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(1-\frac{2 |t|}{\tau})p_\tau (t) \frac{\tau}{2} sinc^2 \frac{\tau \omega}{4 \pi} \frac{\tau}{2} sinc^2 ( \frac{\tau t}{4 \pi} ) 2 \pi (1-\frac{2|\omega|}{\tau})p_\tau (\omega)
Time Domain Fourier Domain
$ x(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} X(j \omega)e^{j \omega t}d \omega $ $ X(j \omega)=\int_{-\infty}^\infty x(t) e^{-j \omega t}d t $
$ 1\ $ $ 2 \pi \delta (\omega) $
$ − 0.5 + u(t)\ $ $ \frac{1}{j \omega}\ $
$ \delta (t) \ $ $ 1\ $
$ \delta (t-c)\ $ $ e − j \omega c $
$ u(t) $ $ \pi \delta(\omega)+\frac{1}{j \omega} $
$ e ^{− bt}u(t) $ $ \frac{1}{j \omega + b} $
$ cos \omega_0 t $ $ \pi [\delta ( \omega + \omega_0 ) + \delta ( \omega - \omega_0 )] $
cos(ω0t + θ) π[e − jθδ(ω + ω0) + ejθδ(ω − ω0)]?
sinω0t jπ[δ(ω + ω0) − δ(ω − ω0)]?
sin(ω0t + θ) jπ[e − jθδ(ω + ω0) − ejθδ(ω − ω0)]?
rect(\frac{t}{\tau}) \tau sinc \frac{\tau \omega}{2 \pi}
\tau sinc \frac{\tau t}{2 \pi} 2πpτ(ω)
   Note: sinc(x) = sin(x) / x ; pτ(t) is the rectangular pulse function of width τ

Note: Source courtesy Wikibooks.org

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva