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Fourier transform

  1. Sep 25, 2014 #1
    Whais is the effect of a multiplication by (-1)^n in the DTFT ??

    In other words, whats is this transform : x(n)* (-1)^n ??
     
  2. jcsd
  3. Sep 25, 2014 #2

    jbunniii

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    If ##X(\omega) = \sum_{n=-\infty}^{\infty} x(n) e^{-i \omega n}##, then
    $$\begin{align}
    \sum_{n=-\infty}^{\infty} x(n) (-1)^n e^{-i \omega n}
    &= \sum_{n=-\infty}^{\infty} x(n) e^{-i \pi n} e^{-i \omega n} \\
    &= \sum_{n=-\infty}^{\infty} x(n) e^{-i (\omega + \pi) n} \\
    &= X(\omega + \pi)
    \end{align}$$
    Note that this is also equal to ##X(\omega - \pi)## due to the ##2\pi##-periodicity of the discrete-time Fourier transform.
     
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