TRU ING DAI HOC SIX PHAM KY THUAT
t h An h PHO HO CHI MINH
KHOA CO KHI CHE TAO MAY
BO MON CO DIEN TU*
DAP AN HOC KY 1, NAM HQC 19-20
Mon: HE THONG TRUYEN DONG SERVO
MS mon hoc: SERV424029
De so: 01; De thi co 2 trang.
Thai gian: 75phut.
Du&c phep sir dung tai lieu
Dap an:
Bail:
a. So1 do ket noi bo dieu khien cho dong co1 (0.75)
Do dpng ca luang cue 2 pha nen sir dung mach cau H tich hap (vi du: L298)
H-Bridee 1
1
P0 1
1OUT1
IN 1
1
1
1
1
1
PI IN 2 OUT2
1
1
1
1
CONTROLLER 1
P2
1
1
1
1OUT3
IN 3
1
1
1
1
1
P3 IN 4 OUT4
1
1
1H-Bridge 2
b. Tinh thoi gian delay (0.75)
r r
Vai van toe ban m ay 100 mm/s, ta tinh dugc van toe cua dong ca:
o) = 600 (RPM)
pps = rpm
~60~ x ppr —600 360
60 X 1.8 = 2000
1
2000 = 0.5 (ms)
c.
Dua vao hinh cau a, ta co bang trang thai xuat xung cua bo dieu khien (0.75d)
P3 P2 PI P0 HEX
LH L H0x05
HLLH0x09
HL H LOxOA
L H H L 0x06
So hieu: BM1/QT-PDBCL-RDTV 1
Tinh so xung (0.75d)
Ban may di chuyen 200 mm tuang duang vdi dong ca quay 20 vong.
Goc budc a - 1.8°, nen so xung de dong ca quay 1 vong la 200 (xung)
Do do, tong so xung can thiet la 4000 xung.
SVviet giai thuat theo cac ket qua tren (Id)
Bai 2: (43)
Cho sa do khoi bo dieu khien vi tri dong ca DC servo nhu hinh:
a. Doc encoder (0.5d)
Do de khong yeu cau nen SV co the chon che do x l, x2 hoac x4. Gia su chon
x2, encoder co dQ phan giai N (CPR)
De dpc encoder che dp x2 ta su diing 1 ngat ngoai, canh len (rising) va canh xuong
(falling). Gia su dung ngat ngoai kenh 0, (EXTI0), ket noi vdi kenh A cua encoder
Chuo'ng trinh (Id)
EXTI0_Handler() {
if ((rising)&(B==0) |(falling)&(B== 1))
count++;
else count--;
if (count>=2*N){
count=0;
poscnt++;
}
else if (count<=-2*N){
count=0;
poscnt—;
i
_________________________________
Tinh goc quay 9:
theta = poscnt*2*pi+count*pi/N;
b. Viet cong thuc tinh tin hieu dieu khien u(t)? Bien doi u(t) thanh cong thuc rdi rac theo
. , . , d6
thdi diem lay mau thu k? (Biet co = — )
Dua vao sa do khoi, ta co:
u(t) = [K p(Gr - 9 )-(o ] K v = r lf ) K(0.75d)
______
Gia su hp dupe lay mau vdi thdi gian lay m§u T, cong thuc rdi rac:
S6 hieu: BM1/QT-PDBCL-RDTV
u(k) = Kp(0r- m ) - G (k)-0(k-1)K(0.753)
c. Viet giai thuat dieu khien vj tri dong co theo cong thuc cau b? Cho u(t) = 100 tuong
duong vdi PWM = 100%.
Sir dung cong thuc cong b de viet giai thuat. Liru y, gia tri tra ve cua giai thuat phai so
sanh vdi gidi han 100. (Id)
Bai 3:
a. (Id ) Do chuyen dpng CCW, ta co:
AV, = -2 X t +1
A*,+J=A *;.+2
AYj =2Yj+l
AYJ+l=AYJ+2
Xo = 5,Yo = 0; Xf=0, Yf=5
Step D D, d2d3AY AY * / Y,
0 0 -9 15 0
1 1 -9 1-8 -9 3 5 1
2 4 -8 4 -5 -9 5 5 2
3 0 -5 9 0 -7 7 4 3
40 -7 7 0 -5 9 3 4
54-5 9 4 -3 11 2 5
61115 12 -1 11 1 5
7 0 0 12 11 1 11 0 5
b. (Id)
S6 hieu: BM1 /QT-PBBCL-RDTV 1