Contoh Soal FFT
Download
Report
Transcript Contoh Soal FFT
Contoh Soal FFT
Contoh Penghitungan FT 1 dimensi
(Gonzalez hlm 90-92)
1 N 1
1 N 1
f
(
x
)
exp[
2
j
ux
/
N
]
f ( x)(cos(2ux / N ) j sin(2ux / N ))]
N x 0
N x 0
contoh: f (0) 2, f (1) 3, f (2) 4, f (3) 4
F (u )
1 N 1
f ( x)(cos(2 0 x / N ) j sin(2 0 x / N ))]
N x 0
1
[ f (0) f (1) f (2) f (3)] 3.25
4
1 3
F (1) x 0 f ( x)(cos(2x / 4) j sin(2x / 4))]
4
1
[2(1 0) 3(0 j ) 4(1 0) 4(0 j )
4
1
1
(2 3 j 4 4 j ) (2 j ) 0.5 0.25 j
4
4
1
1
F (2) [1] 0.25
F (3) [2 j ] 0.5 0.25 j
4
4
F ( 0)
2
1 3
F (1) x 0 f ( x)(cos(2x / 4) j sin(2x / 4))]
4
1
[ f (0)(cos(2.180.0 / 4) j sin(2.180.0 / 4))
4
f (1)(cos(2.180.1 / 4) j sin(2.180.1 / 4))
f (2)(cos(2.180.2 / 4) j sin(2.180.2 / 4))
f (3)(cos(2.180.3) / 4 j sin(2.180.3) / 4)]
1
[2(1 0) 3(0 j ) 4(1 0) 4(0 j )
4
1
[2 0 0 3 j 4 0 0 4 j ]
4
1
1
(2 3 j 4 4 j ) (2 j ) 0.5 0.25 j
4
4
3
Contoh Penghitungan FT
• Hasil penghitungan FT biasanya mengandung bilangan real
dan imajiner
• Fourier Spectrum didapatkan dari magnitude kedua bilangan
tersebut shg|F(u)| = [R 2(u) + I 2(u)]1/2
• Untuk contoh di halaman sebelumnya, Fourier Spectrumnya
adalah sebagai berikut:
• |F(0)| = 3.25 |F(1)| = [(-0.5)2+(0.25)2]1/2 = 0.5590
• |F(2)| = 0.25 |F(3)| = [(0.5)2+(0.25)2]1/2 = 0.5590
4
Rumus FT – 2 dimensi
FT : F (u , v)
1
MN
M 1 N 1
f ( x, y) exp[2 j (ux / M vy / N )]
• Rumus FT 2 dimensi
1
MN
x 0 y 0
M 1 N 1
f ( x, y) cos(2 (ux / M vy / N ) j sin(2 (ux / M vy / N ))
x 0 y 0
contoh soal :
m isalcitrasebagaiberikut :
1 2 3
A
, lakukanTransform asiFourier!
2 1 3
u 0,1,2; v 0,1
F (0,0)
1 2 1
f ( x, y) cos(2 (0.x / 3 0. y / 2)) j sin(2 (0 x / 3 0 y / 2))
3 .2 x 0 y 0
1
[ f (0,0).(cos(0) j sin 0) f (0,1)(cos0 j sin 0)
6
f (1,0).(cos(0) j sin 0) f (1,1)(cos0 j sin 0)
f (2,0).(cos(0) j sin 0) f (2,1)(cos0 j sin 0)]
1
[1(1 0) 2(1 0) 2(1 0) 1(1 0) 3(1 0) 3(1 0)]
6
1
[1 2 2 1 3 3] 12 / 6 2
6
1 2 1
F (0,1)
f ( x, y) cos(2 (0.x / 3 1. y / 2)) j sin(2 (0 x / 3 1y / 2))
3 .2 x 0 y 0
1
[ f (0,0).(cos0 j sin 0) f (0,1)(cos180 j sin 180)
6
f (1,0)(cos0 j sin 0) f (1,1)(cos180 j sin 180)
f (2,0)(cos0 j sin 0) f (2,1)(cos180 j sin 180)]
1
[1(1 0) 2(1 0) 2(1 0) 1(1 0) 3(1 0) 3(1 0)]
6
1
[1 2 2 1 3 3] 0
6
F (1,0)
1 2 1
f ( x, y) cos(2 (1.x / 3 0. y / 2)) j sin(2 (1x / 3 0 y / 2))
3 .2 x 0 y 0
F (1,1)
1 2 1
f ( x, y) cos(2 (1.x / 3 1. y / 2)) j sin(2 (1x / 3 1y / 2))
3 .2 x 0 y 0
F (2,0) ?
F (2,1) ?
M 1 N 1
InversFT : f ( x, y ) F (u , v) exp[2 j (ux / M vy / N )]
u 0 v 0
M t inggi cit ra (jumlah baris)
N lebar cit ra (jumlah kolom)
5