Revision as of 07:15, 8 October 2008 by Huang122 (Talk)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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


Then

$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega $

$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}cos(t)e^{j\omega t}d\omega $

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





$ x(t) = \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}} $

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

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

$ X(\omega)=0 $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett