Page 704 - The Mechatronics Handbook

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and

1
c 0 = --- ∫ T/2 xt() t = A
---.
d
T – T/2 2
This method of determining c k , where applicable, is less laborious than the direct method.
Fourier Series for Real Signals 3
A real signal x(t) that satisfies the Dirichlet conditions can be expanded by the following procedure.
Recall Eq. (23.17) and replace k by −k, we obtain

1
1
1
c k– = 1 ∫ t +T xt()e j2πkf t t d = --- ∫ t +T xt()e j – 2πkf t 0 t d ∗ = c k ∗ (23.18)
0
---
T t T t
1 1
That is, the positive and negative coefficients are complex conjugates of each other for real signals.
For such signals, |c k | has even symmetry and ∠c k has odd symmetry with respect to k = 0. Denote c k =
(a k − jb k )/2, then c −k = (a k + jb k )/2, so that
j2πkf t j – 2πkf t a k – jb k j2πkf t a k + jb k j – 2πkf t
c k e 0 + c k– e 0 = -----------------e 0 + -----------------e 0
2 2
Since c 0 is real and defined as c 0 = a 0 /2, then we can write

∞
xt() = ---- + ∑ [ a k cos ( 2πkf 0 t) + b k sin ( 2πkf 0 t)] (23.19)
a 0
2
k=1
This relationship, which only holds for a real periodic signal x(t), is called the trigonometric Fourier
series expansion. Since

a k – jb k 1 t +T – j2πkf t
1
c k = ----------------- = --- ∫ xt()e 0 t d
2 T t
1

1 t +T
1
d
= --- ∫ xt() cos[ ( 2πkf 0 t) – jsin ( 2πkf 0 t)] t (23.20)
T t 1
Hence
1
a k = 2 ∫ t +T xt()cos ( 2πkf 0 t) t (23.21a)
---
d
T t
1
2
1
b k = --- ∫ t +T xt()sin ( 2πkf 0 t) t (23.21b)
d
T t
1
and
1
a 0 = 2 ∫ t +T xt() t
d
---
T t
1
The third method of Fourier series expansion of real signals is given by
∞
xt() = c 0 ∑ c k cos ( 2πkf 0 t + θ k ) (23.22)
+
k=1
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