三,z变换的基本性质与定理
1、线性若
[ ( ) ( )] ( ) ( )Z T a x n b y n a X z b Y z
ab,为任意常数
[ ( ) ] ( ) xxZ T x n X z R z R
[ ( ) ] ( ) yyZ T y n Y z R z R
m a x (,) m i n (,)x y x yR R R z R R R
则
2、序列的移位若 [ ( ) ] ( ) xxZ T x n X z R z R
[ ( ) ] ( )mZ T x n m z X zm 为任意整数
xxR z R
则
( ) ( ) ( 3 ) ( )x n u n u n X z例：,求
( ) [ ( ) ( 3 )]X z Z T u n u n解：
[ ( )] [ ( 3 )]Z T u n Z T u n
3 1
11
zzzz
zz


3
2
1
( 1 )
z
zz

2
2
1 0zz z
z

3、乘以指数序列若 [ ( ) ] ( ) xxZ T x n X z R z R
[ ( ) ]n zZ T a x n X a
a为任意常数
xxa R z a R
[ ( ) ] ( )n n n
n
Z T a x n a x n z


()
n
n
zzx n X
aa




x x x x
zR R a R z a R
a
则证：
4、序列的线性加权（ z域求导数）
若 [ ( ) ] ( ) xxZ T x n X z R z R
[ ( ) ] ( )dZ T nx n z X zdzxxR z R
2[ ( ) ] [ ( ) ]Z T n x n Z T n n x n
{ [ ( ) ] }
()
d
z Z T nx n
dz
d dX z
zz
dz dz




则同理：
( ) ( ) n
n
X z x n z

证：
() ( ) ( ) ( )nn
nn
dX z d dx n z x n z
dz dz dz




11( ) ( ) ( )nn
nn
x n n z z nx n z




1 [ ( ) ]z Z T n x n
()[ ( ) ]
xx
dX zZ T nx n z R z R
dz
5、共轭序列若 [ ( ) ] ( ) xxZ T x n X z R z R
* * *[ ( ) ] ( )Z T x n X z?xxR z R
* * * *[ ( ) ] ( ) [ ( ) ( ) ]nn
nn
Z T x n x n z x n z




**()Xz? xxR z R
则证：
6、翻褶序列若 [ ( ) ] ( ) xxZ T x n X z R z R
1[ ( )]Z T x n X
z


11
xx
zRR

则
[ ( ) ] ( ) ( )nn
nn
Z T x n x n z x n z


证：
1 1( ) ( ) n
n
x n z X z




1 1 1
xx
xx
R R zz R R


7、初值定理证：因为 x(n)为因果序列
()
l i m ( ) ( 0)
z
xn
X z x

对于因果序列,有
0
( ) ( ) ( )nn
nn
X z x n z x n z




12( 0 ) ( 1 ) ( 2 )x x z x z
l i m ( ) ( 0 )z X z x
8、终值定理设 x(n)为因果序列，且 X(z)=ZT[x(n)]的极点处于单位圆以内（单位圆上最多在 z=1
处可有一阶极点），则：
1l i m ( ) l i m [ ( 1 ) ( ) ]nzx n z X z
11( ) l i m ( ) l i m [ ( 1 ) ( ) ] R e [ ( ) ] znzx x n z X z s X z
[ ( 1 ) ( )] ( 1 ) ( )Z T x n x n z X z
证：利用序列的移位，得
1
1
[ ( 1 ) ( ) ] [ ( 1 ) ( ) ]
l i m [ ( 1 ) ( ) ]
nn
nn
n
m
n
m
x n x n z x n x n z
x m x m z








1 1
li m [ ( 1 ) ( ) ] li m [ ( 1 ) ( ) ] 1
n
m
zn m
z X z x m x m?


l i m { [ ( 0 ) 0 ] [ ( 1 ) ( 0 ) ] [ ( 2) ( 1 ) ]n x x x x x
[ ( 1 ) ( ) ] } l i m [ ( 1 ) ] l i m ( )nnx n x n x n x n
11( ) l i m [ ( 1 ) ( ) ] R e [ ( ) ] zzx z X z s X z
9、有限项累加特性设 x(n)为因果序列，即 x(n)=0,n<0
[ ( ) ] ( ) xZ T x n X z z R
0
[ ( ) ] ( )1
n
m
zZ T x m X z
zm a x,1xzR
则
()xn证,为因果序列
0 0 0
[ ( ) ] [ ( ) ]
nn
n
m n m
Z T x m x m z


0
() n
m n m
x m z



1
1
0
( ) 1 11
m
m
zx m z z
z


1
0
1 ()
1
m
m
x m zz

()1 xz X z z Rz
m a x,1xzR
n
m m=n
0
10、序列的卷积和（时域卷积和）
设 y(n)为 x(n)与 h(n)的卷积和：
( ) [ ( ) ] xxX z Z T x n R z R
( ) [ ( )] ( ) ( )Y z Z T y n X z H z
( ) [ ( ) ] hhH z Z T h n R z R
( ) ( ) * ( ) ( ) ( )
m
y n x n h n x m h n m


m a x (,) m i n (,)x h x hR R z R R
则且
[ ( ) * ( ) ] [ ( ) * ( ) ] n
n
Z T x n h n x n h n z

证：
( ) ( ) n
nm
x m h n m z



( ) [ ( ) ]n
mn
x m h n m z



( ) ( )
( ) ( )
m
m
x m z H z
H z X z

m a x,m i n,x h x hR R z R R
1
L S I
( ) ( ) ( 1 )
( ) ( )
nn
n
h n b u n ab u n
x n a u n

例：已知 系统的单位抽样响应：
，
求系统输入 的响应。
( ) [ ( ) ] [ ( ) ] n zX z Z T x n Z T a u n z aza解：
1( ) [ ( ) ] [ ( ) ( 1 ) ]nnH z Z T h n Z T b u n a b u n
1[ ( ) ] [ ( 1 ) ]nnZ T b u n a Z T b u n
1 z z z aaz z b
z b z b z b


( ) ( ) ( ) zY z X z H z z bzb
( ) ( ) * ( ) [ ( ) ] ( )ny n x n h n I Z T Y z b u n
Re[ ]z
Im[ ]jz
0
b
a
11、序列相乘（ z域复卷积定理）
若
( ) [ ( ) ] xxX z Z T x n R z R
( ) [ ( )] [ ( ) ( )]Y z Z T y n Z T x n h n
( ) [ ( ) ] hhH z Z T h n R z R
( ) ( ) ( )y n x n h n
x h x hR R z R R
11 ()
2 c
zX H v v d v
jv?


m ax,m i n,hh
xx
zzR v R
RR



则且
[ ( ) ( ) ] ( ) ( ) n
n
Z T x n h n x n h n z

证：
11( ) ( )
2
nn
cn
x n H v v dv zj?




1 ( ) ( )
2
nn
cn
dvx n H v v z
jv?



1 ( ) ( )
2
n
c n
z d vH v x n
j v v?





11 ()
2 c
zH v X v dv
jv?


hhR v R
m ax,m i n,hh
xx
zzR v R
RR



x h x hR R z R R
xx
zRR
v
( ) ( ),( ),( ) ( ) ( )nx n u n y n a a w n x n y n例：已知 < 1 求
1
1( ) 1 <
1X z zz解：
2
1
1
1( )
( 1 ) ( 1 )
aY z a z a
az az


11( ) ( )
2 c
zW z Y v X v d v
jv?

2
1
1 1 1 1
2 ( 1 ) ( 1 ) 1c
a
dv
vj av av v
z



1m ax,m i n,
1
zza v a

1,,v v a a z平面极点,c v a?内极点,单阶
11a a z a
1
1( ) R e [ ( ) ]
1vaW z s F v a zaz
( ) [ ( ) ] ( )nw n I Z T W z a u n
12,Parseval定理若 ( ) [ ( ) ] xxX z Z T x n R z R
( ) [ ( ) ] hhH z Z T h n R z R
1x h x hR R R R
* * 1
*
11( ) ( ) ( )
2 cn x n h n X v H v d vjv?



11m ax,m i n,
xx
hh
R v R
RR



则且


*
*
*1
*
1
=,
2
n
n
x h x h
c
Y z Z T y n x n h n z
z
X v H v dv R R z R R
jv?








利用复卷积公式可得


*
1
*1
*
11
2
z
n
c
Y z x n h n
X v H v dv
jv?




则
*,y n x n h n?证 令
* * * Z T h n H z由于
1x h x hR R R R又

* * 1*
11
2
j
c
n
h n v e
x n h n X v H v dv
jv





当 是实序列，且 时则


22 1
2
j
n
x n h n
x n X e d


当=
则
*112 j j jjX e H e d eje
*12 jjX e H e d
*112 j j jjX e H e j e dje