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Waveforms in Trigonometric Form of Fourier Series

Example 6.4

Compute the trigonometric Fourier series of the sawtooth waveform of Figure 6.16.

T

A

−2π −π π
0 2π ωt

−A

Figure 6.16. Sawtooth waveform

Solution:
This waveform is an odd function but has no half−wave symmetry; therefore, it contains sine
terms only with both odd and even harmonics. Accordingly, we only need to evaluate the b n

coefficients. By inspection, the DC component is zero. As before, we will assume that ω = . 1

If we choose the limits of integration from to 2π we will need to perform two integrations since
0
⎧ A 0 << π
t
----t
⎪ π
ft() = ⎨
⎪ A 2A π << 2π
t
----t –
⎩ π
However, we can choose the limits from π– to +π , and thus we will only need one integration
since
A
ft() = ----t – π << π
t
π
Better yet, since the waveform is an odd function, we can integrate from to , and multiply the
π
0
integral by ; this is what we will do.
2
From tables of integrals,
1
x
∫ xsin ax x = ----- sin ax --cos ax (6.36)
-
d
–
a 2 a
Then,

Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−17
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