Chuong 5 Xu Ly Tin Hieu So

7/30/2019 Chuong 5 Xu Ly Tin Hieu So

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192

Chng V

B LC S C C TNH XUNG HU HN, PHATUYN TNH

5.1 Cc b lc s l tng

B lc s l tng c c tnh bin tn s dng ch nht:

1 i thng( )

0 i chn

jKhi d

H eKhi d

[5.1-1]

Trn thc t khng th xy dng b lc s c c tnh bin tns ( )jH e nh vy,

tuy nhin cc b lc s l tng l c s phn tch v tng hp cc b lc s thc t.

5.1.1 B lc thng thp l tng

a nh ngha:B lc thng thp l tng c c tnh bin tn s khi , nh sau:

1 [ , ]( )

0 [ , ] [ , ]

c cj

hp

c c

KhiH e

Khi v

[5.1-2]

c tnh bin tn s ca b lc thng thp l tng hnh 5.1

b. Cc tham s thc ca b lc thng thp l tng

- Tn s ct : cf

- Di thng : f [0, cf ]

- Di chn : f [ cf ,]

( )jlpH e

1

0c c

Hnh 5.1: c tnh bin tn s ca b lc thng thp l tng


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B lc thng thp l tng cho tn hiu s c ph nm trong di tn cf f i qua,

chn khng chi tn hiu s trong di tn cf f i qua.

c. c tnh xung ( )lph n ca b lc thng thp l tng

Xt b lc thng thp l tng pha tuyn tnh ( ) , c tnh tn s ca

n c dng:

[ , ]( )

0 [ , ] [ , ]

j

j c c

hp

c c

e KhiH e

Khi va[5.1-3]

c tnh xung ( )lph n ca b lc trn c xc nh bng IFT:

1( ) ( ) 1 ( ) .

2

j j j n

hp hp lph n IFT H e H e e d

( )1 1 1( ) .2 2 ( )

c

c

c

c

j j n j n

hph n e e d e

j n

sin ( ) sin ( )( )

( ) ( )

c cc

hp

c

n nh n

n n[5.1-4]

Theo [5.1-4], b lc thng thp l tng pha tuyn tnh c c tnh xung ( )lph n dng

hm sin gim dn v 0 khi n . Ti n =0 c:

0 0

sin ( )(0) lim ( ) lim .

( )

cc clp lp

n nc

nh h n

n

c tnh xung ( )lph n t cc i ti n = 0, v ( ) 0lph n ti cc im / cn k , vi k l s

nguyn.

V d 5.1: Hy xc nh v v th c tnh xung ( )lph n ca b lc thng thp l tng pha

khng ( ) 0 , c tn s ct / 3c .

Gii: c tnh xung ca b lc thng thp pha khng l tng:

sin( . / 3)( )

.lp

nh n

n

Theo cng thc trn lp c bng 5.1


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Bng 5.1

n 0 1 2 3 4 5 6 7 8

( )lph n 0,33 0,28 0,14 0 -0,07 -0,05 0 0,04 0,03

Theo cc s liu trn, xy dng c th c tnh xung ( )lph n ca b lc thng thp l

tng pha khng vi / 3c trn hnh 5.2

Nhn xt:c tnh xung ( )lph n ca b lc thng thp l tng l dy chn, i xng

qua trc tung, c di v hn v khng nhn qu, nn khng thc hin c trn thc t.

5.1.2 B lc thng cao l tng

a. nh ngha:B lc thng cao l tng c c tnh bin tn s khi , nh sau:

1 [ , ] [ , ]( )

0 [ , ]

c cj

hp

c c

Khi v H e

Khi

[5.1-5]

th c tnh bin tn s ca b lc thng cao l tng hnh 5.3

( )lph n

0.280.28

0.33

0.140.14

0.040.04

-

-

0 1 2 3 6 9

n

-

-

-

-

-

Hnh 5.2:( )lph n ca b lc thng thp pha khng vi / 3c


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b. Cc tham s thc ca b lc thng cao l tng

- Tn s ct : cf

- Di thng f [ cf ,]

- Di chn : f [0, cf ]

B lc thng cao l tng cho tn hiu s c ph nm trong di tn f > cf i qua, chn

khng cho tn hiu trong di tn f < cf i qua.

c. c tnh xung (hph n ca b lc thng cao l tng

Xt b lc thng cao l tng tuyn tnh () = - c tnh tn s ca n c dng :

[ , ] [ , ]( )

0 [ , ]

j

j c c

hp

c c

e khi vaH e

khi

[5.1-6]

V di thng v dI chn ca b lc thng cao ngc vi b lc thng thp ,nn c th biu

din ( )jhpH e qua ( )jlpH e

nh sau:

( ) 1 ( )j jhp lpH e H e [5.1-7]

Theo [5.1-7] c th tm c c tnh tn s ca b lc thng cao t c tnh ca b lc

thng thp c cung tn s ct .

c tnh xung ( )hph n ca b lc trn c xc nh bng IFT :

1( ) ( ) 1 ( ) .

2

j j j n

hp hp lph n IFT H e H e e d

( )jlpH e

1

0c c

Hnh 5.3: c tnh bin tn s ca b lc thng cao l tng


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1 1( ) .

2 2

j n j n j n

hph n e d e e d

( )1 1 1 1( )2 2 ( )

c

c

j n j n

hph n e ejn j n

sin ( )sin( . )( )

( )

c

hp

nnh n

n n

Hay: sin ( )sin( . )

( )( )

cchp

c

nnh n

n n

[5.1-8]

Vi :1 0sin( ) sin( )

( )0

khi nn nn

o khi nn n

Nn c th vit li [5.1-8] di dng :

sin ( )sin( )( ) ( ) ( ) .

( )

cc chp

c

nnh n n n

n n

[5.1-89]

So snh [5.1-9] vi [5.1-4] c th biu din c tnh xung ca b lc thng cao

qua c tnh xung ca b lc thng thp :

( ) ( ) ( )hp lph n n h n [5.1-10]

Theo [5.1-10] c th tm c c tnh xung ( )hph n ca b lc thng cao t c tnh

xung ( )lph n ca b lc thng thp c cng tn s ct c

c tnh xung ( )hph n ca b lc thng cao l tng l dy chn ,i xng qua trc

tung v t cc i ti n=0 . Khi tn s ct c N th c tnh xung ( ) 0hph kN ti cc

im n=kN, vi k l s nguyn .

V d 5.2:Hy xc nh v v c tnh xung ( )hph n ca b lc s thng cao l tng pha

khng c tn s ct 3c

Gii:C c tnh xung ca b lc thng cao pha khng l tng :

sin( 3)( ) ( ) ( ) ( )hp lp

nh n n h n n

n

Theo cng thc trn v kt qu ca v d 5.1 lp c bng 5.2 :

Bng 5.2


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n 0 1 2 3 4 5 6 7 8

( )hph n 0,33 0,28 0,14 0 -0,07 -0,05 0 0,04 0,03

( )lph n 0,77 -0,28 -0,13 0 0,07 0,05 0 -0,04 -0,03

Theo cc s hiu trn , xy dng c th c tnh xung cu b lc thng cao l

tng pha khng vi 3c trn hnh 5.4

Nhn xt:Theo [5.1-9] , b lc thng cao l tng l h x l s HR khng nhn qu ,

v th khng th thc hin c trn thc t

5.1.3 B lc di thng l tng .

a. nh ngha :B lc di thng l tng c c tnh bin tn s khi , nh sau:

1 2 1 21 [ , ] [ , ]( )

0

c c c cj

hp

khi vaH e

khi nam ngoai cac khoang tren

[5.1-11]

th c tnh bin tn s ca b lc dI thng l tng hnh 5.5.

b. Cc tham s ca b lc dI thng l tng

- Tn s ct : 1 2,c cf f

- Di thng : f [ 1 2,c cf f ]

)( jbp eH

1

0

2c 2c 1c 1c

Hnh 5.5: c tnh bin tn s ca b lc di thng l tng .


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- Di chn : f [0, 1cf ] v [ 2cf ,]

B lc di thng l tng cho tn hiu s c ph nm trong di tn s 1 2c cf f f i

qua , chn khng cho tn hiu ngoi di tn i qua.

c. c tnh xung ( )bph n ca b lc di thng

Xt b lc di thng l tng c pha tuyn tnh ( ) , c tnh tn s ca n

c dng :

1 2 1 2, ,( )0 m ngoi c c khong trn

jj c c c c

bp

e Khi v H e

Khi n [5.1-12]

C th biu din ( )jbpH e qua c tnh tn s

1( )j

lpH e v

2( )j

lpH e ca cc b lc

thng thp l tng c cc tn s ct 1c

v 2c tng ng:

2 1( ) ( ) ( )j j j

bp lp lp H e H e H e [5.1-13]

Theo [5.1-13] c th tm c c tnh tn s ca b lc di thgn c tn s ct 1c

v 2c , tc tnh tn s ca hai b lc thng thp c tn s ct 1c v 2c tng ng.

c tnh xung ( )bph n ca b lc trn c xc nh bng IFT:

2 1

1 1

2 2

2 1

1( ) ( ) ( ) ( ) .

2

1 1( ) . . . .

2 2

sin ( ) sin ( )( )

.( ) .( )

i j j j

bp bp lp lp

j j n j j n

bp

c c

bp

h n IFT H e H e H e e

h n e e d e e d

n nh n

n n

[5.1-14]

2 12 1

sin ( ) sin ( )( )

.( ) .( )

c cc cbp

n nh n

n n[5.1-15]

Hay: 2 1( ) ( ) ( )bp lp lp h n h n h n [5.1-16]

Theo [5.1-16] c th tm c c tnh xung ( )bph n ca b lc di thng theo c tnh

xung 1( )lph n v 2( )lph n ca cc b lc thng thp c tn s ct 1c v 2c tng ng:

V d 5.3: Hy xc nh v v c tnh xung ( )bph n ca b lc s di thng l tng pha

khng c cc tn s ct 1 / 3c v 2 / 2c .


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Gii: C c tnh xung ca b lc di thng pha khng l tng:

2 1

si n( / 2) si n( / 3)( ) ( ) ( )

. .bp lp lp

n nh n h n h n

n n

Theo cng thc trn v kt qu ca cc v d 5.1 lp c bng 5.3:

Bng 5.3

n 0 1 2 3 4 5 6 7 8

2( )lph n 0.50 0.32 0 -0.11 0 0.06 0 -0.04 0

1( )lph n 0.33 0.28 0.14 0 -0.07 -0.05 0 0.04 0.03

( )lph n 0.17 0.04 -0.14 -0.11 0.07 0.01 0 -0.08 -0.03

Theo cc s liu trn, xy dng c th c tnh xung ca b lc di thng l

tng vi

Nhn xt:B lc di thng l tng l h x l s IIR khng nhn qu, v th n

khng th thc hin c trn thc t.

0.1

0.0

0.0

0.0

0.0

-

-

-

-

-

-

-

-

n

hbp(

Hnh 5.6: c tnh xung ca b lc di thng l tng


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5.1.4 B lc di chn l tng

a nh ngha B lc di chn l tng c c tnh bin tn s khi , nh sau:

1 2 1 20 , ,( )1 ng thuc c c khong trn

j c c c c bs

Khi v H e

Khi kh[5.1-17]

th c tnh bin tn s ca b lc di chn l tng nh hnh 5.7

b. Cc tham s thc ca b lc di chn l tng

- Tn s ct: 1 2,c cf f

- Di thng: 1 c20, ,cf f v f

- Di chn : f[fc1 , fc2 ]

B lc di chn l tng khng cho tn hiu s c ph nm trong di tn fc1< f < fc2iqua , cho tn hiu s ngoi di tn s i qua .

c. c tnh xung hbs(n) ca b lc di chn l tng

Xt b lc di chn l tng pha tuyn tnh )( , c tnh tn s ca n c

dng :

trenkhoangthuockhongkhie

khieH

j

ccccj

bs

],[va],[0

)(2121

181.5

C th biu din )( jbs eH qua c tnh tn s )(1j

lp eH v )(2j

lp eH ca b lc thng

thp l tng c tn s ct 1c v 2c nh sau :

)( j

bs eH =1- )(2j

lp eH + )(1j

lp eH 191.5

Hnh 5.7 : c tnh bin tn s ca b lc dichn l tng

)( jbp eH

1

0

2c 2c 1c 1c


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Theo 191.5 c th tm c c tnh tn s ca b lc di chn c cc tn s ct

1c v 2c t c tnh tn s ca 2 b lc thng thp c tn s ct 1c v 2c tng ng

Biu din )( jbs eH qua c tnh tn s )(j

bp eH ca b lc di thng

)(

j

bseH

=1- )(

j

bp eH 201.5

Theo 201.5 c th tm c c tnh tn s ca b lc di chn c cc tn s ct

1c v 2c , t c tnh tn s ca b lc di thng c cc tn s ct tng ng

c tnh xunghbs(n) ca b lc trn c xc nh bng IFT :

deeHeeHIFTnh jjlp

jj

bsbs )()(H-12

1)()( 1lp2

1

1

2

221

21

21)(

c

c

c

c

deedeedenh njjnjjnjbs

1

1

2

2

)()(

)(

1

2

1

)(

1

2

11

2

1)(

c

c

c

c

Ienj

Ienj

Iejn

nh njnjnjbs

)(

)(sin

)(

)(sin)sin( 12)(

n

n

n

n

n

nh ccnbs 211.5

)(

)(sin)(sin)sin()(1

11

2

22

nnn

nnnh

c

cc

c

cc

bs 221.5

Hay : )()()()( 12 nhnhnnh lplpbs 231.5

Hoc : )()()( nhnnh bpbs 241.5

Theo 231.5 c th tm c c tnh xung hbs(n) ca b lc di chn khi

bit c tnh xung )(1j

lp eh v )(2j

lp eh ca b lc thng thp tng ng . Theo 241.5 c

th tm c c tnh xung hbs(n) ca b lc di chn khi bit c tnh xung hbp(n) ca b lcdi thng tng ng .

V d 5.4 : Hy xc nh v v c tnh xung hbs(n) ca b lc s di chn l tng pha

khng c cc tn s ct 1c = 3 v 2c = 2

Gii: C c tnh xung ca b lc di chn pha khng l tng :


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)()()( nhnnh bpbs =

n

n

n

n

n

n 3/sin()2/sin()sin(

Theo cng thc trn v kt qu ca v d 5.3 lp c bng 5.4

Bng 5.4

n 0 1 2 3 4 5 6 7 8

)(nhbp 0,17 -0,04 -0,14 -0,11 0,07 0,01 0 -0,08 -0,03

)(nhbs 0,83 -0,04 0,14 0,11 -0,07 -0,01 0 0,08 0,03

Theo cc s liu trn, xy dng c th c tnh xung ca b lc di chn l

tng vi 1c = 3 v 2c = 2

trn hnh 5.8

Theo biu thc 221.5 v kt qu ca v d 5.4, c nhn xt : B lc di chn l tng l

h x l s IIR khng nhn qu , v th n khng th thc hin trn thc t.

5.1.5 Tham s ca b lc thc t

Tt c b lc s l tng c c tnh bin tn s dng ch nht, nn c tnh

xung ca chng u l dy khng nhn qu c di v hn , v th khng th thc hin

c cc b lc l tng .

hbp

0,83

n

0,14

0,11

0 08

0,14

0,11

0 08

Hnh 5.8 : c tnh xung ca b lc di chn l tng


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c tnh bin tn s ca b lc s thc t thng c nhp nh trong di thng

v di chn , vi 2 bin l sn dc nh hnh 5.9 .

c trng cho b lc thc t , ngi ta s dng cc tham s sau :

1. Loi b lc : Thng thp, thng cao, di thng, di chn

2. Tn s gii hn gii thng c (hay fc).

3. Tn s gii hn di chn p (hay fp).

4. rng di qu cpp (hay pf )

5. nhp nh trong di thng1

. Trong di thng, c tnh bin tn s

)(

j

bp eH phi tha mn iu kin

11 )(j

bp eH 11 251.5

6. nhp nh trong di chn 2 . Trong di chn c tnh bin tn s )(j

bp eH

phi tho mn iu kin : )( jbp eH 2 261.5

B lc thc t c p , 1 v 2 cng nh th dc tuyn bin tn s cng gn ging dng

chnht , nn chn lc tn hiu cng tt.

5.2 Phn tch b lc s fir pha tuyn tnh

5.2.1 c tnh xungh(n)ca b lc s FIR pha tuyn tnh

Cc b lc s FIR c c tnh xung h(n)hu hn, nn hm h thng l :

1

0

)(N

n

nznhzH

p

c

)( jbp eH p

Hnh 5.9 : c tnh bin tn s ca b lc thng thp thc t


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V c tnh xung h(n) hu hn nn b lc FIR lun n nh , c ngha l tt c cc

im cc ca hm h thngH(z) nm trong ng trn n v 1z . c tnh tn s ca

b lc s FIR

)(1

0

)()()( jj

N

n

njeeAenhzH

Trong chng ny ch nghin cu cc b lc s FIR c pha tuyn tnh :

Trong v l cc hng s , v l thi gian truyn lan ca tn hiu qua b lc :

d

d 22.5

Theo 22.5 tt c cc thnh phn tn s ca tn hiu i qua b lc s FIR pha tuyn

tnh u b gi tr nh nhau , v th tn hiu khng b mo dng ph .

V )( jeH tun hon vi chu k 2 nn ch nghin cu c tnh bin tn s

)( jeH v pha )( khi )( hoc )20(

Mt khc nu b lc s c c tnh xung h(n)l dy thc th theo tnh cht ca bin

Fourierc : )()( jj eHeH

v : )()(

Nh vy )( jeH l hm chn v i xng , cn )( l hm l v phn i xng. V

th ,khi c tnh xung h(n) l dy thc th ch cn nghin cu b lc s trong khong

)0(

Theo 12.5 ,c 2 trng hp b lc FIR pha tuyn tnh :

1. )(0

2. )(0

a. Trng hp )(0

Khi trin cng thc Euler, biu din c tnh tn s di dng:

)( 5.2 1


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)sin()cos()()()( jeAeeAeH jjjj

)sin()()cos()()( jjj ejAeAeH 32.5

Mt khc c :

1

0

1

0

)sin()cos().()()(N

n

njN

n

jnjnnhenheH

)sin()()cos().()(1

0

1

0

N

n

N

n

jnnhjnnheH 42.5

T 32.5 v 42.5 c :

1

0

)cos()()cos()(N

n

jnnheA

1

0

)sin()()sin()(N

n

jnnheA

Suy ra :

1

0

1

0

)cos()(

)sin()(

)(N

n

N

n

nnh

nnh

tg

V sin(0) = 0 v cos(0) = 1 , nn c th vit li biu thc trn di dng :

1

1

1

0

)cos()()0(

)sin()(

)(N

n

N

n

nnhh

nnh

tg

T y c 2 trng hp 0 l b lc pha khng , v 0

Trng hp 0 : 0)cos()()0(

)sin()(

)0(1

1

1

0

N

n

N

n

nnhh

nnh

tg

Tc l h(n) 0 khi n = 0, v h(n) = 0 0n . B lc nh vy khng c

ngha thc t v khng th thc hin c, v tn hiu truyn qtua b lc lun b gi tr, cho

d thi gian gi tr l nh nht .

Trng hp 0 0)cos()(

)sin()(

)cos(

)sin()(

1

1

1

0

N

n

N

n

nnh

nnh

tg


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Hay :

1

0

1

1

)sin()()cos()cos()()sin()(N

n

N

n

nnhnnhtg

Vy 0)sin()()cos()cos()()sin()(1

0

1

1

N

n

N

n

nnhnnhtg

Tip tc bin i lng gic s nhn c phng trnh :

)(sin)(1

0

nnhN

n

52.5

Phong trnh dng chui Fouriertrn c mt nghim duy nht ti :

2

1

N 62.5

v : h(n) = h(N-1-n) vi n(0, N-1) 72.5

Theo 72.5 , c tnh xung h(n) ca b lc s FIR pha tuyn tnh khi 0 l dy i

xng.

- khi 0 vNl, goi l b lc s FIR pha tuyn tnh loi 1.

- Khi 0 vNchn goi l b lc s FIR pha tuyn tnh loi 2.

V d 5.5: B lc s FIR pha tuyn tnh c )( , viN=5 v h(0) = -1

h(1) = 1, h(2) =2 . tm v v c tnh xung h(n) ca b lc.

Gii : V 0 vNl nn y l b lc

s FIR pha tuyn tnh loi 1.

Theo 62.5 c : 22

15

2

1

N

Theo 72.5 c: h(n) = h(5-1-n) = h(4-n)

Vy : h(4) = h(0) = -1

h(3) = h(1) = 1

h(2) = 2

c tnh xung h(n)c trc i xng ti n = =2,

th h(n) hnh 5.10.

Hnh 5.10 : h(n) ca b lc FIR

pha tuyn tnh loi 1.

h(n)

n

2

1

-1 1 2 3

h(n)

n

1

-11 2 3

Hnh 5.11 : h(n) ca b lc FIRpha tuyn tnh loi 2.


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V d 5.6: B lc FIR pha tuyn tnh c )( ,viN= 4 v h(0) = -1

h(1) = 1. tm v v c tnh xung h(n)ca b lc .

Gii. V 0 v N chn nn y l b

lc FIR pha tuyn tnh loi 2.

Theo 62.5 c : 5,12

14

2

1

N

Theo 72.5 c: h(n) = h(4-1-n) = h(3-n)

Vy : h(3) = h(0) = -1 h(2) = h(1) = 1

c tnh xung h(n)c trc i xng ti n = = 1,5, th h(n) hnh 5.11.

Nhn xt: B lc s FIR pha tuyn tnh loi 1v loi 2 c c tnh xung h(n)i xng

ging nh cc b lc l tng.

- Tm i xngca h(n)trng vi mu ti n = . Nu N l th l s nguyn v trc

i xng h(n)tr ng vi mu ti n = (N-1)/2. Cn nu N chn th l s thp phn v trc

i xng nm gia hai mu ti n = [(N/2)-1)] v n = (N/2)

b Trng hp )(0

Bng cch bin i tng t nh trng hp trn ,nhn c.

0)(sin)( nh 82.5

Phng trnh Fourier trn c 1 nghim duy nht ti:

2

1

N ;

2

92.5

v : h(n) = -h(N-1-n)vi n )1,0( N 102.5

Theo 102.5 , c tnh xung h(n) ca b lc FIR pha tuyn tnh trong trng hp

0 l dy phn i xng.

- khi 0 vNl, gi l b lc s FIR pha tuyn tnh loi 3.

- Khi 0 vNchn goi l b lc s FIR pha tuyn tnh loi 4.


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V d 5.7 : Cho b lc FIR pha tuyn tnh c

)( , vi N=7 v

h(0) = -1, h(1) = -0,5, h(2) = 1,5 . Tm v v

c tnh xung ca b lc.

Gii: V 0 vNl nn y l b

lc FIR pha tuyn tnh loi 3.

Theo 92.5 c :

32

17

2

1

N

Theo 102.5 c :

h(n) = -h(7-1-n) = -h(6-n)

Vy : h(6) = -h(0) = 1

h(5) = -h(1) = 0,5

h(4) = -h(2) = 1,5

h(3) = 0

c tnh xung h(n)c tm phn i xng ti n = = 3 ,

th h(n) trn hnh 5.12.

V d 5.8: Cho b lc FIR pha tuyn tnh c )( vi N=4 v h(0) = -1, h(1) = 1.

Tm v v c tnh xung ca b lc.

Gii: V 0 vNchn nn y l b

lc FIR pha tuyn tnh loi 4

Theo 92.5 c : 5,12

14

2

1

N

Theo 102.5 c : h(n) = -h(4-1-n) = -h(3-n)

Vy : h(3) = -h(0) = 1

h(2) = -h(1) = -1

c tnh xung h(n)c tm phn i xng ti n= 1,5 , th h(n) trn hnh 5.13.

-

Hnh 5.12: h(n) ca b lcFIR pha tuyn tnh loi 3

n

h(n)

5

-1

-0 52 3 4 6

-0 5

h(n)

n

1

-1

2 3

1

Hnh 5.13: h(n) ca b lcFIR pha tuyn tnh loi 4


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Nhn xt :- B lc s FIR pha tuyn tnh loi 3 v loi 4 c c tnh xung h(n)phn i

xng.

-Tm phn i xng ca h(n)ti ti n = . NuN l th l s nguyn v tm phn

i xng h(n)tr ng vi mu ti n = (N-1)/2 v ti h(n) = 0 . Cn nu N chn th l s

thp phn v tm phn i xng nm gia hai mu ti n = [(N/2)-1)] v n = (N/2)Nh vy, c bn loi b lc FIR pha tuyn tnh c )( :

-B lc loi 1: 0 ,Nl , c tnh xung h(n)i xng.

-B lc loi 1: 0 ,Nchn, c tnh xung h(n)i xng.

-B lc loi 1: 2/ , N l , c tnh xung h(n)phn i xng.

-B lc loi 1: 2/ ,N chn , c tnh xung h(n)phn i xng.

5.2.2 c tn s ca b lc s FIR pha tuyn tnh

Khi h(n) l dy thc th ch cn kho st c tnh tn s )( jeH ca b lc s FIR pha

tuyn tnh trong on :0 .

a c tnh tn s ca b lc FIR pha tuyn tnh loi 1

B lc FIR pha tuyn tnh c )( v N l, c tnh tn

s l : jjN

n

njjeeAenheH

)()()(

1

0

V N l nn khai trin biu thc trn thnh tng ca 3 thnh phn :

1

2

1

2

11

0

)(2

1)()(

N

Nn

nj

NjN

n

njj enheN

henheH

i bin thnh phn th 3, t m = (N-1-n) => n = (N-1-n),

khi

12

1N

n th

12

1N

m , khi n = (N-1) th m = 0 :

0

12

1

)1(2

112

1

0

)1(2

1)()(

Nm

mNj

Nj

N

n

njj emNheN

henheH

o chiu ch s v i li bin ca thnh phn th 3 theo n:


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210

12

1

0

)1(2

112

1

0

)1(2

1)()(

N

n

nNj

Nj

N

n

njj enNheN

henheH

V b lc FIR pha tuyn tnh loi 1 c h(n) =h(N-1-n), nn :

1

21

0

)1(2

1

)(2

1)(

N

n

nNjnj

Nj

j eenheN

heH

112.5

Trong :

n

Njn

Nj

Nj

nNjnj eeeee 21

2

1

2

1

)1(

Hay :

nN

eee

Nj

nNjnj

2

1cos2 2

1

)1(

Do 112.5 c a v dng :

2

112

1

0

2

1

2

1cos)(2

2

1)(

Nj

N

n

Nj

jen

Nnhe

NheH

Hay :

2

1

2

112

1

0 2

1cos)(2

2

1)(

Nj

Nj

N

n

j eenN

nhN

heH

i bin, t

m

Nnn

Nm

2

1

2

1,

khi n = 0 th

2

1Nm , khi

1

2

1Nn th m=1, nhn c :

2

11

2

1

.cos2

12

2

1)(

Nj

Nm

j emmN

hN

heH

i bin m tr v n ,o cn ca tng v thm cos 0. =1 vo s hng u :

2

12

1

1

.cos2

120.cos

2

1)(

Nj

N

n

j ennN

hN

heH


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211

Hay :

eeAennaeH j

Nj

N

n

j )(.cos)()( 21

2

1

0

122.5

Vi cc h s ca chui :

2

1)0(

Nha v

nNhna

2

12)( khi n 1 132.5

T 122.5 , c bin tn s ca b lc FIR pha tuyn tnh loi 1 :

2

1

0

)cos()()(

N

n

j nnaeH 142.5

Vi cc h s a(n) ph thuc vo c tnh xung h(n) theo 132.5 .

c tnh pha :

2

1

2

1)(

NN 152.5

Nhn xt: V cos(0) = 1 nn b loc FIR pha tuyn tnh loi 1 khng th dng lm

b lc c 00)( taieH j , l cc b lc thng cao v di V d 5.9: Hy xc nh

cc c tnh )( v )( jeH ca b lc s FIR pha tuyn tnh loi 1 v d 5.5. th c

tnh xung h(n) ca b lc cho trn hnh 5.14. V c tnh tn s bin )( jeH ca b lc

cho .

Gii :c tnh pha theo 152.5 :

2)(22

15

2

1

N

Theo 142.5 : c c tnh bin tn s :

2

0

)cos()()(n

jnnaeH

Tnh cc h s a(n) theo 132.5 : 2)2(21

)(

h

Nhna

2)1(212

12)1(

h

Nha ; 2)0(2)22(2)2( hha

Theo gi tr cc h s nhn c : )2cos(2)cos(22)( jeH


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212

Trn hnh 5.14 l c tuyn bin tn s )( jbp eH ca b lc FIRpha tuyn tnh

loi 1 cho , y l b lc thng thp .

b. c tnh tn s ca b lc FIR pha tuyn tnh loi 2 .

B lc FIR pha tuyn tnh loi 2 c )( v N chn c tnh tn s l :

jN

n

jnjjeeAenheH

)()()(1

0

VNchn nn khai trin biu thc trn thnh tng ca hai thnh phn :

1

2

12

0)()()(

N

Nn

j

N

n

jj

enhenheH

i bin tng th 2 , v bin i tng t mc 5.2.2.a , nhn c :

2

12

1 2

1cos)()(

Nj

N

n

j ennbeH

162.5

Vi cc h s :

n

Nhnb

2

2)( 172.5

T c c tnh bin tn s ca b lc FIR pha tuyn tnh loi 2 :

2

1 2

1cos)()(

N

n

j nnbeH 182.5

Vi cc h s b(n)ph thuc vo c tnh xung h(n) theo 172.5

Hnh 5.14 : c tnh xung h(n) v c tnh bin tn s)( jbp eH

Ca b lc thng FIR pha tuyn tnh loi 1 v d 5.9

h(n

n

2

1

-

1 2 3

)( jbp eH

0 3,1

-

1,5

-1.57


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213

c tnh pha :2

1

2

1)(

NN 192.5

Nhn xt: khi th 012(2

cos2

1cos

nn

vi mi n

nn 0)( jeH khi .Nh vy ,b lc FIR pha tuyn tnhloi 2 khng th dng

xy dng b lc c dc tnh bin tn s khc 0 ti , l b lc thng cao v b

lc di chn .

V d5.10: Hy xc nh c tnh tn s )( v )( jeH ca b lc s FIR pha tuyn tnh

loi 2 v d 5.6 . th c tnh xung h(n) ca b lc cho trn hnh 5.15. V c tnh bin

tn s )( jeH ca b lc cho .

Gii : Theo 192.5 ] c c tnh pha :

5,1)(5,12

14

2

1

N

Theo 152.5 c dc tnh bin tn s :

2

1 2

1cos)()(

n

j nnbeH

Vi cc h s b(n)c xc nh theo 172.5 :

2)1(2)12(2122)1(

hh

Nhb ; 2)0(2)22(2)2( hhb

Vy: )5,1cos(2)5,0cos(2)( jeH

Trn hnh 5.15 l c tnh bin tn s )( jbp eH ca b lc FIR

Hnh 5.15 : c tnh xung h(n) v c tnh bin tn s)( jbp eH

Ca b lc gii thng FIR pha tuyn tnh loi 2 v d 5.10 .

-3,14

)( j

bp eH

0 3,14-1,57 1,57

h(n)

n

1

-11 2 3


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214

pha tuyn tnh loi 2 cho, y l b lc di thng .

c c tnh tn s ca b lc FIR pha tuyn tnh loi 3

B lc FIR pha tuyn tnh loi 3 c )( v N l, c tnh tn s l :

jN

n

jnjjeeAenheH

)()()(

1

0

VNl nn khai trin biu thc trn thnh tng ca 3 thnh phn :

1

2

1

2

12

1

0

)(2

1)()(

N

Nn

nj

Nj

N

n

njj enheN

henheH

V b lc FIR pha tuyn tnh loi 3 c c tnh xung h(n) phn i xng nn

ti n=(N-1)/2 th h(n)=0 . Do biu thc trn c dng :

1

12

1

12

1

0

)()()(N

Nn

nj

N

n

njjenhenheH

i bin tng th hai, t m=(N-1-n) => n = (N-1-m), nhn c :

0

12

1

)1(

12

1

0

)1()()(N

m

mNj

N

n

njjemNhenheH

i li bin m thnh n v o chiu ch s ca tng th hai :

12

1

0

)1(

12

1

0

)1()()(

N

n

nNj

N

n

njj enNhenheH

V b lc FIR phatuyn tnh loi 3 c h(n) = - h(N-1-n), nn :

12

1

0

)1()()(

N

n

nNjnjj eenheH

Tip tc bin i tng ng mc 5.2.2a , nhn c :

21

22

1

1

)sin()()(

Nj

N

n

j ennceH 202.5


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215

Vi cc h s : )(22

12)( nhn

Nhnc

212.5

T c c tnh bin tn s ca b lc FIR pha tuyn tnh loi 3:

2

1

1)sin()()(

N

n

nj

nnceH

222.5

Vi cc h s c(n)ph thuc vo c tnh xung h(n) theo 212.5 .

c tnh pha :

2

1

2)(

N

Suy ra :2

1

N v

2

232.5

Nhn xet : Vi 0 v th sin(0) = 0 v sin( ) = 0 vi mi n, nn khi

0)( jeH . B lc FIR pha tuyn tnh loi 3 khng th dnh xy dng b lc c c

tnh bin tn s khc 0 ti 0 v l cc b lc thng thp , thng cao v di

chn . Nh vy, b lc FIR pha tuyn tnh loi 3 ch xy dng c b lc di thng .

V d 5.11: Hy xc nh c tnh tn s )( v )( jeH ca b lc s FIR pha tuyn tnh

loi 3 v d 5.7 . th c tnh xung h(n) ca b lc cho trn hnh 5.16 . V c tnh bin

tn s )(j

eH ca b lc cho .

Gii : Theo 232.5 c c tnh pha tn s :

1 7 11,5 ( ) 3

2 2 2

N

Theo [5..2-2] c c tnh bin tn s:

2

1

( ) ( ).sin( . )j

n

H e c n

Vi cc h s c(n) c tnh theo [5.2-21]:

(1) 2. ( 1) 2. (3 1) 2. (2) 2.1, 5 3

(2) 2. (1) 2.0, 5 1

(3) 2. (0) 2.1 2

c h h h

c h

c h

Vy ( ) 3.sin( ) 2.sin(2 ) 2.sin(3 )jH e

1,5

h(n

1

( )jH e


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216

d. c tnh tn s ca b lc FIR pha tuyn tnh loi 4

B lc FIR pha tuyn tnh loi 4 c ( ) v N chn, c tnh tn s l:

1

0

( ) ( ). ( ).N

j j n j j

n

H e h n e A e e

V N chn nn khai trin biu thc thnh tng ca hai phn:

112

0

2

( ) ( ). ( ).

N

Nj j n j n

Nnn

H e h n e h n e

i bin tng th hai, v bin i ta nhn c:

12

2 2

1

1( ) ( ).sin .( ) .

2

NN

jj

n

H e d n n e

[5.2-24]

Vi cc h s: ( ) 2. ( )2

Nd n h n [5.2-25]

T c c tnh bin tn s ca b lc FIR pha tuyn tnh loi 4:

2

1

1( ) ( ).sin .( )

2

N

j

n

H e d n n

[5.2-26]

Vi cc h s d(n) ph thuc vo c tnh xung h(n) theo [5.2-25]


26/60

217

c tnh pha:1

( ) ( )2 2

N

Suy ra:1

2

N

v

2

[5.2-27]

Nhn xt: Vi 0 th sin(0) = 0, khi ( ) 0j

H e

, v th b lc FIR pha tuyn tnh loi 4khng th dng xy dng b lc c c tnh bin tn s khc 0 ti 0 l cc b

lc thng thp v di chn.

V d 5.12:Hy xc nh cc c tnh tn s ( ) v ( )jH e ca b lc s FIR pha tuyn

tnh loi 4 v d 5.8. th c tnh xung h(n) ca blc cho trn hnh 5.2-17. V c tnh

bin tn s ( )jH e ca b lc cho.

Gii: Theo [5.2-27] c c tnh pha:

1 4 11,5 ( ) 1,5

2 2 2

N

Theo [5.2-24] c c tnh bin tn s:

2

1

1( ) ( ).sin .( )

2

j

n

H e d n n

Vi cc h s c(n) c xc nh theo [5.2-25]:

(1) 2. ( 1) 2. (1) 2 ; (2) 2. (2 2) 2. (0) 22

Nd h h d h h

Vy ( ) 2sin(0, 5 ) 2sin(1, 5 )jH e

n

h(n

-1

1

0 2 4

1

( )jH e

0.0

-

3.1.57-

Hnh 5.17: c tnh xung h(n) v c tnh bin tn s

( )jH e

cab lc thng cao FIR pha tuyn tnh loi 4 v d 5.12


27/60

218

Trn hnh 5.17 l c tnh bin tn s ( )jH e cab lc s FIR pha tuyntnh

loi 4 cho, y l b lc thng cao.

Theo dng c tnh bin tn s ( )jH e phn tch trn, cc b lc s FIR pha tuyn

tnh ch lm c cc loi b lc nh sau:

B lc loi 1: Ch lm c b lc thng thp v di chn

B lc loi 2: Ch lm c b lc thng thp v di thng

B lc loi 3: Ch lm c b lc di thng

B lc loi 4: Ch lm c b lc thng cao v di thng

5.3 Tng hp b lc s FIR pha tuyn tnh

5.3.1 Tng hp v thc hin b lc s

a Cc bc tng hp v thc hin b lc s

Vic tng hp v thc hin b lc s phi qua bn bc nh sau :

Bc 1: tng hp b lc s ,tm c tnh xung h(n)N c tnh bin tn s

( )jH e cab lc tho mn ch tiu k thut cho.

Bc 2: Xy dng s cu trc ca b lc s h(n)N tng hp .

Bc 3: Lng t ho v m ho cc h s ca b lc thnh cc t m c di

bng s bt ca tn hiu s .

Bc 4: M phng b lc c tng hp trn my tnh kim tra cc c tnh

ca b lc theo cc ch tiu k thut cho v ti u ho ln cui cctham s ca b lc.

b. Ni dung bi ton tng hp b lc s

tng hp b lc s cn bit :

1. dng ca b lc : thng thp , thng cao , di thng , di chn

2. Cc ch tiu k thut :


28/60

219

- Tn s gii hn di thng ( )c cf .

- Tn sgii hn di chn ( )p pf , hay rng di qu p p c .

- nhp nh ca c tnh biu tn s ( )jH e trong di thng 1 .

- nhp nh ca c tnh biu tn s ( )jH e trong di chn 2 .

- Sai s cho php ( ) ( ) ( ) ( )j j j jN pE e H e H e E e . Trong ( )jH e l c

tnh tn s ca b lc s l tng cng loi b lc cn tng hp .

c. Yu cu tng hp b lc s

B lc s c tng hp phi c c tnh biu tn s ( )jH e tho mn tt c

cc ch tiu k thut cho , vi bc nh nht v cu trc n gin nht c th .

c tnh tn s ( )jH e ca b lc s cn tng hp c xc nh theo biu thc

:1

( )

0

( ) ( ) . ( ) .N

j j n j j

Nn N

H e h n e H e e

Biu thc trn cho thy cc mu ca c tnh xung h(n) Nchnh l cc h s ca

chui Fourier xc nh c tnh tn s ( )jH e ca b lc cn bc ca b lc ph thuc vo

di Nca c tnh xung h(n)N .

Nh vy , tng hp b lc s thc cht l tng hp c tnh xung h(n) Nca b lc

, xc nh dng v di N ca h(n)Nsao cho c tnh biu d tn s ( )jH e tho mn tt c

cc ch tiu k thut cho .

V d , tng hp b lc di thng FIR pha tuyn tnh cn xc nh c tnh

xung h(n)Nsao cho c tnh biu tn s ( )j

NH e ca b lc c dng tng t nh hnh

5.18 , tho mn cc ch tiu k thut :

- Di thng : 1 2c c vi 2 1c c

- Di chn l vng tn sp , vi rng ri qu p p c

phi nh hn gi trin cho php .

- nhp nh ca c tnh biu tn s phi m bo :

Trong di thng: 1 11 ( ) 1jH e .


29/60

220

Trong di chn :2( )

jH e

C ba phng php c bn tng hp b lc s FIR pha tuyn tnh

1. phng php ca s .

2. phng php ly mu tn s3. phng php lp ti u .

5.3.2 Phng php ca s

a. C s ca phng php ca s

V c tnh xung h(n) ca b lc s l tng l dy khng nhn qu v hn , nn

khng th thc hin c . C s ca phng php ca s l lm cho c tnh xung h(n) ca

b lc s l tng tr thnh c tnh xung h(n)Nca b lc s cn tng hp , bng cch hn

ch chiu di ca h(n) v a v dng nhn qu . thc hin iu , nhn c tnh xung

h(n) ca b lc s l tng vi hm ca s w(n)N c dng :

( ) 0 0,( 1)Nw n khi n N

Khi , c tnh xung h(n)Nca b lc cn tng hp l dy nhn qu c di

hu hn :

( ) ( ) ( )N Nh n h n w n [5.3-2]

Nh vy, dng ca c tnh xung cn tng hp h(n)Nph thuc vo c tnh xung

l tng h(n) v dng ca s w(n)N, cn di ca h(n)Nph thuc vo rng N ca ca

s w(n)N .

Cc hm ca s w(n)N thng c s dng l :

( )jH e

p

1c 2c

1

2

Hnh 5.18: Cc tham s ca b lc di thng cn tng hp


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221

- Ca s ch nht ( )c Nw n

- Ca s tam gic ( )T Nw n

- Ca s cosin ( )c Nw n

- Cas hanning ( )

Hn Nw n

- Ca s hamming ( )Hm Nw n

Ngoi ra, ngi ta cn s dng mt s hm ca s khc phc tp hn. Cc hm

ca s trn c gii thiu 4.6 Chng bn, chng u c dng i xng trong min

thi gian, nn c tnh tn s c pha tuyn tnh. Ca s ch nht c c tnh bin ng tn s

vi sn dc hn ( p nh hn), nhng nhp nh 1 v 2 cao hn ca s tam gic. Cc

ca s cosin. Hanning, v Hamming c cc tham s dung ho gia hai loi ca s trn. Ca

s cosin c sn dc nht ( p nh nht), cn ca s Hamming c nhp nh 1 v 2

nh nht.

b. Cc bc tng hp b lc stheo phng php ca s.

Tng hp b lc s FIR pha tuyn tnh loi 1 v loi 2 theo phng php ca s

c thc hin theo cc bc nh sau:

Bc 1: Xc nh pha ( ) v c tnh xung h(n) ca b lc s l tng cng loi

b lc cn tng hp.

1( ) . ( )

2

N

[5.3-3]

Bc 2: Chn ca s hm s w(n)Nv chiu di N ca n. Trong min thi gian

hm ca s w(n)Nc tm i xng ti n = (N - 1)/ 2, nn trong min tn s c tnh tn s

HN(ejw) c pha tuyn tnh dng:

1( ) . ( )

2N

N

[5.3-4]

Bc 3: Xc nh c tnh xung h(n) ca b lc s FIR pha tuyn tnh cn tng

hp theo biu thc:

( ) ( ) ( )N Nh n h n w n [5.3-5]

Bc 4: Xc nh c tnh tn s HN(ejw) ca b lc s FIR pha tuyn tnh cn

tng hp:


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222

( ) ( ) ( ).j j jN N NH e FT h n A e e [5.3-6]

Bc 5: Kim tra c tnh bin tn s ( ) ( )j jN NH e A e c t cc ch tiu

k thut cho ..

Nu t th gim di N vlm li cc bc trn cho dn khi chn li hm cas w(n)N . Sau , thc hin li cc bc trn cho n khi chn c Nmin bin tn s

( )jNH e ca b lc cn tng hp t c tt c cc ch tiu cho .

Tng hp b lc s FIR pha tuyn tnh loi 3 v loi4 theo phng php ca s v

c thc hin theo cc bc nh trn, nhng bc 2 cn chuyn hm ca s w(n) N t

dng i xng sang dng phn i xng . Khi , trong min thi ging w(n)N c dng phn

i xng, tn s ( )jNH e c pha tuyn tnh dng :

1( ) .

2 2N

N

N

[5.3-7]

V d 5.13 : Bng phng php ca s, tng hp b lc thng thp FIR pha tuyn tnh loi

1 c tn s ct / 4 ,c vi N = 9.

a. Dng ca s ch nht ; b. Dng ca s tam gic .

Hy xy dng c tnh bin tn s ( )jNH e , xc nh v so snh cc tham s

1 2, , p , nhn c khi dng hai dng ca s trn .

Gii: - Bc 1, xc nh c tnh pha ( ) v c tnh xung h(n) ca b lc s l tng

cng loi b lc cn tng hp .

c tnh pha tn s :1 9 1

4 ( ) 42 2

N

c tnh xung ca b lc thng thp l tng pha tuyn tnh theo [5.1-4]:

sin ( ) sin ( 4) / 4( )

( ) ( 4)

c

lp

n nh n

n n

a. Bc 2, dng ca s hnh ch nht : 9 9( ) ( )Rw n rect n , vi N = 9

Bc 3, xc nh c tnh xung ca b lc cn tng hp :

9 9.( ) ( ) ( )lp R lp h n W n h n

Tnh 9 9( ) , ( ) , ( )lp R lp h n W n h n , v lp bng 5.5 :


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Bng 5.5

n 0 1 2 3 4 5 6 7 8

( )lph n 0.00 0.08 0.16 0.23 0.25 0.23 0.16 0.08 0.00

9( )R n 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

9( )lph n 0.00 0.08 0.16 0.23 0.25 0.23 0.16 0.08 0.00

Bc 4: xc nh c tnh bin tn s ca b lc s FIR pha tuyn tnh loi 1

theo [5.2-14]:

4

0

( ) ( ) ( )cos( . )j jlpn

H e H e a n n

Vi cc h thng chui c xc nh theo [5.2-13]:

9 9

9 9

9

(0) (4) 0,25 ; (1) 2. (3) 0,46

(2) 2. (2) 0,32 ; (3) 2. (3) 0,46

(4) 2. (0) 0

lp lp

lp lp

lp

a h a h

a h a h

a h

Vy : ( ) 0,25 0,46 ( ) 0,32 (2 ) 0,16 (3 )jlpH e coS coS coS

T , tnh c gi tr ca ( )jlp

H e khi

0, v lp bng 5.6 :

Bng 5.6

0 /8 /4 3/8 /2 5/8 3/4 7/8

( )jH e 1.17 0.95 0.46 0.06 0.07 0.01 0.04 0.01 0.03

Theo s liu ca bng 5.5 , bng 5.6 v tnh cht i xng ca dc tnh bin tn s , xy dng c th c tnh xung 9( )lpH n v dc tnh bin tn s ( )

j

lpH e

ca b lc thng thp cho khi dng ca s ch nht trn hnh 5.19 .

Bc 5, t c tnh bin tn s ( )jlpH e trn hnh 5.19 , xc nh c

cc tham s ca b lc dng ca s chnht :


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Tn s gii hn di chn : 2 / 5p (ly ti im 0 u tin)

rng di qu : 2 / 5 / 4 3 / 20p p c .

nhp nh trong di thng : 1 1 0,46 0,54

nhp nh trong di chn : 2 0,07

b. Bc 2, dng ca s tam gic : 92. 4

( ) 19

r

nw n

Bc 3, xc nh c tnh xung ca b lc cn tng hp :

9 9( ) ( ) ( )lp R lp h n W n h n

Tnh 9 9( ) , ( ) , ( )lp R lp h n W n h n v lp bng 5.7 :

Bng 5.7

n 0 1 2 3 4 5 6 7 8 9

( )lph n 0.00 0.08 0.16 0.23 0.25 0.23 0.16 0.08 0.00 -0.04

Hnh 5.19: 9( )lph n v ( )jlpH e ca v d 5.12 khi dng ca s ch nht

( )jH e

0

0.07

p

4

2

5

0.46

11

2 n

h(n

1

0.16

0

0.2

2 3 4 5 6

0.2

0.2

0.16

0.080.08

0.007 8


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9( )

Tn 0.11 0.33 0.56 0.78 1.00 0.78 0.56 0.33 0.11 0.00

9( )lph n 0.00 0.03 0.09 0.18 0.25 0.18 0.09 0.03 0.00 0.00

Bc 4, xc nh c tnh bin tn s ca b lc FIR pha tuyn tnh loi 1 theo [5.2-

14]:4

0

( ) ( ). os( . )j

n

H e a n c n

Vi cc h s ca chui c xc nh theo [5.2-13]

9 9 9

9 9

9

(0) (4) 0,25 ; (1) 2. (4 1) 2. (3) 0,36

(2) 2. (2) 0,18 ; (3) 2. (1) 0,06

(4) 2. (0) 0

lp lp lp

lp lp

lp

a h a h h

a h a h

a h

Vy : ( ) 0,25 0,36 ( ) 0,18 (2 ) 0,06 (3 )jlpH e coS coS coS

T , tnh c gi tr ca ( )jlpH e khi 0, v lp bng 5.8

Bng 5.8

0 /8 /4 3/8 /2 5/8 3/4 7/8

( )jH e 0.83 0.72 0.46 0.22 0.07 0.04 0.04 0.03 0.03

Theo s liu ca bng 5.7 , bng 5.8 v tnh cht i xng ca c tnh bin

tn s, xy dng c th c tnh xung h ip(n)9v c tnh bin tn s ( )j

lpH e ca

b lc thng thp cho khi dng ca s tam gic tren hnh 5.20.

- Bc 5, trong trng hp ny c tnh bin tn s c dng tim cn. Nu ly tn

s gii hn di chn p ti im 2 bngca s ch nht 2 0,07 , th t c tnh

bin tn s ( )jlpH e trn hnh 5.20 xc nh c cc tham s ca b lc dng ca

s tam gic :

Tn s gii hn di chn : / 2p

rng di qu : / 2 / 4 5 / 20p p c .

nhp nh trong di thng : 1 0,83 0,46 0,37


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nhp nh trong di chn : 2 0,07

Nhn xt :So snh cc tham s ca b lc khi tng hp bng ca s ch nht v

ca s tam gic nh sau :

- rng di qu p ca b lc dng ca s tam gic ln hn ca s ch nht, do

b lc dng c s ch nht c hai sn dc hn v chn lc tt hn dng ca

s tam gic .

- nhp nh trong di thng 1 ca b lc dng ca s tam gic nh hn dng ca s

ch nht, l do v b lc dng ca s tam gic lm suy gim tn hiu trung tm di

thng ln hn ca s ch nht .

-

c tnh bin tn s trong di chn ca b lc dng ca s tam gic c dng timcn, cn dng cas ch nht c dng dao ng .

c. Hin tng Gibbs

c tnh bin tn s ( )jlpH e ca cc b lc s l tng c dng ch nht v c

im gin on loi mt ti tn s ct c . V th c tnh xung h(n) ca chng l dy khng

nhn qu v hn. Khi dng hm ca s hn ch h(n) tr thnh c tnh xung h(n)N nhn

qu v hu hn, th c tnh bin tn s ( )jlpH e ca b lc c tng hp c dng sn

dc vng tn s ct c v pht sinh cc nhp nh c di thng v di chn (xem hnh

5.19 v 5.20), hiu ng gi l hin tng Gibbs .

tm hiu bn cht hin tng Gibbs , xut pht t biu thc [5.3 -2]:

( ) ( ) ( )N Nh n h n W n

n

hlp(

10 2 3 4 5 6

0.1

0.2

0.09

0.030.00

7 8

0.03

0.09

0.1

0.00

( )jH e

0

0.0

p

4

2

5

0.4

0.8

1

2

c

p

Hnh 5.20: 9( )lph n v ( )j

lpH e ca v d 5.12 khi ding


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Theo tnh cht ca bin i Fourier, c tnh tn s ( )jlpH e ca b lc cn tng hp

c th xc nh theo biu thc :

*1( ) ( ) ( )2

j j j

NH e W e H e

,( ) ,1( ) ( ) ( )2

j j j

NH e W e H e d

Trong , ( )jw e l dc tnh tn s ca hm ca s :

( ) ( ) ( )j j jN Ww e FT W n A e e

( )jH e l c tnh tn s ca b lc s FIR pha tuyn tnh l tng :

( ) ( ) ( )j j jN Ww e FT W n A e e

Vi :1

.2

N

Do :

,( ) ,

,

1( ( ) ( ) ( ) .

2

1( ) ( ) ( ). ( )

2

j j j j j

N N H W

j j j j j j

N H W N

H e H e A e A e e d

H e e A e A e d A e e

Vy :,1( ) ( ) ( ) ( ).

2

j j j j

N N H W H e A e A e A e d

[5.3-8]

Theo tnh cht ca php tnh tch phn, c tnh bin tn s ca b lc

c tng hp ( )jlpH e dng [5.3-8] phi lin tc, cho d hm ln ( )jNH e

ca b lc

l tng di du tch phn c gin on loi mt ti tn s ct c . chnh l

nguyn nhn pht sinh hin tng Gibbs, lm cho cc b lc thc t khong th c c

tnh bin tn s dng ch nht, m hai bin tn phi c dng sn dc vi cc bin

ng nhp nh ln cn tn s ct c .

5.3.3 Phng php ly mu tn s

a. C s ca phng php ly mu tn s


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B lc s l tng c c tnh bin tn s ( )jlpH e dng ch nht nn

khng th thc hin c trn thc t. C s phng php ly mu tn s l xp x c

tnh tn s ( )jlpH e ca b lc s cn tng hp theoc tnh tn s ( )jH e ca b lc s

l tng cng loi .

Vic xp x c thc hin bng cch ly mu tn s qua DGT, tc l lm

cho cc mu ca ( )jlpH e v ( )jH e bng nhau ti tn s ri rc 1 ( .2 / )kw K k N :

11

( ) ( )j jN kk

H e H e

Hay:1 1( ) ( )jk jk

NH e H e

[5.3-9]

Bng cch nh vy ti cc im tn s ri rc 1k k , sai s xp x gia

( )jlpH e v ( )jH e bng 0 , cn ti cc tn s gia khong 1k v 1( 1)k th sai s xp x l

hu hn. Sai s xp x s gim nh nu gim tn s ly mu c bn 1 (2 / )w N , iu

tng ng vi tng di N ca c tnh xung h(n)Nca b lc s c tng hp .

Trong min k ca DFT, biu thc [5.3-9] c dng :

( ) ( )NH k H k

Hay : ( ) ( )( ) ( ) ( ) ( ) ( ) ( )j k j ke e

N NN

A k A k A k A k H k H k

Trong , ( )NH k c l mu tn s t c tnh bin tn s ( )jH e ca b lc

l tng cng loi , tc l:

1 c di thng ca b lc l t- ng( ) ( )

0 c di chn ca b lc l t-ngN

Khi thuH k H k

Khi thu


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Hnh 5.21 miu t cch ly mu c

tnh bin tn s ( )jH e ca b lc

di thngl tng c cc tn s ct :

1

2

3 /10 0,94

8 /10 2,51

c

c

Vic ly mu tn s

c thc hin trong mt chu k

0 2 , tng ng vi 0 9k v

N =10 .

c tnh bin tn s

c ly mu10( )K k trn hnh 5.21 c

dng i xng khi 1 , 9k , ng vi

di thng ca b lc l tng ,

10( )K k csu mu gi tr 1 , nm mu

nm ngoi di thng gi tr 0:

10( ) 0,0,1,1,1,0,1,1,1,0K k

Sau khi xc nh c ( ) ( )N NH k A k theo [4.4-2] chng bn c :

( )( ) ( ) . j kN NH k A k e [5.3-11]

Cc b lc s FIR pha tuyn tnh loi mt v loi hai c c tnh xung h(n)N

i xng khi 0 ( 1)n N ,nn ( )k c dng [4.4-12] v [4.4-16] :

1

2

1

1

(0) 2( ) ( 1) ( ) . (2 1)

N

kNN N

k

A kh n A k coS n N N N

[5.3-12]

B lc s FIR pha tuyn tnh loi 1 c N l , theo mc 4.4.2 chng bn ,

A(k)Nphn i xng trong khong 1 ( 1)k N v c tnh xung h(n)Nca b lc s cn

tng hp c xc nh theo [4.4-17] :

1

0 0.94 2.51 3.77 5.34 6.28

( )jH e

0

10( )H k

1 2 3 4 5 6 7 8 910

k

Hnh 5.21: Ly mu 10( ) ( )jH e H k

1


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1

2

21

(0) 2( ) ( 1) ( ) . (2 1)

N

kNN N

k

A kh n A k coS n

N N N

[5.3-13]

B lc s FIR pha tuyn tnh loi 3 v loi 4 c c tnh xung h(n)Nphn i

xng khi 0 ( 1)n N nn ( )k c dng [4.4-20] v [4.4-24] :

( 1).( ) .

2

Nk k

N

[5.3-14]

B lc s FIR pha tuyn tnh loi 3 c N l , theo mc 4.4.4 chng bn A(k) N

phn i xng trong khong 1 ( 1)k N v c tnh xung h(n)Nca b lc s cn tng hp

c xc nh theo [4.4-21]:

1

2

3

1

2( ) ( 1) ( ) .sin (2 1)

N

k

N N

k

kh n A k n

N N

[5.3-15]

B lc s FIR pha tuyn tnh loi 4 c N chn , theo mc 4.4.5 chng bn ,

A(k)Nphn i xng trong khong 1 ( 1)k N v c tnh xung h(n)Nca b lc s cn

tng hp c xc nh theo [4.4-25]

1

2

41

( 1) 2 2( ) ( 1) ( ) .sin (2 1)

N

Nk

N NkN

kh n A A k n

N N N N

[5.3-16]

Mt khc , t cc mu ca DFT H(k)N, c th tm c c tnh tn s ( )j

NH e ca

b lc s cn tng hp theo cng thc ni suy [4.2-25] :

( 1)12

0

sin1 2

( ) ( ) . .

sin2

N kn jj N

N Nn

N

H e H k e kN

N

[5.3-17]

T [5.3-17], i vi b lc s pha tuyn tnh loi 1 v loi 2 c :

1 112

0

sin1 2

( ) ( ) . . . . .

sin2

Njk

N

Nkn jjj N

N Nn

N

H e A k e e e kN

N


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Trong 11

2

0

sin(0,5. ) ( )( ) ( 1) . .

sin2

NNn jj k

Nk

N A kH e e

kN

N

[5.3-18]

Tng t , c tnh tn s ca b lc FIR pha tuyn tnh loi 3 v loi 4 :

11 .2 2

0

sin(0,5. ) ( )( ) ( 1) .

sin2

NNn jj k

Nk

N A kH e e

kN

N

[5.3-19]

T [5.3-18] v [5.3-19] c biu thc xc nh c tnh bin tn s ca c bn loi b lc

s FIR pha tuyn tnh cn tng hp :

1

0

sin(0,5. ) ( )( ) ( 1)

sin2

Nnj k

Nk

N A kH e

kN

N

[5.3-20]

c tnh pha ( ) ca b lc FIR pha tuyn tnh loi 1 v loi 2 :

1( ) ;

2N

N

vi

1

2

N

[5.3-21]

c tnh pha ( ) ca b lc FIR pha tuyn tnh loi 3 v loi 4 :

1( ) ;

2 2N

N

vi

1

2 2

Nva

[5.3-22]

gi sai s khi xp x c tnh tn s ( )jNH e theo c tnh tn sb lc l

tng ( )jH e , ngi ta dng hm sai s ( )jE e :

| ( )jE e | = | ( )jNH e - ( )jH e | [5.3-23]

V trong di thng ( )jH e = 1 , con trong di chn ( )jH e = 0 nn c :

1 ( ) i thng( )

( ) Trong di chn

j

Nj

j

N

H e Trongd E e

H e[5.3-24]

Tm cc tr ca hm sai s ( )jE e , s nhn c gi tr sai s ln nht

( )j MAXE e , v tn s ti im sai s ln nht .


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Cn t c sai s ( )j MAXE e nh hn gi tr cho php :

( )j MAXE e ( )j pE e

[5.3-25]

b. Cc bc tng hp b lc s theo phng php ly mu tn s

Cc bc tng hp c tnh xung h(n)N ca b lc s FIR pha tuyn tnh theophng php ly mu tn s nh sau :

Bc 1 :chn s im ly mu N (chnh l di c tnh xung) .Thc hin ly

mu c tnh bin tn s ( )jH e ca b lc l tng cng loi trong mt chu k

0 2 nhn c c tnh bin tn s ri rc N

H k ca b lc s FIR pha tuyn

tnh cn tng hp .

Bc 2 :xc nh c tnh bin tn s ( )jN

H e ca b lc s FIR pha tuyn tnh

cn tng hp bng biu thc ni suy [5.3-20].

tnh ( )jNH e theo[5.3-20], trc ht cn xc nh A(k)N :

- i vi cc loi b lc loi 1 v loi 4 , t N

H k phi ly A(k)Ni xng trong

khong 1 ( 1)k N . Hn na , b lc loi 4 c A(0)N = 0

- i vi cc loi b lc loi 2 v loi 3 , t N

H k phi ly A(k)Nphn i xng

trong khong 1 ( 1)k N . Hn na , b lc loi 3 c A(0)N = 0 .

Bc 3:Kim tra c tnh bin tn s ( )jNH e c t ch tiu k thut cho

1 2, , ,c hay khng ? Hoc ( )j

MAXE e ( )j pE e

.

Nu t tt c cc ch tiu k thut cho th gim s im ly mu N v thc

hin li cc bc trn cho n khi chn c Nminm bo t tt c cc ch tiu k thut

cho.

Nu khng t th tng s im ly mu N v thc hin li cc bc trn cho nkhi chn c Nmin ( )

j

NH e ca b lc cn tng hp t c tt c cc ch tiu k thut

cho.

Bc 4: Xc nh dc tnh xung h(n)Ncu b lc FIR pha tuyn tnh cn tng hp

: ( ) ( )N Nh n IDFT H k [5.3-26]


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- i vi b lc loi 1 , tm h(n)N c th tnh IDFT theo [5.3-12]




Nu N ln , th c th s dng thut ton FFT tnh IDFT [5.3-26].

V d 5.14: Bng phng php ly mu tn s, hy tng hp b lc di thng FIR pha tuyn

tnh loi 2 c tn s ct 1 23 / 10 , 8 / 10c c .

Gii: - Bc 1: Gi s chn N=10 . Vic ly mu c tnh bin tn s ( )jH e ca b lc

tng di thng trong mt chu k 0 2 c thc hin trn hnh 5.21 vi kt qu

nhn c10( )H k l:

10( ) 0,0,1,1,1,0,1,1,1,0K k

- Bc 2 : xc nh dc tnh bin tn s10( )

jH e ca b lc di thngFIR pha tuyn

tnh cn tng hp . y l b lc loi 2 , nn t10( )H k ly 10( )A k phn i xng

trong khong 1 ( 1)k N :

10( ) 0, 0,1,1,1, 0, 1, 1, 1, 0A k [5.3-27]

Xc nh10( )

jH e theo [5.3-20] v s liu ca 10

A k :

9

1010

0

( )sin(0,5.10 )( ) ( 1)

10sin

2 10

sin(5. ) 1 1 1.10 sin(0,5. 0,2 ) sin(0,5. 0,3 ) sin(0,5. 0,4 )

1 1 1

sin(0,5. 0,6 ) sin(0,5. 0,7 ) sin(0,5. 0,3 )

j k

k

A kH e

k


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234

Theo biu thc trn tnh c gi tr ca10( )

jH e bng 5.9 v xy dng bin

tn s trn hnh 5.22.

Bng 5.9

0.00 0.44 0.94 1.26 1.57 1.88 2.51 2.89 3.14

10 ( )jH e 0.00 0.07 0.45 0.99 1.17 1.00 1.17 0.64 0.00

- Bc 3 : Xc nh sai s ( )j MAXE e gia c tnh bin tn s

10( )jH e ca b lc

s c tng hp v c tnh bin tn s ( )jH e ca b lc s l tng cng loi :

Trong di thng ( )j MAXE e = 1 0,45 = 0,55 ti tn s 1 0,94c

Trong di chn ( )j MAXE e = 0,64 ti tn s 2,89

Gi s tham s trn t ch tiu k thut ca b lc cn tng hp.

- Bc 4: Xc nh c tnh xung10( )H n ca b lc di thng FIR pha tuyn tnh cn

tng hp. y l b lc loi 2nn tm c tnh xung ( )NH n theo biu thc IDFT [5.3-

13] :

4

1010 10

1

(0) 2( ) ( 1) . ( ) . os (2 1)

10 10 10

k

k

A kH n A k c n

Theo [5.3-27] c 10( ) 0, 0,1,1,1, 0, 1, 1, 1, 0A k , nn :

10

2 3 4( ) 0,2 cos (2 1) . os (2 1) . os (2 1)

10 10 10H n n c n c n

Theo biu thc trn tnh c cc mu ca 10( )H n bng 5.10 . th ca c

tnh xung 10( )H n trn hnh 5.22 .

Bng 5.10


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235

N 0 1 2 3 4 5 6 7 8 9

10( )h n 0.10 -0.03 0.00 -0.41 0.34 0.34 -0.41 0.00 -0.03 0.10

c tnh bin d tn s10( )

jH e trn hnh 5.22 c nhp nh ln trong di thng,

vi hai bin tn cha c dc. ci thin cc tham s k thut ca b lc s cn tng hp

theo phng php ly mu tn s , cn c iu chnh gi tr cc mu ca ( )NH k ln cn

tn s ct c khi ly mu tn s .

i vi v d 5.14, thc hin iu chnh cc mu 10(1)A v 10(9)A ca 10( )H k thnh

gi tr khc khng v d chn chng bng 0,5

Tc l, iu chnh 10( ) 0,0,1,1,1,0,1,1,1,0A k thnh

10( ) 0,0,5,1,1,1,0,1,1,1,0,5A k . Do c :

10( ) 0,0,5,1,1,1,0, 1, 1, 1,0,5A k [5.3-28]

10( )h n

0.10 0.01

0.34 0.34

-

-

-

-

n1 2 34 5

6 7 8

9 10

0.00 0.63 1.26 1.88 2.513.14

0.00

0.50

0.10

1.50

10( )jH e

Hnh 5.22: c tnh xung h(n)10 v c tnh bin tn s

10( )jH e

ca b lc dithng tng hp theo phng php ly mu tn s


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Khi dc tnh bin tn s10( )

jH e ca b lc cn tng hp s c thm hai

thnh phn ng vi (1)NA v (9)NA , nn s c dng :

910

10

0

( )sin(0,5.10 ) sin(5. ) 1 1 1 1( ) ( 1)

10 10 sin(0,5. 0,2 ) sin(0,5. 0,3 ) sin(0,5. 0,4 ) sin(0,5. 0,5 )sin

2 10

j k

k

A kH e

k

1 1 1

sin(0,5. 0,6 ) sin(0,5. 0,7 ) sin(0,5. 0,8 )

theo biu thc trn tnh c cc gi tr ca10( )

jH e bng 5.11 v xy dung

c c tnh bin tn s trn hnh 5.23

Bng 5.11

0.00 0.44 0.94 1.26 1.57 1.88 2.51 2.89 3.14

10 ( )jH e

0.00 0.28 0.84 1.00 1.00 1.00 1.99 0.53 0.00

Theo s liu ca bng 5.11 xc nh c sai s ( )j MAXE e gia c tnh bin

tn s10( )

jH e ca b lc s c tng hp v c tnh bin tn s ( )jH e ca b lc

s l tng cng loi :

Trong di thng ( )j MAXE e = 1 0,84 = 0,16 ti tn s 1 3 /10 0,94c

Trong di chn ( )j MAXE e = 0,53 ti tn s 2,89

Sau khi hiu chnh, sai s gia10( )

jH e v ( )jH e trong di thng v di chn

u gim. Trong di thng, dng ca10( )

jH e t nhp nh hn, v hai bn sn dc hn

( gn dng ch nht hn)

Xc nh c tnh xung h(n)10ca b lc sau khi c hiu chnh :

4

1010 10

1

(0) 2( ) ( 1) . ( ) . os (2 1)

10 10 10

k

k

A kH n A k c n

Theo [5.3-28] c A(k)10ca b lc sau khi c hiu chnh c thm mt thnh phn ng vi

A(1)N


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10

2 3 4( ) 0,2 0,5 os (2 1) os (2 1) os (2 1) os (2 1)

10 10 10 10H n c n c n c n c n

Theobiu thc trn tnh c cc mu ca h(n)10 bng 5.12

Bng 5.12

N 0 1 2 3 4 5 6 7 8 9

10( )h n 0.01 -0.09 0.00 -0.35 0.43 0.43 -0.35 0.00 -0.09 0.10

thi ca c tnh xung h(n)10 v dc tnh bin tn s 10( )jH e ca b lc di

thng sau khi hiu chnh trn hnh 5.23

Hnh 5.23: c tnh xung h(n)10 v c tnh bin tn s

10( )jH e

ca b lcdi thng sau khi hiu chnh A(1)10 v A(9)10.

10( )h n

0.01 0.01

0.43 0.43

-

-

-

-

n1 2 34 5

6 7 8

9 10

0.00 0.63 1.26 1.882.513.14

0.00

0.40

0.80

1.20

10( )jH e


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So vi c tnh bin tn s khi cha hiu chnh hnh 5.22 c tnh bin tnn

s trn hnh 5.23 gn ging dng ch nht hn.

Tm li, khi tng hp b lc s theo phng php ly mu tn s , cn lu mt

s im sau y :

- B lc loi 1 c N l h(n)N i xng khi 0 ( 1) , ( )Nn N A k i xng khi

1 ( 1)k N ch lm c b lc thng thp v di thng .

- B lc loi 2 c N chn h(n)Ni xng khi 0 ( 1) , ( )Nn N A k phn i xng khi

1 ( 1)k N ch lm c b lc thng thp v di chn .

- B lc loi 3 c N l h(n)N i xng khi 0 ( 1) , ( )Nn N A k i xng khi

1 ( 1)k N ch lm c b lc di thng .

- B lc loi 4 c N l h(n)

N i xng khi 0 ( 1) , ( )Nn N A k i xng khi1 ( 1)k N ch lm c b lc thng cao v di thng .

- tng hp c b lc c c c dc tnh bin tn gn dng ch nht , khi ly

mu tn s b lc l tng cng loi, cn phi hiu chnh cc mu ca ( )NH k ln

cn tn s ct .

5.3.4 Phng php lp ti u

Phng php lp ti u tng hp c tnh xung h(n)N ca b lc s FIR pha tuyn

tnh nhm p ng cc ch tiu k thut cho theo mt tiu chun ti u nht nh .s dng

phng php lp ti u, s a bi ton tng hp b lc s v dng bi ton lp ti u

Chebyschev , v nhn c h(n)N,phi s dng png php ti u ho lp lp v vic

tnh ton kh phc tp . C th nghin cu chi tit phng php ny cc ti liu tham kho

4, 7 ,9 .

5.4 Thc hin b lc s FIR pha tuyn tnh

Sau khi tng hp c tnh xung h(n)N,ca b lc s FIR pha tuyn tnh, hai bc tip theo

thc hinb lc l xy dng cu trc ca b lc v lng t ho, m ho h s ca b lc .

5.4.1 Cu trc dng ni tip ca b lc s FIR pha tuyn tnh

a. B lc s FIR pha tuyn tnh cu trc chun tc

T c tnh xung h(n)N, ca b lc s FIR pha tuyn tnh, xc nh c hm h

thng ( )NH z ca b lc :


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1

0

( )( ) ( ) .

( )

Nn

N N

n

Y zH z h n z

X z

[5.4-1]

vy:1

0

( ) ( ) . ( )N

n

N

n

Y z h n z X z

[5.4-2]

hay: 1 ( 1)( ) (0) ( ) (1) . ( ) ..... ( ) . ( )NN N NY z H X z H z X z h N z X z

Ly bin i Z ngc li c hai v ca [5.4 -2], nhn c phng trnh sai phn

bckhng m t b lc s FIR pha tuyn tnh :

1

0

( ) ( ) ( )N

N

k

Y n h k x n k

[5.4-3]

Hay: ( ) (0) ( ) (1) ( 1) ..... ( 1) ( 1)N N Ny n h x n h x n h N x n N

Theo cc quan h vo ra khng quy [5.4-2] hoc [5.4-3] , xy dng c cc b

lc c cu trc khng phn hi, vi cc mu ca dc tnh xung h(n)Nchnh l cc h s ca

b lc .

Cu trc chun tc ca b lc s FIR pha tuyn tnh trn hnh 5.24a c xy

dng trc tip t phng trnh sai phn [5.4-3]

thc hin b lc s FIR pha tuyn tnh bng phn cng theo cu trc chun tc

cn cb ghi dch (N-1) nhp , N b nhn , v (N-1 ) b cng s hai li vo . v tr ca bghi dch (N-1) nhp , c th dng (N-1) nh hoc thanh ghi cht s liu . cc b ghi dch ,

b nhn , b cng hoc thanh ghi cht s liu u phi c s bt s l bng s bt ca tn hiu

s .

D

D

D

x(

h(

y(

h(

h(

-

D

h(

h(

h(

D

D

y(

x(


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240

b. B lc s FIR pha tuyn tnh cu trc chuyn v

Khi i v tr ca cc b nhn v b tr trong cu trc chun tc, nhn c cu

trc chuyn v trn hnh 5.24b, trong tc ng x(n) c nhn vi cc h s cab lc,

sau mi cng vgi tr .

thc hin b lc s FIR pha tuyn tnh bng phn cng theo c trc chuyn v

cn c (N-1) nh hoc thanh ghi cht s liu, ngoi ra cn N b nhn v (N-1) b cng.

c. B lc s FIR pha tuyn tnh cu trc ni tng

Cu trc ni tng da trn c s biu din hm h thng H N(z) ca b lcs didng tch ca hm c s bc mt v bc hai . cc h s ca b lc c xc nh theo cc

khng ca HN(z) :

11 1 2

0 0 1 2

1 1 1

( ) ( ). ( . ). ( . . )M LN

n

k k i i i

n k i

H z h n z a a z b b z b z

Theo [5.4-4] xy dng c b lc s FIR pha tuyn tnh c cu trc gm cc tng

bc mt v tng bc hai ni tip nhau nh trn hnh 5.25 .

1Z a0k

a1k

1Z

1Z

b

b1i

b1i

X z

Hnh 5.25: B lc FIR pha tuyn tnh cu trc ni tng

Y z


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241

Vi cu trc ni tng trn hnh 5.25 , c th ch to cc m dun bc mt v bc hai

chun , khi xy dng b lc s ch cn thit lp cc h s c th cho modun v ghp ni tip

chng vi nhau theo dng ca hm h thng [5.4-4].

5.4.2 Cc cu trc dng vng ca b lc s pha tuyn tnh

Cu trc dng vng c xy dng trn c s tnh i xng hoc phn i xng

ca c tnh xung h(n)Nca b lc s FIR pha tuyn tnh

a. Cc cu trc dng vng ca b lc s pha tuyn tnh loi 1

B lc s FIR pha tuyn tnh loi 1c 0 , N l , h(n)N i xng h(n)N =h(N-1-

n)Nvi tm i xng ti n=(N-1)/2 , nn c th a HN(z) v dng :

11

1 2 ( 1 )

0 0

1 ( )( ) ( ). ( )2 ( )

N

N n n N n

N

n n

N Y zH z h n z h z h n z zX z

11

1 2( 1 )2

0

1( ) ( ). ( ) .( ). .( )

2

NN

n N n

n

NY z h X z z h n X z z X z z

[5.4-5]

Theo [5.4-5] , xy dng c cu trc dng vng ca b lc s FIR pha tuyn tnh

loi 1 trn hnh 5.26.

thc hin b lc trn hnh 5.26 bng phn cng , cn s dng (N-1) nh hoc

thanh ghi cht s liu , (N-1) b cng , v ( 1) / 2N b nhn . So vi dng chnh tc , s b

nhn gim gn mt na

X z

Y z

1Z

1Z 1Z

h(0

1( 1)

Nh

( 2)( ). NX z z

( 1)( ). NX z z

h(1


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242

b. Cc cu trc dng vng ca b lc s pha tuyn tnh loi 2

B lc FIR pha tuyn tnhloi 2 c 0 , N chn , h(n)N i xng h(n)N =h(N-1-

n)Nvi tm i xng ti n=(N-1)/2 , nn c th a HN(z) v dng :

11

1 2( 1 )

0 0

( )( ) ( ). ( )

( )

N

Nn n N n

N

n n

Y zH z h n z h n z z

X z

Vy :1

( 1 )

0

( ) ( ) .( ). .( )N

n N n

n

Y z h n X z z X z z

[5.4-6]

Theo [5.4-6], xy dng c cu trc dng vng ca b lc s pha tuyn tnh loi2 trn hnh 5.27 .

X z

Y z

1Z

1Z 1Z

h(0

( 2)

2

Nh

( 2)( ). NX z z

( 1)( ). NX z z

h(1


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243

thc hin b lc trn hnh 5.27 bng phn cng , cn s dng (N -1) nh hoc thanh

ghi cht s liu , (N-1) b cng , v (N/2) b nhn .

c. Cc cu trc dng vng ca b lcs pha tuyn tnh loi 3

B lc s FIR pha tuyn tnh loi 3 c 0 , N l , h(n)N phn i xng h(n)N =-

h(N-1-n)Nnn c th a HN(z) v dng :

11

11 2( 1 )2

0 0

1 ( )( ) ( ). ( )

2 ( )

NNN

n n N n

N

n n

N Y zH z h n z h z h n z z

X z

11

1 2( 1 )2

0

1

( ) ( ). ( ) .( ). .( )2

NN

n N n

n

N

Y z h X z z h n X z z X z z

[5.4-7]

Theo [5.4-6], xy dng c cu trc dng vng ca b lc s pha tuyn tnh loi

3 trn hnh 5.28.

X z

Y z

1Z

1Z 1Z

h(

1( 1)

2

Nh

( 2)( ). NX z z

( 1)( ). NX z z

h(


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244

thc hin b lc trn hnh 5.28 bng phn cng , cn s dng (N -1) nh hoc thanh ghi

cht s liu , (N-1) b cng , v ( 3) / 2N b nhn

d. Cc cu trc dng vng ca b lc s pha tuyn tnh loi 4

B lc s FIR pha tuyn tnh loi 4 c 0 , N chn , h(n)N phn i xng h(n)N

=- h(N-1-n)Nnn c th a HN(z) v dng :

1 1( 1 )

0 0

( )( ) ( ). ( )

( )

N Nn n N n

N

n n

Y zH z h n z h n z z

X z

vy:1

( 1 )

0

( ) ( ) .( ). .( )N

n N n

n

Y z h n X z z X z z

[5.4-8]

5.4.3 Cu trc ca b lc s FIR pha tuyn tnh ly mu tn s

Khi tng hp b lc s theo phng php ly mu tn s, sau khi xc nh c

( )Nh n v DFT ca n:

1

1

0

( ) ( ) .N

jk n

N N

n

H k h n e

vi 12

N

Theo [4.2-22], t ( )Nh n c th tm c hm h thng ( )NH z ca b lc s:

1

1

10

( )(1 )( ) ( )

(1 . )

N NN

N N jkk

H kzH z ZT h n

N e z

[5.4-9]

Hay: 1 21

( ) ( ) ( )NH z H z H z N

[5.4-10]


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245

Trong : 1( ) (1 )NH z z

[5.4-11]

V:1

1 1

2 210 0

( )( ) ( )

(1 )

N NN

kjkk k

H kH z H z

e z

[5.4-12]

Thnh phn ( )NH z gm N khu mc

song song, mi khu l:

12 1

( )( )

(1 )

N

jk

H kH z

e z

[5.4-13]

Theo [5.4-13], 2 ( )kH z c cu trc phn hi vi h s phc 1jke nh hnh 5.30.

iu c ngha l b lc FIR pha tuyn tnh c xy dng theo quan h vo ra quy.

T [5.4-10] c:1 2

1( ) ( ). ( ) ( ). ( ) ( )NY z X z H z X z H z H z

N [5.4-14]

Theo quan h vo ra [5.4-14], c s khi b lc FIR pha tuyn tnh hnh 5.31.

Theo hm h thng [5.4-9], xy dng c s cu trc dng ly mu tn s ca b

lc s FIR pha tuyn tnh hnh 5.32.

1Z

1

jk

e

( )NH k

( )NH k

Hnh 5.30: Khu 2( )kH z

1( )H z 2( )H z( )X z

( )NH k

( )Y z

( )NH k

1/

Hnh 5.31: S khi dng ly mu tn s ca b lc FIR


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246

hiu r hn cu trc ca b lc trn cn tm hiu cc thnh phn 1( )H z v 2( )H z

ca n.

11

1 ( )( ) (1 )

( )

NN

N

z Y zH z z

z X z

1( )H z c mt cc bi bc N ti z = 0 v N khng im phn b u trn vng trn

n v ti cc im 12

0

jkjk N

kz e e

, k = [0 (N-1)]. Do thnh phn 1( )H z l h n nh

v c th tm c:

1( ) ( ).(1 ) ( ) ( ).n NY z X z z X z X z z

Xt thnh phn 2( )H z v cc khu phn hi 2 ( )kH z , theo [5.4-13] c:

1Z

1Z

1Z

1Z

1Z

1Z

( )X z

( )NH k ( )Y z

( )NH k

(0)NH

10je

(1)NH

11je

( 1)NH N

1( 1)j Ne

( ). NX z z

1/

-1

Hnh 5.32: B lc s FIR pha tuyn tnh cu trc c phn hi


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247

1 12 1

( )( ) ( )

(1 )N

k Njk jk

H k zH z H k

e z z e

[5.4-15]

Theo [5.4-5], mi 2 ( )kH z c mt im khng ti z = 0 v mt cc im ti

1

2jk

jk Npkz e e

. Do 2( )H z khng c im bi bc N ti z = 0, v N cc im n ti

1

2jk

jk Npkz e e

vi 0 1k N .

Kt hp 1 2( ) ( ). ( )H z H z H z th cc khng im v cc im ca 1( )H z v 2( )H z s b

tr ht cho nhau.

Tuy nhin trn thc t, khi lng t ho cc h s ca 1( )H z v 2( )H z c th dn

n cc khng im ca 1( )H z v cc im ca 2( )H z lch nhau, lm cho b lc mt n

nh. khc phc iu , ngi ta thng lm cho cc khng im ca 1( )H z v cc im

ca 2( )H z dch vo bn trong vng trn n v mt cht bng cch thay 0 0.k kz r z v

.pk pk z r z vi 1r v 1r , nh trn hnh 5.34.

Re[z]

Im[z]

Re[z]

Im[z]

a. Cc v khng ca 1( )H z b. Cc v khng ca 2 ( )H z

Hnh 5.33: Cc im cc v khng ca 1( )H z 2( )H z


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248

Khi hm h thng H(z) s c dng:

1

1

10

( )1 .

( ) 1 . .

N N NN

jkk

H krr z

H z N e r z

[5.4-16]

Bng thc nghim xc nh c, vi gi tr 12 27(1 2 ) (1 2 )r vn m bo b

lc n nh v khng thay i c tnh tn s.

Mt s vn na cn khc phc l cc h s phc 1jke trong cu trc ca b lc.

trnh phi xy dng b lc vi cc h s nhn l s phc, s dng tnh cht i xng ca

H(k)N khi h(n)Nl dy thc, bin i c 2( )H z v dng:

12

2 21 11

( )(0) 2( ) ( )

1 1

NN

Nk

k

NHHH z H z

z z

[5.4-17]

Vi: 1 1

2 1 -2

1

os ( ) os (k)-k( ) 2. ( )

1 2 os(k )+zk N

c k z c H z H k

z c

[5.4-18]

Trong : ( )( ) ( ) j kN NH k H k e

V: 11 1

0

0 0

(0) ( ) . ( )N N

j n

N N N

n n

H h n e h n

l s thc

21 1.2

0 0

( ) ( ) ( 1) ( )2

NN Nj nnN

N N N

n n

NH h n e h n

l s thc

Theo [5.4-16] c s cu trc cc khu phn hi 2 ( )kH z trn hnh 5.35.

Re[z]

Im[z]

Re[z]

Im[z]

a.V tr cc ca 2( )H z b. V tr cc mi ca 2( )H z

Hnh 5.34: Cc im cc v khng ca 1( )H z


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Hnh 5.35: S cu trc cc khu phn hi 2 ( )kH z

5.4.4 Lng t ho v m ha cc h s ca b lc

Sau khi xy dng c cu trc ca b lc s FIR pha tuyn tnh, thc hin

nhn tn hiu s vi cc h s ca b lc, cn lng t ho v m ho h s thnh hng s

ho nh phn . V d, vi cc cu trc dng chun tc thi cn lng t ho v m ho cc

mu ca c tnh xung h(n)N thnh cc hng s m nh phn c di bng s bt ca tn

hiu s .

Vic lng t ho cch s ca b lc s gy sai s v lam thay i h thng H(z)

cng nh c tnh tn s jH e ca b lc tng hp. Trong mt s trng hp sai s

lng t c th lm mt i n nh hoc lm thay i tn s ca b lc. V d nh lm tn h

n nh khi xy dng cu trc ca b lc theo phng php ly mu tn s.

Gi s h s trc khi lng t ho gi tr lin tc k , sau khi lg t ho n s c

gi tr l vi k l sai s lng t . Gi tr ca k ph thuc vo s bt ca tn hiu s v gi

tr tuyt i ca k .

nh gi nh hng ca sai s lng t h s k n c tnh tn s jH e hochm h thng H(z) ,ngi ta a ra khi nim nhy ring ( )jkS e

v ( )kS z :

( )( )

jj

k

k

H eS e

[5.4-19]

12 os(k )c

1Z

1

os (k)c

1os ( ) kc k


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250

( )( )k

k

H zS z

[5.4-20]

nh gi nh hng sai s lng t ca tt c cc h s k hay c tnh tn s

jH e hoc hm h thng H(z) , ngi ta s dng khi nim nhy tuyt i ( )jkS e v

( )kS z :

( ) ( )j j

k k

k

S e S e

[5.4-21]

( ) ( )k kk

S z S z [5.4-22]

1

Nz

1Z

1Z

H21(z

H21(z

H2(N/2-

X(z

Y(z

1/N H(0)N

H(N/2

Hnh 5.36: S khi ca b lc s vi ly mu tn s


60/60

hoc nhy cu phng ( )jq

S e v ( )qS z

2( ) ( )j j

q k

k

S e S e

[5.4-23]

2( ) ( )q kk

S z S z

[5.4-24]

Nhn xt :

- Nu nhy cng nh th nh hng ca sai s lng t n hm h thng H(z) v

c tnh tn s jH e cng nh .

- Nu chn cu trc b lc thch hp c th lm gim ng k nhy, v th cn tm

cu trc c nhy thp .

- M phng b lc trn my tnh thy c y nh hng ca sais lng t

n cc dc tnh ca b lc v t c nh hng khc phc cc nh hng xu

gy bi sai s lng t. ng thi m phng cho php ti u ho b lc ln cui

cng .

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Chuong 5 Xu Ly Tin Hieu So