Tugas Pengolahan Sinyal Digital 2011 2012 06

8/2/2019 Tugas Pengolahan Sinyal Digital 2011 2012 06

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TUGAS PENGOLAHAN SINYAL DIGITAL

KELOMPOK 6

Oleh :

YANUAR EKA PRAYUDI H1C007012 (1,5,9)

BUDI CAHYADI H1C008055 (4,8,12)

SYAMSUL MA'ARIEF H1C009009 (3,7,11)

YULIASIH H1C009010 (2,6,10)

KEMENTERIAN PENDIDIKAN NASIONAL

UNIVERSITAS JENDERAL SOEDIRMAN

FAKULTAS SAINS DAN TEKNIK

TEKNIK ELEKTRO

2012


2/17

SOAL :

1. given a sequence x(n) for 0n 3 where x(0) = 1, x(1) = 1, x(2) = -1 and x(3) = 0, compute its

DFT X(k).

2. Given a sequence x(n) for 0


3/17

9.

Compute the following window functions for size of 8:

a. Hamming window function

b. Hanning window function

10. Given the following data sequence with a length of 6,x(0) = 0, x(1) = 1, x(2) = 0, x(3) = 1, x(4) = 0, x(5) = 1compute the windowed sequence xw(n) using thea. triangular window function.b. Hamming window function.c. Hanning window function.

11.


4/17

12.

Jawab:

1. X(0) =n=0

3

x n = ( x(0) + x(1) + x(2) + x(3) ) = (1+1-1+0) = 1

X(1)= n=0

3

x n =1

4

( + x 2 e j + )

= x(0) - jx(1) x(2) + jx(3) = 1 j + 1 + 0 = 2-j

X(2)= = ( x 0 x 1 e j + x 2 e j2 + x 3 ej3 )

= x(0) - x(1) + x(2) -x(3)= 1 1 1 + 0 = - 1

X(3) = = ( + x 2 e j3 + )

= x(0) + jx(1) - x(2) - jx(3)

= 1 + j + 1 - 0 = 2 + j

2. N = 4

W4 = ej

2

3X(k) = x(n) W4kn

n=03

= x(n) ej

k n2

n=0

untuk :3

k = 0 = x(0) = x(n) e j0 = x(0) e

j0 + x(1) e j0 + x(2) e

j0 + x(3)


5/17

ej0

n=0

= x(0) + x(1) + x(2) + x(3)= 4 + 3 + 2 + 1= 10

3

k = 1 = x(1) = x(n) ej

n2 = x(0) e

j0 + x(1) + ej

2 + x(2) e

j +

x(3) ej

32

n=0= x(0) - jx(1) + x(2) + jx(3)= 4 j3 2 + j= 2- j2

3k = 2 = x(2) = x(n) e

j n = x(0) e j0 + x(1) e

j + x(2) e j2 + x(3)

ej 3

n=0= x(0) - x(1) + x(2) - x(3)= 4 3 + 2 - 1= 2

3

k = 3 = x(3) = x(n) ej

3n2 = x(0) e

j0 + x(1) ej

32 + x(2) e

j 3

n=0

+ x(3) ej

92

= x(0) + j x(1) - x(2) - jx(3)= 4 + j3 - 2 j

= 2 +j2

dengan menggnakan MATLAB function fft(), maka :X = fft([1 1 -1 0])X = 1.0000 2.0000 - 1.0000i -1.0000 2.0000 + 1.0000i

3. N = 4 dan W4-1 = ej 2 , maka:

x n=1

4Xk W4-nk = x n=1

4Xke

jk n

2

untuk n = 0

x 0=1

4X0ej 0=

1

4X0e

j 0X1e

j 0X2e

j 0X3e

j 0


6/17

=1

4102J2 22J2 =4

untuk n = 1

x 1=1

4 X ke

jk2 =

1

4X0e

j 0X1e

j2X2e

jX3e

j32

=1

4X0jX1X2jX3

=1

410j 2j22j 2j2=3

untuk n = 2

x 2=1

4Xkej k=

1

4X0e

j0X1e

jX2e

j 2X3e

j3

=1

4X0X1X2X3

=1

4102j222j2=2

untuk n = 3

x 3=1

4 Xke

jk3

2 =1

4X0e

j 0X1e

j32 X2e

j 3X3e

j33

2

=1

4X0jX1X2jX3

=

1

4 10j 2j22j 2j2=1

4. Since N = 4 and W4 = ej

2

We have simplified formula :

X(k) = 3

n=0

x nW4kn

= 3

n=0

x n ej

2

This , for k=0

X(0) = 3

n=0

x n ej0 = x(0) ej0 + x(1) ej0 + x(2) ej0 + x(3) ej0

= x(0) + x(1) + x(2) + x(3)

= 4 + 3 + 2 + 1

= 10

for k=1


7/17

X(1) = 3

n=0

x n ej

n2

= x(0) ej0

+ x(1) ej

2

+x(2) ej

+x(3) ej

32

= x(0) - jx(1) - x(2) + jx(3)

= 4 - j3 - 2 + j

= 2 - 2jfor k=2

X(2) = 3

n=0

x ne jn

= x(0) e j0

+ x(1) e j

+x(2) e j2

+x(3) e j3

= x(0) - x(1) + x(2) - x(3)

= 4 - 3 + 2 - 1

= 2

and for k=1

X(1) = 3

n=0

x n ej

3n2

= x(0) ej0

+ x(1) ej

32

+x(2) ej3

+x(3) ej

92

= x(0) + jx(1) - x(2) - jx(3)

= 4 + j3 - 2 - j

= 2 + 2j

DFT Using Matlab :

>> X = fft ([4 3 2 1])

X =

10.0000 2.0000 - 2.0000i 2.0000 2.0000 + 2.0000i

DFT Using matlab with zero padding :

>> X = fft ([4 3 2 1 0 0])

X =

10.0000 3.5000-4.3301i 2.5000-0.8660i 2.0000 2.5000+0.8660i 3.5000+4.3301i

5. Dari soal 4.4 didapatkan: x(0) = 10

X(1) = 3,5 4,3301j

X(2) = 2,5 0,8660jX(3) = 2

X(4) = 2,5 + 0,8660j

X(5) = 3,5 + 4,3301j


8/17

Ditanyakan x(0) dan x(4)

X(0) = 1

6

(x(0) + x(1) + x(2) + x(3) + x(4) + x(5)

=1

6 ( 10 + 3,5 - 4,3301j + 2,5 0,8660j + 2 + 2,5 + 0,8660j + 3,5 + 4,3301j)

= 1

6

(10 + 3,5 + 2,5 + 2 + 2,5 + 3,5)

= 1

6

(24)

= 4

X(4) = 1

6n=0

5

x n

= 1

6

( + + x 3 e2 j + + )

= 1

6

( 10 (3,5 - 4,3301j) (2,5 0,8660j) + 2 (2,5 + 0,8660j) (3,5 +4,3301j))

= 1

6

(0)

= 0

6. a. f =fs

N

=20.000

8000

= 2.5 Hz

b. f max =N

2f = 10.000 Hz

7. Jawab: f=0.5Hz

N=fs

f=

2000

0.5=4000

Since we use the FFT to compute the spectrum, the number of the data points must be a power

of 2, that is,

N=215=32768

And the resulting frequency resolution can be recalculated as

f=fs

N=

2000

32768=0.061Hz

8. Diketahui : N = 4

fs = 100 Hz


9/17

W4 = ej

2

x(0) = -1; x(1)= 2; x(2) = 1; x(3) = 4

Ditanyakan : amplitude spectrum, phase spectrum and power spectrum . . .?

Jawab :

using matlab function we can get DFT X(k)

>> X=fft ([-1 2 1 4])

X =

6.0000 -2.0000 + 2.0000i -6.0000 -2.0000 - 2.0000i

we get X(0) = 6;

X(1) = -2+2j = 2.828


10/17

2 = tan-1 ( |ImagX0

RealX0| ) = tan-1 ( |

0

6| ) = 00

P2 =1

42 |X(2)|

2 =1

16|-6|2 = 2.25

For k = 3, f = k . fs /N = 3 x 100/4 = 75 Hz

A3 =1

4|X(3)| =

1

4x 2.828 = 0.707 ;

3 = tan-1 ( |ImagX0

RealX0| ) = tan-1 ( |

2

2| ) = -1350

P3 =1

42 |X(3)|

2 =1

16|2.828|2 = 0.5

9. N = 8

Whm(0)= 0,54 - 0,46 cos (81

2 x

0

) = 0.0800

Whm(1) = 0,54 0,46 cos (81

2 x 1

) = 0,2532

Whm(2) = 0,54 0,46 cos (81

2 x 2

) = 0,6424

Whm(3) = 0,54 0,46 cos (81

2 x 3

) = 0,9544

Whm(4) = 0,54 0,46 cos (81

2 x 4

) = 0,9544

Whm(5) = 0,54 0,46 cos (81

2 x 5

) = 0,6424

Whm(6) = 0,54 0,46 cos (81

2 x 6 ) = 0,2532

Whm(7) = 0,54 0,46 cos (81

2 x 7

) = 0,0800

Jadi fungsi jendela hamming adalah :W = [ 0,0800 0,2532 0,6424 0,9544 0,9544 0,6424 0,2532 0,0800 ]

a. N=8 dicari fungsi jendela hanning

Whn(0) = 0,5 0,5cos (81

2 x 0

) = 0

Whn(1) = 0,5 0,5cos (81

2 x 1

) = 0,1883

Whn(2) = 0,5 0,5cos (81

2 x 2

) = 0,6113

Whn(3) = 0,5 0,5cos (81

2 x 3

) = 0,9505

Whn(4) = 0,5 0,5cos (81

2 x 4

) = 0,9505

Whn(5) = 0,5 0,5cos (81

2 x 5

) = 0,6113


11/17

Whn(6) = 0,5 0,5cos (81

2 x 6

) = 0,1883

Whn(7) = 0,5 0,5cos (81

2 x 7

) = 0

Jadi fungsi jendela hanning adalah :W = [ 0 0,1883 0,6113 0,9505 0,9505 0,6113 0,1883 0 ]

10. 10. N = 6

a. triangular

wtri (n) = 1-2nN1

N1

wtri (0) = 1-2061

61= 0

wtri (1) = 1-

2161

61 = 0.4

wtri (2) = 1- 2261

61= 0.8

wtri (3) = 1-2361

61= 0.8

wtri (4) = 1-24 61

61= 0.4

wtri (5) = 1-2561

61= 0

windowed sequence :Xw (0) = X (0) x Wtri (0) = 0 x 0 = 0

Xw (1) = X (1) x Wtri (1) = 1 x 0.4 = 0.4

Xw (2) = X (2) x Wtri (2) = 0 x 0.8 = 0

Xw (3) = X (3) x Wtri (3) = -1 x 0.8 = -0.8

Xw (4) = X (4) x Wtri (4) = 0 x 0.4 = 0

Xw (5) = X (5) x Wtri (5) = 1 x 0 = 0

b. Hamming

Whm (n) = 0.54 0.46 cos2 n

N1

Whm (0) = 0.54 0.46 cos20

61= 0.08

Whm (1) = 0.54 0.46 cos21

61= 0.4


12/17

Whm (2) = 0.54 0.46 cos22

61= 0.91

Whm (3) = 0.54 0.46 cos23

61= 0.91

Whm (4) = 0.54 0.46 cos24

61 = 0.4

Whm (5) = 0.54 0.46 cos25

61= 0.08

windowed sequence :

Xw (0) = X (0) x Whm(0) = 0 x 0.08 = 0

Xw (1) = X (1) x Whm (1) = 1 x 0.4 = 0.4

Xw (2) = X (2) x Whm (2) = 0 x 0.91 = 0

Xw (3) = X (3) x Whm (3) = -1 x 0.91 = -0.91

Xw (4) = X (4) x Whm (4) = 0 x 0.4 = 0

Xw (5) = X (5) x Whm (5) = 1 x 0.08 = 0.08

c. Hanning

whn (n) = 0.5- 0.5 cos2 n

N1

whn (0) = 0.5- 0.5 cos20

61= 0

whn (1) = 0.5- 0.5 cos21

61= 0.35

whn (2) = 0.5- 0.5 cos22

61

= 0.9

whn (3) = 0.5- 0.5 cos23

61= 0.9

whn (4) = 0.5- 0.5 cos2 4

61= 0.35

whn (5) = 0.5- 0.5 cos25

61= 0

windowed sequence :

Xw (0) = X (0) x Whn(0) = 0 x 0 = 0

Xw (1) = X (1) x Whn (1) = 1 x 0.35 = 0.35Xw (2) = X (2) x Whn (2) = 0 x 0.9 = 0

Xw (3) = X (3) x Whn (3) = -1 x 0.9 = -0.91

Xw (4) = X (4) x Whn (4) = 0 x 0.35 = 0

Xw (5) = X (5) x Whn (5) = 1 x 0 = 0

11. a. Triangular Window

N = 4, maka:


13/17

Wtri(0) = 12x 041

41= 0

Wtri(1) = 12x 141

41= 0.6667

Similarly, wtri(2) = 0:6667, wtri(3) = 0. Next, the windowed sequence is computed as

xw(0) = x(0) x Wtri(0) = 4 x 0 = 0

xw(1) = x(1) x Wtri(1) = -1 x 0.6667 = -0.6667

xw(2) = x(2) x Wtri(2) = 2 x 0.6667 = 1.3334

xw(3) = x(3) x Wtri(3) = 1 x 0 = 0

Applying DFT Equation (4.8) to xw(n) for k = 0, 1, 2, 3, respectively,

X(K) = xw(0) W4kx0 + xw(1) W4

kx1 + xw(2) W4kx2 + xw(3) W4

kx3

We have the following results:

X(0) = 0.6667X(1) = -1.3334 + j0.6667

X(2) = 2

X(3) = 1.3334 j0.6667

f=1

NT=

1

4 .0.01=25Hz

Applying Equations (4.19), (4.22), and (4.23) leads to

A0 =1

4|X(0)| = 0.1667, = tanh

0

0.6667 = 00

P0 =1

42 |X(0)|

2 = 0.02778

A1 =1

4|X(1)| = 0.3727, = tanh

0.6667

1.3334 = 153.440

P1 =1

42 |X(1)|

2 = 0.1389

A2=1

4|X(2)| = 0.5, = tanh

0

2 = 00

P2 =1

42 |X(2)|

2 = 0.25

A3=1

4|X(3)| = 0.3727, = tanh

0.6667

1.3334 = -26.560

P3 =1

42 |X(3)|

2 = 0.1389

b. Hamming Windows


14/17

N = 4, maka:

Whm(0) = 0.54 0.46 cos2x 0

41=0.08

Whm(1) = 0.54 0.46 cos2x 1

41=0.77

Whm(2) = 0.54 0.46 cos2x 241

=0.77

Whm(3) = 0.54 0.46 cos2x 3

41=0.08

the windowed sequence is computed as

xw(0) = x(0) x Whm(0) = 4 x 0.08 = 0.32

xw(1) = x(1) x Whm(1) = -1 x 0.77 = -0.77

xw(2) = x(2) x Whm(2) = 2 x 0.77 = 1.54

xw(3) = x(3) x Whm(3) = 1 x 0.08 = 0.08


X(K) = xw(0) W4kx0 + xw(1) W4

kx1 + xw(2) W4kx2 + xw(3) W4

kx3


X(0) = 1.17

X(1) = -1.22 + j0.85

X(2) = 2.55

X(3) = -1.22 j0.85

f=1

NT=

1

4 .0.01=25Hz

Applying Equations (4.19), (4.22), and (4.23) we achieve

A0 =1

4|X(0)| = 0.2925, = tanh

0

1.17 = 00

P0 =1

42 |X(0)|

2 = 0.0856

A1 =

1

4 |X(1)| = 3717, = tanh

0.85

1.22 = 145.140

P1 =1

42 |X(1)|

2 = 0.1382

A2=1

4|X(2)| = 0.6375, = tanh

0

2 = 00

P2 =1

42 |X(2)|

2 = 0.4064


15/17

A3=1

4|X(3)| = 0.3717, = tanh

0.85

1.22 = 214.860

P3 =1

42 |X(3)|

2 = 0.1382

c. Hanning Windows

N = 4, maka:

Whn(0) = 0.5 0.5 cos2x 0

41=0

Whn(1) = 0.5 0.5 cos2x 1

41=0.75

Whn(2) = 0.5 0.5 cos2x 2

41=0.75

Whn(3) = 0.5 0.5 cos 2x 341

=0

the windowed sequence is computed as

xw(0) = x(0) x Whn(0) = 4 x 0 = 0

xw(1) = x(1) x Whn(1) = -1 x 0.75 = -0.75

xw(2) = x(2) x Whn(2) = 2 x 0.75 = 1.5

xw(3) = x(3) x Whn(3) = 1 x 0 = 0


X(K) = xw(0) W4kx0 + xw(1) W4kx1 + xw(2) W4kx2 + xw(3) W4kx3


X(0) = 0.75

X(1) = -1.5 + j0.75

X(2) = 2.25

X(3) = -1.5 j0.75

f=1

NT=

1

4 . 0.01=25Hz

Applying Equations (4.19), (4.22), and (4.23) we achieve

A0 =1

4|X(0)| = 0.1875, = tanh

0

0.75 = 00

P0 =1

42 |X(0)|

2 = 0.352


16/17

A1 =1

4|X(1)| = 0.4193, = tanh

0.75

1.5 = 153.440

P1 =1

42 |X(1)|

2 = 0.1758

A2=1

4

|X(2)| = 0.5625, = tanh 0

2.25

= 00

P2 =1

42 |X(2)|

2 = 0.3164

A3=1

4|X(3)| = 0.4193, = tanh

0.75

1.5 = 206.560

P3 =1

42 |X(3)|2 = 0.1758

12. a.

Pada grafik di atas menunjukkan adanya kebocoran spectral karena

amplitudonya tidak kontinyu pada domain waktu

b.


17/17

Comparison of a spectrum without using a window function and a spectrum

using a hamming window of size of 73 sample

Pada grafik di atas menunjukkan adanya kebocoran spectral karena amplitudonya tidak kontinyu

pada domain waktu

Tugas Pengolahan Sinyal Digital 2011 2012 06

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