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( ω +2 πr )
2 =1 .
|
(8.36)
r = −∞
φ ( t )is
( ω )
2 and sampling a function periodizes its Fourier
transform. Equation (8.36) is true if we choose
This is because φ
|
1
a ( ω )=
.
| N n ( ω +2 πr )
2 ) 1 / 2
(
|
r = −∞
Using the above result we can write,
2 n sin n ω/ 2 e −iβω/ 2
φ ( ω )=
,
| N n ( ω +2 πr )
2 ) 1 / 2
|
ω n (
r =
−∞
2 n sin n ω/ 2 e −iβω/ 2
(8.37)
=
,
1
( ω +2 πr ) 2 n ) 1 / 2
ω n 2 n sin n ω/ 2(
r = −∞
e −iβω/ 2
ω n S 2 n ,
=
where
1
( ω +2 πr ) 2 n ,
S 2 n =
(8.38)
r = −∞
d 2 n− 1
1
2 2 n
1
=
dx 2 n− 1 cot x.
(2 n
1)!
Example 1: Linear splines
For linear splines, the order of the polynomial is n =2.Alsowhen n is even,
β =0and β = 1 when n is odd. Hence, from equation (8.38),
1+2cos 2 x
sin 4 x
S 4 ( ω )= 1
48
.
Therefore,
1+2cos 2 x
sin 2 x
S 4 ( ω )=
1
4 3
,
and so,
4 3sin 2 x
1+2cos 2 x .
1
ω 2
φ ( ω )=
Example 2: Cubic splines
For cubic spline, n = 4. Hence, from equation (8.38),
 
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