What happens when we modulate a function in time with a complex exponential?

\begin{align*}
g(n) &= \hat{G}(z) \quad R_1 \lt \abs{z} \lt R_2 \\
h(n) &= z_0^ng(n) \ztp \hat{H}(z) = \mbox{ } ?
\end{align*}

The result must be consistent with our result for the fourier transform version. 

\begin{align*}
\hat{H}(z) &= \sum \limits_{n\in\Z} h(n) z^{-n} \\
\hat{H}(z) &= \sum \limits_{n\in\Z} z_0^ng(n) z^{-n} \\
\hat{H}(z) &= \sum \limits_{n\in\Z} g(n) \lr{\frac{z}{z_0}}^{-n} \\
\hat{H}(z) &= \hat{G}\lr{\frac{z}{z_0}} \quad \quad R_1 \lt \abs{\frac{z}{z_0}} \lt R_2
\end{align*}

The $RoC$ simplifies to $\{ z \st \abs{z_0}R_1 \lt \abs{z} \lt \abs{z_0}R_2\}$, such that $RoC(h) = \abs{z_0}Roc(g)$.

\begin{example}

Given the following transform pair, determine what happens to the poles and zeros.

$$
e^{i\omega_0 n} g(n) \ztp  \hat{G}(e^{-i\omega_0 }z)
$$

We know that the $RoC$ is the same as $RoC(g)$, since $\abs{e^{i\omega}} = 1$. Assume that $z=p$ is a pole of $\hat{G}$. Now we have $e^{-i\omega_0} z = p$ is a pole. This implies that when $z = e^{i\omega_0} p$, we have a pole rotated by $\omega_0$ from the position of the original pole.

Suppose that $\omega_0 = \frac{\pi}{4}$, and $g(n) = u(n)$.
\begin{align*}
u(n) &\ztp \hat{U}(z) \\
u(n) &\ztp \frac{1}{1-z^{-1}} = \frac{z}{z-1}\\
e^{i\omega_0}u(n) &\ztp \hat{U}(e^{-i\omega_0}z) \\
e^{i\omega_0}u(n) &\ztp \frac{e^{-i\omega_0}z}{e^{-i\omega_0}z -1} \\
e^{i\omega_0}u(n) &\ztp \frac{z}{z - e^{i\omega_0}} \\
\end{align*}

%\img{images/roc/rotated/rot.ps}

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Here the $RoC$ stays the same since the magnitudes are equal.

\end{example}

\begin{example}

Determine what happens to the $RoC$ for $z_0 = 2$.

\begin{align*}
h(n) = 2^n u(n) &\ztp \hat{U}\lr{ \frac{z}{2} } \\
h(n) = 2^n u(n) &\ztp \frac{1}{1-2z^{-1}} \\
\end{align*}

The $RoC$ gets scaled:

%\img{images/roc/scaled/scale.ps}

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\end{example}

\begin{example}

Determine what happens to the $RoC$ for $z_0 = 2e^{-i\frac{\pi}{4}}$.

\begin{align*}
f(n) = \lr{2e^{-i\frac{\pi}{4}}}^n u(n) &\ztp \hat{U}\lr{ \frac{z}{2} } \\
f(n) = \lr{2e^{-i\frac{\pi}{4}}}^n u(n) &\ztp \frac{1}{1-\lr{2e^{-i\frac{\pi}{4}}}z^{-1}} \\
\end{align*}

The $RoC$ gets rotated and scaled, a combination of the above two cases.

\end{example}

Modulation by Complex Exponentials