Bài giảng điều khiển số (chương 6)

C.6: CHT LNG IU KHIN
C.6: CHT LNG IU KHIN
H THNG IU KHIN S
H THNG IU KHIN S
6.1. SAI LCH TNH
• nh ngha: Sai lch gia đi lng đu
vào và đi lng đu ra  trng thái xác
lp.
6.2. Kiu (loi) hàm truyn đt
•Kiu (loi) hàm truyn đt bng s lng đim cc bng 1.
10
1
()
1
A
zA
Gz
z
+
=
−
…kiu “1”
10
2
()
A
zA
Gz
z
+
=

…kiu “0”
()( )
10
3
()
10.5
Az A
Gz
zz
+
=
−−
…kiu “1”
10
3
32
()
2.5 2 0.5
Az A
Gz
zzz
+
=
−+−
()( )
10
2
10.5
Az A
zz

+
=
−−
…kiu “2”
6.3. H thng có mt vòng kín
G
h
(z)
(-)
X(z)
Y(z)E(z)
x(kT)
e(kT) y(kT)
lim ( )
t
k
s
ekT
→∞
=
1
1
lim ( )
z
z
E
z
z
→
−

=
1
1()
lim
1()
z
h
z
Xz
z
Gz
→
−
=⋅
+
nh ngha các hng s
•Hng s bc thang
1
lim ( )
bt h
z
K
Gz
→
=
•Hng s bc mt
()
1
1
lim 1 ( )

bm h
z
K
zGz
T
→
=−
•Hng s bc hai
()
2
2
1
1
lim 1 ( )
bh h
z
K
zGz
T
→
=−
Tín hiu đu vào
()
1
z
Xz
z
ρ
⇒=
−

• Tín hiu đu vào
là hàm bc thang:
() .1()
x
kT kT
ρ
=
11
1() 1
lim lim
1() 1()1
tbt
zz
hh
zXz z z
ss
zGz zGzz
ρ
→→
−
−
== ⋅ = ⋅ ⋅
+
+−
1
1
lim
1()1lim()
bt
z

hh
z
s
Gz Gz
ρ
ρ
→
→
==
++
1
bt
bt
s
K
ρ
=
+
Tín hiu đu vào
()
2
()
1
zT
Xz
z
ρ
⇒=
−
• Tín hiu đu vào

là hàm t l bc
mt vi thi gian:
() .()
x
kT kT
ρ
=
()
2
11
1() 1
lim lim
1() 1()
1
tbm
zz
hh
zXz z zT
ss
zGz zGz
z
ρ
→→
−
−
== ⋅ = ⋅ ⋅
++
−
1
1

lim
11 1
( 1) ( 1) ( ) lim( 1) ( )
bm
z
hh
z
s
zzGz zGz
TT T
ρ
ρ
→
→
==
−+ − −
bm
bm
s
K
ρ
=
Tín hiu đu vào
()
2
3
(1)
()
2
1

zz T
Xz
z
ρ
+
⇒=
−
2
() .()
2
xkT kT
ρ
=
• Tín hiu đu vào
là hàm t l bc
hai vi thi gian:
()
2
3
11
1() 11 (1)
lim lim
1() 1()2
1
tbh
zz
hh
zXz z zzT
ss
zGz zGz

z
ρ
→→
−− +
== ⋅ = ⋅ ⋅⋅
++
−
1
2
22
2
22
1
(1)
lim
1
11
lim( 1) ( )
2(1) (1)()
bh
z
h
h
z
z
s
zGz
zzGz
T
TT

ρ
ρ
→
→
+
==
⎡⎤
−
−+ −
⎢⎥
⎣⎦
bh
bh
s
K
ρ
=
Hàm truyn đt G
h
(z)
()()
()
12
()
( ) ; 1; 1,2, ,
hi
n
Mz
Gz z i n
zz zz zz

=∀≠=
− − ⋅⋅⋅ −
•G
h
(z) kiu “0”:
()()
()
()()
()
11
12
12
()
lim ( ) lim
(1)
11 1
bt h
zz
n
bt
n
Mz
KGz
zz zz zz
M
Kconst
zz z
→→
==
−−⋅⋅⋅−

==
−−⋅⋅⋅−
1
bt
bt
s
const
K
ρ
==
+
Hàm truyn đt G
h
(z)
()()
()
12
()
( ) ; 1; 1,2, ,
hi
n
Mz
Gz z i n
zz zz zz
=∀≠=
− − ⋅⋅⋅ −
•G
h
(z) kiu “0”:

()()
()
()()
()
11
12
12
11(1).()
lim( 1) ( ) lim
10.(1)
0
11 1
bm h
zz
n
bm
n
zMz
KzGz
T T zz zz zz
M
K
Tz z z
→→
−
=−=
− − ⋅⋅⋅ −
==
− − ⋅⋅⋅ −
bm

bm
s
K
ρ
=
=∞
Hàm truyn đt G
h
(z)
()()
()
12
()
( ) ; 1; 1,2, ,
hi
n
Mz
Gz z i n
zz zz zz
=∀≠=
− − ⋅⋅⋅ −
•G
h
(z) kiu “0”:
()()
()
()()
()
2

2
22
11
12
2
12
11(1).()
lim( 1) ( ) lim
10.(1)
0
11 1
bh h
zz
n
bh
n
zMz
KzGz
T T zz zz zz
M
K
Tzz z
→→
−
=−=
−−⋅⋅⋅−
==
− − ⋅⋅⋅ −
bh
bh

s
K
ρ
=
=∞
Hàm truyn đt G
h
(z)
()( )
()
2
()
( ) ; 1; 2,3, ,
1
hi
n
Mz
Gz z i n
zzz zz
=∀≠=
−−⋅⋅⋅−
•G
h
(z) kiu “1”:
()( )
()
()
()
11

2
2
()
lim ( ) lim
1
(1)
0. 1 1
bt h
zz
n
bt
n
Mz
KGz
zzz zz
M
K
zz
→→
==
− − ⋅⋅⋅ −
==∞
− ⋅⋅⋅ −
0
1
bt
bt
s
K
ρ

=
=
+
Hàm truyn đt G
h
(z)
()( )
()
2
()
( ) ; 1; 2,3, ,
1
hi
n
Mz
Gz z i n
zzz zz
=∀≠=
−−⋅⋅⋅−
•G
h
(z) kiu “1”:
()( )
()
()
()
11
2
2

11(1).()
lim( 1) ( ) lim
1
1(1)
11
bm h
zz
n
bm
n
zMz
KzGz
TTzzzzz
M
Kconst
Tz z
→→
−
=−=
− − ⋅⋅⋅ −
==
− ⋅⋅⋅ −
bm
bm
s
const
K
ρ
==

Hàm truyn đt G
h
(z)
()( )
()
2
()
( ) ; 1; 2,3, ,
1
hi
n
Mz
Gz z i n
zzz zz
=∀≠=
−−⋅⋅⋅−
•G
h
(z) kiu “1”:
()( )
()
()
()
()
2
2
22
11
2
2

2
11(1).()
lim( 1) ( ) lim
1
1. (1)
1
0
11
bh h
zz
n
bh
n
zMz
KzGz
TTzzzzz
zM
K
Tz z
→→
−
=−=
− − ⋅⋅⋅ −
−
==
−⋅⋅⋅−
bh
bh
s
K

ρ
=
=∞
Hàm truyn đt G
h
(z)
()
()()
2
3
()
( ) ; 1; 3, ,
1
hi
n
Mz
Gz z i n
zzzzz
=∀≠=
−−⋅⋅⋅−
•G
h
(z) kiu “2”:
()
()()
()()
2
11
3

3
()
lim ( ) lim
1
(1)
0. 1 1
bt h
zz
n
bt
n
Mz
KGz
zzzzz
M
K
zz
→→
==
− − ⋅⋅⋅ −
==∞
− ⋅⋅⋅ −
0
1
bt
bt
s
K
ρ
=

=
+
Hàm truyn đt G
h
(z)
()
()()
2
3
()
( ) ; 1; 3, ,
1
hi
n
Mz
Gz z i n
zzzzz
=∀≠=
− − ⋅⋅⋅ −
•G
h
(z) kiu “2”:
()
()()
()()
2
11
3
3

11(1).()
lim( 1) ( ) lim
1
1(1)
0. 1 1
bm h
zz
n
bm
n
zMz
KzGz
TT
zzzzz
M
K
Tz z
→→
−
=−=
−−⋅⋅⋅−
==∞
− ⋅⋅⋅ −
0
bm
bm
s
K
ρ
=

=
Hàm truyn đt G
h
(z)
()
()()
2
3
()
() ; 1; 3, ,
1
hi
n
Mz
Gz z i n
zzzzz
=∀≠=
−−⋅⋅⋅−
•G
h
(z) kiu “2”:
()
()()
()()
2
2
2
22
11

3
2
3
11(1).()
lim( 1) ( ) lim
1
1(1)
11
bh h
zz
n
bh
n
zMz
KzGz
TT
zzzzz
M
Kconst
Tz z
→→
−
=−=
−−⋅⋅⋅−
==
− ⋅⋅⋅ −
bh
bh
s
const

K
ρ
==