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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 | + | <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 {e^{j2t} - 1}{jt\pi} - \frac {1 - e^{-j2t}}{jt\pi} </math> | ||
| + | |||
| + | <math> \Rightarrow ~~~~~~~= \left ( \frac {e^{j2t} + e^{-j2t} - 2}{jt\pi} \right ) </math> | ||
| + | |||
| + | <math> \Rightarrow ~~~~~~~= \left ( \frac {2\cos(2t) -2}{jt\pi} \right ) </math> | ||
| + | |||
| + | <math> \Rightarrow ~~~~~~~= \frac {-(4j\sin^2(t))}{t\pi} </math> | ||
Latest revision as of 19:13, 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 $
$ \Rightarrow ~~~~~~~= \frac {e^{j2t} - 1}{jt\pi} - \frac {1 - e^{-j2t}}{jt\pi} $
$ \Rightarrow ~~~~~~~= \left ( \frac {e^{j2t} + e^{-j2t} - 2}{jt\pi} \right ) $
$ \Rightarrow ~~~~~~~= \left ( \frac {2\cos(2t) -2}{jt\pi} \right ) $
$ \Rightarrow ~~~~~~~= \frac {-(4j\sin^2(t))}{t\pi} $
