Transcript Chapter 8 Discrete (Sampling) System

Chapter 8 Discrete (Sampling) System
8.1 Introduction
8.2 Z-transform
8.3 Mathematical describing of the sampling systems
8.4 Time-domain analysis of the sampling systems
8.5 The root locus of the sampling control systems
8.6 The frequency response of the sampling control
systems
8.7 The design of the “least-clap” sampling systems
Chapter 8 Discrete (Sampling) System
8.1 Introduction
8.1.1 Sampling
Make a analog signal to be a discrete
signal shown as in Fig.8.1 .
x(t) —analog signal .
x*(t) —discrete signal .
8.1.2 Ideal sampling switch —sampler
x(t)
x*(t)
x(t)
t
0
t1 t2 t3 t4 t5 t6
Fig.8.1 signal sampling
Sampler —the device which fulfill the sampling.
Another name —the sampling switch — which works like a
switch shown as in Fig.8.2 .
T
8.1.3 Some terms
1. Sampling period T— the time interval
of the signal sampling: T = ti+1 － ti .
x*(t)
x(t)
0
t
0
Fig.8.2 sampling switch
t
8.1.3 Some terms
2. Sampling frequency ωs — ωs = 2π fs = 2π / T .
3. Periodic Sampling — the sampling period Ts = constant.
4. Variable period sampling — the sampling period Ts≠constant.
5. Synchronous sampling —not only one sampling switch in a
system, but all work Synchronously.
6. Multi-rate sampling.
7. Opportunity(Random) sampling.
We mainly discuss the periodic and synchronous sampling in
chapter 8.
8.1.4 Sampling (or discrete) control system
There are one or more discrete signals in a control system —
the sampling (or discrete) control system. For example the
digital computer control system:
8.1 Introduction
r(t)
e(t)
－
A/D
e*(t)
computer
u*(t)
D/A
u (t)
process
measure
Fig.8.3 computer control system
8.1.5 Sampling analysis
Expression of the sampling signal:
x * ( t )  x( t )   T ( t )  x( t ) 


k 0
k 0
  (t  kT )   x(kT ) (t  kT )
It can be regarded as Fig.8.4:
c(t)
8.1.5 Sampling analysis
T
x(t)
x(t)
×
t
0
x*(t)
x*(t)
δT(t)
＝
t
t
0
0
Modulation
modulating
modulated
signal
pulse(carrier)
wave
Fig.8.4 sampling process
because :  T ( t ) 

C  1
 n T

We have:


k 0
k  
  (t  kT )   C ne
jk s t
1

T

 e jk st   s  2 / T
k  
1 
1 
1 
jk s t
jk s t
  T (t )e dt  T   T (t )e dt  T   T (t )dt  T 
T / 2
0
0

T /2
0
0
8.1.5 Sampling analysis
1
x * ( t )  x( t ) 
T

e
k  
jk s t
Laplace
1
X
*
(
s
)


T
transforma tion

 X ( s  jk s )
k  



 here : X ( s )  L[ x( t )]  x ( t )e  st dt 



0


1
The frequency spectrum of x * ( t ) : X* ( j ) 
T

 X [ j(  k s )]
k  
This means: for the frequency spectrum of x(t) shown in Fig.8.5,
the frequency spectrum of x*(t) is like as Fig.8.6.
X ( j )
 max  max
Fig.8.5
X * ( j )
 max
Filter


 max
s
Fig.8.6
2 s
Only:  s  2max
X ( j ) could be
reproduced
8.1 Introduction
So we have:
8.1.6 Sampling theorem ( Shannon’s theorem)
If the analog signal could be whole restituted from the sampling
signal, the sampling frequency  s must be satisfied :
 s  2 max or T 
here :

 max
 max  the maximu m frequenc y of the analog signal .
T
 sampling period .
 s  sampling frequency ,  s  2 T .
8.1.7 zero-order hold
Usually the controlled process require the analog signals, so
we need a discrete-to-analog converter shown in Fig.8.5.
x*(t)
discrete-to-analog
converter
xh(t)
Fig.8.7
D/A convert
8.1.7 zero-order hold
The ideal frequency response of the D/A converter is shown in
Fig.8.8.
 ( )
A(ω)
To put the ideal frequency response in
ω
practice is difficult, the zero -order hold
is usually adopted.
Fig.8.8
The action of the zero-order hold is shown
x*(t)
x(t)
in Fig.8.9.
xh(t)
The mathematic expression of xh(t) :
xh (t )  x(kT )
kT  t  (k  1)T
The unity pulse response of the zeroorder hold is shown in Fig.8.10.
The transfer function of the zero-order
hold can be obtained from the unity pulse
response:
1  e Ts
G ( s )  L1( t )  1( t  T ) 
s
Fig.8.9
g(t)
t
T
Fig.8.10
8.1 Introduction
8.1.7 zero-order hold
The frequency response of
the zero-order hold, which is
shown in Fig.8.11, is:
G( j ) , G( j )
G( j )
T
0
s
3 s
2 s

G( j )
Fig.8.11
sin(T / 2)

1  e  jT
 G ( j )  T
G ( j ) 

T / 2
j
 G ( j )  (T / 2)
8.2 Z-transform
8.2.1 Definition
Expression of the sampled signal:
Using the Laplace transform:
Define:
ze
Ts

x * ( t )   x( kT ) ( t  kT )
k 0

x * ( s) 
 kTs
x
(
kT
)
e

k 0
8.2 Z-transform
We have the Z-transform:

X ( z )  Z x( t )  Z x * ( t )   x( kT ) z k
k 0
8.2.2 Z-transforms of some
common signals
The Z-transforms of some
common signals is shown in
table 8.1.
8.2.3 characteristics of Ztransform
The characteristics of Ztransform is given in table
8.2.
Table 8.1
x( t )
X ( s)
X (z)
 (t )
1
1
z
z 1
Tz
1( t )
1
t
1
e t
sint
cost
s
s2
1
s 

( z  1) 2
z
z  e T
z sinT
s2   2
s
z 2  2 z cosT  1
z ( z  cosT )
s2   2
z 2  2 z cosT  1
Table 8.2
x( t )
k1 x1 ( t )  k 2 x2 ( t )
x( t  mT )
x( t  kT )
X (z)
k1 X 1 ( z )  k 2 X 2 ( z )
z
m
X (z) 
m 1
 x(iT  mT )z i
i 0
m 1
z m X (z) 
 x(iT )z m i
i 0
 Tz
tx( t )
dX ( z )
dz
e t x( t )
X ( z ) z  zeT
a k x( t )
X ( z) z z
a
Initial value
lim x( t )  lim X ( z )
Final value
lim x( t )  lim( z  1) X ( z )
t 0
t 
z 
z 1
k

Z x1 ( t )  x2 ( t )  Z   x1 ( iT ) x2 ( kT  iT )
Real convolutio n
 i 0

 X1( z) X 2 ( z)
8.2.3 characteristics of Z-transform
Using the characteristics of Z-transform we can conveniently
deduce the Z-transforms of some signals.
Such as the examples shown in table 8.3:
Table 8.3
x( t )
1  e t
te
t
X (z)
z
z
(1  e T ) z




T
z 1 z  e
( z  1)( z  e T )
Tz
( z  1) 2
ak

z  zeT
Tze T
( z  e T ) 2
z
z

z  1 z z z  a
a
t
2
 ( t  mk )
d Tz
T 2 z ( z  1)
 Tz

dz ( z  1) 2
( z  1) 3
z m
8.2 Z-transform
8.2.4 Z-transform methods
1. Partial-fraction expansion approaches
If :
X(s) 
Kn
A(s)
K1
K2


  
( s  a1 )( s  a2 )    ( s  an ) s  a1 s  a2
s  an
n
then : X ( z )  
Ki z
 ai T
z

e
i 1
Example 8.1
 5( s  4) 
5  10 z
15 z
5z
 10 15
Z

Z






 s s  1 s  2 z 1
s
(
s

1
)(
s

2
)
z  e T z  e  2T




2. Residues approaches
n

z 
X ( z )   res  X ( s ) 
  Ri  Residues
Ts 
z  e  s   ai i 1

i 1
n
1
 q 1 
z 
q
Ri 
lim
( s  ai ) X ( s )

q

1
Ts 
s


a
(q  1)！
z

e
i s


For q-order poles of X(s)
8.2.4 Z-transform methods
Example 8.2
 10 
10
z
Z


2
2
 s( s  1)  s( s  1) z  eTs
s 0
T
1
  10
z 

lim


( 2  1)！s1 s  s z  eTs 
10 z  z 2  ze (1  T )


T 2
z 1
ze


8.2.5 Inverse Z-transform
x(kT )  Z 1 X ( z )
 Inverse z- transform
1. Partial-fraction expansion approaches
If :
X(z) 
A(z)
(z  e
n
a1T
)( z  e
then : X ( kT )   K i e ai kT
i 1
a2T
)   (s  e
anT

)
K1z
ze
a1T

K2z
se
a2T
 
8.2.5 Inverse Z-transform
Example 8.3
 2T

z
(
1

e
)  1  z
z 
1
 2kT
x( kT )  Z 

Z


1

e

z 1
 2T
 2T 
ze 

 ( z  1)( z  e ) 
2. Power-series approaches
If :
X(z) 
A(z)
 K 1 z 1  K 2 z  2  K 3 z  3    
B( z )
then : X ( kT )  K 1 ( t  T )  K 2 ( t  2T )  K 3 ( t  3T )    
Example 8.4
3
2

z

2
z
1 
1
x( kT )  Z  3

2
 z  1.5 z  0.5 z 


 Z 1 1  3.5 z 1  4.75 z  2  6.375 z  3    
 1  3.5 ( t  T )  4.75 ( t  2T )  6.375 ( t  3T )    
8.2.5 Inverse Z-transform
3. Residues approaches
n

x( kT )   res X ( z )  z
i 1
k 1
   R  Residues
n
i 1
i

1
 q 1
q
k 1
Ri 
lim
(
z

a
)
X
(
z
)
z
i
(q  1)！z  ai  z q 1

For q-order poles of X(z)
Example 8.5
Z
1


z2
z 2  z k 1
z 2  z k 1
( z  1) 
( z  0.5)


( z  1)( z  0.5)
 ( z  1)( z  0.5)  ( z  1)( z  0.5)
z 1
z 0.5

kT
2  (0.5) T
Chapter 8 Discrete (Sampling) System
8.3 Mathematical modeling of the sampling systems
8.3.1 Difference equation
For a nth-order differential equation:
dn
d n1
d n 2
d
c( t )  a1 n1 c( t )  a2 n 2 c( t )      an1 c( t )  anc( t )
n
dt
dt
dt
dt
 b0
dm
dt m
d m 1
d
r ( t )  b1 m 1 r ( t )      bm 1 r ( t )  bm r ( t )
dt
dt
Make:
d
c( k  1)T   c( kT )  c( k  1)  c( k )  c( k )
c( t ) 


dt
T
T
T
 1th-order Forward difference
8.3.1 Difference equation
d 2c( t )
dt 2
…
Or :
d  dc( t )  2c( k ) c( k )
 


2
dt  dt 
T
T2
c(k  1 )-c(k) c(k  2 )- 2c(k  1 )  c(k)


2
T
T2
 2th-order Forward difference
dc( t ) c( k )  c( k  1)

dt
T
d 2c( t )
dt 2
…

 2c( k )



c ( k )
 1th-order backward difference
T
c( k )
T2
T2
c(k)-c(k  1 ) c(k)- 2c(k  1 )  c(k  2 )


2
T
T2
 2th-order backward difference
8.3.1 Difference equation
A nth-order differential equation can be transformed into a
nth-order difference equation by the backward or forward
difference:
c( k )   1c( k  1)   2c( k  2)       n1c( k  n  1)   nc( k  n)
  0 r ( k )   1r ( k  1)   2 r ( k  2)       m 1r ( k  m  1)   m r ( k  m )
To get the solution of the difference equation is very simple by
the recursive algorithm.
T c(mT )

Example 8.6
e
e
r
For the sampling system
shown in Fig.8.12, Assume:
K = 10, T = 0.5s, r(t) = 1(t)
Determine the output c*(mT).
Solution

T
K (5 s  1)
s( 2 s  1)
Fig.8.12
c
8.3.1 Difference equation
K (5 s  1)
d 2c( t ) dc( t )
de ( t )



2


5
K

Ke
(t )

2
dt
dt
e ( s ) s( 2 s  1)
dt
d 2c( t )
c( k )  2c( k  1)  c( k  2) dc( t ) c( k )  c( k  1)
 2
2
;

2
2
dt
T
dt
T
C ( s)
de  ( t )
e  ( k )  e  ( k  1)
5K
 5K
; Ke ( t )  Ke ( k )
dt
T
 The difference equation :
2T
T2
c( k ) 
4T
T2
c( k  1) 
2
T
c ( k  2) 
2
5 K  KT 
5K 
e (k ) 
e (k  1)
T
T
4T
2
5 KT  KT 2 
5 KT 
c( k ) 
c( k  1) 
c ( k  2) 
e (k ) 
e ( k  1)
2T
2T
2T
2T
5 KT  KT 2 
5 KT 
4T
2
c( k ) 
e (k ) 
e ( k  1) 
c( k  1) 
c ( k  2)
2T
2T
2T
2T
8.3.1 Difference equation
For K = 10, T = 0.5s, we have:
c(k )  11e (k )  10e (k  1)  1.8c(k  1)  0.8c(k  2)
Consider e*(k) = r(k)－c(k) = 1－c(k):
41
4
c( k )  1.75 
c( k  1) 
c ( k  2)
60
60
If c(0) = 0, applying the recursive algorithm we have:
41
4
c(1)  1.75 
c ( 0) 
c( 1)  1.75
60
60
41
4
c( 2)  1.75 
c(1) 
c(0)  0.554
60
60
41
4
c( 3)  1.75 
c ( 2) 
c(1)  1.255
60
60
……
41
4
c( m )  1.75  c( m  1)  c( m  2)
60
60
8.3 Mathematical modeling of the sampling systems
8.3.2 Z-transfer (pulse) function
Definition: Z-transfer (pulse) function — the ratio of the Ztransformation of the output signal versus input signal for the
linear sampling systems in the zero-initial conditions, that is:
G( z ) 
C(z)
R( z )
1. The Z-transfer function of the open-loop system
r(t)
T
G1(s)
R(z)
r(t)
T
G1(s)
R(z)
T
c*(t)
c(t)
G2(s)
G1(z)G2(z)
C(z)
c*(t)
c(t)
G2(s)
G1G2(z)
G1(z) =Z [ G1(s)] G2(z) =Z [ G2(s)]
C(z)
G1G2(z) =Z [ G1(s)G2(s) ]
8.3.2 Z-transfer (pulse) function
2. The z-transfer function of the closed-loop system
r
G(s)
－
c
C(z) 
R( z )G( z )
 GH ( z )  Z G( s ) H ( s )
1  GH ( z )
c
C(z) 
RG( z )
 RG( z )  Z R( s )G( s )
1  GH ( z )
c
C(z) 
R( z )G ( z )
1  G( z ) H ( z )
H(s)
r
－
G(s)
H(s)
r
G(s)
－
H(s)
r
－
G1(s)
G2(s)
c
C (z) 
RG1( z )G2 ( z )
1 G1G2 H ( z )
C(z) 
R( z )G1( z )G2 ( z )
1 G1( z )G2 H ( z )
H(s)
r
－
G1(s)
H(s)
G2(s)
c
8.3.2 Z-transfer (pulse) function
r
－
G1(s)
G2(s)
G3(s)
c
H(s)
C(z) 
r
G1(s)
－
G2(s)
－
c
RG1 ( z )G2 ( z )G3 ( z )
1  G2 ( z )G3 ( z )G1 ( z ) H ( z )
H1(s)
H2(s)
r
－
G1(s)
C(z) 
－
G2(s)
G3(s)
R( z )G1 ( z )G2 ( z )
1  G2 H1 ( z )  G1 ( z )G2 H 2 ( z )
c
H1(s)
H2(s)
RG1 ( z )G2 ( z )G3 ( z )
C(z) 
1  G2 ( z )G3 H1 ( z )  G2 ( z )G1G3 H 2 ( z )
Chapter 8 Discrete (Sampling) System
8.4 Time-domain analysis of the sampling systems
8.4.1 The stability analysis
1. The stability condition
The characteristic equation of the sampling control systems:
1  GH ( z )  0
∵ z  eTs  1  GH (eTs )  0
Suppose: s    j  eTs  eT (  j )  eT  e jT
In s-plane, α need to be negative for a stable system, it means:
eTs  z  eT  1
So we have:
The sufficient and necessary condition of the stability for the
sampling control systems is:
The roots zi of the characteristic equation 1+GH(z)=0 must all
be inside the unity circle of the z-plane, that is: zi  1
8.4.1 The stability analysis
critical stability
The graphic expression of the stability
condition for the sampling control systems
is shown in Fig.8.4.1.
unstable zone
Im
z-plane
1
Re
Stable zone
2. The stability criterion
Fig.8.4.1
In the characteristic equation 1+GH(z)=0, substitute z with
w  1 —— W (bilinear) transformation.
z
w 1
We can analyze the stability of the sampling control systems
the same as we did in chapter 3 (Routh criterion in the w-plane) .
 Proof : suppose w    j , z  x  jy , then :


z  1 x  jy  1  x  1  jy 
x2  y2  1
2y





j
 w    j 

2
2
2
2
z

1
x

jy

1
x

1

jy
(
x

1
)

y
(
x

1
)

y




2
2
2
2


0

x

y

1

0

x

y
1

 ( for the left half of the w-plane )  ( inside the unit circle of the z-plane )










8.4.1 The stability analysis
Example 8.7
1  G( z )  1 
0.632 Kz
0
z  1.368 z  0.368
Determine K for the stable system.
w 1
Solution : make
z
w 1
0.632 Kz
1 2
 0  0 .632 Kw  1.264w  ( 2.736  0.632 K )  0
z  1.368 z  0.368
0 .632 K
2.736  0.632 K
In terms o f the Routh criterion :
1.264
2.736  0.632 K
We have: 0 < K < 4.33.
2
8.4.2 The steady state error analysis
The same as the calculation of the steady state error in Chapter
3, we can use the final value theorem of the z-transform:
e ss  lim ( z  1) E ( z )
z 1
8.4.2 The steady state error analysis
For the stable system shown in Fig.8.4.2
R( z )G( z )
R( z )
E ( z )  R( z )  c( z )  R( z ) 

1  G( z ) 1  G( z )
r
r ( t )  1( t )  R( z ) 
r (t )  t
r (t )  t 2
 R( z ) 
 R( z ) 
G(s)
－
c
Fig.8.4.2
R( z )
e ss  lim ( z  1) E ( z )  lim ( z  1)
z 1
z 1
1  G( z )
 1

*
1  K p
 T

*
K

v
 T2

 K a*
e
z
;
z 1
Tz
( z  1)
K *p  lim G ( z )
z 1
;
2
T 2 z ( z  1)
( z  1) 3
K v*  lim ( z  1)G ( z )
z 1
; K a*  lim ( z  1) 2 G ( z )
z 1
8.4.2 The steady state error analysis
Example 8.8
e
r
－
T
Z.o.h
G (s)
c
K
Z.o.h —Zero-order hold. T  1s G( s ) 
s ( s  5)
1) Determine K for the stable system.
2) If r(t) = 1+t, determine ess=？
Solution
1)
Kz
 KTz
1  e
G( z )  Z 
 s
Ts
K 


s( s  5) 


5  5
25 
 (1  z 1 )
 ( z  1)2 z  1 z  e 5T 

 T 1
 K

2
K
z
 2.2067 z  0.2135
 (1  e ) Z  2




 s ( s  5) 
5 ( z  1)( z  0.0067)
K
K 
K

5  25 
 (1  e Ts ) Z  25 
s
s5
 s


Ts
Kz
8.4.2 The steady state error analysis
The charecteristic equation of the system :
K z 2  2.2067 z  0.2135
1  G( z )  1  
0
5
( z  1)( z  0.0067)
(5 -K ) z 2  ( 2.2067 K  5.0335) z  (0.0335  0.2135 K )  0
w 1
z
 0.9932 w  (9.993  1.573 K )w  (10.067  2.4202 K )  0
w 1
0  K  4.16
2)
2
K
z
 2.2067 z  0.2135
*
K p  lim G ( z )  lim 

z 1
z 1 5
( z  1)( z  0.0067)
2
K
z
 2.2067 z  0.2135
K v*  lim ( z  1)G ( z )  lim  
 0.2 K
z 1
z 1 5
( z  0.0067)
e ss 
1
1
K *p

T
K v*
T
 0
0.2 K
5

K
T1
8.4 Time-domain analysis of the sampling systems
8.4.3 The unit-step response analysis
A( z )
z
suppose : C ( z )   ( z ) R( z ) 

( s  p1 )( s  p2 )    ( s  pn ) z  1
A0 z n Ai z


z  1 i 1 z  pi
n
then : c( kT )  A0   Ai ( pi
kT
)T
Im
i 1
1
Re
Analyzing c(kT) we have
the graphic expression of
C(kT) is shown in Fig.8.4.3.
Fig.8.4.3
Chapter 8 Discrete (Sampling) System
8.5 The root locus of the sampling control systems
The plotting procedure of the root loci of the sampling systems
are the same as that we introduced in chapter 4.
But the analysis of the root loci of the sampling systems is
different from that we discussed in chapter 4 (imaginary axis of
the s-plane ←→ the unit circle of the z-plane).
8.6 The frequency response of the sampling control systems
The analysis and design methods of the frequency response of
the sampling systems are the same as that we discussed respectively in chapter 5 , chapter 6, only making:
w 1
z
and w  jv
w 1
Here: v — the counterfeit frequency
Chapter 8 Discrete (Sampling) System
8.7 The design of the “least-clap” sampling systems
The transition process of the sampling control systems can be
finished in the minimum sampling periods—the “least-clap”
systems.
8.7.1 design of D(z)
For the system shown in
Fig.8.6.1.
We have:
r
e
－
D(z)
G (s)
c
Fig.8.6.1
C(z)
D( z )G( z )
 (z) 

R( z ) 1  D( z )G( z )
E(z)
1
e ( z ) 
 1   (z) 
R( z )
1  D( z )G( z )
1  e ( z )
 (z)
D( z ) 

G ( z )1   ( z ) G ( z )e ( z )
8.7.1 design of D(z)
For the input signal :
r ( t )  1( t )  R( z ) 
r ( t )  t 2  R( z ) 
we have
 ( z )  z 1
1
1 z
1
;
r(t)  t  R( z ) 
Tz 1
(1  z 1 )2
T 2 z 1 (1  z 1 )
(1  z 1 )3
or e ( z )  1  z 1
 r ( t )  1( t )
or e ( z )  (1  z 1 )2  r ( t )  t
1
 ( z )  3 z 1  3 z 2  z 3 or e ( z )  (1  z 1 )3  r ( t )  t 2
2
1  e ( z )
 (z)
D( z ) 

G ( z )1   ( z ) G ( z )e ( z )
 ( z )  2 z 1  z 2
8.7.1 design of D(z)
1
R( z )
1  D( z )G ( z )
1
A( z )
1
 lim (1  z )

z 1
1  D( z )G ( z ) (1  z 1 )v
Proof: E ( z )  e ( z ) R( z ) 
e ss
1
1  z 1
If : e ( z ) 

and B( z )  1
1  D( z )G ( z )
B( z )
then : e ss  lim (1  z 1 ) A( z )  e ss  0
z 1
The transient process of the system response can be ended
in the minimum periods.
The responses of
the “least-clap”
System are shown
in Fig.8.6.2.
c*(t)
c*(t)
c*(t)
1
t
T 2T
t
T 2T 3T
Fig.8.6.2
t
T 2T 3T
8.7.1 design of D(z)
Example 8.9
For the system shown in Fig.8.6.3
1  e Ts
r
e
u
c
Gh ( s ) 
G
(s)
Gh(s)
D(z)
s
－ T
T
10
G( s) 
Fig.8.6.3
s( s  1)
Determine D(z), make the system to be the “least-clap” system
for r(t) = t.
3.68 z 1 (1  0.718 z 1 )
Solution G ( z )  Z Gh ( s )G ( s ) 
1
1
(1  z )(1  0.368 z )
r ( t )  t  R( z ) 
Tz 1
(1  z 1 ) 2

e ( z )  (1  z 1 )2
1  e ( z ) 0.543(1  0.368 z 1 )(1  0.5 z 1 )
D( z ) 

G ( z )e ( z )
(1  z 1 )(1  0.718 z 1 )
8.7 The design of the “least-clap” sampling systems
8.7.2 The realization of D(z)
To illustrate by example 8.10
Example 8.10 To realize D(z) for Example 8.9
Solution
U ( z ) 0.543(1  0.368 z 1 )(1  0.5 z 1 )
D( z ) 

E(z)
(1  z 1 )(1  0.718 z 1 )
U ( z )[(1  z 1 )(1  0.718z 1 )]  E ( z )[0.543(1  0.368z 1 )(1  0.5z 1 )]
U ( z )  0.282z 1U ( z )  0.718z 2U ( z )  0.543 E ( z )  0.471z 1 E ( z )  0.1z 2 E ( z )
u(k )  0.543e(k )  0.471e(k  1)  0.1e(k  2)  0.282u(k  1)  0.718u(k  2)
We can program the computer in terms of above formula to
realize D(z).
Chapter 8 Discrete (Sampling) System
Exercises: p820～ E13.11; E13.14; P13.12; p13.18; AP13.2
In example 8.9, if G ( s ) 
10
s ( s  5)
and respectively for r(t)=1(t); t; t2 .