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Final_Formula_sheet_back[1]

Final_Formula_sheet_back[1] - Consider a feedback system...

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Unformatted text preview: Consider a feedback system. Hhe transfer function of the plant is + e U H E C G + e U o—+ “E C —> G Y _ A) suppose that teh reference input signal is r(t) = 1(t) and that C(s)=K is a proportional (s + 1) controller with gain K. can the steadv state error be arbitrarily decresed by increasing the T I C G G (5) = m controller gain KYES = E = 1 + C G - 1 C (5} = K Given E(S)=R(s)—¥(s) e(oo)=L:i.msE(s) e(oo)=L:i.ms( ][_) (5.1; K 5.,0 5—:0 1+1“; 5 1(5) =13 (s)KG(s) R T (s) _ sCs—IJCS-S) _ . e( )=Lj.ms = ' +1 subsut‘ltl‘m m 5—:0 (1+KG) ‘3 (m) Ho 1+KG) + 5 (5—1) (3.6) E = R— (EKG) 1" E+EKG=R G=m (5.1314 + + — E (1+KG) =R r (t) = 1"” T (s) _ 59.1)(353 E - R Hr (m = i . 1 M _1+KG 5 e(°°)=1,‘_],"o" 1+K( ) scs—1)(s.5) _ _ 52.254 Justifyyou answer 1 T I: ) (s + 1) K s (s — 1) (S + 6) s = — — BymakingKbiggereGanetssmaller e(0°)=[1+K(2:Lo 5(5—1)(S+6) 5(5—1)(5+6)+(5+1)K O ¢2*0¢1 1 (5 + 1) K e(w)=—m=— mugs):— 1+K(T) 1+Kw10 s(s—1)(S+6)+(s+1)l( B) suppose that the reference input signalis r(t)=t*1(t) and that C(s)= k/s is anintergral T {5) = (5 + 1) K controller with gain K. What ist he smallest error can be achieved by increasing the — 6 s + 5 52 + s3 + K s + K controller gain K? _ _ 5+ 1 K e(oo)—I;_ZI)JD'I'ISE(S) G: 1|] T13): ( } R 52+2s+1 S3+552+(K—6)S+K e (on) = L:i.ms( ) 5-:0 1+CG _ 1 1 K ””3““ W i)](;] 3 2 6:; 52.254 5 +55 +(K—6)S+K R e()=Lj_mg — e(oo)=L:i.m ao+a1+ a2 +a3 °° [14m] we H21: 1—)] 4 5 5 ‘ 5‘ — K — 6 > I] r (t) = tw1(t) 5 1 e (on) = LilDTl U 4 L[r(t)]=s—2 4 Sufism“) I] glass 1 _ 1 1 e (on) = 3|] =L — _ _ Bun) 5':I’J°ns[1+(§]l}][132) 0+K(02.::°‘1) K> 4 8(m)_Lm 4(1) e(m)=101K KM] _ 5.,0 1+ (é) G s w 13) Lot 1(1) =9 1(t)S|10W1£alilispnssihlctn specify K so ltattheslaady stall: can: is finit: Lim 339.1111“ 5K 1 an an s(s+2) LiJllK = sac (“2) 5—10 (n+2) ) Lam-(1+1) 1(1) semiannual-mun m. 1H gainKsntha‘Hhesleadystatemrislcssumi Ho 2 . Thasystemis uftypclthu'efinrethe syflemcanmttracktypefl gammy? 1(1) w! E (S) = 11 (5) -Y (I) finile steady stateermr 2(5) = EC:)KG(:) _ . v I 'tutim QIDetmmnccuanIlergamKEnflaatthenmpmhasanavasmutufnu mnmflzanlfl‘kzn E (s) =11 (s) — [I (amt-(s)! msem aunitstqs mam signal n+3“ = n Knosfsng up = at: E(1+KE) - R ~43— i 1CD=';~5 14?.Q‘IIT-E; -+ m(!lP)=Ln[e‘l:;] —an(x,.)-¢‘G—? = 1+xs Substitution ’ (hill-15H2 m =L' E - W = 515! at } 3:: (I) R 31(3) = 1+KG (KG) (Lump): A; “2* (Ln(.15))3 em). Lin: =Lim : I“) = “5 g: m (Ln(.1u))2 5—10 1+1“; sen Ideals) lug) 1+1“ nu, I: ——-—--xz+ (Lid-10))” = -5912 um) Lin! R Lms_1_m[l+}— 1 K R “a 1‘1: “:2; “a 1+K sci-.2) a 22 _ K G _ 5 (9+2) :1 [3+2] _ 5 [5+2] _ 1 1 _ 1 1 fl [5) - 1+KG- _ 1+3 1 + X '— M 3(w1tlgfl s 1.3—1” (F) +13;- 5 ”—1 {;) s (n+2) a (n2) a (mu 5 (5.2) s (“2) K K "n2 .0»)me 1 din 1 (l) 2 =minmme “a 1+ 1 Ho 1 , s(s+ )+K s + 5+ 3 an n :(s.2) 59.2) 1 a( )- +1.11: - = = = . m “01.2) ,_,n ""K‘gia) 2:61. _ 2 —) 2 (.5169) an 2 —) a. 2 (.5169) 1 9346 e (m) a 0+1;i.||:]| 1 1 Founluln " ““9.” x:anZ-ax=1.93462=3.1421 ...
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