1
These changes were made 11/17/2004 to these notes:
1. Added an explanatory note (in pink!) in parentheses on page 2. 2. Corrected total derivatives to partial derivatives in Eq. (4) and preceding equation on page 4
and in Eqs. (32) and (33).
11. Distortion arising from static- field inhomogeneity for Spin-Echo images
11.1. In-plane effect
Assume the usual (spin-warp) SE sequence, first with no static- field inhomogeneity (i.e., 0BD = ):
( ) ( ) ( )2, , , x f y pi x G t y G
f y as t G A x y z e d x d yp g t
r- +
= ò ò
If we add inhomogeneity:
( )
( ) ( )( )2 , ,
,
, , x f y p a f
f y
i x G t y G B x y z t
a
s t G
A x y z e d x d yp g t
r- + + D
=
ò ò (1)
( ) ( ) ( )( )2 , ,, , , x f y p a fi x G t y G B x y z t
f y as t G A x y z e d x d yp g t
r- + + D
= ò ò
We note that there are two terms in the exponent that involve ft . We combine them:
( ) ( ) ( )( )( )2 , ,, , , x a f y pi x G B x y z t y G
f y as t G A x y z e d x d yp g t
r- + D +
= ò ò
Then we divide and multiply by Gx to prepare for a change of integration variable:
( ) ( )( ), ,
2
, , ,
ax f y p
x
B x y zi x G t y G
G
f y as t G A x y z e d x d yp g t
r
æ öæ öD- + +ç ÷ç ÷ç ÷
è øè ø= ò ò
The new integration variable is
( )( ), ,
, , aa
x
B x y zx x y z x
G
D¢ º + (2)
It’s meaning is that, because of a shift of ( ) , ,aB x y zg D from the frequency, xx Gg , at which
they should have been precessing, the spins at ( ), ,ax y z , act as if they are located at , ,ax y z¢ .
After the change of variables, we have
( ) ( )( ) ( )2, , , , , ,x f y pi x G t y G
f y a a
x
xs t G A x x y z y z e d x d y
p g tr
¢- + ¶
¢¶¢ ¢= ò ò (3)
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We insist that the function, ( ), ,ax x y z¢ must exist, which means that Eq. (2) produces a different x¢
for every different x , which in turn means that for fixed values of y and z, ( ), ,ax x y z¢ increases
with x . Given that ( ), ,ax x y z¢ increases with x , we know that the derivative, 0x x¢¶ ¶ > , and is
therefore never equal to zero. This is a reasonable assumption as long as
( ), ,
x
B x y zG
x
¶ D
¶=
Since x x¢¶ ¶ is nonzero, the derivative going the other way exists:
1x x
x x
-¢¶ ¶æ ö
= ç ÷¢¶ ¶è ø
(Note that, in addition to the requirement that x x¢¶ ¶ be nonzero, we are using the fact that this
change of variables involves only one variable. If, in addition to defining the new variable of
integration ( ), ,x f x y z¢ = , we also define ( ), ,ay g x y z¢ = , then the inverse relationship will
exist if and only if 0g x¶ ¶ = . Thus, the inverse would exist, for example, if x x y z¢ = and
y y z¢ = , but not if x x y z¢ = and y x y¢ = .)
Using our definitions for kx and ky, we have
( ) ( )( ) ( )1
2, , , , , ,x yi x k y k
x y a a
xs k k A x x y z y z e d x d y
x
p gr
-¢- +¢¶æ ö ¢ ¢= ç ÷
¶è øò ò
Taking a Fourier transform to calculate the value of the image at the position, x1,y1, we have
( ) ( ){ } ( ) ( )
( )( )
( )( )
1 1
1
1
211 1 1 1
1
1
11 1 1
, , , , ,
, , , ,
, , , , ,
x yi x k y k
x y x y x y
a ax x
y y
a a
i x y s k k x y s k k e d k d k
d xB x x y z y z
d x
d xB x x y z y z
d x
p g
r
r
+ +-
-
¢=
=
-
= =
æ ö¢æ ö ¢= ç ÷ç ÷ç ÷è øè ø
æ ö= ç ÷
è ø
ò òF
where
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( )
( )
1
1
, ,,
, , 11
a
x
x
B x y zx x
G
B x y zd x
d x x G
D= +
¶ D= +
¶
The third expression for ( )1 1,i x y above, tells us that, not only do the spins at , ,ax y z act as if
they were located at 1, ,ax y z, but also their image intensity appears to be divided by 1d x
d x.
The reason for the shift from x to 1x is, as mentioned above, the shift in their frequency caused
by BD . The reason for the change in intensity is more subtle. There are three possible situations:
1 1x
x
¶=
¶: There is no change in the intensity
1 1x
x
¶>
¶: The intensity is decreased. This happens because the signal from the spins is spread out
over a larger portion of the image.
1 1x
x
¶<
¶: The intensity is increased. This happens because the signal from the spins is focused
onto a smaller portion of the image. Discuss Homework 3:
· Significant figures · Confusing x’ and x in number 2 · Simple calculation for number 5 (I counted 5 points off for a non-simple calculation)
Discuss presentations
· Give “classic” papers from which most other references spring: § Forward-reverse: Chang and Fitzpatrick, 1992 § Delayed echo
· Sekihara and/or Feig, 1985 · Sumanaweera, 1993 · Jezzard, 1995
· Show how to find their citations on using Web of Knowledge
We can rewrite the final equation above for ( )1 1,i x y as follows:
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( ) ( )1
11 1 1, , , a
xi x y B x y z
xr
-¶æ ö
= ç ÷¶è ø
or
( ) ( )1
11 1 0 1, , ,
xi x y i x y
x
-¶æ ö
= ç ÷¶è ø
(4)
where ( )0 1,i x y is the image that would be formed if there were no inhomogeneity.
11.2. Through-plane plus in-plane effect.
During slice-selection, spins are excited by varying amounts. The degree of excitement is
measured in terms of the signal-producing part of the magnetization, x yM . There is a frequency
“profile” that describes the size of the signal produced by the spins as a function of the spins’
frequency of precession, n . The function is conveniently written in the form ( )1p n n- , where
1n is the frequency of the RF pulse. The shape of the function ( )1p n n- is determined by the
time-dependence of the RF field 1B in the rotating frame. Determining ( )1p n n- when given
( )1B t requires that the Bloch equation be solved numerically. Determining ( )1B t when given
( )1p n n- requires “inverting the Bloch equation. The relationship between these two functions
is beyond the scope of this course, but for very small flip angles (< 5 degrees), ( )1p n n- is
approximately proportional to a Fourier transform of ( )1B t , which, in turn, is similarly
proportional to an inverse Fourier transform of ( )1p n n- . Therefore, for small angles, since it
is typically desired to excite uniformly a range zG zn gD = D of frequencies, which would
mean that ( )1p n n- equals the boxcar function, ( )1r e c tn n- , ( )1B t is made to approximate
a sinc function, ( )s i nc tp nD . Note: The effects that we are considering are present regardless of
the shape of ( )1p n n- , so at this point we do not need to make an assumption. Below, we will
assume, however, that both the spin density and the spatial derivatives of BD are approximately
constant over the nonzero range of ( )1p n n- .
If we include slice selection explicitly in our expression, Eq. (1), for the signal from a SE sequence, it becomes
( )
( )( ) ( ) ( )( )
1
2 , ,
1
, ,
, , , , ,x f y p f
f y
i x G t y G B x y z t
s t G v
A p x y z x y z e d x d y d zp g t
n n r- + + D
=
-ò ò ò
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where
( ) ( )( )
( )0
1 0 1
, , , ,
,
z
z
x y z B z G B x y z
B z G
n g
n g
= + + D
= +
where have named our selected slice 1z instead of az to for later notational consistency.
Therefore,
( )
( )( )
( )
1
, ,2
1
, ,
, ,, , ,
x f y px
f y
B x y zi x G t y G
G
z
z
s t G z
B x y zA p G z z x y z e d x d y d z
G
p g t
g r
æ öæ öD- + +ç ÷ç ÷ç ÷
è øè ø
=
æ öDæ ö+ -ç ÷ç ÷
è øè øò ò ò
Now we change integration variables for both x and z:
( )( )
( )( )
, ,, ,
, ,, ,
x
z
B x y zx x y z x
G
B x y zz x y z z
G
D¢ º +
D¢ º +
(5)
We insist that the functions, ( ), ,x x y z¢ ¢ and ( ), ,z x y z¢ ¢ both exist. This requirement is
equivalent to the requirement that the Jacobian of the transformation from , ,x y z¢ ¢ to , ,x y z be
nonzero. The Jacobian is
,
x x x
x y z
x y z y y yJ
x y z x y z
z z z
x y z
¢ ¢ ¢¶ ¶ ¶
¶ ¶ ¶¢ ¢æ ö ¶ ¶ ¶
=ç ÷ç ÷ ¶ ¶ ¶è ø¢ ¢ ¢¶ ¶ ¶
¶ ¶ ¶
which in this case becomes
DON’T SHOW EQ. (6). IT IS PART OF THE SOLUTION TO A HOMEWORK PROBLEM!
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1
0 1 0
1
x x x
z z z
B B BG G G
x y zx y z
Jx y z
B B BG G G
x y z
¶ D ¶ D ¶ D+
¶ ¶ ¶¢ ¢æ ö
=ç ÷ç ÷è ø ¶ D ¶ D ¶ D
+¶ ¶ ¶
(6)
Eq. (6) reduces to
1 x z
x y z B BJ G G
x y z x z
¢ ¢æ ö ¶ D ¶ D= + +ç ÷ç ÷ ¶ ¶è ø
.
Note that we can view J as being a function of either x,y,z or x’,y’,z’, since either set of variables can be determined from the other set. Thus,
( ) ( ), , , ,x y z
J J x y z J x y zx y z
¢ ¢æ ö¢ ¢= =ç ÷ç ÷
è ø
It is reasonable to expect this Jacobian to be nonzero as long as
,x
z
BG
x
BG
z
¶ D
¶
¶ D
¶
=
=
The change of variables leads to
( ) ( )( ) ( ) ( )( )( ) ( )
( ) ( )( )( ) ( )
1 1
12
1 1
, , , , , , '
1
, , , , ' , , , , '
, ,
, ,
x f y p
f y z
i x G t y G
x y zx z
x x y z y z x y z
B BG Gx z
s t G z A p G z z x x y z y z x y z
e J x y z d x d y d z
s k k z A p G z z d z e
p g t
r
g r
g
¢ -- +
æ öç ÷ -è ø
¢ ¢ ¢
¶ D ¶ D+ +¶ ¶
¢ ¢ ¢= - ´
¢ ¢ ¢ ¢
æ ö¢ ¢= -ç ÷
ç ÷è ø
ò ò ò
ò ò ò( )2 x yi x k y k
d x d yp g ¢ +
¢
where the arguments for BD are the same as for r . Taking a Fourier transform to calculate the
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value of the image at the position, 1 1,x y, we have
( ) ( ){ } ( ) ( )
( )( )( ) ( )( )
1 1211 1 1 1 1
1 1 1 1 1
1
, , , , , ,
, , , , , ,,
1
x yi x k y k
x y x y x y
z
x z
i x y z s k k x y s k k e d k d k
x x y z y z x y zB p G z z d z
B BG G
x z
p g
rg
+ +-= =
¢ ¢¢ ¢= -
¶ D ¶ D+ +
¶ ¶
ò ò
ò
F
(7)
where, again, the arguments BD are the same as for r .
At this point, we make an approximation. We assume that both r and the derivatives of BD
and are constant over the range of z¢ for which p is nonzero. Thus, over that range,
( ) ( )( ) ( ) ( )( )( ) ( )( ) ( ) ( )( )
( ) ( )( ) ( ) ( )( )
1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1
, , , , , , , , , , , ,
, , , , , , , , , , , ,
, , , , , , , , , , , ,
x x y z y z x y z x x y z y z x y z
B x x y z y z x y z B x x y z y z x y z
x x
B x x y z y z x y z B x x y z y z x y z
z z
r r¢ ¢ =
¢ ¢¶ D ¶ D=
¶ ¶
¢ ¢¶ D ¶ D=
¶ ¶
With this approximation,
( ) ( )( )( ) ( ) ( )( )
( ) ( )( )
( ) ( )( )
1 1 1 1 1 1 1
1 1 1 1
1 1 1 1 1 1 1
0 1 1 1 1 1 1 1
, , , , , ,, ,
1
, , , , , ,
1
, , , , , ,,
1
z
x z
x z
x z
x x y z y z x y zi x y z B p G z z d z
B BG G
x z
x x y z y z x y zC
B BG G
x z
i x x y z y z x y z
B BG G
x z
rg
r
¢ ¢= -¶ D ¶ D
+ +¶ ¶
=¶ D ¶ D
+ +¶ ¶
=¶ D ¶ D
+ +¶ ¶
ò
(8)
where 0i is, as before, the image that would be acquired if there were no inhomogeneity, and
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( )( )
( )( )
1
1
, ,, ,
, ,, ,
x
z
B x y zx x y z x
G
B x y zz x y z z
G
Dº +
Dº +
(9)
Since the transformation in Eqs. (9) is one-to-one, we can rewrite the last of Eqs. (8) as
( ) ( ) ( )( )( )
( ) ( )
1 1 1
0
, , , , , , , ,
, , ,
, , , ,1 x z
i x x y z y x y z z x y z
i x y z
B x y z B x y zG G
x z
=
¶ D ¶ D+ +
¶ ¶
(10)
where we are for notational consistency also using
1y y= (11)
Now we define a new coordinate system rotated about the y axis relative our our current system
by ( )1t a n x zG Gq -º . We designate coordinates that are referred to this system by putting a
: over them. Thus,
1 1 1
1 1 1
c o s s i n ,
s i n c o s ,
c o s s i n ,
s i n c o s
x x z
z x z
x x z
z x z
q q
q q
q q
q q
= +
= - +
= +
= - +
%
%
%
%
where 2 2s i n x x zG G Gq = + and
2 2c o s z x zG G Gq = + . With this transformation, we find
that DON’T SHOW EQ.(12). IT IS PART OF THE SOLUTION TO A HOMEWORK PROBLEM!
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( ) ( )
( )
( )
( ) ( )
1
2 2 2 2
2 2
, , , ,c o s s i n
c o s s i n, ,
, ,
, , ,
x z
x z
z x z x x z
x z
x z x z
B x y z B x y zx x z
G G
x B x y zG G
G G G G G Gx B x y z
G G
x B x y z G G G G
q q
q q
æ öD D æ ö= + + +ç ÷ ç ÷
è øè ø
æ ö= + D +ç ÷
è ø
æ ö+ +ç ÷= + D +ç ÷è ø
= + D +
%
% % % %
% % % %
% % % %
(12)
which becomes
( )1 , ,x x B x y z G= + D %% % % % % (13)
where G% is a simple function of the gradients.
DON’T SHOW EQS.(14) or (15) . THEY ARE PART OF THE SOLUTION TO A
HOMEWORK PROBLEM!
2 2 2 21 1 1x z x z x zG G G G G G G= + = +% (14)
It can easily be shown also that
s i n c o s .z xG G Gq q= =% (15)
These two figures show cute (but maybe not so useful!) geometric relationships among xG , zG ,
and G% :
Furthermore,
q q
zG
G% xG
q 1 xG
1 G% 1 zG
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DON’T SHOW EQ.(16). IT IS PART OF THE SOLUTION TO A HOMEWORK PROBLEM!
( ) ( )
( )
( )
1
2 2 2 2
, , , ,s i n c o s
s i n c o s, ,
, ,
x z
x z
x x z z x z
x z
B x y z B x y zz x z
G G
z B x y zG G
G G G G G Gz B x y z
G G
q q
q q
æ öD D æ ö= - + + +ç ÷ ç ÷
è øè ø
æ ö= + D - +ç ÷
è ø
æ ö- + +ç ÷= + D +ç ÷è ø
%
% %% %
% %% %
(16)
which becomes
1 .z z=% %
Furthermore, DON’T SHOW EQ.(17). IT IS PART OF THE SOLUTION TO A HOMEWORK PROBLEM!
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( ) ( )
2 2 2 2
, , , ,1
1
1
1
x z
x z
x z x z
z x z x x z
x z
B x y z B x y zG G
x z
B x B z B x B zG G
x x z x x z z z
B Bx x z zG G G G
x z x zx z
G G G G G GB
x G G
¶ D ¶ D+ +
¶ ¶
¶ D ¶ ¶ D ¶ ¶ D ¶ ¶ D ¶æ ö æ ö= + + + +ç ÷ ç ÷
¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶è ø è ø
¶ D ¶ D¶ ¶ ¶ ¶æ ö æ ö= + + + +ç ÷ ç ÷
¶ ¶ ¶ ¶¶ ¶è ø è ø
æ ö+ +¶ Dç ÷= + + +ç ÷¶è ø
% %% %
% %% %
% % % %
% %
%
2 2 2 2
2 2 2 2
1 0
1
x x z z x z
x z
z x z x x z
x z
G G G G G GB
z G G
G G G G G GB B
x G G z
BG
x
æ ö+ +¶ Dç ÷- +ç ÷¶è ø
æ ö+ +¶ D ¶ Dç ÷= + + + ´ç ÷¶ ¶è ø
¶ D= +
¶
%
% %
%%
(17)
Using Eq. (13) in Eq. (17), gives
( ) ( ) 1, , , ,
1 x z
B x y z B x y z xG G
x z x
¶ D ¶ D ¶+ + =
¶ ¶ ¶
%
% (18)
Thus, in summary,
( )
( ) ( )
1
1
1
1
, ,,
,
,
, , , ,1 .x z
B x y zx x
G
y y
z z
B x y z B x y z xG G
x z x
D= +
=
=
¶ D ¶ D ¶+ + =
¶ ¶ ¶
% %%% %
%
% %
% %
%
%
(19)
Using Eqs. (19) in Eq. (10) yields
( ) ( )1
11 1 1 0, , , , ,
xi x y z i x y z
x
-¶æ ö
= ç ÷¶è ø
%% % % %% %
% (20)
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which looks surprisingly like Eq. (4), which ignored through-plane distortion.
Again, the intensity can change, but this time it is a function of the derivative in a direction between the x and z axes. 11.3. Echo-Planar Imaging (EPI) We ignore the effects of slice selection for now and again alter Eq. (1), this time to correspond to Spin-Echo EPI. Note that we are assuming a sequence with only forward-going readouts:
( ) ( )( )( )( )
( )( )( )
, ,
1 1
2 , ,
, ,
, , ,
z
x x b p a x y
B x y z
z G
i x G m t y n G B x y z m t n t
a
s m n z A p G z z
x y z e d x d y d zp g t
g
r
D
- D + + D D + D
= + - ´ò
ò ò (21)
and the integral over x and y can be transformed as follows:
( )( )( )( )
( )( ) ( )
( )
( ) ( )( )
e f
e f f
2 , ,
, ,, ,2
, , , ,2
, ,
, ,
, ,
x x b p a x y
a yax x y p
x y p
a ax x y
x y
i x G m t y n G B x y z m t n t
B x y z tB x y zi x m G t y n G
G G
B x y z B x y zi x m G t y n G
G G
x y z e d x d y
x y z e d x d y
x y z e
p g t
p g tt
p g
r
r
r
- D + + D D + D
æ öæ öD Dæ öDç ÷- + D + + Dç ÷ç ÷ ç ÷ç ÷ Dè ø è øè ø
æ öæ öD Dç ÷- + D + +ç ÷ç ÷è ø è ø
=
=
ò ò
ò ò( )f
,yt
d x d y
æ öç ÷ Dç ÷è øò ò
where yGD is the “blip” gradient, which is used to provide the phase encoding between
readouts, and ( )e f f
y y p yG G ttº D D is the “effective” gradient for the purposes of assessing the
effects of inhomogeneity. It is the gradient that would required to produce the same change in
phase by being turned on at a constant rate during ytD as yGD would by being turned on for
only only pt . Since it has so much more time to act, it is much smaller than yGD .
It is important to note here that the time dependence of the blip gradient need not be a square wave. The important thing is the effect caused by the gradient over the time that it is applied. Its time dependence may be of any shape without affecting the image, so long as the integral
( )0
p
y b bG t d tt
Dò
has the desired value. For this more general case,
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( ) ( )e f f
0
1 p
y y b b
y
G G t d tt
t
= DD ò .
(This independence of time shape is true of all phase encoding gradients, including those used for the non-EPI sequences.) As usual, we define spatial frequencies for x and y. For EPI they are as follows:
( )e f f
x x x
y y y
k m G t
k n G t
g
g
= D
= D
Similarly, as before for spin-warp imaging, we can deduce from these expressions that
( )
( )( )
1
1e f f
x x
y y
X G t
Y G t
g
g
-
-
º D
º D
With these definitions, the x,y integral becomes
( )
( ) ( )( )e f f
, , , ,2
, , .
a ax y
x y
B x y z B x y zi x k y k
G Gx y z e d x d y
p
r
æ öæ öæ öD Dç ÷ç ÷- + + +ç ÷ç ÷ç ÷è ø è øè øò ò (22)
Now we transform the integration using
( )( )
( )( )
( )
( )( )
e f f
, ,, ,
, ,, ,
, ,, ,
x
y
z
B x y zx x y z x
G
B x y zy x y z y
G
B x y zz x y z z
G
D¢ º +
D¢ º +
D¢ º +
(23)
The Jacobian for this transformation can be shown to be equal to
1 x y z
x y z B B BJ G G G
x y z x y z
¢ ¢ ¢æ ö ¶ D ¶ D ¶ D= + + +ç ÷ç ÷ ¶ ¶ ¶è ø
(24)
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Using Eqs. (23) and (23) in Eq. (22) and inserting that into the z integration that we last saw in Eq. (21), yields
( )
( )( )( ) ( ) ( ) ( )
1
2
1
, , , , , ' , , , '
1
, ,
x y
x y
i x k y k
zx y z
x x y z y x y z z x y z
B B BG G Gx y z
s k k z
A p G z z d z e d x d yp g
rg
æ öç ÷ ¢ ¢- +è ø
¢ ¢ ¢ ¢
¶ D ¶ D ¶ D+ + +¶ ¶ ¶
=
æ öç ÷¢ ¢ ¢-ç ÷è ø
ò ò ò
Taking the inverse Fourier transform gives
( ) ( ){ } ( ) ( )
( )( )( ) ( ) ( )( )
1 1211 1 1 1 1 1
1 1 1 1 1 1
1
, , , , , , ,
, , , , , , , ,,
1
x yi x k y k
x y x y x y
z
x y z
i x y z s k k z x y s k k e d k d k
x x y z y x y z z x y zB p G z z d z
B B BG G G
x y z
p g
rg
+ +-= =
¢ ¢ ¢¢ ¢= -
¶ D ¶ D ¶ D+ + +
¶ ¶ ¶
ò ò
ò
F
(25)
which is very similar to Eq. (7). Making the same assumptions that we made after Eq. (7) but including changes with respect to y as well as x and z yields
( ) ( )( )( )( ) ( ) ( )( )
( ) ( ) ( )( )
( ) ( ) ( )( )
1 1 1 1 1 1 1 1 1
1 1 1 1
1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1
, , , , , , , ,, ,
1
, , , , , , , ,
1
, , , , , , , ,,
1
z
x y z
x y z
x y z
x x y z y x y z z x y zi x y z B p G z z d z
B B BG G G
x y z
x x y z y x y z z x y zC
B B BG G G
x y z
i x x y z y x y z z x y z
B B BG G G
x y z
rg
r
¢ ¢= -¶ D ¶ D ¶ D
+ + +¶ ¶ ¶
=¶ D ¶ D ¶ D
+ + +¶ ¶ ¶
=¶ D ¶ D ¶ D
+ + +¶ ¶ ¶
ò
(26)
where 0i is, as before, the image that would be acquired if there were no inhomogeneity, and
( )( )
( )( )
( )( )
1
1
1
, ,, ,
, ,, ,
, ,, ,
x
y
z
B x y zx x y z x
G
B x y zy x y z y
G
B x y zz x y z z
G
Dº +
Dº +
Dº +
(27)
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Where these equations are similar to Eqs. (8) and (9). As before, we insist that the Jacobian for our transformation, Eq. (with primes replaced by subscript ones) (24), is nonzero
Since the transformation in Eqs. (9) is one-to-one, we can rewrite the last of Eqs. (8) as
( ) ( ) ( )( )( )
( ) ( ) ( )
1 1 1
0
, , , , , , , ,
, , ,
, , , , , ,1 x y z
i x x y z y x y z z x y z
i x y z
B x y z B x y z B x y zG G G
x y z
=
¶ D ¶ D ¶ D+ + +
¶ ¶ ¶
(28)
As before, we can rotate our coordinate system to make things simpler. By defining a new coordinate system rotated so that
( )
( ) ( ) ( )
( )
( ) ( ) ( )
22 2
1 1 1
1 22 2
1 1 1
1 1 1
x y z
x y z
x y z
x y z
x G y G z Gx
G G G
x G y G z Gx
G G G
+ +=
+ +
+ +=
+ +
%
%
(There are an infinite number of such coordinate systems!), we can show that
( )
( ) ( ) ( )
1
1
1
1
, ,,
,
,
, , , , , ,1 .x y z
B x y zx x
G
y y
z z
B x y z B x y z B x y z xG G G
x y z x
D= +
=
=
¶ D ¶ D ¶ D ¶+ + + =
¶ ¶ ¶ ¶
% %%% %
%
% %
% %
%
%
(29)
where
( ) ( ) ( )
2 2 2
2 2 21 1 1 1 x z z
x y z
x y y z z x
G G GG G G G
G G G G G G= + + =
+ +
% (30)
With these relationships Eq. (28) becomes
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( ) ( )1
11 1 1 0, , , , ,
xi x y z i x y z
x
-¶æ ö
= ç ÷¶è ø
%% % % %% %
% (31)
which is exactly the same as Eq. (20)!
11.4. Ignoring small geometric distortions
Typically, in a spin-warp image z xG G? and in an echo-planar image, ( )e f f
x yG G? and
( )e f f
z yG G? . In fact, in order to carry out EPI in one shot, it is necessary to move very quickly
through k-space, particularly in the x direction, so Gx is very large. As a result, for spin-warp images
( )( )
( )
( )
( )
( ) ( )
1 1
1
1
1
1
1
1
11 1 1 0 1 1
,
, ,, ,
, ,
, ,
, ,
, , , ,
x
x x x x
B x y zx x y z x
G
y x y z y
z x y z z
xJ x y z
x
xi x y z i x y z
x
-
-
» »
Dº +
=
»
¶æ ö» ç ÷
¶è ø
¶æ ö» ç ÷
¶è ø
% %
(32)
and for echo-planar images
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( )
( )( )
( )
( )
( )
( ) ( )
1 1
1
1 e f f
1
1
1
1
11 1 1 0 1 1
,
, ,
, ,, ,
, ,
, ,
, , , ,
y
x y x y
x x y z x
B x y zy x y z y
G
z x y z z
yJ x y z
x
yi x y z i x y z
x
-
-
» »
»
D= +
»
¶æ ö» ç ÷
¶è ø
¶æ ö» ç ÷
¶è ø
% %
(33)
11.5 EPI for back-and-forth readout EPI rarely has forward-going readouts only. Typically the readouts go back and forth. The x,y integral in that case has the same form for even values of n, but for odd values, it looks like this:
( )( )( )( )
( )( ) ( )
( )
( ) ( )( )e f f
2 , ,
, ,, ,2
, , , ,2
, ,
, ,
, ,
x x b p a x y
a yax x y p
x y p
a ax x
x y
i x G m t y n G B x y z m t n t
B x y z tB x y zi x m G t y n G
G G
B x y z B x y zi x m G t y n G
G G
x y z e d x d y
x y z e d x d y
x y z e
p g t
p g tt
p g
r
r
r
- - D + + D D + D
æ öæ öD Dæ öDç ÷- - + D + + Dç ÷ç ÷ ç ÷ç ÷ Dè ø è øè ø
æ öæ öD Dç ÷- - + D + +ç ÷ç ÷è ø è ø
=
=
ò ò
ò ò( )
( )
( ) ( )( )
e f f
e f f
, , , ,2
,
, , .
y y
a ax y
x y
t
B x y z B x y zi x k y k
G G
d x d y
x y z e d x d yp
r
æ öç ÷ Dç ÷è ø
æ öæ öæ öD Dç ÷ç ÷- - + + +ç ÷ç ÷ç ÷è ø è øè ø=
ò ò
ò ò
We alter the signal in this case by reversing the sign of kx for odd values of n, where
( )e f f
y y yk n G tg= D .
As result we have
( ) ( )
( ) ( )( )e f f
, , , ,2
1, , , , ,
a ax y
x y
B x y z B x y zi x k y k
G G
x ys k k z x y z e d x d yp
r
æ öæ öæ öD Dç ÷ç ÷- ± + +ç ÷ç ÷ç ÷è ø è øè ø= ò ò
where the ± is a plus for even n and minus for odd n. This change of sign means that our change of variables for x no longer works. The effect on the image is not at all obvious. It causes a “ghost” image to appear above and below the true image. The effect is called “N-over-2 ghosting”, and it is called that because the ghost appears to be a faint, rough facsimile of the “unghosted” image shifted up N/2 pixels and another version shifted down N/2 pixels. (For those who are interested in the reason for this effect, it is a consequence of the fact that even lines are slightly different from odd lines. The
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mathematical description of this effect involves multiplying in k-space by a function that changes from one to zero and back as we move upward (y-direction) in k-space by one discrete unit. Multiplying by a modulation function in k-space is equivalent to convolving with a kernel in image space, where the kernel is the Fourier transform of the modulation function. The Fourier transform for the modulation function in this space is a kernel that equals zero everywhere except at the origin and at positions that are above or below the origin at a distance of / 2q N± , where q is any integer.
away. The value of the kernel at these non-zero positions is 1 for the even lines; for the odd lines it is one for even values of q and -1 for odd values.) However, since Gx is very large, the effect is negligible, so we can neglect this term. As a result, we can use the approximations given by Eqs. , as if the readout were always forward going.
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