Difference between revisions of "ECE438 Week8 Quiz Q3sol" - Rhea

Revision as of 17:25, 12 October 2010

a) We decimate 3 times to compute DFT. The gain of one path equals to the product of gain of each decimate.

According to the figure. $ \text{ Gain }=(W_N^0)*(-1)*(W_N^2)=-W_N^2 $

b) Only one. In general, there is only one path between each input sample and each output sample.

c) $ x[0]\text{ to }X[2]\text{ : The gain is }1 $

$ x[1]\text{ to }X[2]\text{ : The gain is }W_N^2 $

$ x[2]\text{ to }X[2]\text{ : The gain is }-W_N^0=-1 $

$ x[3]\text{ to }X[2]\text{ : The gain is }-W_N^0 W_N^2=-W_N^2 $

$ x[4]\text{ to }X[2]\text{ : The gain is }W_N^0=1 $

$ x[5]\text{ to }X[2]\text{ : The gain is }W_N^0 W_N^2=W_N^2 $

$ x[6]\text{ to }X[2]\text{ : The gain is }-W_N^0 W_N^0=-1 $

$ x[7]\text{ to }X[2]\text{ : The gain is }-W_N^0 W_N^0 W_N^2=-W_N^2\text{, as in Part (a)} $

$ \text{Since }X[k]=\sum_{n=0}^{N-1}x[n]W_N^{nk} \text{, k=0,1,...,N-1} $

Now

$ \begin{align} X[2] &= \sum_{n=0}^7 x[n]W_8^{2n} \\ &= x[0]+x[1]W_8^2+x[2]W_8^4+x[3]W_8^6+x[4]W_8^8+x[5]W_8^10+x[6]W_8^12+x[7]W_8^14 \\ &= x[0]+x[1]W_8^2+x[2](-1)+x[3](-W_8^2)+x[4](1)+x[5]W_8^2+x[6](-1)+x[7](-W_8^2) \end{align} $

Thus, each input sample contributes the proper amount to the output DFT sample.

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