1
PERTEMUAN 2
Even and odd functions
Half-range expansions
If a function is defined over half the range, say 0 to L, instead of the full range
from - L to L, it may be expanded in a series of sine terms only or of cosine
terms only. The series produced is then called a half range Fourier series.
Conversely, the Fourier Series of an even or odd function can be analysed using
the half range definition.
Even Function and Half Range Cosine Series
An even function can be expanded using half its range from



0 to L or
-L to 0 or
L to 2L
That is, the range of integration = L. The Fourier series of the half range even
function is given by:
for n = 1, 2, 3, ... , where
bn = 0
2
In the figure below, f(t) = t is sketched from t = 0 to t =
.
An even function means that it must be symmetrical about the f(t) axis and
this is shown in the following figure by the broken line between t = - and t
= 0.
It is then assumed that the "triangular wave form" produced is periodic with
Example
3
We are given that
and f(t) is periodic with period 2π.
a) Sketch the function for 3 cycles.
b) Find the Fourier trigonometric series for f(t), using half-range series.
Solution:
a) Sketch:
b) Since the function is even, we have bn = 0.
In this example, L = π.
We have:
4
To find an, we use a result from before:
We have:
When n is odd, the last line gives us
.
5
When n is even, the last line equals 0.
For the series, we need to generate odd values for n. We need to use (2n - 1)
for n = 1, 2, 3,...
So we have:
Check: The graph for the first 40 terms:
Odd Function and Half Range Sine Series
An odd function can be expanded using half its range from 0 to L, i.e. the
range of integration = L. The Fourier series of the odd function is:
Since ao = 0 and an = 0, we have:
6
for n = 1, 2, 3, ...
In the figure below, f(t) = t is sketched from t = 0 to t = , as before.
An odd function means that it is symmetrical about the origin and this is
shown by the red broken lines between t = - and t = 0.
It is then assumed that the waveform produced is periodic of period 2
outside of this range as shown by the dotted lines.
7
Example. Find the Fourier series of the function
Answer. Since f(x) is odd, then an = 0, for
the coefficients bn. For any
We deduce
Hence
, we have
. We turn our attention to
8
Example. Find the Fourier series of the function
Answer. We have
and
We obtain b2n = 0 and
Therefore, the Fourier series of f(x) is
9
Example. Find the Fourier series of the function function
Answer. Since this function is the function of the example above minus the
constant
. So Therefore, the Fourier series of f(x) is
10
Example. Find the Fourier series of
Answer. Since L = 2, we obtain
for
. Therefore, we have
11
http://aurora.phys.utk.edu/~forrest/papers/fourier/
Example. Perderetkan
8
0  x  2
8
2  x  4
f(x) = 
( Periode 4 ,
L=2)
Penyelesaian :
Y
8
-4
-2
0
X
-8
a0 

4
2
1
1
f(x) dx  (  8. dx 

20
2 0
2
4

1  8x - 8x





2
0
2

 0
4
2 8. dx
)

1
( 16 - 32  16 )
 
2

2
4
12
an 
1
nx
1
nx
f(x) cos
dx  (  8.cos
dx 

20
2
2 0
2
4
2
  2
nx  2
 4 
sin
2  0
  n
2 8. cos
nx
dx )
2



nx  4
 2
sin
 n
2  2

( sin n  0 ).
 0
bn 
-
4
1
nx
1
nx
nx
f(x) sin
dx  (  8.sin
dx -  8. sin
dx

20
2
2 0
2
2
2
4
  - 2
nx  2

 4 
cos



2 0
  n
8

( - cos n  1  1 n
16

( 1 - cos n ) 
n
2
4
2
nx  4
cos
n
2  2



cos n )
32
, ( n ganjil ,
n
Jadi f(x) =
32

cos n   (-1)n )
(2n-1)x
sin
2
(
)

(
2n
1)
n1

πx
32
(
π
Atau dapat ditulis : f(x) =
Example. Perderetkan
periode 2π )
a0 


1

:
2

2
1
0 f(x) dx

1

2

2
3
0 (  - x ). dx
(2 2  2 2 )  0
5 πx
sin
f(x) = π – x ,
1
1 2  2

x
x
 
2  0
1
3 πx
sin
sin

2
5
 ..... )
0 < x < 2π , (
13
an 

1

1

2
0
f(x) dx 
1

2
0 (  - x ).cos
nx

dx
 x
-1
2
cos n
1
sin nx
n
0 (  - x ).cos nx dx
0
-1
cos nx
n2

 2
1   x
1
sin nx cos nx 

  n
 0
n2

bn 


1
n 2
1

1

(1  1)  0
2
0
f(x) dx 
1

2
0 (  - x ).sin
nx

2
dx
  x sin nx
0 (  - x ).sin nx dx
-1
1   x
1
cos nx sin nx 

2
  n

n
cos nx
n
 2

-1
0
0
-1
n
2
sin nx
1
2
( 2 ) 
n
n

Jadi f(x) = π – x =
2

n1
sinnx
n
Atau dapat ditulis sebagai berikut :
f(x) = π - x = 2 (
Example. Perderetkan f(x) = ex
Half range Sinus.
Penyelesaian :
sinx sin2x sin3x sin4x



 ...
1
2
3
4
untuk
0 < x < π ,
dalam
14
L
bn 


2
nπx
f(x)sin
dx

L0
L
2

2

L

0

2 -1
1
1
e sinnxdx  ( e xcosnx  2 e x sin nx - 2
 n
n
n
x
n2
1
1
( - e x cos nx  2 e x sin nx)
2
n 1 n
n


 e sin nx dx )
x
0
e
e
x
x
sin nx
- cos nx
n

e
0
x
- sin nx
n2
2  n


(1 - e  cos n )
2

  n 1

f(x)  e x 
2

(
1  e
1  e
1  e
sin
x

2
sin
2x

3
sin3x  .... )
12  1
22 1
32 1
Example. Perderetkan f(x) = sin x ,
range Cosinus
0 < x < π , ke dalam Half
Penyelesaian :
a0 

2

-2


 sin x dx 
0
( -1 - 1 ) 
-2


cos x 
0
sin x
cos x
4

- sin x
cos nx
sin nx
n
- cos nx
n2
an 
2


 sin x . cos
0
nx
2
dx 



 sin x cos nx dx
0
 sin x sin nx cos x cos nx 1 



sin
x
cos
nx
dx


n
n2
n 2 0


2

n
2  sin x sin nx cos x cos nx 
 2

0
(n  1)  
n
n2

2


15
n 2 2 - cos n - 1
1 - 2 cos n  1
 2
(
)
(
) , ( 0 utk n ganjil )
2
2
(n  1) 
n
(n  1) 
1

-4
((2n) 2  1)
f(x)
atau 
2 4
-
 
2

-
n genap

1
 (2n)
n1
2
1
cos2nx
4 cos2x
cos2x cos6x cos8x
( 2
 2


 ...
 2 1
4 1 6 2 1 8 2 1
Soal – soal
Untuk soal 1 s/d 4 , bila f(x) = x
1. Perderetkan f(x) dalam deret fourier cosinus untuk 0 < x < π
2. Perderetkan f(x) dalam deret fourier sinus untuk 0 < x < π
3. Perderetkan f(x) dalam deret fourier cosinus untuk 0 < x < 1
4. Perderetkan f(x) dalam deret fourier sinus untuk 0 < x < 1
5. Perderetkan f(x) = 2x – 1 ,
0 < x < 1 dalam deret sinus
6 . Perderetkan f(x) = 2x – 1 ,
0 < x < 1 dalam deret cosinus
TERIMA KASIH