tugas servo+stability range

TUGAS
SISTEM PENGENDALIAN MODERN
DESIGN SERVO AND STABILITY RANGE
Disusun oleh :
Kartini
02311640000041
DEPARTEMEN TEKNIK FISIKA
INSTITUT TEKNOLOGI SEPULUH NOPEMBER
SURABAYA
2019
Plant yang digunakan adalah vehicle suspension (1/4 vehicle)
Sistem memliki persamaan seperti berikut :
Ẋ= Ax + Bu
Y= Cx
Dengan matriks A, B, dan C untuk plant tersebut adalah :
0
0
0
1
−25.31 −0,14
A = [−4252500 0
]
−47.76 1
46.55 0
−1832 0
1551.7 0
0
6.517
B= [
]
−46.55
−1551.7
C = [0 0 1
0]
DESIGN SERVO
Menggunakan n = 4, sehingga sinyal kontrol u sebagai berikut :
u = - (k2x2 + k3x3 + k4x4 ) + k1 ( r- x1)
= - Kx + k1 r
dengan : K = [𝑘1
𝑘2
𝑘3
𝑘4 ]
State feedback matriks K didapat dari matlab :
K = [−5.9461𝑥104
−30.375 −3.529 −0.0214]
Unit-Step Response dari desain sistem :
0
0
0
1
−25.31
−0,14
A-BK =[−4252500 0
]−47.76 1
46.55 0
−1832 0
1551.7 0
0
[ 6.517 ] [−5.9461𝑥104 −30.375 −3.529 −0.0214]
−46.55
−1551.7
0
1
0
0
−37691,69 197,95
−2,309 −0,000126
=[
]
−2767869,9 −1413,97
−169,05 0,0009
−92264312,2 −47133,5 −7308,53 −33,3
Persamaan State :
ẋ = [A – BK]x + Bk1r
ẋ1
𝒙1
0
1
0
0
0
ẋ2
−37691,69 197,95
−2,309 −0,000126 𝒙2
6.517
[ ]= [
] [ ]+ [
]k r
ẋ3
−2767869,9 −1413,97
−169,05 0,0009 𝒙3
−46.55 1
𝒙4
ẋ4
−92264312,2 −47133,5 −7308,53 −33,3
−1551.7
ẋ1
𝒙1
0
1
0
0
0
ẋ2
−37691,69 197,95
−2,309 −0,000126 𝒙2
−0.0388
[ ]= [
] [ ]+ [
]r
ẋ3
−2767869,9 −1413,97
0.2768
−169,05 0,0009 𝒙3
𝒙4
ẋ4
−92264312,2 −47133,5 −7308,53 −33,3
9.2266
Persamaan output :
y = [0
𝒙1
𝒙2
0 1 0] [𝒙 ]
3
𝒙4
u(∞)= -Kx(∞) + k1 r(∞)
= - Kx(∞) + k1 r
Grafik unit-step response
Sehingga :
u(∞) = -[−5.9461𝑥104
= -[−5.9461𝑥104
𝑥1 (∞)
𝑥2 (∞)
+ −5.9461𝑥104 𝑟
−30.375 −3.529 −0.0214]
𝑥3 (∞)
[ 𝑥4 (∞) ]
𝑟
0
−30.375 −3.529 −0.0214] [0] + −5.9461𝑥104 𝑟 = 0
0
Pada saat steady state, sinyal kontrol u menjadi 0
STABILITY RANGE
State Equation system :
0
0
0
1
−25.31 −0,14
−4252500
0
A=[
]
−47.76 1
46.55 0
−1832 0
1551.7 0
Liapunov Function :
V(x) = xT Px
Untuk menganalisa stabilitas, digunakan matriks Q, positive-semidefinitereal symmetric :
0
Q = [0
0
0
0
0
0
0
0
0
0
0
0
0]
0
1
P didedifinisikan sebagai : AT P + PA = - Q
𝑝11 𝑝12 𝑝13 𝑝14
−4252500 46.55 1551.7
𝑝
𝑝
𝑝
𝑝
0
0
0
] [𝑝 21 𝑝22 𝑝23 𝑝24 ]+
0 −25.31
31
32
33
34
−47.76 −1832
𝑝41 𝑝42 𝑝43 𝑝44
0 −0,14
1
0
𝑝11 𝑝12 𝑝13 𝑝14
0
0
0
1
0
𝑝21 𝑝22 𝑝23 𝑝24 −4252500 0 −25.31 −0,14
[𝑝
] = [0
𝑝32 𝑝33 𝑝34 ] [ 46.55 0
31
0
−47.76 1
𝑝41 𝑝42 𝑝43 𝑝44
0
−1832 0
1551.7 0
0
1
[
0 0 0
0 0 0]
0 0 0
0 0 −1
−4252500𝑝21 + 46.55𝑝31 + 1551.7𝑝41 + (−4252500𝑝12 + 46.55𝑝13 + 1551.7𝑝14 ) = 0
−4252500𝑝21 + 46.55𝑝31 + 1551.7𝑝41= - (−4252500𝑝12 + 46.55𝑝13 + 1551.7𝑝14)
𝑝21 = −𝑝12
𝑝31 = −𝑝13
𝑝41 = −𝑝14
𝑝11 − 4252500𝑝22 + 46.55𝑝23 + 1551.7𝑝24 = 0
−25.31𝑝21 − 47.76 𝑝31 − 1832𝑝41 − 4252500𝑝32 + 46.55𝑝33 + 1551.7𝑝34 = 0
−0,14 𝑝21 + −4252500𝑝42 + 46.55𝑝43 + 1551.7𝑝44 = 0
−4252500𝑝22 + 46.55𝑝32 + 1551.7𝑝42 + 𝑝11 = 0
𝑝23 = 𝑝32
𝑝24 = 𝑝42
𝑝12 + 𝑝21 = 0
−25.31𝑝22 − 47.76 𝑝32 − 1832𝑝42 + 𝑝31 = 0
−0,14 𝑝22 + 𝑝23 + 𝑝41 = 0
𝑝11
𝑝21
P = [𝑝
31
𝑝41
𝑝12
𝑝22
𝑝32
𝑝42
𝑝13
𝑝23
𝑝33
𝑝43
𝑝14
𝑝24
𝑝34 ]
𝑝44
Untuk P masing-masing > 0 maka sistem adalah stabil