Page 271 - Numerical Analysis Using MATLAB and Excel

Chapter 6 Fourier, Taylor, and Maclaurin Series

Therefore,

1 4A ⎛ 1 1 1 ⎞ A 4A ∞ 1
ft() = --a – ------- cos + -- cos 3t + ------ cos 5t + ------ cos 7t + … = ---- – ------- ∑ ----- cos nt
-
t
-
2 0 π 2 ⎝ 9 25 49 ⎠ 2 π n 2
n = odd
2.
ft() 2A
π
A -------t
ωt
⁄
⁄
0 π 2 π 3π 2 π
This is an even function; therefore, the series consists of cosine terms only. There is no half−
wave symmetry and the average (DC component) is not zero.

⁄
⁄
) [
1 Area 2 × ( A 2 ⋅ ( π 2 ⁄ ) ] + Aπ 3A ( π 2 ) ⋅ 3A
-
Average = --a = ----------------- = ---------------------------------------------------------------- = --------------------------- = -------
2 0 Period 2π 2π 4
⁄
2 π 2 2A 2 π
a = --- ∫ -------tcos nt t + --- ∫ Acos nt t (1)
d
d
n
π
π
π
⁄
and with 0 π 2
1
∫ xcos ax x = ----- cos ax + x - ax = ----- cos( 1 ax + axsin ax )
-- sin
d
a 2 a a 2
(1) simplifies to
⁄
π 2
4A 1 2A π
a = ------- ----- cos( nt + ntsin nt ) + ------- sin nt ⁄
n
π 2 n 2 nπ π 2
0
4A ⎛ nπ nπ nπ ⎞ 2A ⎛ nπ⎞
= ----------- cos ------ + ------ sin ------ – 1 0 + ------- sin nπ – sin ------ ⎠
–
nπ ⎝
2 2⎝
⎠
n π 2 2 2 2
and since sin ntπ = 0 for all integer , n
4A nπ 2A nπ 4A 2A nπ 4A ⎛ nπ ⎞
a = ----------- cos ------ + ------- sin ------ – ----------- – -------sin ------ = ----------- cos ------ – 1 ⎠
2 2⎝
n
2 2
2 2
n π 2 nπ 2 n π nπ 2 n π 2
------- 0 –(
)
)
,
for n = 1 a =, 1 4A 1 = – 4A , 2 a = --------- –( 4A 1 1 = – 2A
------- for n =
–
-------
2
π 2 π 2 4π 2 π 2
4A 4A – 4A
)
,
for n = 3 a =, 3 --------- 0 –( 1 = – --------- for n = 4 a =, 4 ----------- 11–( ) = 0
2 2
9π 2 9π 2 7 π
We observe that the fourth harmonic and all its multiples are zero. Therefore,
6−54 Numerical Analysis Using MATLAB® and Excel®, Third Edition
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