Ask
Login
0
11kviews
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.
written 8.2 years ago by gravatar for teamques10 teamques10 ★ 68k modified 2.8 years ago by gravatar for sagarkolekar sagarkolekar ★ 11k
discrete time signal processing
ADD COMMENT EDIT
1 Answer
0
1.8kviews
written 8.2 years ago by gravatar for teamques10 teamques10 ★ 68k

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 $
ADD COMMENT EDIT
Please log in to add an answer.