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<math>\begin{align} | <math>\begin{align} | ||
\mathcal{X}(\omega)&= \sum_{n=-\infty}^{\infty} 5^{-|n+2|}e^{-j\omega n}\\ | \mathcal{X}(\omega)&= \sum_{n=-\infty}^{\infty} 5^{-|n+2|}e^{-j\omega n}\\ | ||
| − | &= \sum_{n=-\infty}^{-3} 5^{(n+2)}e^{-j\omega n} | + | &= \sum_{n=-\infty}^{-3} 5^{(n+2)}e^{-j\omega n} + \sum_{n=-2}^{\infty}e^{-j\omega n} \\ |
&= 25 \sum_{n=-\infty}^{-3} \left(\frac{1}{5}e^{j\omega}\right)^{-n} + e^{2j\omega}\sum_{n=-2}^{\infty} (5e^{j\omega})^{-(n+2)} \\ | &= 25 \sum_{n=-\infty}^{-3} \left(\frac{1}{5}e^{j\omega}\right)^{-n} + e^{2j\omega}\sum_{n=-2}^{\infty} (5e^{j\omega})^{-(n+2)} \\ | ||
&= 25 \sum_{n=3}^{\infty} \left(\frac{1}{5}e^{j\omega}\right)^{n} + e^{2j\omega}\sum_{n=0}^{\infty} (5e^{j\omega})^{-n}\\ | &= 25 \sum_{n=3}^{\infty} \left(\frac{1}{5}e^{j\omega}\right)^{n} + e^{2j\omega}\sum_{n=0}^{\infty} (5e^{j\omega})^{-n}\\ | ||
Revision as of 15:27, 20 March 2011
Homework 7 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)&= \sum_{n=-\infty}^{\infty} 5^{-|n+2|}e^{-j\omega n}\\ &= \sum_{n=-\infty}^{-3} 5^{(n+2)}e^{-j\omega n} + \sum_{n=-2}^{\infty}e^{-j\omega n} \\ &= 25 \sum_{n=-\infty}^{-3} \left(\frac{1}{5}e^{j\omega}\right)^{-n} + e^{2j\omega}\sum_{n=-2}^{\infty} (5e^{j\omega})^{-(n+2)} \\ &= 25 \sum_{n=3}^{\infty} \left(\frac{1}{5}e^{j\omega}\right)^{n} + e^{2j\omega}\sum_{n=0}^{\infty} (5e^{j\omega})^{-n}\\ &= 25\cdot \frac{\left(\frac{1}{5}e^{j\omega}\right)^{3}}{1-\frac{1}{5}e^{j\omega}} + \frac{e^{2j\omega}}{1-\frac{1}{5}e^{-j\omega}}\\ &=\frac{e^{3j\omega}}{5-e^{j\omega}} + \frac{5e^{2j\omega}}{5-e^{-j\omega}} \end{align} $
To verify our answer using the table, we first write:
$ x[n]=5^{(n+2)}u[-n-3]+5^{-(n+2)}u[n+2]=\frac{1}{5}\left(\frac{1}{5}\right)^{-(n+3)}u[-(n+3)]+5^{-(n+2)}u[n+2] $.
Now, using the time reversal property (for the first term) and the time shift property of the DTFT (for both terms), we get:
$ \begin{align} \mathcal{X}(\omega)&=\frac{e^{3j\omega}}{5}\left(\frac{1}{1-\frac{1}{5}e^{j\omega}}\right)+\frac{e^{2j\omega}}{1-\frac{1}{5}e^{-j\omega}} \\ &=\frac{e^{3j\omega}}{5-e^{j\omega}}+\frac{5e^{2j\omega}}{5-e^{-j\omega}} \end{align} $.
