If X(k) is the N-point DFT of a sequence x(n), then circular time shift property is that N-point DFT of x((n-l))N is X(k)e-j2πkl/N.

If X(k) is the N-point DFT of a sequence x(n), then circular time shift property is that N-point DFT of x((n-l))N is X(k)e-j2πkl/N. Correct Answer True

According to the circular time shift property of a sequence, If X(k) is the N-point DFT of a sequence x(n), then the N-pint DFT of x((n-l))N is X(k)e-j2πkl/N.

Related Questions

{a(n)} is a real-valued periodic sequence with a period N. x(n) and X(k) form N-point Discrete Fourier Transform (DFT) pairs. The DFT Y(k) of the sequence
$$y\left( n \right) = \frac{1}{N}\sum\limits_{r = 0}^{N - 1} {x\left( r \right)} x\left( {n + r} \right)$$      is
If X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n)?