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FFT of the Given Signal

We assume for now that $w(n)$ is a rectangular windowing function from $-\ensuremath{\frac{N-1}{2}}$ to $\ensuremath{\frac{N-1}{2}}$. (We will consider other windowing functions in a section below.) The DFT of our rectangular windowed $y(n)$ may be written

$\displaystyle Y(k)$ $\textstyle =$ $\displaystyle \sum_{n=-\ensuremath{\frac{N-1}{2}}}^{\ensuremath{\frac{N-1}{2}}}e^{i \alpha n^2}
e^{-i2\pi k n/K }$ (18)
  $\textstyle =$ $\displaystyle \sum_{n=-\ensuremath{\frac{N-1}{2}}}^{\ensuremath{\frac{N-1}{2}}}\exp(i( \alpha n^2
-2\pi k n/ K)).$ (19)

where $K$ is the length of the optionally zero padded transform. (If no zero padding is used, $K=N$.) Using Euler's identity ( $e^{i\theta}=\cos(\theta)+i\sin(\theta)$), it is easy to see that the real and imaginary parts of $Y(k)$ are, respectively,
$\displaystyle \Re{Y(k)}$ $\textstyle =$ $\displaystyle \sum_{n=-(N-1)/2}^{(N-1)/2}\cos( \alpha n^2
-2\pi k n/ K)$ (20)
$\displaystyle \Im{Y(k)}$ $\textstyle =$ $\displaystyle \sum_{n=-(N-1)/2}^{(N-1)/2}\sin( \alpha n^2
-2\pi k n/ K).$ (21)



Aaron S. Master 2002-10-17