Page 231 - Numerical Analysis Using MATLAB and Excel

Chapter 6 Fourier, Taylor, and Maclaurin Series

A
)
b = ------ 12– +( nπ 1 = 0
n
as expected, since the square waveform has half−wave symmetry.

For n = odd , (6.21) reduces to
A 4A
)
b = ------ 1 +( nπ 2 + 1 = -------
n
nπ
and thus
b = 4A
-------
1
π
4A
b = -------
3π
3
-------
b = 4A
5
5π
and so on.
Therefore, the trigonometric Fourier series for the square waveform with odd symmetry is

4A ⎛ 1 1 ⎞ 4A 1
-
-
ft() = ------- sin ωt + --sin 3ωt + --sin 5ωt + … = ------- ∑ -- sin nωt (6.27)
-
π ⎝ 3 5 ⎠ π n
n = odd
It was stated above that, if the given waveform has half−wave symmetry, and it is also an odd or
an even function, we can integrate from to π 2⁄ , and multiply the integral by . We will apply
4
0
this property to the following example.

Example 6.2
Compute the trigonometric Fourier series of the square waveform of Example 1 by integrating
from to π 2⁄ , and multiplying the result by . 4
0
Solution:

Since the waveform is an odd function and has half−wave symmetry, we are only concerned with
the odd b n coefficients. Then,

⁄
1 π 2 4A π 2 4A ⎛ π ⎞
⁄
d
b = 4--- ∫ 0 ft()sin nt t = ------- – ( nπ cos nt 0 ) = ------- – cos n--- + 1 ⎠ (6.28)
nπ ⎝
n
π
2
For n = odd , (6.28) becomes

6−14 Numerical Analysis Using MATLAB® and Excel®, Third Edition
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