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Thread: Hilbert transform problem

  1. #1
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    Hilbert transform problem

    ok , my bad , here is the post again ..


    i was trying to come with an easier way of finding the Hilbert transform of a function $\displaystyle \ x(t)$ , and here is what i did :

    starting with Fourier transform of x :

    $\displaystyle X(f)=\int^{\infty}_{-\infty}\ x(t) \ e^ {\ -i2\pi ft} \ dt$

    now

    $\displaystyle \ e^ {\ -i2\pi ft} = \frac{1}{2\pi i}\oint\frac{e^ {\ -i2\pi
    fz}}{z-t} \ dz$

    where the contour encloses t .

    "this is done by cauchy's integral formula"

    =>

    $\displaystyle X(f) = \frac{1}{2\pi i}\int^{\infty}_{-\infty}\ x(t) \oint\frac{e^ {\ -i2\pi fz}}{z-t} \ dz \
    dt = \frac{1}{2\pi i} \oint\ e^ {\ -i2\pi fz}\int^{\infty}_{-\infty}\frac{x(t)}{z-t}\ dt \ dz$

    now

    $\displaystyle \pi\hat{x}(z)=\int^{\infty}_{-\infty}\frac{x(t)}{z-t}\ dt$

    where

    $\displaystyle \hat{x}(z)$ is the Hilbert transform of x .

    =>

    $\displaystyle
    \ X(f)= \frac{1}{2i}\oint\hat{x}(s)\ e^ {\ -i2\pi fs}\ ds$

    "z is replaced with s for conventional reasons"

    now the program is to relate the last integral to the inverse laplace (or fourier !!) transform of$\displaystyle \hat{x}(s)$, by choosing the suitable contour(s) , and here is where i'm stuck !! so any help is appreciated .
    Last edited by mmzaj; May 17th 2008 at 04:28 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    You are going to have to type these out on your own or find a better way to copy and paste them. There is a good intro thread to learn basic LaTeX here.

    -Dan
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    i think it's readable now !!
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  4. #4
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    Quote Originally Posted by mmzaj View Post
    ok , my bad , here is the post again ..


    i was trying to come with an easier way of finding the Hilbert transform of a function $\displaystyle \ x(t)$ , and here is what i did :

    starting with Fourier transform of x :

    $\displaystyle X(f)=\int^{\infty}_{-\infty}\ x(t) \ e^ {\ -i2\pi ft} \ dt$

    now

    $\displaystyle \ e^ {\ -i2\pi ft} = \frac{1}{2\pi i}\oint\frac{e^ {\ -i2\pi
    fz}}{z-t} \ dz$

    where the contour encloses t .

    "this is done by cauchy's integral formula"

    =>

    $\displaystyle X(f) = \frac{1}{2\pi i}\int^{\infty}_{-\infty}\ x(t) \oint\frac{e^ {\ -i2\pi fz}}{z-t} \ dz \
    dt = \frac{1}{2\pi i} \oint\ e^ {\ -i2\pi fz}\int^{\infty}_{-\infty}\frac{x(t)}{z-t}\ dt \ dz$

    now

    $\displaystyle \pi\hat{x}(z)=\int^{\infty}_{-\infty}\frac{x(t)}{z-t}\ dt$

    where

    $\displaystyle \hat{x}(z)$ is the Hilbert transform of x .

    =>

    $\displaystyle
    \ X(f)= \frac{1}{2i}\oint\hat{x}(s)\ e^ {\ -i2\pi fs}\ ds$

    "z is replaced with s for conventional reasons"

    now the program is to relate the last integral to the inverse laplace (or fourier !!) transform of$\displaystyle \hat{x}(s)$, by choosing the suitable contour(s) , and here is where i'm stuck !! so any help is appreciated .
    Here's an easier way:

    1. You know that $\displaystyle \hat{x}(z) = x(z) * \frac{1}{\pi z}$ where * denotes the convolution operator, right?

    2. $\displaystyle FT\left[ \frac{1}{\pi z} \right] = - i \, \text{sgn} (f)$ where sgn is the signum function (see PlanetMath: signum function).

    3. From the multiplication to convolution theorem: $\displaystyle FT\left[ \hat{x}(z) \right] = - i \, X(f) \, \text{sgn} (f)$.

    4. Therefore: $\displaystyle \hat{x}(z) = -i \, FT^{-1} \left[X(f) \, \text{sgn} (f) \right]$.
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  5. #5
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    thanx a lot , that really helped .
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