Biomedical Engineering Reference
In-Depth Information
Applying the condition of orthogonality, the right side of the equation becomes (4.7).
t
2
j
(
t
)
x
(
t
)
dt
=
λ
(4.7)
a
j
j
t
1
Note that when the indexes of the basis function are equal,
j
=
n
, the resulting integral
will be some value, lambda (4.8);
t
2
j
(
t
)
n
(
t
)
dt
=
λ
(4.8)
j
t
1
otherwise, if indexes of the basis function are not equal,
j
n
, the integral will equal
zero. The equation may be rewritten to solve for the coefficients as shown in the following
equation:
=
t
2
1
λ
j
(
t
)
x
(
t
)
dt
=
a
j
(4.9)
j
t
1
Recall the general equation for an energy or average power signal as shown in the fol-
lowing equation:
t
2
x
2
(
t
)
dt
E
=
(4.10)
t
1
By substituting the general basis function representation of a signal (4.11) into (4.10);
N
x
(
t
)
=
a
n
n
(
t
)
(4.11)
n
=
0
the energy equation (4.10), may be expressed as (4.12):
t
2
t
2
t
2
N
N
x
2
(
t
)
dt
E
=
=
x
(
t
)
a
n
n
(
t
)
dt
=
a
n
n
(
t
)
x
(
t
)
dt
(4.12)
n
=
0
n
=
0
t
1
t
1
t
1
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