DTFT is periodic with period 2π

The DTFT of a sequence $x[n]$ is

$$X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n]\,e^{-j\omega n}$$

Derivation: Replace $\omega$ by $\omega + 2\pi$:

$$\begin{aligned} X(e^{j(\omega+2\pi)}) &= \sum_{n=-\infty}^{\infty} x[n]\,e^{-j(\omega+2\pi)n} \\ &= \sum_{n=-\infty}^{\infty} x[n]\,e^{-j\omega n}\,e^{-j2\pi n} \end{aligned}$$

Since $n$ is an integer, $e^{-j2\pi n} = \cos(2\pi n) - j\sin(2\pi n) = 1$. Therefore

$$X(e^{j(\omega+2\pi)}) = \sum_{n=-\infty}^{\infty} x[n]\,e^{-j\omega n} = X(e^{j\omega})$$

More generally $X(e^{j(\omega+2\pi k)}) = X(e^{j\omega})$ for any integer $k$. So the DTFT is periodic in $\omega$ with period $2\pi$.

Why it happens: The discrete-time exponentials $e^{j\omega n}$ and $e^{j(\omega+2\pi)n}$ are the same sequence, because $n$ takes only integer values. Frequencies $2\pi$ apart cannot be told apart in discrete time. This is unlike continuous time, where $e^{j\omega t}$ are all different for different $\omega$.

Consequences:

• Only one period, usually $-\pi \le \omega < \pi$ (or $0 \le \omega < 2\pi$), needs to be computed or plotted.
• Low frequencies are near $\omega = 0, \pm 2\pi, \dots$; the highest frequency is $\omega = \pm\pi$.
• The inverse DTFT integrates over one period only: $x[n] = \frac{1}{2\pi}\int_{2\pi} X(e^{j\omega})e^{j\omega n}d\omega$.

Example: For $x[n] = a^n u[n]$, $X(e^{j\omega}) = \frac{1}{1-ae^{-j\omega}}$, which repeats every $2\pi$ since $e^{-j\omega}$ does.