Consider the casading system:

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% \includegraphics{images/cascade.ps}
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\begin{center}
\begin{pspicture}(0,-3)(8,3)

\rput(0,0){$x(t)$}
\rput(8.2,0){$y(t)$}
\rput(2.75,0){$f(t)$}
\rput(5.25,0){$g(t)$}

\psframe(1,-1.5)(7,1.5)
\psframe(2,-0.5)(3.5,0.5)
\psframe(4.5,-0.5)(6,0.5)

\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(0.5,0)(2,0)
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\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6,0)(7.75,0)

\end{pspicture}
\end{center}

To find $h$, simply let $x$ = $\delta$. When $x$ first enters the system, it goes through $f$ and becomes $f$. Then we pass $f$ through the system $g$. Thus $y$ is the convolution of $f$ and $g$:

$$ h(t) = (f*g)(t) $$

To find $H$, let $x(t) = e^{i \omega t}$. When $x$ first enters the system, it goes through $F$ and becomes $F(\omega)e^{i\omega t}$ since the system is linear. Then the signal enters $G$ and becomes $F(\omega)G(\omega)e^{i\omega t}$. Therefore,

$$H(\omega) = G(\omega)F(\omega)$$

We can take note that convolution in the time domain corresponds to multiplication in the frequency domain:

\begin{align*}
h(t) = (f*g)(t) &\iff H(\omega) = F(\omega)G(\omega) \\ 
\end{align*}

\begin{example}
Consider the system
%$$ x(t) \to \fbox{ $\to \fbox{f(t)} \to \fbox{$\delta(t-T)$}\to$ } \to y  $$

\begin{center}
\begin{pspicture}(0,-3)(8,3)

\rput(0,0){$x(t)$}
\rput(8.2,0){$y(t)$}
\rput(2.75,0){$f(t)$}
\rput(5.25,0){$\delta(t-T)$}

\psframe(1,-1.5)(7,1.5)
\psframe(2,-0.5)(3.5,0.5)
\psframe(4.5,-0.5)(6,0.5)

\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(0.5,0)(2,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(3.5,0)(4.5,0)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(6,0)(7.75,0)

\end{pspicture}
\end{center}

Let $x$ be $\delta$. When $x$ first enters the system it becomes $f(t)$. Then  $f(t)$ enters the system given by $\delta(t-T)$, thus we get the convolution 
$$ g(t) = \lr{f(t) * \delta(t-T)} $$

We know can determine $G(\omega)$ because $\delta(t-T)$ in the time domain corresponds to $e^{-i\omega T}$ in the frequency domain:

$$ G(\omega) = F(\omega)e^{-i\omega T}  $$
\end{example}

\begin{example}
Given $f(n)$ and $g(n)$, find $H(\omega)$ and $h(n)$:
\begin{align*}
f(n) &= \sum \limits_{k=0}^{\infty}\alpha^k\delta(n-kN) \\
g(n) &= \delta(n) - \alpha\delta(n-N) \\
\end{align*}

Lets find $G(\omega)$ first, since it is simpler: 
\begin{align*}
G(\omega) &= \sum \limits_{n\in\Z} \lr{\delta(n) - \alpha\delta(n-N)}e^{-\omega n} \\
G(\omega) &=  e^{-i\omega 0} - \alpha e^{-i\omega N} \\
G(\omega) &=  1 - \alpha e^{-i\omega N} \\
\end{align*}

This gives us some insight into $F(\omega)$:
\begin{align*}
F(\omega) &= \sum \limits_{n \in \Z} \lr{\sum \limits_{\ell=0}^{\infty} \alpha^\ell \delta(n-\ell N)}e^{-i\omega n} \\
F(\omega) &= \sum \limits_{n \in \Z} \lr{\delta(n) + \alpha \delta(n-N) + \alpha^2 \delta(n-2N) + \cdots }e^{-i\omega n} \\
F(\omega) &= \sum \limits_{n \in \Z}\delta(n)e^{-i\omega n} + \alpha \sum \limits_{n \in \Z}\delta(n-N)e^{-i\omega n} + \alpha^2 \sum \limits_{n \in \Z}\delta(n-2N)e^{-i\omega n} + \cdots \\
F(\omega) &= 1 + \alpha e^{-i\omega N} + \alpha^2 e^{-i\omega 2N} + \cdots  \\
F(\omega) &= \sum \limits_{k \in \Z_{\oplus}} \alpha^k e^{-i\omega kN} \\
F(\omega) &= \sum \limits_{k \in \Z_{\oplus}} \lr{\alpha e^{-i\omega N}}^k \\
F(\omega) &= \frac{1}{1 - \alpha e^{-i\omega N}}
\end{align*}

Now we can solve for $H(\omega)$:
\begin{align*}
H(\omega) &= F(\omega)G(\omega) \\
H(\omega) &=  \frac{1}{1 - \alpha e^{-i\omega N}} \lr{1 - \alpha e^{-i\omega N}} \\
H(\omega) &= 1
\end{align*}
This implies that $h(t) = \delta(t)$.

\end{example}


Cascade (series)