Homework 7 Solutions, ECE301 Spring 2011 Prof. Boutin

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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}5^{-(n+2)}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)]+\left(\frac{1}{5}\right)^{(n+2)}u[n+2] $.

Using the time reversal property (for the first term), the time shift property (for both terms), the appropriate pair from the table, and the linearity of the FT, 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} $.

Question 2

We choose the period from $ (-\pi,\pi) $ (which corresponds to the part of the signal for k=0) to compute the inverse DT Fourier transform of the given signal:

$ \begin{align} x[n]&=\frac{1}{2\pi}\int_{-\pi}^{\pi} \mathcal{X}(\omega)e^{j\omega n}d\omega \\ &=\frac{1}{2\pi}\int_{-\pi}^{\pi} \left[ \delta (\omega) +\pi\delta\left(\omega -\frac{\pi}{2}\right)+\pi\delta\left(\omega +\frac{\pi}{2}\right) \right] d\omega \\ &=\frac{1}{2\pi}\int_{-\pi}^{\pi} \delta (\omega) d\omega +\frac{1}{2}\int_{-\pi}^{\pi}\delta\left(\omega -\frac{\pi}{2}\right) d\omega+\frac{1}{2}\int_{-\pi}^{\pi}\delta\left(\omega +\frac{\pi}{2}\right) d\omega \\ &=\frac{1}{2\pi} + \frac{1}{2}e^{-j\frac{\pi}{2}n}+\frac{1}{2}e^{j\frac{\pi}{2}n} \\ &=\frac{1}{2\pi} + \cos\left(\frac{\pi}{2}n\right) \end{align} $

Question 3

a) First, we find the FT of $ x[n] $ and $ h[n] $: $ \mathcal{X}(\omega)=\frac{1}{1-\frac{3}{4}e^{-j\omega}} $ $ \mathcal{H}(\omega)=\frac{1}{1-\frac{1}{2}e^{-j\omega}} $

Then,

$ \mathcal{Y}(\omega)=\mathcal{H}(\omega)\mathcal{X}(\omega)=\frac{8}{(4-3e^{-j\omega})(2-e^{-j\omega})} $

Now, we need to write out the partial fraction expansion.

$ \frac{1}{(4-3e^{-j\omega})(2-e^{-j\omega})}=\frac{A}{4-3e^{-j\omega}}+\frac{B}{2-e^{-j\omega}} $

Multiplying both sides by $ (4-3e^{-j\omega})(2-e^{-j\omega}) $, we get:

$ A(2-e^{-j\omega})+B(4-3e^{-j\omega})=1 $

Comparing both sides and solving, we have that:

$ A=\frac{3}{2} $, $ B=-\frac{1}{2} $.

Then, we have that:

$ \mathcal{Y}(\omega)=\frac{12}{4-3e^{-j\omega}}-\frac{4}{2-e^{-j\omega}}=\frac{3}{1-\frac{3}{4}e^{-j\omega}}-\frac{2}{1-\frac{1}{2}e^{-j\omega}} $

Using DTFT pairs, we get:

$ y[n]=3\left(\frac{3}{4}\right)^nu[n]-2\left(\frac{1}{2}\right)^nu[n]=\left[3\left(\frac{3}{4}\right)^n-2\left(\frac{1}{2}\right)^n\right]u[n] $