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<math> \therefore x_2(t) = \frac {1}{2\pi} \int_{-\infty}^{\infty} X_2(j\omega) e^{jt\omega}\,d\omega </math>
 
<math> \therefore x_2(t) = \frac {1}{2\pi} \int_{-\infty}^{\infty} X_2(j\omega) e^{jt\omega}\,d\omega </math>
  
<math> \Rightarrow = \frac {1}{2\pi} \int_{0}^{2} 2 e^{jt\omega}\,d\omega + \frac {1}{2\pi} \int_{-2}^{0} (-2) e^{jt\omega}\,d\omega </math>
+
<math> \Rightarrow ~~~~~= \frac {1}{2\pi} \int_{0}^{2} 2 e^{jt\omega}\,d\omega + \frac {1}{2\pi} \int_{-2}^{0} (-2) e^{jt\omega}\,d\omega </math>

Revision as of 19:58, 1 July 2008

Solution to Prob 4.4b

Its given that X(jw) =

$ ~2, ~~0 \le \omega \le 2 $

$ -2, ~~-2 \le \omega < 0 $

$ ~0, ~~|\omega| > 2 $

$ \therefore x_2(t) = \frac {1}{2\pi} \int_{-\infty}^{\infty} X_2(j\omega) e^{jt\omega}\,d\omega $

$ \Rightarrow ~~~~~= \frac {1}{2\pi} \int_{0}^{2} 2 e^{jt\omega}\,d\omega + \frac {1}{2\pi} \int_{-2}^{0} (-2) e^{jt\omega}\,d\omega $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva