0
$\begingroup$

I am trying to find the minimum-phase system of the transfer function $H(z) = \frac{2 + 3.125z^{-2}}{1-0.9z^{-1}+0.81z^{-2}}$.

I know I need to find and remove the allpass (basically reflect the two zeroes on the imaginary axis inside the unit circle?), and I've gotten so far as this: $H(z) = 2\frac{1 + \frac{3.125}{2}z^{-2}}{1-0.9z^{-1}+0.81z^{-2}} = 2\frac{1 + \frac{3.125}{2}z^{-2}}{1-0.9z^{-1}+0.81z^{-2}} \frac{\frac{3.125}{2} + z^{-2}}{1 + \frac{3.125}{2}z^{-2}}=,$

and then I'm not really sure where to go next. Is my thinking correct thus far? What step should I do next?

Thank you.

Edit: This is the kind of thing I'm trying to do, as per the example in my book: allpass and minimum phase example

|improve this question
$\endgroup$
  • 1
    $\begingroup$ Your math is wrong as is... did you type something wrong? $\endgroup$ – CMDoolittle Sep 28 '15 at 5:04
  • $\begingroup$ Yeah, it probably is... I'm a little confused on how this works. I've added the example from my book, to better illustrate what I'm trying to do. $\endgroup$ – enpi Sep 28 '15 at 8:20
3
$\begingroup$

The allpass part must have its zeros outside the unit circle, so it gets the zeros of your system, with the corresponding poles inside the unit circle. You get the correct pole locations by simply swapping the coefficients of the numerator polynomial:

$$H_a(z)=\frac{2+3.125z^{-2}}{3.125+2z^{-2}}\tag{1}$$

The minimum phase part is obtained by taking the system's poles and cancelling the poles of the allpass part:

$$H_m(z)=\frac{3.125+2z^{-2}}{1-0.9z^{-1}+0.81z^{-2}}\tag{2}$$

The transfer function of the original system is given by

$$H(z)=H_a(z)\cdot H_m(z)\tag{3}$$

|improve this answer
$\endgroup$
  • $\begingroup$ how does equation 2 cancel the poles of the allpass part, it looks identical to the original H? $\endgroup$ – panthyon Sep 28 '15 at 15:18
  • $\begingroup$ @panthyon: No, it's not identical to the original $H(z)$, its numerator is different (swapped coefficients). And the numerator equals the denominator of allpass part, hence canceling the allpass poles. $\endgroup$ – Matt L. Sep 28 '15 at 15:23
  • $\begingroup$ aha! reading comprehension error on my part. thanks for clarifying. $\endgroup$ – panthyon Sep 28 '15 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.