can't it be implemented as an unstable lattice? $\endgroup$ – robert bristow-johnson Mar 18 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 at 9:27
  • $\begingroup$ well, i know i can do it for a 2nd-order lattice. $\endgroup$ – robert bristow-johnson Mar 18 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 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 at 21:59

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