Question

Find N-point DFT of x[n]= $z8nP0DHzB+IMQAAAABJRU5ErkJggg==$         n=0,1,…,N-1

X[n] = $wP8wOd6LBLk0gAAAABJRU5ErkJggg==$

Using the periodicity of the complex exponentials, we can write x[n] follows:

X[n] = $X5iKbLiiSH8AAAAASUVORK5CYII=$

The DFT coefficients are

              9N/2 k=0

X[k]= N/4                          k=2 and k=-2

                0                              else

Answer 1

2nn Given x[n] = 4 + cos? , and n = 0,1,2,....., N - 1 N

1+ cos20 NOTE : cos?(0) =

e-jø + ejo cos(0) =

+1 = ux : (uw) sos +I

$\mathrm{=\:\:\frac{9+cos\left ( \frac{4\pi \,n}{N} \right )}{2}}$

+ 6 + Na #3 +0] *

$\mathrm{=\frac{9}{2}+\left [ \frac{e^{-j\frac{4\pi \,n}{N} }+e^{j\frac{4\pi \,n}{N} }}{4} \right ] }$

2. е де (0) + . ein (-2) + . ein (2)

ev " " " ਡਫ)

N-1 N – Point DFT of x[nis defined as X(k) = xne-in

IDFT of X(k) is defined as x[n] - 12 2X(k) Nk

( * * * * -- ) : - [ax :

Hence by comparison DFT coefficients of x[nare,

for k= -2 9N for k= 0 X(k) = { À for k= 2 0 otherwise