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 “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
→
−

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
⇒=
−

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

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

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

=∀≠=
− − ⋅⋅⋅ −
•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
→→
==
−−⋅⋅⋅−

()()
()
()()
()
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

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

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

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
==
class="bi xb7 ydd w24 h53"

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

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
=

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
=

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