FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD 2π 661L
=5
2 π{[
−cos[
2 π( 1
2 +n)]( 1
2 +n)−cos[
2 π( 1
2 −n)]
( 1
2 −n)]−[
−cos 0
( 1
2 +n)−cos 0
( 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
and so onbn=
1
π∫ 2 π05 sinθ
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{[
sin 2π( 1
2 −n)( 1
2 −n) −sin 2π( 1
2 +n)( 1
2 +n)]−[
sin 0
( 1
2 −n)−sin 0
( 1
2 +n)]}Whennis both odd and even,bn=0 since sin (−π),
sin 0, sinπ, sin 3π,...are all zero. Hence the Fourier
series for the rectified sine wave,i=5 sinθ
2is given by:f(θ)=a 0 +∑∞n= 1(ancosnθ+bnsinnθ)i.e. i=f(θ)=10
π−20
3 πcosθ−20
(3)(5)πcos 2θ−20
(5)(7)πcos 3θ−···i.e. i=20
π(
1
2−cosθ
( 3 )−cos 2θ
( 3 )( 5 )−cos 3θ
( 5 )( 7 )−···)Now try the following exercise.Exercises 240 Further problems on 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
3sin 3x+1
5sin 5x+···)⎤⎥
⎥
⎦- 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+···]- For the waveform shown in Fig. 69.6 deter-
mine (a) the Fourier series for the function