# 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?

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

• 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. Mar 13 '20 at 20:02
• @robertbristow-johnson OK! After I’ve finished my tea. About 30 minutes.
– Peter K.
Mar 13 '20 at 20:24
• i'll tweak it a little, @PeterK. just obvious stuff. i just didn't want to take time to deal with the content. Mar 13 '20 at 21:29

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.

• why can't it be implemented as an unstable lattice? Mar 18 '19 at 2:59
• @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). Mar 18 '19 at 9:27
• well, i know i can do it for a 2nd-order lattice. Mar 18 '19 at 19:36
• @robertbristow-johnson: Give it a try then, I would be curious to see if it can be done for the given transfer function. Mar 18 '19 at 21:48
• 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. Mar 18 '19 at 21:59