DTFT of (1/2)^(−n)u[−n−1]

Simplify the signal: $\left(\frac12\right)^{-n} = 2^n$, and $u[-n-1] = 1$ for $n \le -1$. So

$$x[n] = 2^n \text{ for } n \le -1, \quad 0 \text{ for } n \ge 0$$

i.e. $x[n] = \{\dots, \frac18, \frac14, \frac12\}$ ending at $n = -1$. It decays as $n \to -\infty$, so it is absolutely summable and the DTFT exists.

DTFT:

$$\begin{aligned} X(e^{j\omega}) &= \sum_{n=-\infty}^{-1}2^n e^{-j\omega n} \quad (\text{put } m = -n) \\ &= \sum_{m=1}^{\infty}2^{-m}e^{j\omega m} = \sum_{m=1}^{\infty}\left(\tfrac12 e^{j\omega}\right)^m \\ &= \frac{\frac12 e^{j\omega}}{1 - \frac12 e^{j\omega}}, \qquad \left|\tfrac12e^{j\omega}\right| < 1 \end{aligned}$$

Answer:

$$X(e^{j\omega}) = \frac{0.5\,e^{j\omega}}{1 - 0.5\,e^{j\omega}} = \frac{e^{j\omega}}{2 - e^{j\omega}}$$

Magnitude: $|X(e^{j\omega})| = \frac{1}{\sqrt{5 - 4\cos\omega}}$, giving $1$ at $\omega = 0$ (equal to $\sum x[n] = \frac12 + \frac14 + \dots = 1$) and $\frac13$ at $\omega = \pi$.

Phase: $\angle X(e^{j\omega}) = \omega + \tan^{-1}\frac{\sin\omega}{2 - \cos\omega}$.

(Check with z-transform: $-a^n u[-n-1] \leftrightarrow \frac{1}{1-az^{-1}}$, $|z|<|a|$. With $a = 2$: $X(z) = -\frac{1}{1-2z^{-1}} = \frac{z}{2 - z}$, ROC $|z|<2$, which contains the unit circle. Putting $z = e^{j\omega}$ gives the same result.)