Controls Formula Sheet Test 2

# Controls Formula Sheet Test 2 - V Problem 1(15 pts Using...

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Unformatted text preview: V Problem 1 (15 pts). Using proportional feedback control, control designers have obtained a closed-loop system with the unit impulse response shown by the following ﬁgure. When the gain Ku = 10, , the system is on the verge of instability. Determine the PID controller paranfeters according to the ultimate gain method. ' ‘ 1.2 1.4 4:3 I I _.. I Tim. (coo) 1: Problem 1 — DesiginID Controller Problem 2 (15‘pts) Consider a system with transfer function (3 + 100)(s + 500) ms) = (s2 + s + 1)(s + 4000)‘ Compute the overshoot and settling time of the step response of the system. Problem 3 Consider the unit feedback system shown in Figure 2. Suppose that the controller is a proportional controller with gain K. Let G'(s) = (a) (10 pts). Show that feedback system is stable if K is positive. (b) (10 pts). Let r(t) = (l + t) 1(t). Determine. controller gain K so that the steady state error is less that 0001? V ' ~ (c) (10 pts). Let r(t) = t2 1(t). Show that it is not possible to specify K so that the steady state error is ﬁnite. L' ' Problem 4 Consider a system shown in Figure 3. Let the reference input be 'r(t) = a1(t) where a is a constant. Let the disturbance signal be d(t) = w 1(t) where w is an unknown constant. Let the controller be C(s) = Let G(s) = (a) (10 pts). Show that E(s§ = ﬁ—G—R— EEC—GD where R, D are the Laplace transforms of\r(t), d(t) respectively. ’ V (b) (10 pts). Suppose that K > 0. Show that the feedback system is stable. (c) (20 pts). Suppose that K > 0. Show that the steady state error will be 0 for any constant a and any unknown constant 11). / J 10km \0 ‘ Pmon 1f 1 m2 K=oxku=ko§ nJiPw—ms:T»=%Pu:0“‘>"s { ’ID Ca<s>- [(sme\(s+§oo\]/ (sixtwﬂkﬂoom E WM cm : mambo“ [moomtsusm] =v (RH/(SHE) u) : Dug ‘tu (Amps C76): \/(54_\$+\\ : Lani /L51+ ‘Zﬁahg+ LORI) a 56leﬂow Du" :\ 1 on:\ 1) HQWMZ‘ ) T: 05 it WWONoJr M%M« ‘qrf/W “V hue r i ‘ : ‘L .~ LLE Mp : €\«(DS)IVI‘-&05)l S » "Lb/ I = (Jug " (mg I v i i J 3 l i a i 1 :73) m V) % (‘55 Ya) - mag) b, GL6)? ‘l [slam] 3 E ! i 3 i E ¢ ﬂ Q - i \ 3 l : Rs) : P73) ’ Hkménb) ! aw»: {3,3 gm —— {:3 3k W.) =\< Tm \s We Lawn l i g f: imam Edi-«3:? :(z—E \JK:C1LCE+I) =24? QiE+CCqE+rbé7 E E E E g 1 X‘ 7 11 \$ : V‘QX‘BKSHH » 'k 62 ‘x k i 1° 5 '“ A“ " \$1 +s+k S. o E ‘ \+k(€) SKQFDB (30 X l, \ x=k mkath‘msﬁM \4>o I M rm rm» 1m 96) :ﬂrm : mew—s2) 93) "' (1(a) ‘- (3(5) L256) 3* kﬁﬂ kaq as): ﬂaw/AR : Q ekookllw S \Q 2 H145 mun gm» ' man x up“ 399“ ‘k 7 \‘g 5063 40.0w :3 [2 (00¢! 2? k>\0€o SW0 #1 \+[:(.S;D g (a) rm —. ALA-10:) \ a /k) 1 % ‘ =EU+C6A+V>L1 «,1 Ezgvgb (‘1) sly‘k' ‘ U Hch mac ' s, \ oi ‘ b : Q—E (A. k j W Q- LIZ/(Hm) 4» (fa/(HGGWI Wm \00 9%, w swap 2 CC? R + D t +lez \\ 1w; HCG. ‘ 1+ CG: ; (1(3):‘</s 3 we): '/(su) . U5.» cemgb k 6(SHHCC—lflw‘ﬁb k ‘ ‘ \{§)(%:‘ 9 4 g + k, LO ’5 ’4 5(0) ﬂow % 3 D15) Em : 9 _ __Ge_, {3 1+ Cc] Hm ‘ a to _, ebb“) ~ 52%. SI: ~_ in S § 5L» 9"?“ s+l /' x “’ i " )‘VC ‘é‘OQ g;— é‘Ob K ~ :7 = L» 06%“? V by» ‘st3 ‘5‘?!) 31—6S4k. €40 314\$H< 0.0 0 13 (3‘?ch ‘Hacﬁ QCOQ \S We“? AQ.‘)QJAC\QLJ 5W ck + w Q(m)‘0 M96 Q (Sm/«A mks MXW @534.ka L3 - ##ﬁ\$ﬁﬂ%*%lﬁwﬂkw¥€¥%kéw‘ﬂyﬁwmkxgﬁv§£\63¥{c \,c 1 (u 1 g u g xv 1 9 "UMG (943 Twig Plikpzoq/(m; TI: L/m : T1 : 2”.) To: 050,) L: {mew/(sgqxgmﬂ Q [1+ (Hem) / w 9081(0351 : (ks+\<>/ (834812 M406 +¥> \ 2 \ é L2 08% ﬂak-mm JTE *- Mbakz O-loh‘ ; Tx :3???) ~. IBM/g 1w:1/{squﬁgghzsh smqawi) "Atkirm's HRH ~ " } 3) xtiq Z ‘: 2:3— W‘I‘Q’\\ O: C) : K ‘- C m 1 mg Le + £31 512’th A 1%: ~®-* xx mu 2 m new ji®v5°e\$kﬁﬁ3 VWXLCwa/w MWQ“ <sx, Wm, A: Magma m NW Ammwi w {we WQLQ W = i 4‘ 3 wk‘k‘bawbkm ® (1%“th 3(‘{:f> (it? (Lu/«awe? (-m—‘qka Why. “1:0 W ‘\o www.me 4w» “3% ‘62 (\G‘ﬁ 6E ‘ sigma 51%an 6142mm r‘ SQMIL 5 (£6) 94% MmMmm-mwk (ﬁlm, Q, M . A. bx WV/W w Cl" ‘1‘) ‘MCLL‘IVL N vet-v“ *2 \&{o .r {62> E ow> as»: aw Msuzfusnw A )9: : \’ (H: I, ) e (cawas’c + <6». can) Wm V 2m“ 04 :wm‘rf1 @W’Z o\ (9 :6“ ° sob—2 Rf i: ” \- ‘ Tan WM . VLwa > MI“. f ~ Wtw 3:1. . ~ \_,71. . E JD" 1 L33; : v i 410 ‘11: Jﬁs /w~ AdAm/Q/Lgaw ' I ' We): \/ _\,_ ‘ s1+ a\$-+¥~ K (93" k wak Cl 93a mec 7523 95k». cm \4\ .;1®Lt\}>é [DE/o % bk: 6.2; :9. ‘ , a r (a . 7g . I: /; 2EKE ‘e 5 Deb<z—F'E’T<\ a>0 ; k>o LE &;1; (-3 @1116»: 0“ (DV (TE 5 ’TZ-T’ ~ 0. II anew-"2E u U? a 3 ;,ﬁ «200 be {4 3 7 ;_ 2% 4 ..ZT‘ , m, _ A? 6 K) A Q-k’m" .g ~72 3; gag ~« LB 1 m. \$2 + 2 fuju§ 4‘ \JEWI 10, G2 0t 2 3 o? m M ’ AMI/V“ 15PM '1 {be/4M Ox 1 ...
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Controls Formula Sheet Test 2 - V Problem 1(15 pts Using...

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