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Trouble Understanding the Fourier Integral

  1. Feb 17, 2008 #1
    [SOLVED] Trouble Understanding the Fourier Integral

    I am having a bit of trouble understanding how to use the Fourier Integral Formula.

    The integral is [tex]\int_{0}^{\infty} [A_{\omega}cos(\omega x) + B_{\omega}sin(\omega x)]d\omega[/tex]


    [tex]A_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty} f(\xi)cos(\omega\xi)d\xi[/tex]
    [tex]B_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty} f(\xi)sin(\omega\xi)d\xi[/tex]

    The example in the book doesn't help because f(x) is 1 for -1<=x<=1, and 0 for |x|>1. This doesn't clear up my confusion as to what the integral is supposed to look like with [tex]f(\xi)[/tex] inserted. What is [tex]f(\xi)[/tex]? Is it the function given, f(x)? Like if I had f(x) = sin(x) on [-2, 2], would A_w look like:

    [tex]A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega\xi)d\xi[/tex],

    with sin(x) being treated as a constant that can be brought outside the integral, or am I supposed to make A_w be

    [tex]A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega x)dx[/tex]?

    This is where I am confused.
  2. jcsd
  3. Feb 17, 2008 #2
    Your book gives: f(x) = 1 for values between -1 and 1. Lets just say you wanted to integrate this function:

    eg. [tex] \int_{-\infty}^{\infty}f(x)dx [/tex]
    This integral would reduce to: [tex] \int_{-1}^{1}dx [/tex] since it is zero outside of x=-1 and x=1 and is unity otherwise.

    Note: You could also integrate this function as:
    [tex] \int_{-\infty}^{\infty}f(\xi)d\xi[/tex]
    and you would get the same thing.

    The [itex] \xi [/itex] is a dummy variable.

    For example, imagine I wanted to define a quantity that was the area of an arbitrary rectangle. Lets call this number [tex] R_{AB} [/tex] where A and B are the lengths of the rectangle respectively.

    Well I could simply define [tex] R_{AB} = R \times B[/tex] or I could also write it as:

    [tex] R_{AB} = \int_0^A \int_0^B d\xi_1 d\xi_2 [/tex] Does that make sense? Its kinda like a loop.
    Last edited: Feb 17, 2008
  4. Feb 17, 2008 #3
    I don't think I fully answered your question:

    If you were to integrate:

    [tex] \int_{-\infty}^{\infty} f(\xi)cos(\omega \xi) d\xi[/tex]

    Lets let f(x) = 1 for x= 0 to 1 and f(x)=0 otherwise. Thus, your integral would become:

    [tex] A = \int_{a}^{b} (1)cos(\omega \xi) d\xi[/tex]
    [tex] A = \frac{\sin b \omega }{\omega }-\frac{\sin a \omega }{\omega} [/tex]
  5. Feb 17, 2008 #4


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    Apparently you are confused over basic definitions. The notation [itex]f(\xi)[/itex] refers to the value that the function f takes when you substitute [itex]x = \xi[/itex], just like f(0) means: the value of f in x = 0, and f(12.4872) means: the value of f in x = 12.8472.
    So if you put that into the formula you gave, you get
    [tex]A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(\xi)cos(\omega\xi)d\xi[/tex].
  6. Feb 17, 2008 #5
    Yeah that's what I was having trouble understanding. Pretty simple actually, but it was late at night and my brain wasn't working. I don't know why I didn't see that in the first place. Thanks.
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