Pertemuan
5
Transformasi Fourier
Transformasi Fourier Sinus dan Fourier Cosinus
11 July 2017
Kalkulus Lanjut
1
Dengan menggunakan rumus V
:
V. Transformasi Sinus Fourier :
Fs f(x)  Fs (  ) 
2

f(  ) sin d


0
f(x) 
2

F (  ) sinx d


( inverse )
s
0
1 . Cari Transformasi Sinus Fourier dari f(x) = 1/x
Penyelesaian
Fs (  ) 
11 July 2017

2 sinx
dx 

0 x

2 sinx
dx

 0 x
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
2
 2


2
LihatsoalII -1 )
2
2 Carilah f(x) bila diketahui :
1


0 f(x) cos x dx  
0
0 x  1

F(  ) 
2
 1

f(x) cos x dx


0
f(x) 
2

F(  ) cos x d


0

2
1

0
2

(1 -  ) cos x d
2 1- 
1



(1

)
cos

x
d


sin

cos

x

 0
  x
x2

2
1

11 July 2017
2  cos x
1



  x2
x2
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



1
0




2 (1 - cos x )
x 2
3
3. Cari f(x) bila diketahui :
1



f(x)
sin

x
dx

2
0

 0
0  1
1  2
  2
Peyelesaian : Ruas kiri dan kanan dikalikan dengan

1


2

f(x) sin x dx  2

 0


0

11 July 2017
2

2

2

didapat :
0  1
1  2
  2
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4
Fs(f(x) = Fs(α) =
f ( x) 
2




1


2


0

2
0  1

2
1  2

  2
Fs ( ) sin  d
0
1
2

2
   sin x d  2  sin x d 
0
1

11 July 2017
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5
2 1
41


  - cos x  -  cos x
 x
0 x

1

2
1
2 1
1 4  1
1

cos
x

cos
2x
cos
x

  x
x    x
x

2

 1  cos x - 2 cos 2x 
x
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6

cos x d
 -x

e ,
0  2  1
2
4 Buktikan
Bukti :
x  0
f ( x)  e  x
misalkan
Dari rumus VI diperoleh :
f ( x) 
2


11 July 2017



-u
cos

x
d

e
cos

u
du

e
(
e


 cos u du 
-u
0
2

-x
0


0
cos x d
-x

e
 2 1
0

atau

0
1
 2 1
cos x d
 -x

e
2
2
 1
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7
5 Cari Transformasi sinus Fourier dari
Jawab :
Fs f(x)   Fs (  ) 

2
2


e


- ax
f(x) =
e  ax

 f(x) sin x
dx
0
sin x dx
0

e - ax
 a2  2
2


 0
2 
1

.


  a 2   2



11 July 2017

 - a sin x -  cos x

2 
 
  a 2   2 
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8
6 Cari Transformasi cosinus Fourier dari
f(x) =
Jawab
e
 ax
Fc f(x)  F( ) 
2

f(x)cosx dx


0



11 July 2017
2


- ax
e
 cosx dx
0
e- ax
 a2  2
2

 - a cosx   sinx

2 
a

0

 
a 2   2 
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


 0
2 
a 
  a 2   2 
9
7 . Cari Transformasi cosinus Fourier dari :

x
f ( x)   1- x

0
Fc f(x)  F( ) 
2
0  x  1/2
1/2  x  1
x 1

f(x)cosx dx


0

2

11 July 2017
1/ 2
 x cosx dx 
0
2
1
 (1 - x)cosx dx
 1/ 2
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10

2  x sinx cosx


  
2
 1/ 2
2  (1 - x) sinx cosx

 

 

2
 0
2  sin(  / 2) cos(  / 2) 1

 2

2
 
2





2  - cos sin (/2) cos(  / 2)




2
  
2
2

2  - cos 1 2 cos(  / 2)
 2

2
  

2
11 July 2017
1

 1/2
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


11



11 July 2017
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