[gate-users] ECAT scanner sinogram output

[Schmidtlein, Ross/Sloan-Kettering Institute]

Hi Uday,

I think you have the general picture.  Ring pairs add some complications
I forgot to mention.  

Let suppose that we have a two ring system, with rings i and j.  If we
look at just ring j or ring j we have the ii and jj sinograms which have
N/2 by N/2 dimension with interlacing (I forgot to mention last time
that radial LOR's with |r| > N/4 are often truncated from the sinogram,
this is done because of depth of penetration issues that can strongly
degrade the reconstruction of these LOR's).  In this case (ii or jj)
sinograms an LOR originating from say the top of the ring or the bottom
of the ring both travel the same path.  However, if we next look at the
ring pair ij if the LOR at the top of the sinogram goes from i to j
instead of j to i then the two LOR's paths are different.  This means
that for a ring pair i != j will have N projections planes rather than
N/2 and would thus create an N by N/2 sinogram.  The convention has been
to split these up into two sinograms ij and ji.  I'll attach some Latex
I wrote up that describes this procedure.

ross


The LORs are rearranged (binned) into a  Michelogram format (developed
by Christian Michel~\cite{3DPET:1998}).  Specifically, the LORs are
binned into a block matrix with indice dimensions in the $\phi$,
\emph{r}, and $\theta$ directions, where $\phi$ = azimuthal, \emph{r} =
radial, and $\theta$ = ring difference bins.

The azimuthal and radial binning of the LORs are performed by using the
properties of the crystal and ring number sets.  The azimuthal binning
of the LORs is achieved by using the property of this set, i.e., that
for each projection plane the sum of the crystal numbers (C\#) for a
coincidence pair will always sum to a unique set of numbers.  Each
projection plane's set contains two unique numbers that are offset from
one another by the total number of crystals in the scanner's ring (i.e.
in the GE Advance/Discovery LS scanner $N=672$, thus \emph{sum} = C\#
and C\# + 672).  The index \emph{i} for a given $\phi$ can then be
expressed as

\begin{equation}
i = \left[ \left( Crystal_{1} + Crystal_{2} \right) + Const. \right], 
\end{equation}

\noindent where the \emph{Const.} is an arbitrary offset which satisfies
the boundary condition $\phi = 0$.  In the case of interlacing, $i \in
[0,N/2)$, and $\phi_i$ is therefore given by

\begin{equation}
\phi_i = \frac{2 \pi i}{N}.
\end{equation}

Radial binning uses the property of LORs, which is that the radial
distance from the center of the scanner is related to the index
difference of the crystals that make up an LOR.  In terms of indexing,
the index $j$ for $r$ can be expressed as,

\begin{equation}
\left| j \right| = N/2 - \left| Crystal_{1} - Crystal_{2} \right|, 
\end{equation}

\noindent with $j \in [0,N/2)$ for the case of interlacing.  The sign of
\emph{j} is determined by the position of the LOR relative to $\phi_i$:

\begin{equation} j = \left\{ \begin{array}{ll} 
               +\left|j\right| & \mbox{if } crystal_{1,2} \in \left[i, i
+ N/2 \right] \\ 
               -\left|j\right| & {otherwise}\end{array} 
      \right. .
\end{equation}

\noindent This is equivalent to stating that an LOR for a crystal pair
in the right-half plane, with respect to the LOR's projection plane,
will have a positive index, whereas the left-half plane will have a
negative index.  Thus for a scanner of radius $R$, this results in the
radial distance $r_j$ being expressed by,

\begin{equation}
r_j = R \sin \left[\pi \left( \frac{j + 1/2}{N} \right) \right] . 
\end{equation}


\noindent This is equivalent to stating that an LOR for a crystal pair
in the right-half plane, with respect to the LOR's projection plane,
will have a positive index, whereas the left-half plane will have a
negative index.

Lastly, the ring difference, or the index of the axial angle $\theta$,
is simply given by the ring difference as determined above in
equation~\ref{eq:ring_num}.  Here coincidence pairs with $\theta > 0$
are stored as $(ring_1, ring_2)$ while those with $\theta < 0$ are
stored as $(ring_2, ring_1)$.  They are assigned in the following
manner,
 
\begin{equation}
(ring_1, ring_2) ~\forall~ crystal_1 \in \left[i - N/4, i + N/4 \right]
, 
\end{equation}

\noindent and

\begin{equation}
(ring_2, ring_1) ~\forall~ crystal_2 \in \left[i - N/4, i + N/4 \right]
. 
\end{equation}

\noindent The crystal that lies in the lower-half plane with respect to
the $i^{th}$ projection plane will define the first index.

In practice, the coincidence events described in this section are
interlaced from $\phi$ into $r$.  This has the effect of halving the
number of $\phi$ indices and doubling the number of $r$ indices, $i \in
[0,N) \rightarrow [0,N/2)$, and $j \in [0,N/4) \rightarrow [0,N/2)$.
The addition of interlacing somewhat complicates matters, but the
general character of the equations remains the same.
Figure~\ref{fig:sinogram_binning} illustrates the indexing strategy used
to convert the detector and ring number pairs into sinogram counts.

\begin{figure}
\includegraphics [scale=.37,angle=0]{./figures/sinogram_binning.eps}
\footnotesize{\caption{A simple example that shows the sinogram indexing
for two projection planes of a scanner with 16 detectors with
interlacing from $\phi$ into $r$.}\label{fig:sinogram_binning}}
\end{figure}




-----Original Message-----
From: gate-users-bounces at lphe1pet1.epfl.ch
[mailto:gate-users-bounces at lphe1pet1.epfl.ch] On Behalf Of Uday Tipnis
Sent: Monday, February 28, 2005 2:31 PM
To: gate-users at lphe1pet1.epfl.ch
Subject: [gate-users] ECAT scanner sinogram output


Ross,

Thank you very much for your reply. I now understand how each of the 
sinogram slices is N/2 x N/2 in dimension. But if there are m crystal 
rings then there would be (m-1)*(m-1) pairs of crystal rings. I tried to

simulate just one crystal ring but this is not possible. I need at least

two rings and that generates 4 slices. I still don't understand how m*m 
slices are generated. Could you please help?

Thanks,
Uday

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