f(t)=∫∞−∞F(ω)eiωtdωF(ω)=12π∫∞−∞f(t)e−iωtdt
Periodic Function
What does the transform F(ω) look like for a periodic function? Let f(t) be periodic: f(t) = ∑nf0(t−nT) where f0 is a time-limited function defined over the interval [0,T].
Compute F(ω):
\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.
