If $x(n) (↔)^{DFT}$ X(k),then

$x(n-l) (↔)^{DFT} X(k).e^{\frac{-j 2πkl}{N}}$

Proof:We have IDFT equation

$x(n)=\frac{1}{N} ∑_{k=0}^{N-1} X(k).e^{\frac{-j 2πkn}{N}}$

$IDFT [X(K).e^{\frac{-j 2πkl}{N}} ]=1/N ∑_{k=0}^{N-1}X(k).e^{\frac{-j 2πkl}{N}}.e^{\frac{j 2πkn}{N}}$

$1/N ∑_{k=0}^{N-1}X(K).e^{\frac{-j (2πk(n-l))}{N}}$

Comparing with DFT equation we have

$x(n-l) \lt-^{DFT}\gt X(K).e^{\frac{-j 2πkl}{N}}$