082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

23.5 Fourier series 741
y
4
– 3π –– –π π –2
π –2
2
–
π
3π ––2
t
Figure23.17
Graph forExample 23.17.
inside. The integral thus reduces to
a n
= 1 π
∫ π/2
−π/2
4cosntdt
= 4 π
∫ π/2
−π/2
cosntdt
[ sinnt
] π/2
= 4 π n
−π/2
= 4 [sin nπ (
nπ 2 −sin − nπ 2
= 8
nπ sinnπ 2
)]
We obtaina 1
= 8 π ,a 2 =0,a 3 =−8 3π , etc. We finda 0
using Equation (23.3), again
[
integrating over − T ]
2 ,T :
2
a 0
= 2 T
= 1 π
= 4
∫ T/2
−T/2
∫ π/2
−π/2
f(t)dt
4dt = 4 π [t]π/2 −π/2
Similarly, tofind the Fourier coefficients,b n
, we use Equation (23.5):
b n
= 2 T
∫ T/2
−T/2
which reduces to
∫ π/2
f(t)sin 2nπt
T
dt
b n
= 1 4sinntdt
π −π/2
= 4 [ ] −cosnt π/2
π n
−π/2
= 4 [
−cos nπ ]
nπ 2 +cosnπ 2
= 0