Editing Minimum phase (section)

== Frequency analysis ==
{{unsourced section|date=October 2023}}

=== 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.

=== Continuous-time frequency analysis ===

Analysis for the continuous-time case proceeds in a similar manner, except that we use the [[Laplace transform]] for frequency analysis. The time-domain equation is
<math display="block">
 (h * h_\text{inv})(t) = \delta(t),
</math>
where <math>\delta(t)</math> is the [[Dirac delta function]]{{snd}} the identity operator in the continuous-time case because of the sifting property with any signal {{math|''x''(''t'')}}:
<math display="block">
 (\delta * x)(t) = \int_{-\infty}^\infty \delta(t - \tau) x(\tau) \,d\tau = x(t).
</math>

Applying the [[Laplace transform]] gives the following relation in the [[s-plane]]:
<math display="block">
 H(s) H_\text{inv}(s) = 1,
</math>
from which we realize that
<math display="block">
 H_\text{inv}(s) = \frac{1}{H(s)}.
</math>

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

=== Relationship of magnitude response to phase response ===
{{See also|Kramers–Kronig relations#Magnitude (gain)–phase relation}}
A minimum-phase system, whether discrete-time or continuous-time, has an additional useful property that the [[natural logarithm]] of the magnitude of the frequency response (the "gain" measured in [[neper]]s, which is proportional to [[Decibel|dB]]) is related to the phase angle of the frequency response (measured in [[radian]]s) by the [[Hilbert transform]]. That is, in the continuous-time case, let
<math display="block">
 H(j\omega)\ \stackrel{\text{def}}{=}\ H(s)\Big|_{s=j\omega}
</math>
be the complex frequency response of system {{math|''H''(''s'')}}. Then, only for a minimum-phase system, the phase response of {{math|''H''(''s'')}} is related to the gain by
<math display="block">
 \arg[H(j\omega)] = -\mathcal{H}\big\{\log\big(|H(j\omega)|\big)\big\},
</math>
where <math>\mathcal{H}</math> denotes the Hilbert transform, and, inversely,
<math display="block">
 \log\big(|H(j\omega)|\big) = \log\big(|H(j\infty)|\big) + \mathcal{H}\big\{\arg[H(j\omega)]\big\}.
</math>

Stated more compactly, let
<math display="block">
 H(j\omega) = |H(j\omega)| e^{j\arg[H(j\omega)]}\ \stackrel{\text{def}}{=}\ e^{\alpha(\omega)} e^{j\phi(\omega)} = e^{\alpha(\omega) + j\phi(\omega)},
</math>
where <math>\alpha(\omega)</math> and <math>\phi(\omega)</math> are real functions of a real variable. Then
<math display="block">
 \phi(\omega) = -\mathcal{H}\{\alpha(\omega)\}
</math>
and
<math display="block">
 \alpha(\omega) = \alpha(\infty) + \mathcal{H}\{\phi(\omega)\}.
</math>

The Hilbert transform operator is defined to be
<math display="block">
 \mathcal{H}\{x(t)\}\ \stackrel{\text{def}}{=}\ \hat{x}(t) = \frac{1}{\pi} \int_{-\infty}^\infty \frac{x(\tau)}{t - \tau} \,d\tau.
</math>

An equivalent corresponding relationship is also true for discrete-time minimum-phase systems.