Chapter 8 Discrete (Sampling) System 8.1 Introduction 8.2 Z-transform 8.3 Mathematical describing of...
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Chapter 8 Discrete (Sampling) System 8.1 Introduction 8.2 Z-transform Mathematical describing of the sampling syst 8.4 Time-domain analysis of the sampling systems The root locus of the sampling control syste 8.6 The frequency response of the samp ling control systems he design of the “least-clap” sampling syst
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Transcript of Chapter 8 Discrete (Sampling) System 8.1 Introduction 8.2 Z-transform 8.3 Mathematical describing of...
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 systems8.7 The design of the “least-clap” sampling systems
Chapter 8 Discrete (Sampling) System
Make a analog signal to be a discrete signal shown as in Fig.8.1 .
t
x(t)
0 t1 t2 t3 t4 t5 t6
Fig.8.1 signal sampling
x(t)x*(t)
x(t) —analog signal .x*(t) —discrete signal .
8.1.2 Ideal sampling switch —sampler
Sampler —the device which fulfill the sampling. Another name —the sampling switch — which works like a switch shown as in Fig.8.2 . T
t
x*(t)
0 t
x(t)
0
Fig.8.2 sampling switch
8.1.3 Some terms1. Sampling period T— the time interval of the signal sampling: T = ti+1 － ti .
8.1 Introduction8.1.1 Sampling
8.1.3 Some terms2. 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:
A/D D/Acomputer process
measure
r(t) c(t)e(t)
－e*(t) u*(t) u (t)
Fig.8.3 computer control system
8.1.5 Sampling analysis
Expression of the sampling signal:
8.1 Introduction
)()()()()()()(*
00
kTtkTxkTttxttxtx
kkT
It can be regarded as Fig.8.4:
0 t
x*(t)
t
x(t)
0
T
t
δT(t)
0
× ＝
modulating pulse(carrier)
modulated wave
Modulation signal
Fig.8.4 sampling process
x(t) x*(t)
Tdtt
Tdtet
Tdtet
TC
TeT
eCkTttbecause
Ttjk
T
T
T
tjkTn
sk
tjk
k
tjkn
kT
ss
ss
1)(
1)(
1)(
1
/2 1
)()( :
0
0
0
0
2/
2/
0
8.1.5 Sampling analysis
We have:
ks
st
ks
Laplace
tiontransformak
kjXT
jX*txofspectrumfrequencyThe
dtetxtxLsXhere
jksXT
sXeT
txtxtsjk
)]([1
)( :)(*
)()]([)( :
)(1
)(* 1
)()(*
0
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.
)( jX
maxmax
Fig.8.5Fig.8.6
)(* jX
max
max s
s2
Filter Only: max2 s
)( jX could be
reproduced
8.1.5 Sampling analysis
If the analog signal could be whole restituted from the sampling signal, the sampling frequency must be satisfied : s
maxmax 2
Tors
. 2 ,
.
. :
s
max
Trequencysampling f
eriodsampling pT
alnalog signy of the am frequencthe maximuhere
s
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.
discrete-to-analog converter
x*(t) xh(t) Fig.8.7D/A convert
So we have:
8.1.6 Sampling theorem ( Shannon’s theorem)
8.1 Introduction
x*(t)x(t)
xh(t)
Fig.8.9
The action of the zero-order hold is shown in Fig.8.9.
The unity pulse response of the zero-order hold is shown in Fig.8.10.
The mathematic expression of xh(t) :
TktkTkTxtxh )1( )()(
The transfer function of the zero-order hold can be obtained from the unity pulse response:
s
eTttLsG
Ts
1)(1)(1)( T
Fig.8.10
t
g(t)
ω
A(ω) )(
Fig.8.8
To put the ideal frequency response inpractice is difficult, the zero -order hold is usually adopted.
8.1.7 zero-order hold The ideal frequency response of the D/A converter is shown in Fig.8.8.
The frequency response of the zero-order hold, which is shown in Fig.8.11, is:
8.1.7 zero-order hold
)2/()(2/
)2/sin()(1
)(TjGT
TTjG
j
ejG
Tj
)( ,)( jGjG
s
s2 s3
T
0
Fig.8.11
)( jG
)( jG
8.1 Introduction
8.2 Z-transform8.2.1 DefinitionExpression of the sampled signal: )()()(*
0
kTtkTxtxk
Using the Laplace transform:
0
)()(*k
kTsekTxsx
Define: Tsez
We have the Z-transform:
0
)()(*)()(k
kzkTxtxZtxZzX
8.2.2 Z-transforms of some common signals
The Z-transforms of some common signals is shown in table 8.1.
8.2 Z-transform
1cos2
)cos(cos
1cos2
sinsin
1)1(
1
11)(1
11)(
)()()(
222
222
22
Tzz
Tzz
s
st
Tzz
Tz
st
ez
z
se
z
Tz
st
z
zst
t
zXsXtx
Tt
Table 8.1
8.2.3 characteristics of Z- transform
The characteristics of Z-transform is given in table 8.2.
)()(
)()()()(nconvolutio Real
)()1lim()(lim value Final
)(lim)(lim value Initial
)()(
)()(
)()(
)()()(
)()()(
)()()()(
)()(
21
02121
1
0
1
0
1
0
22112211
zXzX
iTkTxiTxZtxtxZ
zXztx
zXtx
zXtxa
zXtxedz
zdXTzttx
ziTxzXzkTtx
zmTiTxzXzmTtx
zXkzXktxktxk
zXtx
k
i
zt
zt
a
zz
kzez
t
m
i
imm
m
i
im
T
Table 8.2
m
a
zz
k
T
T
zez
t
T
T
T
zmkt
z
zzT
z
Tz
dz
dTzt
az
z
z
za
ez
Tze
z
Tzte
ezz
ze
ez
z
z
ze
zXtx
T
t
)(
)1(
)1(
)1(
1
)()1(
))(1(
)1(
11
)()(
3
2
22
22
Table 8.3
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:
8.2.3 characteristics of Z-transform
n
iTa
i
n
n
n
iez
zKzXthen
as
K
as
K
as
K
asasas
A(s)X(s)If
1
2
2
1
1
21
)( :
)())(( :
Example 8.1
TT ez
z
ez
z
z
z
sssZ
sss
sZ
2
515
1
10
2
5
1
1510
)2)(1(
)4(5
2. Residues approaches
of X(s)r polesFor q-ordeez
zsXas
sqR
ResiduesRez
zsXreszX
Tsq
iq
q
asi
n
ii
as
n
iTs
i
i
)()(
lim)1(
1
)()(
1
1
11
！
8.2 Z-transform8.2.4 Z-transform methods
1. Partial-fraction expansion approaches
22
10
22
)1(
1
10
10lim
)12(
1
)1(
10
)1(
10
T
T
Tsss
Ts
ez
Tzez
z
z
ez
z
ssez
z
ssssZ
！
8.2.5 Inverse Z-transform
transformInverse z-zXZkTx )()( 1
1. Partial-fraction expansion approaches
n
i
kTai
TaTaTaTaTa
i
n
eKkTXthen
es
zK
ez
zK
esezez
A(z)X(z)If
1
21
)( :
)())(( :
2121
Example 8.2
8.2.4 Z-transform methods
kTTT
T
eez
z
z
zZ
ezz
ezZkTx 2
21
2
21 1
1))(1(
)1()(
2. Power-series approaches
)3()2()()( :
)( :
321
33
22
11
TtKTtKTtKkTXthen
zKzKzKzB
A(z)X(z)If
Example 8.4
)3(375.6)2(75.4)(5.31
375.675.45.31
5.05.1
12)(
3211
23
231
TtTtTt
zzzZ
zzz
zzZkTx
Example 8.3
8.2.5 Inverse Z-transform
3. Residues approaches
of X(z)r polesFor q-ordezzXazzq
R
ResiduesRzzXreskTx
kqiq
q
azi
n
ii
n
i
k
i
)()(
lim)1(
1
)()(
11
1
11
1
！
8.2.5 Inverse Z-transform
Example 8.5
1
1221 )1(
)5.0)(1()5.0)(1(
z
k
zzz
zz
zz
zZ
5.0
12
)5.0()5.0)(1(
z
k
zzz
zz
T
kT
)5.0(2
8.3 Mathematical modeling of the sampling systems
8.3.1 Difference equation
For a nth-order differential equation:
)()()()(
)()()()()(
11
1
10
12
2
21
1
1
trbtrdt
dbtr
dt
dbtr
dt
db
tcatcdt
datc
dt
datc
dt
datc
dt
d
mmm
m
m
m
nnn
n
n
n
n
n
Make:
fferenceForward di1th-orderT
kckckc
T
kTcTkctc
dt
d
)(
T
)()1(
)()1()(
Chapter 8 Discrete (Sampling) System
fferenceForward dith-orderT
c(k))c(k)-c(k
T
)-c(k)c(kT
kc
T
kc
dt
tdc
dt
d
dt
tcd
2
1221
)()(
)()(
22
22
2
2
2
Or :
ifferencebackward d1th-orderT
kckckc
dt
tdc
)(
T
)1()()(
ifferencebackward dth-orderT
)c(k)c(kc(k)-
T
)c(k)-c(kT
kc
T
kc
dt
tcd
2
2121
)()(
)(
22
22
2
2
2
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:
)()1()2()1()(
)()1()2()1()(
1210
121
mkrmkrkrkrkr
nkcnkckckckc
mm
nn
8.3.1 Difference equation
To get the solution of the difference equation is very simple by the recursive algorithm.Example 8.6
)12(
)15(
ss
sK
r e e
c
)(mTc
Fig.8.12
T
T
K = 10, T = 0.5s, r(t) = 1(t) Determine the output c*(mT).
For the sampling systemshown in Fig.8.12, Assume:
Solution
8.3.1 Difference equation
)1(5
)(5
)2(2
)1(4
)(2
222
keT
Kke
T
KTKkc
Tkc
T
Tkc
T
T
: ionence equatThe differ
)1(2
5)(
2
5)2(
2
2)1(
2
4)(
2
keT
KTke
T
KTKTkc
Tkc
T
Tkc
)2(2
2)1(
2
4)1(
2
5)(
2
5)(
2
kcT
kcT
Tke
T
KTke
T
KTKTkc
)()(
5)()(
2 )12(
)15(
)(
)(2
2
tKedt
tdeK
dt
tdc
dt
tcd
ss
sK
se
sC
)()( ;)1()(
5)(
5
)1()()( ;
)2()1(2)(2
)(2
22
2
kKetKeT
kekeK
dt
tdeK
T
kckc
dt
tdc
T
kckckc
dt
tcd
If c(0) = 0, applying the recursive algorithm we have:
8.3.1 Difference equation
)2(8.0)1(8.1)1(10)(11)( kckckekekc
For K = 10, T = 0.5s, we have:
Consider e*(k) = r(k) － c(k) = 1 －c(k): )2(
60
4)1(
60
4175.1)( kckckc
……
75.1)1(60
4)0(
60
4175.1)1( ccc
554.0)0(60
4)1(
60
4175.1)2( ccc
255.1)1(60
4)2(
60
4175.1)3( ccc
)2(60
4)1(
60
4175.1)( mcmcmc
8.3.2 Z-transfer (pulse) function
Definition: Z-transfer (pulse) function — the ratio of the Z-transformation of the output signal versus input signal for the linear sampling systems in the zero-initial conditions, that is:
)(
)()(
zR
zCzG
1. The Z-transfer function of the open-loop system
TG1(s)
r(t)G2(s) c(t)
c*(t)
G1(z)G2(z)R(z) C(z)
G1(s) G2(s)
T T
r(t)c(t)
c*(t)
G1G2(z)R(z) C(z)
G1G2(z) =Z [ G1(s)G2(s) ]
8.3 Mathematical modeling of the sampling systems
G1(z) =Z [ G1(s)] G2(z) =Z [ G2(s)]
G(s)r c
－H(s)
rG2(s)
c－
G1(s)
H(s)
r－
G2(s)c
G1(s)
H(s)
r c－
G(s)
H(s)
r－
cG(s)
H(s)
)()()( )(
)()()( sHsGZzGH
zGH
zGzRzC
1
)()()( )(
)()( sGsRZzRG
zGH
zRGzC
1
)()(
)()()(
zHzG
zGzRzC
1
)(
)()()(
zHGG
zGzRGzC
21
21
1
)()(
)()()()(
zHGzG
zGzGzRzC
21
21
1
8.3.2 Z-transfer (pulse) function
2. The z-transfer function of the closed-loop system
r
－G1(s)
H2(s)
G2(s) c
H1(s)
－ G3(s)
r－
G3(s)c
G2(s)
H(s)
G1(s)
r
－G2(s)
cG1(s)
H2(s)
H1(s)
－
)()()()(1
)()()()(
132
321
zHzGzGzG
zGzGzRGzC
)()()(1
)()()()(
22112
21
zHGzGzHG
zGzGzRzC
)()()()(1
)()()()(
2312132
321
zHGGzGzHGzG
zGzGzRGzC
8.3.2 Z-transfer (pulse) function
Chapter 8 Discrete (Sampling) System8.4 Time-domain analysis of the sampling systems8.4.1 The stability analysis
The characteristic equation of the sampling control systems:0)(1 zGH
0)(1 TsTs eGHez∵
Suppose: TjTjTTs eeeejs )( In s-plane, α need to be negative for a stable system, it means:
1 TTs eze
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: 1iz
1. The stability condition
1
Re
Imz-plane
Stable zone
Fig.8.4.1
The graphic expression of the stability
condition for the sampling control systems
is shown in Fig.8.4.1.
2. The stability criterionIn the characteristic equation 1+GH(z)=0, substitute z with
1
1
w
wz —— W (bilinear) transformation.
We can analyze the stability of the sampling control systems the same as we did in chapter 3 (Routh criterion in the w-plane) .
)( ) (
101 0
)1(
2
)1(
1
1
1
1
1
1
1
: , , :
2222
2222
22
z-planele of the unit circinside theethe w-planofft halffor the le
yxyx
yx
yj
yx
yx
jyx
jyx
jyx
jyx
z
zjw
thenjyxzjwsupposeProof
8.4.1 The stability analysis
unstable zone
critical stability
0368.0368.1
632.01)(1
2
zz
KzzG
Determine K for the stable system.Solution :
0)632.0736.2(264.16320 0368.0368.1
632.01
2
KwKw.
zz
Kz
K
KK.
nh criteriof the RoutIn terms o
632.0736.2
264.1
632.0736.26320
:
We have: 0 < K < 4.33.
1
1
w
wzmake
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:
)()1(lim1
zEzez
ss
8.4.1 The stability analysis
Example 8.7
8.4.2 The steady state error analysis
)(1
)(
)(1
)()()()()()(
zG
zR
zG
zGzRzRzczRzE
G(s)r c
－e
Fig.8.4.2
For the stable system shown in Fig.8.4.2
*
2
*
*
11
1
1
)(1
)()1(lim)()1(lim
a
v
p
zzss
K
T
K
TK
zG
zRzzEze
)()1(lim ;)1(
)1()( )(
)()1(lim ;)1(
)( )(
)(lim ;1
)()(1)(
2
1
*3
22
1
*2
1
*
zGzKz
zzTzRttr
zGzKz
TzzRttr
zGKz
zzRttr
za
zv
zp
Z.o.h —Zero-order hold.)5(
)( 1
ss
KsGsT
2) If r(t) = 1+t, determine ess= ？1) Determine K for the stable system.
Solution
Example 8.8
8.4.2 The steady state error analysis
52555)1(
)5()1(
)5(
1)(
2
2
s
K
s
K
s
KZe
ss
KZe
ss
K
s
eZzG
Ts
Ts
Ts1)
)0067.0)(1(
2135.02067.2
5
2515
)1(5)1(
2
1
521
zz
zzK
ez
Kz
z
Kz
z
KTzz
T
T
r－ G (s)
cZ.o.hT
e
0)4202.2067.10()1.5739.993(0.9932 1
1
0)2135.00335.0()0335.52067.2()5(
0)0067.0)(1(
2135.02067.2
51)(1
:
2
2
KwKww
wz
KzKz-K
zz
zzKzG
m the systequation ofteristic eThe charec
8.4.2 The steady state error analysis
16.40 K
2)
KK
T
K
T
Ke
Kz
zzKzGzK
zz
zzKzGK
vpss
zzv
zzp
5
2.00
1
1
2.0)0067.0(
2135.02067.2
5lim)()1(lim
)0067.0)(1(
2135.02067.2
5lim)(lim
1T**
2
11
*
2
11
*
8.4 Time-domain analysis of the sampling systems8.4.3 The unit-step response analysis
n
i
T
kT
ii
n
i i
i
n
pAAkTcthen
pz
zA
z
zA
z
z
pspsps
zAzRzzCsuppose
10
1
0
21
)()( :
1
1)())((
)()()()( :
Fig.8.4.3
Analyzing c(kT) we have the graphic expression of C(kT) is shown in Fig.8.4.3.
Im
Re1
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 respect-ively in chapter 5 , chapter 6, only making:
jvww
wz
and 1
1
Here: v — the counterfeit frequency
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.
For the system shown in Fig.8.6.1.
We have:
8.7.1 design of D(z)
Chapter 8 Discrete (Sampling) System
r－ G (s)
cD(z)
Fig.8.6.1
e
)()(1
1)(1
)(
)()(
)()(1
)()(
)(
)()(
zGzDz
zR
zEzezGzD
zGzD
zR
zCz
)()(
)(1
)(1)(
)()(
zzG
z
zzG
zzD
e
e
8.7.1 design of D(z)
31
1122
21
1
1
)1(
)1()()(
)1()( ;
1
1)()(1)(
:
z
zzTzRttr
z
TzzRtr(t)
zzRttr
signalinputtheFor
havewe
2
2
1)(
)(
)(1)(
ttr
ttr
ttr
31e
321
21e
21
1e
1
)1()( 33)(
)1()( 2)(
1)( )(
zzorzzzz
zzorzzz
zzorzz
)()(
)(1
)(1)(
)()(
zzG
z
zzG
zzD
e
e
8.7.1 design of D(z)
Proof:
0 )()1(lim :
1)( )(
1
)()(1
1)( :
)1(
)(
)()(1
1)1(lim
)()()(1
1)()()(
1
1
1
11
1
ssz
ss
e
vzss
e
ezAzethen
zBandzB
z
zGzDzIf
z
zA
zGzDze
zRzGzD
zRzzE
periods. minimum the in
ended be can response system the of processtransient The
The responses of the “least-clap” System are shown in Fig.8.6.2.
Fig.8.6.2
1c*(t)
t
T 2T
c*(t)
t
T 2T 3T
c*(t)
t
3T2TT
8.7.1 design of D(z)
Example 8.9For the system shown in Fig.8.6.3
)1(
10)(
1)(
sssG
s
esG
Ts
h
Determine D(z), make the system to be the “least-clap” system for r(t) = t.
)718.01)(1(
)5.01)(368.01(543.0
)()(
)(1)(
)1()( )1(
)()(
)368.01)(1(
)718.01(68.3)()()(
11
11
2121
1
11
11
zz
zz
zzG
zzD
zzz
TzzRttr
zz
zzsGsGZzG
e
e
e
h
r－
G (s)c
Gh(s)D(z)
e
TT
Fig.8.6.3
u
Solution
8.7.2 The realization of D(z)
8.7 The design of the “least-clap” sampling systems
To illustrate by example 8.10
Example 8.10 To realize D(z) for Example 8.9
)718.01)(1(
)5.01)(368.01(543.0
)(
)()(
11
11
zz
zz
zE
zUzDSolution
)]5.01)(368.01(543.0)[()]718.01)(1)[(( 1111 zzzEzzzU
)(1.0)(471.0)(543.0)(718.0)(282.0)( 2121 zEzzEzzEzUzzUzzU
)2(718.0)1(282.0)2(1.0)1(471.0)(543.0)( kukukekekeku
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
)5(
10)(
sssGIn example 8.9, if
and respectively for r(t)=1(t); t; t2 .
