0
420views
Given a sequence $x(n)$ for $0 \leq n \leq 3$, where $x(0)=1, x(1)=3, x(2)=3$, and $x(3)=4$. Evaluate its DFT $X(k)$.
0
23views

Solution:

Since $N=4, W_4=\mathrm{e}^{-\mathrm{j} \pi / 2}$, then using:

$X(k)=\sum_{n=0}^3 x(n) W_4^{k n}=\sum_{n=0}^3 x(n) e^{-j \frac{\pi k n}{2}}\\$

Thus, for $k=0$

\begin{aligned} & X(0)=\sum_{n=0}^3 x(n) e^{-j 0}=x(0) e^{-j 0}+x(1) e^{-j 0}+x(2) e^{-j 0}+x(3) e^{-j 0} \\\\ &=x(0)+x(1)+x(2)+x(3) \\\\ &=1+2+3+4=10 \\\\ & \text { for } k=1 \\\\ & X X(1)=\sum_{n=0}^3 x(n) e^{-j \frac{\pi n}{2}}=x(0) e^{-j 0}+x(1) e^{-j \frac{\pi}{2}}+x(2) e^{-j \pi}+x(3) e^{-j \frac{3 \pi}{2}} \\\\ &=x(0)-j x(1)-x(2)+j x(3) \\\\ &=1-j 2-3+j 4=-2+j 2 \\\\ & \text { for } k= 2 \\\\ & X(2)=\sum_{n=0}^3 x(n) e^{-j \pi n}=x(0) e^{-j 0}+x(1) e^{-j \pi}+x(2) e^{-j 2 \pi}+x(3) e^{-j 3 \pi} \\\\ &=x(0)-x(1)+x(2)-x(3) \\\\ &=1-2+3-4=-2 \\\\ & \text { and for } k=3 \\\\ & X(3)=\sum_{n=0}^3 x(n) e^{-j \frac{3 \pi n}{2}}=x(0) e^{-j 0}+x(1) e^{-j \frac{3 \pi}{2}}+x(2) e^{-j 3 \pi}+x(3) e^{-j \frac{j \pi}{2}} \\\\ &=x(0)+j x(1)-x(2)-j x(3) \\\\ &=1+j 2-3-j 4=-2-j 2 \\ \end{aligned}\\