This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ME 375 FINAL EXAM
Friday, May 4, 2007 Division Shaver 11:30 / Meckl 2:30 (circle one) Name So (wife A Instructions ('1) This is a closed book examination, but you are allowed three 8.5x11 crib sheets. (2) You have two hours to work all six problems on the exam. (3) Use the solution procedure we have discussed: what are you given, what are you asked to
ﬁnd, what are your assumptions, what is your solution, does your solution make sense. You
must show all of your work to receive ﬂy credit. (4) Circle or boxin your answers. (5) You must write neatly and should use a logical format to solve the problems. You are
encouraged to really “think” about the problems before you start to solve them. Please write your name in the top righthand corner of each page. (6) A table of Laplace transform pairs and properties .of Laplace transforms is attached at the end
of this exam set. Problem N0. 1 (30) Problem No. 2 (30) I l I"
Problem No. 3 (30) ‘
Problem No. 4 (40) Problem No. 5 (30) Problem No. 6 (30) TOTAL (***/200) May 4, 2007 2 Name PROBLEM #1: (30 points) For the following transfer functions: (i) determine the poles (ii) compute time constant, natural frequency and/or damping‘ratio (iii) indicate which time response (1, 2, 3, or 4) below represents the unit step response of each
transfer function. ' 25 ' 10 /D ,
a ——————— 2 b __,___.__ = __ )
() s2+3s+25 C ) () 52+11s+10 (5+!)(MIO) (4
. —— 2 ’J‘“ , , _ __
f4“. 5 ~ 14;)? rain. 5— l/ (O I
was—Fl: 7.": l/T1=Z;szc.. 230,23 air0.3 (cw£— (3) (aw—3— (I) s2+6s+25 s+3
.6 __
Falls: 5 = _ g i); I” (L: J 3
‘="31‘j4' ”c‘SLJze. (An: S— r]:
Z‘Jun =b=> ﬁsb'B 1 Step Response 1 4 Step Response
0.9
1.2
0.5
0.7 1 '
11>
0.6 3 0.8
a 0.5 “E:
E .
0.4 < 0.6
0.3 0.4
0.2
0.2
0.1 _
O , . 0 _ . ..
O 0.2 0.4 0.6 10.8 1 1.2 1.4 1.5 1.5 2 O 0.5 1 1.5 2 2.5 3, 3.5 4
Time (sec) , ‘ Time (sec)
Step Response ' ' . Ste Res onse
1A ‘ 1 p p Amplitude
O .u
'm x i»: A,‘ .0
(D . .0
§, 0.2 May 4, 2007 ' ‘_ 3 Name PROBLEM #2: (30 points) The transfer function for a mechanical system is given by: __i:(3~_)__ 40(s+10) 0(3) " U(s) ', (s+1)(s2 +‘4s+400) (a) Write down expressions for 0060),  GO" (0)1, and A G(,1' co). \
Q
(I
2
\J
n
E
'0
+
\
O May 4, 2007 4 Name PROBLEM #2 (cont! >
(b) Find the magnitude and phase of G(ja)) at co = 10 and 50 rad/sec. car—mfg IG‘QW = 0/86,
469/0) = ~4é,9° z O.82_ NA “150‘ {6997/ = 0.0/9 ff:
4 ‘4 0
1mm: 4w~“5~ 7‘“ 57> L3:
‘ /80°"/‘av\—‘ 22/0:
469573) " 4941?” = ’3 224 m1 (0) Find the steady—state output yss(t) of the system to an input of u(t) = 3 + sin(10t) + 23in(5 0t). (ACé) = 3/§(0)( + /CCJ‘/0)/sfn(lof+ (@9109 W 7 + a/GCJ'J‘OJ/s/‘KﬁbHéégny ULU‘) = 3 + OJBG 5.‘n(/06—O.82)
+0.038 51th (5.0% “ 3. 22,4) May 4, 2007 5 Name PROBLEM #3: (30 points) A computer chip with added heat sink is represented in the schematic diagram below: CP.PP.TP The mass density of the chip material is represented by pp, op represents the specific heat of the
chip material, Tp is the temperature of the chip, dp is the height of the chip, and Ap is its top
surface area. The heat sink has height d5. Assume that only convective heat transfer (with
convective coefﬁcient 11) occurs to the surroundings, which are at ambient temperature TA.
Conductive heat transfer (with thermal conductivity 0:) occurs between the chip and the heat
sink. Ignore any heat transfer through the sides and bottom of the chip and the sides of the heat
sink. ‘ The chip is turned on and begins generating heat at a rate given by q. Derive a differential
equation that describes the time response of the chip temperature T p as function of q and T A. Conservarhhn as; W7 ; #Laj‘ gator«.4 ‘: MSW/«ital " M 64:14 May 4, 2007 6 Name PROBLEM #4: (40 points) Consider the following feedback control system: Where: C(s) = (s + 10)(s + 40) and P(s) = lOO/(s2 + 35 +100)
What is the transfer function: a) from Rto Y, GR(s) (ME 11> = 0) b) _frotho Y, GD(s) (ml— e = o )
Find the steadystate output y(t) when: c) r(t) = unit step, d(t) = 0 d) r(t) = O, d(t) = unit step (LP
/+ ItP Ce C5) = W s"'+3¥+/°° ____________,_______________________
/+ (co + (:1!o)(.s+4o) [on 514,354,400 J‘Hp 331. [$0 Gk“? = ,L__g__2.<_,:2__P—. 51435 4 (00 + loo 4 (m (MNXH‘FO) May 4, 2007 7 Name W
66/) WWI" ﬂ 3 O/ yeahaw (4°ch diagram: FD“) ' , Y0) ma 4.» +L.. 61; U) = YLS) ___ PG) 13(5)
:08) [123220) :— /+ Rs) + “99(5)
/ /.,. 7’0)
G©C;) : ___/o—o—_——— /0 0 11143: + l°° ‘— ' 133
/+ ___—”m 4. (:+ “X“ «9/00 sit35+ /‘°+ /°° + /°° (“"36” ) 5‘44““ “Tm
oo
1 (S) 1’ ___ ___L______._ (
CD /o/;‘+ 53035440/20o J.
(C) (AR ﬁnd Ua‘wz 7124M , s ‘
(‘6) : (I‘M 5 Y6) 2 (AM sGéLQég = 61:; 6% (S)
a" 5‘90 {#0 s
= 4—0 000 = 20°
) ‘3“ H) Zia—o: 29! J— She 53° 5.30
<
_ m ____ __—
‘Jssm ‘ $0722? 402 May 4, 2007 8 Name PROBLEM #5: (30 points) (a) Indicate the pole locations for the following root locus that correspond to a %OS 5 0.5% _ ‘.... j. u
E
v
3
a
I
n
I
I" 1;.. l n
An...— .u.. ....L.... ‘ 4 ' . C r'
—~_.——‘—_.._‘I..m_4!~=——~ ._. n. ”3‘.“ m mni. . ,_ an _ __ .r l_._,__..__..c_..___u.._..._.j 46 14 —12  8 6 4 _ 2 “WW/ ‘1
%OS = /00 9, 50,5” frj/rj" a (LooS May 4, 2007 9 Name (b) Indicate the pole locations for the following root locus that correspond to a 2% settling
time _<_ 0.45 seconds. 7 8 I 5
«3
l
l
A N O (c) Indicate the pole locations for the following root locus that correspond to both
requirements.
8. i
q
l
1
1
l c
n
I
I 1
I
1
l
l
l
1
I J .
bvv
. o :......r........ May 4, 2007 10 Name PROBLEM #6: (30 points) Consider the following feedback control system: Y(S)
«“7» where P(s) = (of/(s2 +_2gcons + of), with can = 2 rad/s, Q = 0.1 R(s) Design a PD controller C(s) so that the resulting closedloop system has damping ratio of 0.7 and
natural frequency of 4 rad/s. >1ch __ 0P : W
KO) ‘ /+ C? {1'4 045+ 4
“A /+ (W
51+0.4S+4
: 4C kgsJ— KP) 4(“454— Ko) :2 M Sat 0.4” 4 + 4(1415+K,o) 3+ (0.4+ 4K4)s+(4+4/4) gw Zjuwca we: LJLI—a. ju. 094 coca “H M~C¢V dawn? radio (6 W ireﬁmw, , NS/aﬂJWVW.
71W 0.4+ 414 = 21cm)“ = 2(o.?)(4) = so 414 :52. 3 l/ék=/.3{ “4/40 == wcf = /¢ ...
View
Full Document
 Fall '10
 Meckle

Click to edit the document details