The sampling theorem is what relates DT to CT. Consider a system where we multply the input signal $x(t)$ by a pulse train $q(t)$.

%\img{images/sampling/mult/justboxes.ps}
%\img{images/sampling/mult/justfunction.ps}

When we multiply the signals, we get the original signal scaled by the pulse train. The figure below shows the case where $A=1$:

%\img{images/sampling/mult/both.ps}

In essence we are tossing out values from the signal. That is the critical thing that happens when you sample a signal. How much can we afford to throw out and still recover a signal? This is what the sampling theroem tells us. Lets start by looking at the frequency spectrum of this signal when we have a bandlimited input signal given by the following spectra:

%\img{images/triangles/voice/triangle.ps}

We know that multiplication in the time domain corresponds to convolution in the frequency domain:

$$
X_q(\omega) = \frac{1}{2\pi} (X * Q)(\omega)
$$

What is $Q(\omega)$? Why not just plug in $q(t)$ into the CTFT?

\begin{align*}
Q(\omega) &= \int_{-\infty}^{\infty} q(t) e^{-i\omega t}dt \\
\end{align*}

No periodic signal can be plugged in to the analysis equation to get a fourier transform---it doesn't have finite energy and it is not absolutely integrable. So, we have got to go through a round about way. For periodic signals we talk about energy in a period. We have to use something else to figure out what the transform is. In general, no periodic signal can directly go into the CTFT.


To bypass this, we can :


\begin{enumerate}
\item Express $q(t)$ in terms of its fourier series coefficients.

\begin{align*}
q(t) &= \sum \limits_{k\in\Z} Q_k e^{ik\omega_s t} \quad \mbox{where } \omega_s = 2\pi/T \\
\end{align*}

\item Then use the CTFT on these terms.

\begin{align*}
Q(\omega) &= 2\pi \sum \limits_{k\in\Z} Q_k \delta(\omega - k\omega_s) \\
\end{align*}

\end{enumerate}

Lets derive the CTFT for this periodic signal:


We first equate the CFS synthesis with the CTFT synthesis equations:
\begin{align*}
q(t) &= \sum \limits_{k \in \Z} Q_k e^{ik\omega_s t} = \frac{1}{2\pi} \int_{\R} Q(\omega) e^{i\omega t} d\omega \\
\end{align*}

Then we can solve:

\begin{align*}
\sum \limits_{k \in \Z} Q_k e^{ik\omega_s t} &= \frac{1}{2\pi} \int_{\R} Q(\omega) e^{i\omega t} d\omega \\
\sum \limits_{k \in \Z} Q_k e^{ik\omega_s t} &= \frac{1}{2\pi} \int_{\R} \sum \limits_{k\in \Z} A_k \delta(\omega - k\omega_s)  e^{i\omega t} d\omega \\
\sum \limits_{k \in \Z} Q_k e^{ik\omega_s t} &= \sum \limits_{k\in \Z} \frac{A_k}{2\pi} \int_{\R} \delta(\omega - k\omega_s)  e^{i\omega t} d\omega \\
\sum \limits_{k \in \Z} Q_k e^{ik\omega_s t} &= \sum \limits_{k\in \Z} \frac{A_k}{2\pi} e^{ik\omega_s t}  \\
\end{align*}

Now we can equate the coefficients:

\begin{align*}
\sum \limits_{k \in \Z} Q_k e^{ik\omega_s t} &= \sum \limits_{k\in \Z} \frac{A_k}{2\pi} e^{ik\omega_s t}  \\
Q_k &= \frac{A_k}{2\pi} \\
2\pi Q_k &= A_k \\
\end{align*}

This tells us what the fourier transform is:

\begin{align*}
Q(\omega) &= 2\pi \sum \limits_{k\in \Z} Q_k \delta(\omega - k\omega_s) \\
\end{align*}


Now lets find the fourier series coefficients:

\begin{align*}
Q_k &= \frac{1}{T} \int_{\langle T \rangle} q(t) e^{-ik\omega_st}dt \\
Q_k &= \frac{A}{T} \int_{-\Delta/2}^{\Delta/2} e^{-ik\omega_st}dt \\
Q_k &= \frac{A}{T} \left. \frac{e^{-ik\omega_st}}{-ik\omega_s} \right|^{\Delta/2}_{-\Delta/2} \\
Q_k &= \frac{A}{T} \frac{2\sin( k \omega_s \Delta/2)}{k\omega_s} \\
Q_k &= \frac{A\sin( k \omega_s \Delta/2)}{k\pi} \\
\end{align*}

% image of Q_k as  kronecker deltas
%\img{images/sampling/samples/qk/graph.ps}

Now that we know $Q_k$, we can determine $Q(\omega)$:

\begin{align*}
Q(\omega) &= 2\pi \sum \limits_{k\in\Z} Q_k \delta(\omega - k\omega_s) \\
Q(\omega) &= 2\pi \sum \limits_{k\in\Z} \frac{A\sin( k \omega_s \Delta/2)}{k\pi} \delta(\omega - k\omega_s) \\
\end{align*}


% image of Q(\omega) as dirac deltas  spaces by \omega_s
% \img{images/sampling/samples/qw/graph.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-5,-1)(5,3)
    
\rput(4,1.79990000199998){$Q(\omega)$}

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

% begin x-axis labels:
\psline(-4.2, -0.1)(-4.2, 0.1)
\rput(-4.2, -0.3){$-7\omega_s$}
\psline(-3.6, -0.1)(-3.6, 0.1)
\rput(-3.6, -0.3){$-6\omega_s$}
\psline(-3, -0.1)(-3, 0.1)
\rput(-3, -0.3){$-5\omega_s$}
\psline(-2.4, -0.1)(-2.4, 0.1)
\rput(-2.4, -0.3){$-4\omega_s$}
\psline(-1.8, -0.1)(-1.8, 0.1)
\rput(-1.8, -0.3){$-3\omega_s$}
\psline(-1.2, -0.1)(-1.2, 0.1)
\rput(-1.2, -0.3){$-2\omega_s$}
\psline(-0.6, -0.1)(-0.6, 0.1)
\rput(-0.6, -0.3){$-\omega_s$}
\psline(0, -0.1)(0, 0.1)
\rput(0, -0.3){$0$}
\psline(0.6, -0.1)(0.6, 0.1)
\rput(0.6, -0.3){$\omega_s$}
\psline(1.2, -0.1)(1.2, 0.1)
\rput(1.2, -0.3){$2\omega_s$}
\psline(1.8, -0.1)(1.8, 0.1)
\rput(1.8, -0.3){$3\omega_s$}
\psline(2.4, -0.1)(2.4, 0.1)
\rput(2.4, -0.3){$4\omega_s$}
\psline(3, -0.1)(3, 0.1)
\rput(3, -0.3){$5\omega_s$}
\psline(3.6, -0.1)(3.6, 0.1)
\rput(3.6, -0.3){$6\omega_s$}
% end x-axis labels:


\fileplot[linewidth=1.0pt,linestyle=dashed,linecolor=blue!40]{sinc.dat}

%dirac
\psline[linewidth=1.25pt]{->}(-3.6, 0)(-3.6, 0.16534747163524)

%dirac
\psline[linewidth=1.25pt]{->}(-3, 0)(-3, -0.0698538745497315)

%dirac
\psline[linewidth=1.25pt]{->}(-2.4, 0)(-2.4, -0.3113014402612)

%dirac
\psline[linewidth=1.25pt]{->}(-1.8, 0)(-1.8, -0.184383518039522)

%dirac
\psline[linewidth=1.25pt]{->}(-1.2, 0)(-1.2, 0.422164487844469)

%dirac
\psline[linewidth=1.25pt]{->}(-0.6, 0)(-0.6, 1.16504885745903)

%dirac
\psline[linewidth=1.25pt]{->}(0, 0)(0, 1.49990000199998)
\rput(0, 1.7){$(2\pi Q_0)$}

%dirac
\psline[linewidth=1.25pt]{->}(0.6, 0)(0.6, 1.16504885745903)

%dirac
\psline[linewidth=1.25pt]{->}(1.2, 0)(1.2, 0.422164487844469)

%dirac
\psline[linewidth=1.25pt]{->}(1.8, 0)(1.8, -0.184383518039522)

%dirac
\psline[linewidth=1.25pt]{->}(2.4, 0)(2.4, -0.3113014402612)

%dirac
\psline[linewidth=1.25pt]{->}(3, 0)(3, -0.0698538745497315)

%dirac
\psline[linewidth=1.25pt]{->}(3.6, 0)(3.6, 0.16534747163524)

\end{pspicture}
\end{center}


Now we are ready to perform the convolution:

$$
X_q(\omega) = \frac{1}{2\pi} (X * Q)(\omega)
$$

Recall that the spectrum of our input signal was a triangle. What happens when the triangle is convolved with scaled and shifted impulses? We get scaled and shifted triangles:

% image of X_q(\omega) with the signal X convolved with Q
%\img{images/sampling/samples/xq/axis.ps}

Recoverability is based on not overlapping replicated spectra. The sampling frequency $\omega_s$ is sufficiently large such that when I replicate the triangle on adjacent multiples of fundamental frequencies they don't overlap.

%\img{images/sampling/impulse/triangles/zoom2/impulse.ps}
%\img{images/sampling/impulse/triangles/zoom/threetris2.ps}

Notice that $D \leq \omega_s - D \Rightarrow \omega_s \leq 2D$, where $2D$ is the bandwidth of the spectra. We refer to bandwidth in this course as the total footprint of the signal on the omega axis. Some books use half of this value. They would say you need two times the bandwidth to recover the original signal. Note that $2D \leq \omega_s$ is sufficient, but not necessary. Compressed sampling is an area of research that studies this concept.

Sampling using a pulse train