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Find IDFT of the sequence X (K) = (5,0,1-j,0,1,0,1+j,0)
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Solution:

We have,

$\mathrm{x}(\mathrm{n})=\frac{1}{N} \sum_{k=0}^{N-1} X(\mathrm{k}) e^{j 2 \pi k n / N} n=0,1, \ldots ., N-1\\$

For $\mathrm{N}=8$

$\mathrm{x}(\mathrm{n})=\frac{1}{8} \sum_{k=0}^{N-2} X(\mathrm{k}) \mathrm{e}^{j \pi n / 4} n=0,1, \ldots .7\\$

For $\mathrm{n}=0 ; \mathrm{x}(0)=\sum_{k=0}^7 X(\mathrm{k})=\frac{1}{8}[5+0+1-\mathrm{j}+0+1+\mathrm{j}+0]=1$

For $\mathrm{n}=1 ; \mathrm{x}(1)=\frac{1}{8} \sum_{k=0}^7 X(\mathrm{k}) \mathrm{e}^{j \pi k / 4}=\frac{1}{8}[5+-(1-\mathrm{j})(\mathrm{j})+1(-1)+(1+\mathrm{j})(-\mathrm{j})]=6 / 8=0.75$

For $\mathrm{n}=2 ; \mathrm{x}(2)=\frac{1}{8} \sum_{k=0}^7 X(\mathrm{k}) \mathrm{e}^{\mathrm{j} \pi k / 2}=\frac{1}{8}[5+(1-\mathrm{j})(-1)+1(1)+(1+\mathrm{j})(-1)]=4 / 8=0.5$

For $\mathrm{n}=3 ; \mathrm{x}(3)=\frac{1}{8} \sum_{k=0}^7 X(\mathrm{k}) \mathrm{e}^{\mathrm{j} 3 \pi k / 4}=\frac{1}{8}[5+(1-\mathrm{j})(-\mathrm{j})+1(-1)+(1+\mathrm{j})(\mathrm{j})]=2 / 8=0.25$

For $\mathrm{n}=4 ; \mathrm{x}(4)=\frac{1}{8} \sum_{k=0}^7 X(\mathrm{k}) \mathrm{e}^{j 5 \pi k / 4}=\frac{1}{8}[5+(1-\mathrm{j})(1)+1(1)+(1+\mathrm{j})(1)]=8 / 8=1$

For $\mathrm{n}=5 ; \mathrm{x}(5)=\frac{1}{8} \sum_{k=0}^7 X(\mathrm{k}) \mathrm{e}^{j 5 \pi k / 4}=\frac{1}{8}[5+(1-\mathrm{j})(\mathrm{j})+(1)(-1)+(1+\mathrm{j})(-\mathrm{j})]=6 / 8=0.75$

For $\mathrm{n}=6 ; \mathrm{x}(6)=\frac{1}{8} \sum_{k=0}^7 X(\mathrm{k}) \mathrm{e}^{j 3 \pi k / 2}=\frac{1}{8}[5+(1-\mathrm{j})(-1)+1(1)+(1+\mathrm{j})(-\mathrm{j})]=4 / 8=0.5$

For $\mathrm{n}=7 ; \mathrm{x}(7)=\frac{1}{8} \sum_{k=0}^7 e^{j 7 \pi k / 4}=\frac{1}{8}[5+(1-\mathrm{j})(-\mathrm{j})+1(1)+(1+\mathrm{j})(\mathrm{j})]=2 / 8=0.25$

$x(n)=\{1,0.75,0.5,0.25,1,0.75,0.5,0.25\}$