Page 475 - Advanced engineering mathematics

13.5 Phase Angle Form 455

8

6

4

2

0
–2 0 2 4 6
x
FIGURE 13.17 f (x) in Example 13.16.

and

2 3 2nπx 9
2
b n = x sin dx =− .
3 0 3 nπ
The Fourier series of f (x) is
∞
9 1 2nπx 2nπx
3 + cos − sin .
nπ nπ 3 3
n=1
We may think of this as the Fourier series of f (x) on the symmetric interval [−3/2,3/2].How-
2
ever, keep in mind that f (x) is not x on this interval. We have f (x) = x on 0 ≤ x < 3, hence
2
2
also on [0,3/2]. But from Figure 13.17, f (x) = (x + 3) on [−3/2,0).
This Fourier series converges to
⎧
⎪9/4 for x =±3/2,
⎪
⎪
9/2 for x = 0,
⎨
⎪(x + 3) 2 for −3/2 < x < 0,
⎪
⎪
x for 0 < x < 3/2.
⎩ 2
For the phase angle form, compute
9 √

2
2
2
c n = a + b = 1 + n π 2
n
n
n π 2
2
and

−9/nπ
δ n = arctan − = arctan(nπ) = 0.
2
9/n π 2
The phase angle form of the Fourier series of f (x) is
∞
9 √ 2nπx

2
2
3 + 1 + n π cos .
2
n π 2 3
n=1
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October 14, 2010 14:57 THM/NEIL Page-455 27410_13_ch13_p425-464