Revision as of 14:10, 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} $

To verify our answer using the table, we first write:

$ x(t)=e^{-3|t|}=e^{3t}u(-t)+e^{-3t}u(t) $.

From the table:

$ \mathfrak{F}\left\{e^{-3t}u(t)\right\}=\frac{1}{j\omega+3} $.

Using time-reversal property from the table:

$ \mathfrak{F}\left\{e^{3t}u(-t)\right\}=\frac{1}{-j\omega+3} $.

Now, since the Fourier transform (FT) is linear, we have that:

$ \begin{align} \mathcal{X}(\omega)&=\mathfrak{F}\left\{e^{-3t}u(t)\right\}+\mathfrak{F}\left\{e^{3t}u(-t)\right\} \\ &=\frac{1}{j\omega+3}+\frac{1}{-j\omega+3} \\ &=\frac{6}{\omega^2+9} \end{align} $.

HW5

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