Editing Hilbert transform (section)

== Relationship with the Fourier transform ==
The Hilbert transform is a [[Multiplier (Fourier analysis)|multiplier operator]].{{sfn|Duoandikoetxea|2000|loc=Chapter 3}} The multiplier of {{math|H}} is {{math|1=''σ''<sub>H</sub>(''ω'') = −''i'' sgn(''ω'')}}, where {{math|sgn}} is the [[sign function|signum function]]. Therefore:

<math display="block">\mathcal{F}\bigl(\operatorname{H}(u)\bigr)(\omega) = -i \sgn(\omega) \cdot \mathcal{F}(u)(\omega) ,</math>

where <math>\mathcal{F}</math> denotes the [[Fourier transform]]. Since {{math|1=sgn(''x'') = sgn(2{{pi}}''x'')}}, it follows that this result applies to the three common definitions of <math> \mathcal{F}</math>.

By [[Euler's formula]],
<math display="block">\sigma_\operatorname{H}(\omega) = \begin{cases}
 ~~i = e^{+i\pi/2} & \text{if } \omega < 0\\
 ~~                      0 & \text{if } \omega = 0\\
  -i = e^{-i\pi/2} & \text{if } \omega > 0
\end{cases}</math>

Therefore, {{math|H(''u'')(''t'')}} has the effect of shifting the phase of the [[negative frequency]] components of {{math|''u''(''t'')}} by +90° ({{frac|{{pi}}|2}}&nbsp;radians) and the phase of the positive frequency components by −90°, and {{math|''i''·H(''u'')(''t'')}} has the effect of restoring the positive frequency components while shifting the negative frequency ones an additional +90°, resulting in their negation (i.e., a multiplication by −1).

When the Hilbert transform is applied twice, the phase of the negative and positive frequency components of {{math|''u''(''t'')}} are respectively shifted by +180° and −180°, which are equivalent amounts. The signal is negated; i.e., {{math|1=H(H(''u'')) = −''u''}}, because

<math display="block">\left(\sigma_\operatorname{H}(\omega)\right)^2 = e^{\pm i\pi} = -1 \quad \text{for } \omega \neq 0 .</math>