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If a system is given by a transfer function in the $z$ domain that has all poles and zeros inside the unit circle except for a factor of $z^{-1}$ in the denominator (pole at infinity), can it still be considered minimum phase? If not, how would I create an all pass system to neutralize this pole and obtain a minimum phase representation of the system without also adding a zero at infinity (which i assume would also cause the system to not be min-phase?

The transfer function:

$$H(z) = \frac{1 - \frac 12 z^{-1}}{z^{-1}}$$

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  • $\begingroup$ It would be helpful if you could write down the transfer function. I think there's a problem with your system, because the way you describe it, it's not causal and stable. $\endgroup$ – Matt L. Mar 18 '14 at 20:40
  • $\begingroup$ H(z) = (1 - (1/2)*z^-1)/(z^-1) $\endgroup$ – user3361675 Mar 18 '14 at 21:02
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If I see it correctly, your transfer function is

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

This system is not a minimum-phase system because its pole is at infinity (as you've already pointed out). You could use a trivial all-pass filter with impulse response $h[n]=\delta[n-1]$, i.e. a simple delay by one sample to change your original (non-causal) transfer function to a stable and causal transfer function

$$H_m(z)=z^{-1}H(z)=1-\frac{1}{2}z^{-1}$$

which is also minimum-phase, because it has a pole at $z=0$ and a zero at $z=1/2$. This means that all the system's poles and zeros are inside the unit circle and the minimum-phase requirement is met.

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No, to the first question. A pole at infinity means the region of convergence does not include infinity. This means that the system is not causal. But the definition of a minimum phase system usually includes the requirement that the system be causal.

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