\begin{align}
f(t) = &\int_{-\infty}^{\infty} {F(\omega) e^{i \omega t} d{\omega }}\\
F(\omega) = \frac 1 {2 \pi} &\int_{-\infty}^{\infty}{f(t) e^{-i \omega t} dt}
\end{align}
Periodic Function
What does the transform $F(\omega)$ look like for a periodic function? Let $f(t)$ be periodic: $f(t)$ = $ \sum_n {f_0(t-nT) } $ where $f_0$ is a time-limited function defined over the interval $[0, T]$.
Compute $F(\omega)$:
\begin{align}
F(\omega) &= \frac 1 {2 \pi} \int_{-\infty}^{\infty}{ \sum_{n=-\infty}^{\infty} f_0(t-nT) e^{-i \omega t} {dt} }\\
&= \sum_{n=-\infty}^{\infty} \frac 1 {2 \pi} \int_{-\infty}^{\infty}{ f_0(t-nT) e^{-i \omega t} {dt} } \\
&= \sum_{n=-\infty}^{\infty} \frac 1 {2 \pi} \int_{-\infty}^{\infty}{ f_0(t) e^{-i \omega t}e^{-i \omega nT} {dt}} \\
&= \bbox[ border:2px solid yellow ]{\left( \sum_{n=-\infty}^{\infty} e^{-i \omega nT} \right)} \bbox[ border:2px solid cyan]{\left( \frac 1 {2 \pi} \int_0^T{ f_0(t) e^{-i \omega t}{dt} } \right)}
\end{align}
The yellow box is ${\Omega} \sum_{n=-\infty}^{\infty} \delta (\omega-n \Omega)$, where $\Omega = \frac {2 \pi} T$
So we end up with
\begin{align*}
F(\omega) &= \Omega \sum_{n=-\infty}^{\infty} \delta (\omega-n \Omega) F_0(\omega) \\
&= \Omega \sum_{n=-\infty}^{\infty} \delta (\omega-n \Omega) F_0(n \Omega)
\end{align*}
$F(\omega)$ is just a sampled version of the transform of $f_0(t)$.
Sampled Function
Working the other way now, let's transform a sampled function.
Let $F$ be the transform of $f$:
$$
F(k) = \frac 1 {2 \pi} \int_{-\infty}^{\infty}{f(x) e^{-i k x} dx}
$$
$f(x)$ is a sampled version of $g(x)$:
$$
f(x) = \sum_{n=-\infty}^{\infty} \delta(x - nL) g(x)
$$
The result is immediate:
$$
F(k) = \frac 1 {2 \pi} \int_{-\infty}^{\infty}{ \left( \sum_{n=-\infty}^{\infty} \delta(x - nL) \right) g(x) e^{-i k x} dx}
$$
This looks like a transform of a product, which resolves to a convolution product of the individual transforms. Since $\sum_{n=-\infty}^{\infty} \delta(x - nL) \iff \bbox[ border:2px solid yellow ]{\frac {\Omega} {2 \pi} \sum_ {n=-\infty}^{\infty}\delta(k - n\Omega)}$, we can write
\begin{align}
F(k) &= \int_{-\infty}^{\infty} { \left( \frac {\Omega} {2 \pi} \sum_{n=-\infty}^{\infty} \delta(k - k^{'} - n\Omega) \right) G(k^{'}) dk^{'} } \label{ISI} \\
&= \frac {\Omega} {2 \pi} \sum_{n=-\infty}^{\infty} G(k - n \Omega)
\end{align}
This is a set of shifted copies of the transform of $g$, which might be a mess unless $g$ is zero outside the interval controlled by the sample rate $\Omega$. Although, we will return to this later with an interesting example of when the overlap works out nicely.
Periodic, Sampled Function
Finally, let's work out the transform of a periodic, sampled function.
\begin{align}
F(\omega) &= \frac {1}{2 \pi} \int f(t) e^{-i \omega t} {dt} \\
&=\frac {1}{2 \pi} \int \sum_{m=-\infty}^{\infty} \sum_{n=0}^{L-1} d_n \delta(t-mP-nT) e^{-i \omega t} {dt}\\
&=\frac {1}{2 \pi}\sum_m \sum_{n=0}^{L-1} d_n e^{-i \omega m P} e^{-i \omega n T}\\
&=\frac {1}{2 \pi}\bbox[ border:2px solid yellow ]{\sum_m e^{-i \omega m P}} \sum_{n=0}^{L-1} d_n e^{-i \omega n T} \\
&= \frac {1}{P}\sum_m \delta(\omega - \frac {m 2 \pi}{P}) \sum_{n=0}^{L-1} d_n e^{-i \omega n T} \\
&= \sum_m \delta(\omega - \frac {m 2 \pi}{P}) \frac {1}{P} \sum_{n=0}^{L-1} d_n e^{-i 2 \pi m n \frac {T}{P} } \\
&= \sum_m \delta(\omega - m \Omega) \frac {\Omega}{2 \pi} \sum_{n=0}^{L-1} d_n e^{-i 2 \pi m \frac {n} L } \\
&= \sum_m \delta(\omega - m \Omega) c_m
\end{align}
We used $\Omega P = 2 \pi$ and $P \equiv T L$, and defined
\begin{align*}
c_m &= \frac {\Omega}{2 \pi} \sum_{n=0}^{L-1} d_n e^{-i 2 \pi m \frac {n} L } \\
&= \left({1 \over T}\right) \left({1 \over L} \sum_{n=0}^{L-1} d_n e^{-i 2 \pi m \frac {n} L }\right)
\end{align*}
In general, the transform of a periodic function is discrete, the transform of a discrete function is periodic, and the transform of a periodic and discrete function is periodic and discrete.
The Nysquist Criterion for Avoiding ISI
Returning to equation \eqref{ISI}, consider the following system
$$ y(t) = \sum_n c_n h(t-nT) $$
Sampled at intervals $t-kT=0$, this looks like
\begin{align*}
y(kT) &= \sum_n c_n h(kT-nT) \\
\hat y(k) &= \sum_n c_n \hat h(k-n)
\end{align*}
We would like to investigate the situation where $\hat y(k)$ is just $c_k$. That is, $\hat h(k)$ are all zero with one exception: $\hat h(k) = \delta_k$. So let's look at the Fourier analysis of a discrete version of $h(t)$:
\begin{align*}
\sum_{n=-\infty}^{\infty} \delta(t - nT) h(t) &= \int \frac {\Omega} {2 \pi} \sum_{n=-\infty}^{\infty} H(\omega - n \Omega) e^{i \omega t} {d\omega} \\
\delta(t) &= \int \frac {\Omega} {2 \pi} \sum_{n=-\infty}^{\infty} H(\omega - n \Omega) e^{i \omega t} {d\omega}
\end{align*}
So we must have
$$
\Omega \sum_{n=-\infty}^{\infty} H(\omega - n \Omega) = 1
$$
In the study of communication in bandlimited channels, this is known as the Nyquist Criterion for the elimination of inter-symbol interference.
