-
Notifications
You must be signed in to change notification settings - Fork 4
Commit
This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository.
Merge pull request #13 from kashefy/dt
density transformation makeover
- Loading branch information
Showing
26 changed files
with
2,127 additions
and
312 deletions.
There are no files selected for viewing
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -0,0 +1,55 @@ | ||
|
||
\section{The ICA problem} | ||
|
||
\begin{frame}{\secname} | ||
|
||
Let $\vec s = (s_1, s_2,...,s_N)^\top$ denote the concatenation of independent sources | ||
and $\vec x \in \R^N$ describe our observations. $\vec x$ relates to $\vec s$ through a | ||
\emph{linear transformation} $\vec A$: | ||
|
||
\begin{equation} | ||
\label{eq:ica} | ||
\vec x = \vec A \, \vec s | ||
\slidesonly{\hspace{3cm}\text{(the ICA problem)}\hspace{-4cm}} | ||
\end{equation} | ||
|
||
We refer to $\vec A$ as the \emph{mixing matrix}\notesonly{ and Eq.\ref{eq:ica} as the \emph{ICA problem}}. | ||
Solving the ICA problems is to recover $\vec s$ from only observing $\vec x$. | ||
|
||
\end{frame} | ||
|
||
\subsection{Example scenario} | ||
|
||
\begin{frame}{\subsecname} | ||
|
||
\begin{center} | ||
\includegraphics[width=0.7\textwidth]{img/setting} | ||
\notesonly{ \captionof{figure}{Example scenario} } | ||
\end{center} | ||
|
||
\notesonly{ | ||
|
||
Two speakers are placed in a room and emit signals $s_1$ and $s_2$. | ||
The speakers operate independent of one another. | ||
Two microphones are placed in the room and start recording. | ||
The first microphone is placed slightly closer to speaker 2, while | ||
the second microphone is placed slightly closer to speaker 1. | ||
$x_1$ and $x_2$ denote the recordings of the first and second microphone respectively. | ||
When we listen to the recordings we expect to hear a mix of $s_1$ and $s_2$. | ||
Since microphone 1 was placed closer to speaker 2, when we only listen to $x_1$ we hear more of $s_2$ than $s_1$. | ||
The opposite can be said when we listen only to $x_2$. | ||
|
||
Acoustic systems are linear. This means that $x_1$ is a superposition of \emph{both} sources $s_1$ and $s_2$. | ||
We will assume here that the contribution of a source $s_i$ | ||
to an observation $x_j$ is inversely proportional to the distance between the source and the microphone. | ||
The distance-contribution relationship is \emph{linear}. We don't need this to be any more realistic. | ||
|
||
If we had a measurement of the distance between each microphone and each speaker, | ||
we would tell exactly what the contribution of each of $s_1$ and $s_2$ is to each recorded observation. | ||
If we know the exact contribution of a source to an observation, we can look at both observations and recover each source in full. | ||
} | ||
|
||
\pause | ||
This is what ICA tries to solve, except that it does not have any knowledge about the spatial setting. It is blind. | ||
|
||
\end{frame} |
Oops, something went wrong.