Fourier series for periodic functions of period 2π 615
=5
2 π{[
−cos[
2 π( 1
2 +n)]
( 1
2 +n)−cos[
2 π( 1
2 −n)]
( 1
2 −n)]−[
−cos0
( 1
2 +n)−cos0
( 1
2 −n)]}Whennis both odd and even,
an=5
2 π{[
1
( 1
2 +n)+1
( 1
2 −n)]−[
− 1
( 1
2 +n)−1
( 1
2 −n)]}=5
2 π{
2
( 1
2 +n)+2
( 1
2 −n)}=5
π{
1
( 1
2 +n)+1
( 1
2 −n)}Hence
a 1 =5
π[
1
3
2+1
−^12]
=5
π[
2
3−2
1]
=− 20
3 πa 2 =5
π[
1
5
2+1
−^32]
=5
π[
2
5−2
3]
=− 20
( 3 )( 5 )πa 3 =5
π[
1
7
2+1
−^52]
=5
π[
2
7−2
5]
=− 20
( 5 )( 7 )π
andsoonbn=1
π∫ 2 π05sinθ
2sinnθdθ=5
π∫ 2 π0−1
2{
cos[
θ(
1
2+n)]−cos[
θ(
1
2−n)]}
dθfrom Chapter 40=5
2 π[
sin[
θ( 1
2 −n)]
( 1
2 −n) −sin[
θ( 1
2 +n)]
( 1
2 +n)] 2 π0=5
2 π{[
sin2π( 1
2 −n)
( 1
2 −n) −sin2π( 1
2 +n)
( 1
2 +n)]−[
sin0
( 1
2 −n)−sin0
( 1
2 +n)]}Whennis both odd and even,bn=0sincesin(−π),
sin0, sinπ,sin3π,...are all zero. Hence the Fourier
series for the rectified sine wave,i=5sinθ
2is given by:f(θ )=a 0 +∑∞n= 1(ancosnθ+bnsinnθ)i.e. i=f(θ )=10
π−20
3 πcosθ−20
( 3 )( 5 )πcos2θ−20
( 5 )( 7 )πcos3θ−···i.e. i=
20
π(
1
2−
cosθ
( 3 )−
cos2θ
( 3 )( 5 )−
cos3θ
( 5 )( 7 )−···)Now try the following exerciseExercises228 Further problemson Fourier
series of periodic functions of period 2π- Determine the Fourier series for the periodic
function:
f(x)={
− 2 , when −π<x< 0+ 2 , when 0<x<πwhich is periodic outside this range of
period 2π.
⎡
⎢
⎢
⎣f(x)=8
π(
sinx+1
3sin3x+1
5sin5x+···)⎤
⎥
⎥
⎦- For the Fourier series in Problem 1, deduce a
series for
π
4at the point wherex=π
2
[
π
4= 1 −1
3+1
5−1
7+···]