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\author{Don H. Johnson\\[4pt]
Electrical \& Computer Engineering Department\\
Rice University\\
\url{dhj@rice.edu}
}
\title{Representation of Signals}

\begin{document}
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\maketitle

\section{Introduction}\label{sec:intro}

In simple terms, a signal $\signal{t}$ can be decomposed as a superposition---a sum---of simpler signals in most cases of interest.
In particular, we can define a \emph{basis} as a family of functions $\{\phi_k(\cdot)\}$ that represents a signal according to
\[
\signal{t} = \sum_{k=1}^{\infty} a_k\phi_k(t)\; ,
\]
where $a_k$ is a sequence of scalars defined both by the basis and the signal.

In simple terms, a signal $\signal{t}$ can be decomposed as a superposition---a sum---of simpler signals in most cases of interest.
In particular, we can define a \emph{basis} as a family of functions $\{\phi_k(\cdot)\}$ that represents a signal according to
\begin{equation}\label{eq:expansion}
\signal{t} = \sum_{k=1}^{\infty} a_k\phi_k(t)\; ,
\end{equation}
where $a_k$ is a sequence of scalars defined both by the basis and the signal.

When the basis consists of an orthonormal set of functions, the coefficients are found by calculating
\[
a_k = \int \!\signal{t} \phi_k(t)\,dt
\]

\section{Frames}\label{sec:frames}
As we saw in \S\ref{sec:intro}, signals can be represented by an orthonormal basis%
\footnote{``Orthonormal'' means $\int \!\phi_i(t)\phi_j(t)\,dt = \delta_{ij}$.}
as in \equationname~\eqref{eq:expansion}.
In this case, the expansion fits ``just right:''
it produces exactly the precise number of terms needed to represent the signal without error.
A \emph{frame} is ``fat:''
it is over-complete in that it typically employs more terms than needed~\citep{Donoho-06,Baraniuk-07,Candes-Romberg-Tao-05}.
Consequently, its basis functions do \emph{not} form an orthonormal set.
Reasons for choosing a frame rather than a basis are
\begin{itemize}
\item
The non-unique representation means that one can choose the representation best suited to alternate criteria without sacrificing accuracy.
\begin{itemize}
\item
De-noising can be performed.
\item
Sparse representations (i.e., those having the fewest terms) are (almost) always possible.
\end{itemize}
\item
Confuses the opposition.
\end{itemize}

\section{Really cool stuff}
First, some large operators both in text:
$\iiint\limits_{\mathcal{Q}} f(x,y,z)\,dx\,dy\,dz$ and $\prod_{\gamma\in\Gamma_{\tilde{C}}} \partial(\tilde{X}_{\gamma})$;
and in display mode.
\begin{equation}\label{eq:impressive}
\iiiint\limits_{Q}f(w,x,y,z)\,dw\,dx\,dy\,dz 
\le \oint_{\bm{\partial Q}} f'\left(\max\left\{\frac{\|w\|}{|w^2+x^2|}; \frac{\|z\|}{|y^2+z^2|}; \frac{\|w\oplus z\|}{\|x\oplus y\|}\right\}\right)
\precapprox \biguplus_{\mathbb{Q}\Subset\bar{\bm{Q}}}\left[f*\left(\frac{\left\lmoustache\mathbb{Q}(t)\right\rmoustache}{\sqrt{1-t^2}}\right)\right]_{t=\alpha}^{t=\vartheta}
\end{equation}
This impressive result~\eqref{eq:impressive} is way too long for the line.

For $x$ in the open interval $]-1,1[$ the infinite sum in \equationname~\eqref{eq:sum} is convergent;
however, this does not hold throughout the closed interval $[-1,1]$.
\begin{equation}\label{eq:sum}
(1-x)^{-k} = \begin{cases}
1 + \sum_{j=1}^{\infty} (-1)^{j} \fancycomb{n}{k}x^{j}& 
k\in \mathbb{N}; k\ne0\\

1 & k=0\\

\text{undefined} & k \not\in \mathbb{N}
\end{cases}
\end{equation}

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