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\begin{center}
{\Large \textbf{Representation of Signals}}\\
Don H. Johnson\\
\today
\end{center}

\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
\begin{equation}\label{eq:expansion}
\signal{t} = \sum_{k=1}^{\infty} a_k\phi_k(t)\; ,
\end{equation}
where, finally, $a_k$ 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.
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}
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