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Appendix.lyx
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Appendix.lyx
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#LyX 2.0 created this file. For more info see http://www.lyx.org/
\lyxformat 413
\begin_document
\begin_header
\textclass article
\begin_preamble
\numberwithin{figure}{section}
\numberwithin{table}{section}
\end_preamble
\use_default_options true
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\language english
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\end_header
\begin_body
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
appendix
\end_layout
\end_inset
\end_layout
\begin_layout Section
Appendix
\end_layout
\begin_layout Subsection
Data and sequences
\end_layout
\begin_layout Standard
Summary of the acquisition protocols used in this thesis.
\begin_inset Newline newline
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
noindent
\end_layout
\end_inset
\series bold
101 and 117 gradients at MRC-CBU
\series default
.
3T scanner (TIM Trio, Siemens) with a 12 channel coil, using Siemens advanced
diffusion work-in-progress sequence, and STEAM
\begin_inset CommandInset citation
LatexCommand cite
key "merboldt1992diffusion,MAB04"
\end_inset
as the diffusion preparation method.
The field of view was
\begin_inset Formula $240\times240\, mm^{2}$
\end_inset
, matrix size
\begin_inset Formula $96x96$
\end_inset
, and slice thickness
\begin_inset Formula $2.5mm$
\end_inset
(no gap).
55 slices were acquired to achieve full brain coverage, and the voxel resolutio
n was
\begin_inset Formula $2.5\times2.5\times2.5\, mm{}^{3}$
\end_inset
.
Two sampling schemes were considered: a
\begin_inset Formula $102$
\end_inset
-point half grid acquisition(TR=
\begin_inset Formula $8,200\, ms$
\end_inset
, TE=
\begin_inset Formula $69\, ms$
\end_inset
) with a maximum b-value of
\begin_inset Formula $4,000\, s/mm^{2}$
\end_inset
, and a single shell acquisition using
\begin_inset Formula $118$
\end_inset
non-collinear gradient directions (TR=
\begin_inset Formula $7,000\, ms$
\end_inset
, TE=
\begin_inset Formula $47ms$
\end_inset
) and a b-value of
\begin_inset Formula $1,000\, s/mm^{2}$
\end_inset
.
The two acquisition schemes were matched for total acquisition time (
\begin_inset Formula $14\, min\,37s$
\end_inset
), voxel resolution, and bandwidth.
\begin_inset Newline newline
\end_inset
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
noindent
\end_layout
\end_inset
\series bold
101 gradients at MRC-CBU
\series default
.
3T scanner (TIM Trio, Siemens) with a 32 channel coil, using Siemens advanced
diffusion work-in-progress sequence, and STEAM
\begin_inset CommandInset citation
LatexCommand cite
key "merboldt1992diffusion,MAB04"
\end_inset
as the diffusion preparation method.
The field of view was
\begin_inset Formula $240\times240\, mm^{2}$
\end_inset
, matrix size
\begin_inset Formula $96\times96$
\end_inset
, and slice thickness
\begin_inset Formula $2.5\, mm$
\end_inset
(no gap).
55 slices were acquired to achieve full brain coverage, and the voxel resolutio
n was
\begin_inset Formula $2.5\times2.5\times2.5\, mm{}^{3}$
\end_inset
.A 102-point half grid acquisition with a maximum b-value of
\begin_inset Formula $4,000\, s/mm^{2}$
\end_inset
was used.
The total acquisition time was
\begin_inset Formula $14\, min\,21s$
\end_inset
with TR=
\begin_inset Formula $8,200ms$
\end_inset
and TE=
\begin_inset Formula $69ms$
\end_inset
.
\begin_inset Newline newline
\end_inset
\series bold
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
noindent
\end_layout
\end_inset
\series bold
257 gradients at EPFL
\series default
.
3T scanner (TIM Trio, Siemens) with a 32 channel coil.
The field of view was
\begin_inset Formula $210\times210\, mm^{2}$
\end_inset
, matrix size
\begin_inset Formula $96\times96$
\end_inset
, and slice thickness
\begin_inset Formula $3\, mm$
\end_inset
.
44 slices were acquired and the voxel resolution was
\begin_inset Formula $2.2\times2.2\times3.0\, mm{}^{3}$
\end_inset
.A 258-point half grid acquisition scheme with a maximum b-value of
\begin_inset Formula $8011\, s/mm^{2}$
\end_inset
(DSI515) was used.
The total acquisition time was
\begin_inset Formula $34\, min$
\end_inset
with TR=
\begin_inset Formula $8200\, ms$
\end_inset
and TE=
\begin_inset Formula $165\, ms$
\end_inset
.
\end_layout
\begin_layout Subsection
The Cosine Transform
\begin_inset CommandInset label
LatexCommand label
name "sub:The-cosine-transform"
\end_inset
\end_layout
\begin_layout Standard
Here we provide a simple derivation of the cosine transform:
\begin_inset Formula $\int_{0}^{\infty}\cos(st)g(t)dt$
\end_inset
where
\begin_inset Formula $g(t)$
\end_inset
defined on
\begin_inset Formula $[t,\infty)$
\end_inset
.
\end_layout
\begin_layout Standard
Let
\begin_inset Formula $f(t)$
\end_inset
be an even function,
\begin_inset Formula $f(t)=f(-t)$
\end_inset
, defined for
\begin_inset Formula $-\infty<t<\infty$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray*}
F(s) & = & \int_{-\infty}^{\infty}f(t)e^{its}dt\\
& = & \int_{0}^{\infty}f(t)e^{its}dt+\int_{-\infty}^{0}f(t)e^{its}dt\\
& = & \int_{0}^{\infty}f(t)e^{its}dt-\int_{-\infty}^{0}f(-t)e^{-its}dt\\
& = & \int_{0}^{\infty}f(t)e^{its}dt+\int_{0}^{\infty}f(t)e^{-its}dt\\
& = & \int_{0}^{\infty}f(t)[e^{its}+e^{-its}]dt\\
& = & \int_{0}^{\infty}f(t)[\cos(its)+i\sin(its)+\cos(its)-i\sin(its)]dt\\
& = & 2\int_{0}^{\infty}f(t)cos(st)dt
\end{eqnarray*}
\end_inset
\begin_inset Formula $\noindent$
\end_inset
\begin_inset space ~
\end_inset
In the third row above we replaced
\begin_inset Formula $t\rightarrow-t$
\end_inset
in the second integral.
If we want to express the Fourier transform of a symmetric function as
an integral over the whole space we have
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $F(s)=\int_{-\infty}^{\infty}f(t)\cos(st)dt$
\end_inset
.
\end_layout
\begin_layout Subsection
Fourier Transform of
\begin_inset Formula $P(\mathbf{r})r^{2}$
\end_inset
\begin_inset CommandInset label
LatexCommand label
name "sub:Fourier-transform-ofPr2"
\end_inset
\end_layout
\begin_layout Standard
From Fourier analysis we know that if
\begin_inset Formula $E(\mathbf{q})$
\end_inset
is the Fourier transform function of
\begin_inset Formula $P(\mathbf{r})$
\end_inset
then
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray*}
\mathfrak{F\mathrm{(xP(\mathbf{r}))}} & = & i\frac{\partial E(\mathbf{q})}{\partial\mathbf{q}_{x}}\\
\mathfrak{F\mathrm{(x^{2}P(\mathbf{r}))}} & = & -\frac{\partial^{2}E(\mathbf{q})}{\partial\mathbf{q}_{x}^{2}}
\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
\align left
where
\begin_inset Formula $\mathfrak{F\mathrm{()}}$
\end_inset
is the Fourier transform.
By writing the second equation for
\begin_inset Formula $y$
\end_inset
and
\begin_inset Formula $z$
\end_inset
and summing them all together we obtain
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray*}
\mathfrak{F\mathrm{(r^{2}P(\mathbf{r}))}} & = & -\frac{\partial^{2}E(\mathbf{q})}{\partial\mathbf{q}_{x}^{2}}-\frac{\partial^{2}E(\mathbf{q})}{\partial\mathbf{q}_{y}^{2}}-\frac{\partial^{2}E(\mathbf{q})}{\partial\mathbf{q}_{z}^{2}}=-\nabla^{2}E(\mathbf{q})
\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection
Radial projection of a symmetric function
\begin_inset CommandInset label
LatexCommand label
name "sub:Radial-projection-of-symmetric"
\end_inset
\end_layout
\begin_layout Standard
Let
\begin_inset Formula $f:\mathbb{R}^{3}\rightarrow\mathbb{R}$
\end_inset
be a symmetric function with the 3D Fourier transform function
\begin_inset Formula $\hat{f}(\mathbf{q})$
\end_inset
and
\begin_inset Formula $\hat{\mathbf{u}}$
\end_inset
be an arbitrary unit vector.
We will show that
\begin_inset Formula $\intop_{0}^{\infty}f(r\hat{\mathbf{u}})dr=\frac{1}{8\pi^{2}}\int\int_{\hat{\mathbf{u}}^{\bot}}\hat{f}(\mathbf{q})qdqd\phi$
\end_inset
where
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $\hat{\mathbf{u}}^{\bot}$
\end_inset
is the plane perpendicular to
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\color inherit
\begin_inset Formula $\hat{\mathbf{u}}$
\end_inset
.
\end_layout
\begin_layout Standard
Without loss of generality, we align
\begin_inset Formula $\hat{\mathbf{u}}$
\end_inset
with the z-axis having
\begin_inset Formula $\hat{\mathbf{z}}=\hat{\mathbf{u}}$
\end_inset
.
Using the Dirac delta function (make use of Lebesgue integral) we can now
write
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray*}
\int_{0}^{\infty}f(r\hat{\mathbf{z})}dr & = & \int_{0}^{\infty}f(0,0,z)dz\\
& = & \frac{1}{2}\int\int\int_{\mathbb{R}^{3}}f(x,y,z)\delta(x)\delta(y)dxdydz
\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
The factor
\begin_inset Formula $1/2$
\end_inset
is required because we need the integral only in the positive half of the
z-axis, and the function is symmetric.
Let us define
\begin_inset Formula $g(x,y,z)\equiv\delta(x)\delta(y)$
\end_inset
.
For the two functions
\begin_inset Formula $f,g:\mathbb{R}^{3}\rightarrow\mathbb{R}$
\end_inset
with Fourier transform functions
\begin_inset Formula $\hat{f}(\mathbf{q})$
\end_inset
and
\begin_inset Formula $\hat{g}(\mathbf{q})$
\end_inset
, Parseval's theorem states that
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray*}
\int\int\int_{\mathbb{R}^{3}}f(x,y,z)g^{*}(x,y,z)dxdydz & =\quad\quad\quad\quad\quad\quad\quad\quad
\end{eqnarray*}
\end_inset
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\lang british
\begin_inset Formula
\begin{eqnarray*}
(2\pi)^{-3}\int\int\int_{\mathbb{R}^{3}}f(q_{x},q_{y},q_{z})\hat{g}^{*}(q_{x},q_{y},q_{z})dq_{x}dq_{y}dq_{z}
\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Furthermore,
\begin_inset Formula $\hat{g}(q_{x},q_{y},q_{z})=2\pi\delta(q_{z})$
\end_inset
and replacing it in the above equations leads to
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray*}
\int_{0}^{\infty}f(r\hat{\mathbf{z})} & = & \int\int\int_{\mathbb{R}^{3}}\frac{1}{2}f(x,y,z)g(x,y,z)dxdydz\\
& = & \frac{1}{2(2\pi)^{3}}\int\int\int_{\mathbb{R}^{3}}f(q_{x},q_{y},q_{z})2\pi\delta(q_{z})dq_{x}dq_{y}dq_{z}\\
& = & \frac{1}{8\pi^{2}}\int_{-\infty}^{\infty}\hat{f}(q_{x},q_{y},0)dq_{x}dq_{y}
\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection
The Tensor in GQI
\begin_inset CommandInset label
LatexCommand label
name "sub:Tensor-in-GQI"
\end_inset
\end_layout
\begin_layout Standard
We now apply this formulation under the assumption that the diffusion voxel
can be represented by a single tensor model.
Eq.
\begin_inset space ~
\end_inset
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:W"
\end_inset
can be written in the form
\begin_inset Formula
\begin{eqnarray}
S(\mathbf{q}) & = & S_{0}exp(-b\mathbf{q}^{T}D\mathbf{q})\label{eq:Tensor}
\end{eqnarray}
\end_inset
\end_layout
\begin_layout Standard
where
\begin_inset Formula $S_{0}$
\end_inset
is the image when b-value is equal to 0,
\begin_inset Formula $b$
\end_inset
is the b-value for a specific direction and
\begin_inset Formula $D$
\end_inset
is a
\begin_inset Formula $3x3$
\end_inset
matrix, known as the diffusion tensor.
Then from Eq.
\begin_inset space ~
\end_inset
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Q2S_complex"
\end_inset
and
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:Tensor"
\end_inset
the Fourier transform of
\begin_inset Formula $S$
\end_inset
is equal to
\end_layout
\begin_layout Standard
\align center
\begin_inset Formula
\begin{eqnarray}
Q(\mathbf{R}) & = & \int S_{0}\exp(-b\mathbf{q}^{T}D\mathbf{q})\exp(-j2\pi\mathbf{q}\cdot\mathbf{R})d\mathbf{q}\label{eq:FourierW}
\end{eqnarray}
\end_inset
\end_layout
\begin_layout Standard
The same equation in its triple integral form can be written as
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
Q(R) & =S_{0} & \iiint\exp(-b\sum_{i=1}^{3}q_{i}^{2}\lambda_{i}-j2\pi\sum_{i=1}^{3}q_{i}R_{i})dq_{1}dq_{2}dq_{3}\nonumber \\
& =S_{0} & \iiint\prod_{i=1}^{3}exp(-bq_{i}^{2}\lambda_{i}-j2\pi q_{i}R_{i})dq_{1}dq_{2}dq_{3}\nonumber \\
& =S_{0} & \prod_{i=1}^{3}\int\exp(-bq_{i}^{2}\lambda_{i}-j2\pi q_{i}R_{i})dq_{i}\nonumber \\
& =S_{0} & \prod_{i=1}^{3}\int\exp(-b\lambda_{i}[q_{i}^{2}+\frac{j2\pi R_{i}}{b\lambda_{i}}q_{i}])dq_{i}\nonumber \\
& =S_{0} & \prod_{i=1}^{3}\int\exp\{-b\lambda_{i}[(q_{i}+\frac{j\pi R_{i}}{b\lambda_{i}})^{2}+\frac{\pi^{2}R_{i}^{2}}{b^{2}\lambda_{i}^{2}}]\}\nonumber \\
& =S_{0} & \prod_{i=1}^{3}\int\exp\{-b\lambda_{i}(q_{i}+\frac{j\pi R_{i}}{b\lambda_{i}})^{2}\}exp\{-\frac{\pi^{2}R_{i}^{2}}{b\lambda_{i}}\}\label{eq:NextQ}
\end{eqnarray}
\end_inset
\end_layout
\begin_layout Standard
In that stage we can make use of the formula
\begin_inset Formula $\int\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp(-\frac{(x-\mu)^{2}}{2\sigma^{2}})dx=1$
\end_inset
.
Now we can see that
\begin_inset Formula $b\lambda_{i}=1/2\sigma^{2}$
\end_inset
and
\begin_inset Formula $\mu$
\end_inset
corresponds to
\begin_inset Formula $\mu=-jR_{i}/b\lambda_{i}$
\end_inset
.
Therefore, Eq.
\begin_inset space ~
\end_inset
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:NextQ"
\end_inset
can now be written as
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
Q(\mathbf{R}) & =S_{0} & \prod_{i=1}^{3}\sqrt{\frac{\pi}{b\lambda_{i}}}\exp(-\frac{\pi^{2}R_{i}^{2}}{b\lambda_{i}})\nonumber \\
& =S_{0} & \left(\frac{\pi}{b}\right)^{3/2}\frac{1}{\sqrt{\prod_{i=1}^{3}\lambda_{i}}}\exp(-\frac{\,\,\pi^{2}}{b}\mathbf{R}^{T}D^{-1}\mathbf{R})\label{eq:TensorQ}
\end{eqnarray}
\end_inset
\begin_inset Formula $\noindent$
\end_inset
\begin_inset space ~
\end_inset
where
\begin_inset Formula $D$
\end_inset
is the diffusion tensor.
We can replace the displacement vector
\begin_inset Formula $\mathbf{R}$
\end_inset
with a scalar value
\begin_inset Formula $L$
\end_inset
and a unit vector
\begin_inset Formula $\hat{u}$
\end_inset
i.e.
\begin_inset Formula $\mathbf{R}=L\hat{\mathbf{u}}$
\end_inset
and from Eq.
\begin_inset space ~
\end_inset
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:TensorQ"
\end_inset
we can replace
\begin_inset Formula $\frac{2\pi^{2}}{b}\mathbf{\hat{u}}^{T}D^{-1}\mathbf{\hat{u}}$
\end_inset
with
\begin_inset Formula $k$
\end_inset
and
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\lang british
\begin_inset Formula $S_{0}\left(\frac{\pi}{b}\right)^{3/2}\frac{1}{\lambda_{1}\lambda_{2}\lambda_{3}}$
\end_inset
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\color inherit
\lang english
with
\begin_inset Formula $\alpha$
\end_inset
.
Using that last change of variables we can now write
\begin_inset Formula
\begin{eqnarray}
\psi_{Q}(\mathbf{r},\mathbf{\hat{u}}) & = & \intop_{0}^{L_{\Delta}}Q(\mathbf{r},L\mathbf{\hat{u}})dL\nonumber \\
& = & \alpha\intop_{0}^{L_{\Delta}}\exp(-L^{2}\frac{k}{2})dL\label{eq:TensorQ2}
\end{eqnarray}
\end_inset
\end_layout
\begin_layout Standard
Setting
\begin_inset Formula $m=\sqrt{k}L$
\end_inset
and using the derivation for the error function Eq.
\begin_inset space ~
\end_inset
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:TensorQ2"
\end_inset
illustrates the remarkable result that we can calculate analytically the
spin ODF for Gaussian diffusion using the cumulative distribution function
\begin_inset Formula $CDF$
\end_inset
.
\begin_inset Newline newline
\end_inset
\begin_inset Formula
\begin{eqnarray}
\psi_{Q}(\mathbf{\hat{u}}) & = & \frac{\alpha}{\sqrt{k}}\intop_{0}^{\sqrt{k}L_{\Delta}}e^{-m^{2}/2}dm\\
& = & \alpha\sqrt{\frac{2\pi}{k}}\left[CDF(\sqrt{k}L_{\Delta})-\frac{1}{2}\right]\label{eq:spinodf_cdf}
\end{eqnarray}
\end_inset
\end_layout
\begin_layout Standard
This can be used as a check to compare the approximated or sampled spin
ODF that is derived in
\begin_inset CommandInset citation
LatexCommand cite
key "Yeh2010"
\end_inset
with Eq.
\begin_inset space ~
\end_inset
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:spinodf_cdf"
\end_inset
for the case of gaussian diffusion.
\end_layout
\begin_layout Standard
What is also very interesting is to try to derive what the normalization
factor should be for the spin ODF in Eq.
\begin_inset space ~
\end_inset
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:TensorQ2"
\end_inset
.
Because calculating a spherical integral from Eq.
\begin_inset space ~
\end_inset
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:spinodf_cdf"
\end_inset
seems at the moment very complicated we first work with the simpler gaussian
diffusion ODF derived by Tuch
\begin_inset CommandInset citation
LatexCommand cite
key "Tuch2004"
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
\psi_{p_{\Delta}}=\frac{1}{Z}\sqrt{\frac{\pi\tau}{\mathbf{u}^{T}D^{-1}\mathbf{u}}}\label{eq:tuchs_gaussian_odf}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
\frac{Z}{\sqrt{\pi\tau}} & = & \iintop_{S^{2}}(\mathbf{u}^{T}D^{-1}\mathbf{u})^{-\frac{1}{2}}d\mathbf{u}\label{eq:spherical_integral_Z}
\end{eqnarray}
\end_inset
\end_layout