DTFT of u[n−2] − u[n−6]

$x[n] = u[n-2] - u[n-6]$ is a rectangular sequence that equals 1 for $n = 2, 3, 4, 5$ and 0 elsewhere:

$$x[n] = \{0, 0, 1, 1, 1, 1\} \quad (n = 0 \text{ to } 5)$$

DTFT:

$$\begin{aligned} X(e^{j\omega}) &= \sum_{n=2}^{5} e^{-j\omega n} = e^{-j2\omega} + e^{-j3\omega} + e^{-j4\omega} + e^{-j5\omega} \end{aligned}$$

Closed form (geometric series of 4 terms, first term $e^{-j2\omega}$, ratio $e^{-j\omega}$):

$$\begin{aligned} X(e^{j\omega}) &= e^{-j2\omega}\,\frac{1 - e^{-j4\omega}}{1 - e^{-j\omega}} \\ &= e^{-j2\omega}\,\frac{e^{-j2\omega}\left(e^{j2\omega} - e^{-j2\omega}\right)}{e^{-j\omega/2}\left(e^{j\omega/2} - e^{-j\omega/2}\right)} \\ &= e^{-j2\omega}\,e^{-j3\omega/2}\,\frac{2j\sin 2\omega}{2j\sin(\omega/2)} \\ &= e^{-j7\omega/2}\,\frac{\sin 2\omega}{\sin(\omega/2)} \end{aligned}$$

Answer:

$$X(e^{j\omega}) = \frac{\sin(2\omega)}{\sin(\omega/2)}\,e^{-j3.5\omega}$$

• Magnitude $|X| = \left|\frac{\sin 2\omega}{\sin(\omega/2)}\right|$, maximum value $4$ at $\omega = 0$ (= number of ones), zeros at $\omega = \pm\pi/2, \pm\pi$.
• Phase is linear, $-3.5\omega$ (plus $\pi$ jumps where the sine ratio is negative), because the pulse is symmetric about $n = 3.5$.

(Alternative: $u[n] \leftrightarrow \frac{1}{1-e^{-j\omega}} + \pi\sum_k\delta(\omega-2\pi k)$ with time shifting gives the same result, since the impulse terms cancel.)