Digital Signal Processing Reference
In-Depth Information
M
The resultant magnetic moment
depends upon the first term and the second
term of the RHS of the (
2.16
) independently. The resultant
(
t
)
M
(
t
)
due to the second
term ends up with
M
x
(
components as described in the Eqs. (
2.9
)-
(
2.11
). Note that the
z
-component of the resultant vector is constant due to the second
term. Now the magnetic moment
t
),
M
y
(
t
)
and
M
z
(
t
)
M
due to the first term is obtained as follows.
Rewriting (
2.15
) with only first term of the RHS as
(
t
)
→
dM
(
t
)
M
E
=
γ
(
t
)
×
(
(
t
))
(2.16)
dt
⎡
⎤
i
j
k
M
E
⎣
⎦
⇒
⇒
(
)
×
(
)
=
M
x
(
t
)
M
y
(
t
)
M
z
(
t
)
t
t
E
x
(
t
)
E
y
(
t
)
0
dM
x
(
t
)
=−
γ
E
y
(
t
)
M
z
(
t
)
=−
γ
E
0
sin
(
−
γ
B
0
t
+
θ)
M
0
cos
(α)
(2.17)
dt
dM
y
(
t
)
=
γ
E
x
(
t
)
M
z
(
t
)
=
γ
E
0
cos
(
−
γ
B
0
t
+
θ)
M
0
cos
(α)
(2.18)
dt
dM
z
(
t
)
=
γ(
M
x
(
t
)
E
y
(
t
)
−
M
y
(
t
)
E
x
(
t
))
(2.19)
dt
Solving the Eqs. (
2.17
)-(
2.19
) as described in Sect.
2.1
, we get the following.
dM
x
(
t
)
j
dM
y
(
t
)
+
=
γ
E
0
M
0
cos
(α)(
−
sin
(
−
γ
B
0
t
+
θ)
dt
dt
+
jcos
(
−
γ
B
0
t
+
θ))
(2.20)
dM
xy
(
t
)
e
j
(
−
γ
B
0
t
+
θ)
⇒
=
j
γ
E
0
M
0
cos
(α)
(2.21)
dt
)
=
−
E
0
M
0
cos
(α)
e
j
(
−
γ
B
0
t
+
θ)
K
⇒
M
xy
(
t
,
B
0
where
K
is the constant.
As
M
xy
(
0
)
=
M
0
sin
(α)
cos
(φ)
+
jM
0
sin
(α)
sin
(φ)
=
M
0
sin
(α)
e
(
j
φ)
we get
K
as follows.
−
E
0
M
0
cos
(α)
e
(
j
θ)
K
M
xy
(
0
)
=
B
0
⇒
−
E
0
M
0
cos
(α)
cos
(θ)
K
=
M
0
sin
(α)
cos
(φ)
B
0
Also,
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