1Chng

4 Ly mu Lng

t

- M ha

Th.S

ng

Ngc

KhoaKhoa in - in T

2Nhc li

Hu ht cc h thng thng tin ngy nay u ldng thng tin s.

u im ca thng tin s: tin cy cao do ckh nng khi phc d liu, linh hot (lp trnhc), d dng lu tr,

Bc quan trng u tin trong qu trnh thngtin s: l qu trnh bin i ngun tin tc thnhmt dng tng thch vi h thng thng tin s c th truyn i.

3S h

thng

thng

tin s

bng

gc (baseband)

M

haPhtiu chxungLy mu Lng

t

Gii

iu ch/Tch

sng

Knh

truyn

ThuLc

thngthp Gii m

Dng

xungChui bit

nh

dng

nh

dng

Tin tc s.

Tin tcvn bn.

Tin tcanalog

Tin tcvn bn

Tin tcanalog

Tin tcdng

s.

Ngun

ch

4nh

dng

tn

hiu tng t

chuyn mt tin tc dng tng t sang mt dng thch hp vi h thng thng tin s, cn thc hin cc bc nh sau: Ly mu (sampling) Lng t (quantization) M ha (Encoding) Truyn d liu di nn (baseband transmission)

5Qu trnh ly mu

Qu trnh ly mu

Bt k tn hiu no cng c th phn tch thnhtng ca cc tn hiu dng sin do vy ta s btu vi vic nghin cu mt tn hiu sin n

6Qu trnh ly mu (tt)

V d qu trnh ly mu

7Qu trnh ly mu (tt)

Cho tn hiu c dng hnh sin x(t) = sin(2f0t) Tn hiu trn c ly mu vi tn s fs=1/Ts

fs: tn s ly mu Ts: chu k ly mu

Suy ra

c gi l tn s tn hiu sau khi ly mu

( ) ( ) ( )

=

== n

TTn

ffnTfnxnTx ss

ss0

00 2sin2sin2sin

sff0

8Qu trnh ly mu (tt)

Nu bt u ly mu ti thi m t = 0, chng tas c mu ti nhng thi im 0, Ts, 2Ts, 3Ts,

=

=

=

=

0

0

0

0

2sin)(

22sin)2(

2sin)(

02sin)0(

TnTnTx

TTTx

TTTx

Tx

ss

ss

ss

#

9Qu trnh ly mu (tt)

Tn s ly mu bao nhiu l ph hp?

-1.1

0

-1.1

0

-1.1

0

Ts=T/10

Ts=T/4

Ts=3/4T

10

Qu trnh ly mu (tt)

Cho hai tn hiu

C hai tn hiu c ly mu tn s fs = 40Hz

Hai tn hiu khc nhau nhng cho cng mt ktqu, nguyn nhn l tn s ly mu khng ph hp

( )( ) Hzfttx

Hzfttx50)100cos(10)20cos(

22

11

====

( ) ( )( ) ( )

=

+=

=

=

=

2cos

22cos

40100cos

2cos

4020cos

22

11

nnnnnxnTx

nnnxnTx

s

s

11

Qu trnh ly mu (tt)

Mt cch tng qut, cc tn hiu c tn s khcnhau kfs, s c tn hiu sau khi ly mu gingnhau

( ) ( ) ( )( ) ( ){ } ( )

=

+=+=

==

nffn

ffkn

ffnxtkfftx

nffnxtftx

ss

s

ss

s

00202

0101

2cos22cos2cos

2cos2cos

12

B ly mu l tng

x Zero-orderhold hO

(t)

=

=n

)nTt()t(s

xa

(t) xS

(t) xQ

(t)

Sample-and-Hold System

-3T -2T -T 0 T 2T 3T 4T 5T

xS

(t)xa

(t)

13

nh l ly mu

c th biu din chnh xc tn hiu x(t) bi ccmu x(n) th Tn hiu x(t) phi c gii hn bng thng ngha l

tn ti tn s cc i (fmax) Tn s ly mu fs 2fmax

fs = 2fmax: tc Nyquist

: tn s Nyquist

Do vy tn hiu phi i qua b lc thng thp

2sf

14

nh l ly mu (tt)

Oversampling tn s ly mu cao hntc Nyquist

Undersampling tn s ly mu thphn tc Nyquist

Critical sampling tn s ly mu bngvi tc Nyquist

15

nh l ly mu (tt)

Tn hiu thoi c bng thng 3.4KHz. Do vy trong in thoi s, tn s ly muthng c chn l 8KHz.

Trong m nhc, tn hiu cht lng cao cbng thng 20KHz. Do vy, a CD nhcc lu vi tn s ly mu l 44.1KHz.

16

V d 4-1

Mt ng dng m thanh c: Tn s ly mu: 44.1Khz Kch thc mu: 16 bits

Tn hiu sau khi ly mu c lu vo aCD (700MB). Xc nh thi lng ti ac lu trong a.

17

Bi tp 4-1

Cho tn hiu lin tc

Xc nh x(n) vi fs = 5000Hz Xc nh fs tn hiu c ly mu ng Vi fs = 5KHz, tm x2(t) chng ph vi x1(t) sau khi

ly mu

( ) ( ) ( ) ( )ttttx 12000cos106000sin52000cos31 ++=

18

Ph ca tn hiu ly muXXCC

(j(j))

N--

NN

11

22/T/TS(jS(j))

0 S-S

SN-N-S

nhnh

hhngng

ccaa

chchngng

phph

SS

> 2> 2

NN

SS

= 2= 2

NN

SS

< 2< 2

NN

PhPh

ccaa

ttnn

hihiuu

nguynnguyn

ththyy

PhPh

ccaa

hhmm

llyy

mmuu

PhPh

ccaa

ttnn

hihiuu

sausau

khikhi

llyy

mmuu

XXSS

(j(j))

19

Ph ca tn hiu ly mu

Nhn xt: Ph ca tn hiu ly mu c dng lp tun

hon ph ca tn hiu trc khi ly mu vcch nhau fs

Ph ca tn hiu ly mu trn c th cto thnh t mt trong cc tn hiu c tn s f+ mfs

Nu tn s ly mu khng tha nh l lymu s xy ra hin tng chng ph.

20

Hin tng

aliasing (chng

ph)

Lc

thng

thp

Tc

Nyquist

aliasing

21

Hin tng

aliasing (chng

ph)

22

Ly mu

Min thi gian Min tn s)()()( txtxtxs = )()()( fXfXfX s =

|)(| fX)(tx

|)(| fX

|)(| fX s)(txs

)(tx

23

Bi tp 4-2

Cho tn hiu lin tc

Xc nh x(n) vi fs = 1.5KHz Xc nh fs tn hiu c ly mu ng Vi fs = 5KHz, tm xa(t) chng ph vi x(t) sau khi

ly mu

( ) ( ) ( ) ( )ttttx 3000cos2000sin21000cos34 +++=

24

Khi phc tn hiu

Nu tc ly mu ln hn tc Nyquist, cth khi phc li tn hiu gc bng mt b lcthng thp c tn s ct bng fm.

Trong min thi gian, tn hiu khi phc s ltch chp ca cc mu ca tn hiu gc vihm sinc (sinc(t) = Sa(t)):

( )

n s

ss T

nTtcnTx sin

25

Ph

ca

tn

hiu ly mu (tt)

26

V d minh ha

Tn hiu thoi

Tn hiu phi thoi

27

V d minh ha

Tn hiu sau b lc thp(SSB=Fs/4=11.025 kHz)

Tn hiu ly mu

Fs/4(new Fs=11.025 kHz)

Tn hiu gc(Fs=44.1 kHz)

28

Qu trnh lng t

Qu trnh lng t s biu din gi tr ly mubi B bit nh phn [b1, b2, , bB]

QuantizerTn

hiu ly muTn

hiu lng

t

B bits/sample

.

.

.

B : s bit ca mt mu

c

2B

mc lng

t

R : tm o ton thang ca b bin i ADC

rng

lng

tBRQ 2=

29

Qu trnh lng t

B bin i A/D lng cc c cc gi trlng t nm trong khong

B bin i A/D n cc c cc gi tr lngt nm trong khong

( ) QRnxR Q 22

( ) QRnxQ 0

30

B lng t lng cc

31

B lng t n cc

32

Sai s trong lng t

33

Sai s trong lng t

Sai s lng t e(nT) = xQ(nT) x(nT) Khi x nm gia hai mc lng t, ty theo x

nm na trn hay na di ca khonglng t m s c lm trn ln hay xung

Sai s lng t22QeQ

12Qerms =

34

Bi tp 4-3

Trong mt ng dng m thanh, tn hiuc ly mu tn s 44KHz. B ADC cR=10V. Xc nh s bit B sai s lng t hiu

dng thp hn 40uV Xc nh sai s hiu dng tht s. Tnh tc bit theo bps

35

Bi tp 4-4

Cho mt tn hiu hnh sin c tn s 1KHz v bin 1 volt, tn hiu ny c lymu vi tn s 8KHz. Tn hiu sau khi ly mu c a n b lng t lng cc, u, 3 bit, R = 2V. Xc nh chui bit sau khi lng t trong mt chu k?

36

V d minh ha

tTs: chu

k

ly mu

x(nTs

): gi

tr

cc

muxq

(nTs

): gi

tr

lng

t

Bin

gii

Mc lng

t

111 3.1867

110 2.2762

101 1.3657

100 0.4552

011 -0.4552

010 -1.3657

001 -2.2762

000 -3.1867

T

mPCM 110 110 111 110 100 010 011 100 100 011Chui PCM

amplitudex(t)

01234567m7

m6m5

m4

m3

m2m1

m0

37

B chuyn i D/A

Qu trnh bin i D/A bao gm hai khichnh sau: B khi phc bc thang (Mch gi) B lc ng ra

Hold circuit Post-filterya

(t)yS

(t)yQ

[n]

38

B chuyn i D/A

Tn hiu trc khi a vo b khi phc bcthang phi c bin i t gi tr nh phnsang gi tr ri rc.

M ha n cc

M ha lng cc

( ) ( )BBQ bbbRny +++= 2...22 2211

( )

+++= 212...22 22

11

BBQ bbbRny

39

Bi tp 4-5

Sau khi lng t mt chui cc mu bngb lng t 3 bit c R=2V ta c chuibit sau y

Xc nh gi tr cc mu khi s dng Lng t n cc Lng t lng cc

40

B chuyn i D/A

B khi phc bc thang s cho ra tn hiutng t nhng cha phng do c thnhphn cao tn

t

Ys

(t)

41

B chuyn i D/A

So snh p ng tn s ca b lc ltng v b khi phc bc thang

B

lc l tngT

-2/T -/T 0 /T 2/T

B

khi

phc bc

thang

42

B chuyn i D/A

B lc post-filter c tc dng b phnchnh lch gia hai b khi phc. png tn s ca b lc post-filter nh sau

-/T 0 /T

1

|Hr~(j)|

43

B chuyn i D/A

44

Lng

t

ha

u v khng u

Lng t ha u: Khng c thng tin thng k bin v c tnh tng quan ca

tn hiu vo. Khng s dng nhng c tnh lin quan n ngi dng. Thch ng vi nhng bin thin nh ca tn hiu v khng cn

tinh chnh cc thng s u vo. D thc hin ng dng trong x l tn hiu, ha v iu khin qu trnh

Lng t ha khng u: S dng thng tin thng k ca tn hiu vo tinh chnh cc

thng s ca b lng t Cho t s SNR tt hn lng t ha u vi cng s mc

lng t. Cc khong lng t c chia khng u v bin trong ding ca b lng t nhng c t s SNR bng nhau.

ng dng ch yu trong tn hiu ting ni.

45

Lng t ha khng u

c thc hin bng cch nn tn hiu trc khi lngt ha u.

my thu, phi din ra qu trnh ngc li, gi l gintn hiu, trnh mo dng tn hiu thu.

Nn

(compress) Lng

t

Knh

truynGin

(expanding)

My

pht My

thu

)(ty)(tx )( ty )( tx

x

)(xCy = x

y

46

Thng

k

bin

tn

hiu ting

ni

i vi ting ni, cc tn hiu nh thng xut hinnhiu hn cc tn hiu ln.

Nu dng lng t ha u s dn n t s thpi vi cc tn hiu b v cao i vi cc tn hiu ln.

iu chnh khong cch lng t s ci thin t s SNR trn ton di ng ca tn hiu.

0.0

1.0

0.5

1.0 2.0 3.0Bin

chuNn ha ca

tn

hiu ting

ni

H

m

m

t

x

c

s

u

t

qNS

qNS

47

Lng t ha u i vi tn hiu m thanh

Vi cc tn hiu c m lng ln hay nh, cng sut nhiu lng t u c gi tr nh nhau.

Nh vy SQNR (Signal to Quantization Noise Ratio) s c gi tr nh i vi cc tn hiu m lng nh v ngc li i vi cc tn hiu m lng ln.

Cn ci thin t s SQNR cho cc tn hiu nh.

48

Lng t ha u i vi tn hiu m thanh

Cn xc nh di bin thin ca cng sut tn hiu vo sao cho trong khong , t s SQNR l chp nhn c i vi ngi s dng.

000

111

001

volts

OK

qu ln vi cc m thanh nh

qu nh vi cc m thanh ln

49

Di ng ca tn hiu

nh ngha: l t s cng sut ti a ca tn hiu so vi cng sut ti thiu ca tn hiu sao cho t s SQNR l chp nhn c.

T s SQNR ph thuc cng sut tn hiu, v cng sut nhiu lng t l khng i v bng 2/12.

Vi tn hiu sin c lng t ha u bng n bit th:

SQNR

6n

+ 1.8 (dB) Cng thc ny cng gn ng i vi tn hiu

m thanh.

50

V d v di ng ca tn hiu

Nu SQNR ti thiu chp nhn c l 30dB, hy tnh di ng ca tn hiu sin lng t ha u 8 bit?

Gii:Dy = (6n

+ 1.8)

30 = 6x8 + 1.8

30 =

= 19.8 dBGi tr ny qu nh i vi tn hiu thoi.

Bi tp: lm li v d trn vi nu dng 12 bit lng t u.

51

Lng t ha khng ng u

tng t s SQNR i vi cc tn hiu c m lng nh, cn gim sai s lng t bng cch tng s mc lng t.

x(t)

t001111

52

V d minh ha

Lng t ha u

1-bit Q.

2-bits Q.

3-bits Q.

4-bits Q.

Lng t ha khng u

1-bit Q.

2-bits Q.

3-bits Q.

4-bits Q.

53

Truyn tn hiu di nn (baseband)

truyn tin tc qua knh truyn vt l, cc t m PCM (codewords) c chuyn thnh cc dng xung (waveforms).

Mi xung mang mt k hiu trong tp M k hiu c th c.

Mi k hiu tng ng vi bit trong chui m PCM.

C nhiu cch khc nhau truyn cc k hiu. l cc phng php iu ch xung.

Mn 2log=

54

Cc

phng

php

iu ch

xung

Chui xung PCM c truyn trc tiptheo mt trong cc dng sau: Nonreturn to Zero-Level (NRZ-L) Nonreturn to Zero Inverted (NRZI) Multilevel (Bipolar AMI) Manchester Differential Manchester B8ZS

55

Nonreturn

to Zero-Level (NRZ-L)

Hai mc in p khc nhau cho hai mclogic 0 v 1

Gi tr in p truyn khng mc 0V V d: -5V miu t mc logic 1 v +5V miu t mc logic 0.

56

Nonreturn to Zero Inverted

Truyn hai mc in p khc khng vo ti mc logic 1.

Invert on 1

0 v

0 v

+ve

-ve

-ve

+ve

1 0 1 1 0 0 1 1 1 0 1 0 0 0 0 0 0 0 1 1 1 0 0 1 0Bi

tp

57

Bipolar-AMI

Bipolar-AMI (Alternate Mark Inversion) Khng c tn hiu truyn (0 volt) m t bit 0 Bit 1 c miu t bi in p dng hay in p m. Gi tr xung m v dng xen k nhau H thng s mt ng b khi khng c tn hiu trong

thi gian di.

1 0 1 1 0 0 1 1 1 0 1 0 0 0 0 0 0 0 1 1 1 0 0 1 0Bi

tp

58

Manchester

Miu t gi tr bit ngay gia chu k xung Nguyn tc:

T thp cao: miu t bit 1 T cao thp: miu t bit 0Chu k

bit

High

Low

1 0 1 1 0 0 1 1 1 0 1 0 0 0 0 0 0 0 1 1 1 0 0 1 0Bi

tp

59

Differential Manchester

Nguyn tc: Truyn bit 0: c s thay i v mc truyn ti u chu k bit. Truyn bit 1: khng c s thay i v mc truyn ti u chu

k bit.

1 0 1 1 0 0 1 1 1 0 1 0 0 0 0 0 0 0 1 1 1 0 0 1 0Bi

tp

60

B8ZS

B8ZS (Bipolar with 8-Zeros Substitution) Ging nh bipolar-AMI nhng khc phcc trng hp truyn nhiu bit 0 lin tip

Nu xut hin 8 bit 0 v xung cui cngtrc l xung dng th 8 bit 0 c mha thnh 000+-0-+

Nu xung cui cng l xung m th 8 bit 0c m ha thnh 000-+0+-

61

B8ZS

xung cui cng l xung m v 8 bit 0 c mha thnh 000-+0+-

1 0 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 1 0Bi

tp

62

Cc dng xung PCM

La chn cc dng xung PCM da trn cc tiu ch sau: c tnh ph (ph mt cng sut v hiu

sut s dng bng thng). Kh nng ng b bit. Kh nng pht hin li. Tnh chng nhiu v giao thoa. phc tp v chi ph thit k h thng.

63

Ph cc tn hiu PCM

64

Cc dng iu ch xung M-ary

Cc dng iu ch xung M-ary khc bao gm: iu ch bin xung M-ary (PAM) iu ch v tr xung (PPM) iu ch rng xung (PWM hay PDM)

iu bin xung PAM: L tn hiu a mc, mi mc bin ng vi 1 k

hiu M-ary (tng ng bng n = log2M bit PCM). Vi cng tc d liu, phng php ny yu cu

bng thng nh hn PCM. Vi cng mc cng sut trung bnh ca xung,

phng php PCM d pht hin d liu hn PAM.

65

V d iu ch PAM

-0.5 0 0.5 1 1.5 2 2.5 3 3.5 4-6

-4

-2

0

2

4

6

8

time

AnalogPAM

Matlab

Demo: pamm.m

66

Cc dng iu ch xung M-ary

Pulse Amplitude Modulation (PAM): A m(t) Pulse Time Modulation (PTM): Td m(t) Pulse Width Modulation (PWM): T m(t)

t

T

Ts

TdA PAMPTM

PWM

67

Cu hi?

Chng 4Ly mu Lng t - M haNhc liS h thng thng tin s bng gc (baseband)nh dng tn hiu tng tQu trnh ly muQu trnh ly mu (tt)Qu trnh ly mu (tt)Qu trnh ly mu (tt)Qu trnh ly mu (tt)Qu trnh ly mu (tt)Qu trnh ly mu (tt)B ly mu l tngnh l ly munh l ly mu (tt)nh l ly mu (tt)V d 4-1Bi tp 4-1Ph ca tn hiu ly muPh ca tn hiu ly muHin tng aliasing (chng ph)Hin tng aliasing (chng ph)Ly muBi tp 4-2Khi phc tn hiuPh ca tn hiu ly mu (tt)Vi du minh hoaVi du minh hoaQu trnh lng tQu trnh lng tB lng t lng ccB lng t n ccSai s trong lng tSai s trong lng tBi tp 4-3Bi tp 4-4V d minh haB chuyn i D/AB chuyn i D/ABi tp 4-5B chuyn i D/AB chuyn i D/AB chuyn i D/AB chuyn i D/ALng t ha u v khng uLng t ha khng uThng k bin tn hiu ting niLng t hoa u i vi tin hiu m thanhLng t hoa u i vi tin hiu m thanhDai ng cua tin hiuVi du v dai ng cua tin hiuLng t hoa khng ng uVi du minh hoaTruyn tin hiu dai nn (baseband)Cc phng php iu ch xungNonreturn to Zero-Level (NRZ-L)Nonreturn to Zero InvertedBipolar-AMIManchesterDifferential ManchesterB8ZSB8ZSCac dang xung PCMPh cac tin hiu PCMCac dang iu ch xung M-aryVi du iu ch PAMCac dang iu ch xung M-aryCu hi?