There are cases where you want to use digital signal processing on a continuous time signal. So you have got to convert the input signal to discrete time signal, do the processing in discrete time, then finally reconstruct the continuous time signal. We do DT processing of CT signals. This allows us to use digital circuits.

% \img{images/conversions/processing/dtpofct.ps}

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%frame
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\psframe(3.5,-0.25)(2.5,0.25)


%frame
\rput(5,0){$D/C$}
\psframe(5.5,-0.25)(4.5,0.25)

%text
\rput(-0.5,0){$x(t)$}

%text
\rput(6.5,0){$y(t)$}

%text
\rput(1,-1.2){$T_s$}

%text
\rput(5,-1.2){$T_r$}

%text
\rput(2,0.2){$x_d(n)$}

%text
\rput(4,0.2){$y_d(n)$}


% wire with arrow 
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Where $T_s$ is the sampling period, $T_r$ is the reconstruction period, C/D means CT to DT conversion, and D/C means DT to CT conversion. We also may want to utilize analog cicuits to process digital signals. 

% \img{images/conversions/processing/ctpofdt.ps}

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%frame
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%frame
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\psframe(3.5,-0.25)(2.5,0.25)


%frame
\rput(5,0){$C/D$}
\psframe(5.5,-0.25)(4.5,0.25)

%text
\rput(-0.5,0){$x_d(n)$}

%text
\rput(6.5,0){$y_d(n)$}

%text
\rput(1,-1.2){$T_r$}

%text
\rput(5,-1.2){$T_s$}

%text
\rput(2,0.2){$x(t)$}

%text
\rput(4,0.2){$y(t)$}


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So far, we have are talked about taking a continuous time signal, sampling it with an impulse train so we get $x_g(t)$.  Then there is a box that converts the signal, let's call this the ``Dirac to Kronecker'', and puts out an $x_d(n)$. If we look closely at the C/D box, we have the following system:

% \img{images/conversions/ctod/cd.ps}

\begin{center}
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%frame
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\rput(5,0){to}
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%text
\rput(-0.5,0){$x(t)$}

%text
\rput(7,0){$x_d(n)$}

%text
\rput(2,-1.2){$g(t)=\sum\limits_{k=-\infty}^{\infty}\delta(t-kT_s)$}

%text
\rput(3.25,0.2){$x_g(n)$}


% wire with arrow 
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Lets start out relating the discrete time signal $x_d(n)$ to the continuous time signal $x(t)$. When we look at the system, we start with $x(t)$ and multiply that by an impulse train $g(t)$ to obtain $x_g(t)$. 

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

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\begin{pspicture}(-5,-1.5)(6,2.5)

    
% function name
\rput(4,1.7){$x_g(t)$}


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%dirac
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%dirac
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% x-axis


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\psline(4, -0.1)(4, 0.1)
\rput(4, -0.3){$4T_s$}
% end x-axis labels:


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\end{center}

We pass this signal into the ``Dirac to Kronecker'' box. From there we get our discrete time signal.

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

\begin{center}
\begin{pspicture}(-5,-1.5)(6,2.5)


    
% function name
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\rput(-1, 0.8){$(x(-T_s))$}

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\psline[linewidth=1.25pt]{-*}(0, 0)(0, 0.152)
\rput(0, 0.4){$(x(0))$}

%dirac
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\rput(1, -1){$(x(T_s))$}

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\psline[linewidth=1.25pt]{-*}(2, 0)(2, -0.4816)

%dirac
\psline[linewidth=1.25pt]{-*}(3, 0)(3, -0.064)

%dirac
\psline[linewidth=1.25pt]{-*}(4, 0)(4, 1.2032)
% x-axis


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\end{pspicture}
\end{center}

Notice the relationship between $x_d(n)$ and $x(t)$:

\begin{align*}
x_d(n) &= \left. x(t) \right|_{t = nT_s} \\
x_d(n) &= x(nT_s) \\
\end{align*}

Lets express $x_g(t)$ in terms of $x_d(n)$. This will give us intuition for plotting $X_d(\Omega)$. Looking at the two plots above for $x_g(t)$ and $x_d(n)$, we can see that $x_g(t)$ is just a bunch of shifted and scaled deltas. Each delta has a strength of $x(nT_s) = x_d(n)$.

$$
x_g(t) = \sum \limits_{n \in \Z} x(nT_s) \delta(t - nT_s) = \sum \limits_{n \in \Z} x_d(n) \delta(t - nT_s) 
$$

Now lets solve for the fourier transform of $x_g(t)$ when in terms of $x_d(n)$:

\begin{align*}
x_g(t) &= \sum \limits_{n \in \Z} x_d(n) \delta(t - nT_s)  \\
\ftrans{x_g(t)} &= \sum \limits_{n \in \Z} x_d(n) e^{-i\omega nT_s}  \\
X_g(\omega) &= \sum \limits_{n \in \Z} x_d(n) e^{-i (\omega T_s) n}  \\
X_g(\omega) &= X_d(\omega T_s) \\
\end{align*}

This gives us the relation:

$$
X_g(\omega) = \left. X_d(\Omega) \right|_{\Omega = \omega T_s} 
$$

Note that we are using $\Omega$ to denote radians per sample and $\omega$ radians per second. We can check if the units make sense:

$$
\Omega \frac{rad}{sample} = \omega \frac{rad}{sec} T_s \frac{sec}{sample} 
$$


Here is a way to solve for the freqency response $X_d(\Omega)$ explicity by first solving for $X_g(\omega)$ and then using the relation $X_g(\omega) = \left. X_d(\Omega) \right|_{\Omega = \omega T_s}$:

\begin{align*}
x_g(t) &= x(t)g(t) \\
\ftrans{x_g(t)} &= \frac{1}{2\pi} \lr{X(\omega) * G(\omega)} \\
X_g(\omega) &= \frac{1}{2\pi} \lr{\infint X(\omega - \Omega) G(\Omega) d\Omega} \\
X_g(\omega) &= \frac{1}{2\pi} \lr{\infint X(\omega - \Omega) 2\pi \sum \limits_{k\in\Z} G_k \delta(\Omega - k\omega_s)d\Omega} \\
X_g(\omega) &= \sum \limits_{k\in\Z} G_k \infint X(\omega - \Omega) \delta(\Omega - k\omega_s)d\Omega \\
X_g(\omega) &= \sum \limits_{k\in\Z} G_k X(\omega - k\omega_s) \\
X_g(\omega) &= \frac{1}{T_s} \sum \limits_{k\in\Z} X(\omega - k\omega_s) \\
\end{align*}


If you look at the relation between $X_d(\Omega)$ and $X_g(\omega)$,

\begin{align*}
X_g(\omega) &= \left. X_d(\Omega) \right|_{\Omega = \omega T_s} \\
X_d(\Omega) &= \left. X_g(\omega) \right|_{\omega = \Omega/T_s} \\
\end{align*}


this relation can help us solve explicity for $X_d(\Omega)$:

\begin{align*}
X_d(\omega T_s) &= X_g(\omega) = \frac{1}{T_s} \sum \limits_{k\in\Z} X(\omega - k\omega_s) \\
X_d(\Omega) &= X_g\lr{\frac{\Omega}{T_s}} = \frac{1}{T_s} \sum \limits_{k\in\Z} X\lr{\frac{\Omega}{T_s} - k\omega_s} \quad \mbox{let } \Omega = \omega T_s \\
X_d(\Omega) &= \frac{1}{T_s} \sum \limits_{k\in\Z} X\lr{\frac{\Omega}{T_s} - k\omega_s} \\
X_d(\Omega) &= \frac{1}{T_s} \sum \limits_{k\in\Z} X\lr{\frac{\Omega - 2\pi k}{T_s}} \\
\end{align*}

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

If you look at either the calculations or the graph, you will see that all we did was multiply $\omega$ by $T_s$, hence, $\Omega = \omega T_s$. So each replica of the original spectrum, which was at multiple of $\omega_s$, now comes down to be at a multiple of $2\pi$. That is a big relief because the DTFT has to be periodic with period $2\pi$. Scaling is done on the horizontal axis, the frequency axis---not an the vertical axis. The vertical peak is still $\frac{1}{T_s}$. This result is frequency rescaling. Looking at the plot of $X_g(\omega)$, you can see this very clearly; we have only multiplied the $\omega$ axis by $T_s$:

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

If we want to relate $X_d(\Omega)$ to $X(\omega)$, we find that

$$
X_d(\Omega) = \left. \frac{1}{T_s} X(\omega) \right|_{\omega=\frac{\Omega}{T_s}} \quad \forall \abs{\Omega} \lt \pi
$$

We cannot forget to limit $\Omega$ when writing the signal in this form. The sampling creates more replicas, so we must limit the domain to allow this definition.

\begin{align*}
X_d(\Omega) &= \frac{1}{T_s} \sum \limits_{k\in\Z} X\lr{\frac{\Omega - 2\pi k}{T_s}} \\
X_d(\Omega) &= \frac{1}{T_s} X\lr{\frac{\Omega}{T_s}} \quad \mbox{(restricted domain)}\\
\end{align*}

Here we can see that we are effectively scaling the frequency and the amplitude. 

Overview of Sampling

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){$x(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(-2.7,0)(-0.25,0)
% out to y
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% multiplier
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\psline(0.175,0.175)(-0.175,-0.175)

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\rput(4.75, -0.3){$\frac{\omega_s}{2}$}

\psframe(2.65, -0.75)(6, 1)


% impulses
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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}

8 Week 8