Page 236 - Numerical Analysis Using MATLAB and Excel

Waveforms in Trigonometric Form of Fourier Series

integral by .
4
By inspection, the DC component is zero. From tables of integrals,

1
x
∫ xsin ax x = ---- sin ax --cos ax (6.39)
–
-
d
a 2 a
Then,
⁄
⁄
⁄
4 π 2 2A 8A π 2 8A ⎛ 1 t ⎞ π 2
2 ∫
d
d
-
b = --- ∫ -------tsin nt t = ------- tsin nt t = ------- ----- sin nt – -- cos nt ⎠
2 ⎝
n
π
2
0 π π 0 π n n
0 (6.40)
π
π
π
8A
⎛
⁄
= ---------- sin( 8A nt – ntcos nt ) π 2 = ---------- sin n--- – n--- cos n--- ⎞
2 ⎠
2 ⎝
2
2
n π 2 0 n π 2 2
We are only interested in the odd integers of , and we observe that:
n
π
cos n--- = 0
2
For odd integers of , the sine term yields
n
⎧ 1 for n = 1 5 9 … then b =,,, , -----------
8A
⎪ n 2 2
π ⎪ n π
sin n--- = ⎨
2 ⎪ 8A
⎪ – 1 for n = 3 7 11 … then b =,, , , n – -----------
2 2
⎩ n π
Thus, the trigonometric Fourier series for the triangular waveform with odd symmetry is
( n – 1 )
8A ⎛ 1 1 1 ⎞ 8A ---------------- 1
2
2 ∑
ft() = ------- sin ωt --sin 3ωt + ------ sin 5ωt – ------ sin 7ωt + … = ------- – ( 1 ) ----- sin nωt (6.41)
-
–
2 ⎝
⎠
π 9 25 49 π n 2
n = odd
Example 6.6
A half−wave rectification waveform is defined as
⎧ sin ωt 0 < ωt < π
ft() = ⎨ (6.42)
⎩ 0 π ω < t < 2π
Express ft() as a trigonometric Fourier series. Assume ω = . 1
Solution:
The waveform for this example is shown in Figure 6.18.
Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−19
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