What if we have a signal that upsamples another signal by dialating it in the time domain?

$$ h(n) = \begin{cases} g\lr{\frac{n}{N}} & n\bmod N=0 \\ 0 & \mbox{otherwise} \\ \end{cases} $$

The input signal becomes dialated in the time domain by a factor of $N$---the system inserts $N-1$ zeros between samples.

\begin{align*}
\hat{H}(z) &= \sum \limits_{n\in\Z} h(n) z^{-n} \\
\hat{H}(z) &= \sum \limits_{n\in N\Z} g\lr{\frac{n}{N}} z^{-n} \\
\end{align*}

I want to mention that we performed a \emph{dilation} of the integers. A dilation of a set $X$ is a subset $cX = \{ cx \st x \in X \}$, and $c \in \Z$. Here we have successfully dilated the integers such that we are summing over the set $\{Nn \st n \in \Z \}$.

\begin{align*}
\hat{H}(z) &= \sum \limits_{mN\in N\Z} g(m) z^{-mN} \quad \mbox{ let } m = \frac{n}{N} \\
\hat{H}(z) &= \sum \limits_{mN\in N\Z} g(m) z^{-mN} \\
\hat{H}(z) &= \sum \limits_{m\in \Z} g(m) z^{-mN} \\
\end{align*}

The reason that we can now let $m$ sum over all integers is because every $mN$ element in the set $N\Z$ is factoring out $m$ implicitly, so the $N$ drops out of our limits in the sum.
 
\begin{align*}
\hat{H}(z) &= \sum \limits_{m\in \Z} g(m) \lr{z^{N}}^{-m}\\
\hat{H}(z) &= \hat{G}(z^N) \\
\end{align*}

Lets look at the $RoC$ for a dilated signal. 

\begin{align*}
RoC(g) &: R_1 \lt \abs{z} \lt R_2 \\
RoC(h) &: R_1 \lt \abs{z^N} \lt R_2 \\
RoC(h) &: \sqrt[n]{R_1} \lt \abs{z} \lt \sqrt[n]{R_2} \\
\end{align*}

We find that $RoC(h) = \sqrt[n]{RoC(g)}$. But what happens to the poles and the zeros? Assume $z=p$ is a pole of $\hat{G}(z)$, then $\hat{H}(z)$ has $N$ poles at the $N^{th}$ roots of $p$. For demonstrative purposes, assume that $N = 3$, and $\frac{1}{8}$ is a pole of $\hat{G}(z)$. 

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\begin{align*}
z = \frac{1}{8} = \frac{1}{8}e^{i2\pi k}
\end{align*}

We can take any number and multiply it by 1. Then we can take the root.

\begin{align*}
z &= \sqrt[3]{\frac{1}{8}e^{i2\pi k}} \\
z &= \frac{1}{2}e^{i\frac{2\pi}{3} k} \\
\end{align*}

This gives us the set $\{ \frac{1}{2}, \frac{1}{2}e^{i2\pi/3}, \frac{1}{2}e^{i4\pi/3} \}$. Notice that the $RoC$ has been pushed out to $\frac{1}{2}$ and now we have 3 roots:

%\img{images/roc/dilated/dilated.ps}

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(1) If $g(n)$ is stable, what can you say about $H(\omega)$? 


If $g(n)$ is stable, then $G(\omega) = \left. \hat{G}(z) \right|_{z = e^{i\omega}}$, and it is absolutely summable. We aren't adding any new values of $G$ to $H$. We are simply interspersing the same values of $G$, which implies that $H$ is also absolutely summable. We are inserting zeros between adjacent samples in $h$, so we don't change its stability.

$$ \sum \limits_{n \in Z} \abs{g(n)} \lt \infty \Rightarrow \sum \limits_{n \in Z} \abs{h(n)} \lt \infty $$

$H(\omega)$ is also well-defined.

\begin{align*}
\hat{H}(z) &= \hat{G}(z^N) \\
H(\omega)  &= \hat{G}(e^{i\omega N}) = G(\omega N) \\
H(\omega)  &=  G(\omega N) \\
\end{align*}

Dilation