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Lecture 3: The MR Signal Equation
• We have solved the Bloch equation and examined– Precession
– T2 relaxation
– T1 relaxation
• MR signal equation– Understand how magnetization changes phase due to
gradients
– Understand MR image formation or MR spatial encoding
Complex transverse magnetization: mm is complex.
m =Mx+iMy
Re{m} =Mx Im{m}=My
This notation is convenient:It allows us to represent a two
element vector as a scalar.
Re
Im
m
Mx
My
Inhomogeneous Object - Nonuniform field.Now we get much more general.
Before, M(t) was constant in space and so no spatial variable was needed to describe it.
Now, we let it vary spatially.
k
j
i
z
y
x
r
We also let the z component of the B field vary in space.
),B(γ),ω(Let
γBω Recall,
k)B(B),(B
oo
o
trtr
,trtr
This variation gives us a new phase term, or perhaps a more general one than we have seen before.
),,,(),,,(),(
ion,magnetizat erseFor transv
),()(
tzyxiMtzyxMtrm
trMtM
yx
Transverse Magnetization Equation
We can number the terms above 1-41. - variation in proton density
2. - spatial variation due to T2
3. - rotating frame frequency- “carrier” frequency of signal
4.
- spins will precess at differential frequency according to the gradient they experience (“see”).
ttrTto τreerMtrm
0iω)(/ )d,ω(iexp )(),( o2
)(rM o
)(/ 2 rTte
te oiω
tτr
0)d,ω(iexp
),ω( r
Transverse Magnetization Equation
Phase is the integral of frequency.
tτr
0)d,ω(iexp
),ω( r
- spins will precess at differential frequency according to the gradient they “see”.
t
τr0
)d,ω(
Thus, gradient fields allow us to change the frequency and thus the phase of the signal as a function of its spatial location.
Let’s study this general situation under several cases, from the specific to the general.
The Static Gradient Field
In a static (non-time-varying) gradient field, for example in x,
xrrGx xxx γG)B(γ)ω(
B
Plugging into solution (from 2 slides ago)
ttrTto τreerMtrm
0iω)(/ )d,ω(iexp )(),( o2
txeerMtrm trTto x
iω)(/ Gγiexp )(),( o2
and simplifying yields
The Static Gradient Field
Now, let the gradient field have an arbitrary direction. But the gradient strength won’t vary with time.
k)GGGB()(B o zyxr zyx
Using dot notation,
k)GB()(B o rr
Takes into account the effects of both the static magnetic and gradient fields on the magnetization gives
treerMtrm trTto r
iω)(/ Gγiexp )(),( o2
Time –Varying Gradient Solution (Non-static)
. We can make the previous expression more general by allowing the gradients to change in time. If we do, the integral of the gradient over time comes back into the above equation.
ttrTto τreerMtrm
0iω)(/ d)(Giexp )(),( o2
Receiving the signalLet’s assume we now use a receiver coil to “read” the transverse
magnetization signal We will use a coil that has a flat sensitivity pattern in space. Is this a good assumption?
dxdydzreerMtSx y z
ttrTtoreceive τ
0iω)(/ d)(Giexp )()( o2
- Yes, it is a good assumption, particularly for the head and body coils.
Then the signal we receive is an integration of
over the three spatial dimensions.
),( trm
),( trm
Receiving the signal (2)
3) Demodulate to baseband.i.e.
dxdydzreerMtSx y z
ttrTtoreceive τ
0iω)(/ d)(Giexp )()( o2
zoslice thickness z
2/
2/),,(),(
zz
zzo
odzzyxmyxm
Let’s simplify this:
1) Ignore T2 for now; it will be easy to add later.
2) Assume 2D imaging. (Again, it will be easy to expand to 3 later.)
tett oωireceive )(S)S( S(t) is complex.
Receiving the signal (3)
3) Demodulate to baseband.i.e. tett oωi
receive )(S)S( S(t) is complex.
Then,
dxdyrγx,ymtSx y
tτ
0d)(Giexp )()(
end goal
Gx(t) and Gy(t)
Now consider Gx(t) and Gy(t) only.
dxdyyγ
xγ
x,ymtS
t
x y
t
τ
τ
0 Y
0 x
)d(G2
2iexp
)d(G2
2iexp )()(
Deja vu strikes, but we push on!
Let t
xxτγ
tk 0)d(G
2)(
t
yyτγ
tk 0)d(G
2)(
dxdyykxkx,ymtS yx y
x 2iexp2iexp )()(
kxand ky related to t
Let t
xxτγ
tk 0)d(G
2)(
t
yyτγ
tk 0)d(G
2)(
dxdyykxkx,ymtS yx y
x 2iexp2iexp )()(
As we move along in time (t) in the signal, we simply change where we are in kx, ky
dxdyex,ymkkMx y
ykxkyx
yx ][2i
)(),(
(dramatic pause as students inch to edge of seats...)
kxand ky related to t (2)
x,y are in cm, kx,ky are in cycles/cm or cycles/mm
kx, ky are reciprocal variables.x y
))(),(()( tktkMtS yx
Remarks:1) S(t), our signal, are values of M along a trajectory in F.T. space.
- (called k-space in MR literature)
2) Integral of Gx(t), Gy(t), controls the k-space trajectory.
3) To image m(x,y), acquire set of S(t) to cover k-space and apply inverse Fourier Transform.
Spatial Encoding Summary Consider phase.
)(i),( x,y,teyxm
Receiver detects signal
planeyx
x,y,tR dxdyeyxmtS
,
)(i),()(
What is ? )(x,y,t
dt
tdt
)()(
Frequency
t
dttyxx,y,t0
')',,()(
is time rate of change of phase.
dt
tyxdtyx
),,(),,(
So
Spatial Encoding Summary
So t
dttyxx,y,t0
')',,()(
But ),,γB()( tyxx,y,t
Expanding, ])(G)(Gγ[B)( o ytxtx,y,t yx
We control frequency with gradients.
ydxdt
ydxx,y,t
ty
txo
tyx
00
0 o
)(Gγ)(Gγ
)(G)(GBγ)(
So,
dxdydyGxGieyxmtSx y
t
yxti
r ]))()(([exp(),()(0
0
Complex Demodulation
• The signal is complex. How do we realize this?
• In ultrasound, like A.M. radio, we use envelope detection, which is phase-insensitive.
• In MR, phase is crucial to detecting position. Thus the signal is complex. We must receive a complex signal. How do we do this when voltage is a real signal?
Complex Demodulationsr(t) is a complex signal, but our receiver voltage can only read a singlevoltage at a time. sr(t) is a useful concept, but in the real world, wecan only read a real signal. How do we read off the real and imaginary components of the signal, sr(t)?
titir
tir
eetts
etsts0
0
)()()(
)()(
Complex receiver signal ( mathematical concept)
The quantity s(t) or another representation, )()( tiet
is what we are after. We can write s(t) as a real and imaginary component, I(t) and Q(t)
))(sin()()(
))(cos()()(
)()()( )(
tttQ
tttI
tiQtIet ti
Complex Demodulation
An A/D converter reading the coil signal would read the real portion of sr(t), the physical signal sp(t)
))(cos()()(
)}(Re{)(
0 tttts
tsts
p
rp
Complex Demodulation
sp(t)
X
X
cos(t)
Inphase: I(t)
Quadrature: Q(t)
sin(t)
Low Pass Filter
Low Pass Filter
))(cos()(
)}(Re{)(
0 ttt
tsts rp
Let’s next look at the signal right after the multipliers. We canconsider the modulation theorem or trig identities.
Complex Demodulation
sp(t)
X
X
cos(t)
Inphase: I(t)
Quadrature: Q(t)
sin(t)
Low Pass Filter
Low Pass Filter
))](2cos())()[cos((2
1)(
)cos())(cos()()(
0int
00int
ttttts
ttttts
ermediatei
ermediatei
))](sin())(2)[sin((2
1)(
)sin())(cos()()(
0int
00int
ttttts
ttttts
ermediateq
ermediateq
))(cos()(
)}(Re{)(
0 ttt
tsts rp
Complex Demodulation
sp(t)
X
X
cos(t)
Inphase: I(t)
Quadrature: Q(t)
sin(t)
Low Pass Filter
Low Pass Filter
))(cos()(2
1)( tttI
))(sin()(2
1)( tttQ
Complex Demodulation
The low pass filters have an impulse response that removes thefrequency component at 2We can also consider them to have gain to compensate for the factor ½.