Editing Analytic signal (section)

==Definition==
[[File:Analytisches-signal-uebertragungsfunktion-en.svg|thumb|250px|Transfer function to create an analytic signal]]
If <math>s(t)</math> is a ''real-valued'' function with Fourier transform <math>S(f)</math> (where <math>f</math> is the real value denoting frequency), then the transform has [[Hermitian function|Hermitian]] symmetry about the <math>f = 0</math> axis:
:<math>S(-f) = S(f)^*,</math>
where <math>S(f)^*</math> is the [[complex conjugate]] of <math>S(f)</math>.
The function:
:<math>
\begin{align}
S_\mathrm{a}(f) &\triangleq
\begin{cases}
2S(f), &\text{for}\ f > 0,\\
S(f), &\text{for}\ f = 0,\\
0, &\text{for}\ f < 0
\end{cases}\\
&= \underbrace{2 \operatorname{u}(f)}_{1 + \sgn(f)}S(f) = S(f) + \sgn(f)S(f),
\end{align}
</math>
where
*<math>\operatorname{u}(f)</math> is the [[Heaviside step function]],
*<math>\sgn(f)</math> is the [[sign function]],

contains only the ''non-negative frequency'' components of <math>S(f)</math>. And the operation is reversible, due to the Hermitian symmetry of <math>S(f)</math>:
:<math> 
\begin{align}
S(f) &=
\begin{cases}
\frac{1}{2}S_\mathrm{a}(f), &\text{for}\ f > 0,\\
S_\mathrm{a}(f), &\text{for}\ f = 0,\\
\frac{1}{2}S_\mathrm{a}(-f)^*, &\text{for}\ f < 0\ \text{(Hermitian symmetry)}
\end{cases}\\
&= \frac{1}{2}[S_\mathrm{a}(f) + S_\mathrm{a}(-f)^*].
\end{align}
</math>

The '''analytic signal''' of <math>s(t)</math> is the inverse Fourier transform of <math>S_\mathrm{a}(f)</math>:
:<math>\begin{align}
s_\mathrm{a}(t) &\triangleq \mathcal{F}^{-1}[S_\mathrm{a}(f)]\\
&= \mathcal{F}^{-1}[S (f)+ \sgn(f) \cdot S(f)]\\
&= \underbrace{\mathcal{F}^{-1}\{S(f)\}}_{s(t)} + \overbrace{
\underbrace{\mathcal{F}^{-1}\{\sgn(f)\}}_{j\frac{1}{\pi t}} * \underbrace{\mathcal{F}^{-1}\{S(f)\}}_{s(t)}
}^\text{convolution}\\
&= s(t) + j\underbrace{\left[{1 \over \pi t} * s(t)\right]}_{\operatorname{\mathcal{H}}[s(t)]}\\
&= s(t) + j\hat{s}(t),
\end{align}</math>
where
*<math>\hat{s}(t) \triangleq \operatorname{\mathcal{H}}[s(t)]</math> is the [[Hilbert transform]] of <math>s(t)</math>;
*<math>*</math> is the binary [[convolution]] operator;
*<math>j</math> is the [[imaginary unit]].

Noting that <math>s(t)= s(t)*\delta(t),</math> this can also be expressed as a filtering operation that directly removes negative frequency components''':'''

:<math>s_\mathrm{a}(t) = s(t)*\underbrace{\left[\delta(t)+ j{1 \over \pi t}\right]}_{\mathcal{F}^{-1}\{2u(f)\}}.</math>

===Negative frequency components===
Since <math>s(t) = \operatorname{Re}[s_\mathrm{a}(t)]</math>, restoring the negative frequency components is a simple matter of discarding <math>\operatorname{Im}[s_\mathrm{a}(t)]</math> which may seem counter-intuitive. The complex conjugate <math>s_\mathrm{a}^*(t)</math> comprises ''only'' the negative frequency components. And therefore <math>s(t) = \operatorname{Re}[s_\mathrm{a}^*(t)]</math> restores the suppressed positive frequency components.  Another viewpoint is that the imaginary component in either case is a term that subtracts frequency components from <math>s(t).</math> The <math>\operatorname{Re}</math> operator removes the subtraction, giving the appearance of adding new components.