# factoring poles / zeros: off by constant gain compared with textbook

(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?

• 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 Jan 20, 2019 at 23:23
• 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 $\left(1-2z^{-1}\right)=0$ Jan 20, 2019 at 23:30
• 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.. Jan 20, 2019 at 23:42
• i would suggest multiplying both numerator and denominator by a sufficient power of $z$ so that all of the unfactored terms of $z^n$ have a non-negative power, $n$, and that when factored, all of the factors have $z$, not $z^{-1}$. Jun 20, 2019 at 19:48

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.

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...

• 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? Jan 21, 2019 at 1:28
• 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. Jan 21, 2019 at 1:32
• ok... thanks... i'll chew on that for awhile.. Jan 21, 2019 at 1:41

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.