DE TUI CUOI KY HQC KY I (2023 - 2024)
Mon: TOAN tfNG DyNG CO KHI (Dai Tra)
Ma mon hoc: AMME131529
Be so/Ma de: 01 Be thi co 02 trang
Thai gian: 90 phut Ngay thi 18/12/2023
SV chi duac phep sit dung mot td A4 chep tay.
TRUING DAI HQC SU' PHAM KY THU AT
t h An h PHO HO CHI MINH
KHOA CO KHI CHE TAO MAY
BO MON CO DIEN Til
Cau 1: (3,0 diem)
Mo hinh toan hoc cua mot he thong ca khi duac mo ta bai phuang trinh vi phan nhu sau:
mx"(t) + bx'(t) + kx(t) = r(t)
Bi6t khoi luang m = 1/9{kg), he so giam chan b = 4^9 (^Ns/rnj, do cung 16 xo k = 5^9 ^N/rnj
Vai vi tri va van toe ban dau la .t(0) = 0, i(0) = 0, va tin hieu dau vao r(t) — t, 0 < t < 3
t T 2, t ^ 3
a. Tim bien doi Laplace R(s)
b. Tim bien doi Laplace X(s)
c. Tim tin hieu dao dong x(t)
Cau 2: (2,0 diem) Cho phuang trinh vi phan va dieu kien ban dau nhu sau:
— = ~t [v 2 +y), t € [1,2], 2/(1) = - 2
Hay tim gia tri y(t) bSng phuang phap Runge-Kutta bac 3 vai buac tinh la h — 0.2 (tinh chinh xac
den 4 chu so thap phan).
Cau 3: (2,0 diem) Gia tri dien ap V {volt) do duac cua mot cam bien do khoang each duac cho nhu
bang ben duai:
D(cm) 20 40 50 60 70 90
V(volt) 2.55 1.55 1.25 1.05 0.88 0.7
Hay uac luang gia tri dien ap F(volt) cua cam bien 6 khoang each 42 cm theo phuang phap noi suy
Newton cho da thuc bac 3 (tinh chinh xac den 4 chit so thap phan).
Cau 4: (3,0 di6m) Cho h? phuang trinh dai so tuyen tinh sau:
- x 1 + 2x2 + 2x3 = 0
2xx + 2x2 + /3x3 = 1 (1)
2xx + a x 2 + 2x3 - - 1
a. Viet lai he phuang trinh (1) duai dang Ax — B , vai x = x2 x3]T, B = [0 1 — i f . Tim
dieu kien cua a, 0 de ton tai ma tran AT1
b. Tinh ma tran A"1 tu do suy ra gia tri x = [x^ x2 x j 1, biet A la ma tran doi xung va 0 = — 1
c. Tim tri rieng va vec ta rieng cua ma tran A 6 cau b.
HET.
Ghi chu: Can bo coi thi khong duac giai thich de thi.
So hieu: BM1/QT-PBT-RBTV/02 Lan soat xet: 02 Ngay hieu luc: 15/5/2020 Trang: 1/2
Chu£n dau ra cua hpc phSn (ve kien thirc) Ngi dung kieni tra
[CLOl]: Co kha nang ap dung cac phuang phap bien doi Laplace va cac
phuang phap so gan dung (Euler, Euler cai tien, Runge-Kutta) de giai cac
bai toan phuong trinh vi phan dugc mo hinh hoa tu cac he thong co khi,
co dien tir.
Cau 1
Cau 2
[CL03]: Co kha nang ap dung cac phuang phap ngi suy da thuc
(Lagrange, Newton, Spline) de uro'c lutnig gia tri dap ung cua he thong
ky thuat.
Cau 3
[CL04]: Co kha nang ap dung cac kien thuc ve ma tran nhu djnh thuc
cua ma tran, ma tran nghich dao, tri rieng, vec to rieng,..., de giai dugc
cac bai toan ky thuat lien quan den dinh thuc ma tran.
Cau 4
Ngayl4 thang 12 nam 2023
Tru’d'ng bo in on
So hi?u: BM1/QT-PBT-RBTV/02 Lan soat xet: 02 Ngay hieu lure: 15/5/2020 Trang: 2/2
TRUC5NG DAI HQC SU' PHAM KY THUAT
THANH PHO HO CHI MINH
KHOA CO KHI CHE TAO MAY
BO MON CO D1EN Ttf
DAP AN BE THI CUOI KY HQC KY I
(2023- 2024)
Mon: TOAN iTNG DIJNG CO KHI
Ma mon hoc: AMME131529
Bes6/M ade:01 De thi co 01 trang
Thai gian: 90 phut
Bugc phep su dyuig mot to A4 chep tay.
Noi dung va diem dap an
Dap an D i e m
Cau 1: (3 diem)
Ta co: x"(t) + 4x'(t) + 5x(t) = 9r(t)
L + 4 x' + 5 x 1 = ( s 2 + 4 s + 5 j X (s) 0.25
r(f) = t[l - H(t - 3)] + (t + 2)H (t - 3)
- t + 2H{t - 3)
R(s) = \ + - e ~ 3'
s s
0.25
0.25
= + ( s 2 + 4 s + 5 ) x ( s ) = 9
1
^ X(s) = 9
• I
1 2 _j,
T + - e 3s
s s
-3s
s 2 (s2 + 4 s + 5 j s (s2 + 4 s + 5 j
1 1
s 2 ( s 2 + 4 s + 5 ) 5
_ 1
~ 5
1 4 1
2 5 s
1 4 Y
s'2 t 2 5 A
1
H
-----
2 5
1
H
-----
2 5
= 9 ( / + J )
4 s + 1 1
s 2 + 4 s + 5
(s + 2)
+ 3-
^s + 2j + 1 ^s + 2j + 1
=>■ L 1 { / } — —t — — + — e 2t c o s t + — e 2t s i n /
1 1 5 2 5 2 5 2 5
2 c~35 — 2
1 s + 4 p -3»
s ( s 2 + 4 s + 5 j 5s s 2 + 4 s + 5
0.25
0.25
0.25
0.25
0.25
• J
s 4 " 2
-2-
s ^s + 2j + 1 ^s + 2j + 1
-3s
L~l{J) = - [ l - e - 2i+6 c o s (t - 3 ) - 2 e _2(+fi s i n ( i - 3 )] t f ( / - 3 )
0.25
0.25
0.25
{l} + L~l {j})
9 3 6 3 6 _ 2t . , 2 7 _ 2( .
= —t
-----------
1
-------
e c o s H
--------
e su it
5 2 5 2 5 2 5
+ — [ l - e ~ 2(+0 c o s (t - 3 ) - 2 e _2<+6 s i n (t - 3 )j H{t - 3 ) 0 . 2 5
So hieu: BM1/QT-PBT-RBTV/02 Lan soat xet: 02 Ngay hieu luc: 15/5/2020 Trang: 1/4
Cau 2: (2 diem)
y' = \ ( y 2 + 2/), t e [1,2], y( 1) = - 2 ,h = 0.2
Su dung cong thuc RK bac 3
f(t,y) = -t (y2 + y), h = 0.2 =>n = ^ = 5, 0.25
^0 — ^i+l — + ^-2
t 1 = ¥ M , ) = 0 .2 i (j ,J + y ()
k2 = h f
l
h 1 ,
t> + o ^ + o ki °'2 t + o l( ( y< + ° -5fci)2 + + °-5k!
k3 = hf (t + /i, 2/j — fcj + 2^2) = 0.2 ^ ^ ^ ^ (^i — ^1 + 2^2) + 2/; ~ ^ + 2£^
V<+l = J/<+ ^ K + 4 ^ + fc3]
>at A;, — yl, A;2 = B,k3 — C,y. — y, t. = x de bam may dnh
1 0-2/ 2 , \
1 = — (2/ +2/)
0.25
0.25
0.25
0.2 y + 0.5v4j + y + 0.5^4
x 4- 0.1
= 7 ^ { ( « - a + 2B)’ + v~a + 2B
= y + i[A + \B + C]
= x + 0.2
ing ket qua:
it.
lVi
01-2
11.2 -1.7131
21.4 -1.5545
31.6 -1.4537
4 1.8 -1.3839
52 -1.3328
Id
(Dung
moi gia
tri dupe
0.2d)
Cau 3: (2.0 d)
a) Bai vi D — 42 E (40,50], ta chon 4 diem de noi suy da thuc bac 3 nhu sau:
D(cm) 20 40 50 60 70 90
V(volt)2.55 1.55 1.25 1.05 0.88 0.7
S6 hieu: BM1/QT-PBT-RBTV/02 Lan soat xet: 02 Ngay hieu lire: 15/5/2020 Trang: 2/4