Page 249 - Numerical Analysis Using MATLAB and Excel

Chapter 6 Fourier, Taylor, and Maclaurin Series

Example 6.9
Compute the exponential Fourier series for the square waveform of Figure 6.22 below. Assume
that ω = . 1

T
A

π 2π
ωt
0

−A
Figure 6.22. Waveform for Example 6.9
Solution:
This is the same waveform as in Example 6.1, and as we know, it is an odd function, has half−
wave symmetry, and its DC component is zero. Therefore, the C n coefficients will be imaginary,

C = 0 for n = even , and C = 0 . Using (6.95) with ω = 1 , we get
0
n
1 2π – jnt 1 π – jnt 1 2π – jnt
C = ------ ∫ ft()e t d = ------ ∫ Ae t d + ------ ∫ – Ae t d
2π
n
0 2π 0 2π π
and for n = , 0
1 π – 0 2π – 0 A
)
)
C = ------ ∫ Ae d + ∫ – ( A e d t = ------ π –( 2π + π = 0
t
2π
0
0 π 2π
as expected.
For n ≠ , 0
1 π – jnt 2π – jnt 1 A – jnt π – A – jnt 2π
C = ------ ∫ Ae t d + ∫ – Ae t d = ------ --------e + --------e
n
2π
0 π 2π – jn 0 – jn π
1 A – jnπ A – jn2π – jnπ A – jnπ – jn2π – jnπ (6.102)
)
------ -------- e ( = – 1 + ----- e ( – e ) = ------------ 1 –( e + e – e )
2π – jn jn 2jπn
= ------------ 1 +( A e – jn2π – 2e – jnπ ) ------------ e ( = A – jnπ – 1 ) 2
2jπn 2jπn

– jnπ
For n = even , e = 1 ; then,

C n ------------ e ( = A – jnπ – 1 ) 2 = ------------ 1 –( A 1 ) 2 = 0 (6.103)
n = even 2jπn 2jπn

6−32 Numerical Analysis Using MATLAB® and Excel®, Third Edition
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