What if we multiply with an impulse train instead of a pulse train? Not that this is ideal because in real life you cannot generate these impulses. Normally, signals that can be generated approximate the pulse trains. Consider a system where we multply the input signal $x(t)$ by an impulse train $g(t)$.


% \img{images/sampling/impulse/impulse/impuls.eps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-5,-1)(5,3)
    
% function name
\rput(4,1.7){$g(t) = \sum \limits_{k=-\infty}^{\infty} \delta(t-kT)$}

\rput(-4.4, 0.5){$\cdots$}
\rput(4.4, 0.5){$\cdots$}

%dirac
\psline[linewidth=1.25pt]{->}(-4, 0)(-4, 1)

%dirac
\psline[linewidth=1.25pt]{->}(-3, 0)(-3, 1)

%dirac
\psline[linewidth=1.25pt]{->}(-2, 0)(-2, 1)

%dirac
\psline[linewidth=1.25pt]{->}(-1, 0)(-1, 1)

%dirac
\psline[linewidth=1.25pt]{->}(0, 0)(0, 1)
\rput(0,1.3){(1)}

%dirac
\psline[linewidth=1.25pt]{->}(1, 0)(1, 1)

%dirac
\psline[linewidth=1.25pt]{->}(2, 0)(2, 1)

%dirac
\psline[linewidth=1.25pt]{->}(3, 0)(3, 1)

%dirac
\psline[linewidth=1.25pt]{->}(4, 0)(4, 1)


% x-axis
\psline{->}(-4.4, 0)(4.9, 0)
\rput(4.4,-0.3){$t$}

% begin x-axis labels:
\psline(-4, -0.1)(-4, 0.1)
\rput(-4, -0.3){$-4T$}
\psline(-3, -0.1)(-3, 0.1)
\rput(-3, -0.3){$-3T$}
\psline(-2, -0.1)(-2, 0.1)
\rput(-2, -0.3){$-2T$}
\psline(-1, -0.1)(-1, 0.1)
\rput(-1, -0.3){$-T$}
\psline(0, -0.1)(0, 0.1)
\rput(0, -0.3){$0$}
\psline(1, -0.1)(1, 0.1)
\rput(1, -0.3){$T$}
\psline(2, -0.1)(2, 0.1)
\rput(2, -0.3){$2T$}
\psline(3, -0.1)(3, 0.1)
\rput(3, -0.3){$3T$}
\psline(4, -0.1)(4, 0.1)
\rput(4, -0.3){$4T$}
% end x-axis labels:
\end{pspicture}
\end{center}

How does this impules train $g(t)$ relate to pulse train $q(t)$? 

$$
g(t) = \lim_{\Delta \to 0} q(t)
$$

% \img{images/sampling/impulse/multiplied/impulse.ps}

%% image of multiplication


%%% graph of x_g(t) and x(t)

What is the fourier transform of the impulse train?

%% image of impulse train
%% image of new impulse train

\begin{align*}
G(\omega) &= 2\pi \sum \limits_{k\in\Z} G_k \delta(\omega - k\omega_s) \\
G(\omega) &= 2\pi \sum \limits_{k\in\Z} \frac{1}{T} \delta(\omega - k\omega_s) \\
G(\omega) &= \frac{2\pi}{T} \sum \limits_{k\in\Z} \delta(\omega - k\omega_s) \\
G(\omega) &= \omega_s \sum \limits_{k\in\Z} \delta(\omega - k\omega_s) \\
\end{align*}

% \img{images/sampling/impulse/ftrans/impulse.ps}

\begin{center}
\begin{pspicture}(-5,-1)(5,3)
    
% function name
\rput(3.5,1.7){$G(\omega) = \omega_s \sum \limits_{k=-\infty}^{\infty} \delta(\omega - k\omega_s)$}

\rput(-4.4, 0.5){$\cdots$}
\rput(4.4, 0.5){$\cdots$}

%dirac
\psline[linewidth=1.25pt]{->}(-4, 0)(-4, 1)

%dirac
\psline[linewidth=1.25pt]{->}(-3, 0)(-3, 1)

%dirac
\psline[linewidth=1.25pt]{->}(-2, 0)(-2, 1)

%dirac
\psline[linewidth=1.25pt]{->}(-1, 0)(-1, 1)

%dirac
\psline[linewidth=1.25pt]{->}(0, 0)(0, 1)
\rput(0,1.3){$\left(\frac{2\pi}{T}\right)$}

%dirac
\psline[linewidth=1.25pt]{->}(1, 0)(1, 1)

%dirac
\psline[linewidth=1.25pt]{->}(2, 0)(2, 1)

%dirac
\psline[linewidth=1.25pt]{->}(3, 0)(3, 1)

%dirac
\psline[linewidth=1.25pt]{->}(4, 0)(4, 1)


% x-axis
\psline{->}(-4.4, 0)(4.9, 0)
\rput(4.6,-0.3){$\omega$}

% begin x-axis labels:
\psline(-4, -0.1)(-4, 0.1)
\rput(-4, -0.3){$-4\omega_s$}
\psline(-3, -0.1)(-3, 0.1)
\rput(-3, -0.3){$-3\omega_s$}
\psline(-2, -0.1)(-2, 0.1)
\rput(-2, -0.3){$-2\omega_s$}
\psline(-1, -0.1)(-1, 0.1)
\rput(-1, -0.3){$-\omega_s$}
\psline(0, -0.1)(0, 0.1)
\rput(0, -0.3){$0$}
\psline(1, -0.1)(1, 0.1)
\rput(1, -0.3){$\omega_s$}
\psline(2, -0.1)(2, 0.1)
\rput(2, -0.3){$2\omega_s$}
\psline(3, -0.1)(3, 0.1)
\rput(3, -0.3){$3\omega_s$}
\psline(4, -0.1)(4, 0.1)
\rput(4, -0.3){$4\omega_s$}
% end x-axis labels:

\end{pspicture}
\end{center}


Convolution of a function with a shifted dirac shifts the function by an amount equal to the shift of the dirac and scales it by the strength of the dirac. We have found that periodic signals have impulsive fourier transforms where impulses are uniformly spaced. Note that the strength of the specta is $\frac{2\pi}{T} = \omega_s$, and the periodicity and spacing is $\omega_s$. Recall that the impulse train has coefficients that don't depend on $k$. All frequencies equally contibute. Now we can find the convolution of the original input signal $x(t)$ and $g(t)$:

%% image of X_g(\omega)
% \img{images/sampling/impulse/triangles/impulse.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-5,-1)(5,3)
    
% function name
\rput(4,1.7){$X_g(\omega) = \frac{1}{2\pi}(X * G) (\omega)$}

\rput(-4.5, 0.5){$\cdots$}
\rput(4.5, 0.5){$\cdots$}


%fill plot for triangle.dat
\pscustom[fillstyle=solid,fillcolor=blue!40,linestyle=none]{
	\fileplot[linewidth=0pt]{triangle.dat}
}
\fileplot[linewidth=1.5pt]{triangle.dat}

\rput(0,1.4){$\left(\frac{1}{T}\right)$}

% x-axis
\psline{->}(-4.4, 0)(4.9, 0)
\rput(4.6,-0.3){$\omega$}

% begin x-axis labels:
\psline(-4, -0.1)(-4, 0.1)
\rput(-4, -0.3){$-4\omega_s$}
\psline(-3, -0.1)(-3, 0.1)
\rput(-3, -0.3){$-3\omega_s$}
\psline(-2, -0.1)(-2, 0.1)
\rput(-2, -0.3){$-2\omega_s$}
\psline(-1, -0.1)(-1, 0.1)
\rput(-1, -0.3){$-\omega_s$}
\psline(0, -0.1)(0, 0.1)
\rput(0, -0.3){$0$}
\psline(1, -0.1)(1, 0.1)
\rput(1, -0.3){$\omega_s$}
\psline(2, -0.1)(2, 0.1)
\rput(2, -0.3){$2\omega_s$}
\psline(3, -0.1)(3, 0.1)
\rput(3, -0.3){$3\omega_s$}
\psline(4, -0.1)(4, 0.1)
\rput(4, -0.3){$4\omega_s$}
% end x-axis labels:

\end{pspicture}
\end{center}

We can simply do a low pass filter to recover the signal from its samples. But we cannot over look the fact that we must sample at a certain rate to recover the signal. Lets take a closer look at the signal and its boundaries:

% \img{images/sampling/impulse/triangles/zoom/threetris.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-5,-1)(5,3)
    
% function name
\rput(2.5,1.3){$X_g(\omega) = \frac{1}{2\pi}(X * G) (\omega)$}

\rput(-2.6, 0.5){$\cdots$}
\rput(3, 0.5){$\cdots$}


%fill plot for triangle.dat
\pscustom[fillstyle=solid,fillcolor=blue!40,linestyle=none]{
	\fileplot[linewidth=0pt]{1.dat}
	\fileplot[linewidth=0pt]{2.dat}
	\fileplot[linewidth=0pt]{3.dat}
}
\fileplot[linewidth=1.5pt]{1.dat}
\fileplot[linewidth=1.5pt]{2.dat}
\fileplot[linewidth=1.5pt]{3.dat}

\rput(0,1.3){$\left(1/T_s\right)$}

% x-axis
\psline{->}(-2.7, 0)(3, 0)
\rput(3.1,-0.2){$\omega$}

% critical points

\rput(-0.5,-0.2){$-D$}
\rput(0.5,-0.2){$D$}
\psline(-0.5, -0.05)(-0.5, 0.05)
\psline(0.5, -0.05)(0.5, 0.05)

\rput(-2.5,-0.2){$-\omega_s - D$}
\rput(-1.5,-0.2){$-\omega_s + D$}
\psline(-2.5, -0.05)(-2.5, 0.05)
\psline(-1.5, -0.05)(-1.5, 0.05)

\rput(2.5,-0.2){$\omega_s + D$}
\rput(1.5,-0.2){$\omega_s - D$}
\psline(2.5, -0.05)(2.5, 0.05)
\psline(1.5, -0.05)(1.5, 0.05)

% begin x-axis labels:
\psline(-2, -0.05)(-2, 0.05)
\rput(-2, -0.2){$-\omega_s$}
\psline(-1, -0.05)(-1, 0.05)
\psline(0, -0.05)(0, 0.05)
\rput(0, -0.2){$0$}
\psline(1, -0.05)(1, 0.05)
\psline(2, -0.05)(2, 0.05)
\rput(2, -0.2){$\omega_s$}

\end{pspicture}
\end{center}

We must have $D \leq \omega_s - D$. This implies the same condition, $2D \leq \omega_s$. This is called the \emph{Nyquist Criterion}. If you have a bandlimited signal, you can sample it at a rate of the bandwidth or faster and still recover the signal. In conclusion, a set of sufficient conditions to recover a signal $x$ from its samples $x(nT)$ are:

\begin{enumerate}
\item $x$ is bandlimited
\item $\omega_s = \frac{2\pi}{T} \leq 2D$
\end{enumerate}

Note that in reality you want to make room for the rolloff of the low pass filter, since the Nyquist rate assumes an ideal low pass filter. What happens if you don't sample fast enough?

% \img{images/sampling/impulse/triangles/aliased/impulse.ps}

Notice that the higher frequency content gets ``folded'' into the lower frequency range. This phenomenon is called aliasing.

Ideal Sampling with an impulse train