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Math Help - Fourier Transform

  1. #1
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    Fourier Transform

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    I was asked to find the fourier transform of x(n). Could someone explain whats happening in this answer?

    Thanks
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  2. #2
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    Quote Originally Posted by hammer View Post
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    I was asked to find the fourier transform of x(n). Could someone explain whats happening in this answer?

    Thanks
    I presume the problem is in simplifying X(\omega)?

    X(\omega) = \sum_{n = -4}^{\infty} \left ( \frac{1}{4} \right )^ne^{-j \omega n}

    Define m = n + 4. Then
    X(\omega) = \sum_{n = -4}^{\infty} \left ( \frac{1}{4} \right )^ne^{-j \omega n} = \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^{m - 4}e^{-j \omega (m - 4)}

    = \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m \cdot \left ( \frac{1}{4} \right )^{-4}e^{-j \omega m }e^{-(-4)j \omega }

    So
    X(\omega) = \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m e^{-j \omega m }4^4e^{4j \omega }

    Now, note that the last two factors are independent of m:
    X(\omega) = 4^4e^{4j \omega } \cdot \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m e^{-j \omega m }

    Note also that the summation is the summation of an infinite geometric series. The sum of any geometric series is
    \sum_{m = 0}^{\infty} ar^m = \frac{a}{1 - r}

    Thus
    \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m e^{-j \omega m } = \sum_{m = 0}^{\infty} \left ( \frac{e^{-j \omega}}{4} \right )^m = \frac{1}{1 - \frac{e^{-j \omega}}{4}}

    So finally we get:
    X(\omega) = 4^4e^{4j \omega } \cdot \frac{1}{1 - \frac{e^{-j \omega}}{4}}

    X(\omega) = \frac{4^4e^{4j \omega }}{1 - \frac{e^{-j \omega}}{4}}

    -Dan
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    Quote Originally Posted by hammer View Post
    ImageShack - Hosting :: questionyw0.jpg

    I was asked to find the fourier transform of x(n). Could someone explain whats happening in this answer?

    Thanks
    This is a Discrete Fourier Transform, X(\omega) (on the second line) is the transform of x(n)
    by definition of the DFT (and that u(n)=0 for n<0, and 1 otherwise).

    The rest is just simplification using the sum of a geometric series formula.

    ZB
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