#20 (11/06/2023)

Note

Fourier coefficients (integrals) are pseudo-symmetric with respect to m or ω as

C−m
=
Cm
 
(1)
F−ω
=
F(ω)
 
(2)
Proof.
Just use the definition.

Power spectrum is symmetrical, i.e.
P(−ω) = P(ω).
(3)
Proof.
Just use the definition.

Nyquist-Shannon Sampling theorem

sampling.jpg
A Fourier transformable function, f(x), can be completely represented by its discrete values at x = …−3, 2, 1, 0, 1, 2, … by
f(x)
=
…+ f(−1) sin π(x + 1)
π(x + 1)
+f(0) sin πx
πx
+f(1) sin π(x − 1)
π(x − 1)
+ …
=
∞
∑
k=−∞ 
 f(k) sin π(x−k)
π(x−k)
(4)
if f(x) does not contain higher frequency components than π, i.e.
F(ω) = 0     |ω| > π.
Proof As the bandwidth of f(x) is restricted between −π < ω < π, the inverse Fourier transform of f(x) is

f(x) = 1
2 π
 ⌠
⌡ 		∞

−∞ 
 F(ω) ei ωx dω = 1
2 π
 ⌠
⌡ 		π

−π 
 F(ω) ei ωx dω,
(5)
where
F(ω) = ⌠
⌡ 		∞

−∞ 
 f(x) e−iωx dx.
As F(ω) exists only in the interval [−π, π], it can be expressed by the Fourier series as

F(ω) = ∞
∑
k=−∞ 
 Ck ei k ω,
(6)
where
Ck
=
1
2 π
 ⌠
⌡ 		π

−π 
 F(ω) e−i k ω dω
=
1
2 π
 ⌠
⌡ 		π

−π 
 F(ω) ei ω(−k) dω
=
f(−k),
(7)
where we used Eq. (5). By substituting Eq. (7) into Eq.(6), one gets
F(ω) = ∞
∑
k=−∞ 
 f(−k) ei k ω = ∞
∑
k=−∞ 
 f(k) e− i k ω.
(8)
Finally, using Eq. (8), f(x) is expressed as

f(x)
=
1
2 π
 ⌠
⌡ 		π

−π 
 ⎛
⎝ 		∞
∑
k=−∞ 
 f(k) e−i k ω⎞
⎠ 		ei ωx d ω
=
1
2 π
 ∞
∑
k=−∞ 
 ⎛
⎝ 		⌠
⌡ 		π

−π 
 e i ω(x − k) dω⎞
⎠ 		f(k)
=
1
2π
 ∞
∑
k=−∞ 
 ⎛
⎝ 		2 sin π(x − k)
(x − k)
⎞
⎠ 		f(k)
=
∞
∑
k=−∞ 
 f(k) sin π(x − k)
π(x − k)
 .
(9)
This conclusion can be generalized as
f(x) = ∞
∑
k=−∞ 
 f⎛
⎝ 		1
2W
 k⎞
⎠
sin ⎛
⎝ 		2 πW ⎛
⎝ 		x− 1
2W
 k⎞
⎠ 		⎞
⎠
2 πW ⎛
⎝ 		x− 1
2W
 k⎞
⎠
,
(10)
if
F(ω) = 0,     |ω| ≥ W.
(11)
Derivation

Sampling Rate
Period
Definition
1
(−π, π)
(angular frequency)
2π
(− 1
2
 , 1
2
 )
(frequency)
π
W
(−W, W)
(frequency)
1
2 W
(−2 πW, 2 πW)
(angular frequency)
(12)
1
The sampling theorem indicates that if the bandwidth of f(x) is limited to [−W, W], f(x) can be completely reconstructed by sampling the value of f(x) with the interval of τ = 2 W.


Examples: Human ears can hear frequencies up to 22 kHz. Therefore, the CD sample rate is 44.1 kHz. Copper phone lines pass frequencies up to 4 kHz, hence, phone companies sample at 8 kHz.

Parseval's theorem for Fourier series

The Fourier and inverse Fourier transform formulas are given as

f(x)
=
∞
∑
m=−∞ 
 Cm ei m x,
(13)
Cm
=
1
2 π
 ⌠
⌡ 		π

−π 
 f(x) e−i m x dx,
(14)
from which
f(x)
f(x)
 
 = ∞
∑
m=−∞ 
 ∞
∑
n=−∞ 
 Cm
Cn
 
 ei m x e−i n x
(15)
Integrating the both sides gives
⌠
⌡ 		π

−π 
 f(x)
f(x)
 
 dx
=
∞
∑
m=−∞ 
 ∞
∑
n=−∞ 
 Cm
Cn
 
 ⌠
⌡ 		π

−π 
 ei m x e−i n x dx
=
∞
∑
m=−∞ 
 ∞
∑
n=−∞ 
 Cm
Cn
 
 ⌠
⌡ 		π

−π 
 ei (m − n) x dx
=
2 π ∞
∑
m=−∞ 
 |Cm|2,
(16)
where
⌠
⌡ 		π

−π 
 eimx e−inx dx = ⎧
⎪
⎨
⎪
⎩
0
m ≠ n
2π
m=n
was used.
Hence, the Parseval identityfor Fourier series is stated as
⌠
⌡ 		π

−π 
 {f(x)}2 dx = 2 π ∞
∑
m=−∞ 
 |Cm|2,
(17)
or

1
2π
 ⌠
⌡ 		π

−π 
 {f(x)}2 dx = ∞
∑
m=−∞ 
 |Cm|2.
(18)
The left hand side represents the average power of f(x) while the right hand side represents the power of f(x).
Example
parseval1.jpg

Cm
=
1
2 π
 ⌠
⌡ 		π

−π 
 f(x) e−i m x dx
=
1
2 π
 ⎛
⎝ 		⌠
⌡ 		0

−π 
 (−1) e−i m x dx + ⌠
⌡ 		π

0 
 (+1) e−i m x dx⎞
⎠
=
1
π
 ⎛
⎝ 		(−1)m − 1
m
⎞
⎠ 		i,
(19)
so

|Cm|2 = ((−1)m − 1)2
π2 m2
 ,
(20)
or
|Cm|2 = ⎧
⎪
⎨
⎪
⎩
4
π2
 1
(2 m − 1)2
m = ±1, ±3, ±5, …
0
m = 0, ±2, ±4, ±6, …
(21)
Hence, Eq. (18) becomes
2 π = 2 π ∞
∑
m=−∞ 
 4
π2
 1
(2 m − 1)2
(22)
from which one obtains
1
12
 + 1
32
 + 1
52
 + 1
72
 + … = π2
8
 .
(23)

Footnotes:

1
ω = 2 πf
where ω is an angular frequency (unit: rad/time) and f is a (regular) frequency (unit: Hz).

File translated from TEX by TTH, version 4.03.
On 07 Nov 2023, 19:56.