Difference between revisions of "ECE438 HW8 Solution" - Rhea

Revision as of 12:53, 10 November 2010

Solution to Question 1 of HW5

Solution to Question 2 of HW5

a)

$ x[n]=6\delta[n]+5 \delta[n-1]+4 \delta[n-2]+3 \delta[n-3]+2 \delta[n-4]+\delta[n-5]\,\! $

Using the DFT formula,

$ \begin{align} X[k] &=\sum_{n=0}^{5}\left(6\delta[n]+5 \delta[n-1]+4 \delta[n-2]+3 \delta[n-3]+2 \delta[n-4]+\delta[n-5]\right)e^{-j\frac{2\pi}{6}kn} \\ &= 6 + 5e^{-j\frac{2\pi}{6}k} + 4e^{-j\frac{2\pi}{6}2k} + 3e^{-j\frac{2\pi}{6}3k} + 2e^{-j\frac{2\pi}{6}4k} + e^{-j\frac{2\pi}{6}5k} \\ \end{align} $

b)

We use the inverse-DFT formula to obtain $ y_6[n] $

$ y_6[n]=\frac{1}{N}\sum_{k=0}^{k=5} W_6^{-2k} X[k] e^{j\frac{2\pi}{6}nk} = \frac{1}{N}\sum_{k=0}^{k=5} X[k] e^{j\frac{2\pi}{6}(n+2)k} $

If you compare this with the inverse-DFT of $ X[k] $

$ x_6[n]=\frac{1}{N}\sum_{k=0}^{k=5}X[k] e^{j\frac{2\pi}{6}nk} $

then, you will notice that $ y_6[n]=x_6[(n+2)\text{mod}6] $. Thus, it becomes

$ y_6[n]=4\delta[n]+3 \delta[n-1]+2\delta[n-2]+\delta[n-3]+6 \delta[n-4]+5\delta[n-5]\,\! $

(Producting $ W^{-2k} $ to $ X[k] $ yields circular-shifting to the left by 2 in the periodic discrete-time signal)

c)

$ h[n]=\delta[n]+\delta[n-1]+\delta[n-2]\,\! $

computing the circular convolution with $ x[n] $ and $ h[n] $,

$ \begin{align} y[n] =& x[n]\circledast_6 h[n] \\ =& \quad \{\quad 6,\quad 5,\quad 4,\quad 3,\quad 2,\quad 1\} \\ & +\! \{\quad 1,\quad 6,\quad 5,\quad 4,\quad 3,\quad 2\} \\ & +\! \{\quad 2,\quad 1,\quad 6,\quad 5,\quad 4,\quad 3\} \\ =& \quad \{\quad 9,\;\;12,\;\;\!15,\;\;12,\quad 9,\quad 6\} \\ =& 9\delta[n]+12\delta[n-1]+15\delta[n-2]+12\delta[n-3]+9\delta[n-4]+6\delta[n-5] \\ \end{align} $

d)

In order for the periodic repetition (with period N) of the usual convolution between x[n] and h[n] to be the same with the N-point circular convolution,

$ N \geq L+M-1 $ where L is the length of x[n] and M is the length of h[n].

Therefore, $ N\geq8 $.

Solution to Question 3 of HW5

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