Page 270 - Numerical Analysis Using MATLAB and Excel

Solutions to End−of−Chapter Exercises

6.13 Solutions to End−of−Chapter Exercises

1.

ft()
A A
----t
π

ωt
– 2π – π 0 π 2π
This is an even function; therefore, the series consists of cosine terms only. There is no half−
0
wave symmetry and the average (DC component) is not zero. We will integrate from to π
and multiply by . Then,
2
2 π A 2A π
2∫
d
d
a = --- ∫ ----tcos nt t = ------- tcos nt t (1)
n
π
0 π π 0
From tables of integrals,
1
∫ xcos ax x = ----- cos ax + x - ax
d
-- sin
a 2 a
and thus (1) becomes
2A 1 t π 2A 1 t 1
⎛
⎛
-
a = ------- ----- cos nt + -- sin nt ⎞ = ------- -----cos nπ + -- sin ntπ – ----- – 0 ⎞
-
2 ⎝
2 ⎝
n
π n 2 n ⎠ π n 2 n n 2 ⎠
0
and since sin ntπ = 0 for all integer , n
2A ⎛ 1 1 ⎞ 2A
–
a = ------- -----cos nπ -----– 2 ⎠ = ----------- cos( nπ 1 ) (2)
2 ⎝
n
2 2
π n 2 n n π
We cannot evaluate the average 12⁄( ) a ⁄ 0 from (2); we must use (1). Then, for n = , 0
1 2A π A t 2 π A π 2
⋅
-
⋅
--a = --------- 2∫ tt = ----- ---- = ----- -----
d
2 0 2π π 2 2 π 2 2
0
0
or
⁄
⁄
)
( 12 ⁄ a = A2
0
We observe from (2) that for n = even , a n = even = 0 . Then,
–
4A
4A
,
,
,
----------- for n =,
for n = 1 a =, 1 – 4A , 3 a = – 4A 5 a =, 5 – ----------- for n = 7 a = -----------
------- for n =
3
3
2 2
2 2
2 2
π 2 3 π 5 π 7 π
and so on.
Numerical Analysis Using MATLAB® and Excel®, Third Edition 6−53
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