The continuous-time impulse, Dirac Delta:

%\begin{figure}[htp]
%\centering
%\includegraphics{images/dirac/dirac.ps}
%\caption{ The Dirac Delta }
%\label{ fig:dirac }
%\end{figure}

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The (1) in the figure represents the area under the impulse, or the \emph{strength} of the impulse. It is not the height as in discrete-time, as the hieght of the Dirac Delta is infinite.

A definition, but not very useful one, is

$$ \delta(t) = \begin{cases} \infty & t=0 \\ 0 & \mbox{otherwise} \end{cases} $$

Lets look at some others that may prove to be more useful. Consider the unit area, where its minimum $t$ value is $-\Delta/2$, maximum $t$ value is $\Delta/2$, and a height of $1/\Delta$. This can be defined more concretely as 

$$
\delta_\Delta(t)=
\begin{cases}
0 & t<-\frac{\Delta}{2} \\
\frac{1}{\Delta} & -\frac{\Delta}{2}\leq t \leq\frac{\Delta}{2} \\
0 & t\gt\frac{\Delta}{2} \\
\end{cases}
$$

%\begin{figure}[htp]
%\centering
%\includegraphics{images/dirac/dirac-unitarea.ps}
%\caption{ The Unit Area }
%\label{ fig:dirac-unitarea }
%\end{figure}

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Then $\delta(t) = \lim_{\Delta \to 0}\delta_\Delta(t)$. The unit area goes to an infinite height, zero width, but its area does not change. 

Lets consider a triangle whose base spans $t$ from $-\Delta$ to $\Delta$, and whose height is $1/\Delta$. This can be defined as
$$
\delta_\Delta(t)=
\begin{cases}
0 & t\lt -\frac{\Delta}{2} \\
\frac{1}{\Delta^2}t+\frac{1}{\Delta} & -\frac{\Delta}{2}\leq t \lt 0 \\
-\frac{1}{\Delta^2}t+\frac{1}{\Delta} & 0\leq t \leq\frac{\Delta}{2} \\
1 & t\gt\frac{\Delta}{2} \\
\end{cases}
$$

%\begin{figure}[htp]
%\centering
%\includegraphics{images/dirac/dirac-triangle.ps}
%\caption{ A triangle with an area of 1 }
%\label{ fig:dirac-triangle }
%\end{figure}

{\bf A triangle with an area of 1}


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Again, $\delta(t) = \lim_{\Delta \to 0}\delta_\Delta(t)$. Lets consider the gaussian probability density function, where $\delta_\Delta(t) = \frac{1}{\sqrt{2\pi\Delta^2}}e^{-t^2/2\Delta^2}$.

%\begin{figure}[htp]
%\centering
%\includegraphics{images/dirac/dirac-gaussian.ps}
%\caption{ Gaussian Probability Density Function }
%\label{ fig:dirac-gaussian }
%\end{figure}



The standard deviation is $\Delta$, the maximum height at 0 is $\frac{1}{\sqrt{2\pi\Delta^2}}$. In this case $\int_{-\infty}^{\infty}\delta_\Delta(t)dt=1$, and again, $\delta(t) = \lim_{\Delta \to 0}\delta_\Delta(t)$.


The truth is, we don't care about what the delta \emph{is}, we care about what it does to other functions when it comes into contact with them.


Take an arbitrary function $f(t)$ and multiply it with the unit area definition $\delta_\Delta(t)$. We would be left with the area underneath $f(t)$ where it comes into contact in the section of the domain in $\pm\Delta/2$. The height is then $f(t)/\Delta$. If $\Delta$ is sufficiently small, then we can represent the height as $f(0)/\Delta$ from $-\Delta/2$ to $\Delta/2$. Thus, the area is equal to $f(0)$.

$$ \int_{-\infty}^{\infty}f(t)\delta_\Delta(t)dt = f(0)$$

Lets take the limit of this integral as $\Delta$ approaches 0:
\begin{align*}
\int_{-\infty}^{\infty}f(t)\delta_\Delta(t)dt &= f(0) \\
\lim_{\Delta \to 0}\int_{-\infty}^{\infty}f(t)\delta_\Delta(t)dt &= f(0) \\
\int_{-\infty}^{\infty}f(t)\lim_{\Delta \to 0}\delta_\Delta(t)dt &= f(0) \\
\int_{-\infty}^{\infty}f(t)\lim_{\Delta \to 0}\delta_\Delta(t)dt &= f(0) \\
\int_{-\infty}^{\infty}f(t)\delta(t)dt &= f(0)
\end{align*}
The value of the function is picked up wherever the impulse is active. This is a sampling property of the Dirac Delta. We can generalize this formula slightly: 

$$ \int_{-\infty}^{\infty}f(t)\delta(t-t_0)dt=f(t_0) $$

The new resulting strength on a graph would be ($f(t_0)$). This is called the \emph{sifting property} of the Dirac Delta.

\begin{problem}
What are the values of these integrals?

\begin{enumerate}
\item $\int_{-\infty}^{\infty}\cos(\omega_0t)\delta(t-T)dT$
\item $\int_{-\infty}^{\infty}\delta(t)e^{i\omega_0 t}dt$
\end{enumerate}

\end{problem}

\begin{solution}
These can be fairly simple to integrate:

\begin{enumerate}
\item $\int_{-\infty}^{\infty}\cos(\omega_0t)\delta(t-T)dT = \cos(\omega_0T)$
\item $\int_{-\infty}^{\infty}\delta(t)e^{i\omega_0 t}dt = 1$
\end{enumerate}
\end{solution}

$$ x(t) = \int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)d\tau $$
$$ \delta(t) \to \fbox{S} \to h(t)  $$
This gives us the continuous-time convolution integral:
$$ y(t) = \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau $$