When cascading to systems $g$ and $h$, what happens if we pass in a signal $x(n) = z^n$?

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Let $f = g * h$. If we follow the path of the signal piecewise, we can first solve the system for $q(n)$, and then pass $q(n)$ through $h$ to determine $y$.

\begin{align*}
q(n) &= x(n) * g(n) \\
q(n) &= \sum \limits_{k=-\infty}^{\infty} g(k) x(n-k) \\  
q(n) &= \sum \limits_{k=-\infty}^{\infty} g(k) z^{n-k} \\  
q(n) &= z^n \sum \limits_{k=-\infty}^{\infty} g(k) z^{-k} \\  
q(n) &= z^n \hat{G}(z) \\  
\end{align*}

Now the second part of the system:

\begin{align*}
y(n) &= q(n) * h(n) \\
y(n) &= z^n \hat{G}(z) * h(n) \\
y(n) &= \sum \limits_{k=-\infty}^{\infty} h(k) z^{n-k} \hat{G}(z) \\
y(n) &= z^n \hat{G}(z) \sum \limits_{k=-\infty}^{\infty} h(k) z^{-k} \\
y(n) &= z^n \hat{G}(z) \hat{H}(z) \\
\end{align*}

We find that convolution in the time domain is multiplication in the transform domain---a relationship similar to what you knew from the fourier transform:

\begin{nicebox}
$$
f(n) = (g * h)(n) \ztp \hat{F}(z) = \hat{G}(z) \hat{H}(z)
$$
\end{nicebox}

\begin{proof}

\begin{align*}
f(n) &= (g*h)(n) \\
\hat{F}(z) &= \sum \limits_{n\in\Z}(g*h)(n)z^{-n} \\
\hat{F}(z) &= \sum \limits_{n\in\Z} \sum \limits_{k\in\Z}g(k)h(n-k)z^{-n} \\
\hat{F}(z) &= \sum \limits_{k\in\Z} g(k) \sum \limits_{n\in\Z}h(n-k)z^{-n} \quad \mbox{swap summations}\\
\hat{F}(z) &= \sum \limits_{k\in\Z} g(k) \sum \limits_{m\in\Z}h(m)z^{-(m+k)} \quad \mbox{let } m = n-k\\
\hat{F}(z) &= \sum \limits_{k\in\Z} g(k) z^{-k} \sum \limits_{m\in\Z}h(m)z^{-m} \\
\hat{F}(z) &= \hat{G}(z) \hat{H}(z) \\
\end{align*}

For the system $f = g * h$, the $RoC(g) \cap RoC(h) \subset RoC(f)$. If $\hat{G}(z)$ and $\hat{H}(z)$ are rational and there are no pole-zero cancellations, then $RoC(g) \cap RoC(h) = RoC(f)$.

\end{proof}

\begin{example}

Two systems are in cascade, $\hat{G}$ and $\hat{H}$. Find the overall $RoC$ of the system.

\begin{align*}
\hat{G}(z) = \frac{1}{1 - \frac{1}{2}z^{-1}} \quad \mbox{(causal)} \\
\hat{H}(z) = \frac{z - \frac{1}{2}}{z + \frac{1}{3}} \quad \mbox{(causal)} \\
\end{align*}

First find $\hat{F}$ and solve for the poles and zeros:

\begin{align*}
\hat{F}(z) &= \hat{G}(z)\hat{H}(z) \\
\hat{F}(z) &= \lr{\frac{z}{z - \frac{1}{2}}} \lr{\frac{z - \frac{1}{2}}{z + \frac{1}{3}}} \\
\hat{F}(z) &= \frac{z\lr{z - \frac{1}{2}}}{\lr{z - \frac{1}{2}}  \lr{z + \frac{1}{3}}} \\
\hat{F}(z) &= \frac{z}{\lr{z + \frac{1}{3}}} \\
\end{align*}

So we have a pole at $-\frac{1}{3}$ and a zero at 0.

\begin{align*}
\hat{F}(z) = \frac{z}{\lr{z + \frac{1}{3}}} \iztp -\frac{1}{3}^n u(n) \quad \mbox{ where } \abs{z} \gt \frac{1}{3}
\end{align*}

Looking at the regions of convergence we find that the pole-zero cancellation changed the region of convergence.

\begin{align*}
RoC(g): &\frac{1}{2} \lt \abs{z} \\
RoC(h): &\frac{1}{3} \lt \abs{z} \\
RoC(f): &\frac{1}{2} \lt \abs{z} \\
\end{align*}

In this case $RoC(f) \gt RoC(g) \cap Roc(h)$.



In general, when given two systems in cascade, there is a good method for solving for the system function without calculating it explicitly. However, you must be given the input/output pair. For example, given a system

\begin{align*}
x \to &\fbox{h} \to y \\
y(n) = (x*h)(n) &\ztp \hat{Y}(z) = \hat{X}(z)\hat{H}(z) \\ 
\end{align*}

We can use the relation $\hat{H}(z) = \frac{\hat{Y}(z)}{\hat{X}(z)}$. You should be able to determing the system, but $RoC(x) \cap RoC(h)$ must have non-trivial overlap. This is because $RoC(y) \supseteq RoC(x) \cap RoC(h)$.

\end{example}

Systems in Cascade