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Waveforms in Trigonometric Form of Fourier Series

T
A

π 2π ωt
0

−A

Figure 6.14. Square waveform for Example 6.1

Solution:

The trigonometric series will consist of sine terms only because, as we already know, this wave-
form is an odd function. Moreover, only odd harmonics will be present since this waveform has
half−wave symmetry. However, we will compute all coefficients to verify this. Also, for brevity,
we will assume that ω = . 1

The coefficients are found from
a
i
1 2π 1 π 2π A π 2π
)
(
d
d
d
a = --- ∫ 0 ft()cos nt t = --- ∫ 0 Acos nt t + ∫ π – ( A cos nt t = ------ sin nt – sin nt π ) (6.24)
n
π
nπ
π
0
)
–
= ------ sin( A nπ 0 – sin n2π + sin nπ = ------ 2sin( A nπ – sin n2π )
nπ nπ
and since is an integer (positive or negative) or zero, the terms inside the parentheses on the
n
second line of (6.24) are zero and therefore, all coefficients are zero, as expected, since the
a
i
square waveform has odd symmetry. Also, by inspection, the average (DC ) value is zero, but if
we attempt to verify this using (6.24), we will get the indeterminate form 00⁄ . To work around
this problem, we will evaluate a 0 directly from (6.12). Then,
1 π 2π A
)
)
–
d
a = --- ∫ At + ∫ – ( A t = ---- π –( 0 2π + π = 0 (6.25)
d
π
0
0 π π
The b i coefficients are found from (6.14), that is,
1 2π 1 π 2π A π 2π
)
b = --- ∫ 0 ft()sin nt t = --- ∫ 0 Asin nt t + ∫ π – ( A sin nt t = ------ – ( nπ cos nt + cos nt π ) (6.26)
d
d
d
n
π
π
0
)
= ------ –( A cos nπ ++ cos 2nπ – cos nπ = ------ 12cos–( A nπ + cos 2nπ )
1
nπ nπ
For n = even , (6.26) yields
Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−13
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