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Math Help - Request for Explanation of Fourier Transform

  1. #1
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    Request for Explanation of Fourier Transform

    The introduction to the Laplace transform chapter in my DE textbook says (I'm paraphrasing some of this):

    In general, integral transforms address the question: How much is a given function y(t) "like" a particular standard function?

    For example, if y(t) represents a radio signal and we want to compare it to sin \omega t with frequency \frace \omega {2 \pi}, just integrate:

    \int_{-N}^N y(t)\; sin (\omega t)\; dt.

    If y(t) goes up when sin(\omega t) goes up, and goes down when sin(\omega t) goes down, then the integral will be large, because y(t) is very much "like" sin(\omega t). So far, this makes perfect sense.

    Then, it says you can compare y(t) to e^{-zt}, where z=s+i \omega is Complex, using:
    \int_{-N}^N y(t)\; e^{-zt}\; dt

    Which can be split into

    \int_{-N}^N y(t)\; e^{-st}\; dt\,\, and \mathbf{\int_{-N}^N y(t)\; e^{-i \omega t}\; dt }.

    The last (bold) of those integrals is called the Fourier Transform (which is what I really need to learn). Then the book says:

    The Fourier Transform's value at a particular value of \omega is a measure of the extent to which y(t) oscillates with a frequency of \frac \omega {2 \pi}.
    Can someone explain how that integral compares y(t) to an oscillation of \frac \omega {2 \pi}?

    I am aware of Euler's equation:
    e^{it}=cos(t)+i \cdot sin(t)

    If I apply it to just the Fourier integral, I get:

    \int_{-N}^N y(t)\; e^{-i \omega t}\; dt\, = \int_{-N}^N y(t)\; \cos (\omega t)\; dt + i \int_{-N}^N y(t)\; \sin (\omega t)\; dt.

    Unless the \sin( \omega t) is always zero, I don't see how this can make that comparison.

    Thanks for your explanations.
    Jeff
    Last edited by MSUMathStdnt; August 7th 2011 at 12:03 PM. Reason: screwed up the title
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  2. #2
    A Plied Mathematician
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    Re: Request for Explanation of Fourier Transform

    Off-hand, I would say that e^{-i\omega t} oscillates around the unit circle in the complex plane. So that's the oscillation being described. Does that answer your question?
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  3. #3
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    Re: Request for Explanation of Fourier Transform

    Quote Originally Posted by MSUMathStdnt View Post
    ....

    If I apply it to just the Fourier integral, I get:

    \int_{-N}^N y(t)\; e^{-i \omega t}\; dt\, = \int_{-N}^N y(t)\; \cos (\omega t)\; dt + i \int_{-N}^N y(t)\; \sin (\omega t)\; dt.

    Unless the \sin( \omega t) is always zero, I don't see how this can make that comparison.

    Thanks for your explanations.
    Jeff
    You want to know how like a sinusoid of arbitary phase your signal is. To do that you need to first find the in phase and quadrature components which are the two terms you have above.

    CB
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