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Find IDFT of the sequence X (K) = {1,0,1,0}
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Solution:

We have,

$y(n)=\frac{1}{N} \sum_{k=0}^{N-1} Y(k) e^{\frac{j 2 \pi n n}{N}}, \quad n=0,1 \ldots . . N-1\\$

For $\mathrm{N}=4$

$y(n)=\frac{1}{4} \sum_{k=0}^3 Y(k) e^{\frac{j 2 \pi k n}{4}}, \quad n=0,1,2,3\\$

For $\mathrm{n}=0$

\begin{aligned} y(0) & =\frac{1}{4} \sum_{k=0}^3 Y(k) \\\\ & =\frac{1}{4}[y(0)+y(1)+y(2)+y(3)] \\\\ & =\frac{1}{4}[1+0+1+0] \\\\ & =0.5 \end{aligned}

For $\mathrm{n}=1$

\begin{aligned} y(1) & =\frac{1}{N} \sum_{k=0}^3 y(k) e^{\frac{j \pi k}{2}} \\\\ & =\frac{1}{4}\left[y(0)+y(1) e^{\frac{j \pi}{2}}+y(2) e^{j \pi}+y(3) e^{\frac{j 3 \pi}{2}}\right] \\\\ & =\frac{1}{4}[1+0+\cos \pi+j \sin \pi] \\\\ & =\frac{1}{4}[1+0-1+0]=0 \end{aligned}

For $n=2$

\begin{aligned} y(2) & =\frac{1}{N} \sum_{k=0}^3 y(k) e^{\frac{j 2 \pi k}{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] \\\\ & =\frac{1}{4}[1+0+1+0]=0.5\\ \end{aligned}

For $\mathrm{n}=3$

\begin{aligned} y(3) & =\frac{1}{4}\left[y(0)+y(1) e^{\frac{j 3 \pi}{2}}+y(2) e^{j 3 \pi}+y(3) e^{\frac{j 9 \pi}{2}}\right] \\\\ & =\frac{1}{4}[1+0+\cos 3 \pi+j \sin 3 \pi] \\\\ & =\frac{1}{4}[1+0-1+0]=0\\ \end{aligned}

$y(k)=\{0.5,0,0.5,0\}$