Ch2_Lay Mau Va Khoi Phuc Tin Hieu

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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2

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

5/22/2010

t = nTs

Tn hiu vo

x(t)

Tn hiu ri rc

xs(t)


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3


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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5


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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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.

5/22/2010


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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:

5/22/2010

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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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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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.

5/22/2010

Bi i X l hi


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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.

5/22/2010

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

Bi i X l t hi


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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:

5/22/2010

3

2 3 2 2 2

25log ( ) log ( / 2) log log 0.322

/ 2 20s s

ff f octave

f

= = =

Bi i X l t hi


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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)

5/22/2010

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

= =

Bi i X l t hi


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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)

5/22/2010

Bi ging: X l s tn hiu


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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.

5/22/2010

0 fs/2

| ( ) |PREH f

f

1

-fs/2



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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)

5/22/2010

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

5/22/2010

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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2.2 B tin lc (tt):b. Khi dng b tin lc l tng

Ph ca tn hiu sau b tin lc X(f).



LdB = 0 dB

5/22/2010

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



LdB = AdB(f = 8 Khz) = -55 dB

5/22/2010

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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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:

5/22/2010

0 4

| ( ) |H f

f (Khz)

35dB/octave

-4



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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)

5/22/2010

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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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

5/22/2010



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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:

5/22/2010

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

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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:

5/22/2010

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

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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:

5/22/2010

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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g g

27


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:

5/22/2010

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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g g

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2.4.2 B khi phc bc thang: to xp xhnh thang

M t:

Qu trnh khi phc:

5/22/2010

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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g g

29


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

5/22/2010



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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).

5/22/2010



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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]

5/22/2010

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]

5/22/2010

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 = + + + +



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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.

5/22/2010

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.

5/22/2010

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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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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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.

5/22/2010

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.

5/22/2010

( )8

1| ( ) | , :

1 0.1a

X f f Khz

f

=

+

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Ch2_Lay Mau Va Khoi Phuc Tin Hieu