a) Calculate DFT x [k] using result of (a), Calculate DFT of following

$x_1(n) = \begin{cases} 1 & n=0 \\ 0 & 1\leq n \leq 4\\ 1 5\leq n \leq 7 . \end{cases} \hspace {2cm} x_2(n) = \begin{cases} 0 & 0\leq n \leq 1 \\ 1 & 2\leq n \leq 5\\ 1 6\leq n \leq 7 . \end{cases}$

Mumbai University > EXTC > Sem 6 > Discrete Time Signal Processing

Marks : 10

Year : MAY 2015

$x (n) = {1, 1, 1, 1, 0, 0, 0, 0}$ To find x [k] By DFT, $\sum\limits_{n=0}^{N-1}x(n).W_N^{nk} $ For $x (n), N = 8\ x[k]=x[0]w_N^0+x[1]w_N^{K}+x[2]w_N^{2K}+x[3]w_N^{3K}+0+0+0+0\ \therefore x[k]=1+w_N^{K}+w_N^{2K}+w_N^{3K}$ At $k = 0, x [0] = 1 +1 +1 + 1 = 4$ At $ k = 1, x [1] = 1 + w^1_8 + w^2_8 + w^3_8$\\ = 1 + (0.707 – j0.707)+ (-j) +(-0.707 – j0.707)\\ = 1 – j2.414$

At $k =2, x [2] = 1 + w^1_8 + w^4_8 + w^6_8 \\ = 1 + (-j) + (-1) + (-j) \\ = 0 $

At $k = 3, x [3] = 1 + w^3_8 + w^6_8 + w^9_8\\ = 1 + (-0.707 – j0.707) + (j) + (0.707 – j0.707) \\ =1 – j1.414$

At $k = 4, x [4] = 1 – 1 + 1 – 1 = 0$

By symmetry about $k=\dfrac N2=\dfrac 82=4$

$x [k] = x* [-k] = x*[N - k] \\ x [k] = x* [8 - k] \\ k [5] = x* [3] = 1 + j1.414 \\ k [6] = x* [2] = 0\\ k [7] = x* [1] = 1 + j 2.414 \\ x [k] = \{4, 1 – j2.414, 0, 1 – j1.414, 0, 1 + j1.414, 0, 1 + j2.414\} $

Now,

$X_1 (n) = \{1, 0, 0, 0, 0, 1, 1, 1\} \\ X_1 (n) = x (-n)$

By DFT property of time reversal,

$x_1[k]=x[-k]\\ x_1=x[k].e^{-j2\pi k/N}$

At $k=0 \space x_1[0]=x[0].e^0=4$

At $k=1\space x_1[1]=x[1].e^{-j2\pi/8}= (1-j2.414).e^{-j\pi/4}=-1-2.414j$

At $k=2\space x_1[2]=x[2].e^{-jj\pi/2}=0$

At $k=3\space x_1[3]=x[3].e^{-j6\pi/8}=-1.7069+j0.2927$

At $k=4\space x_1[4]=0$

At $k=5\space x_1[5]=x[5].e^{-j10\pi/8}=-1.7069-j0.2927$

At $k=6\space x_1[6]=x[6].e^{-j12\pi/8}=0$

At $k=7\space x_1[7]=x[7].e^{-j7\pi/4}=-1+2.414j \\ x_1 [k] = \{4, -1 – j2.414, 0, -1.70697 + j.0.927, 0, -1.7069 – j0.2927, 0, -1+2.414j\}$

Now,

$x_2 (n) = \{0, 0, 1, 1, 1, 1, 1, 1\}$