Difference between revisions of "ECE438 Week13 Quiz Q3sol" - Rhea

Revision as of 11:08, 17 November 2010

Solution to Q3 of Week 13 Quiz Pool

a)

$ x[n]=\delta[n]\,\! $

Using the N-point DFT formula,

$ X[k]=\sum_{n=0}^{N-1}\delta[n]e^{-j\frac{2\pi}{N}kn} = 1 $

b)

We use the N-point inverse-DFT formula to obtain $ y[n] $

$ \begin{align} y[n] & =\frac{1}{N}\sum_{k=0}^{N-1} \left( W_N^{k} + W_N^{2k} + W_N^{3k} \right) X[k] e^{j\frac{2\pi}{N}nk} \\ & =\frac{1}{N}\sum_{k=0}^{N-1} W_N^{k}X[k] e^{j\frac{2\pi}{N}nk} + \frac{1}{N}\sum_{k=0}^{N-1} W_N^{2k}X[k] e^{j\frac{2\pi}{N}nk} + \frac{1}{N}\sum_{k=0}^{N-1} W_N^{3k}X[k] e^{j\frac{2\pi}{N}nk} \\ & =\frac{1}{N}\sum_{k=0}^{N-1} X[k] e^{j\frac{2\pi}{N}(n-1)k} + \frac{1}{N}\sum_{k=0}^{N-1} X[k] e^{j\frac{2\pi}{N}(n-2)k} + \frac{1}{N}\sum_{k=0}^{N-1} X[k] e^{j\frac{2\pi}{N}(n-3)k} \quad \text{where} \;\; W_N=e^{-j\frac{2\pi}{N}} \\ \end{align} $

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

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

then, you will notice that $ y[n]=x_N[(n-1)\text{mod}N] + x_N[(n-2)\text{mod}N] + x_N[(n-3)\text{mod}N] $. Thus, it becomes

$ y[n]=\delta[n-1]+\delta[n-2]+\delta[n-3] \quad n=0,1,2,...,N-1\,\! $ and it is periodic with $ N $.

(Producting $ W^{k} $ to $ X[k] $ yields circular-shifting to the right by 1 in the periodic discrete-time signal)

c)

Let fix $ N=5 $ and $ h[n]=\delta[n]+2\delta[n-1]+3\delta[n-2] $

computing 5-point circular convolution with $ y[n] $ and $ h[n] $,

$ \begin{align} z[n] =& y[n]\circledast_5 h[n] \\ =& \quad \{\quad 0,\quad 1,\quad 1,\quad 1, \quad 0\} \\ & +\! \{\quad 0,\quad 0,\quad 2,\quad 2, \quad 2\} \\ & +\! \{\quad 3,\quad 0,\quad 0,\quad 3, \quad 3\} \\ =& \quad \{\quad 3,\quad 1,\quad 3,\quad 6, \quad 5\} \\ =& 3\delta[n]+\delta[n-1]+3\delta[n-2]+6\delta[n-3]+5\delta[n-4] \\ \end{align} $

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