Digital Signal Processing Reference
In-Depth Information
Ae
jωt
. Therefore,
solved assuming that
x
takes the form
x
=
dx
dt
=
jAωe
jωt
d
2
x
dx
2
=−
ω
2
Ae
jωt
which are substituted back into (6-19), so the coefficient
A
can be solved for:
+
j
b
k
qE
0
m
−
ω
2
Ae
jωt
m
Aωe
jωt
m
Ae
jωt
e
jωt
+
=
A
+
j
b
k
m
qE
0
m
−
ω
2
m
ω
+
=
qE
0
/m
A
=
−
ω
2
+
j(b/m)ω
+
k/m
Therefore,
(qE
0
/m)e
jωt
x
=
Ae
jωt
=
−
ω
2
+
j(b/m)ω
+
k/m
Dropping the time-harmonic function and rearranging yields
qE
0
/m
(k/m
−
ω
2
)
+
j(b/m)ω
x
=
(6-20)
The natural (resonant) frequency of the oscillator is defined when,
ω
2
=
k
/
m
,
which allows the definition of
ω
0
:
k
m
ω
0
=
(6-21)
Therefore, (6-20) can be simplified:
qE
0
/m
ω
0
−
ω
2
x
=
(6-22)
+
j(b/m)ω
From equation (6-2), the electric dipole moment is
q
2
E
0
/m
p
=
q x
=
(6-23)
ω
0
−
ω
2
+
j(b/m)ω
and subsequently, the polarization vector is
N(q
2
E
0
/m)
ω
0
−
ω
2
P
=
N p
=
+
j(b/m)ω
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