# How to find lattice coefficients of all pole IIR filter with filter coefficient $k_m$ is 1?

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?

• @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). – Matt L. Mar 18 at 9:27
• 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. – robert bristow-johnson Mar 18 at 21:59