Line 9: Line 9:
 
&= \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\[e^{-(j\omega +3)t}\right\]^{\infty}_0 \\
+
&= \frac{1}{-j\omega +3} \left[e^{(-j\omega +3)t}\right]_{-\infty}^0 - \frac{1}{j\omega +3} \left\[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} \\
 
&= \frac{j\omega +3}{\omega^2 +9}+\frac{-j\omega +3}{\omega^2 + 9} \\
 
&= \frac{j\omega +3}{\omega^2 +9}+\frac{-j\omega +3}{\omega^2 + 9} \\

Revision as of 13:51, 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} $


HW4

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Ryne Rayburn