Page 238 - Numerical Analysis Using MATLAB and Excel

Waveforms in Trigonometric Form of Fourier Series

Next, we can evaluate all the a n coefficients, except , from (6.44).
a
1
⁄
a
First, we will evaluate to obtain the DC value. By substitution of n = 0 , we get a = 2A π
0
0
Therefore, the DC value is
1 A
-
--a = ---- (6.45)
2 0 π
We cannot use (6.44) to obtain the value of ; therefore, we will evaluate the integral
a
1
A π
d
a = ---- ∫ sin tcos t t
π
1
From tables of integrals, 0
1
∫ ( sin ax cos ax x = ------ sin( 2a ax ) 2
)
)
(
d
and thus, π
a = ------ sin( A t ) 2 = 0 (6.46)
1
2π
0
,,,,
From (6.44) with n = 2 3 4 5 … , we get
A cos 2π + 1 ⎞ 2A
⎛
a = ---- ------------------------- ⎠ = – ------- (6.47)
π ⎝
2
3π
2
2 )
(
1 –
(
a = A cos 3π + 1 ) 0 (6.48)
---------------------------------- =
3
π 1 –( 3 ) 2
We see that for odd integers of , n 0 . However, for n = even , we get
na =
(
A cos 4π + 1 ) 2A
a = ---------------------------------- = – --------- (6.49)
4
π 14 ) ( – 2 15π
(
2A
a = A cos 6π + 1 ) – --------- (6.50)
---------------------------------- =
6
π 16 ) ( – 2 35π
(
2A
a = A cos 8π + 1 ) – --------- (6.51)
---------------------------------- =
8
π 18 ) ( – 2 63π
and so on.
Now, we need to evaluate the b n coefficients. For this example,
1 2π A π A 2π
d
d
b = A--- ∫ ft()sin nt t = ---- ∫ sin tsin nt t + ---- ∫ 0sin nt t
d
n
π
0 π 0 π π
Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−21
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