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(From Schaum's DSP outline, 2nd edition, problem 5.32)

Book says factor it and extract H(z) from the factored product:

$$ H(z)H(z^{-1})= \frac{ \frac{5}{4} - \frac{1}{2}z - \frac{1}{2}z^{-1} }{ \frac{10}{9}- \frac{1}{3}z - \frac{1}{3}z^-1 } $$

ok.. no problem... so i convert to the form used by the book for nearly every problem:

$$ H(z)H(z^{-1})= \left( \frac{z^{-1}}{z^{-1}} \right) \left( \frac{ \frac{5}{4} - \frac{1}{2}z - \frac{1}{2}z^{-1} }{ \frac{10}{9}- \frac{1}{3}z - \frac{1}{3}z^-1 } \right) $$

$$ H(z)H(z^{-1})= \left( \frac{ - \frac{1}{2} + \frac{5}{4}z^{-1} - \frac{1}{2}z^{-2} }{ \frac{1}{3} + \frac{10}{9}z^{-1}- \frac{1}{3}z^{-2} } \right) $$

Then I apply quadractic formula to determine the zeros and poles:

$$zeros = \left\{ \frac{1}{2}, 2 \right\}$$ $$poles = \left\{ \frac{1}{3}, 3 \right\}$$

Then I rewrite $H(z)H^{-1}(z)$ in factored form:

$$ H(z)H(z^{-1})=\frac{\left(1-\frac{1}{2}z^{-1}\right)(1-2z^{-1})}{\left(1-\frac{1}{3}z^{-1}\right)(1-3z^{-1})} $$

At this point I realize that I have a different answer from the book which says, the answer at this point should be:

$$ H(z)H(z^{-1})=\frac{\left(1-\frac{1}{2}z^{-1}\right)(1-\frac{1}{2}z)}{\left(1-\frac{1}{3}z^{-1}\right)\left(1-\frac{1}{3}z\right)} $$

I think... well... ok.. it should be equivalent to my version... i can rewrite my equation to get it to match the book.. so i try that:

$$ H(z)H(z^{-1})=\frac{\left(1-\frac{1}{2}z^{-1}\right)(-2z^{-1})(1-\frac{1}{2}z)}{\left(1-\frac{1}{3}z^{-1}\right)(-3z^{-1})(1-\frac{1}{3}z)} $$

$$ H(z)H(z^{-1})=\left(\frac{2}{3}\right)\frac{\left(1-\frac{1}{2}z^{-1}\right)(1-\frac{1}{2}z)}{\left(1-\frac{1}{3}z^{-1}\right)(1-\frac{1}{3}z)} $$

At this point i'm scratching my head wondering why it doesn't match because its off by a different gain? Wondering why the gain doesn't matter when picking out H(z) from the factorization of $H(z)H(z^{-1})$? Is this good practice to write it the other way around? which way is correct?

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  • $\begingroup$ Its like you only look at one pole and one zero to recreate $H(z)$ and ignore the other pole and zero, then define $H(z^{-1})$ from $H(z)$....then, it matches the book $\endgroup$ – Bill Moore Jan 20 at 23:23
  • $\begingroup$ what i don't understand is $\left(1-\frac{1}{2}z\right) \ne \left(1-2z^{-1}\right)$. if you set z =1. it doesn't equate. but individually the roots are the same.$\left(1-\frac{1}{2}z\right)=0$ verses $\left(1-2z^{-1}\right)=0$ $\endgroup$ – Bill Moore Jan 20 at 23:30
  • $\begingroup$ or maybe its like this... if you see $H(Z)H(z^{-1})$ then you just automatically know that your zeros and also your poles come in reciprocal pairs so you need to use the form $(1-(zeropole)^{-1}z)$ on the first zero/pole and $(1-(zeropole)z^{-1})$ on the second zero/pole... that's just the rule... no idea... it just works.. $\endgroup$ – Bill Moore Jan 20 at 23:42
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Spectral factorization is not unique, unless you specify further constraints.

So let your given $H(z)H(z^-1)$ be (taking from your derivation) :

$$ \begin{align} H(z)H(z^{-1})&= \frac{ \frac{5}{4} - \frac{1}{2}z - \frac{1}{2}z^{-1} }{ \frac{10}{9}- \frac{1}{3}z - \frac{1}{3}z^-1 } \tag{1}\\ &= \frac{\left(1-\frac{1}{2}z^{-1}\right)(1-2z^{-1})}{\left(1-\frac{1}{3}z^{-1}\right)(1-3z^{-1})} \end{align} $$

Then all of the following are valid possibilities that satisf Eq.(1)

$$ H_1(z) =\frac{1-\frac{1}{2}z^{-1}}{1-\frac{1}{3}z^{-1}} $$

$$ H_2(z) =\frac{1-2z^{-1}}{1-3z^{-1}} $$

$$ H_3(z) =\frac{1-\frac{1}{2}z^{-1}}{1-3z^{-1}} $$

$$ H_4(z) =\frac{1-2z^{-1}}{1-\frac{1}{3}z^{-1}} $$

So you must have further constraints that make one of them as a unique answer. Such as minimum-phase, maximum phase, causal, stable, bothe etc...

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  • $\begingroup$ that's an interesting point of view...so basically what you are saying is that reciprocating a real pole/zero has the same magnitude response $|H(e^{j\omega})|$. but $X[z^{-1}] <-> x[-n]$ is time-reversal and anti-casual? $\endgroup$ – Bill Moore Jan 21 at 1:28
  • $\begingroup$ No. What I say is given $G(z)=H(z)H(z^{-1})$ there are many possible $H(z)$ that yield the same $G(z)$. Each $H(z)$ will have different pole zero combinations and hence their magnitudes, in general, will not be same, except those of reciprocal ones. $\endgroup$ – Fat32 Jan 21 at 1:32
  • $\begingroup$ ok... thanks... i'll chew on that for awhile.. $\endgroup$ – Bill Moore Jan 21 at 1:41
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for consistency with the book, i'm just going to make up the following rule:

if $\alpha$ is a pole or zero of $H(z)$, then it must be factored as: $\left(1-\alpha z^{-1}\right)$

if $\alpha$ is a pole or zero of $H(z^{-1})$, then it must be factored as: $\left(1-\frac{1}{\alpha}z\right)$

if $\alpha$ is a pole or zero of $H\left( \frac{1}{\alpha^*}\right)$, then it must be factored as: $\left(z^{-1} - \alpha^*\right)$

At this point, i'm not really sure what the casual, min-max phase, stable, etc... implication are for always choosing this way of factoring... it just follows what the book always does.

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You got the poles and zeros right, but you ignored the scaling. Your factorization

$$H(z)H(z^{-1})=\frac{\left(1-\frac{1}{2}z^{-1}\right)(1-2z^{-1})}{\left(1-\frac{1}{3}z^{-1}\right)(1-3z^{-1})}\tag{1}$$

is not equal to the original function

$$H(z)H(z^{-1})= \frac{ \frac{5}{4} - \frac{1}{2}z - \frac{1}{2}z^{-1} }{ \frac{10}{9}- \frac{1}{3}z - \frac{1}{3}z^{-1} }\tag{2}$$

This is easy to see by looking at the constants corresponding to the terms with no powers of $z$ in the numerator and denominator in $(1)$ (they're all $1$), and by comparing them to the corresponding terms in $(2)$ (here we need to look at the constants of the $z$ terms, which are $-\frac12$ and $-\frac13$).

So if you take those constant into account you end up with

$$\frac{-\frac12}{-\frac13}\cdot\frac{\left(1-\frac{1}{2}z^{-1}\right)(1-2z^{-1})}{\left(1-\frac{1}{3}z^{-1}\right)(1-3z^{-1})}=\frac{\left(1-\frac{1}{2}z^{-1}\right)(z^{-1}-\frac12)}{\left(1-\frac{1}{3}z^{-1}\right)(z^{-1}-\frac13)}\tag{3}$$

If you multiply the numerator and the denominator of $(3)$ by $z$ then you end up with the same result as in the given solution. Note that in this case your dividing by $z$ had to be undone, so you could have skipped that step.

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