Assumption: F(ω)exits only between [−l, l].
The formula for inverse Fourier transform is
f(x)
=
1
2π
 ⌠
⌡ 		∞

−∞ 
 F(ω) ei ωx d ω
=
1
2π
 ⌠
⌡ 		l

−l 
 F(ω) ei ωx d ω
(1)
As F(ω) exists only between [−l, l], it can be expressed by Fourier series as:
F(ω) = ∞
∑
k=−∞ 
~
F
 
k 
 ei k [(π)/(l)] ω
(2)
with its Fourier coefficient as
~
F
 
k 
 = 1
2l
 ⌠
⌡ 		l

−l 
 F(ω) e−i k [(π)/(l)] ω dω
(3)
Combining Eq.(1) with Eq.(2) yields
f(x)
=
1
2π
 ⌠
⌡ 		l

−l 
 F(ω) ei ωx d ω
=
1
2π
 ⌠
⌡ 		l

−l 
 ⎛
⎝ 		∞
∑
k=−∞ 
~
F
 
k 
 ei k [(π)/(l)] ω⎞
⎠ 		ei ωx d ω
=
1
2π
 ∞
∑
k=−∞ 
~
F
 
k 
 ⌠
⌡ 		l

−l 
 ei k [(π)/(l)] ωei ωxd ω
=
1
2π
 ∞
∑
k=−∞ 
~
F
 
k 
 2 lsin (k π+lx)
kπ+lx
(4)
On the other hand, combining Eq.(1) with Eq.(3) yields

~
F
 
k 
=
1
2l
 ⌠
⌡ 		l

−l 
 F(ω) e−i k [(π)/(l)] ω dω
=
π
l
 ⎛
⎝ 		1
2π
 ⌠
⌡ 		l

−l 
 F(ω) ei ω(−k π/l) dω⎞
⎠
=
π
l
 f⎛
⎝ 		− π
l
 k⎞
⎠ 		.
(5)
Therefore, Eq.(4) can be written as

f(x)
=
1
2π
 ∞
∑
k=−∞ 
 π
l
 f⎛
⎝ 		− π
l
 k⎞
⎠ 		2 lsin (k π+lx)
kπ+lx
=
∞
∑
k=−∞ 
 f⎛
⎝ 		− π
l
 k⎞
⎠ 		sin (k π+lx)
kπ+lx
=
∞
∑
k=−∞ 
 f⎛
⎝ 		π
l
 k⎞
⎠ 		sin (−k π+lx)
−kπ+lx
 .
(6)
If the interval is set as (−2 πW, 2 πW), choose l = 2 πW to rewrite Eq.(6) as
f(x) = ∞
∑
k=−∞ 
 f⎛
⎝ 		1
2W
 k⎞
⎠
sin ⎛
⎝ 		2 πW ⎛
⎝ 		x− 1
2W
 k⎞
⎠ 		⎞
⎠
2 πW ⎛
⎝ 		x− 1
2W
 k⎞
⎠
,
(7)
which implies that f(x) can be expressed with a sampling rate of 1/(2 W).

File translated from TEX by TTH, version 4.03.
On 06 Nov 2022, 23:32.