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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!

gerrgheiser
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  • wait, so you're just looking for factorizations to $(1+\alpha_i D^{-n})$ of $a + bD^{-1}+cD$? Your $\alpha_i$ are just the inverse of the zeros of that polynomial, and finding those zeros is equivalent to finding the zeros of a quadratic function (multiply $a + bD^{-1}+cD=0$ with $D$ on both sides, yields trivially solvable quadratic form). – Marcus Müller Mar 21 '17 at 07:43
  • I'm wanting to factorize it into a form of $c(1+\alpha D)(1+\alpha D^{-1})$, where $c$ is just a constant. This way each half of the factorization is the conjugate of the other. – gerrgheiser Mar 21 '17 at 13:23

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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:

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

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...

msm
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    ahhh, that makes perfect sense. I think I was trying to work toward that, but was missing something. Thanks a bunch! – gerrgheiser Mar 21 '17 at 16:22
  • I still didn't get it how you do the nominator part of your equation I guess at this point I am so confused that someone has to tell me step by step I am taking course of static signal processing. and we have to do a lot of spectral factorization in the Yule-Walker method. X(D)=.3477(1+.2588D)(1+.2588D−1)(1+.9D)(1+.9D−1)=cY(D)Y∗(D−∗). my question where I am stuck is [B(z)B(1/z)]=0.61-0.14Z−1+0.14Z After spectral factorization B(z)= 0.76(1-0.24Z^-1)(1-0.24Z) p(e^jw)=0.58*|1-0.24e^-jw|^2/|1-0.5e^-jw|^2 – Hamood Faisal Aug 14 '20 at 13:04