Biomedical Engineering Reference
In-Depth Information
So, to begin solving this problem, you first need to multiply the denominator by 2/2, as
shown by (4.32), (4.33), and (4.34).
sin
nw
0
t
a
2
A
T
2
2
e
−
j
nw
0
t
a
α
=
(4.32)
2
n
nw
0
/
2
Then, multiply by
t
a
/
t
a
.
sin
nw
0
t
a
2
A
T
t
a
t
a
e
−
j
nw
0
t
a
α
=
(4.33)
2
n
nw
0
2
sin
nw
0
t
a
2
⎡
⎤
⎣
⎦
At
a
T
e
−
j
nw
0
t
a
α
=
(4.34)
2
n
nw
0
t
a
2
Since
w
0
=
2
π
f
0
and
f
0
=
1
/
T
, then
w
0
=
(2
π
)
/
T
. Thus, alpha can be rewritten
as (4.35).
sin (
e
−
j
2
n
π
t
a
At
a
T
π
nt
a
/
T
)
α
=
(4.35)
2
T
n
n
π
t
a
/
T
The complete series is written as (4.36) or equations in (4.37).
∞
n
e
jnw
0
t
x
(
t
)
=
α
or
(4.36)
n
=−∞
⎧
⎨
⎡
⎤
⎫
⎬
∞
At
a
T
sin (
n
π
t
a
/
T
)
⎣
⎦
e
−
j
2
n
π
t
a
e
j
n
2
π
t
x
(
t
)
=
2
T
T
n
π
t
a
⎩
⎭
n
=−∞
T
At
a
T
sin(
n
e
−
j
2
n
π
(
t
−
t
2
)
∞
π
t
a
/
T
)
x
(
t
)
=
(4.37)
T
π
/
n
t
a
T
n
=−∞
Keep in mind that the signal is being represented in terms of sinusoids having fre-
quencies that are multiples of the fundamental frequency 1
n
, give
the magnitude and phase of these sinusoids, which constitute an approximate frequency-
domain description of the explicit time-domain signal. The Fourier series is by far the
most commonly used orthogonal representation.
/
T
. The coefficients,
α
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