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Waveforms in Trigonometric Form of Fourier Series

4A 4A
)
b = ------- – +( nπ 0 1 = ------- (6.29)
n
nπ
as before, and thus the series is the same as in Example 1.

Example 6.3

Compute the trigonometric Fourier series of the square waveform of Figure 6.15.
Solution:

This is the same waveform as in Example 6.1, except that the ordinate has been shifted to the
right by π 2⁄ radians, and has become an even function. However, it still has half−wave symme-
try. Therefore, the trigonometric Fourier series will consist of odd cosine terms only.

T
A

π / 2 3π / 2

ωt
0 π 2π

−A

Figure 6.15. Waveform for Example 6.3

Since the waveform has half−wave symmetry and is an even function, it will suffice to integrate
from to π 2⁄ , and multiply the integral by . The a n coefficients are found from
4
0
⁄
⁄
1 π 2 4 π 2 4A π 2 4A ⎛ π ⎞
⁄
(
a = 4 --- ∫ 0 ft()cos nt t = --- ∫ 0 Acos nt t = ------- sin nt 0 ) = ------- sin n--- (6.30)
d
d
nπ ⎝
2 ⎠
n
π
π
nπ
We observe that for n = even , all a n coefficients are zero, and thus all even harmonics are zero
as expected. Also, by inspection, the average (DC ) value is zero.
π
For n = odd , we observe from (6.30) that sin n --- , will alternate between +1 and 1– depending
2
on the odd integer assigned to . Thus,
n
a = ± 4A (6.31)
-------
n
nπ
,,,
For n = 1 5 9 13 , and so on, (6.30) becomes
Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−15
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