M. Lustig, EECS UC Berkeley
Principles of MRIEE225E / BIO265
Lecture 10
Instructor: Miki LustigUC Berkeley, EECS
M. Lustig, EECS UC Berkeley
Mathematical Description of MRI
• Plan:1) Derive Math for each element2) Put together : e.g., the BLOCH equation3) Solve the Bloch eqn. for special cases
a) Excitation CH. 6 (later)b) Reception CH. 5 (first)
i) Derive k-space (AGAIN!!!)ii) Pulse sequenceiii)Sampling
M. Lustig, EECS UC Berkeley
The Bloch Equation
• In the rotating frame:
d ~M
dt= �� ~B ⇥ ~M � M
x
i+My
j
T2� M
z
�M0
T1k
~r = [x, y, z]T
~B = � ~G · ~rk +B1xi+B1y j
M. Lustig, EECS UC Berkeley
The Bloch Equation
• In Matrix Form :2
4M
x
My
Mz
3
5 =
2
40 � ~G · ~r ��B1y
�� ~G · ~r 0 �B1x
�B1y ��B1x 0
3
5
2
4M
x
My
Mz
3
5+
+
2
4� 1
T20 0
0 � 1T2
00 0 � 1
T1
3
5
2
4M
x
My
Mz
3
5+
+
2
4001T1
3
5M0
M. Lustig, EECS UC Berkeley
Bloch Equation
• Combined (rotating frame)
• T1 is the source of all signals!• Magnetization distribution unknown• Can probe by changing B1 and G
2
4M
x
My
Mz
3
5 =
2
64� 1
T2� ~G · ~r ��B1y
�� ~G · ~r � 1T2
�B1x
�B1y ��B1x � 1T1
3
75
2
4M
x
My
Mz
3
5+
2
4001T1
3
5M0
M. Lustig, EECS UC Berkeley
Solving the Bloch Equation
Two Special Cases:• Reception
– Data acquisition– Spatial encoding– Explicit solutions! (Today)
• Excitation– Non linear problems– No general solution– Many solutions for special cases (Ch. 6)
B1x = 0, B1y = 0
B1x 6= 0, B1y 6= 0
M. Lustig, EECS UC Berkeley
Reception
• Magnetization is a function of and t.Also, assume single, constant T1 and T2.
• We would like to resolve this using a time varying gradient
~r
~M(~r, t) =h~Mx
(~r, t), ~My
(~r, t), ~Mz
(~r, t)iT
~G(t) = [Gx
(t), Gy
(t), Gz
(t)]T
2
4M
x
My
Mz
3
5 =
2
64� 1
T2� ~G · ~r 0
�� ~G · ~r � 1T2
00 0 � 1
T1
3
75
2
4M
x
My
Mz
3
5+
2
4001T1
3
5M0
M. Lustig, EECS UC Berkeley
Reception
• With no B1(t) :
• With no B1(t) :
• Decouples into two independent eqns.
2
4M
x
My
Mz
3
5 =
2
64� 1
T2� ~G(t) · ~r 0
�� ~G(t) · ~r � 1T2
00 0 � 1
T1
3
75
2
4M
x
My
Mz
3
5+
2
4001T1
3
5M0
M. Lustig, EECS UC Berkeley
Reception
M. Lustig, EECS UC Berkeley
Solution for Reception
• Mz: (2) is already solved!
M. Lustig, EECS UC Berkeley
Solution for Reception
• Mxy: (1)M
xy
=
✓� 1
T2� i� ~G(t) · ~r
◆M
xy
) 0 = Mxy
+
✓1
T2+ i� ~G(t) · ~r
◆M
xy
M. Lustig, EECS UC Berkeley
Solution for Reception
• Can be solved using an integrating factor
y(x)M(x) =
ZQ(x)M(x) + C
M. Lustig, EECS UC Berkeley
Differential Equations 101
• Given an ODE
• Multiply with integrating factor M(x)
• So,
• And,
y
0 + p(x)y = Q(x)
M(x)y0 +M(x)p(x)y = M(x)Q(x)
(M(x)y)0 = M(x)Q(x)
Mxy
=
✓� 1
T2� i� ~G(t) · ~r
◆M
xy
) 0 = Mxy
+
✓1
T2+ i� ~G(t) · ~r
◆M
xy
M. Lustig, EECS UC Berkeley
Solution for Reception
• Can be solved using an integrating factor
et
T2+i�~r·
R t0
~G(⌧)d⌧
Mxy
+
✓1
T2+ i� ~G(t) · ~r
◆M
xy
= 0
M. Lustig, EECS UC Berkeley
Solution for Reception
• Integrate:
Mxy
(~r, t)et
T2+i�~r·
R t0
~
G(⌧)d⌧ = Mxy
(~r, 0)
⇣M
xy
et
T2+i�~r·
R t0
~
G(⌧)d⌧⌘0
= 0
Mxy
(~r, t) = Mxy
(~r, 0)e�t
T2 e�i~r·R t0 �
~
G(⌧)d⌧
~k(t) =�
2⇡
Z t
0
~G(⌧)d⌧
M. Lustig, EECS UC Berkeley
Solution for Reception
• Define a Spatial frequency vector:~k(t) = [k
x
(t), ky
(t), kz
(t)]T
M. Lustig, EECS UC Berkeley
Solution for Reception
• Transverse Magnetization Mxy is then:
• Received Signal is proportional to Mxy integrated over volume:
Mxy
(~r, t) = Mxy
(~r, 0)e�t
T2 e�i2⇡~k(t)·~r
Fourier kernel
s(t) =
Z
~
R
Mxy
(~r, 0)e�t
T2 e�i2⇡~k(t)·~rd~r
Also a function of r!M. Lustig, EECS UC Berkeley
Signal Equation
• Assume that T2 is LARGE
• So,
s(t) =
Z
~
R
Mxy
(~r, 0)e�i2⇡~k(t)·~rd~r
M. Lustig, EECS UC Berkeley
k-Space
M. Lustig, EECS UC Berkeley
k-Space Trajectories