Stale and prove frequency shifting property of DFT.
1 Answer

If $x(n) \stackrel{\text { DFT }}{\longleftrightarrow} \times(\mathrm{k})$


$e^{j \frac{2 n L n}{N} x(n)} \stackrel{\text { DFT }}{\longleftrightarrow} \times(k-l)$

Proof: By definition of DFT

$x(k)=\sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi n k}{N}}$

Then DFT of x(n - l) is,

$D F T[x(n-l)]=\sum_{n=0}^{N-1} x(n-l) = e^{-j} {\frac{2 \pi n k}{N}}$

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$\operatorname{DFT}[x(n-1)]=\sum_{p=1}^{N-1} x(p) e^{-j \frac{2 \pi}{N}(p+l) k}$

$D F T[x(n-e)]=\left[\sum_{p=0}^{N-1} x(p) e^{-j \frac{2 \pi p k}{N}}\right] e^{-j}\frac{2 \pi \mu c}{N}$

$=X(k) e^{-j 2 \pi \mu c / N}$

By definition of DFT.

$x(k)=\sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi n k}{N}}$ $0 \leq k \leq N-1$

$\therefore$ DFT of $e^{j \frac{2 \pi \ell n}{N} x(n)}$ is

$\operatorname{DFT}\left[e^{j \frac{2 \pi l n}{N}} x(n)\right] =\sum_{n=0}^{N-1} e^{j \frac{2 \pi l n}{N} \cdot x(n) e^{j \frac{2 \pi n k}{N}}}$

$=\sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi}{N}(k-1) n}$

$D F T\left[e^{j \frac{2 \pi L n}{N}} x(n)\right]=X(k-l)$

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