Page 272 - Numerical Analysis Using MATLAB and Excel

Solutions to End−of−Chapter Exercises

3A 4A ⎛ 1 1 ⎞
-
-
ft() = ------- – ------- cos + -- cos 2t + -- cos 3t + …
t
4 π 2 ⎝ 2 9 ⎠
3.
ft()
A

0 π 2π ωt
This is neither an even nor an odd function and has no half−wave symmetry; therefore, the
series consists of both cosine and sine terms. The average (DC component) is not zero. Then,

1 2π – jnωt
C = ------ ∫ 0 ft()e ( d ωt )
2π
n
and with ω = 1

1 2π – jnt 1 π – jnt 2π – jnt A π – jnt
C = ------ ∫ 0 ft()e t d = ------ ∫ 0 Ae t d + ∫ π 0e t d = ------ ∫ 0 e t d
2π
2π
2π
n
The DC value is
A π 0 A π A
C = ------ ∫ 0 e td = ------t 0 = ----
0
2π
2π
2
For n ≠ 0
A
A
C = ------ ∫ π e – jnt t d = ---------------e – jnt π ------------ 1 –( = A e – jnπ )
2π
n
0 – j2nπ 0 j2nπ
Recalling that
e – jnπ = cos nπ – jsin nπ

for n = even , e – jnπ = 1 and for n = odd , e – jnπ = – 1 . Then,

A
)
C n = even = ------------ 1 –( j2nπ 1 = 0
and
A A
C n = odd ------------ 1 –[ = j2nπ – ( 1 ] ) = --------
jnπ
By substitution into the expression

ft() = … + C e – j2ωt + C e – jωt + C + C e jωt + C e j2ωt + …
–
–
2
1
2
1
0
we find that
Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−55
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