Page 247 - Numerical Analysis Using MATLAB and Excel

Chapter 6 Fourier, Taylor, and Maclaurin Series

Then, (6.88) is written as

ft() = … + C e – j2ωt + C e – jωt + C + C e jωt + C e j2ωt + … (6.92)
2
–
1
–
2
1
0
We must remember that the C i coefficients, except C 0 , are complex and occur in complex conju-
gate pairs, that is,
C – n = C ∗ (6.93)
n
We can derive a general expression for the complex coefficients C n , by multiplying both sides of
(6.92) by e – jnωt and integrating over one period, as we did in the derivation of the a n and b n
coefficients of the trigonometric form. Then, with ω = , 1

2π 2π 2π
–
jt –
∫ ft()e – jnt t d = … + ∫ C e – j2t – jnt t d + ∫ C e e jnt t d (6.94)
e
1
–
–
2
0 0 0
2π 2π
jt –
+ ∫ C e – jnt t d + ∫ C e e jnt t d
0
1
0 0
2π 2π
+ ∫ C e j2t – jnt t d + … + ∫ C e jnt – jnt t d
e
e
n
2
0 0
We observe that all the integrals on the right side of (6.97) are zero except the last. Therefore,
2π 2π 2π
∫ ft()e – jnt t d = ∫ C e jnt – jnt t d = ∫ C t = 2πC n
e
d
n
n
0 0 0
or
1 2π – jnt
C = ------ ∫ 0 ft()e t d
n
2π
and, in general, for ω ≠ , 1
1 2π – jnωt
C = ------ ∫ ft()e ( d ωt ) (6.95)
n
2π
0
or
1 T – jnωt
C = --- ∫ 0 ft()e ( d ωt ) (6.96)
n
T
We can derive the trigonometric Fourier series from the exponential series by addition and sub-
traction of the exponential form coefficients C n and C – n . Thus, from (6.89) and (6.90),
1
C + C – n = -- a –( 2 - n jb + a + jb ) n
n
n
n
6−30 Numerical Analysis Using MATLAB® and Excel®, Third Edition
Copyright © Orchard Publications