Revision as of 08:06, 8 October 2008 by Huang122 (Talk)

Let x(t)= $ cos(t) $


Then

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

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

$ X(\omega) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{-j\omega t}dt $

$ X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt) $

$ X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt) $

$ X(\omega)={\left. \frac{e^{jt(1-\omega)}}{j(1-\omega)}\right]_{-\infty}^{\infty}} + {\left. \frac{e^{-jt(1+\omega)}}{-j(1+\omega)}\right]_{-\infty}^{\infty}} $

$ X(\omega)={\left.\frac{(1+\omega)e^{jt(1-\omega)}-(1-\omega)e^{-jt(1+\omega)}}{j(1-\omega^2)}\right]_{-\infty}^{\infty}} $

$ X(\omega)={\left.\frac{2e^{-\omega}(1+\omega)cos(t)}{j(1-\omega^2)}\right]_{-\infty}^{\infty}} $

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