| Line 6: | Line 6: | ||
<math>\begin{align} | <math>\begin{align} | ||
| − | \mathcal{X}(\omega)&= \int_{-\infty}^{infty} e^{-3|t|}e^{-j\omega t} dt \\ | + | \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^{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 \\ | &= \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 | + | &= \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} \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} \\ | ||
Revision as of 14:54, 2 March 2011
Homework 5 Solutions, ECE301 Spring 2011 Prof. Boutin
Students should feel free to make comments/corrections or ask questions directly on this page.
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} $
