Revision as of 17:23, 8 October 2008 by Chen18 (Talk)

Specify a signal x(t)

$ x(t)=cos(8 \pi t)e^{-t^{2}} $

Fourier Transform of x(t)

$ \begin{align} X(\omega) &=\int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty} cos(8 \pi t)e^{-t^{2}}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t}-e^{-j8\pi t}}{2}e^{-t^{2}}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t-t^2}-e^{-j8\pi t-t^2}}{2}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{t(j8\pi -t)}-e^{-t(j8\pi +t)}}{2}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{(t-4\pi j)^2+16\pi ^2}-e^{-(t-4\pi j)^2-16\pi ^2}}{2}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{(t-4\pi j)^2+16\pi ^2}}{2}e^{-j\omega t}dt -\int_{-\infty}^{\infty}\frac{e^{-(t-4\pi j)^2-16\pi ^2}}{2}e^{-j\omega t}dt \end{align} $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang