Homework 1, ECE438, Fall 2011, Prof. Boutin

Question 1-5

(1) $ x(t)=\frac{Wt}{\pi t} $

Compute CTFT of $ x(t) $. Instead of using CTFT formula, we solve it by using inverse CTFT and comparison.

$ \begin{align} x(t) &= \frac{sin(Wt)}{\pi t} = \frac{e^{jWt}-e^{-jWt}}{2j\pi t}\ (*)\\ x(t) &= \int\limits_{-\infty}^{\infty}X(f)e^{j2\pi ft}df\ (**) \end{align} $

Since

$ \int\limits_{-\frac{W}{2\pi}}^{\frac{W}{2\pi}}1\cdot e^{j2\pi ft}df = \frac{e^{jWt}-e^{-jWt}}{2j\pi t} $

Therefore,

$ X(f) = rect(\frac{\pi f}{W}) = \left\{\begin{array}{ll}1, & \text{ if }|t|<\frac{W}{2\pi},\\ 0, & \text{else.}\end{array} \right. \ $

Question 6

$ \text{a)} \;\; \text{General Relation for the decimation with a factor of } D \,\! $.

$ \text{Let } X(w) = \mathcal{F}(x[n]) $

$ \begin{align} Y(w) &= \sum_{n=-\infty}^{\infty} y[n]e^{-jwn} = \sum_{n=-\infty}^{\infty} x[3n]e^{-jwn} \\ &= \sum_{m=-\infty, m=Dk}^{\infty} x[m]e^{-j\frac{wm}{D}} = \sum_{m=-\infty}^{\infty} x[m] \left( \sum_{k=-\infty}^{\infty} \delta[m-Dk] \right) e^{-j\frac{wm}{D}} \\ &= \sum_{m=-\infty}^{\infty} x[m] \left( \frac{1}{D} \sum_{k=0}^{D-1} e^{j\frac{2\pi}{D}km} \right) e^{-j\frac{wm}{D}} = \sum_{k=0}^{D-1} \frac{1}{D} \sum_{m=-\infty}^{\infty} x[m]e^{j\left(w-\frac{2\pi}{D}k\right)m} \\ &= \sum_{k=0}^{D-1} \frac{1}{D} X\left(\frac{w-2\pi k}{D}\right) \\ \end{align} $

Replacing D with 5 would be the answer.

$ \text{b)} \;\; \text{General Relation for the upsampling with a factor of } L \,\! $.

$ \begin{align} Z(w) &= \sum_{n=-\infty}^{\infty} z[n]e^{-jwn} \\ &= \sum_{n=-\infty}^{\infty} \left( \sum_{k=-\infty}^{\infty} x[k] \delta[n-kL] \right) e^{-jwn} \\ &= \sum_{k=-\infty}^{\infty} x[k] \sum_{n=-\infty}^{\infty} \delta[n-kL] e^{-jwn} = \sum_{k=-\infty}^{\infty} x[k] e^{-jwkL} \\ &= \sum_{k=-\infty}^{\infty} x[k] e^{-jLwk} = X(Lw) \\ &\end{align} $

Since $ X(w) $ is periodic with $ 2\pi $, $ Z(w)=X(Lw) $ is periodic with $ 2\pi/L $.

Replaing L with 7 would be the answer.

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