2
$\begingroup$

My question is about an example in Adaptive Filter Theory, Haykin 4th ed (refer figures 1.8 and 1.10 of Haykin). We want to determine the region for asymptotic stationarity of an AR process in the 2D space of $\rho_1$ and $\rho_2$. Process model (characteristic equation): $$1+a_1z^{-1}+a_2z^{-2}=0$$ For asymptotic stationarity, the system must be stable, i.e. the roots must be inside the unit circle in the z-plane. This means the parameters must be constrained to be in the following region (topic of a previous post). $$a_2<1$$ $$a_1+a_2+1>0$$ $$a_1-a_2-1<0$$ We get the following from the Yule-Walker equations. $$\rho_1=\frac{-a_1}{1+a_2}$$ $$\rho_2=-a_2+\frac{a_1^2}{1+a_2}$$ where $\rho_i=\frac{r[i]}{r[0]}$

Using the last two limits on the parameters ($a_1+a_2+1>0$ and $a_1-a_2-1<0$, sorry, cross referencing may not be working, so I have to resort to quoting!), I can derive the following limits on the region in $\rho_1\rho_2$-space: $-1<\rho_1<1$.

Question: The limit on $\rho_2$ as derived by Haykin is $\rho_1^2<\frac{1}{2}(1+\rho_2)$. However I get the opposite inequality: $\rho_1^2>\frac{1}{2}(1+\rho_2)$. What did I do wrong? Derivation below.

Substitute $\rho_1$ in $\rho_2$: $$\rho_2=-a_2+\frac{a_1^2}{1+a_2}=-a_2-a_1\rho_1$$ Use $-a_1=\rho_1(1+a_2)$ and $a_2<1$ $$\rho_2=-a_2+(1+a_2)\rho_1^2=\rho_1^2+a_2(\rho_1^2-1)<\rho_1^2+(\rho_1^2-1)$$ Therefore $$\rho_2<2\rho_1^2-1\Rightarrow\rho_1^2>\frac{1}{2}(1+\rho_2)$$

$\endgroup$
2
  • $\begingroup$ I get the same result if that makes you feel any better ;) $\endgroup$
    – Jdip
    Commented Jul 28, 2023 at 20:46
  • $\begingroup$ Consider the possibility that it is just a typo, probably in the LaTEx where >ge got mis-typed as \le by Haykin's secretary, and Haykin missed it during the proof-reading of the mss. $\endgroup$ Commented Aug 1, 2023 at 18:23

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.