FT2

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Find the Fourier transform of f(t) = \begin{cases}1&|t|<1\\0&|t|>1\end{cases}\,

F(\omega) = \int_{-1}^1 e^{-i\omega t}\,dt = \frac{e^{-i\omega t}}{-i\omega}\Bigg|_{-1}^1 = \frac{2}{\omega} \cdot \frac{e^{i\omega}-e^{-i\omega}}{2i} = \frac{2\sin(\omega)}{\omega},\,\,\,\omega\ne 0\,

When \omega=0\,, e^{-i\omega t}=1\,

In this case,

F(0) = \lim_{\omega\rightarrow 0} F(\omega) = 2\,.

Main Page : Partial Differential Equations : Fourier Transforms

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