Time-shifting property of DTFS

Statement

If $x[n]$ is periodic with period $N$ and $x[n] \overset{FS}{\longleftrightarrow} a_k$, then

$$x[n - n_0] \overset{FS}{\longleftrightarrow} a_k\, e^{-jk(2\pi/N)n_0}$$

A shift in time multiplies each coefficient by a linear phase term; the magnitudes $|a_k|$ do not change.

Proof

$x[n - n_0]$ is also periodic with period $N$. Its coefficients are

$$b_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n - n_0]\,e^{-jk(2\pi/N)n}$$

Put $m = n - n_0$, so $n = m + n_0$. As $n$ runs over one period, $m$ also runs over one period:

$$\begin{aligned} b_k &= \frac{1}{N}\sum_{m=\langle N\rangle}x[m]\,e^{-jk(2\pi/N)(m + n_0)} \\ &= e^{-jk(2\pi/N)n_0}\left[\frac{1}{N}\sum_{m=\langle N\rangle}x[m]e^{-jk(2\pi/N)m}\right] \\ &= e^{-jk(2\pi/N)n_0}\,a_k \end{aligned}$$

Interpretation

• $|b_k| = |a_k|$: the magnitude spectrum is unchanged by a shift.
• $\angle b_k = \angle a_k - k\frac{2\pi}{N}n_0$: the phase changes linearly with $k$.

Example: $x[n] = \cos(2\pi n/N)$ has $a_{\pm1} = 1/2$. Then $x[n-1]$ has $b_{1} = \frac{1}{2}e^{-j2\pi/N}$ and $b_{-1} = \frac{1}{2}e^{j2\pi/N}$, still of magnitude $1/2$.