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Math Help - Fourier transform

  1. #1
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    Fourier transform

    Hi there. I'm starting with the Fourier transforms, and I'm having some trouble with my first exercise on this topic.
    The problem says: Given f(x)=H(x)-H(x-l) (H(x) is the Heaviside unit step function).
    a) Consider the odd extension for f and find its Fourier integral representation.
    b) Using the previous incise calculate the value for \int_0^{\infty}\frac{1-cos (\omega l)}{\omega}\sin (\omega l) d\omega

    Well, for a) I think I should get the fourier transfor for the sign function, I don't know if this is right, but anyway I've defined the function like this:
    g(x)=\begin{Bmatrix}{ 0}&\mbox{ if }& x>l\\1 & \mbox{if}& 0<x<l\\-1 & \mbox{if}& -l<x<0\\0 & \mbox{if}& x<-l\end{matrix}

    Then the Fourier integral representation:
    g(x)=\displaystyle\frac{1}{2\pi}\displaystyle\int_  {-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}g(x)e^{-i \omega x}dx e^{i \omega x}d\omega

    But g(x) is odd, then:

    g(x)=\displaystyle\frac{1}{2\pi}\displaystyle\int_  {-\infty}^{\infty} 2 \displaystyle\int_{0}^{\infty}g(x)(-i)\sin(\omega x)}dx e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{0}^{\infty}g(x)  \sin(\omega x)}dx e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty} \left (  \displaystyle\int_{0}^{\infty} \sin(\omega x)}dx -  \displaystyle\int_{-\infty}^{0} \sin(\omega x)}dx\right ) e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty} \left ( \cos(\omega x)} |_0^{\infty} -   \cos(\omega x)}|_{-\infty}^{0} \right ) e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty} \left ( \frac{2}{\omega}-\displaystyle\lim_{c \to{+}\infty}{}\cos{\omega c}-\displaystyle\lim_{c \to{-}\infty}{}\cos{\omega c}}\right ) e^{i \omega x}d\omega

    And the thing is that the integral for the cosine diverges as I see it, but I'm probably doing something wrong.

    Help please
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Ulysses View Post
    Hi there. I'm starting with the Fourier transforms, and I'm having some trouble with my first exercise on this topic.
    The problem says: Given f(x)=H(x)-H(x-l) (H(x) is the Heaviside unit step function).
    a) Consider the odd extension for f and find its Fourier integral representation.
    b) Using the previous incise calculate the value for \int_0^{\infty}\frac{1-cos (\omega l)}{\omega}\sin (\omega l) d\omega

    Well, for a) I think I should get the fourier transfor for the sign function, I don't know if this is right, but anyway I've defined the function like this:
    g(x)=\begin{Bmatrix}{ 0}&\mbox{ if }& x>l\\1 & \mbox{if}& 0<x<l\\-1 & \mbox{if}& -l<x<0\\0 & \mbox{if}& x<-l\end{matrix}
    Assuming that l>0 you should have:

    g(x)= \left\{ \begin{array}{ll}1,&0\le x<l \\ 0,&\text{otherwise}\end{array} \right.

    What you give is the odd extension (more of less).

    Then the Fourier integral representation:
    g(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(x)e^{-i \omega x}dx e^{i \omega x}d\omega

    But g(x) is odd, then:

    g(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty} 2 \int_{0}^{\infty}g(x)(-i)\sin(\omega x)}dx e^{i \omega x}d\omega
    OK to this point

    .................. =\frac{-i}{\pi}\int_{-\infty}^{\infty}\int_{0}^{\infty}g(x)\sin(\omega x)}dx e^{i \omega x}d\omega
    You have lost the 2, and this last integral becomes (putting the 2 back in) OK found it below now fixed:

    .................. =\frac{-i}{\pi}\int_{-\infty}^{\infty}\int_{0}^{l}\sin(\omega x)}dx e^{i \omega x}d\omega
    Last edited by CaptainBlack; June 11th 2011 at 10:52 AM.
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  3. #3
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    I canceled it out with the 2 in the denominator, are you sure that's wrong? I've found one mistake, it's pretty silly. I was making for -infinity to zero, but I shouldn't use that part. I see now, I shouldn't take from zero to infinity, I must use l. Thanks
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Ulysses View Post
    I canceled it out with the 2 in the denominator, are you sure that's wrong? I've found one mistake, it's pretty silly. I was making for -infinity to zero, but I shouldn't use that part. I see now, I shouldn't take from zero to infinity, I must use l. Thanks
    You are correct about the 2.

    CB
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