Fourier Series - (12.1)(12.2)
1. Periodic Functions:
A function f is a periodic function with period 2p if fŸx 2p fŸx .
Example Sketch the graph of function.
a. fŸx b. gŸx -2
-1
x
if 0 t x 1
2"x
if 1 t x t 2
sinŸx if 0 t x t =
0
if = t x t 2=
with a period of 2
with a period of 2=.
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
1
2
3
-6
4
-4
-2
0
y fŸx 2
4
6
8
10
12
y gŸx 2. Fourier Series of a periodic function with period 2p :
Functions sinŸbx and cosŸbx are periodic functions with a period of 2= . The periods of
b
=x
=x
cos p , sin p
are 2p, and periods of
n=x
cos n=x
p , sin p
2p
are n where n is a positive integer. A linear combination of
=x
=x
=x
1, cos =x
p , sin p , cos 2 p , sin 2 p , . . . ,
also has a periodic of 2p.
Let fŸx be defined a function on the interval "p, p with period 2p. Then the Fourier series of fŸx is
defined by
.
n=x
fŸx a 0 ! a n cosŸ n=x
p b n sinŸ p 2
where for n 1, . . . ,
n1
a 0 1p
p
; "p fŸx dx,
b n 1p
a n 1p
p
; "p fŸx cosŸ n=x
p dx
p
; "p fŸx sinŸ n=x
p dx
Note that:
a. If fŸx is an even function, then fŸx sinŸ n=x
p is an odd function and b n 0 for all n u 1 and
1
a 0 2p
p
; 0 fŸx dx,
a n 2p
p
; 0 fŸx cosŸ n=x
p dx,
.
fŸx a 0 ! a n cosŸ n=x
p " a cosine series
2
n1
b. If fŸx is an odd function, then fŸx cosŸ n=x
p is an odd function and a n 0 for all n u 0 and
b n 2p
p
; 0 fŸx sinŸ n=x
p dx,
.
fŸx ! b n sinŸ n=x
p " a sine series
n1
c. The derivation is based on the following property: functions
2=x
2=x
=x
1, cos =x
p , sin p , cos p , sin p , . . . ,
are orthogonal on "p, p , i.e.,
p
p
p
=nx p
; "p cos =nx
p dx n= sin p | "p n= ¡2 sinŸ=n ¢ 0
p
; "p sin
=nx dx 0
p
p
=nx
p
=n sin =m 0 if n p m
cos =mx
dx 2p n sin =n cos =m2 " m cos
2
p
=Ÿn " m p
=nx
p
sin =mx
dx 0
p
p
=nx
p
=n cos =m 0 if n p m
sin =mx
dx "2p n cos =n sin =m2 " m sin
p
=Ÿn " m 2 ; "p cos
; "p cos
; "p sin
Convergence of a Fourier Series:
U
U
Let f and f be piecewise continuous on the interval "p, p , that is, let f and f be continuous except at a
finite number of points in the interval and have only finite discontinuities at these points. Then the Fourier
series of f on the interval converges to fŸx at a point of continuity. At a point of discontinuity, the Fourier
series converges to the average
1 lim fŸx lim fŸx .
xva 2 xva "
Example Find Fourier series of the periodic function fŸx "k
if " = x 0
k
if 0 x =
with a period of 2=.
1
0.5
-6
-4
0
-2
2
x
4
6
-0.5
-1
fŸx "k
if " = x 0
k
if 0 x =
2
p =. Since fŸx is odd, a n 0, for n u 0.
n
2 ; = fŸx sinŸnx dx 2 ; = k sinŸnx dx " 2k cos =n " 1 " 2k Ÿ"1 " 1
bn =
= 0
=
=
n
n
0
4k .
when n is even, cosŸn= 1, and then b n 0; and when n is odd, cosŸn= "1 and then b n n=
0
4k
n=
bn .
fŸx !
m0
n 2m
n 2m 1
1
1
4k
sinŸŸ2m 1 x 4k
= sinŸx 3 sinŸ3x 5 sinŸ5x . . . .
Ÿ2m 1 =
Let k 1.
4 sinŸx F 1 Ÿx =
4 ¡sinŸx F 2 Ÿx =
4 ¡sinŸx F 3 Ÿx =
1 sinŸ3x ¢
3
1 sinŸ3x 1 sinŸ5x ¢
5
3
1
0.5
-6
-4
-2
0
2
x 4
6
-0.5
-1
y fŸx , F 1 Ÿx , F 2 Ÿx , F 3 Ÿx ,
Example Find Fourier series of the periodic function fŸx x if 0 x 1
1 " x if 1 x 2
, with a period of 2.
3
1
0.5
-2
-1
0
1
2
-0.5
-1
p 1. Since fŸx 1 " Ÿx 2 "x " 1, for "1 x 0. the function fŸx can also be written as:
fŸx a0 an bn 1
1
x if 0 x 1
"Ÿx 1 if " 1 x 0
1
0
; "1 fŸx dx ; "1 "Ÿx 1 dx ; 0 xdx 0
1
0
; "1 fŸx cosŸn=x dx ; "1 "Ÿx 1 cosŸn=x dx ; 0 x cosŸn=x dx
1
n 2m
0
"4
= 2 Ÿ2m 1 2
2 Ÿ"1 cosŸ=n 2 Ÿ"1 Ÿ"1 n =2n2
=2n2
n 2m 1
, m 0, 1, . . .
1
0
; "1 fŸx sinŸn=x dx ; "1 "Ÿx 1 sinŸn=x dx ; 0 x sinŸn=x dx
1 Ÿ1 " cosŸn= 1 Ÿ1 " Ÿ"1 n =n
=n
0
2
=Ÿ2m 1 n 2m
n 2m 1
, m 0, 1, . . .
.
fŸx a 0 !¡a n cosŸn=x b n sinŸn=x ¢
2
n1
.
!
m0
2
"4
cosŸŸ2m 1 =x sinŸŸ2m 1 =x =Ÿ2m 1 = 2 Ÿ2m 1 2
2 sinŸ=x F 1 Ÿx " 42 cosŸ=x =
=
F 2 Ÿx F 1 Ÿx " 4 2 cosŸ3=x 2 sinŸ3=x 3=
9=
F 3 Ÿx F 2 " 4 2 cosŸ5=x 2 sinŸ5=x 5=
25=
4
1
0.8
0.6
0.4
0.2
-3
-2
-1
0
-0.2
1
2
x
3
-0.4
-0.6
-0.8
-1
y fŸx , F 1 Ÿx , F 2 Ÿx , F 3 Ÿx Example Find Fourier series of the periodic function fŸx 0 if " = x 0
2
, with a period of
cosŸx if 0 t x =
2
=.
0.8
0.6
0.4
0.2
-1.5
-1
-0.5
0
0.5
x
1
1.5
y fŸx , " = t x t =
2
2
p =
2
=/2
2
=
=/2
n=x
2
a0 =
; "=/2 fŸx dx 2
an =
; "=/2 fŸx cos
=
2
=/2
;0
2
cosŸx dx =
2
dx =
=/2
;0
2 cos =n
cosŸx cosŸ2nx dx " =
"1 4n 2
n
2Ÿ"1 n1
2 Ÿ"1 "=
"1 4n 2
=Ÿ"1 4n 2 5
2
bn =
=/2
; "=/2 fŸx sinŸ2nx dx 2
=
=/2
;0
2 " sin =n 2n
cosŸx sinŸ2nx dx =
"1 4n 2
4n
=Ÿ"1 4n 2 .
fŸx a 0 ! a n cosŸ2nx b n sinŸ2nx 2
n1
.
2Ÿ"1 n1
4n
cosŸ2nx sinŸ2nx =Ÿ"1 4n 2 =Ÿ"1 4n 2 1 !
=
n1
1 2
=
=
1 2
=
=
.
!
n1
Ÿ"1 n1
2n
cosŸ2nx sinŸ2nx "1 4n 2
"1 4n 2
1 cosŸ2x 2 sinŸ2x " 1 cosŸ4x 4 sinŸ4x 1 cosŸ6x 6 sinŸ6x ". . .
3
3
15
15
35
35
1
0.8
0.6
0.4
0.2
-3
-2
-1
0
1
x 2
3
y fŸx , " = t x t =
2
2
6