DTFT of sgn[n] and u[n]

The definition used here is $\mathrm{sgn}[n] = 1$ for $n \ge 0$ and $-1$ for $n < 0$, so that $u[n] = \frac{1}{2}(1 + \mathrm{sgn}[n])$. (A note at the end covers the version with $\mathrm{sgn}[0] = 0$.)

 sgn[n]
   1             *  *  *  *
                 |  |  |  |
 ----+--+--+--+--+--+--+--+--- n
     |  |  |  |
  -1 *  *  *  *
    -4 -3 -2 -1  0  1  2  3

DTFT of sgn[n]

$\mathrm{sgn}[n]$ is not absolutely summable, so take it as a limit of decaying sequences ($0 < a < 1$):

$$\mathrm{sgn}[n] = \lim_{a\to 1}\left[a^n u[n] - a^{-n}u[-n-1]\right]$$

First part:

$$\sum_{n=0}^{\infty}a^n e^{-j\omega n} = \frac{1}{1 - ae^{-j\omega}}$$

Second part, with $m = -n$:

$$\sum_{n=-\infty}^{-1}a^{-n}e^{-j\omega n} = \sum_{m=1}^{\infty}\left(ae^{j\omega}\right)^m = \frac{ae^{j\omega}}{1 - ae^{j\omega}}$$

Hence

$$S_a(e^{j\omega}) = \frac{1}{1 - ae^{-j\omega}} - \frac{ae^{j\omega}}{1 - ae^{j\omega}}$$

Let $a \to 1$. In the second term, divide the numerator and denominator by $e^{j\omega}$:

$$\frac{e^{j\omega}}{1 - e^{j\omega}} = \frac{1}{e^{-j\omega} - 1} = -\frac{1}{1 - e^{-j\omega}}$$

Therefore

$$\boxed{\mathrm{sgn}[n] \xleftrightarrow{DTFT} \frac{2}{1 - e^{-j\omega}}, \quad \omega \ne 2\pi k}$$

Since $\mathrm{sgn}[n]$ has zero average value, there is no impulse at $\omega = 0$.

Magnitude and phase: $1 - e^{-j\omega} = e^{-j\omega/2}\cdot 2j\sin(\omega/2)$, so

$$\frac{2}{1 - e^{-j\omega}} = \frac{e^{j\omega/2}}{j\sin(\omega/2)}, \qquad |S(e^{j\omega})| = \frac{1}{|\sin(\omega/2)|}$$

The phase is $\frac{\omega}{2} - \frac{\pi}{2}$ for $0 < \omega < \pi$ (and $\frac{\omega}{2} + \frac{\pi}{2}$ for $-\pi < \omega < 0$).

DTFT of u[n]

$$u[n] = \frac{1}{2} + \frac{1}{2}\mathrm{sgn}[n]$$

The constant $1$ has DTFT $2\pi\sum_k\delta(\omega - 2\pi k)$. (Its inverse DTFT over one period is $\frac{1}{2\pi}\int 2\pi\delta(\omega)d\omega = 1$.) So

$$\frac{1}{2} \xleftrightarrow{DTFT} \pi\sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)$$

By linearity:

$$\boxed{U(e^{j\omega}) = \frac{1}{1 - e^{-j\omega}} + \pi\sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)}$$

Check

• As $a \to 1$, $a^n u[n] \to u[n]$ and $\frac{1}{1 - ae^{-j\omega}} \to \frac{1}{1 - e^{-j\omega}}$. The impulses account for the DC value of $u[n]$.
• Accumulation property: $u[n] = \sum_{k=-\infty}^{n}\delta[k]$, which gives $\frac{1}{1 - e^{-j\omega}}\cdot 1 + \pi\cdot 1\cdot\sum_k\delta(\omega - 2\pi k)$, the same result.
• This is the DT counterpart of $u(t) \leftrightarrow \frac{1}{j\omega} + \pi\delta(\omega)$.

Note: if $\mathrm{sgn}[n]$ is defined with $\mathrm{sgn}[0] = 0$, it equals the version above minus $\delta[n]$. Its DTFT is then $\frac{2}{1 - e^{-j\omega}} - 1 = \frac{1 + e^{-j\omega}}{1 - e^{-j\omega}}$. Using $u[n] = \frac{1}{2}(1 + \mathrm{sgn}[n] + \delta[n])$ gives the same $U(e^{j\omega})$.