BGXLTH_CLC_chuongI-II _ Tin Hieu Lien Tuc_SV

Bin i tn hiu v ng dng

Phm Th Ngc Yn

2012

B mn

K thut o&

Tin hc Cng nghip

Phm Th Ngc Yn - 20122

Ni dung mn hc

Chng 1 : Gii thiu chung

Chng 2 : Tn hiu v h thng lin tc

Chng 3 : Bin i tn hiu lin tc thnh tn hiuri rc - Khi phc tn hiu

Chng 4 : Tn hiu v h thng ri rc

Chng 5 : Php bin i Fourier ri rc

Chng 6 : ng dung: Lc s

Q&R


Ti liu tham kho

1. L. Rabiner and B. Gold, "Theory and

Application of Digital Signal Processing",

Prentice Hall, New-Jersey, 1997

2. Jacques Max & Jean-Louis Lacoume.

"Mthodes et techniques de traitement du

signal et applications aux mesures

physiques". Masson, 1996.

3. Phm Th Ngc Yn. "X l tn hiu -Tn hiu

, h thng tuyn tnh, lc s v DSP " - NXB

KHKT, 2010

4. Nguyn Quc Trung. "X l tn hiu v lc

s", NXB KHKT. Tp 1 (1999), tp 2 (2001)


Chng I: Gii thiu chung

Hnh 1-1

S tng qut mt

knh truyn tin

Ngun to

thng tin

Thng tin

To tn hiu

Tn hiu

H thng truyn tn hiuNhiu

Tn hiu

Nhn tn hiu

Thng tin

Khai thc

Thng tin

1.1. Tn hiu v phn loi tn hiu


Tn hiu l mt i lng vt l c th, c nhng quilut bin thin c th, mang theo nhng thng s phn

nh thuc tnh cn nghin cu ca i tng o.

Xt tn hiu nh l hm ca mt bin c lp(thi gian- x(t))




Tn hiu s l tn hiu c biu din bng mt dy s. X ltn hiu s bao hm mi php x l cc dy s c cthng tin cn thit nh phn tch, thay i, tng hp, m ho ...

X l tn hiu thu

c (o, lu gi, sa i ...)

Tn hiu tng t

Tn hiu s

Tn hiu s

Ri rc ho

Tn hiu ri rc

c cu

tha hnh

Qu trnh

vt lCm bin

Ly mu

A/D

D/A

H thng iu khin

s (My tnh)

Thng tin Thng tin Tn hiu tng t

Hnh 1-2 : V d iu khin qu trnh cng nghip bng h thng s


Khi nim tn hiu v nhiu ch l tng i v ph thuc vo mc ch ca ngi s dng.


NhiuCc hin tng lm nh hng n qu trnh thu nhn tn hiu


2. Phn loi :

Theo thi gian,

Theo c tnh nng lng,

Theo c tnh ph,

Theo tn hiu lin tc hoc ri rc.



a) Biu din tn hiu theo thi gian

Tn hiu vt l

Tn hiu tin nh Tn hiu ngu nhin

T/h chu k T/h khng chu k Dng Khng dng

Hnh Chu k Gi Chuyn Egodic Khng

sin phc tp chu k tip egodic



b) Biu din tn hiu theo c tnh nng lng

Tn hiu c nng lng ton phn hu hn

Tn hiu c cng sut trung bnh hu hn

)(2

dttx

)(1

lim02/

2/

2

T

TT dttx

T



c) Biu din tn hiu theo c tnh ph

Phn loi theo phn b nng lng hoc cngsut tn hiu theo hm tn s (ph tn hiu)

Vng tn s F = Fmax- Fmin (Hz) c gi l rng bng (di) tn ca tn hiu

Tn s trung bnh Ftb=( Fmax+ Fmin)/2

Fmin Fmax

Tn s

Phn b

ph

F



Tn hiu c di tn hp: F/Ftb nh (Fmax Fmin)

Tn hiu c di tn rng: F/Ftb ln (Fmax>> Fmin)

Cc tn hiu c di tn hp c phn loi theo Ftb Ftb


d) Biu din tn hiu theo tn hiu lin tc hoc ri rc

Theo bin thi gian, c th phn thnh tn hiu thi gianlin tc (tn hiu lin tc) v tn hiu thi gian ri rc (tn hiu

ri rc hay tn hiu ly mu)

Bin ca tn hiu cng c th lin tc hoc ri rc(lng t) T hp ca 2 bin c 4 dng tn hiu

Tn hiu c bin v thi gian lin tc (t/h tng t): x(t) Tn hiu c bin ri rc v thi gian lin tc (t/h lng

t ho): xq(t)

Tn hiu c bin lin tc v thi gian ri rc (t/h lymu): x(nTe)

Tn hiu c bin ri rc v thi gian ri rc (t/h logic):xq(nTe) (thng c s dng trong my tnh )




Cc cng vic chnh ca x l tn hiu :

a. Chun b (to tn hiu)

Tng hp tn hiu

iu ch tn hiu

M ho tn hiu

b. Can thip vo tn hiu

Lc tn hiu

Tch tn hiu ra khi nhiu

Nhn dng tn hiu

Phn tch tn hiu

o cc thng s c trng ca tn hiu



1.2. u im v nhc im ca x l s tn hiu

chnh xc cao

Sao chp trung thc nhiu ln

Tnh bn vng

Cng ngh ngy cng hon thin

Tnh linh hot v mm do

Tnh nng cao

Tc v gi thnh

Thi gian thit k.

Vn di hu hn



1.3. Phm vi ng dng

X l nh: nhn dng, hot hnh, mt ngi my

Thit b o lng iu khin: phn tch ph, iu khin

v tr v tc , gim n, nhiu, nn d liu, o lnga cht ..

X l ting ni, m thanh: nhn dng ting ni, ngini, tng hp ting ni, m thanh s ...

Qun s: truyn thng bo mt, x l tn hiu rada,

dn ng tn la ...

Sinh hc v in t y t: qut nh, hnh nh no ,

in tim ...


Ni dung mn hc

Chng 1 : Gii thiu chung

Chng 2 : Tn hiu lin tc

Chng 3 : Bin i tn hiu lin tc thnh tn hiuri rc - Khi phc tn hiu

Chng 4 : Tn hiu v h thng ri rc

Chng 5 : Php bin i Fourier ri rc

Chng 6 : ng dung: Lc s

Q&R


Chng II : Tn hiu lin tc

Hai bi ton thng gp: Phn tch h thng: tm y(t) ca h thng bit x(t) v h(t) Tng hp h thng: tm cu trc ca h thng h(t) bit x(t) v y(t)

M

M

M1N

N

N10dt

)t(yda...

)t(d

)t(dya

dt

)t(xdb...

)t(d

)t(dxb)t(xb)t(y

Gii phng trnh vi phn bc cao m t h thng

Y()=b0X()+b1(j)X()+....+bN(j)NX()+a1(j)Y()+....+aM(j)

MY()

2.1. Nguyn tc c bn x l tn hiu lin tc



2.2. Bin i Fourrier ca tn hiu lin tc bt k -

Tch phn Fourrier

1. nh ngha

X() = R() + jI() = A() ej()

1)-(2 dtetxX tj

)()(

2)-(2 2

1

deXtx tj

)()(

x(t) X()

Phn tch tnh cht ph ca x(t) ???



2. iu kin tn ti ca php bin i Fourier

x(t) l mt hm gii hn Tch phn ca x(t) t - ti phi l gi tr xc nh Gi tr cc i v cc tiu cng nh cc gi tr ginon ca x(t) phi l gi tr xc nh

3. Mt s tnh cht c bn Tnh tuyn tnh Dch chuyn theo thi gian Dch chuyn theo tn s o hm theo thi gian o hm theo tn s Nhn chp trong min thi gian Nhn chp trong min tn s nh l Parseval



Tnh cht ca php nhn chp

Tnh giao hon

Tnh phn b

Tnh kt hp

Nhn t trung tnh (phn b Dirac)

d)t(y)(x)t(y)t(x



Bi tp v d1:

Tm bin i Fourrier v v ph ca hm ca s CN

t 0

t x(t)

A

BT V d 2: Tm p ng tn s Y() ca tn hiu sau

t khi e

t khi 0)t(y

t10j



BT V d 4:

Gii li bi ton nhn chp:

Cho x(t) v h(t) c dng:

Tnh y(t) bng phng php tn s

1t vi

1t vi

0t vi

0t vi

0

1)t(h

0

1)t(x

j

1)()X(

0 tkhi 0

0 tkhi 1))t(u( )t(x

BT V d 3: Hy chng minh cp bin i Fourrier



2.3. Hm Delta v xung Durac

t

t

0

2

1

)(tf )(lim)( 0 tft

)sin(

2

1

dteF tj)( D()=lim0 F()=1

(t-t0)= (t0-t) biu din hm Delta ti t= (t-t0)

)()()(

)()()()(

)()()()()(

00

000

000

ttxtttx

tttxtttx

txdttxttdttxtt

2(t ) khng

nh ngha.



VD: Tnh y(t)=x(t)*h(t) ca cc hm sau:


3)-(2

1

000 )sin()cos()(k

kk tkbtkaatx


2.4. Bin i Fourrier ca tn hiu lin tc chu k - Chui

Fourrier

Cch biu din 1 pha trn trc to

4)-(2

1k

2/T

2/Tok

2/T

2/Tok

2/T

2/T0

dt)tksin()t(xT

2b

dt)tkcos()t(xT

2a

dt)t(xT

1a

nh l : Nu hm x(t) lin tc v tun hon theo bin c lp t vi chu k

tun hon T th x(t) c th khai trin thnh chui Fourier l t hp tuyn

tnh ca cc hm iu ho c tn s k0, trong o=2 F=2/T.



)a

barctan(ba

)tkcos(ca)t(x

k

kk

2k

2k

1kk0k0

v c

k

Vit cch khc:

Ph tn ca x(t) l ph vch

Nu t k= 1/2.(ak-j bk) 22

1 22 kkkk

cba

dte)t(xT

1

5)-(2 e)t(x

2/T

2/T

tjk

k

k

tjk

k

o

o

vi

0

-k

k

T

kX

2

)62()(2)( 0

Biu din 2 pha trn trc to

CM t (2-6) cth tm c (2-5)



BT V d:

1. Tm k v hm ph X() ca tn hiu

x(t)=cos(2Ft)=cos(0t)

2. Hy chng minh cp bin i Fourier phn f v r (ph lc)

3. Bi tp ng dng nh l Parseval



2.5. H thng lin tc

Tnh cht tuyn tnh: ax1(t)+bx2(t) ay1(t)+by2(t)

Tnh bt bin theo thi gian: x(t) y(t); x(t-) y(t-)

Tnh nhn qu: h(t)=0 t



1. Biu din h thng bng p ng xung h(t)

Tn hiu vo l (t) tn hiu ra l h(t)

Tn hiu vo l x(t) tn hiu ra y(t) = x(t)* h(t)

Gii thch cch tnh y(t)???



Tnh cht ca php nhn chp

Tnh giao hon : x * y = y * x

Tnh phn b : x * (y+z) = (x * y) + (y * z)

Tnh kt hp : x * (y * z) = (x * y) * z

Nhn t trung tnh (phn b Dirac): x * = * x = x

d)t(y)(x)t(y)t(x



Cch tnh php nhn chp

Tnh h(t-) bng cch tnh i xng h()

h(-), dch chuyn h(-) mt khong t1, ta s c

h(t1-)

Tnh tch x() . h(t1-)

Tnh tch phn ca ca tch trn theo bin .

Gi tr ca tch phn ny l gi tr ca y(t) ti

thi im t1.

Cho t1 thay i t - cho n + , tnh

c tt c cc gi tr khc ca y(t).

dthxty )()()( 11

dthxthtxty )()()()()(



Bi tp

1. Cho x(t) v h(t) c dng:

Hy tnh y(t) theo 2 phng php: hnh hc v bin iFourrier

1t vi

1t vi

0t vi

0t vi

0

1)t(h

0

1)t(x



2. Biu din h thng bng p ng tn s H()

h(t) H()

y(t)=x(t)*h(t) Y()=X().H()

y(t)=x(t).h(t) Y()=1/2 [X()*H()]



3. Biu din h thng bng im cc v im khng

7)-(2 M

M

N

N

jaja

jbjbb

X

YH

)(...1

)(...

)(

)()(

1

10

8)-(2 M

M

N

N

papa

pbpbbpH

)(...1

)(...)(

1

10

9)-(2

M

i

iM

N

i

iN

NM

NN

ppa

zpb

ppppppa

zpzpzpbpH

1

1

121

121

)(

)(

))...()((

))...()(()(

T cc v tr ca im cc (p) v im khng (z)

trn mt phng p (hay s) c th nhn bit c tnhcht ca h thng



Cch c tnh cht h thng t imcc v im khng

Nu cc im cc lun nm v pha

tri ca trc tung, cc im khng c th

nm mt v tr bt k trn mt phng, h

thng lun lun n nh.

Nu =0, tt c cc gi tr ca p u

nm trn trc tung.

Nu khng c cc im khng nm 1/2 mt phng phi, h thng c pha tithiu, tc l vi mt bin bt k cho trc, thi gian truyn ()=-d()/d()

lun nh nht c th i vi tt c cc tn s.

Nu cc im khng ch nm 1/2 mt phng phi, im cc ch nm 1/2mt phng tri, hm truyn t H()=1 vi mi h thng c di thng mi

tn s (cho tt c cc tn s i qua).

i vi 1 h thng thc t c th thc hin c, s im M lun ln hn hocbng N. Tc l s im cc phi ln hn hoc bng s im khng ca h thng.



2.6. Hm tng quan v mt ph tn hiu

1. Nhc li cng sut v nng lng ca tn hiu

Cng sut tc thi

Cng sut trung bnh ca tn hiu trong khong thi gian

T0

Nng lng ton phn ca tn hiu

Cng sut tc thi tc ng qua li gia hai tn hiu

Cng sut trung bnh tc ng gia hai tn hiu trong

khong thi gian T0 Cng sut trung bnh tc ng gia hai tn hiu c

di v hn



2. Nng lng ca tn hiu trong min tn s, mt ph

nng lng

Mt ph nng lng Mt ph tn hiu ca hai tn hiu Nng lng ca tn hiu trong di tn s quanh mt tn

s c bn 0

Nng lng ton phn ca tn hiu trong min tn s



3. Hm tng quan v mt ph tn hiu

Hm t tng quan :

Hm h tng quan :

Quan h vi mt ph tn hiu

Hm tng quan ca tn hiu tun hon cng l mthm tun hon

dt)t(x)t(x)(R *xx

dt)t(y)t(x)(R *xy

Hy CM

Tnh hm t tng

quan ca x(t)=cos(0t)



2.7. Nhiu

1. Ngun nhiu

a. Ngun gy nhiu ngoi:

Nhiu mi trng t nhin

Nhiu nhn to

b. Ngun gy nhiu ni b hay "nhiu trong":

Nhiu gy bi xung in

Nhiu nn : gm nhiu nhit v nhiu ht



Nhiu nhit: Ngun gc xut hin ca cc in p nhiu trong cc

mch th ng

Gy bi hiu ng Johnson: b2 = 4k.T.R.fk: hng s Boltzmann

R: in tr ()

T: nhit (0K)

f: di thng ca in tr gy nhiu

Ti mt di tn cho trc, mt ph ca nhiu nhit l hng s: B(f)=B0 vi B0=1/2 k.T

Nhiu " ht " : Xut hin trong cc mch tch cc

Pht sinh bi dng cc ht mang in tnh chy qua liin p to thnh dng in

Nhiu nhit v nhiu ht thng c gi l mtdng n trng (c gi tr trung bnh =0)



Nhiu nhit v nhiu ht thng c gi l mtdng n trng (c gi tr trung bnh =0) nhiu nn

Rt kh loi tr nhiu nn.

Cc loi nhiu ngoi v trong c th c loi tr hoc b lmsuy gim ng k khi s dng mng chn (xung in).



2. T s tn hiu/nhiu

c trng cho s suy gim ca mt tn hiu no

c tnh ho mt h thng truyn tn hiu:

T s tn hiu/nhiu

Biu din theo dB

b

s

P

P

bi

bi

nhiu nh trung sut cng

hiu tn nh trung sut cng

)P

P(log10)(log10

b

s1010dB

3. Tch tn hiu tun hon x(t) b chm trong nhiu b(t):

Phng php hm tng quan

Gi thit s(t)=x(t)+b(t), x(t) v b(t) hon ton c lp



2.6. Lc cc tn hiu tng t

1. Ca s thi gian (ca s hu hn)

Thut ton cho php thc hin vic trch tn hiu, can thip

vo tn hiu hoc thay i bin tn hiu

Ca s thi giane(t) s(t)

s(t) = e(t).f(t)

S() =1/2 [E()*F()]



nh hng ca hm ca s

Khi dng ca s thi gian, ph ca tn

hiu c b nh hng khng?

Kim chng vi hm ca s hu hn CN c rng 2 (- n+ ) v tn hiu vo

e(t)=cos(0t)



2. Lc tn s

Thut ton cho php trch cc thnh phn ph tn hiu,

can thip vo ph tn hiu hoc lm gim mt phn hay

ton phn ph tn hiu.

d)t(h)(e)t(h)t(e)t(s

Lc tn sE() S() S() = E().F()

s(t) = e(t)f(t)= e(t)h(t)

Nu e(t) v h(t) nhn qu:

t

0

t

0

d)(h)t(ed)t(h)(e)t(h)t(e)t(s

Gii thch



ng dng tnh p ng ca b lc theo chui

h(t) = h1(t) * h2(t) *....* hn(t)

)(H)(Hn

1i

i

Gt c n b lc mc ni tip vi nhau, mi b lc c ctrng bi p ng xung hi(t) v hm truyn t Hi()

C th thay th n b lc ny bng mt b lc tngng c p ng xung

iu kin thchin c php tnhtng ng?



Bi tp ng dng

1. Cho mt b lc thng thp c thit k nhhnh BT2-1

a. Vit phng trnh vi phn ca mch

b. Tnh hm truyn t ca mch

c. Tnh p ng xung ca mch

d. Gi thit tn hiu vo e(t) c dng nh hnhBT2-2. Hy tnh tn hiu ra s(t) tng ng.

2. Cho mt b lc thng cao c thit k nhhnh BT2-3

a. Vit phng trnh vi phn ca mch

b. Tnh hm truyn t ca mch

c. Tnh p ng xung ca mch

d. Gi thit tn hiu vo e(t)=at.u(t) vi t0.

Hy tnh tn hiu ra s(t) tng ng.

BT2-1

BT2-2

BT2-3

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BGXLTH_CLC_chuongI-II _ Tin Hieu Lien Tuc_SV