ELEG 479
Lecture #12
Magnetic Resonance (MR) Imaging
Mark Mirotznik, Ph.D.
Associate Professor
The University of Delaware
Physics of Magnetic Resonance
Summary
 Protons and electrons have a property called spin that results in them
looking like tiny magnets.
 In the absence of an external magnetic field all the magnetic dipole are
oriented randomly so we get zero net magnetic field when we add them
all up.
=
No Net
Magnetization
Random
Orientation
Physics of Magnetic Resonance
Summary
 When we add a large external magnetic field we can get the protons to
line up in 1 of 2 orientations (spin up or spin down) with a few more per
million in one of the orientations than the other. This produces a net
magnetization along the axis of the applied magnetic field.
Bo 2 2
Mo 
PD
4kT
 When we add a large external magnetic field we cause a torque on the
already spinning proton that causes it to precess like a top around the
applied magnetic field. The frequency is precesses , called its Larmor
frequency., is determined from Larmor’s equation.
Physics of Magnetic Resonance
Summary from Last Lecture
The net magnetization vector is the sum of all of these little
magnetic moments added together. This is what we measure.
z
z


z
z
z


z

 xy
y

z




x
z
Net Magnetization Vector


 xy
 xy
y
y
x
y
x
z
z

 xy
 xy
x
y


x
N


M (t )    n ( xn , yn , zn , t )
n 1


M (t )  M xy (t )  M z zˆ
Physics of Magnetic Resonance
Summary from Last Lecture
 However since all the spinning protons are precessing out of
phase with each other this results in zero net magnetization in
the transverse plane.
=0


M (t )  M xy (t )  M z zˆ
This is bad news since

M xy (t )
is where are signal comes from!
Somehow we need to get these
guys spinning together!
Physics of Magnetic Resonance
Summary from Last Lecture
To get them to all spin together we add a RF field whose frequency is the same as the Larmor
resonant frequency of the proton and is oriented in the xy or transverse plane.
B1
RF Excitation
time
B1
Physics of Magnetic Resonance
Summary from Last Lecture
z
B1

M
a
x

M xy
y
a B1Dt
Tip Angle
Amplitude of RF
Pulse
Time of Application
of RF Pulse
Physics of Magnetic Resonance
Summary from Last Lecture
z
e
1
B (t )

M
a

M xy
y
x
e
1
B (t ) = envelope of
In general

a    B (t )dt
0
e
1
the RF signal
Physics of Magnetic Resonance
Summary from Last Lecture
 To get the signal out we place a coil near the sample. A timevarying transverse magnetic field will produce a voltage on the
coil that can be digitized and stored for processing.
d
V (t )   M xy (t )
dt
V (t )  K  o  M o  sin( a )
recall
Bo 2 2
Mo 
PD
4kT
and
o    Bo
Relaxation Processes
Physics of Magnetic Resonance
Relaxation
 After the RF field is removed over time the spin system will
return back to it’s equilibrium state due to several relaxation
processes.
Physics of Magnetic Resonance
Relaxation
 After the RF field is removed over time the spin system will
return back to it’s equilibrium state due to several relaxation
processes. These are:
(1) Spin-Spin relaxation (also called the T2 relaxation): Due to
random processes in which neighboring proton spins
effect each other spin system will lose coherence and Mxy
will decay. This is an irreversible process.
(2) Spin-Lattice relaxation (also called T1 relaxation): Due to
another random process the Mz will begin to recover back
to it’s original equilibrium state. Also irreversible.
(3) T2* relaxation: Due to inhomogenities in the external Bo
field Mxy will decay much faster than T2. This is a
reversible process.
T1 Relaxation
T2 Relaxation (FID)
T2 Decay
RF
RF
*
T2 Relaxation
T2* Decay: Dephasing due to field
inhomogeneity
z'
Mxy
y'
=0
x'
T2* relaxation is dephasing of transverse magnetization
too but it turns out to be reversible
Animation of T2* Dephasing
Spin- Echo
Spin Echo
Summary of Relaxation Processes
MRI Image
 An MRI image is determined by two things
 three intrinsic properties of the tissue. These are:
T1, T2 and Pd. (two relaxation time constants and
the density of protons)
 the details of the external magnetic fields (Bo, B1
and the gradient magnetics (have not talked about
these yet)). How they are configured and how we
turn them on and off (pulse sequence) effects what
the image looks like.
By varying the pulse sequence we can control which
of the intrinsic properties to emphasize in the image.
Tissue Contrast
TR
180 degree
RF pulses
0.5TE
90 degree
RF pulses
0.5TE
0.5TE
0.5TE
TE
TE
White matter
T1=813 ms
T2=101 ms
TR
180 degree
RF pulses
CASE I: TR>>T1 , TE<T2
What do we measure?
90 degree
RF pulses
TE
TE
White matter
T1=813 ms
T2=101 ms
TR
180 degree
RF pulses
90 degree
RF pulses
CASE I: TR>>T1 , TE<T2
Short TE means that the signal
has not decayed much due to
T2relaxation.
 Long TR means that by the next
pulse the system is back at
equilibrium (Mz due to T1
relaxation has fully recovered)
 So what are we measuring?
TE
TE
White matter
T1=813 ms
T2=101 ms
TR
180 degree
RF pulses
90 degree
RF pulses
TE
CASE I: TR>>T1 , TE<T2
Short TE means that the signal
has not decayed much due to
T2relaxation.
 Long TR means that by the next
pulse the system is back at
equilibrium (Mz due to T1
relaxation has fully recovered)
 So what are we measuring?
PD Weighted imaging!
White matter
T1=813 ms
T2=101 ms
TE
TR
180 degree
RF pulses
CASE II: TR>>T1 , TE~T2
90 degree
RF pulses
TE
TE
White matter
T1=813 ms
T2=101 ms
TR
180 degree
RF pulses
90 degree
RF pulses
CASE II: TR>>T1 , TE~T2
 TE on the order of T2means that
the signal is proportional to the
T2relaxation constant.
 Long TR means that by the next
pulse the system is back at
equilibrium (Mz due to T1
relaxation has fully recovered)
 So what are we measuring?
TE
TE
White matter
T1=813 ms
T2=101 ms
TR
180 degree
RF pulses
90 degree
RF pulses
CASE II: TR>>T1 , TE~T2
 TE on the order of T2means that
the signal is proportional to the
T2relaxation constant.
 Long TR means that by the next
pulse the system is back at
equilibrium (Mz due to T1
relaxation has fully recovered)
 So what are we measuring?
T2 Weighted imaging!
TE
TE
White matter
T1=813 ms
T2=101 ms
TR
180 degree
RF pulses
90 degree
RF pulses
TE
CASE III: TR~T1 , TE<T2
What do we measure?
TR
180 degree
RF pulses
90 degree
RF pulses
TE
CASE III: TR~T1 , TE<T2
 TE is shorter than T2means that the
signal is not heavily weighted on the
T2relaxation constant.
 TR on the order of T1 means that by
the next pulse the system is not back at
equilibrium (Mz due to T1 relaxation has
not fully recovered)
 So what are we measuring?
TR
180 degree
RF pulses
90 degree
RF pulses
CASE III: TR~T1 , TE<T2
 TE is shorter than T2means that the
signal is not heavily weighted on the
T2relaxation constant.
 TR on the order of T1 means that by
the next pulse the system is not back at
equilibrium (Mz due to T1 relaxation has
not fully recovered)
 So what are we measuring?
T1 Weighted imaging!
TE
Tissue Contrast
Summary
TE
PD weighted
T1weighted
T2 weighted
TR
TE<T2 (short TE)
TR>>T1 (long TR)
TE<T2 (short TE)
TR~T1 (short TR)
TE~T2 (long TE)
TR>>T1 (long TR)
Bloch Equations
Full Bloch equation including relaxation



( M x (t) x̂  M y (t) ŷ) ( M z (t )  M o )ẑ
dM
 M (t )  γB(t) 

dt
T2
T1
precession,
RF excitation
transverse
magnetization

B (t)  Bo  B1 (t )
includes Bo and B1
longitudinal
magnetization
Example: Solve for the transverse
components of M after a 90 degree pulse.



( M x (t) x̂  M y (t) ŷ) ( M z (t )  M o )ẑ
dM
 M (t )  γB(t) 

dt
T2
T1
 1

M x (t) 

 M y (t ) Bo  M z (t ) B y (t )   T2
 M x (t ) 

d

  1


M
(
t
)

γ
M
(
t
)
B

M
(
t
)
B
(
t
)

M
(t)
y
x
o
z
x
y

  T


dt 
2
M x (t ) B y (t )  M y (t ) Bx (t ) 
 M z (t ) 


 1
 M z (t) - M o 
 T1

Example: Solve for the transverse
components of M after a 90 degree pulse.
 1

M x (t) 

 M y (t ) Bo  M z (t ) B y (t )   T2
 M x (t ) 

d

  1


M
(
t
)

γ
M
(
t
)
B

M
(
t
)
B
(
t
)

M
(t)
y
x
o
z
x
y

  T


dt 
2
M x (t ) B y (t )  M y (t ) Bx (t ) 
 M z (t ) 


 1
 M z (t) - M o 
 T1

After 90 degree pulse the RF field is shut down and only Bo is non-zero
 1

M x (t) 

 M x (t ) 
 M y (t ) B0   T2

d
  γ - M (t ) B    1 M (t) 
M
(
t
)
y
0

 x
 T2 y

dt 
 M z (t ) 

 
0

1
 M z (t) - M o 
 T1

Example: Solve for the transverse
components of M after a 90 degree pulse.
After 90 degree pulse the RF field is shut down and only Bo is non-zero
 1

M x (t) 

 M x (t ) 
 M y (t ) B0   T2

d
1





M
(
t
)

γ
M
(
t
)
B

M
(t)
y
0

 x
 T2 y

dt 
 M z (t ) 

 
0

1
 M z (t) - M o 
 T1

Initial conditions for 90 degree pulse: M (0)  0
x
M y ( 0)  M o
M z ( 0)  0
Example: Solve for the transverse
components of M after a 90 degree pulse.
After 90 degree pulse the RF field is shut down and only Bo is non-zero
 1

M x (t) 

 M x (t ) 
 M y (t ) B0   T2

d
1





M
(
t
)

γ
M
(
t
)
B

M
(t)
y
x
0
y


  T

dt 
2
 M z (t ) 

 
0

1
 M z (t) - M o 
 T1

M x (t )  M o sin( ot )e
Solutions
 t
M y (t )  M o cos(o t )e
M z (t )  M o (1  e
 t
T1
)
T2
 t
T2
Example: Solve for the transverse
components of M after a 90 degree pulse.
After 90 degree pulse the RF field is shut down and only Bo is non-zero
 1

M x (t) 

 M x (t ) 
 M y (t ) B0   T2

d
1





M
(
t
)

γ
M
(
t
)
B

M
(t)
y
x
0
y


  T

dt 
2
 M z (t ) 

 
0

1
 M z (t) - M o 
 T1

M xy (t )  M x (t )  jM y (t )
Solutions
M xy (t )  M o e
 j o t
M z (t )  M o (1  e
e
 t
 t
T1
T2
)
Example: Solve for the transverse
components of M after an arbitrary flip
angle (a)
 1

M x (t) 

 M y (t ) Bz  M z (t ) B y (t )   T2
 M x (t ) 

d

  1


M
(
t
)

γ
M
(
t
)
B

M
(
t
)
B
(
t
)

M
(t)
y
x
z
z
x
y

  T


dt 
2
M x (t ) B y (t )  M y (t ) Bx (t ) 
 M z (t ) 


 1
 M z (t) - M o 
 T1

After an arbitrary RF pulse the RF field is shut down and only Bo is
non-zero
 1

M x (t) 

 M x (t ) 
 M y (t ) B0   T2

d
  γ - M (t ) B    1 M (t) 
M
(
t
)
y
0

 x
 T2 y

dt 
 M z (t ) 

 
0

1
 M z (t) - M o 
 T1

Example: Solve for the transverse
components of M after an arbitrary flip
angle (a)
After an arbitrary RF pulse the RF field is shut down and only Bo is
non-zero
 1

M x (t) 

 M x (t ) 
 M y (t ) B0   T2

d
  γ - M (t ) B    1 M (t) 
M
(
t
)
y
0

 x
 T2 y

dt 
 M z (t ) 

 
0

1
 M z (t) - M o 
 T1

Initial conditions for 90 degree pulse:
M x (0)  0
M y (0)  M o sin( a )
M z (0)  M o cos(a )
Example: Solve for the transverse
components of M after an arbitrary flip
angle (a)
 1

M x (t) 

 M x (t ) 
 M y (t ) B0   T2

d
1





M
(
t
)

γ
M
(
t
)
B

M
(t)
y
x
0
y


  T

dt 
2
 M z (t ) 

 
0

1
 M z (t) - M o 
 T1

M xy  M x  jM y
M xy (t )  M xy (0  )e  jBot e t T2
Solutions
 M o sin( a )e  jot e t T2
M z (t )  M o (1  e
t
 M o (1  e
T1
t
)  M z (0)e
T1
t
T1
)  M o cos(a )e
t
T1
Solve full Bloch equation with only B=Bo
Solution for transverse components Mx and My

 jBo t t T2

 jo t t T2
M xy (t )  M xy (0 )e
 M xy (0 )e
e
e
M xy (0 )  M z (0 ) sin( a )

M z (t )  M o (1  e

t
T1
)  M o cos(a )e
t
T1
Where a is the flip angle after RF excitation
Signal Detection
Signal Detection via RF coil
Signal Detection via RF coil
Transverse magnetization at t=0.
s (t )  A  
z max
z min
y max
xmax
y min
xmin
 
M xy ( x, y, z ,0  )e  jBot e t / T2 ( x , y , z ) dxdydz
 Coils oriented as shown above will only respond to changes in the
transverse magnetic field (this is what we want)
 Assuming the magnetic fields are homogenous the signal will be a
weighted integration of all the protons within the coil.
 The waiting will be based on the total magnetization at location x,y,z at
the start of the pulse (Mxy(x,y,z,0)) and the tissue decay time T2(x,y,z)
 This is not an image!!
Signal Detection via RF coil
s (t )  A  
z max
z min
s (t )  A  
z max
z min
s (t )  e
 j o t
y max
xmax
y min
xmin
y max
xmax
y min
xmin
 
 
M xy ( x, y, z ,0  )e  jBot e t / T2 ( x , y , z ) dxdydz
M xy ( x, y, z ,0  )e  jot e t / T2 ( x , y , z ) dxdydz
z max y max xmax

 A     M xy ( x, y, z ,0  )e t / T2 ( x , y , z ) dxdydz


z min y min xmin
After demodulation:
so (t )  A  
z max
z min
y max
xmax
y min
xmin
 
M xy ( x, y, z ,0  )e t / T2 ( x , y , z ) dxdydz
Creating an Image
Creating an Image
To create an
image using NMR
we need to figure
out a way to
encode the
proton spins
spatially in three
dimensions.
But how?
Frequency and Phase Are Our Friends in MR Imaging

q= t
q
The spatial information of the proton pools contributing
MR signal is determined by the spatial frequency and
phase of their magnetization.
Gradient Coils
z
z
z
y
y
x
x
X gradient
y
Y gradient
x
Z gradient
Gradient coils generate spatially varying magnetic field
so that spins at different location precess at frequencies
unique to their location, allowing us to reconstruct 2D
or 3D images.
Gradient Coils
Purpose: Spatially alter magnitude of B0 (not direction)
Sounds generated during imaging due to mechanical stress within gradient coils.
Vector Notation

G  Gx aˆ x  G y aˆ y  Gz aˆ z

r  x aˆ x  y aˆ y  z aˆ z
 
Bz ( x, y, z )  Bo  G  r
Larmor frequency within a gradient field

G  Gx aˆ x  G y aˆ y  Gz aˆ z

r  x aˆ x  y aˆ y  z aˆ z
 
Bz ( x, y, z )  Bo  G  r
 

 (r )   Bo  G  r


Slice Selection
Slice Selection Gradient
BG
Coil 2
Coil 1
Helmholtz Coils
Z-Gradient Fields
 ( z )   Bo  Gz  z 
 ( z )   0    Gz  z
By adding a z-gradient field we cause a variation in the
resonant frequency from head to toe.
Example
A sample is put inside a 1.5T magnet. A zgradient of 3 gauss/cm is applied. If we
wish to image a 2 ft in length section of a
person what is the range of resonant
frequencies we will encounter?
Example
A sample is put inside a 1.5T magnet. A z-gradient of
3 gauss/cm is applied. If we wish to image a 2 ft in
length section of a person what is the range of resonant
frequencies we will encounter?
 
Bz ( x, y, z )  Bo  G  r  Bo  Gz  z
Bzmin
Bzmax
 3 
 1.5  
 z
 10000 
 3 
 1.5  
  12  2.54  1.490856 T
 10000 
 3 
 1.5  
  12  2.54  1.509144 T
 10000 
Example
A sample is put inside a 1.5T magnet. A z-gradient of
3 gauss/cm is applied. If we wish to image a 2 ft in
length section of a person what is the range of resonant
frequencies we will encounter?
 3 
B  1.5  
  12  2.54  1.490856 T
 10000 
 3 
max
Bz  1.5  
  12  2.54  1.509144 T
 10000 
 ( z )    B( z ),   42.58 MHz / tesla
 min  42.58 1.490856  63.48 MHz
min
z
 max  42.58 1.509144  64.26 MHz
A Field Gradient Makes the Larmor
Frequency Depend upon Position
1.500 T
1.501 T
B0
63,480,000 Hz
64,260,000 Hz
Z
B(Z)  B o  G Z * Z
(z)   B(z)
Gradient in Z
Slice Selection
(-)
62 MHz
63 MHz
64 MHz
G
65 MHz
66 MHz
(+)
Slice Selection
How do we determine the slice width and center?
z
z-gradient
Bo
(slice center)
z
After z selection
gradient and excitation
Dz (slice width)
x
Determining slice thickness
Resonance frequency range as the result of sliceselective gradient:
 ( z )   Bo  Gz  z 
 min  Bo
 max  Bo
 z min 
, z max 
Gz
Gz

 max  min   Bo  Bo
 Dz  z max  z min 
Gz
D
Dz 
Gz
Changing slice thickness
There are two ways to do this:
(a) Change the slope of the slice selection gradient
(b) Change the bandwidth of the RF excitation pulse
D
Dz 
G z
Both are used in practice, with (a) being more popular
Example
Suppose we wish to have a slice thickness
of 2 mm and we are using a z-gradient of
1.0 G/cm ? What range of RF frequencies
should we use?
Example
Suppose we wish to have a slice thickness of 2 mm and
we are using a z-gradient of 1 G/cm ? What range of RF
frequencies should we use?
D
Dz 
Gz
D
0.2cm 
 1 
42.58  

 10000 
4
 D  8.516 10 MHz  851.6 Hz
Selecting different slices
 ( z )   Bo  Gz  z 
 min  Bo
 max  Bo
 z min 
, z max 
Gz
Gz
z max  z min  max  min   2Bo
z

2
2Gz

 max  min  / 2  Bo   o
z

Gz
Gz
  o
z
G z
Selecting different slices
In theory, there are two ways to select different slices:
(a) Change the position of the zero point of the slice
selection gradient with respect to isocenter
(b) Change the center frequency of the RF to correspond
to a resonance frequency at the desired slice
  o
z
G z
Option (b) is usually used as it is not easy to change the
isocenter of a given gradient coil.
RF Excitation (RF Pulse)
Fo
FT
0
Fo Fo+1/ t
t
Time
Frequency
Fo
t
Fo
FT
DF= 1/ t
RF Excitation (RF Pulse)
Fo
t
sin( D  t ) j 2 t
s(t )  AD 
e
D  t
FT
A
D
1
2
 v
S ( )  A  rect (
)
D
v2  v1
v
2
RF Excitation: Flip angle
Fo
FT
A
D
t
1
 v
S ( )  A  rect (
)
D
sin( D  t ) j 2 t
s(t )  AD 
e
D  t
Envelope of the pulse
a 
2
t max
e
B
 1 (t )dt
t min
v2  v1
v
2
RF Excitation: Flip angle
Fo
FT
A
D
t
1
 v
S ( )  A  rect (
)
D
sin( D  t ) j 2 t
s(t )  AD 
e
D  t
a 
v2  v1
v
2
t max
t max
t min
t min
e
B
 1 (t )dt  
2
 j 2 t
s
(
t
)

e
dt

RF Excitation: Flip angle
Fo
FT
A
D
t
1
2
 v
S ( )  A  rect (
)
D
sin( D  t ) j 2 t
s(t )  AD 
e
D  t
v2  v1
v
2

sin( D  t )
a    B (t )dt    AD 
dt
D  t
t min

t max
e
1
 v
zz
a  A  rect (
)  A  rect (
)
D
Dz
RF Excitation: Flip angle (truncated sinc)
 p / 2
p /2
sin( D  t ) j 2 t
t
~
s (t )  AD 
e
 rect ( )
D  t
p
p
2
a 


p
sin( D  t )
AD 
dt
D  t
2
zz
a  A p  rect (
) * sin c( p Gz ( z  z ))
Dz
A potential problem
Wait a minute! Dr. M I remember you telling us that if
the magnetic field varied from place to place
(inhomogeneous) then we would get rapid dephasing
of spins. Since some spins are spinning faster than
others they quickly get out of phase. That was the
whole reason behind the spin echo stuff!
Won’t that happen again?
Wait a minute! Dr. M I
remember you telling us that if
the magnetic field varied from
place to place (inhomogeneous)
then we would get rapid
dephasing of spins. Since some
spins are spinning faster than
others they quickly get out of
phase. That was the whole
reason behind the spin echo
stuff! Won’t that happen again?
Why yes it will! It is
called gradient dephasing
Good question!
Spinning slow
Spinning fast
Why yes it will! It is
called gradient
dephasing. It will quickly
kill our signal much faster
than T2 or even T2*
Any ideas on how to
get around this?
Spinning slow
Spinning fast
Localization in xy plane
Lets Start with a Simple Flat Person
(only xz plane)
z
Bo
x
Lets Start with a Simple Flat Person
(only xz plane)
z
z-gradient
After z selection
gradient and excitation
Bo
x
Lets Start with a Simple Flat Person
Frequency Encoding Mathematical Analysis
Recall that the signal we measure is given by:
s (t )  A  
z max
z min
y max
xmax
y min
xmin
 
M xy ( x, y, z ,0  )e  jBot e t / T2 ( x , y , z ) dxdydz
Now we have selected only a single slice in z (z=zo) and we have
no y dependence (flat person)
s(t )  e
 j 2 o t



A  M xy ( x,0  )e t / T2 ( x ) dx
After demodulation (envelope detection)
so (t )  s (t )  e
j 2 o t

  A  M xy ( x,0  )e t / T2 ( x ) dx

Lets Start with a Simple Flat Person
Frequency Encoding Mathematical Analysis
After demodulation (envelope detection)
so (t )  s (t )  e
Let
j 2 o t

  A  M xy ( x,0  )e t / T2 ( x ) dx

f ( x)  A  M xy ( x,0 )et / T2 ( x)

so (t )   f ( x)dx

This is what we want to
image (called the
effective proton density)
Lets Start with a Simple Flat Person
(frequency encoding using x-gradient)
z
After z selection
gradient and excitation
Bo
x
Lets Start with a Simple Flat Person
Frequency Encoding Mathematical Analysis
Now lets apply a gradient in the x direction (Gx)

s(t )   A  M xy ( x,0  )e  j 2  o Gx x t e t / T2 ( x ) dx

s(t )  e
 j 2 o t



A  M xy ( x,0  )e  j 2Gx x t e t / T2 ( x ) dx
After demodulation (envelope detection)

s o (t )   A  M xy ( x,0  )e  j 2Gx x t e t / T2 ( x ) dx

f ( x)  A  M xy ( x,0 )et / T2 ( x)

s o (t )   f ( x)e  j 2Gx x t dx

What does this look like?
Lets Start with a Simple Flat Person
Frequency Encoding Mathematical Analysis
After demodulation (envelope detection)

s o (t )   f ( x)e  j 2Gx x t dx

Let
u  Gxt

u
F (u )  s o (
)   f ( x)e  j 2ux dx

Gx
The received signal is related to the Fourier transform (THIS IS THE KEY!)
Lets Start with a Simple Flat Person
Frequency Encoding Mathematical Analysis
After demodulation (envelope detection)

s o (t )   f ( x)e  j 2Gx x t dx

f ( x)  A  M xy ( x,0 )et / T2 ( x)
Let u  Gxt

u
F (u )  s o (
)   f ( x)e  j 2ux dx

Gx
We can now find our image as a function of x by taking
an inverse Fourier Transform
A Simple Example of Spatial
Encoding with Frequency Encoding
Constant
Magnetic
Field
w/o encoding
Varying
Magnetic
Field
w/ encoding
A Simple Example of Spatial
Encoding with Frequency Encoding
Frequency
Decomposition
Decays faster than T2*
Extend this to a full 3D person
Extend this to a full 3D person
After slice selection we need to image in xy plane
y
x
Spatial Encoding in xy plane
Frequency Encoding Mathematical Analysis
Now lets apply a gradient in the x direction (Gx)

s (t ) 




A  M xy ( x, y,0  )e  j 2  o Gx x t e t / T2 ( x , y ) dxdy
After demodulation (envelope detection)

so (t ) 




A  M xy ( x, y,0  )e  j 2Gx xt e t / T2 ( x , y ) dxdy
f ( x, y)  A  M xy ( x, y,0 )et / T2 ( x, y )
s o (t )  





Effective proton density
f ( x, y )e  j 2Gx x t dxdy
Spatial Encoding in xy plane
Frequency Encoding Mathematical Analysis
After demodulation (envelope detection)

so (t ) 




A  M xy ( x, y,0  )e  j 2Gx xt e t / T2 ( x , y ) dxdy
f ( x, y)  A  M xy ( x, y,0 )et / T2 ( x, y )
s o (t )  




f ( x, y )e  j 2Gx x t dxdy

Let u  Gxt

u

F (u, v  0)  s o (t 
)    f ( x, y )e  j 2ux dxdy
 
Gx
Spatial Encoding in xy plane
Frequency Encoding Mathematical Analysis
s o (t )  




f ( x, y )e  j 2Gx x t dxdy

Let u  Gxt

u

F (u, v  0)  s o (t 
)    f ( x, y )e  j 2ux dxdy
 
Gx
Corresponds to a single
line or trajectory in the
uv plane
Spatial Encoding in xy plane
Frequency Encoding Mathematical Analysis
Now lets apply gradients in both the x direction (Gx) and y
direction (Gy)
After demodulation (envelope detection)

so (t ) 




A  M xy ( x, y,0  )e  j 2Gx xt e
 j 2G y yt t / T2 ( x , y )
e
dxdy
Let
f ( x, y)  A  M xy ( x, y,0 )et / T2 ( x, y )
s o (t )  





f ( x, y )e
Effective proton density
 j 2G x x t  j 2G y y t
e
dxdy
Spatial Encoding in xy plane
Frequency Encoding Mathematical Analysis
Now lets apply gradients in both the x direction (Gx) and y
direction (Gy)
s o (t )  


Let


f ( x, y )e
 j 2G x x t  j 2G y y t
e
dxdy

u  Gx t , v  G y t
F (u , v)  





f ( x, y )e  j 2uxe  j 2vx dxdy
Spatial Encoding in xy plane
Frequency Encoding Mathematical Analysis
Let
u  Gx t , v  G y t
F (u , v)  




f ( x, y )e  j 2uxe  j 2vx dxdy

Polar Scanning
Gradient Echo
(A brief detour)
Ts/2
Spatial Encoding in xy plane
Gradient Echo Mathematical Analysis
Spins will dephase very quickly (quicker than T2*) due to the
gradient fields.
After negative x-gradient
 ( x, t )     Gx xdt  Gx xt   p ,  p  t   p  Ts 2
t
p
 ( x, p  Ts 2 )  Gx x Ts 2
Spatial Encoding in xy plane
Gradient Echo Mathematical Analysis
After negative x-gradient
 ( x, t )     Gx xdt  Gx xt   p ,  p  t   p  Ts 2
t
p
 ( x, p  Ts 2 )  Gx x Ts 2
Now impose positive x-gradient for Ts
 ( x, t )  Gx x
Ts
2

t

 p Ts 2
Gx xdt  Gx x
Ts
T
 Gx x t   p  s 
2
2

 ( x, t )  Gx xTs  Gx xt   p ,  p  Ts 2  t   p  3Ts 2
Spatial Encoding in xy plane
Gradient Echo Mathematical Analysis
Now impose positive x-gradient for Ts
 ( x, t )  Gx xTs  Gx xt   p ,  p  Ts 2  t   p  3Ts 2
 ( x, t   p  Ts )  Gx xTs  Gx x p  Ts   p   0
Phase Encoding
Pulse Repetition
Image Contrast
Image Quality
Field of View and Resolution in MRI
Fourier Plane
Spatial Domain
Dx
Du
Dv
Vcoverage
FOVy
Dy
Ucoverage
FOVx
Nyquist Sampling Theorem: Review
• Assume we have a continuous signal with maximum
frequency of fmax
• To avoid aliasing we must sample the signal at a
sampling frequency of fs>=2 fmax
• The sampling interval T=1/ fs
• fmax<=1/(2T)
Sampling in MRI
• Slice selection direction: sampling in z-direction
Slice thickness (Dz) controlled by RF excitation
bandwidth (D)
To avoid aliasing 1  2 f max, z  Dz  1
Dz
2 f max, z
Where fmax,z is the highest spatial frequency in along
the z-axis
Sampling in MRI
• Within each slice: sampling in xy plan
 We sample in the Fourier domain (u,v)
(called k-space in MRI literature, kx=u, ky=v)
 Rectilinear Scan
Du depends on sampling interval T during
readout (ADC)
Du depends on sampling interval during this time
Sampling in MRI
• Within each slice: sampling in xy plan
 We sample in the Fourier domain (u,v)
(called k-space in MRI literature, kx=u, ky=v)
 Rectilinear Scan
Dv depends on spacing between phase
encoding
Dv depends on the integrated phase shift here
Sampling in MRI
• Within each slice: sampling in xy plan
 We sample in the Fourier domain (u,v)
(called k-space in MRI literature, kx=u, ky=v)
 Polar Scan
Angle scan depends on steps in Gy/Gx
Angle scan depends on steps in Gy/Gx
Sampling in MRI
• Within each slice: sampling in xy plan
 We sample in the Fourier domain (u,v)
(called k-space in MRI literature, kx=u, ky=v)
 Polar Scan
Rho spacing depends on sampling interval T
during readout
rspacing depends on sampling interval during this time
Dv
X-gradient relates dimension x with Larmor freq  by
 ( x)   o  Gx x
To avoid aliasing only frequency given below are measured
o 
fs
f
  ( x)   o  s
2
2
 
fs
f
 Gx x  s
2
2
X-gradient relates dimension x with Larmor freq  by
 ( x)   o  Gx x
To avoid aliasing only frequency given below are measured
fs
f
  ( x)   o  s
2
2
fs
fs
 
x
2Gx
2Gx
o 
 
fs
f
 Gx x  s
2
2
Field of view in the x-direction (FOVx) is thus given by
FOVx  xmax  xmin
FOVx 
fs
fs
fs
1
 (
)

2Gx
2Gx
Gx GxT
1
1
FOVx 

GxT Du
 Dependant on the phase encoding gradient Gy. The amount of
phase change is given by
D  Dv  DG yTPE
 Field of view in y
FOVy  ymax  ymin
1
1


GyTPE Dv
 While FOV is limited by the sampling interval in the UV plane
(Fourier plane) the resolution is limited by the total extent of
the UV plane being sampled. If we ignore high spatial frequency
content we will have lower resolution (blur our image). Since
MRI scans cover only a finite area of the Fourier space we can
expect a finite resolution.
 Fourier space coverage in MRI
U cov erage  N x Du  N xGxT
Vcov erage  N y Dv  N yDGyTPE
 Fourier space coverage in MRI
U cov erage  N x Du  N xGxT
Vcov erage  N y Dv  N yDGyTPE
 Outside of this range we assume the contributions to be
zero. This is equivalent to passing the actual image through
a low-pass filter in the uv plane whose transfer function is
given by
u
v
H (u, v)  rect (
)  rect (
)
U cov erage
Vcov erage
 In the spatial domain this is then given by the point spread
function (PSF)
h( x, y )  U cov erage Vcov erage  sin c(U cov erage  x)  sin c(Vcov erage  y )
 In the spatial domain this is then given by the PSF
h( x, y )  U cov erage Vcov erage  sin c(U cov erage  x)  sin c(Vcov erage  y )
 The full width half max (FWHM) resolution is given by the
width of the sinc function’s main lobe
FWHM x 
FWHM y 
1
U cov erage
1
Vcov erage
1
1


N x Du N xGxT

1
1

N y Dv N yDG yTPE
 Increasing the U,V (coverage area in Fourier space)
reduces blurring.
Dx 
FOVx
1
1


M
MDu U cov erage
Dy 
FOV y
N

1
1

NDv Vcov erage
Field of View and Resolution in MRI
Spatial Domain
Fourier Plane
Du
Dx
Dv
Vcoverage
FOVy
Dy
Ucoverage
FWHM x 
FWHM y 
FOVx
1
U cov erage
1
Vcov erage
1
Du
1
FOV y 
Dv
FOVx 