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7/31/2019 Ch2_Lay Mau Va Khoi Phuc Tin Hieu
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Bi ging: Xl s tn hiu
5/22/20101
Chng 2
LY MU V KHI PHC TN HIU
Ni dung:
2.1 Ly mu tn hiu2.2 B tin lc2.3 Lng t ha2.4 Khi phc tn hiu tng t2.5 Cc b bin i ADC v DAC
Bi tp
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.1 Ly mu tn hiu: Qu trnh bin i tn hiu lin tc thnh cc mu tn hiu ri rc theo thi gian.
2.1.1 Nguyn l ly mu:
trong: Ts: chu kly mu [giy]
fs = 1/Ts: tn s ly mu [Hz] hay tc ly mu [mu/giy]
La chn hp l gi trca fs l vn quan trng:
fsphi ln biu din y tnh cht ca tn hiu.
fs qu ln s yu cu cao vphn cng, tn b nh,vv
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t = nTs
Tn hiu vo
x(t)
Tn hiu ri rc
xs(t)
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.1.2 M t qu trnh ly mu:M t min thi gian M t min tn s
5/22/2010
X0
X0/Ts
1/Ts
t
x(t)
0
s(t)
0 T 2T 3T 4T 5T t
t
xs(t)
0
X(f)
fM-fM 0
0 fs
( )S f
2fs-fs-2fs
f0 fs
( )S
X f
2fs-fs-2fs
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2.1.2 M t qu trnh ly mu (tt):
Quan h gia ng vo ng ra ca b ly mu:
Trong min thi gian:
Trong min tn s:
Nhn xt:
Qu trnh ly mu to ph rng v hn nhng tun hon vi chu kfS.
Ngha l, ph ca xs(t) chnh l ph ca x(t) v cc lp litn s fs,2fs,vv
5/22/2010
( ) ( ) ( ) ( ) ( )s s s
n
x t x t s t x n T t n T
=
= =
1( ) ( ) * ( ) ( )
S s
nS
X f X f S f X f nfT
=
= =
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.1.3 nh l ly mu Nyquist: Dng b lc thng thp
khi phc tn hiu.
khi phcng th:
trong: fs: tc Nyquist
fs/2: tn s Nyquist
[-fs/2; fs/2]: khong Nyquist.
Nhvy, tcc mu ta c th khi phc ling tn hiu ban u, khi lymu phi chn tc ly mu ln hn hay t nht l bng hai ln thnh phn tns cao nht c trong tn hiu tng t.
nh l Nyquist xcnh gii hn di ca fs.
5/22/2010
0 fs 2fs-fs-2fs
0 0 20-0-20
0 fs
( )SX f
2fs-fs-2fs
2s M
f f
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.1.3 nh l ly mu Nyquist (tt):V d1: Cho tn hiu sau:
x(t) = 4 + 2cos2 t + 6cos8 t (t:ms)
Xc nh gi tr hp l ca fs ?Li gii:
Xc nh cc thnh phn tn s:
f1 = 0 Khz; f2 = 1 Khz; f3 = 4 Khz
Thnh phn tn s cao nht:
fM = max{f1, f2, f3} = f3 = 4Khz.
Chn gi tr fs da vo nh l ly mu Nyquist:
fs P2fM = 2x4 = 8 Khz.
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.1.3 nh l ly mu Nyquist (tt): Gii hn trn ca fs:
Gi s Tp: thi gian xl mi mu dliu (ty thuc vo phn cng).
fp = 1/Tp: tcxl mi mu.
gi trcc mu khng chng ln nhau th:
Tm li, tm gi tr ca fs:
Tc ly mu c trng cho mt vi ng dng:
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s pf f
2 s pf f f
Lnh vc fM fsThoi 4 Khz 8 Khz
Audio 20 Khz 40 Khz
Video 4 Mhz 8 Mhz
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.1.4 Hin tng Alias (Chng ln ph):xy ra khinh l ly mu Nyquist khng tha, tc l: fs < 2fM.
Cc tn hiu c tn s khc nhau c biu din bi cc mu nhnhau khng phn bitc.
V d2: Hai tn hiu c tn s ln lt l: 10 Hz v 90Hzc ly mu tcfs = 100 Hz s cho tp mu nhnhau.
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.1.4 Hin tng alias (tt):nh hng ca hin tng alias c th hin khi khi phc.
Gi sb khi phc l b lc thng thp l tng, tn s ct fc= fs/2.
Khi, tn s khi phc:(*)
(chn m sao cho thnh phn tn s nm trong khong Nyquist: [-fs/2; fs/2])
V d3: Hai tn hiu c tn s ln lt l: f1= 10 Hz v f2= 90Hzc ly mu tc fs = 100 Hz s c ph nhsau. Sau khi khi phc, ta thu c hai thnhphn tn s: 10 Hz v -10Hz.
5/22/2010
mod ; 0, 1, 2,....a s sf f f f mf m= = =
0 fs/2 fs-fs/2-fs f
10 Hz-90 Hz110 Hz
90 Hz-10 Hz
-110 Hz
i i i
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2.1.4 Hin tng alias (tt):
V d3(tt):
Kt qu c c dng cng thc (*), nhsau:
f1a = f1 mod fs = 10mod100 =10 - 0x100 = 10 Hz
f1a = f2mod fs = 90mod100 = 90 - 1x100 = -10 Hz
V d4: Cho tn hiu sau: xa(t) = sin200t (t:giy)
Xcnh tn hiu khi phc ya(t) trong hai trng hp:
a. Tn s ly mu fs = 120 Hz.
b. Tn s ly mu fs = 240 Hz.
Li gii: a. - Khong Nyquist: [-60 Hz, 60Hz]
- Tn s khi phc: fa = f mod fs = 100mod120 =100 - 1x120 = -20 Hz
- Tn hiu thu c: ya(t) = sin2fat = - sin40tkhi phc sai do
khng tha nh l ly mu Nyquist.
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Bi i X l hi
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.1.4 Hin tng alias (tt):V d4: Cho tn hiu m thanh sau:
xa(t) = 2cos10t + cos30t + cos50t + cos90t (t:ms)
Tn hiu ca qua h thng DSP:
a. Xcnh tn hiu x(t)?
b. Xcnh tn hiu khi phc ya(t)?
Bit rng, b tin lc c p ng tn s nhsau.
B qua nh hng ca pha.
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ya(t)
HPRE (f)Tn hiu vo
xa(t)
ADC
x(t)
DSPKhi phcl tngx(n) y(n)
= x(n)
fs= 40 Khz
0 20
| ( ) |PREH f
f
Suy hao
60dB/octave
0
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Chng 2 LY MU V KHI PHC TN HIU (tt)
V d4 (tt):Li gii:
Xcnh cc thnh phn tn s trong tn hiu vo:
f1 = 5 Khz; f2 = 15 Khz; f3 = 25 Khz; f4 = 45 Khz Xcnh tn hiu ng ra b tin lc x(t)
Dng tn hiu x(t):
x(t) = 2|H(f1)|cos(2f1t) + |H(f2)|cos(2f2t) + |H(f3)|cos(2f3t) + |H(f4)|cos(2f4t)
Xcnh cc suy hao bin do b tin lc:
V f1, f2< 20 Khz, nn: |H(f1)| = |H(f2)| =1;
Xcnh |H(f3)| v |H(f4)|:
Khong cch tf3, f4n fs/2 theo octave:
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3
2 3 2 2 2
25log ( ) log ( / 2) log log 0.322
/ 2 20s s
ff f octave
f
= = =
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Chng 2 LY MU V KHI PHC TN HIU (tt)
V d4 (tt):
Mc suy hao so vi di thng (tnh theo dB)f3: (0.322 octave )x (60 dB/octave) = 19.3 dB
f4: (1.17 octave )x (60 dB/octave) = 70.2 dB
Tnh suy hao:
Thay cc gi trvo, ta c:
x(t) = 2cos(10t) + cos(30t) + 0.1084.cos(50t) + 3.09.10-4cos(90t)
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42 4 2 2 2
45log ( ) log ( / 2) log log 1.17
/ 2 20s
s
ff f octave
f
= = =
19.3/20
3| ( ) | 10 0.1084H f= =
70.2/20 4
4| ( ) | 10 3.09 10H f
= =
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Chng 2 LY MU V KHI PHC TN HIU (tt)
Xcnh tn hiu khi phc ya(t):
Tn hiu c dng:
ya(t) = 2cos(2f1at) + cos(2f2at) + 0.1084cos(2f3at) + 3.09.10-4cos(2f4at)
Xcnh cc thnh phn tn s sau b khi phc l tng:
Khong Nyquist: [-20 Khz, 20Khz]
Cc thnh phn:
f1a = f1 mod fs = 5mod40 = 5 Khz
f2a = f2mod fs = 15mod40 = 15 Khz
f3a = f3 mod fs = 25mod40 = -15 Khz
f4a = f4 mod fs = 45mod40 = 5 Khz
Thay vo, ta c:
ya(t) = 2cos(10t) + cos(30t) + 0.1084cos(-30t) + 3.09.10-4cos(10t)
= 2,003.cos(10t) + 1,1084.cos(30t) (t:ms)
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.2 B tin lc: b lc tng tthng thp dng gii hn ph tn hiu ng vochng hintng chng ln ph.
B tin lc l tng:
b lc thng thp l tng c tn s ct fc= fs/2.
loi b tt c cc thnh phn tn s ln hn fs/2
khng xy ra hin tng chng ln ph.
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0 fs/2
| ( ) |PREH f
f
1
-fs/2
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.2 B tin lc (tt): B tin lc thc t:
b lc thng thp c p ng nhhnh v.
khng loi b hon ton cc thnh phn tn s ln hn fs/2 hin tng chng ln ph vn xy ra nhng gim nhiu.
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2.2 B tin lc (tt): Cch la chn cc thng s cho b tin lc thc t:
Chn tn s ct di thng fpass sao cho di thng [-fpass; fpass] cha trn vntm gi trquan tm [-fM; fM].
Chn tn s ct di chn fstop v suy hao di chn Astop sao cho ti thiunh hng ca hin tng alias.
Suy hao ca b lc (theo dB):
(f0: tn s trung tm ca b lc)
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stop s passf f f=
10
0
( )( ) 20 log( )
dB H fA fH f=
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2.2 B tin lc (tt):V d 5: Cho tn hiu tng t x(t) c ph nh sau:
Tn hiu c ly mu fs = 12 Khz.
Xc nh mc chng ln ph:a. Khi khng dng b tin lc.
Ph ca tn hiu sau khi ly mu Xs(f).
Mc chng ln ph vo vng tn hiu quan tm [-4 Khz; 4 Khz] l:
LdB = AdB(f = 8 Khz) = -15 dB
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0 4
| ( ) |aX f
f (Khz)
-15dB/octave
-4
0 4
| ( ) |sX f
f (Khz)
-15dB/octave
-4 8 12 16 -8-12-16
0
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.2 B tin lc (tt):b. Khi dng b tin lc l tng
Ph ca tn hiu sau b tin lc X(f).
Ph ca tn hiu sau khi ly mu Xs(f).
Mc chng ln ph vo vng tn hiu quan tm [-4 Khz; 4 Khz] l:
LdB = 0 dB
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0 4 6
| ( ) |X f
f (Khz)
-15dB/octave
-6 -4
0 4 6
| ( ) |sX f
f (Khz)
-15dB/octave
-6 -4 8 12 16 -8-12-16
0
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2.2 B tin lc (tt):c. Khi dng b tin lc thc t c p ng nh sau:
Ph ca tn hiu sau b tin lc X(f) s
c suy hao ngoi di thng l:15 dB + 40 dB = 55 dB
Ph ca tn hiu sau khi ly mu Xs(f).
Mc chng ln ph vo vng tn hiu quan tm [-4 Khz; 4 Khz] l:
LdB = AdB(f = 8 Khz) = -55 dB
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0 4
| ( ) |H f
f (Khz)
-40dB/octave
-4
0 4
| ( ) |sX f
f (Khz)
-55dB/octave
-4 8 12 16 -8-12-16
0
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.2 B tin lc (tt):d. mc chng ln ph vo di tn quan tm nh hn 50 dB.
Xc nh cc thng s ca b lc?
Chn fpass: fpass = 4 Khz Chn fstop: fstop = fs fpass = 12 4 = 8 Khz
Chn Astop:
Ta c: LdB = AdB(fstop) + Xa(fstop)
Suy ra: Astop = LdB - Xa(fstop)
> 50 15 = 35 dB.
C th chn dng ca b tin lc nhhnh v:
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0 4
| ( ) |H f
f (Khz)
35dB/octave
-4
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.3 Lng t ha (Quantization): qu trnh xp xgi trcc mu ri rcchuyn mt tp cc mu ri rc c
s gi trrt ln thnh mt tp c s gi trt hn.
V tr ca khi lng t ha trong h thng:
Hai ki
u l
ng t
ha: Kiu lm trn (rounding)
Kiu ct bt (truncation)
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Chuyn i ADC (Analog to Digital Conversion)
Lymu
Tn hiu t ngra b tin lc
x(t)
Lng tha
xs(t)
M ha
nh phn B bitxsQ(t)
fs
n khiDSP
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.3 Lng t ha (tt) : c tnh ca b lng tha th hin qua quan h ng vo - ng ra.
V d 6: B lng tha u (uniform quantizer) 3 bit.
Dng lng cc Dngn cc
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2.3 Lng t ha (tt) : Vi b lng tha c tm ton thang R, biu din B bit2B mc lng t.
rng lng t:
Sai s lng t:
hay:
Sai s lng t(quantization error) hay nhiu lng t(quantization noise):bin ngu nhin c phn bu, cc trng bng sai s hiu dng:
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2 BR =
( ) ( ) ( )sQ se t x t x t =
2
12
rm se e
= =
sQ se x x=
0
( )p e
e-/2 /2
1/
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g g
25
Chng 2 LY MU V KHI PHC TN HIU (tt)
2.3 Lng t ha (tt) : Ts SNR ca b lng tha:
(lut 6 dB trn bit)
Nhn xt: B ADC tng thm 1 bit ts SNR tng thm 6 dB.
S bit cng nhiu th nhiu lng tcng nh.
Ts SNR khng ph thuc vo bin tn hiu.V d7: H thngin thoi s: fs=8 Khz; biu din 8 bit/mu; R = 10.
Li gii: Sai s lng thiu dng:
Tc bit:
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6 [ ]SNR B dB=
8
102 11.3 ( )12 12 2 12
B
rm s
Re m V= = = =
. 8 ( / ) 8( / sec) 64sB f bit sam ple sam ple kbps= =
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g g
26
Chng 2 LY MU V KHI PHC TN HIU (tt)
2.4 Khi phc tn hiu tng t : chuyn dng tn hiu ri rc sang dng tn hiu tng t.
Quan h gia ng vo v ng ra:
vi:
Suy ra:
Vy:
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B khi phcTn hiu
y(t)
Tn hiu ng ra
ya(t)
( ) ( ) * ( ) ( ') ( ') 'ay t y t h t h t t y t dt
= = ( ) ( ) ( )
s s
n
y t y nT t nT
=
=
( ) ( ') ( ) ( ' ) 'a s sn
y t h t t y nT t nT dt
=
=
( ) ( ) ( )a s sn
y t y n T h t n T
=
=
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27
Chng 2 LY MU V KHI PHC TN HIU (tt)
2.4.1 B khi phc l tng: b lc thng thp l tng c tn s ct fc= fs/2.
M t:
Qu trnh khi phc:
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0fs/2
| ( ) |POSTH f
f
Ts
-fs/2
h(t)
1
0 t
sin( ) s
s
f th tf t
=
( ) ( ) ( ) ( )aY f H f Y f X f = =
sin ( )( ) ( )
( )
s sa s
n s s
f t nTy t y nT
f t nT
=
=
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.4.2 B khi phc bc thang: to xp xhnh thang
M t:
Qu trnh khi phc:
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0 Ts t
1( ) ( ) ( )sh t u t u t T =
( ) ( )[ ( ) ( )]a s s s sn
y t y nT u t nT u t nT T
=
= f
0
Ts
|H(f)|
fs 2fs-fs-2fs
sinsj fTs
s
s
fTT e
fT
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.4.3 B hu lc: l b lc thng thp, nm ngay sau b khi phc bc thang.
dng loi b cc thnh phn ph nh cn st li sau b khi phc bc thang.
Min thi gian:
Min tn s
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2.4 Khi phc tn hiu tng t : Nhn xt:
Cch khi phc dng b khi phc l tng l khng thc t.
Ng ra sau khi hu lc gn ging ng ra b khi phc l tng
b khi phc bc thang+ b hu lc~ b khi phc l tng.
tng cht lng chuyn i DAC, dng thm b cn bng c p ng tns: HEQ(f) = 1/H(f).
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2.6 Cc b chuyn i ADC v DAC:2.6.1 B chuyn i DAC B bit:
S khi:
Cc loi chuyn i:
Dng nhphn n cc (unpolar natural binary):
V d 8: b = (0,0,,0) xQ = 0 [V]
b = (0,0,,1) xQ = R.2-B = Q [V]
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LSB
xQ(t)B chuyn iDAC
Ng vo
B bit
b1
b2b3
b4
bB
MSB
l mt trongs 2B gi tr
mc lng t
trong tm tonthang R
1 2 31 2 32 2 2 .... 2
BQ Bx R b b b b
= + + + +
1 2 3[ , , ..., ]Bb b b b b=
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2.6.1 B
chuyn
i DAC B bit (tt) Dng nhphn offset lng cc (polar offset binary):
tm gi trbdch i R/2.
Dng b 2 (twos complement): (ly b bit c trng s ln nht)
V d 9: Mt b chuyn i DAC: B=4 bit; R = 10 V. Dliu: b = [1 0 0 1]
Dng 1: xQ = 10[1x2-1+0x2-2+0x2-3+1x2-4]
=10x[1/2+1/16] = 5.625 [V]
Dng 2: xQ = 10[1x2-1+0x2-2+0x2-3+1x2-4- 0.5] = 0.625 [V]
Dng 3: xQ = 10[0x2-1+0x2-2+0x2-3+1x2-4- 0.5] = - 4.375 [V]
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1 2 3
1 2 32 2 2 .... 2 0.5B
Q Bx R b b b b = + + + +
1 2 3
1 2 32 2 2 .... 2 0.5B
Q Bx R b b b b = + + + +
Bi ging: Xl s tn hiu
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Chng 2 LY MU V KHI PHC TN HIU (tt)
2.6.2 B
chuyn
i ADC: S khi:
BADC tc cao (flash ADC):
V d 10: B flash ADC 2 bit.Hnh v minh ha khi gi tr
mu ng vo Vin = 3V
th ng ra 10.
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B bit
ng ra
LSB
Mu d liuvo x B chuyn i
ADC
b1
b2
b3
b4
bB
MSB
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Chng 2 LY MU V KHI PHC TN HIU (tt)2.6.2 B chuyn i ADC (tt):
BADC xp x lin tip (Successive Approximation ADC):
S khi:
Nguyn tc hotng:
Tt c cc bit trong thanh ghi (SAR) c khinggi tr[0,0,.,0]. Ln lt cc bitc bt ln kim tra, btu tbit b1 (MSB).
Trong mi ln bt bit, SAR bi gi trsang DAC. DAC to ra xQ. B so snh sxcnh ng ra c=0 hay 1. Nu c = 1 bitc ginguyn, ngc li bt v 0.
Sau B ln kim tra, SAR gigi trng b=[b1,b2,,b3]gi ra output.
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x
DAC
+ Thanh ghi
xp x lin tip
xQ
c=0/1
B bit
ng ra
b1 b2 b3 b4 bB
b1 b2 b3 b4 bB
_B so snh:
x PxQ: c = 1
x < xQ: c = 0
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Chng 2 LY MU V KHI PHC TN HIU (tt)
V d 11: BADC xp x lin tip: tm ton thang R =10V; m ha B = 4 bit.Lng tha kiu ct bt; DAC dng loi chuyn i nhphn offset.
Xcnh gi trng ra khi mu ng vo x = 3.5V.
Li gii: Lp bng hot ng nh sau:
Kiu offset:
Ln lt bt v test cc bit:
b=1000: xQ = 10(1/2-1/2)
= 0
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Chng 2 LY MU V KHI PHC TN HIU (tt)
Bi tp:
2.1 (bi 3.3.1 trang 100) Cho tn hiu m thanh sau:
xa(t) = sin10t + sin20t + sin60t + sin90t (t:ms)
Tn hiu ca qua b tin lc, c ly mu tc fs = 40 Khz, v sau ca qua mch khi phc l tng. Xcnh tn hiu khi phc ya(t)?
a. Khng dng b tin lc.
b. Dng b tin lc l tng c tn s ct 20 Khz.
c. Dng b tin lc c p ng tn s nhsau.
B qua nh hng ca pha.
2.2(bi 3.3.2 trang 101) Khong tn s quan tm trong tn hiu ting ni l
[0; 3.4 Khz]. Bn ngoi khong ny tn hiu suy gim dB/decade.Tn hiunyca qua b tin lc c p ng phngn fM, ri suy gimdB/decade. Hy chng t rng, mc chng ln ph vo di tn quan tm
nh hn A dB th tc ly mu ti thiu l: fs = fM+10A/(+)
fM.
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0 20
| ( ) |PREH f
f
Suy hao
48dB/decade0
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Bi tp:
2.3 (bi 3.3.3 trang 101) Mt tn hiu tng tc di tn quan tm [0,20Khz],v c phc m t nhsau
Tn hiu c ly mu tc fs. Ngi ta mun mc chng ln ph vo ditn quan tm phi nh hn 60 dB. Hy xcnh gi trca fs tha mn yu
cu trn nu khng dng b tin lc.2.4 (bi 3.5.2 trang 102) Mt tn hiu tng tsau khi qua b tin lcc ly
mu tc fs = 8 Khz. Tn hiu s sau c lc dng b lc s thngthp l tng fc= 1 Khz. Tn hiu s ng ra ca n mch khi phc
hnh thang rin b hu lc. Hy xcnh cc thng s ca b hu lcmc ph nh c gim t hn 40 dB.
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( )8
1| ( ) | , :
1 0.1a
X f f Khz
f
=
+
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