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Bi ging: X l s tn hiuChng 5

BIN I ZNi dung:5.1 Bin i Z 5.1.1 nh ngha bin i Z 5.1.2 Cc tnh cht ca bin i Z 5.1.3 Gin cc-khng 5.2 Bin i Z ngc 5.2.1 Phng php phn tch thnh chui ly tha 5.2.2 Phng php phn tch thnh phn thc s cp 5.3 Phn tch h thng dng bin i Z Bi tp

5/22/2010

1

Bi ging: X l s tn hiuChng 55.1 Bin i Z: l php chuyn tn hiu sang min Z thun tin trong phn tch, x l. bin i Z c vai tr nh php bin i Laplace trong mch tng t. c dng tnh ton p ng ca h thng LTI, thit k cc b lc,vv... 5.1.1 nh ngha: Bin i Z ca mt tn hiu ri rc x(n):+

BIN I Z

X (z) =K hiu:

n =

x(n) z nhay:

(z: bin phc)

Z x(n) X ( z )

X ( z ) = Z [ x ( n) ]

Vng hi t ca bin i Z (ROC: Region Of Convergence) ROC l tp hp nhng gi tr ca Z lm cho X(z) c gi tr hu hn.

ROC =5/22/2010

{z

| X (z) }2

Phi ch r ra khi ni n bin i Z.

Bi ging: X l s tn hiuChng 55.1 Bin i Z (tt): V d 1: Xc nh bin i z ca cc tn hiu sau a. x(n) = {1,2,5,7,0,1} b. x(n) = anu(n) c. x(n) = -anu(-n-1) d. x(n) = Li gii: a. T nh ngha: X(z) = z2 + 2z + 5 + 7z-1+ z-3 ; b. Ta c: ROC: z 0; z anu(n) bnu(-n-1)ReZ

BIN I Z (tt)

ROC

ImZ

-1

0

a

1

X (z) =Nu: |az-1||a| th:

X (z) =5/22/2010

1 1 a z 1

ROC: |z| > |a|3

Bi ging: X l s tn hiuChng 55.1 Bin i Z (tt): c. Ta c: BIN I Z (tt)+ 1 1 +

X (z) =

n =

x(n) z

n

=

n =

a zn

n

=

n=

a

n

z = ( a 1 z ) nn n =1

Nu: |a-1z|| z |> 0.5 X(z) = 4z 1 10.5z2

5/22/2010

7

Bi ging: X l s tn hiuChng 5c. Vi phn trong min Z: BIN I Z (tt) 5.1.2 Cc tnh cht ca bin i Z (tt):

dX ( z ) x(n) X ( z ) nx(n) z dzV d 4: Tm bin i Z ca tn hiu sau: Vit li x(n):

x (n) = na n u ( n) x (n) = nx1 (n), x1 (n) = a n u ( n)

p dng cp bin i c bn:n

p dng tnh cht trn:

1 x1 (n) = a u ( n) X 1 ( z ) = , | z |>| a | 1 1 az

dX1(z) d 1 az1 ; | z |>| a | X(z) =z =z = 2 1 dz dz 1az (1az1 )5/22/2010

8

Bi ging: X l s tn hiuBIN I Z (tt) 5.1.2 Cc tnh cht ca bin i Z (tt): d. Tch chp:

Chng 5

x1 ( n ) X 1 ( z ) x ( n ) = x1 ( n ) * x 2 ( n ) x ( z ) = X 1 ( z ) X 2 ( z ) x2 ( n ) X 2 ( z )chuyn i php tch chp trong min thi gian sang php nhn thng thng trong min Z thun tin trong phn tch h thng. V d 5: Tnh tch chp ca hai tn hiu sau:

x1 (n) = {1, 2,1};Ta c:

x2 (n) = u ( n) u (n 6)ROC: z 0;

X1(z) = 1- 2z-1 + z-2;

X2(z) = 1+ z-1 + z-2 + z-3 + z-4 + z-5; ROC: z 0; p dng tnh cht trn: X(z) = X1(z)X2(z) = (1- 2z-1 + z-2)(1+ z-1 + z-2 + z-3 + z-4 + z-5) = 1- z-1 - z-6 + z-7 Suy ra:5/22/2010

x(n) = {1,-1,0,0,0,0,-1,1}

9

Bi ging: X l s tn hiuChng 5e. o thi gian: BIN I Z (tt) 5.1.2 Cc tnh cht ca bin i Z (tt):

x(n) X ( z) x(n) X ( z1 )V d 6: Tm bin i Z ca tn hiu sau:

1 x ( n) = u ( n) 3 n 1 y (n) = x ( n) = u (n) = 3n u (n) t: 3 p dng cp bin i c bn: 1 y ( n) Y ( z ) = , | z |> 3 1 1 3zp dng tnh cht trn:

n

1 ; | z | 1

Chia a thc c dng ly tha:13

5/22/2010

Bi ging: X l s tn hiuChng 5Li gii: Chia a thc c dng ly tha: BIN I Z (tt) 5.2.1 Phng php khai trin thnh chui ly tha (tt):

X (z) =

1 1 1 .5 z 1 + 0 .5 z 2

= 1+

3 1 7 2 z + z + ..... 2 4Khng cho dng biu thc khp kn ca x(n)

Suy ra gi tr chui x(n):

3 7 x ( n ) = 1, , , ... 2 4

5.2.2 Phng php khai trin thnh cc phn thc s cp: Biu din X(z) thnh dng sau: Lc :

X (z) =

ak =0

N

k

X k (z)

trong : Xk(z) l cc biu thc c bin i Z ngc xk(n) bit.

x(n) =

ak =0

N

k

xk (n )14

5/22/2010

Bi ging: X l s tn hiuChng 5BIN I Z (tt) 5.2.2 Phng php khai trin thnh cc phn thc s cp (tt): V d 9: Tm bin i Z ngc ca tn hiu sau:

X (z) =Li gii:

1 1 1 .5 z1

+ 0 .5 z

2

,

R O C :| z | > 1

a v dng tng cc phn thc s cp:

X (z) =

1 1 1 .5 z 1 + 0 .5 z 2

1 (1 z 1 )(1 0 .5 z 1 ) 2 1 = 1 z 1 1 0 .5 z 1 =

Mc khc,p dng cp bin i Z c bn:

1 ,| Z |> 1 u (n) 1 1 1 z , | z |>| a | a nu (n) 1 1 az 1 (0.5) n u ( n ) ,| z |> 0.5 1 1 0.5 z Suy ra:5/22/2010

x(n) = 2u (n) (0.5) n u (n)

15

Bi ging: X l s tn hiuChng 5Gi s X(z) c dng hu t: BIN I Z (tt) Phng php a v tng cc phn thc s cp:

N ( z 1 ) X ( z) = D( z 1 )

Trng hp 1: (bc t s nh hn mu s) xt 2 kh nng D(z) ch c cc nghim thc n, tc l c th biu din:

N ( z 1 ) N ( z 1 ) = X ( z) = 1 D( z ) (1 p1 z 1 )(1 p2 z 1 )(1 p3 z 1 )....... A3 A1 A2 = + + + .... 1 1 1 1 p1 z 1 p2 z 1 p3 ztrong , cc h s c xc nh nh sau:

Ai = (1 p i z 1 ) X ( z ) 5/22/2010

z = pi16

Bi ging: X l s tn hiuChng 5BIN I Z (tt) V d 9: Tm bin i Z ngc ca tn hiu sau:

2 2 .0 5 z 1 X (z) = 1 2 .0 5 z 1 + z 2Biu din thnh tng cc phn thc s cp:

A1 A2 2 2.05 z 1 2 2.05 z 1 X ( z) = = = + 1 2.05 z 1 + z 2 (1 0.8 z 1 )(1 1.25 z 1 ) (1 0.8 z 1 ) (1 1.25 z 1 )Xc nh cc h s:

A1 = (1 0.8 z ) X ( z ) 1 1

z = 0.8

2 2.05 z 1 = =1 1 1 1.25 z z =0.8

2 2.05 z 1 A2 = (1 1.25 z ) X ( z ) z =1.25 = =1 1 0.8 z 1 z =1.25 Cc bin i Z ngc c th c:

(0.8)n u(n) + (1.25)n u(n), | z |>1.25 x(n) = (0.8)n u(n) (1.25)n u(n 1), 1.25 >| z |> 0.8 (0.8)n u(n 1) (1.25)n u(n 1), | z |< 0.8

5/22/2010

17

Bi ging: X l s tn hiuChng 5BIN I Z (tt) Phng php a v tng cc phn thc s cp: D(z) c cc nghim thc bi, tc l c th biu din:

N(z1) N(z1) X(z) = 1 = D(z ) (1 pz1)(1 p2z1)...(1 pk z1)h...... 1 Ak Ak A A A 1 2 hk 1 2 = + + + +...+ +... 1 1 1 1 2 1 h 1 pz 1 p2z 1 p3z (1 p3z ) (1 p3z ) 1trong , cc h s c xc nh nh sau:

Ai = (1 p i z 1 ) X ( z )

z = pi

;i kz= pk

1 dh j (1 p k z1)h X (z) Ajk = (h j)! dzh j 5/22/2010

; j =1,..., h18

Bi ging: X l s tn hiuChng 5BIN I Z (tt) Phng php a v tng cc phn thc s cp: Trng hp 2: (bc t s bng bc mu s)

N ( z 1 ) N ( z 1 ) = X ( z) = 1 D( z ) (1 p1 z 1 )(1 p2 z 1 )(1 p3 z 1 )....... A3 A1 A2 = A0 + + + + .... 1 1 1 1 p1 z 1 p2 z 1 p3 ztrong , cc h s c xc nh nh sau:

A0 = [ X ( z )]z =0 ; Ai = (1 p i z 1 ) X ( z )

z = pi

V d 10: Tm tt c cc bin i Z ngc c th c ca X(z):

1 + z + 1 0 z 2 X (z) = 0 .2 5 + z 25/22/2010 19

Bi ging: X l s tn hiuChng 5BIN I Z (tt) Biu din thnh tng cc phn thc s cp:

A1 A2 1 + z + 10 z 2 10 + z 1 z 2 X ( z) = = = A0 + + 0.25 + z 2 1 0.25 z 2 1 0.5 z 1 1 + 0.5 z 1Xc nh cc h s:

A0 = [ X ( z ) ]

z =0

10 z 1 z 2 = =4 2 0.25 z z =0z = 0.5

A1 = (1 0.5 z 1 ) X ( z ) A2 = (1 + 0.5 z 1 ) X ( z ) Suy ra:

=4

z =0.5

=2

X ( z) = 4 +

4 2 + 1 0.5 z 1 1 + 0.5 z 1

Cc bin i Z ngc c th c:

4 (n) + 4(0.5)n u(n) + 2(0.5)n u(n); | z |> 0.5 x(n) = 4 (n) 4(0.5)n u(n 1) 2(0.5)n u(n 1); | z |> 0.5 5/22/2010

20

Bi ging: X l s tn hiuChng 5BIN I Z (tt) Phng php a v tng cc phn thc s cp: Trng hp 3: (bc t s ln hn mu s) Chia t s cho mu s a v dng:

N ( z 1 ) R( z 1 ) 1 = Q( z ) + X ( z) = 1 D( z ) D( z 1 )Vic tm bin i Z ngc ca Q(z) l d dng, cn vi a thc cn li dng trng hp 1. V d 11: Tm tt c cc bin i Z ngc c th c ca X(z):

6 + z 5 X (z) = 1 0 .2 5 z 2Biu din thnh tng cc phn thc s cp:

6 + z 5 6 + 1 6 z 1 1 3 X (z) = = 16 z 4 z + 2 1 0 .2 5 z 1 0 .2 5 z 2Xc nh cc h s: (tng t trng hp 1)..21

5/22/2010

Bi ging: X l s tn hiuChng 5BIN I Z (tt) 5.3 Phn tch h thng dng bin i Z: Xt h thng ri rc c p ng xung h(n). Bin i Z ca p ng xung c gi l hm truyn (transfer function) ca h thng. Hm truyn ca h thng ri rc:

H (z) =Quan h gia ng vo- ng ra:

n =

+

h(n) z

n

H(z) thng c s dng m t v phn tch h thng ri rc

H thngTn hiu vo

ri rc H

Tn hiu ra

x(n) X(z)5/22/2010

y(n)=h(n)*x(n) Y(z)=X(z)H(z)22

Bi ging: X l s tn hiuChng 5Tnh n nh v nhn qu: Nhn qu: H thng LTI nhn qu: h(n) = 0, n3. p ng xung ca h thng:

Lc ny, h thng s khng n nh do ROC khng cha vng trn n v.5/22/2010 24

1 h(n) = u (n) + 2.3n u (n) 2

n

Bi ging: X l s tn hiuChng 5Bi tp: 5.1 (bi 8.1.3 trang 311) 5.2 (bi 8.2.1 trang 312) 5.3 (bi 8.2.2 trang 312) 5.3 (bi 8.2.3 trang 312) 5.4 (bi 8.2.9 trang 313) 5.5 (bi 8.2.11 trang 313) 5.6 (bi 8.3.6 trang 315) 5.7 (bi 8.3.9 trang 315) 5.8 (bi 8.4.1 trang 315) 5.9 (bi 8.5.2 trang 316) 5.10 (bi 8.5.3 trang 316)25

BIN I Z (tt)

5/22/2010