0
$\begingroup$

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?

|improve this question
$\endgroup$
1
$\begingroup$

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.

|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.