Line 1: Line 1:
<math>\mathcal{X}(\omega) = \frac{\frac{1}{2j}}{(2 - j4 + jw)^{2}} - \frac{\frac{1}{2j}}{(2 + j4 - jw)^{2}}</math>
+
<math>\mathcal{X}(\omega) = \left\{ {\begin{array}{*{20}c}
 +
  {1,} & {-2 \le  \omega \le 0} \\
 +
  { -1,} & {0 \le g(x) \ge 2} \\
 +
  {0,} & {| \omega| > 2} \\
 +
\end{array}} \right.</math>
  
<math>\ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}\,d\omega </math>
 
  
<math>= \frac{1}{2\pi}\int_{-\infty}^{\infty}(\frac{\frac{1}{2j}}{(2 - j4 + jw)^{2}} - \frac{\frac{1}{2j}}{(2 + j4 - jw)^{2}})e^{j\omega t}\,d\omega</math>
+
<math>\ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}\,d\omega </math>
  
by looking at the table on p. 329 of the book some observations can be made:
+
<math>\ x(t)=\frac{1}{2\pi}\int_{-2}^{0}e^{j\omega t}\,d\omega + \frac{1}{2\pi}\int_{0}^{2}-e^{j\omega t}\,d\omega </math>
  
the form <math>\frac{1}{(2 - j4 + jw)^{2}}</math> means that there is a term of the form <math>\ te^{-2t}</math>
+
<math>x(t)=\frac{1}{2\pi}(\frac{1}{jt} - \frac{e^{-j 2 t}}{jt}) + \frac{1}{2\pi}(-\frac{e^{j 2 t}}{jt} + \frac{1}{jt}) </math>

Revision as of 15:20, 8 October 2008

$ \mathcal{X}(\omega) = \left\{ {\begin{array}{*{20}c} {1,} & {-2 \le \omega \le 0} \\ { -1,} & {0 \le g(x) \ge 2} \\ {0,} & {| \omega| > 2} \\ \end{array}} \right. $


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

$ \ x(t)=\frac{1}{2\pi}\int_{-2}^{0}e^{j\omega t}\,d\omega + \frac{1}{2\pi}\int_{0}^{2}-e^{j\omega t}\,d\omega $

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

Alumni Liaison

Followed her dream after having raised her family.

Ruth Enoch, PhD Mathematics