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The Hilbert-transform

The Hilbert transform

Though it's use is frequent in signal processing, it does have a significance in understanding tomographic image reconstruction, the Hilbert transform.
. The $ \mathfrak{H} Hilbert transform is defined as:
$\mathfrak{H}g\left (x \right )=\frac{1}{\pi}\int_{-\infty }^{\infty }\frac{g\left ( y\right )}{x-y}dy

The definition looks simple, to evaluate the integral looks a lot harder as the denominator harbours a singularity. It means we have to take the integral in a Cauchy principal value, so
$ \mathfrak{H}g\left ( x \right )=\underset{\varepsilon \rightarrow 0}{\lim} \frac{1}{\pi}\int_{-\infty }^{t-\varepsilon  }\frac{g\left ( y \right )}{x-y}dy+\int_{t+\varepsilon  }^{\infty }\frac{g\left ( y \right )}{x-y}dy
Existence of such a limit can easily imagined, as for function g=1 the function 1/x can be integrated in Principa Value, being odd the result is 0, since the range of the integral is symmetric.

To ease the evaluation of the integral

  • note that the Hilbet transfrom is a convolution with function 1/x
  • Fourier transform then inverse Fourier transform the expression

$\mathfrak{H}g\left ( x \right )=g\left ( x \right )*\frac{1}{\pi x}=\mathfrak{F}^{^{-1}}\mathfrak{F}\left \{ g\left ( x \right )*\frac{1}{\pi x} \right \}=\mathfrak{F}^{^{-1}}\left \{ \mathfrak{F}\left [ g\left ( x \right ) \right ] \mathfrak{F}\left [ \frac{1}{\pi x} \right]\right \}
Let us evaluate the Fourier transform of 1/x:
$\mathfrak{F}\left [ \frac{1}{x} \right ]=-\frac{1}{\pi}\int_{-\infty }^{\infty }\frac{e^{-i k x}}{x}dx=-\frac{1}{\pi}\int_{-\infty }^{\infty }\frac{\cos( kx)-i\sin\left ( kx \right )}{x}dx

The function cos is even, divided by the odd "x" function the first term is again zero.
We have the next two terms remaining:
$ -\frac{2i}{\pi}\int_{0}^{\infty }\frac{\sin\left (  kx \right )}{x}dx \mid k< 0
$ \frac{2i}{\pi}\int_{0}^{\infty }\frac{\sin\left (  kx \right )}{x}dx \mid k>0

$ \int_{0}^{\infty }\frac{\sin\left (  kx \right )}{x}dx=\int_{0 }^{\infty }sinc\left (  kz \right )}dz=\frac{\pi}{2}

The result is, then:

$\mathfrak{F}\left [ \frac{1}{x} \right ]=i sgn(k)
where sgn is the sign function.
$\mathfrak{H}g\left ( x \right )=\mathfrak{F}^{^{-1}}\left \{ \mathfrak{F}\left [ g\left ( x \right ) \right ] i sgn(k)\right \}

This result, regarding the numerical evaluation technique is a lot simpler then the application of the basic definition, since the digital Fourier transform and its implementation technique (FFT) is a routinely applied, accessible and fast.

Our results also shows, that if we apply the Hilbert transform twice on the same function we obtain:
$\mathfrak{H}\mathfrak{H}g\left ( x \right )=\mathfrak{F}^{^{-1}}\left [\left \mathfrak{F}[\mathfrak{F}^{^{-1}}\left \{ \mathfrak{F}\left [ g\left ( x \right ) \right ] i sgn(k)\right \}  \right ]i sgn(k)  \right ]=-g\left ( x \right )

thus the inverse of the Hilbert-transfrom is -apart from a sign- is itself.
As an illustration we have prepared the 2D Hilbert transform of an image:

Image Image


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