Derive DTFT; DTFT periodic with 2π

Derivation of the DTFT

1. Periodic extension: let $x[n]$ be aperiodic and of finite length ($-N_1 \le n \le N_2$). Form $\tilde{x}[n]$ by repeating $x[n]$ with a period $N$ large enough to avoid overlap. Then $\tilde{x}[n] = x[n]$ over one period, and $\tilde{x}[n] \to x[n]$ as $N \to \infty$.
2. DTFS of the extension: with $\omega_0 = 2\pi/N$,

$$a_k = \frac{1}{N}\sum_{n=\langle N\rangle}\tilde{x}[n]e^{-jk\omega_0 n} = \frac{1}{N}\sum_{n=-\infty}^{\infty}x[n]e^{-jk\omega_0 n}$$

The second form holds because $x[n] = 0$ outside the chosen period. 3. Envelope: define $X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$. Then $a_k = \frac{1}{N}X(e^{jk\omega_0})$. 4. Synthesis: using $1/N = \omega_0/2\pi$,

$$\tilde{x}[n] = \sum_{k=\langle N\rangle}a_k e^{jk\omega_0 n} = \frac{1}{2\pi}\sum_{k=\langle N\rangle}X(e^{jk\omega_0})e^{jk\omega_0 n}\omega_0$$

5. Limit $N \to \infty$: $\omega_0 \to d\omega$, $k\omega_0 \to \omega$, and the sum over $N$ terms covers $N\omega_0 = 2\pi$, so it becomes an integral over $2\pi$. Also $\tilde{x}[n] \to x[n]$.

This gives

$$X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}, \qquad x[n] = \frac{1}{2\pi}\int_{2\pi}X(e^{j\omega})e^{j\omega n}d\omega$$

Proof that the DTFT is periodic with period 2π

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

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, $X(e^{j(\omega+2\pi r)}) = X(e^{j\omega})$ for any integer $r$.

Reason: discrete-time exponentials whose frequencies differ by $2\pi$ are identical sequences. Only $0 \le \omega < 2\pi$ (or $-\pi \le \omega < \pi$) needs to be considered. Low frequencies lie near $0, \pm2\pi, \dots$ and the highest frequency is $\omega = \pm\pi$.