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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