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Find DFT of $x(n)=2^n$ using the 8-point DIT-FFT algorithm.
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written 3.1 years ago by
prajapatijaimin
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Solution:
$ \begin{aligned} x[n] &=2^n \\\\ &=\{1,2,4,8,16,32,64,128\} .\\ \end{aligned}\\ $
$ \begin{aligned}\\ &x[k]=\sum_{n=0}^{N / 1} x[n] W_N^{k n} \\\\ &W_8^0=e^{-j 2 \pi / 8 \cdot 0}=0,1 . \cdot N-1 \\\\ &W_8^1=e^{-j 2 \pi / 8 \cdot 1}=0.707-j 0.707 \\\\ &W_8^2=e^{-j 2 \pi / g^2 \cdot 2}=-j \\\\ &W_8^3=e^{-j 2 \pi …
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