Page 274 - Numerical Analysis Using MATLAB and Excel

Solutions to End−of−Chapter Exercises

A π 2⁄ – jnt A – jnt π 2⁄ A – jnπ 2 jnπ 2
⁄
⁄
C = ------ ∫ – π 2 e t d = ---------------e – π 2 = --------------- e ( – j2nπ – e )
j2nπ
2π
n
–
⁄
⁄
⁄
⁄
–
jnπ 2
jnπ 2
e
A
–
A e
⎞
⎛
⁄
⁄
= ------------ e ( A jnπ 2 – e – jnπ 2 ) = ------ -------------------------------------- = ------ sin nπ
------
j2nπ nπ ⎝ j2 ⎠ nπ 2
and we observe that for n = even , C = 0
n
For n = odd , C n alternates in plus (+) and minus (−) signs, that is,
A
,,
,
C = ------ if n = 159 …
nπ
n
A
,,
,
C = – ------ if n = 3711 …
n
nπ
Thus,
A ⎛ A jnωt ⎞
ft() = ---- + ∑ ± ------e
2 ⎝ nπ ⎠
n = odd
,,,
where the plus (+) sign is used with n = 1 5 9 … and the minus (−) sign is used with
,
,,
n = 3 7 11 … . We can express ft() in a more compact form as
A ( n – 1 ⁄ A jnωt
)
2
ft() = ---- + ∑ – ( 1 ) ------e
2 nπ
n = odd
6.
ft() 2A 1
-------t –
A π
⁄
– π 2 π 2
⁄
– π 0 π ωt
−A
We will find the exponential form coefficients C n from
1
C = ------ ∫ π ft()e – jnt t d
n
2π
From tables of integrals – π
∫ xe d x = e ax 1 )
ax
------- ax –(
a 2
Then,
Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−57
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