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The first five points of eight point DFT of real valued signal are $\{0.25, 0.125 -j0.3018, 0, 0.125-j0.0150, 0\}$. Determine the remaining three points.
1 Answer
written 7.5 years ago by |
The given DFT is $\{0.25, 0.125 -j0.3018, 0, 0.125-j0.0150, 0\}$
Now, here
$X[0]= 0.25, \\ X[1]= 0.125 -j0.3018, \\ X[2]= 0, \\ X[3]= 0.125-j0.0150 , \\ X[4]= 0$
By Symmetry Property of DFT,
$$X* [k ] = X[ N-k ]$$
Or
$$X* [ N-k ] = X[ k ]$$
Here, $N = 8, \\ X* [k ] = X[8-k ] \\ X[ 5 ] = X*[ 8 - 5 ] = X* [3] \\ = 0.125+ j0.0150 \\ X[ 6 ] = X*[ 8 - 6 ] = X* [2 ] \\ = 0\\ X[ 7 ] = X*[ 8 - 7 ] = X* [ 1 ] \\ = 0.125 + j0.3018 $