DAI HQC SIT PHAM KY THUAT IT.HCM Mon thi: DIEU KHIEN TUDONG
KHOA CO KHI CHE TAO MAY Ma mon hoc: AUCO330329
B 0 m6n CO DIEN Ttr Da s6 01; De thi co 02 trang
HQC KY 1 ,2019-2020 Thai gian: 75 phut
DAP AN
^ ^ ^ G3(GlG2+G2+Gt)
td “n m 1 + G2G3H x0 ,5
Neu SVsir dung Graph va cho ket qua dung van cham du diem
Cau 2 a. VeQDNS
(3 .5 a) Phuong trinh dac tnrng he kin co dang:
1+ K S + l = 0
s(s - 2)
He ha co hai cue la pi = 0; P2 = 2; va 1 zero z = -1. (n=2, m =l)
0 ,5
=> Quy dao nghiem he kin g6m hai nhanh xudt phat tCr cac cue khi K=0.
Khi K—>co, 1 nhanh tien ve zero -1 va 1 nhanh tien toi oo theo tiem can xac
dinh boi:
Goc giua cac tiem can va true thuc
(2/ — l)zr (2/-1W
a,. =
-
------
-
— =
-
------
-— = n
n -m 2-1
0 ,5
r
QDN doi xim g qua true thiec
Xac dinh diem tach/nhap cua QDN
Tu phuong trinh dac tinh, ta co:
Y _ s(s - 2)
s + l
„ dK s2 + 2 s-2
Suy ra: = .
ds (s + l)2
dK
Dodo: — = 0 <=> s2+ 2 s-2 = 0
ds
Giai ra ta dupe hai nghiem si = —2,73 va S2 = 0,73
Ca 2 nghiem deu phu hop la diSm tach/nhap vi thoa dieu kien tong so cue
va zero ben phai no la mot so le. 0 .5
Giao diem cua QDN voi true ao:
Thay s = jeo vao phuong trinh dac tinh ta dupe:
(jeo)2 + (K -2)joo+K = 0
f—co2 + K = 0
<=>i[K -2 = 0
" K = 2
<=> r
co = V2 ;
Vay cac giao diem can tim la s = ±j%/2 0 ,5
+jj-T
Do thi cua QDN:
- Ve dung bien dang quy dao nghiem so
- Dat dung cac gia tri tren quy dao nghiem so 0,5
Root Locus
2
1.5
' f ' 1
(fl
"O
8 0.5
3
e-
E -1
' 0.945 ■ '
0.974
'099.
-0 997 ...J
o V ../ 0.82 ' 0.68 0.4
'
& 5 4 3 j ? t :
' 0:997 " " i"
10:99
: ... - —i' ' "
0 974 -
01945
—:
---------
1—
------
0.9 . 0.82 0.66 0:4
----:
-------
J
____
:
____
l
_
:
____
:
__
l
__
:
_____
t.;., iix*.
__
i
_____
-3 -2 -1 0
Real Axis (seconds'1)
Dua tren quy dao nghiem so
0<K<2 He thong khong on dinh
K=2 He thong o bien gidi on dinh
K>2 He thong on dinh
SV su dung each khac va cho dung dap so van cham du diem (Id)
Cau 3:
(3.5 d) a. Su dung bo dieu khien PI 0.25
Do ngo vao la ham bac thang nen de he co sai so xac lap bSng 0 thi trong
ham truyen GCG phai co it nhat 1 khau tich phan ly tuong. Do ham truyen 0.5
cua doi tuong khong co khau tich phan nen bo dieu khien su dung phai la
bo PI.
b. Gc = Kp =13; r(t) = l(t)
r G<-G 26
G k ( s) 26
1 + GCG (s + l)(s + 3) + 26 s2 + 4s + 29
Taco: Y(s) = Gk^ = . 26 _ A + Bs + C
s s(s +4s + 29) s sz + 4s + 29
Giai ra, ta co: A - 26/29; B = -26/29; C = -104/29
c 261 26 (s + 2) + 0.4x5
Suy ra Y(s) =
-----------
-p
-----
----------=r
29 s 29 [(s + 2)2 +52]
Vay ta co ham qua do:
y(t) = L_1 [ Y (s)] = ^ (e"2t cos 5t + 0.4e“2t sin 5t)
_ 26 26729 _2t( 5 2 . _ ^
=> y(t) =
---------------
e -7=cos5t + -j= sin 5 t
29 145 1,729 729
J
m 26 26^29 _2t . ,_4 . M 2
=>y(t) “ 29 145 e sm(5t + (p), voi cos (p -0.5
Taco y(oo) = — :=>e(oo) = l — — = — = 0.1034
29 29 29
0.25
c. Su dung bo dieu khien PI: Gc = 13+ —
s
f K 2
Phuong trinh dac tnrng he kin: 1 + 13 + — ---------------= 0
l s y (s + l)(s + 3)
Rut gon ta co: s3 + 4s2 + 29s + 2KX = 0
Thanh lap bang Routh, ta co dieu kien de he on dinh: 0 < K i< 58
Khi Ki = 2, da thuc dac tnrng co dang:
A(s) = s3 + 4s2 + 29s + 4
SV chung minh A(s) = 0 co nghipm phuc (phan th\rc am) hoac tinh he so
suy giam 0 < £, < 1 de suy dang dap ung ngo ra co dao dong tit din.
0.5
0.5
Vo Lam Chirong