written 21 months ago by |
Solution:
Let us assume $N=L=4$.
$ \begin{aligned}\\ & \text { We have } X(k)=\sum_{n=0}^{N-1} x(n) e^{-j 2 \pi n k / N} \quad k=0,1, \ldots, N-1 \\\\ & \qquad \begin{aligned}\\ X(0)=\sum_{n=0}^3 x(n) & =x(0)+x(1)+x(2)+x(3) \\\\ & =1+1+0+0=2\\ \end{aligned}\\ \end{aligned}\\ $
$ \begin{aligned} X(1)=\sum_{n=0}^3 x(n) e^{-j \pi n / 2} & =x(0)+x(1) e^{-j \pi / 2}+x(2) e^{-j \pi}+x(3) e^{-j 3 \pi / 2} \\\\ & =1+\cos \frac{\pi}{2}-j \sin \frac{\pi}{2} \\\\ & =1-j \end{aligned} $
$ \begin{aligned} X(2)=\sum_{n=0}^3 x(n) e^{-j \pi n} & =x(0)+x(1) e^{-j \pi}+x(2) e^{-j 2 \pi}+x(3) e^{-j 3 \pi} \\\\ & =1+\cos \pi-j \sin \pi \\\\ & =1-1=0\\ \end{aligned} $
$ \begin{aligned} X(3)=\sum_{n=0}^3 x(n) e^{-j 3 n \pi / 2} & =x(0)+x(1) e^{-j 3 \pi / 2}+x(2) e^{-j 3 \pi}+x(3) e^{-j 9 \pi / 2} \\\\ & =1+\cos \frac{3 \pi}{2}-j \sin \frac{3 \pi}{2} \\\\ & =1+j \\\\ X(k) & =\{2,1-j, 0,1+j\}\\ \\ y(n) & =\frac{1}{N} \sum_{k=0}^{N-1} Y(k) e^{j 2 \pi n k / N} \quad n=0,1, \ldots N-1 \\\\ y(0) & =\frac{1}{4} \sum_{k=0}^3 Y(k) \quad n=0,1,2,3 \\\\ & =\frac{1}{4}[y(0)+y(1)+y(2)+y(3)] \\\\ & =\frac{1}{4}[1+0+1+0] \\\\ & =0.5\\ \end{aligned}\\ $
$ \begin{aligned} & y(1)=\frac{1}{N} \sum_{k=0}^3 Y(k) e^{j \pi k / 2} \\\\ & y(1)=\frac{1}{4}\left[Y(0)+Y(1) e^{j \pi / 2}+Y(2) e^{j \pi}+Y(3) e^{j 3 \pi / 2}\right] \\\\ & =\frac{1}{4}[1+0+\cos \pi+j \sin \pi+0] \\\\ & =\frac{1}{4}[1+0-1+0]=0 \\\\ & y(2)=\frac{1}{4}\left[Y(0)+Y(1) e^{j \pi}+Y(2) e^{j 2 \pi}+Y(3) e^{j 3 \pi}\right] \\\\ & =\frac{1}{4}[1+0+\cos 2 \pi+j \sin 2 \pi+0] \\\\ & =\frac{1}{4}[1+0+1+0]=0.5 \\\\ & y(3)=\frac{1}{4}\left[Y(0)+Y(1) e^{j 3 \pi / 2}+Y(2) e^{j 3 \pi}+Y(3) e^{j 9 \pi / 2}\right] \\\\ & =\frac{1}{4}[1+0+\cos 3 \pi+j \sin 3 \pi+0] \\\\ & =\frac{1}{4}[1+0+(-1)+0]=0 \\\\ & y(n)=\{0.5,0,0.5,0\} \\\\ & \end{aligned} $