Advances in PID Control Part 1 pot
RECENT ADVANCES IN ROBUST CONTROL – NOVEL APPROACHES AND DESIGN METHODSE Part 1 pdf
...
equilibrium points. Hence, the control law is chosen in following form:
2
11 1
ab
ukxkx ,
2
222
ac
ukxkx (2 .14 )
Hence, the system (2 .13 ) with set control (2 .14 ) is:
2
1
11 1 1
2
2
22 ... kx
, (3 .10 )
and obtaining new control system
12
213
22
34
2
41 11
111
1
,
,
,
.
u
xx
kk
xxx
mm
xx
kk
xx xkx
mmm
...
RECENT ADVANCES IN ROBUST CONTROL – NOVEL APPROACHES AND DESIGN METHODSE Part 2 potx
... ⎣⎦
⎣⎦⎣⎦
(20)
From (18 ), (19 ) and (20) and by using the separation lemma (Shi & al, 19 92)), we finally obtain:
1
2
2
0
0
D
T
M
T
⎡
⎤
≤
⎢
⎥
⎣
⎦
( 21)
Where:
11 1 1
11 111
11
11 1 1
tttt tt
ij ... B
−−Δ
⎡
⎤
⎢
⎥
Σ=
−−Δ
⎢
⎥
⎣
⎦
,
11 11
2
11 11
0
0
ttt
i
j
i
iij
XAXKB
AX BK X
⎡⎤
Δ+ Δ
⎢⎥
Σ=
Δ+Δ
⎢⎥
⎣⎦
and
1
3
2
0
0
Δ
⎡
⎤
Σ=
⎢
⎥
Δ
⎣
⎦
Let
1
11 1 11 2
, XPXP
−
==....
RECENT ADVANCES IN ROBUST CONTROL – NOVEL APPROACHES AND DESIGN METHODSE Part 3 pot
... - 0.70i
11 111 11
()
aa bb
AHE BHEK−++
-2.58 +0 .10 i -2.58- 0 .10 i -2.58 + 0 .10 i -2.58 - 0 .10 i
2222222
()
aa bb
AHE BHEK−++
-3.09 +0.54i -3.09-0.54i -3.09 + 0.54i -3.09 - 0.54i
11 1 11 1bb
AGCHEK+− ... Control – Novel Approaches and Design Methods
52
()
2
12 2
22
11
2 212
2
1
2
2
1
11
() cos( ())- () 1 ()
1( ) 1( )
1
() 1 sin( ()) -1. 5 ()-3 ()
1( )
cos (...
RECENT ADVANCES IN ROBUST CONTROL – NOVEL APPROACHES AND DESIGN METHODSE Part 6 potx
... [1, 3] , b ∈ [0.05, 0 .1] , d ∈ [0.3, 1]
0
3 (1 0.05 )
()
(1 0.35)
s
Ps
s
−
=
+
2
.WΔ<
1
1.8
()
2.80
s
Ws
s
+
=
and
32
2
32
2(0.0074 0.333 1. 5 51 1) (.000 01 1)
()
3(0.0049 0.246 1. 157 1)
sss ...
where η is a strictly positive constant.
The control discontinuity can be found from the above inequality:
11 1
11 1
11 1
() (1) ()()sgn()
() (1) () ()
() ( 1) ( )
es...
RECENT ADVANCES IN ROBUST CONTROL – NOVEL APPROACHES AND DESIGN METHODSE Part 4 ppt
... following 3
rd
order reduced model:
-0.6 910 1. 3088 -3.8578 -0.76 21
( ) -1. 3088 -0.6 910 -1. 5 719 ( ) -0 .11 18 ( )
0 0 -0.3697 0.4466
xt xt ut
⎡⎤⎡⎤
⎢⎥⎢⎥
=+
⎢⎥⎢⎥
⎢⎥⎢⎥
⎣⎦⎣⎦
0.00 61 0.02 61 0. 011 1 ...
pp. 13 13 -13 19, 19 68.
[15 ] S. Haykin, Neural Networks: A Comprehensive Foundation, Macmillan Publishing
Company, New York, 19 94.
[16 ]
W. H. Hayt, J. E. Kemmerly, a...
RECENT ADVANCES IN ROBUST CONTROL – NOVEL APPROACHES AND DESIGN METHODSE Part 5 pdf
... Processing, Vol.
36, No. 7, pp. 11 41- 115 1.
Recent Advances in Robust Control – Novel Approaches and Design Methods
12 2
Fig. 3. Two-layer product WNN structure.
4 .1 Input layer
11 11 1 1
; ...
Fig. 17 . Bode plot of the PID controller.
10
3
10
4
10
5
10
6
-10 0
-50
0
50
magnitude [db]
10
3
10
4
10
5
10
6
-250
-200
-15 0
-10 0
-50
0
frequency [...
RECENT ADVANCES IN ROBUST CONTROL – NOVEL APPROACHES AND DESIGN METHODSE Part 9 docx
... ⎝⎠ ⎦
1
,
1
ˆ
1
n
n
n
n
α
+
+
+
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎛⎞
⎢
⎥
⎜⎟
⎣
⎝⎠⎦
(28)
where
1
1
2
2
12
1
1
1
1
.
10 .00
ˆ
1. 00
ˆ
ˆ
ˆ
ˆ
.1
.
ˆˆ
.10
ˆ
ˆ
12
11
ˆˆˆ
.1
11
nn
n
n
nn
n
k
k
nn
nn
k
nn
nn
α
χ
χ
αα
χ
ααα
−−
+
+
−
=
−
−−
−
−
⎡⎤
⎢⎥
⎛⎞
⎢⎥
⎜⎟
⎢⎥
⎝⎠
⎡
⎤
⎡⎤
⎢⎥
⎢
⎥
⎢⎥
⎢⎥
⎢
⎥
⎢⎥
⎢⎥
⎢
⎥
⎢⎥
⎛⎞ ... or (22), are given by using the affine transformatio...
Tài liệu Project Planning and Control Part 1 pptx
... 11 5.76 is to divide it by
1. 05 × 1. 05 × 1. 05 or 1. 157, for
11 5.76
1. 05 × 1. 05 × 1. 05
=
11 5.76
1. 157
= 10 0.
If instead of dividing the 11 5.76 by 1. 157, it is multiplied by the inverse of
1. 157, ... were invested in a bank earning an interest of 5%
The value in 1 year would be 10 0 × 1. 05 = 10 5
The value in 2 years would be 10 0 × 1. 05 × 1. 05 = 11 0.25...