Chuong 4 Lay Mau - Luong Tu - Ma Hoa
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Transcript of Chuong 4 Lay Mau - Luong Tu - Ma Hoa
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1Chng
4 Ly mu Lng
t
- M ha
Th.S
ng
Ngc
KhoaKhoa in - in T
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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.
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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)
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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
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6Qu trnh ly mu (tt)
V d qu trnh ly mu
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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
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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
#
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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
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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
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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
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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)
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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
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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
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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.
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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.
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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 ++=
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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))
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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.
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20
Hin tng
aliasing (chng
ph)
Lc
thng
thp
Tc
Nyquist
aliasing
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21
Hin tng
aliasing (chng
ph)
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22
Ly mu
Min thi gian Min tn s)()()( txtxtxs = )()()( fXfXfX s =
|)(| fX)(tx
|)(| fX
|)(| fX s)(txs
)(tx
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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 +++=
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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
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25
Ph
ca
tn
hiu ly mu (tt)
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26
V d minh ha
Tn hiu thoi
Tn hiu phi thoi
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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)
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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=
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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
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30
B lng t lng cc
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31
B lng t n cc
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32
Sai s trong lng t
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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 =
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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
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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?
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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
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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]
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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
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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
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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)
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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)|
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43
B chuyn i D/A
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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.
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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=
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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.
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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+-
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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.
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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.
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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
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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?