Suppose we pass our sampled signal $x_g(t)$ through a low pass filter. 

% \img{images/sampling/lowpass/sys.ps}

\begin{center}
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% in from x
\rput(-3.2,0){$x(t)$}
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% out to y
\rput(2,0.3){$x_g(t)$}
\rput(8.6,0){$y_g(t)$}
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% impulses
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\end{center}

We know that $Y_g(\omega) = H(\omega)X_g(\omega)$, and hence, $y_g(t) = (h * x_g)(t)$. Suppose $H(\omega)$ filters out values that are outside of $(-\omega_s/2, \omega_s/2)$ and has a gain of $T_s$. What is $h(t)$?

\begin{align*}
h(t) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} H(\omega) e^{-i\omega t} d\omega \\
h(t) &= \frac{1}{2\pi} \int_{-\omega_s/2}^{\omega_s/2} T_s e^{-i\omega t} d\omega \\
h(t) &= \frac{T_s}{2\pi} \left. \frac{e^{-i\omega t}}{-it} \right|_{-\omega_s/2}^{\omega_s/2} \\
h(t) &= \frac{1}{\omega_s} \frac{e^{-i(\omega_s/2) t} - e^{i(\omega_s/2) t}}{-it} \\
h(t) &= \frac{1}{\omega_s} \frac{2\sin((\omega_s/2) t)}{t} \\
h(t) &= \frac{\sin(\frac{\omega_s}{2}t)}{\frac{\omega_s}{2}t} \\
\end{align*}

Recall the relation between $x_g(t)$ and $x_d(n)$, $x_g(t) = \sum \limits_{n\in\Z} x_d(n) \delta(t - nT_s)$. 

\begin{align*}
(h * x_g)(t) &= \infint h(\tau) x_g(t - \tau) d\tau \\
(h * x_g)(t) &= \infint h(\tau) \sum \limits_{n\in\Z} x_d(n) \delta(t - \tau - nT_s) d\tau \\
(h * x_g)(t) &= \sum \limits_{n\in\Z} x_d(n) \infint h(\tau) \delta(t - \tau - nT_s) d\tau \\
(h * x_g)(t) &= \sum \limits_{n\in\Z} x_d(n) h(t - nT_s) \\
\end{align*}

In this case, $y_g(t) = (h * x_g)(t) = x(t)$. This is because we pass it through a low pass filter. This means that $x$ is being expressed as a linear combination of its samples and its shifted impulse responses $h(t)$, where $h(t)$ is a sinc function. Our input signal $x$ was bandlimited. We sampled it at or above the Nyquist rate. If you take a bandlimited signal and sample it at or above the Nyquist rate, you can represent that signal as a linear combination of shifted sincs. Before when doing fourier analysis, we could represent functions as a linear combination of complex exponentials. Now that we are dealing with the space of bandlimited signals (in this case from $-\omega_s/2$ to $\omega/2$), we can represent functions as a linear combination of sincs.

% \img{images/sampling/sincs/impulseresp/graph.ps}

Notice that this sinc function has a very special property. It is 1 at 0, and is 0 at every other crossing. That is, any time it crosses the horizontal axis at $t = kT_s$ such that $t \neq 0$, the function is zero. We can shift the sinc to the right and or the left, but at any of these sample points are zero except the sinc whose peak is at that value. This allows us to take this as a basic function and replicate it where we took our samples from $x(t)$:

% \img{images/sampling/sincs/sincs/multiplied/g3.ps}

Notice that all but the peak are at zero crossings of the sample rate.

% \img{images/sampling/sincs/sincs/multiplied/g2.ps}
% \img{images/sampling/sincs/sincs/multiplied/g1.ps}

The final result is the addition of the sinc functions:

% \img{images/sampling/sincs/sincs/multiplied/graph.ps}

We have used the fact that these sincs form an orthogonal basis---in the space of bandlimited signals. It is hard to prove directly that these form an orthogonal basis, but you can do so by first showing the the inner product in the time domain is equal to $\frac{1}{2\pi}$ times the inner product in the frequency domain:

\begin{align*}
\langle x,y\rangle  &\bydef \infint x(t) y^*(t) dt \\
\langle X,Y\rangle  &\bydef \infint X(\omega) Y^*(\omega) d\omega \\
\end{align*}

Want to show that $\langle x,y\rangle  = \frac{1}{2\pi} \langle X,Y\rangle $.

\begin{proof}
\begin{align*}
\langle x,y\rangle  &= \infint x(t) y^*(t) dt \\
\langle x,y\rangle  &= \infint x(t) \lr{\frac{1}{2\pi} \infint Y(\omega) e^{i\omega t} d\omega}^* dt \\
\langle x,y\rangle  &= \frac{1}{2\pi} \infint x(t)  \infint Y^*(\omega) e^{-i\omega t} d\omega dt \\
\langle x,y\rangle  &= \frac{1}{2\pi} \infint Y^*(\omega) \lr{\infint x(t) e^{-i\omega t} dt} d\omega \\
\langle x,y\rangle  &= \frac{1}{2\pi} \infint Y^*(\omega) X(\omega) d\omega\\
\langle x,y\rangle  &= \frac{1}{2\pi} \langle X,Y\rangle \\
\end{align*}

Now to show orthogonality of the sinc functions:

\begin{align*}
\phi_k(t) = sinc(t-k) \bydef \frac{\sin(\pi(t-k))}{\pi(t-k)} \\
\end{align*}

Want to show $\langle \phi_k, \phi_\ell\rangle  = c \delta(k-\ell) \mbox{ } \exists c \in \Z\setminus \{0\}$

\begin{align*}
\langle \Phi_k, \Phi_\ell\rangle  &= \infint \Phi_k(\omega) \Phi_\ell^*(\omega) d\omega \\
\langle \Phi_k, \Phi_\ell\rangle  &= \infint \lr{\Phi_0(\omega) e^{i\omega k}} \lr{\Phi_0(\omega) e^{i\omega \ell}}^* d\omega \\
\langle \Phi_k, \Phi_\ell\rangle  &= \infint \abs{\Phi_0(\omega)}^2 e^{i\omega k} e^{-i\omega \ell} d\omega \\
\langle \Phi_k, \Phi_\ell\rangle  &= \int_{-\pi}^{\pi} e^{i\omega k} e^{-i\omega \ell} d\omega \\
\langle \Phi_k, \Phi_\ell\rangle  &= 2\pi \delta(k-\ell) \\
2\pi\langle \phi_k, \phi_\ell\rangle  &= 2\pi \delta(k-\ell) \quad \mbox{ (by proof above)} \\
\langle \phi_k, \phi_\ell\rangle  &= \delta(k-\ell) \\
\end{align*}

Therefore, $\phi_k \perp \phi_\ell \mbox{ } \forall k \neq \ell$.

\end{proof}

Time Domain Representation of a Low Pass Filter