spectral factorization method So I'm trying to figure out a method to do spectral factorization, where we take some function $X(D) = (1+cD)(1+cD^{-1})$. I believe this is used in equaliation I can do this if I just plugging in numbers until i get something that works, but that doesn't seem like the best way to go about it. Lets look at the following example:

$$X(D) = \frac{.19}{(1+.9D)(1+.9D^{-1})}+\frac{1}{10} = \frac{.371+.09D^{-1} +.09D}{(1+.9D)(1+.9D^{-1})}$$

The bottom part is easy, because it's already factorized into the form we want, so it's the top part that's the hard part. I can come up with the following answer but trying random numbers until I start getting close to what I want:

$$ X(D) =  \frac{.3477(1+.2588D)(1+.2588D^{-1})}{(1+.9D)(1+.9D^{-1})} = cY(D)Y^*(D^{-*})$$
where $Y(D) = \frac{(1+.2588D)}{(1+.9D)}$ and $c= .3477$.

I've tried searching for a method of doing this, but I just can't find one that makes sense and that works every time.

I've tried getting it in the form of $D(D^2 + bD + c)$ where the $D^2$ term has a coefficient of 1, and $b$ and $c$ are constants, and then try factoring that into two parts, but it just doesn't seem to be working out for me. I'd appreciate any advice!
The question is not so clear. I am not sure what other types of examples you consider. In this particular case, if you didn't get the hint notice that:

$$\begin{align}
c(1+\alpha D)(1+\alpha D^{-1}) &amp;=  (c+\alpha^2c)+\alpha cD^{-1}+\alpha cD \\&amp;=0.371+0.09D^{-1} +.09D
\end{align}$$

which gives you
$$\left\{\begin{matrix}
c(1+\alpha^2) = 0.371\\
\alpha c=0.09
\end{matrix}\right.
\Rightarrow \frac{1+\alpha^2}{\alpha}=\frac{0.37}{0.09}$$
and you will actually have two solutions

$$\left\{\begin{matrix}
\alpha=3.8634, c=    0.0233\\
\alpha=0.2588, c=  0.3477
\end{matrix}\right.
$$

See if the second answer is also valid...