DTFT of a rectangular pulse

The signal is a rectangular pulse of $2N_1 + 1$ samples, all equal to 1, centred at $n = 0$.

DTFT

$$X(e^{j\omega}) = \sum_{n=-N_1}^{N_1}e^{-j\omega n}$$

Put $m = n + N_1$ ($m = 0$ to $2N_1$) and sum the geometric series:

$$\begin{aligned} X(e^{j\omega}) &= e^{j\omega N_1}\sum_{m=0}^{2N_1}e^{-j\omega m} = e^{j\omega N_1}\frac{1 - e^{-j\omega(2N_1+1)}}{1 - e^{-j\omega}} \\ &= \frac{e^{j\omega(N_1+\frac{1}{2})} - e^{-j\omega(N_1+\frac{1}{2})}}{e^{j\omega/2} - e^{-j\omega/2}} \end{aligned}$$

The second line comes from multiplying the numerator and denominator by $e^{j\omega/2}$. Using $e^{j\theta} - e^{-j\theta} = 2j\sin\theta$:

$$\boxed{X(e^{j\omega}) = \frac{\sin\left(\omega\left(N_1 + \frac{1}{2}\right)\right)}{\sin(\omega/2)}}$$

At $\omega = 0$ the value is $2N_1 + 1$ (by L'Hôpital, or as the number of samples).

Remarks

• $X(e^{j\omega})$ is real and even, because $x[n]$ is real and even.
• It is the DT counterpart of the sinc function and is periodic with period $2\pi$.
• The first zeros are at $\omega = \pm\frac{2\pi}{2N_1+1}$. A wider pulse gives a narrower main lobe.
• Example: for $N_1 = 2$, $X(e^{j\omega}) = \frac{\sin(2.5\omega)}{\sin(0.5\omega)}$, with peak 5 and first zeros at $\pm 2\pi/5$.