Chap.07 Echantillonnage Et Reconstruction Des Signaux Analogiques

8/10/2019 Chap.07 Echantillonnage Et Reconstruction Des Signaux Analogiques

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Filtre Q N

AP

x(t)xa(t) xe(t) yq(t) y(t)y[n]x[n]Filtre

Te CAN

Te

P

Temps

continu discret

Amplitude

continue

discrte

t

x(t)

Signal analogique

(a)

t

xe(t=nTe)

Signal chantillonn

(b)

n

xq[n]

Signal numrique

(c)

t

xq(t)

Signal numrique maintenu

(d)

A

N

N

A

Te

Te

Q


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x(t)

xe(t= nTe)

Te

xq[n]

xq(t)

Te = 0.5 [msec]

x(t) =U0exp(t/) (t) U0= 1 [V] = 1 [ms]

x(t)

xq[n]

xe[n] xq[n] q[n]

0 5.12 [V]

28 = 256


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xe[n] xq[n]

q[n] q[n]

0

0.02

0.04

0.06

5.12

5.10

5.08

5.

11

0.

05

0.

03

0.

01

0

1

2

3

256

255

254

x

[V]

xqq

(a)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 103

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

temps sec

(b)

x(t)

x(t)

Te(t) Te

xe(t)

xe(t) =x(t) Te(t)

x(t)


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t

x(t)

xe(t) = x(t) .Te(t)

t

Te(t)

t

Te

1

t

x(t)

Te

A

B


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Te

x(t)

Te(t)

X(jf)

x(t)

D(jf)

xe(t) =x(t) Te(t) Xe(jf) =X(jf) D(jf)

Te(t)

t

Te

1

0

D(jf) = fe(f)

f

fe

1/Te

0

Te

1

Te(t) Te

fe(f)

fe= 1/Te

1/Te

Te(t)

Te(t) =+

k=

D(jk) exp (+j2 kfet) avec fe= 1

Te

D(jk)

Te(t)

D(jk) 1Te

+Te/2Te/2

(t) exp (j2 kfet) dt= 1Te

0+0

(t) 1 dt= 1Te


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D(jf) = 1

Tefe(f)

X(jf)

D(jf)

fe

X(jf)

fe

Xe(jf) =X(jf) D(jf) = 1Te

+m=

X(j(fm fe))

X(jf)

fe

t

x(t)

xe(t)

tTe

f

X(f)

f

Xe(f)

+fe-fe

x(t)

f0 = 3 [kH z] fe= 8 [kH z]

x(nTe)

x(t)

m fef0 = 3,5,11,13,19,

fe/2 = 4 [kH z]

f0 = 3 [kH z]


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0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

1

0.5

0

0.5

1

temps [ms]

x(t),x(nT

e)

15 10 5 0 5 10 15

0

0.1

0.2

0.3

0.4

0.5

fe

+fe

frquence [kHz]

|Xe

(jf)|

fe>2 f0

fe

x(t)

fe/2

fN=fe/2

fe/2

fapp= m fe f0, m = 0


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xe(t)

tTe

t

x(t)

xe(t)

tTe

f

X(f)

f

Xe(f)

+fe-fe

xe(t)

tTe

Xe,k(f)

-fe

f

Xe(f) = Xe,k(f)

+fe-fe

0 2 4 6 8 10 12 14 16 18 20

1

0.5

0

0.5

1


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0 2k 4k = fe/2

6k

10k

14k

fe= 8k

16k

12k

-2k-4k

-8k

-16k

-6k

4 8 f

[kHz]

fe/2-fe -fe/2 fe

X(f)

62-2-6

fe/2< f


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0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

1

0.5

0

0.5

1

temps [ms]

x(t),x(nT

e)

15 10 5 0 5 10 15

0

0.1

0.2

0.3

0.4

0.5

fe

+fe

frquence [kHz]

|Xe

(jf)|

fapp =|f0 fe| = 3 [kH z]

fN = fe/2 = 4 [kH z]


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+m fe f0 5 m fe f0 5

fe/2 =

4 [kH z]

fapp= 3 [kH z]

T0 = 1 [ms]

f0 =1 [kH z]

fe= 12.8 [kH z]

fe = 12.8 [kH z] f0 = 1 [kH z]

fapp= m fe k f0

m= 1

fapp=

12.8

(1, 3, 5, 7, 9, 11, 13, 15,

)

m= 1, k= 1,

fN=fe/2 = 6.4 [kH z]

T0 = 1 [ms] t= 0.2 [ms]

fe= 16 [kH z]

f0= 1 [kH z] 1/t= 5 [kH z]

fe


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2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

0

0.5

1

x

(t),xe

(t)

20 15 10 5 0 5 10 15 20

0

0.1

0.2

X(jf)

20 15 10 5 0 5 10 15 20

0

0.1

0.2

0.3

X(j(f

kfe))

20 15 10 5 0 5 10 15 20

0

0.1

0.2

0.3

+fe/2f

e/2

X

e(jf)

t, f


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A = 10 V

= 0.2 msec

Te= /2 = 0.1 msec

f = 0

f=fc

x(t) =A exp(t/) (t)

X(jf) =A

1 +j2 f

xe(t)

Xe(jf) = 1

Te

+k=

X(j(f k fe))

= 1

Te

+k=

A

1 +j2 (f k fe)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 103

0

0.2

0.4

0.6

0.8

1

temps [sec]

x(t)x[n]

10 8 6 4 2 0 2 4 6 8 100

0.5

1

1.5

2

x 104

frquence [kHz]

X(f)X

e(f)


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Xe(jf)

Te = /2

fc= 796 [Hz]


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fe>2 fmax

Te 2n1

1 +

fc fe

fc

2m>

2n12

fc fe

fc

2m

>

2n12

fc fe

fc

m> 2n1

fe > fc

1 +

2n11/m

m

n

fe/fc

n

1

2

fe/fc

12

fe (3 5) fc


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0 1 2 3 4 5 6 7 8 9 10

8

6

4

2

0

2

4

6

8

temps

amplitude

x(t)

g(t) =sin( fe t)

( fe t)

t= n Te


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x[n]

xa(t) =+

n=x[n] sin( fe (t n Te))

( fe (t

n Te))

3 2 1 0 1 2 30.5

0

0.5

1

Interpolateur idal

3 2 1 0 1 2 30.5

0

0.5

1

3 2 1 0 1 2 30.5

0

0.5

1

temps [Te]

x[n]

x[n]

h(t)

Te


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0 50 100 150 200 250 3000.2

0

0.2

0.4

0.6

0.8

1

1.2Interpolateur dordre zro

0 50 100 150 200 250 3000.2

0

0.2

0.4

0.6

0.8

1

1.2Interpolateur idal

temps

h(t) =

1 si 0

t < Te

0 sinon

H(jf)

h(t)

H(jf) =Tesin( f Te)

( f Te) exp(j f Te)

fe/2

xs(t)

fe/2


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5 4 3 2 1 0 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1

(t)

5 4 3 2 1 0 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1

temps [Te]

h(t)

5 4 3 2 1 0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

frquence [fe]

module

Te x

CNAidal

5 4 3 2 1 0 1 2 3 4 5

1

0.5

0

0.5

1

phase/

frquence [fe]


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Systme

numrique

x[n] y[n]x(t) NA

AN

ys(t)FAR FLx0(t) y(t)

x0(t)

Xe(f)

X0(f)

0 10 20 30 400.05

0

0.05

0.1

0.15

0.2

Signal x0(t)

0 10 20 30 400.05

0

0.05

0.1

0.15

0.2

temps [Te]

Signal x[n]

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

Spectre de x0(t)

X0(f)

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

frquence [fe]

Spectre de x[n]

X

e

(f)

x(t)

Xe(f)


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fc X0(f)

0 10 20 30 400.05

0

0.05

0.1

0.15

0.2

Signal filtr x(t)

0 10 20 30 400.05

0

0.05

0.1

0.15

0.2

temps [Te]

Signal x[n]

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

Spectres

FAR

X0(f)X(f)

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

frquence [fe]

Spectre Xe(f)

X0(f)

y[n]

Ys(f) = Y(f) B(f)

y[n]

ys(t)

fe/2

1/B(f)

f

fe/2


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0 10 20 30 400.05

0

0.05

0.1

0.15

0.2

Signal y[n]

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

Spectre de y[n]

Y(f)

0 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1

1.2

Bloqueur

temps [Te]

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

frquence [fe]

Spectre du bloqueur

B(f)

0 10 20 30 400.05

0

0.05

0.1

0.15

0.2

Signal ys(t)

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

Ys(f) = Y(f) B(f)

0 10 20 30 400.05

0

0.05

0.1

0.15

0.2

temps [Te]

Signal y(t)

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1.2

frquence [fe]

Spectres

FL

X(f)Y(f)


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Y(f) X(f)


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fe= 16 [kH z]

f

20 15 10 5 0 5 10 15 20

0

0.05

0.1

f [kHz]

X(jf)[V/H

z]

xa(t) = cos(2 1000 t)

T0 f0

xa(t) Xa(jf)

xa(t) Te =T0/4

Xe(jf) x[n] Xe(jf)

Te = 3 T /4

Te=T0/2

A = 10 [V]

T0 = 1 [msec]

t= T0/4 Te=T0/20

x(t)

xe(t)

X(jf)

Xe(jf)

X(jf)

X

e(jf)

f= 3 [kH z]

Xe(+j3) =X(+j3) + X(j17) + X(+j23) +


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A= 5 [V]

T0 = 1 [msec] fe= 8 [kH z]

x(t)

xe(t)

X(jk) = (1)k+1 A/(jk), k= 0

X(jf)

Xe(jf)

X(jf)

Xe(jf) f= 1 [kH z]

xa(t) = 2 cos(100 t) + 5 sin

250 t +

6

4 cos(380 t) + 16 sin

600 t +

4

fe

fe = 3 fe,min

xe(t)

xa(t) = cos(2 240 t) + 3 cos

2 540 t +6

fN=fe/2

fapp

x(n)

fc=fe/2 ya(t)

T0= 1 [ms]

A = 1 [V]

fe = 9.8 [kH z]


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0 1 2 3 4 5 6 7 8 9 10

1

0.5

0

0.5

1

Signal chantillonn xe(t)

temps [ms]

x(t)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 540

30

20

10

0Spectre thorique (o) et spectre d au repliement spectral ()

frquence [kHz]

|X(jf)|[dB]

f0 = 1

fN

x(t) =eat

(t)

fe

Xe(jf) = 1

a +j2f +

+k=1

2 (a +j2f)

(a +j2f)2 + (2 kfe)2

y(n) =19

m=0h(m) x(n m)


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1

2

1

2

x(t) = 4 cos(2 300 t) 2 cos(2 900 t) [V]

1

2

104

x(t)

A=

f0= 8

f fc

Ar f=fc

1

2

fc

f=f

c

fe= 13.7 fc


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x(t)

A = 1 0 [

]

f =

1 [

]

1 [

]

10 [

]

x(t), xe[n], xq(t)

e(t)


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Chap.07 Echantillonnage Et Reconstruction Des Signaux Analogiques

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Transcript of Chap.07 Echantillonnage Et Reconstruction Des Signaux Analogiques