Biomedical Engineering Reference
In-Depth Information
From trigonometric identities for half-angles, the preceding result simplifies to
ð
t
Þ¼
x
0
B
x
0
þ
y
0
B
y
0
¼
x
0
B
1
2
B
1
þ
y
0
B
1
2
0
1
B
cos 2
o
t
sin 2
o
t
ð
16
:
60e
Þ
2
which, with low-pass filtering to eliminate the components at twice the Larmor frequency, leaves
ð
t
Þ¼
x
0
B
1
2
0
1
B
ð
16
:
60f
Þ
a component of
B
1
that appears to be stationary (not time varying) in the rotating frame, as
expected.
Flip Angles and Decay
In MRI, the ability to control the position of the magnetization vector is important.
A practical configuration to accomplish this magnetic precession is shown in Figure 16.34
for rotated coordinates. First, the magnetization vector is shown aligned with the static field
in Figure 16.34a. Second, a radio frequency (rf) magnetic field
x
0
-axis at the
B
1
along the
Larmor frequency,
v
L
, is applied as a tone burst (a gated sinusoid) of pulse time duration,
M
0
(usually initi-
t
p
. The application of this pulse creates a force to pull the magnetization
ally aligned in an equilibrium position along the
z
-axis) down into a precession angle
j
from the
z
-axis, called the flip or tip angle,
t
p
¼
B
1
2
B
0
1
t
p
j ¼
g
g
ð
16
:
61
Þ
The precessing magnetization vector can also be broken down into its Cartesian compo-
nents in the rotated frame, as shown in Figure 16.34b. In addition to the component along
the
z
0
-axis,
0
x
0
-y
0
plane with a radial magnitude,
M
z
0
, there are two components in the
0
M
x
0
y
0
. It is convenient to look at the complicated resulting changes in time separately along
the two components,
0
z
0
and
0
x
0
y
0
, even though they occur simultaneously.
M
M
EXAMPLE PROBLEM 16.13
Find the pulse length necessary to achieve a phase rotation of p for the field calculated in
Example Problem 16.10. Repeat for a coil that generates a field 100 times greater.
Solution
From Eq. (16.60), solve for
pg
0
, and Table 16.2.
t
p
using Eq. (16.56),
g ¼
2
10
6
T from Example Problem 16.10, and
10
6
MHz/T. Solve for
B
1
¼
B
f
¼
8
g ¼
42.58
t
p
in
Eq. (16.61):
'
p
t
p
¼
Þ
¼
T
Þ
¼
2
:
936 msec
g
0
ð
B
1
=
2
10
6
10
6
2
p
2
p
42
:
58
ð=
s
T
Þð
4
For the second case, the pulse is smaller by a factor of 100, or 29.36 microseconds.
