TUGAS BAB VI (Kedua) PENGOLAHAN SINYAL DIGITAL KELOMPOK 11

RIZKY SUKMA FATMAWATI IWAN SETIAJI

H1C009020 H1C009021

GILANG KURNIAWAN P P H1C009026

Kementerian Pendidikan dan Kebudayaan Nasional Universitas Jenderal Soedirman Fakultas Sains dan Teknik Jurusan Teknik Program Studi Teknik Elektro Purwokerto 2012

TUGAS PSD 6.10 6.13 6.16

Answer : y(n) = 0.5 x(n) + 0.5 x(n-2) Y(z) = 0.5 X(z) + 0.5z-2 X(z) Y(z) = (0.5 + 0.5z-2) X(z) H(z) =Y z = 0.5 + 0.5z-2 X z

a. H e j

= 0.50.5 j2 = 0.5 + 0.5 cos(2 ) - j0.5 sin(2 )

b. Magnitude Respon Frekuensi : | H e j | =

0.50.5 cos 2 2 0.5 sin 220.5 sin 2 0.50.5 cos 2

Phase respon Frekuensi ( H e j ) = tan-1 (

Tabel perhitungan respon frekuensi (radians)

f= 2 fs(Hz) 0 1000 2000 3000 4000

| H e j |

| H e j |dB

< H e j 00 - 45 0 - 90 0 - 135 0 - 180 0

0 0.25 0.5 0.75

1.000 0.968 0.877 0.731 0.54

0 dB - 0.282 dB - 1.14 dB - 2.721 dB - 5.35 dB

c. Tipe filter yang baik digunakan adalah LPF karena respon phase linier

Answer : a. y(n) Y(z) H(z) H( e j ) = 0.5 x(n) + 0.5x (n-1) = 0.5 X(z) + 0.5 z-1 X(z) =Y z = 0.5 + 0.5 z-1 X z

= 0.5 + 0.5 e j = 0.5 + 0.5 cos ( ) - j 0.5 sin ( )

Respon Magnitude : |H( e j )| = maka

0.50.5 cos 20.5 sin = 0 , |H( e j )| = 0.50.5 cos 020.5 sin0 = 1 = 0 dB = 0.5 , |H( e j )| = 1 , |H( e j )| = =

0.50.5 cos 0.520.5sin 0.5 0.50.5 cos 120.5sin 1

= 0.707 = -3.012 dB

= 0 = - dB Tipe Filternya adalah LPF b. y(n) Y(z) = 0.5 x(n) - 0.5x (n-1) = 0.5 X(z) - 0.5 z-1 X(z)

H(z) H( e j )

=

Y z = 0.5 - 0.5 z-1 X z

= 0.5 - 0.5 e j = 0.5 - 0.5 cos ( ) + j 0.5 sin ( )

Respon Magnitude : |H( e j )| = maka

0.50.5 cos 20.5 sin 2 = 0 , |H( e j )| = 0.50.5 cos 020.5 sin0 2 = = 0.5 , |H( e j )| = 1 , |H( e j )| = = = =

0.50.5 cos 0.520.5sin 0.52 0.50.5 cos 120.5sin 12

Tipe Filternya adalah HPF c. y(n) Y(z) H(z) H( e j ) = 0.5 x(n) + 0.5x (n-2) = 0.5 X(z) + 0.5 z-2 X(z) =Y z = 0.5 + 0.5 z-2 X z

= 0.5 + 0.5 e j2 = 0.5 + 0.5 cos ( 2 ) - j 0.5 sin ( 2 )

Respon Magnitude : |H( e j )| = maka

0.50.5 cos2 20.5sin 2 = 0 , |H( e j )| = 0.50.5 cos 2 . 020.5 sin 2 . 0= 1 = 0 dB = 0.5 , |H( e j )| = = 1 , |H( e j )|

0.50.5 cos2 . 0.520.5 sin 2 .0.5= 0.877 = -1.14 dB =

0.50.5 cos 2 . 120.5sin 2 . 1

= 0.54 = -5.35 dB Tipe Filternya adalah LPF, bandstop filter

Answer : a. Direct Form I a1 y(n) = -0.5 | b0 = 1 | b1 = 2 = x(n) + 2x (n-1) + 0.5 y(n-1)

b.

Direct form II y(n) = w(n) + 2w(n-1)

TUGAS PSD 6.12 6.15 6.18 6.12.a mencari H (z )y (n)= x (n)2.cos( ) x (n1)+ x (n2)+2 . cos ( ) 2 dengan =0.8 dan =60oY (z ) X ( z)

rumus H ( z )=

masukan nilai =0.8 dan =60o kedalam persamaano o 2

y (n)= x (n)2.cos(60 ) x (n1)+ x ( n2)+ 2(0.8) .cos (60 )0.8 y (n)= x (n)2.(0.5) x( n1)+ x (n2)+ 2(0.8) .(0.5)0.8 y (n)= x (n) x(n1)+ x (n2)+(1.6) .(0.5)( 0.64) y (n)= x (n) x(n1)+ x (n2)+(0.8)(0.64) y (n)= x (n) x(n1)+ x (n2)+ 0.16 Y ( z)=X ( z ) X (z ) z1+ X ( z ) z 2 +0.16 Y (z )=( X ( z)( 1z 1 + z 2 ))+0.16 Y ( z )0.16=( X ( z )(1 z1+ z2)) Y ( z )0.16 =1 z1+ z2 X (z) Y (z) =(1z 1 + z 2)0.16 X ( z) H (z )=0.160.16z1 +0.16z22

6.12.b plot zero dan pole mengalikan H (z )z 2H (z )=0.160.16z1 +0.16z2 H (z )=(0.160.16z +0.16z ) z1 2 2

2 H (z )=0.16z 0.16z+ 0.16 menghitung pole dengan rumus

b b24.a.c 2.a

0.16 0.162 4.(0.16).(0.16) 2.(0.16)

6.12.c 6.12.d

6.12.e

6.15.a mencari H (z ) persamaanH ( z )= y (n)=0.5x(n)0.7y (n1)0.1y (n2) Y (z) X ( z)

maka Y ( z)=0.5X(z )0.7Y( z )z 1 0.1Y(z ) z2

Y ( z)+0.7Y ( z )z 1 +0.1Y (z ) z 2=0.5X( z) Y ( z)(1+0.7z1 +0.1z2)=0.5X( z )H (z )= H ( z )= H (z )= Y (z ) 0.5 = X ( z) (1+0.7z1 +0.1z2) 0.5 2 xz 1 2 (1+0.7z +0.1z ) 0.5z 2 (z + 0.7z+ 0.1) z z 12

6.15.b mencari respon unit stepx (n)=u (n) maka Y ( z)=H (z ) X (z ) Y ( z)= Y ( z)= 0.5z z ( ) 2 (z +0.7z +0.1) z 1 0.5z z (z +0.5)( z +0.2)(z 1)2 2

X ( z)=

2 Y (z ) 0.5z A B C = = + + z (z +0.5)(z + 0.2)( z1) (z +0.5) ( z+ 0.2) ( z 1)

dimana

0.5z 2 A=(z + 0.5)( )[ z=0.5] ( z + 0.5)( z +0.2)(z1) 0.5z 2 A=( )[ z=0.5] ( z + 0.2)( z1) 0.5(0.5)2 A=( ) (0.5+0.2)((0.5)1) 0.5(0.25) A= (0.3)(1.5) 0.125 A= 0.45 A=0.278 B=(z + 0.2)( 0.5z 2 )[ z =0.2] ( z +0.5)(z +0.2)(z 1) 0.5z 2 B=( )[ z=0.2] ( z +0.5)(z1) 0.5(0.2)2 B=( ) (0.2+0.5)((0.2)1) 0.5(0.04) B= (0.3)(1.2) 0.02 B= 0.36 B=0.05552

C=( z1)(

0.5z )[ z=1] (z + 0.5)( z+ 0.2)( z1) 0.5z 2 C= [ z =1] (z + 0.2)( z +0.5) 0.5(1)2 C= (1+0.2)(1+0.5) 0.5 C= (1.2)(1.5) 0.5 C= 1.8 C =0.277

makaY (z ) 0.278 0.0555 0.277 = + + z (z +0.5) ( z +0.2) ( z1) Y ( z)= 0.278z 0.0555z 0.277z + + ( z +0.5) (z + 0.2) ( z 1)

mengubah persamaan diatas dengan invers z transform untuk respon unit stepy (n)=0.278(0.5)n u (n)+0.0555(0.2)n u (n)+0.277u (n)

6.18.a perbedaan dari ketiganya ialah dimanaH (z )=10.5z(1 ) maka =0.5 H (z )=10.7z(1 ) maka =0.7 H (z )=10.9z(1 ) maka =0.9

6.18.b komponen emphasis dengan frekuensi tertinggi adalah H ( z )=10.9z(1 ) karena =0.9 dimana nilai

dipilih nilai yang paling mendekati 1. TUGAS PSD 6.11 6.14 6.17

6.11.

a..

y (n)= x (n)0.5y ( n2) Y ( z)=X (z )0.5z2 Y ( z) Y ( z)+ 0.5z2 Y ( z)= X ( z) Y (z ) 1 H (z )= = X ( z ) 1+ 0.5z2 1 j H (e )= 1+ 0.52j 1 H (e j )= 1+ 0.5cos (2 )0.5sin(2 )Magnitude response frekuensi:

b.

H (e

j

)=

Phase response frekuensi:

1 (1+ 0.5cos(2 ))2+ ( 0.5sin(2 ))2 (0.5sin (2 )) ) (1+ 0.5cos(2 ))

( H (e j ))=tan1(c.

6.14. a.

b.

y (n)=0.5x( n)+ 0.2y (n1) 1 Y ( z)=0.5X(z )+ 0.2z Y ( z ) Y (z) 0.5 H ( z )= = X ( z ) 10.2z1 0.5z z Y ( z)= z 0.2 z0.5 Y (z) 0.5z = z ( z0.2)( z0.5) Y (z) A B = + z z0.2 z0.5 0.5z 0.5z A= =0.333 B= =0.833 z 0.5 z =0.2 z 0.2 z =0.5

Y ( z)=

0.333z 0.833z + z0.2 z 0.5 y (n)=0.333(0.2)n u (n)+ 0.833(0.5) n u (n) 10.9z 0.1z 1 2 1+ 0.3z 0.04z1 2

6.17. a.

H (z )=

direct-form I y(n)=x(n)-0.9(n-1)-0.1x(n-2)-0.3y(n-1)+0.04y(n-2)

b.

direct-form II: w(n)=x(n)-0.3(999999n-1)+0.04x(n-2) y(n)=w(n)-0.9w(n-1)-0.1w(n-2)

c. cascade relization: section 1:

w 1=x ( n)0.4w1 (n1) y 1 (n)=w1 (n)w1 (n1)

section 2:

w 2 (n)= y1 (n)+ 0.1w2 (n1) y (n)=w2 (n)+ 0.1w 2 (n1)

y 1 (n)=2.5x( n) w 2 (n)= x (n)0.4w 2 (n1) y 2 (n)=2.1w2 (n)

d. paralel realization

w 3 (n)= x( n)+ 0.1W 3 (n1) y 3 (n)=3.6w 3 ( n) y (n)= y 1 ( n)+ y2 (n)+ y 3 (n)