| Line 12: | Line 12: | ||
&= \frac{1}{-j\omega +3} \left\[1-0 \right\] - \frac{1}{j\omega +3} \left\[0-1\right\] \\ | &= \frac{1}{-j\omega +3} \left\[1-0 \right\] - \frac{1}{j\omega +3} \left\[0-1\right\] \\ | ||
&= \frac{1}{-j\omega +3} + \frac{1}{j\omega +3} \\ | &= \frac{1}{-j\omega +3} + \frac{1}{j\omega +3} \\ | ||
| − | &= \frac{j\omega +3}{\ | + | &= \frac{j\omega +3}{\omega^2 +9}+\frac{-j\omega +3}{\omega^2 + 9} \\ |
&= \frac{6}{\omega^2+9} | &= \frac{6}{\omega^2+9} | ||
\end{align} | \end{align} | ||
Revision as of 14:48, 2 March 2011
Homework 5 Solutions, ECE301 Spring 2011 Prof. Boutin
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Question 1
$ \begin{align} \mathcal{X}(\omega)&= \int_{-\infty}^{infty} e^{-3|t|}e^{-j\omega t} dt \\ &= \int_{-\infty}^0 e^{3t}e^{-j\omega t} dt + \int_0^{\infty} e^{-3t}e^{-j\omega t} dt \\ &= \int_{-\infty}^0 e^{(-j\omega +3)t} dt + \int_0^{\infty} e^{-(j\omega +3)t} dt \\ &= \frac{1}{-j\omega +3} \left\[e^{(-j\omega +3)t}\right\]_{-\infty}^0 - \frac{1}{j\omega +3} \left\[e^{-(j\omega +3)t}\right\]^{\infty}_0 \\ &= \frac{1}{-j\omega +3} \left\[1-0 \right\] - \frac{1}{j\omega +3} \left\[0-1\right\] \\ &= \frac{1}{-j\omega +3} + \frac{1}{j\omega +3} \\ &= \frac{j\omega +3}{\omega^2 +9}+\frac{-j\omega +3}{\omega^2 + 9} \\ &= \frac{6}{\omega^2+9} \end{align} $
