Page 241 - Numerical Analysis Using MATLAB and Excel

Chapter 6 Fourier, Taylor, and Maclaurin Series

obtain the DC value. By substitution of n = 0 , we get

a = 4A
-------
0
π
Therefore, the DC value is
1 2A
-
--a = ------- (6.58)
2 0 π
From (6.57) we observe that for all n = odd , other than n = 1 , a = . 0
n
To obtain the value of , we must evaluate the integral
a
1
1 π
a = --- ∫ sin tcos t t
d
π
1
From tables of integrals, 0
1
∫ ( sin ax cos ax x = ------ sin( 2a ax ) 2
(
)
)
d
and thus,
1
a = ------ sin( 2π t ) 2 π = 0 (6.59)
1
0
For n = even , from (6.57) we get
(
– 2A cos 2π + 1 ) 4A
a = ----------------------------------------- = – ------- (6.60)
2
π 2 –( 2 1 ) 3π
(
– 2A cos 4π + 1 ) 4A
a = ----------------------------------------- = – --------- (6.61)
4
π 4 –( 2 1 ) 15π
(
4A
----------------------------------------- =
a = – 2A cos 6π + 1 ) – --------- (6.62)
6
π 6 –( 2 1 ) 35π
(
– 2A cos 8π + 1 ) 4A
a = ----------------------------------------- = – --------- (6.63)
8
π 8 –( 2 1 ) 63π
and so on. Then, combining the terms of (6.58) and (6.60) through (6.63) we get

⎧
cos
cos
8ωt
cos
4ωt
6ωt
------- –
ft() = 2A 4A cos 2ωt ------------------ + ------------------ + ------------------ + … ⎬ ⎫ (6.64)
------- ------------------ +
⎨
π π ⎩ 3 15 35 63 ⎭
Therefore, the trigonometric form of the Fourier series for the full−wave rectification waveform with
even symmetry is
6−24 Numerical Analysis Using MATLAB® and Excel®, Third Edition
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