Bab 7 Filter LTI

5/26/2018 Bab 7 Filter LTI

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Sistem LTI sebagai filter

Karakteristik filter ideal Low Pass Filter (LPF)

High Pass Filter (HPF)

Band Pass Filter (BPF)

Transformasi LPF HPF

Digital resonator

Notch filter

Comb filter

All-pass filter

Digital sinusoidal filter


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Karakteristik filter ideal

)(X)(H)(Y

)(H Weighting function

Spectral shaping function

Filter

Sistem LTI)(X)(H )(Y

)n(x)n(h

)n(y

)n(x)n(h)n(y


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Constant gain pada passband

Filter Ideal :

Zero gain pada stopband

Respon fasa linier

lainnya0

Ce)(H 21

nj o

onje)(CX

)(X)(H)(Y

)nn(Cx)n(y o

on)(


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Prinsip dasar penempatan pole-zero pada bidang z :

Penguatan frekuensi pole di dekat lingkaran satu

Filter stabil semua pole harus di dalam lingkaran satu

Koefisien filter nyata pole kompleks harus konjugate

Pengurangan frekuensi zero di dekat lingkaran satu

N

1k

1

k

M

1k

1

k

ok

N

1k

k

kM

0k

k

)zp1(

)zz1(

bza1

zb

)z(H

1)(Hboo


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Low Pass Filter (LPF)

9,0aaz1

b)z(H

1

o1

1,0a1b1)(H0 ooo

ae

eb)(H

az

zb)z(H

j

j

o1o1


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9,0e

e1,0

ae

eb)(H

j

j

j

j

o1

0MN9,0p0z

cos8,181,1

1,0

sinj)9,0(cos

e1,0

)(H

j

1

)cos8,181,1log(1020

cos8,181,1log201,0log20)(HdB1

9,0cos

sintg)(H 11


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e1,0)(H

j

j

1


8/53

05,02

a1b9,0a

az1

z1b)z(H o1

1

o2

ae

1eb)(H

az

1zb)z(H

j

j

o2o2

0MN9,0p1z

cos8,181,1

cos2205,0

sinj)9,0(cos

sinj)1(cos05,0)(H2

9,0cos

sintg

1cos

sintg)(H 112


9/539,0e

1e05,0)(H

j

j

2


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)(H1

)(H2


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Contoh Soal 8.1

Tentukan respon frekuensi dari LPF dua pole :

Jawab :

21

o

)az1(

b)z(H

2j

2j

o

)ae(

eb

)(H

2

2

o

)az(

zb

)z(H

dengan : 2

1

4/H1)0(H

2

2

o2

o )a1(b1)a1(

b)0(H


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a4/sinj4/cos

)a1()(H

)ae(

eb)(H

2

2j

2j

o

2

1a2a

2

1

)a1(

25,0j)a25,0(

)a1()(H

2

2

2

2

2

22

a2a1

)a1(

2

1)(H

2

1)(H

46,0b32,0aa2a12)a1( o22

2j

2j

)32,0e(

e46,0)(H


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cos32,0cos

2sintg22)(H 12

22 )cos64,01024,1(

46,0

sinj)32,0(cos

46,0)(H


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High Pass Filter (HPF)

9,0aaz1

z1

2

a1)z(H

1

1

3

9,0e

1e05,0)(H

j

j

3


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9,0p1z 9,0e

1e05,0)(H

j

j

3

cos8,181,1

cos2205,0

sinj)9,0(cos

sinj)1(cos05,0)(H3

9,0cos

sintg

1cos

sintg)(H 113


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9,0e

1e05,0)(H

j

j

3


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Band Pass Filter (BPF)

2

1

9

4H

Harus ada satu atau lebih pole konjugate

Terletak di dekat lingkaran satu

Didekat frekuensi tengah pass band

Contoh Soal 8.2Rancang BPF dua pole dengan frekuensi tengah /2,

berharga nol pada frekuensi 0 dan 2, serta :


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Jawab :

)jrz)(jrz(

)1z)(1z(G)z(H

1z,1z,rep 212

j

12

)rz(

)1z(G)z(H22

2

1r1

G2

2H

r1

2G

2H

1e2)re(

)1e(G)(H

22

2j22j

2j


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98sinj

98cose

94

)re(

)1e(

2

r1

)re(

)1e(G)(H

2j

22j

2j2

22j

2j

2

r1G1

r1

G2 2

2

9

8sinj)r

9

8(cos

9

8

sinj)19

8

(cos

2

r1

9

4H

2

2

88


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9

8

sinj)r9

8

(cos

9

8sinj)1

9

8(cos

2

r1

9

4H

2

2

2

1

)9

8cosr2r1(

9

8cos22

4

)r1(

9

4

H 24

222

7,0rr88,1r1)r1(94,1

22422

)7,0z(

)1z(15,0

)z7,01(

)z1(15,0)z(H

2

2

2

2


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)7,0e(

)1e(15,0)(H

2j

2j


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Transformasi LPF HPF>

)(H)(H lphp

)n(h)1()n(h)n(h)1()n(h)e()n(h

hp

n

lp

lp

n

lp

nj

hp

2

Frekuensi rendah

20

Frekuensi tinggi


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M

0k

k

N

1k

k )kn(xb)kn(ya)n(y

N

1k

k)(j

k

M

0k

k)(j

k

lp

ea1

eb

)(H

N

1k

kj

k

M

0k

kj

k

lp

ea1

eb

)(H


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N

1k

kj

k

k

M

0k

kj

k

k

hp

ea)1(1

eb)1(

)(H

M

0k

k

kN

1k

k

k)kn(xb)1()kn(ya)1()n(y

Koefisien pada suku ganjil diganti tandanya


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Contoh Soal 8.3

Ubah LPF yang dinyatakan dengan persamaan beda :

)n(x1,0)1n(y9,0)n(y menjadi HPF

Jawab :

)n(x1,0)1n(y9,0)n(y

jhp

e9,011,0)(H

jlp e9,01

1,0)(H


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)(Hhp )(Hlp

Di it l R t


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Digital Resonator

Bentuk khusus dari BPF dua pole

Sepasang pole konjugate di dekat lingkaran satu

Magnituda besar disekitar o(frekuensi resonansi)

Dapat ditambah satu atau dua zero

zero di titik asal (z = 0)

zero di z = - 1 dan z = 1

1r0rep oj

2,1


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2

o

2

2

o

221

o

o

1j1j

o

1

2

1

1

o

rzcosr2z

zb

zrzcosr21b

)zre1)(zre1(

b

)zp1)(zp1(

b)z(H

oo


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)pz)(pz(

zb)z(H

21

2

o

)()(2)(H

)(U)(Ub)(H

21

21

o

)pe)(pe(

eb)(H

2

j

1

j

2j

o

o

2

oo 2cosr2r1)r1(b1)(H


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8,0r

3/o

95,0r

)pz)(pz(

zb)z(H

21

2

o


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)pz)(pz(

)1z(b

)pz)(pz(

)1z)(1z(b)zp1)(zp1(

)z1)(z1(b)z(H

21

2

o

21

o

1

2

1

1

11

o

)()()()()(H

)(U)(U

)(V)(Vb)(H

2121

21

21o


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)pz)(pz(

)1z)(1z(b)z(H

21

o

8,0r 95,0r

3/o

Notch Filter


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Notch Filter

Kebalikan dari digital resonator

Sepasang zero konjugate di lingkaran satu pada o

Magnituda kecil sekali (nol) disekitar o

Dapat ditambah dua pole konjugate

1r0rep oj2,1

oj

2,1 ez


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)zzcos21(b)ze1)(ze1(b

)zz1)(zz1(b)z(H

21

oo

1j1j

o

1

2

1

1o1

oo

)zrzcosr21(

)zzcos21(b

)zre1)(zre1(

)ze1)(ze1(b

)zp1)(zp1(

)zz1)(zz1(b)z(H

221o

21

o

o

1j1j

1j1j

o

1

2

1

1

12

11

o2

oo

oo


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)zz1)(zz1(b)z(H 121

1o1

)z(H1 4/o

)zp1)(zp1(

)zz1)(zz1(

b)z(H 121

1

1

2

1

1

o2

)z(H2

Comb Filter


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Comb Filter

Notch filter yang lebih umum

Magnituda nol pada berbagai frekuensi secara periodik

M

0k

)kn(x1M

1)n(y

Moving Average (FIR) Filter :

1

)1M(M

0k

k

z1

z1

1M

1z

1M

1)z(H

)1M(11


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1

)(

z1

z1

1M

1)z(H

j

)1M(j

e1

e1

1M

1)(H

)ee(e)ee(e

1M1)(H

2/j2/j2/j

2/)1M(j2/)1M(j2/)1M(j

2sin

21Msin

1M

e)(H

2/Mj


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2sin

2

1Msin

1M

e

)(H

2/Mj

k2

1M0)(H

M,,3,2,1k,1M

k2


39/53

1

)1M(

z1

z1

1M

1)z(H

M = 10


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Comb filter yang lebih umum :

M

0k

k

z)k(h)z(H

M

0k

kL

L

L z)k(h)z(Hzz

)L(He)k(h)(HM

0k

jkL

L


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5L


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2

Lsin

2

1MLsin

1M

e

)(H

2/LMj

L

1

)1M(

z1

z1

1M

1)z(H

L

)1M(

L

z1

Lz1

1M

1)z(H


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L

)1M(

L z1

Lz1

1M

1

)z(H

M = 10

L = 3


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All Pass Filter

k

z)z(H1)(H

1aza

za

)z(H oN

0k

k

k

N

0k

kN

k

N

N

1

1

N1N

1

1

1NN

zaza1

zzazaa)z(H


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N

0k

k

k

N

0k

kN

k

za

za

)z(H

N

0k

k

kza)z(A

)z(A)z(Az)z(H

1N

1)z(A

)z(Az

)z(A

)z(Az

)z(H)z(H)(H 1

N1N12

All Pass Filter

o

o

z

1pzz

ap


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p

z

+

+

2z

1z

1p

2p

oo

o

j

1

j

1

er

1z

rep

o

o

j

2

j2

er

1z

rep

a

1z

ap

p

1z

Single pole single zero filter


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1

1

1

1

1z6,01

)z6,0

11(6,0

z6,01z6,0)z(H

Single polesingle zero filter

Two pole

two zero filter

49,0r

)zrcosr21(

)zcosr2r()z(H

o

22

o

2

o

2

2


48/53

1

1

1z6,01

z6,0)z(H

)zrcosr21(

)zcosr2r()z(H

22

o

2

o

2

2

)(H)(H 21

)(H1

)(H2

Komputasi dari fungsi respon frekuensi


49/53

p g p

N

1k

k

j

M

1k

k

j

)MN(joN

1k

kj

k

M

1k

kj

k

o

)pe(

)ze(

eb

)ep1(

)ez1(

b)(H

)(jkk

j)(jkk

j kk eUpeeVze

)(U)(U)(U

)(V)(V)(Vb)(H

N21

M21o

)]()([

)()()MN(b)(H

N1

M1o


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Re (z)

Im (z)

j

e kz

k

j ze

0

2/

2/

kp

k

j pe

Interpretasi Geometrik


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)(j

kk

j ke)(Upe

Re (z)

Im (z)

kz

)(j

kk

j ke)(Vze

kp

kkVk

U

k


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Contoh Soal 8.4

Tentukan respon frekuensi dari

Jawab :

1z8,01

1)z(H

1b1N1M8,0p0z o

8,0e

e)(Hez

j

jj

8,0z

z

)z8,01(z

z

)z8,01(

1)z(H

11

j


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8,0cos

sintg)(H 1

22o

sin64,0cos6,1cos

1

)(V

)(Ub)(H

sinj)8,0(cos

sinjcos

8,0e

e)(H

j

j

cos6,164,1

1)(H

)()()MN(b)(H o

Documents

Bab 7 Filter LTI