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Unformatted text preview: Name: EML 4312, MID TERM 1 ,
Time: 1 hr 50 minutes (8:20 — 10:10 pm.)
Instructions: points will be awarded for completeness and clarity Problem 1. [2+2+2+2+2 :10 points]
Are the following statements true or false? Provide reasons. A correct answer without a correct
explanation will not fetch you any credit. 1. “The roots of the equation 50:3 + bx + c = 0, where b and c are ﬁxed positive numbers, have
negative real parts.” answer: FALSE
reason: A Y‘Qeﬁﬁﬁﬁda @on&4'+'on "foy all r0513 “+0 Lawn MiamiHQ V‘eal Park ;S M+ W GDL‘ﬁiEA'Dq/rl WM—i b1 PauliWC. Fur/pl +M
(LekH{uani col 731’s 0. Hana, +l~L Met/$5M C0n&l'+\‘w 1‘s
VCOiA'i(d. 3—3 2. “The transfer function H = ——2————————7, where a = 2 i 0.2 and b = 1 :1: 0.2, describes'a BIBO
s + as + b
stable system.” answer: TRUE ' ' '
reason: F0), a Scam! Or4pr po\«anow‘ial MC sV—iaH—b; 0, l " vacuum/‘3 amd Sm—Hib'ien‘i" +13.» an you—h +o 139, GM 44¢. LHPI MI}. “>0; lo)0, which \‘S. +ywr. in \—lm‘s W. W1. haw. Pal0A in +w. LHF, a)wa was W; isng slabn'lz’ria. 3. “Let the output y(t) of a system with H (s) r: %% as the transfer function be decomposed as y(t) = yu(t) + y1_c.(t), where yu(t) is the part due to the input and “C(75) is the part due to
the initial conditions. Even if the system is such that yu(t) is bounded for every bounded input
u(t), it is still possible for y1,C_(t) to be unbounded for certain initial conditions.” answer: F A L 3 E reason: ‘
Y0): .NJJ.) Nut”) . . r. W, 1.6.
Do) US) + N9 , New Mn Conﬁawiz —§v _, N _‘
“lei—“=71 132:”) ,. g< “Wm B ; Pgs’ouu PMS
ks—aﬂsapty '(9—Pn) Oct __ _' T '(I
"' j ( Z > +0xoh‘ov‘ .eagFMg!‘o“
—‘ 4
'ﬂ = or; .9. 1° 7
\=: +‘rs’r \nw 940.040 H is new» 5W (a— an+ {MPL‘WA +w's) I 426% PA
7 W WA‘H—GI‘ I‘mx‘i‘fﬂd thrl'i'ovﬂ m Name: 4. “If + sin(2t)y(t) + y(t) = u(t), then the transfer function from U(s) to Y(s) is given by
1 77
s2 + sin(2t)s + 1‘
answer: » FAtSc reason: TRANSFER FUNCTWN; own defined 0‘43 4w —kuL in van‘an’} 935+Wb, WW” 4*“ Eﬂrué‘x‘iWMA + 9M 2t + g :u a (ﬁzxcw‘w .& iiwu. VWAgriAg view. 5. “When a system whose transfer function is H120 is forced by the input cos(10t), the steady state '
output at time t is «i2: cos(10t + 450)”. ‘ answer: L; E reason: Name: Problem 2. (ABET) [15 points]
A mass is suspended by a non—linear spring ,
against gravity, Where the spring force (f8) is re—
lated to displacement d by f3 :2 Kd3. 1. Find a differential equation relating the po—
sition y(t) to the external force f Call
this equation 1. 2. What is the equilibrium position of the
mass? (Assume f (t) E 0 to ﬁnd the equi—
librium.) 3. Find the linear differential equation for the
displacement of the mass from its equi—
librium position (the linear approximation
to equation 1) when that displacement is
small. 1—)me ‘l'lML was; is al— Fosﬁ{W 3(a) +w. ﬂx+m$fcm oﬁl “HA2... Spring is 3({)« Apphaina waon‘s ‘ Laws, wn. Ogd i;
via. :: m3 +4Ch— Kai—1033 —— EOL. 4.. @ V11
‘ ' TI 2) L£+ \60 he HI\L Qﬂ’uiln‘hvimw‘ Pug—haw, Than Mtge“ 3) ‘3: \a—‘ag M JV“ deviﬁvi'ibn +013” +l/u. gngh’hviwvv a. a +
.5) \ém): T31“ +~ao ' Pvlﬁfna —\—\~is Vliwivnﬂ EaL (L, K“: 3” ; ~ 3 3 if ////// ~
M”; = swat +40.) ._ kw 441342..) = wg +1400 ~ 11 p
(at owicv‘ +W “1
A$$U\wi‘~a 'q is “wall, so wi— wchEﬂMdz‘r +020» m q GS?
\z
/ J‘s}
V W bf. Qd, "HA9. W—hon A; , W 1/ a: .  ﬂ 3) WW3 :wg+—§(tl_.naﬁ3k~b_3)35 wa_ K03; LA A: N  .~ 1,; m D, E. appr oxhrwki'l}
:) "mg + 3 kVS 65%)“3 \d = J“) K .‘ Thus 5 'HN. We ‘
\\ +9 Eq/ 1 who.” q I; 5M4“. Name: Problem 3. [10 points]
What is the transfer function from R(s) to Y(s) in the feedback loop shown? Sketch the poles
of this transfer function in the complex plane.
Find the output Signal (time—function) y(t) when I
7*(t) = 1(t). (Assume all initial conditions are 0, and express each transfer function in the form
518m + b28m_1 + ‘ ' ' + bm5 + bm+l _1 , m < s”+als” ++an_1s+an \
. 9+ $S+t
Tm'n g—gﬂv —§an "3" '{VOYV‘ +0 IL5 Mr” K/MN Name: Problem 4. ViiU [10 points]
The relationship between the output y(t) and input u(t) and disturbance ,ei‘ft) of a system is given by 5y 2 10u(t) + w(t). 1. What is the transfer function from U (s) to Y(s)? What is the transfer function from W(s) to
Y(s)? Draw block diagrams to Show relationships between U(s), W(8), and Y(s). 2. Deﬁne tracking error 6(t) $— 7~(t) — y(t), where 7°(t) is a reference signal that the output y(t) is
required to follow. A controller C(s) (Whose input is E (s) and Whose output is U(s)) is to be
designed. Draw the feedback loop. 3. Is it possible to design a Proportional (P) controller to make the closed loop transfer function
from R(s) to Y(s) BIBO stable? Provide reasons. ’ To {and —H~. ﬁansfer ﬁneHum, ML Cam asst/mu all 0PM Amp 3/ Eu) : K(5)— Yrs); 1’/ Tl; ﬂaw» R’l‘DY .H = 6(5).2.‘é:, . 166) 91L ma, [Pﬁieon‘h‘dmv’
W l + C(S.9\‘l$v s"4— LCCS) ﬂ—{L‘
v = C5) 7:.  t
we: kf‘ 1m .5) UM: 19,5(9 > C K? MM???
‘. HRV = 2"“? ‘HM‘s l; m!" 3,1be slum, *0» all {MAMA 03 k? V ,
S + 1“? $ im+ Pa>sfblx l Name: Problem 5. a [10 pt}
Consider the feedback control system shown on , the right. 1. Design the PD controller gains (i.e., choose
K p and KB) for closed loop transfer func—
tion from R(s) to Y(s) BIBO stable. +0 we. 2. If KP > 4 and KD > 0, is the steady state error e(t) equal to 0 when the reference sig ‘ nal is r(t) = 1(25)? (Assume the initial con—
ditions are 0 for this part). 3. Design the controller gains so that in addi—
tion to stability, both the closed loop poles
have their real parts less than —1. _ ' ,D Z X l ;.
l/anr‘ W ,kp+ KDS V. , Far [also s—l—atita, To wan. m'ym (Pde 12“; W .4) '1 9:!“ wayon kp>2_ and Gwen kg $0 W’r Jug—MWQ “3 3"“? 3‘ @ Name: Problem 6. [10 points]
The relationship between the temperature inside a house T(t), the ambient temperature T a, and the furnace control u is given by d_T(Q dt = Kw) — 5(T(t) — Ta) where K is an unknown constant, and the control signal u(t) can take only two possible values: +1
(furnace on) or 0 (furnace 0H). Time t is measured in hours. With the furnace on, the temperature
of the house T(t) increases from Ta to Ta + 15 in 12 minutes. What is the value of K 7 511+“ we: “ﬂea—Ta
=35: Ker—~63 —_—> \b—lszkw SY(©———k3(®—L‘3’\fm = KUW em: Kuw W  ‘
5+5 5+5“ w. om 4m +M+ MW “4031 (=> 1103235), we) UAW?“ "$70w was = Tagr“: +0 “1%) 45+”) “m = \s, . ' 41‘ wt
Find KEV—ii) ‘bwv‘ (D omd MAE ‘HMS M—lclwa O "'l aw: fifeé) v Jrﬂrcél = 7‘ Ks+ 5) S 5 5+6 5+5 5:0 5
5 $:—.S 5 Name: Problem 7.
The Bode plot of a transfer function 0(5) is
shown on the right. Bode diagram O gnitude (dB)
:‘u
0 I
.1;
O 1. If this system is driven by an input Signal
u(t) = 2sin(10t + 300), what is the output CU
y(t) at steady state? 2 1o
2. If you are given the following two possible jg
choices for G (s), which one do you think is: g
more likely, and why? 3; "
(D
(U
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s + 2 s + 1 Frequency (rad/s)
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" GUM“Ae —§w 045:0 IS 1., sine). +W~ W“
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 Fall '07
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