Difference between revisions of "HW5.3 Xujun Huang ECE301Fall2008mboutin" - Rhea

Revision as of 08:20, 8 October 2008

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

Then

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

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

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

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

@@ Line 8: / Line 8: @@
 <math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}cos(t)e^{j\omega t}d\omega</math>
-<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{j\omega t}d\omega</math>
+<math>x(t)=\frac{1}{2\pi}cos (t)\int_{-\infty}^{\infty}e^{j\omega t}d\omega</math>
+<math>x(t)=\frac{1}{2\pi}cos (t){\left.\frac{e^{j\omega t}}{j(t}\right]_{-\infty}^{\infty}}</math>
-----
-<math>x(t) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{-j\omega t}dt</math>
-<math>X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt)</math>
-<math>X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt)</math>
-<math>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}}</math>
-<math>X(\omega)={\left.\frac{(1+\omega)e^{jt(1-\omega)}-(1-\omega)e^{-jt(1+\omega)}}{j(1-\omega^2)}\right]_{-\infty}^{\infty}}</math>
-<math>X(\omega)={\left.\frac{2e^{-\omega}(1+\omega)cos(t)}{j(1-\omega^2)}\right]_{-\infty}^{\infty}}</math>
-<math>X(\omega)=\frac{(1+\omega)2e^{-\omega}}{j(1-\omega^2)}{\left.cos(t)\right]_{-\infty}^{\infty}}</math>
-<math>X(\omega)=\frac{(1+\omega)2e^{-\omega}}{j(1-\omega^2)}{\left.cos(t)\right]_{-\pi}^{\pi}}</math>
-<math>X(\omega)=0</math>