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Fourier Transform

$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $

$ x(t)=te^{-6t-6}u(t-6)\,\ $

$ X(\omega)=\int_{-\infty}^{\infty}t^2 u(t-1) e^{-j\omega t}dt \; = \int_{1}^{\infty}t^2 e^{-j\omega t}dt $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010