Derive the DTFT pair

The DTFT pair is derived by treating an aperiodic sequence as the limit of a periodic sequence whose period $N \to \infty$, and then using the discrete-time Fourier series.

Step 1: Build a periodic extension

Let $x[n]$ be of finite duration, non-zero only for $-N_1 \le n \le N_2$. Form $\tilde{x}[n]$ by repeating $x[n]$ with period $N$, where $N$ is large enough that the copies do not overlap. Over one period $\tilde{x}[n] = x[n]$, and $\tilde{x}[n] \to x[n]$ as $N \to \infty$.

 x[n]:        ..0 0 [###] 0 0..
 x~[n]: [###] . . . [###] . . . [###]
        |<--- N --->|

Step 2: DTFS of the periodic extension

With $\omega_0 = 2\pi/N$:

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

Choose the period so that it contains $-N_1 \dots N_2$. There $\tilde{x}[n] = x[n]$, and $x[n] = 0$ outside, so the sum can run over all $n$:

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

Step 3: Define the envelope

Define

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

Then $a_k = \frac{1}{N}X(e^{jk\omega_0})$. The DTFS coefficients are samples of this continuous envelope, spaced $\omega_0$ apart.

Step 4: Substitute back

Using $\frac{1}{N} = \frac{\omega_0}{2\pi}$:

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

Step 5: Take the limit N → ∞

• $\omega_0 \to d\omega$, and $k\omega_0 \to \omega$ (a continuous variable).
• The sum over $N$ consecutive $k$ covers a frequency range of $N\omega_0 = 2\pi$, so it becomes an integral over an interval of length $2\pi$.
• $\tilde{x}[n] \to x[n]$.

The result is the DTFT pair:

$$\boxed{x[n] = \frac{1}{2\pi}\int_{2\pi}X(e^{j\omega})e^{j\omega n}d\omega \quad \text{(synthesis)}}$$ $$\boxed{X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \quad \text{(analysis)}}$$

Remarks

• $X(e^{j\omega})$ is continuous in $\omega$ and periodic with period $2\pi$, since $e^{-j(\omega+2\pi)n} = e^{-j\omega n}$. That is why the synthesis integral covers only $2\pi$.
• Convergence: the analysis sum converges if $\sum|x[n]| < \infty$ (absolutely summable) or $\sum|x[n]|^2 < \infty$ (finite energy, convergence in mean square).
• Example: $x[n] = a^n u[n]$ with $|a| < 1$ gives $X(e^{j\omega}) = \frac{1}{1-ae^{-j\omega}}$.