midterm1_sol - Name: EML 4312, MID TERM 1 , Time: 1 hr 50...

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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 Miami-HQ V‘eal Park ;S M+ W GDL‘fiiEA'Dq/rl- WM—i- b1 Pauli-WC. Fur/pl +M (Lek-H-{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—i-aH—b; 0, l " vacuum/‘3 amd Sm—H-ib'ien‘i" +13.» an you—h -+o 139, GM 44¢. LHPI MI}. “>0; lo)0, which \‘S. +y-wr. in -\—lm‘s W. W1. haw. Pal-0A in +w. LHF, a)wa was W; isng s-l-abn'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 4-w- —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— wchEfl-Mdz‘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. Vii-U [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) $— 7~(t) — y(t), where 7°(t) is a reference signal that the output y(t) is required to follow. A controller C(s) (Whose input is E (s) and Whose output is U(s)) is to be designed. Draw the feedback loop. 3. Is it possible to design a Proportional (P) controller to make the closed loop transfer function from R(s) to Y(s) BIBO stable? Provide reasons. ’ To {and —H~. fiansfer fine-Hum, ML Cam asst/mu all 0PM Amp 3/ Eu) : K(5)-— Yrs); 1’/ Tl; flaw» R’l‘DY .H = 6(5).2.‘é:, .- 166) 91L ma, [Pfiieon‘h‘dmv’ W l + C(S.9\‘l$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—l-szkw SY(©———k3(®—L‘3’\fm = KUW em: Kuw- W - ‘ 5+5 5+5“ w. om 4m +M+ MW “4031 (=> 11032-35), we) UAW?“ "$70w was = Tag-r“: +0 “1%) 45+”) “m = \s, . ' 41‘ wt Find KEV—ii) ‘bwv‘ (D omd MAE ‘HMS M—lclwa O "'l aw: fife-é) v Jrflrcél = 7‘ Ks+ 5) S 5 5+6 5+5 5:0 5 5 $:—.S 5 Name: Problem 7. The Bode plot of a transfer function 0(5) is shown on the right. Bode diagram O gnitude (dB) :‘u 0 I .1; O 1. If this system is driven by an input Signal u(t) = 2sin(10t + 300), what is the output CU y(t) at steady state? 2 1o 2. If you are given the following two possible jg choices for G (s), which one do you think is: g more likely, and why? 3; " (D (U ‘C —'100 i 1 1 EL 10‘2 10“ 10° ’ 10‘ 102 s + 2 s + 1 Frequency (rad/s) (MS) ————> Lam) ya) {13> oar 1% mm =2 1 SW (Wt—F 50")» “* 5+”A‘6 35"? " /"’ if _ ‘ ~ ° lo) We“ all?) a 2 ( QM (1045+;0 4" 3 \__kaarm{/ . \f/ Fm» +w. Rm pm, q {5}») = _ 20 d6 ()0 a 2 => mom Mob—w a» Maw?! Log.» 2—90" (“PW‘M’M‘D 2 o .— ° , M. Siam? mm, W}: TS MO t 40) 5 " GUM-“Ae- —§w 045:0 IS 1., sine). +W~ W“ “l ‘l‘w‘ Edi Pldl‘ SW43 0d— 0 0H5 M" “M 90w “('Vfl‘LVLLan L ,4} ' l I l~1 94W 0444i \ ' d‘w—H N 1:7»??? __l_.$ F, _l.. 94W (1944' w '- ~ L Jun-H, TM PW WM ADM M'l Ol-I‘S-Hvsfi/wm,‘ W.Qflvx -§‘-_—f OinQ gim— $‘W‘PJL lac’l'lA will ham; 4% SW Wm. .> 3:“ B ...
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midterm1_sol - Name: EML 4312, MID TERM 1 , Time: 1 hr 50...

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