Transcript Chapter 3

Chapter 3 the z-Transform
3.1 definition
3.2 properties of ROC
3.3 the inverse z-transform
3.4 z-transform properties
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3.1 definition

X ( z) 

x[ n] z
n
, z  re
j
n  
Rx  | z | Rx 
Figure 3.2
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
H ( z) 
 h[n]z
n
: system
function
n  

X e
j
X z  
  X z  |
| z|1,that is z e


 x[n] re
n  
n
j



 ( xnr
j
n
)e
 jn

 FT xnr
n

n  

the condition for convergence：
 | x[n]r
n
| 
n  
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EXAMPLE:
x1[n ]  a u[n ], x2 [n ]   a u[ n  1]
n
X ( z) 
n
1
1  az
1
,
ROC is | z || a | and
| z || a | seperately
ROC takes the
poles as its
boundary
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3.2 properties of ROC
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1. finite  duration : 0  [] | z | [ ]
2.right  side : Rx  | z | []
3.left  side : 0  [] | z | Rx 
4.two  side : Rx   Rx  , then , Rx  | z | Rx 
Rx   Rx  , then
no
ROC
5.causal : Rx  | z | , include 
6.stable : include | z | 1

prove : if
system is stable , then
| h[n] |  ,
n  

| H ( z ) || z |1||

h[ n]  e
 jn

|
n  

| h[ n] |  | e
n  
 jn

|
| h[n] |  
n  
7.causal & stable :
poles are all in unit circle
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3.3 the inverse z-transform
1.inspection method
2.partial fraction expansion
3.power series expansion
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EXAMPLE:
X ( z )  1  0.5 z
1
 0.25 z
2
n0
 1

 0 .5
x[ n]  
0.25
 0

n 1
n2
other
EXAMPLE:
X ( z) 

1
(1  2 z
B1
1  2z
1

B1  (1  2 z
x[ n] 
4
3
1
)(1  0.5 z
1
,0.5 | z | 2
)
B2
1  0.5 z
1
1
) X ( z ) | z  2  4 / 3, B2  (1  0.5 z
n
( 2) u[  n  1]) 
1
1
) X ( z ) | z  0.5  1 / 3
n
0.5 u[ n]
3
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3.4 z-transform properties
x[ n] 
 X ( z )
ROC  Rx
z
x1[ n] 
 X 1 ( z )
ROC  Rx1
z
x2 [ n ] 
 X 2 ( z )
ROC  Rx 2
z
z
1.ax1[n]  bx2[n] 
 aX1( z )  bX 2 ( z )，ROC  Rx1  Rx2
z
2.x[n  n0 ] 
 z
n
n0
X ( z )，ROC  Rx
z
3.z0 x[n] 
 X ( z / z0 )，ROC  Rx | z0 |
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4.nx[ n] 
  z
z
dX ( z )
ROC  Rx
dz
5.x [ n]  X ( z )
*
*
ROC  Rx
*
Re x[ n] 
X ( z)  X ( z )
Imx[n] 
X ( z)  X ( z )
*
*
*
ROC  Rx
2j
z
x [  n] 
 X (1 / z )
z
ROC  Rx
2
6.x[n] 
 X (1 / z )
*
*
*
*
ROC  1 / Rx
ROC  1 / Rx
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7.x[n] is causal
x[0]  lim X ( z )
z 
lim x[ n]  lim ( z  1) X ( z )
n 
z 1
right  lim zX ( z )  X ( z )  lim ( Z [ x[ n  1]  Z [ x[ n]])
z 1
z 1

 lim [  ( x[n  1]  x[n])z ]
n
z 1 n  1
n
 lim [ lim
 ( x[m  1]  x[m])z
m
]
z 1 n  m  1
 lim ( x[0]  x[1]  x[1]  x[0]       x[ n  1]  x[n])
n 
 left
z
8.x1[n]  x2[n] 
 X1( z ) X 2 ( z )，ROC  Rx1  Rx2
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N
a
M
k
 b x[n  k ]
y[ n  k ] 
k
k 0
k 0
M
H ( z) 
Y ( z)

X ( z)

bk z

ak z
k
k 0
N
k
k 0
for
FIR , ak  0, for
k 0
bk  h[ k ]
EXAMPLE:
H ( z) 
2 z
1
1  0.5 z
z
2
3
z
4
y[n]  2 x[n]  x[n  1]  x[n  3]  0.5 y[n  2]  y[n  4]
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Relation between H(z) and frequency responce:
10
0
-10
-20
0
0.2
0.4
0.6
0.8
Normalized Angular Frequency (´p rads/sample)
1
0.2
0.4
0.6
0.8
Normalized Angular Frequency (´p rads/sample)
1
600
Phase (degrees)
B=[2,1,0,-1]
A=[1,0,0.5,0,-1]
freqz(B,A)
Magnitude (dB)
20
400
200
0
-200
0
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H(z)
jω
H(e )
h[n]
convolution(FIR’s realization)
difference equation(IIR’s realization)
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summary:
3.1 the z-transform
3.2 properties of ROC
right-sides sequence: inside the circle
left-sides sequence: outside the circle
finite-duration sequence: the entire z-plane
two-sided sequence: a ring
causal sequence: including infinite
stable sequence: including the unit circle
3.3 the inverse z-transform:
inspection method
partial fraction expansion
power series expansion
3.4 z-transform properties
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Keys and difficulties：
ROC;
the convolution property;
the relationship among system function, the
impulse response, frequency response and difference
equation;
exercises: 3.37 3.38
（these can be solved without contour integral method）
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