BIN I Z NGHCH

1. Tm tt l thuyt 0

0 0( ) ( ) ( )n

x n n u n n z X z

0 0

0

1

0( ) ( ) ( )n nm

m n

x n n x m z z z X z

2. Mt vi v d

2.1. V d 1:

Cho h thng LTI vi p ng xung n v () = 0.5(). Tm p ng () ca h thng vi u vo l () = ().

Gii p:

( ) ( ) ( )Y z X z H z

( )1 0.5

z zY z

z z

( )

( 1)( 0.5) 1 0.5

Y z z A B

z z z z z

1 2 0.5 10.5 1

z zA z B z

z z

( ) 2 1 2( )

1 0.5 1 0.5

Y z z zY z

z z z z z

( ) 2 ( ) 0.5 ( )ny n u n u n

2.2. V d 2:

Xt h thng LTI vi hm biu din tng quan ng nhp v ng xut nh sau:

() 0.5( 1) = ()

Kch thch h thng vi tn hiu () = (). Bit (1) = 0, xc nh p ng () ca h thng.

Gii p:

11 1 1 1 1

1

( 1) ( ) ( ) ( 1) ( )m

m

y n y m z z z Y z y z z z Y z

1( ) 0.5 ( )1

zY z z Y z

z

( )

( 0.5)( 1)

Y z z

z z z

( ) 2 ( ) 0.5 ( )ny n u n u n

3. Bi tp cng c l thuyt

3.1. S dng bin i Z tnh p ng xung n v ca h thng :

( ) ( 2) ( )y n y n x n , vi y(-2) = y(-1) = 0

3.2. Xt h thng c

(2 3)( )

( 1)( 2)

z zH z

z z

Vi ROC |z|>2

tm h(n).

3.3. Xt h thng c :

(2 3)

( )( 1)( 2)

z zH z

z z

Vi ROC |z| 2

b. () =1

2+9+0.7

c. () =

362+116, || > 2

Gi :

C th phn tch H(z) bng cch s dng Scilab thng qua cc bc thc hin sau:

Biu din H(z) thnh a thc theo 1.

V d:

() =1

2 3 + 2=

2

1 31 + 22

() =2

1 3 + 22

Vi = 1

Biu din a thc H(z) trong Scilab

--> s = poly(0,'s')

--> h_z_inv = s^2 / (1 - 3*s + 2*s^2)

--> pf = pfss(h_inv_z, 1/%eps)

pf(1) =

1

-----

- 1 + s

pf(2) =

- 0.25

------

- 0.5 + s

pf(3) =

0.5

Thay = 1 , thu c

() = (1) + (2) + (3) =1

1 + 1

0.25

0.5 + 1+ 0.5

T y c th suy ra c h(n).

5. Bi tp thm

5.1. Tm bin i Z ngc ca cc tn hiu nhn qu sau:

a. 21

1

5.05.11

5.11)(

zz

zzX

b. az

azzX

1

11)(

c. 21 25.01

1)(

zzzX

d. 21 3103

1)(

zzzX

5.2. Tm tt c cc tn hiu (c th c) m c bin i Z nh sau:

a. 2132

1)(

zzzX

b. 21

21

441

21)(

zz

zzzX

c. )3)(2.0)(3.0(

122)(

2

zzz

zzzX

5.3. S dng bin i Z tnh tng chp ca x1(n) * x2(n)

a. x1(n) = {1, 1, 1, 1} v x2(n) = {1, 1, 1, 1} b. x1(n) = {1, 2, 3, 4, 5} v x2(n) = {1, 1, 1} c. x1(n) = (1/5)nu(n) v x2(n) = 2nu(n) d. x1(n) = nu(n) v x2(n) = 2nu(n-1)

5.4. Tm bin i Z ngc:

a. X(z) = log(1-2z), |z| < b. X(z) = log(1-2z-1), |z| >

Gi : S dng tnh cht )(

)()(

zd

zdXznnx Z

5.5. Tnh tng chp ca cc cp tn hiu sau s dng bin i Z mt pha

a. x1(n) = {1, 1, 1, 1, 1} v x2(n) = {1, 1, 1} b. x1(n) = {1, 2, 3, 4} v x2(n) = {4, 3, 2, 1} c. x1(n) = (1/2)nu(n) v x2(n) = (1/3)nu(n)

5.6. Cho phng trnh sai phn

y(n) 0.7y(n-1) = x(n)

a. Tm H(z) b. Tm h(n) c. Tm y(n) nu x(n) = u(n)

5.7. Cho phng trnh sai phn

y(n) 0.5y(n-1) = x(n) + x(n-1)

a. Tm h(n) b. Tm p ng xung bc n v

5.8. Tm gi tr cui cng ca h(n) vi:

h(n) = (0.5)nu(n)

6. Bi tp t gii

6.1. ( ) 10 | | 0.50.5

zH z z

z

6.2. ( ) | | 0.5( 1)( 0.5)

zH z z

z z

6.3. 1

( ) | | 2( 0.3)( 2)

H z zz z

6.4. 2 2

( ) 0.1 | | 3( 3)( 2)( 0.1)

z zH z z

z z z

6.5. 2 2

( ) | | 2( 3)( 2)( 0.1)

z zH z z

z z z

6.6. 1

( ) | | 0.5( 0.5)( 0.5)

zH z z

z z

6.7. 2

1( ) 0.3 | | 0.5

( 0.5) ( 0.3)

zH z z

z z