Editing Minimum phase (section)

=== Discrete-time frequency analysis ===

Performing frequency analysis for the discrete-time case will provide some insight. The time-domain equation is
<math display="block">
 (h * h_\text{inv})(n) = \delta(n).
</math>

Applying the [[Z-transform]] gives the following relation in the ''z''&nbsp;domain:
<math display="block">
 H(z) H_\text{inv}(z) = 1.
</math>

From this relation, we realize that
<math display="block">
 H_\text{inv}(z) = \frac{1}{H(z)}.
</math>

For simplicity, we consider only the case of a [[rational function|rational]] [[transfer function]] {{math|''H''(''z'')}}. Causality and stability imply that all [[pole (complex analysis)|poles]] of {{math|''H''(''z'')}} must be strictly inside the [[unit circle]] (see [[BIBO stability#Discrete-time signals|stability]]). Suppose
<math display="block">
 H(z) = \frac{A(z)}{D(z)},
</math>
where {{math|''A''(''z'')}} and {{math|''D''(''z'')}} are [[polynomial]] in {{mvar|z}}. Causality and stability imply that the [[zero (complex analysis)|poles]]{{snd}} the [[Root of a function|roots]] of {{math|''D''(''z'')}}{{snd}} must be strictly inside the [[unit circle]]. We also know that
<math display="block">
 H_\text{inv}(z) = \frac{D(z)}{A(z)},
</math>
so causality and stability for <math>H_\text{inv}(z)</math> imply that its [[pole (complex analysis)|poles]]{{snd}} the roots of {{math|''A''(''z'')}}{{snd}} must be inside the [[unit circle]]. These two constraints imply that both the zeros and the poles of a minimum-phase system must be strictly inside the unit circle.