TR U IN G DAI HOC S I/ PHAM KY THUAT
THANH PHO HO CHI M1NH
KHOA CO KHI CHE TAO MAY
BO MON CO BIEN TU*
BE THI HOC KY 1, NAM HOC 2019-2020
Mon: HE THONG TRUYEN DONG SERVO
Ma mon hoc: SERV424029
Be so: 01; Be thi co 2 trang.
Thai gian: 75 phut.
Dirac ph e p sir du ng tai lieu
Bai 1: (4d) Cho mot co cau bao gbm: 1 dong cobudc luong cue (bipolar), goc bude a - 1.8°
truyen dong mot ban may dich chuyen thong qua true vitme co budc vit P = 10 (mm), hanh
trinh 400 (mm).
a. Ve so do ket noi bo dieu khien cho dong co?
b. Van tdc tinh tien cua ban may la 100 mm/s, tlnh thdi gian delay khi xuat xung dieu
khien dong co?
c. Cho bang thu tu pha cua dong
co nhu hinh, viet giai thuat
dieu khien dich chuyen cua
ban may 1 khoang 200 (mm)
vdi van tdc nhu cau b (su dung
so do ket noi dieu khien cau a)
SW ITCH IN G SEQUENCE
CCW STEP A A B B
1+-+-
2+- - +
3-+-+
4-+ + -
1+-+-
Bai 2: (4d)
Cho so dd khoi bo dieu khien vi tri dong co DC servo nhu hinh:
a. Mb ta va vidt giai thuat doc encoder, tu do tinh goc quay 6 cua dong co?
b. Viet cong thuc tinh tin hieu dieu khien u(t)? Bien doi u(t) thanh cong thuc rdi rac theo
thdi diem lay mau thu k? (Biet co = ——)
dt
c. Viet giai thuat dieu khien vi tri dong co theo cong thuc cau b? Cho u(t) = 100 tuong
duong voi PWM = 100%.
Bai 3: (2d) Su dung giai thuat DS de noi suy 1 cung tron d goc phdn tu I, theo ngirffc chieu
kirn ddng ho (CCW), R=5.
a. Lap bang tinh cac gia tri can thiet theo luu do giai thuat?
b. Ve ket qua chay duoc theo bang cau a?
Ghi chu: Can bo coi thi khong duac giai thick de thi.
So hieu: BM 1 /QT-PBBCL-RDTV 1
