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The first points of 8-point DFT of real valued sequence are{0.25,0.125-j0.3018,0,0.125-j0.0518,0} Find the remaining three points.
1 Answer
written 6.2 years ago by |
By Symmetry property we have,
$X(k)=X^* (N-k)$
We have
X(0)=0.25
X(1)=0.125-j0.3018
X(2)=0
X(3)=0.125-j0.0518
X(4)=0
$k=5,X(5)=X^* (8-5)$
=$X^* (3)$
=0.125+j0.0518
$k=6,X(6)=X^* (8-6)$
$=X^* (2)$
=0
$k=7,X(7)=X^* (8-7)$
$=X^* (1)$
=0.125+j0.3018