Z变换

advertisement
Ch7 Representation Signals by Using DiscreteTime Complex Exponentials: The Z-Transform
(用离散时间复指数信号表示信号:Z变换)
Ch7.1 Introduction(引言)
(一)使用Z变换分析信号
•The Z-Transform (Z变换)
•Properties of Z-Transform ( Z变换的性质)
•Inversion of the Z-Transform ( Z反变换)
(二)使用Z变换分析系统
•Solving Differential Equations With Initial Conditions
(系统响应求解)
•The Transfer Function (系统函数)
Ch7.2 the Z-Transform
(Z变换)
•
•
•
•
Definitions (定义)
Regions of Convergence (收敛域)
z plane (z平面)
Zeros and Poles (零极点)
The Z-Transform(Z变换)
双边z变换
X ( z) 


x[n]z n
n  
z反变换
1
n 1
x[n] 
X
(
z
)
z
dz

2πj
离散信号可表示为不同频率复指数ej n的加权叠加,
权重正比于X(z) 。
符号表示:
正变换:X(z)=Z{x[n]}
反变换: x[n] =Z-1{X(z)}
z
x[n] 

X ( z)
Regions of Convergence(收敛域)
收敛域: 双边Z变换存在的条件


x[n] r n  
n  
Regions of Convergence (ROC):使上式成立的所有r值。
Regions of Convergence(收敛域)
Illustration of a signal that
has a
z-transform, but does not
have a DTFT.
(a) An increasing
exponential signal for which
the DTFT does not exist.
(b) The attenuating factor r–n
associated with the ztransform. (c) The modified
signal x[n]r–n is absolutely
summable, provided that r >
, and thus the z-transform
of x[n] exists.
The z-Plane(z 平面)
The z-plane. A point z = rej is located at a distance r– from the
origin and an angle  relative to the real axis.
The z-Plane(z 平面)
The unit circle, z = ej, in the z-plane.
Z变换与DTFT变换的关系
1)当收敛域包含单位圆时,Z变换和DTFT均存
在。 X ( e j )  X ( z ) j
z e
2)当收敛域不包含单位圆时,Z变换和DTFT
均不存在。
Zeros and Poles(零极点)
Ex7.2: Determing the Z-transform of the signal x[n] = nu[n]
Depict the ROC and the locations of poles and zeros of in
the z-plane.
Solution:

X ( z)   a z
n 0
n n
1

1  az1
ROC : z  a
Locations of poles and zeros of x[n] = nu[n] in the
z-plane. The ROC is the shaded area.
基本信号的Z变换
 [n]


1
ROC : z  0
n
a u[n]

1
1
1  az
ROC : z  a
u[n]

1
1  z 1
ROC : z  1
nu[n]
nanu[n]
Z
Z
Z

Z
Z

z
1
1  z 
ROC : z  1
1  az 
ROC : z  a
1 2
z 1
1 2
Ch7.4 properties of Laplace Transform
(Z变换的性质)
z
x[n] 

X ( z) with ROC Rx
z
y[n] 

Y ( z) with ROC Ry
1. Linearity(线性特性)
Z
ax[n]  by[n] 

aX ( z)  bX ( z) with ROC at least Rx  Ry
2. Time Reversal (时域翻转)
1
x[n]  X ( )
z
Z
with ROC
1
Rx
3. Time Shift(时移特性)
Z
x[n  n0 ] 

z n0 X ( z)
with ROC Rx , except possibly z  0 or z  
4. Multiplication by a Exponential Sequence
(指数加权性质)

 x[n]  X ( )
n
Z
z
with ROC  Rx
5. Convolution(卷积特性)
x[n]* y[n] 
 X ( z)Y ( z) with ROC at least Rx  Ry
Z
6. Differentiation in the z-domain (z域微分特性)
dX ( z )
nx[n]   z
dz
Z
6. Differentiation in the z-domain (z域微分特性)
dX ( z )
nx[n]   z
dz
Z
Ex:
1
L[u (t )] 
s
d 1  1
L[tu (t )]   ( )
2
s
ds s
2
d 1
L[t u (t )]   ( 2 )  3
ds s
s
n!
n!
n  t
n
L[t e u(t )] 
L[t u(t )]  n 1
n 1
(
s


)
s
2
Ch7.5 Inversion of z-transform
(由X(z)求x(n))
bM z  M    b1 z 1  b0
X ( z) 
aN z  N    a1 z 1  a0
1
 c0  c1 z    cM  N z
当MN时存在
( M  N )
B( z )

A( z )
真分式
 B( z ) 
x(n)  c0 (n)  c1 (n  1)    cM  N  (n  (M  N ))  Z 

A
(
z
)


1
if
then
Where,
Ex:
kN
B( z )
k1
k2



1
1
A( z ) 1  r1 z
1  r2 z
1  rN z 1
 B( z ) 
n
n
n
F 

(
k
r

k
r



k
r
1 1
2 2
N N )u (n)

 A( z) 
1
B( z )
ki  (1  ri z )
A( z )
5 1
5 z
6
X ( z) 
1 1 1 2
1 z  z
6
6
1
z 1 
 1 
1
x[n]  4  u[ n]    u[n]
 2 
 3
n
n
1
ri
, 求x[n]
Ex: Find the inverse z-transform of
1
H ( z) 
1  2 z 1 1  3z 1

Solution:


A
B
H ( z) 

1
1  2z
1  3z 1
A  (1  2z 1 )H ( z) z 2  2
B  (1  3z 1 )H ( z) z 3  3
2
3
H ( z) 

1
1  2z
1  3z 1
1)
z 3
2) 2  z  3
3)
z 2
h[n]  (2n1  3n1 )u[n]
n1
n1
h[n]  2 u[n]  3 u[n 1]
n1
n1
h[n]  2 u[n 1]  3 u[n 1]
Ch7.6 The Transfer Function(系统函数)
x[n]
X(z)
h[n]
H(z)
yzs[n]=x[n]*h[n]
Yzs(z)=X(z)H(z)
系统函数:系统在零状态条件下,输出的拉氏变换式
与输入的拉式变换式之比,记为H(z)。
Z [ y zs ( z )] Yzs ( z )
H ( z) 

Z [ x( z )]
X ( z)
The Transfer function and Difference Equation
(系统函数与差分方程)
Ex: Find the transfer function of the LTI system descriped
by the difference equation
y[n] + 0.7 y[n-1] +0.1 y[n-2] = 2 x[n] + x[n-1]
Solution:
1
2
1
[1  0.7 z  0.1z ]Y ( z)  (2  z ) X ( z)
Y ( z)
2  z 1
H ( z) 

X ( z) 1  0.7 z 1  0.1z 2
Causality and Stability(因果性与稳定性)
The relationship between the location of a pole and the impulse response
characteristics for a causal system. (a) A pole inside the unit circle contributes
an exponentially decaying term to the impulse response. (b) A pole outside the
unit circle contributes an exponentially increasing term to the impulse response.
Causality and Stability(因果性与稳定性)
A system that is both stable and causal must have all its poles inside
the unit circle in the z-plane, as illustrated here.
Causality and Stability(因果性与稳定性)
离散时间LTI系统BIBO稳定的充分必要条件是


h[n]  
n  
因果系统在s域有界输入有界输出(BIBO)的充要
条件是系统函数H(z)的全部极点位于 z平面单位圆内。
Ex: A causal system has the transfer function
1
2 z
H ( z) 
1  0.7 z 1  0.1z 2
Find the impulse response. Is the system stable?
Solution :
极点z= 0.5,在 z平面单位圆内;
极点z=0. 2, 在z平面单位圆内。
The system is stable(稳定)。
Ch7.10 The Unilateral z-Transform (单边Z变换)
1. Definition:

单边z变换
X ( z )   x[n]z n
n 0
符号表示:
x[k ]  X ( z)
zu
Ch7.10.1 Properties of Unilateral z-Transform
(单边Z变换的性质)
Time Shift(时移特性)
x[n]  X ( z)
zu
zu
x[n 1] 
x[1]  X ( z) z 1
1
x[n  2]  x[2]  x[1]z  X ( z) z
zu
2
1
x[n  k ]  x[k ]  x[k  1]z   
zu
x[1]z k 1  x[1]z 1  X ( z ) z k
Ch7.10.2 Solving Difference Equations with
Initial Conditions(利用Z变换分析系统响应)
解差分方程
时域差分方程
时域响应y[k]
z
z
反
变
换
变
换
z域代数方程
z域响应Y(z)
解代数方程
Solving Difference Equations with Initial
Conditions(利用Z变换分析系统响应)
y[n]  a1 y[n 1]  a2 y[n  2]  b0 x[n]  b1 x[n 1] n  0
已知初始状态为y[-1], y[-2],求y[n]。
求解步骤:
1) 经Z变换将时域差分方程变换为Z域代数方程
2) 求解z域代数方程,求出Yzi(z), Yzs(z)
3) Z反变换,求出响应的时域表示式
Solving Diffrential Equations with Initial
Conditions(利用拉普拉斯变换分析系统响应)
Y ( z )  a1 z 1Y ( z )  a1 y[1]  a2 z 2Y ( z )  a2 y[2]  a2 y[1]z 1
 b0 X ( z )  b1 z 1 X ( z )
b0  b1 z 1
 a1 y[1]  a2 y[2]  a2 y[1]z 1
Y ( z) 

X ( z)
1
2
1
2
1  a1 z  a2 z
1  a1 z  a2 z
Yzi(z)
Yzs(z)
y[n]  yzs [n]  yzi [n]  Z 1 Yzi ( z)  Yzs ( z)

Ex: Use the z-transform to determine the output of a system
y[n]  0.7 y[n  1]  0.1y[n  2]  7 x[n]  2 x[n  1] n  0
in response to input x[n] = u[n] . The initial conditions
are y[-1]=-26, y[-2]=-202.
Solution:Taking the unilateral z-transform of both sides of
the difference equation(对差分方程取z变换可得)



Y ( z )  0.7 z 1Y ( z )  y[1]  0.1 z 2Y ( z )  y[2]  y[1]z 1

 7 X ( z )  2 z 1 X ( z )
(0.7  0.1z 1 ) y[1]  0.1y[2]
7  2 z 1
Y ( z) 

X ( z)
1
2
1
2
1  0.7 z  0.1z
1  0.7 z  0.1z
2  2.6 z 1
7  2 z 1
Y ( z) 

1
1
(1  0.2 z )(1  0.5z ) (1  0.2 z 1 )(1  0.5z 1 )(1  z 1 )
Yzi(z)
Yzs(z)
Ex: Use the z-transform to determine the output of a system
y[n]  0.7 y[n  1]  0.1y[n  2]  7 x[n]  2 x[n  1] n  0
in response to input x[n] = u[n] . The initial conditions
are y[-1]=-26, y[-2]=-202.
 0.5
5
12.5
7  2 z 1



Yzs ( z ) 
1
1
1
1
1
1
1

0
.
2
z
1

0
.
5
z
1

z
(1  0.2 z )(1  0.5z )(1  z )
yzs [k ]  Z 1{Yzs ( z)}  [0.5(0.2)n  5(0.5)n 12.5]u[n]
 10
12
2  2.6 z 1


Yzi ( z ) 
1
1
1
1
1

0
.
2
z
1

0
.
5
z
(1  0.2 z )(1  0.5 z )
yzs [k ]  Z 1{Yzi ( z)}  [10(0.2)n  12(0.5)n ]u[n]
y[n]  yzs [n]  yzi [n]
Download