The frequency response is:

$$H(z) = 2-7z^{-1}+7z^{-2}-2z^{-3}$$

I see that it has $3$ zeros: $z_{01} = \frac 12$, $z_{02} = 2$, and $z_{03} = 1$; and $3$ poles in: $$z_x = 0$$

Now, I have to write it like:

$$H(z) = H_{\rm min}(z) H_{\rm max}(z) H_{\rm uc}(z)$$

where $H_{\rm min}(z)$ is the minimum-phase frequency response, $H_{\rm max}(z)$ is the maximum-phase frequency response and $H_{\rm uc}(z)$ only has zeros on $\lvert z\rvert=1$.

  • For $H_{\rm uc}(z)$, I have one zero on $\lvert z\rvert=1$, so: $$H_{\rm uc}(z) = z-1$$ Is that OK?

  • For the minimum-phase frequency, I have all the poles/zeros that are inside the unit circle, so: $$H_{\rm min}(z) = \frac{z-1/2}{z^3}$$

  • For the maximum-phase frequency, I have all the poles/zeros that are outside the unit circle, so: $$H_{\rm max}(z) = z-2$$

    But I also know that I can find $H_{\rm max}(z)$ as: $$H_{\rm max}(z) = H_{\rm min}\left(z^{-1}\right) z^{-M_i}$$ where $M_i$ is the quantity of zeros of $H_{\rm min}(z)$. So: $$H_{\rm max}(z) = \frac{z^{-1} - 1/2}{z^{-3}} = z^3 \left(z^{-1} - 1/2\right) = z^2 \left(z-1/2\right)$$

    So I have $2$ different expressiones for the same $H_{\rm max}(z)$. What am I doing wrong?

  • $\begingroup$ Check your zeros: there's no zero at $z=3$. $\endgroup$
    – Matt L.
    Commented Jun 20, 2016 at 16:20
  • $\begingroup$ That's true. It's fixed now :) $\endgroup$
    – Euler
    Commented Jun 20, 2016 at 17:55

1 Answer 1


So let's check that: $$H(z) = H_{\rm min}(z) H_{\rm max}(z) H_{\rm uc}(z) \tag{1}$$ where: $$ H_{\rm min}(z) = \frac{z-1/2}{z^3}\tag{2} $$ and $$H_{\rm max}(z) = z-2\tag{3}$$ and $$H_{\rm uc}(z) = z-1\tag{4}$$

Substituting $(2)$, $(3)$, and $(4)$ into $(1)$:

\begin{align} H_{\rm min}(z) H_{\rm max}(z) H_{\rm uc}(z) &= \frac{z-1/2}{z^3} (z-2) (z-1)\\ &= \frac{z-1/2}{z^3} \left(z^2 - 3z + 2\right)\\ &= \frac{1}{z^3} \left(z^3 - 3z^2 + 2z - \frac{1}{2}\left(z^2 - 3z + 2\right)\right) \\ &= \frac{1}{z^3} \left( z^3 - \frac{7}{2} z^2 + 5 z - 1\right) \\ &\not= H(z) \end{align}

So something is wrong!

Try: $$ H_{\rm min}(z) = 1-1/2z^{-1}\tag{2A} $$ and $$H_{\rm max}(z) = 1-2z^{-1}\tag{3A}$$ and $$H_{\rm uc}(z) = 2(1- z^{-1})\tag{4A}$$

Now: \begin{align} H_{\rm min}(z) H_{\rm max}(z) H_{\rm uc}(z) &= \left(1-1/2z^{-1}\right) \left(1-2z^{-1}\right) \left(1-z^{-1}\right) 2\\ &= \left(1 - \frac{5}{2} z^{-1} + z^{-2}\right) \left(1-z^{-1}\right) 2\\ &= 2 - 5 z^{-1} + 2z^{-2} - \left(2z^{-1} - 5 z^{-2} + 2 z^{-3}\right)\\ &= 2 - 7 z^{-1} + 7 z^{-2} - 2z^{-3}\\ &= H(z) \end{align}

I believe your interpretation of Oppenheim and Schafer (OS) is incorrect. And I can find no reference that says the poles have to be outside the unit circle for a maximum phase system. This one and the one I referenced in the comments both only mention the zero locations.

And this is what my copy of OS says:

No mention of pole locations

which also only mentions zero locations.

I believe exercise 5.63 is in error. The definition used in the body of the book is:

enter image description here

which again does not mention pole locations.

Also, having $$ H_{\rm max}(z) = H_{\rm min}\left(z^{-1}\right) z^{-M_i} $$ ensures that $H_{\rm max}$ is causal for FIR $H_{\rm min}$.

So you have two options: either the exercise 5.69 definition is wrong, or the relationship between $H_{\rm max}$ and $H_{\rm min}$ is wrong.

  • $\begingroup$ But why did you choose $H_{uc}(z)=2(1-z^{-1})$ instead of $H_{uc}(z) = 1-(z^{-1})$ ?? Both have a zero at $z=1$ and a pole in $z=0$. What's the difference? How did you choose that scale factor? ---------------------------- By definition, $H_{max}(z)$ has all the poles and zeros outside the unit circle, and you choose a zero at $z=2$ and a pole at $z=0$ (inside the unit circle). Why? $\endgroup$
    – Euler
    Commented Jun 20, 2016 at 20:09
  • $\begingroup$ Your choice of $H_{uc}$ or the other components neglects a gain of two. I chose to include it on $H_{uc}$; you could just as validly split it between $H_{max}$ and $H_{min}$ as a $\sqrt{2}$ factor. I suppose I tend to neglect poles at zero and $\infty$. That will mean your $H_{max}$ will not be causal. $\endgroup$
    – Peter K.
    Commented Jun 20, 2016 at 20:27
  • $\begingroup$ How did you know that the scalar factor has to be $2$? I still don't understand why did you choose a pole inside the unit circle for $H_{max}(z)$ $\endgroup$
    – Euler
    Commented Jun 20, 2016 at 20:30
  • 1
    $\begingroup$ Your definition of maximum phase is bogus. Where do you get it from? This page gives a more usual definition. This definition says nothing about pole positions; just zeros. $\endgroup$
    – Peter K.
    Commented Jun 20, 2016 at 20:33
  • 1
    $\begingroup$ I think that the exercise 5.63 is wrong. Thanks Peter! $\endgroup$
    – Euler
    Commented Jun 21, 2016 at 1:55

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