Given a continuous signal \(x\left ( t\right ) \), and we want to sample it. One way to sample it, is to multiply \(x\left ( t\right ) \) it with a pulse train called \(p\left ( t\right ) \), where the time between each pulse is \(T\) sec (hence sampling frequency is \(F_{s}=\frac{1}{T}\)). Let \(x_{p}\left ( t\right ) \) be the generated signal. Notice that \(x_{p}\left ( t\right ) \) will only have values at those time locations where the pulses from the pulse train are located.
The diagram below explains the above. It is from the book ”signals and systems” by Oppenheim and Willsky:
Hence we have
\[ p\left ( t\right ) ={\sum \limits _{n=-\infty }^{\infty }}\delta \left ( t-nT\right ) \]
And the sampled signal is
\begin{align*} x_{p}\left ( t\right ) & =x\left ( t\right ) p\left ( t\right ) \\ & =x\left ( t\right ){\sum \limits _{n=-\infty }^{\infty }}\delta \left ( t-nT\right ) \end{align*}
The goal now is to find the spectrum of \(x_{p}\left ( t\right ) \) and compare it to spectrum of \(x\left ( t\right ) \)
Since \(x_{p}\left ( t\right ) =x\left ( t\right ) p\left ( t\right ) \), then applying Fourier transform we obtain \begin{equation} X_{p}\left ( \omega \right ) =\frac{1}{2\pi }X\left ( \omega \right ) \circledast P\left ( \omega \right ) \tag{1} \end{equation}
Where \(X\left ( \omega \right ) \) is Fourier transform of \(x\left ( t\right ) \), and \(P\left ( \omega \right ) \) is Fourier transform of \(p\left ( t\right ) \). This relation is from convolution properties where multiplication in time domain, becomes convolution in frequency domain, with \(\frac{1}{2\pi }\) factor added.
Now we need to find \(P\left ( \omega \right ) \). But the pulse train is periodic? so we really should be talking about Fourier series coefficients here, not Fourier transform? as Fourier transform is meant for aperiodic signals? Yes, Fourier transform is meant to be used for aperiodic signals, but there is a trick to come up with a Fourier transform for also a periodic signal such as the pulse train here. To apply this trick, first recall that for a periodic signal, it can be approximated by weight sum of complex exponential
\begin{equation} p\left ( t\right ) ={\sum \limits _{n=-\infty }^{\infty }}c\left ( n\right ) e^{j\omega _{0}nt}\tag{2} \end{equation}
Where \(c\left ( k\right ) \) (the weights) are called the Fourier coefficients, and these can be found using
\[ c\left ( n\right ) =\frac{1}{T}{\int \limits _{-T/2}^{T/2}}p\left ( t\right ) e^{j\omega _{0}nt}dt \]
Where \(T\) is the fundamental period of \(p\left ( t\right ) \) and \(\omega _{0}=\frac{2\pi }{T}\). Hence for the pulse train \(p\left ( t\right ) \) we have
\begin{align*} c\left ( n\right ) & =\frac{1}{T}{\int \limits _{-T/2}^{T/2}}\delta \left ( t\right ) e^{j\omega _{0}nt}dt\\ & =\frac{1}{T}{\int \limits _{-T/2}^{T/2}}\delta \left ( t\right ) dt \end{align*}
Hence
\begin{equation} c\left ( n\right ) =\frac{1}{T} \tag{3} \end{equation}
The above is all good and well, but we are back the question of how to find Fourier transform \(P\left ( \omega \right ) \) for the periodic signal \(p\left ( t\right ) \)? Ok, here is the trick. We go backward. We ask, what function of \(P\left ( \omega \right ) \) has as its inverse Fourier transform an aperiodic signal \(\tilde{p}\left ( t\right ) \) whose Fourier series approximation is \(p\left ( t\right ) \) as defined in (2) above? i.e. we want to solve for \(P\left ( \omega \right ) \) in
\begin{align} \tilde{p}\left ( t\right ) & =\frac{1}{2\pi }{\int \limits _{-\infty }^{\infty }}P\left ( \omega \right ) e^{j\omega t}d\omega \nonumber \\{\sum \limits _{n=-\infty }^{\infty }}c\left ( n\right ) e^{j\omega _{0}nt} & =\frac{1}{2\pi }{\int \limits _{-\infty }^{\infty }}P\left ( \omega \right ) e^{j\omega t}d\omega \tag{4} \end{align}
Looking at the above long enough, we ask, what should \(P\left ( \omega \right ) \) be to make RHS be the same as LHS? Let us try \begin{equation} P\left ( \omega \right ) ={\sum \limits _{n=-\infty }^{\infty }}2\pi c\left ( n\right ) \delta \left ( \omega -n\omega _{0}\right ) \tag{5} \end{equation} and see if this does it. Plug the above into (4) we have
\begin{align*}{\sum \limits _{n=-\infty }^{\infty }}c\left ( n\right ) e^{jn\omega _{0}t} & =\frac{1}{2\pi }{\int \limits _{-\infty }^{\infty }}\left ({\sum \limits _{n=-\infty }^{\infty }}2\pi c\left ( n\right ) \delta \left ( \omega -n\omega _{0}\right ) \right ) e^{j\omega t}d\omega \\ & ={\sum \limits _{n=-\infty }^{\infty }}c\left ( n\right ) \left [{\int \limits _{-\infty }^{\infty }}\delta \left ( \omega -n\omega _{0}\right ) e^{j\omega t}d\omega \right ] \\ & ={\sum \limits _{n=-\infty }^{\infty }}c\left ( n\right ) \left [{\int \limits _{-\infty }^{\infty }}\delta \left ( \omega -n\omega _{0}\right ) e^{jn\omega _{0}t}d\omega \right ] \\ & ={\sum \limits _{n=-\infty }^{\infty }}c\left ( n\right ) \left [ e^{jn\omega _{0}t}{\int \limits _{-\infty }^{\infty }}\delta \left ( \omega -n\omega _{0}\right ) d\omega \right ] \\ & ={\sum \limits _{n=-\infty }^{\infty }}c\left ( n\right ) e^{jn\omega _{0}t} \end{align*}
That is it!, hence we conclude that \(p\left ( t\right ) \) has Fourier transform given by (5) above. Now that we found \(P\left ( \omega \right ) \) we go back to (1) and write
\begin{align*} X_{p}\left ( \omega \right ) & =\frac{1}{2\pi }X\left ( \omega \right ) \circledast P\left ( \omega \right ) \\ & =\frac{1}{2\pi }X\left ( \omega \right ) \circledast{\sum \limits _{n=-\infty }^{\infty }}2\pi c\left ( n\right ) \delta \left ( \omega -n\omega _{0}\right ) \end{align*}
But we found that \(c\left ( n\right ) =\frac{1}{T}\), hence above simplifies to
\begin{align*} X_{p}\left ( \omega \right ) & =\frac{1}{T}X\left ( \omega \right ) \circledast{\sum \limits _{n=-\infty }^{\infty }}\delta \left ( \omega -n\omega _{0}\right ) \\ & =\frac{1}{T}{\sum \limits _{n=-\infty }^{\infty }}X\left ( \omega -n\omega _{0}\right ) \end{align*}
Hence we see that the spectrum of the sampled signal \(x_{p}\left ( t\right ) \) is a scaled and shifted version of the spectrum of the signal \(x\left ( t\right ) \) that was sampled. Also we see that many copies of \(X\left ( \omega \right ) \) exist, each one is centered at integer multiples of \(\omega _{0}\) (the radial sampling frequency). The above result immediately leads also to determining what value of \(\omega _{0}\) should be to prevent copies of \(X\left ( \omega \right ) \) to overlap with each others. We see that \(\omega _{0}\) must be larger than twice the bandwidth of \(X\left ( \omega \right ) \), hence Nyquist theory.
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