Page 244 - Numerical Analysis Using MATLAB and Excel

Alternate Forms of the Trigonometric Fourier Series

ft()
3

2

1
ω = 1
t
π/2 π 3π/2 2π
Figure 6.21. Waveform for Example 6.8
Solution:
The given waveform has no symmetry; thus, we expect both cosine and sine functions with odd

and even terms present. Also, by inspection the DC value is not zero.
We will compute the a n and b n coefficients, the DC value, and we will combine them to get an
expression in the form of (6.63). Then,

⁄
⁄
1 π 2 1 2π 3 π 2 1 2π
a = --- ∫ 0 3 ()cos nt t +d --- ∫ π 2 1 ()cos nt t = ------ sin nt 0 + ------ sin nt π 2
d
n
π
π
nπ
nπ
⁄
⁄
3 π 1 1 π 2 π (6.70)
= ------ sin n--- + ------ sin n2π ------ sin– n--- = ------ sin n ---
nπ 2 nπ nπ 2 nπ 2
We observe that for n = even , a = . 0
n
For n = odd ,
a = 2 (6.71)
---
π
1
and
2
a = – ------ (6.72)
3
3π
The DC value is
⁄
1 1 π 2 1 2π 1 π 2 2π
⁄
(
--a = ------ ∫ 3 () t + ------ ∫ 1 () t = ------ 3t + t )
-
d
d
⁄
2 0 2π 2π 2π 0 π 2
⁄
0 π 2
(6.73)
1 ⎛ 3π π⎞ 1 3
)
= ------ ------ + 2π – --- = ------ π +( 2π = -- -
2π ⎝ 2 2 ⎠ 2π 2
The b n coefficients are
Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−27
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