We know $x(n) = \alpha u(n) \ftp \frac{1}{1-\alpha e^{-i\omega}}$ where $\abs{\alpha} \lt 1$. What about a translated version of $x(n)$? We know that $x(n - n_0) \ftp e^{-i\omega n_0}X(\omega)$, so we can determine that 

$$ x(n-n_0) = \alpha^{n-n_0}u(n-n_0) \ftp \frac{e^{-i\omega n_0}}{1-\alpha e^{-i\omega}} $$

Any shifted version of this signal will produce a signal that is rational in $z=e^{i\omega}$.

\begin{example}

Given an impulse response $g(n) = -2^nu(-n-1)$, find the impulse response $G(\omega)$.



First note that $u(-n-1) = \begin{cases} 1 & -n-1\geq0 \\ 0 & \mbox{otherwise} \end{cases}$, which translates to $u(-n-1) = \begin{cases} 1 & n\leq-1 \\ 0 & \mbox{otherwise} \end{cases}$. Then we can start to solve for the frequency response:

\begin{align*}
G(\omega) &= \sum \limits_{n=-\infty}^{\infty}-2^n u(-n-1) e^{-i\omega n} \\
G(\omega) &= \sum \limits_{n=-\infty}^{-1}-2^n e^{-i\omega n} \\
G(\omega) &= -\sum \limits_{n=-\infty}^{-1}\lr{2 e^{-i\omega }}^n \\
G(\omega) &= -\sum \limits_{m=1}^{\infty}\lr{2 e^{-i\omega }}^{-m} \quad \mbox{let }m=-n\\
G(\omega) &= -\sum \limits_{m=1}^{\infty}\lr{\frac{1}{2} e^{i\omega }}^m \\
G(\omega) &= -\sum \limits_{m=0}^{\infty}\lr{\frac{1}{2} e^{i\omega }}^{m+1} \quad \mbox{shift limits}\\
G(\omega) &= -\frac{1}{2}e^{i\omega}\sum \limits_{m=0}^{\infty}\lr{\frac{1}{2} e^{i\omega }}^{m} \\
G(\omega) &= -\frac{1}{2}e^{i\omega} \frac{1}{1-\frac{1}{2}e^{i\omega}} \\
G(\omega) &= \frac{1}{1-2e^{-i\omega}} \\
\end{align*}

Note that we could have solved for $g(-n) = -\frac{1}{2}^nu(n-1)$, and then used the DTFT property that $g(-n) \ftp G(-\omega)$ to solve the problem.

\end{example}

Rational DTFTs