Controls Formula Sheet Test 2b

Controls Formula Sheet Test 2b - Desigt a PE controller...

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Unformatted text preview: Desigt a PE) controller using quarter decay rate method for a system with the foll0wing step response. td=5,H=10,;’=35 PIDoontmllerisgivenas D = +L+tdS)wm‘ 1.2 1.2 t,s ‘ K=—R—L—19*5 35 MT, 2L=2*5=10.TD =O.5L=0.5*5=2.5 2 0. 84 s+1000 (s2 + s + l)(s + 10000) the overshoot and settling time of the step response of the system. (lfint. Truncate the system 3 a second order system by eliminating the non-dominant poles and zeros) Consider a system with transfer functions S) = . Compute @ Use approximation. S+lOOO=lOOO, 5+10000=10000 1000 g 10 1 Due to linearalily (s2 + 3+1)10000 We have 00) a 52 + 5+1 5 4w" @sider a system shown in Figure 1 Let = . Let the disturbance sigtal be0020 d(t) = Wl(t) wherew is an unknown constant. Let the controller be C(S) = 1. Let s l sl+r+l G(.t) = 4a) ShowthatE(S):_ 1 R_ G 1+CG 1+CG respectively. NotethatE=R~Y5L 2R—E=G(CE+D):>R=E+CGE+DG:>R—DG=(1+CG)E R G 1+CG—1+CG 4b) Show that the steady state error will be 0 for any unknown constant w D where R, D are the Laplace transforms of r(t), d(t) :>E= e(ao) = limHu SE '_.. _ A- ! PIS+G A E=R—Y,Y=G( 5+0) R—E=G(CE+D):>E= R — DG 1+GC 1+GC 4b)Y=G(C(R—Y)+D):>Y=CGR+GD—CGY :>Y= CG R+ G D 1+CG 1+CG K 1 l “ z ‘ '2 k 2 Y: SS +s+l R+ S +s+l Djr: R+ D 1+K 1 l k l 3(sz+s+l)+K s(sz+s+l)+K _T 7 +,. is +s+l ssz+s+l s +s2+s+K y+s +s+K 3 2 Therefore, the poles ofthe system are the roots of equation S + s + S ‘l' K =0 (R3,l,l),(R2,l,K),(Rl,l -k,0),(R0,K???_?) By the Routh-Hanzwix Criteria, all the poles will bein the lfl-half-plane ifand only if l-K>0, K>O) 0<K<l Therefore the system is stable it‘and only if 0<k<l c) e(oo) = limH0 sE 1 w . SR . sap . S; . SEC =llmHo -hm,so =1ImHo K -hm.so K 1+CG 1+CG 1+_G 1+_G S s =lim S ' SW0 —o—o=o J-w ‘llms—w - S + KG S + KG '-\ Consider the unit feedback system shown in Fig 2. Suppose that the controller is a proportional controller with gain K Let 0(5) = l . s(s + 2) 3a) Let r([) : (l + [)1([). Determine controller gain K so that the steady state error is 155 than 2? KG R NotethatES =R —Y :R___R=__ () (S) (s) 1+KG, 1+KG where S2 1 4m)=lim,_,os 1 1 1 l+k—- r + R = L(R(t)) = EJ— sofore(00)<2.weneed2<2:>K>l 2 3b) Let r(t) = t . Show that it is not possible to specify K so that the steady state error is finite. . . . l . k . 5"” lth0 SKG = lth0 SK — = 11mm0 — 'SWWW s(s + 2) s + 2 2 constant, the system is oftype 1. Therefore, the system can not beoftype 2. 3c) Detemtine controller gainKsothat the output hasantwershoot ofno more than 15% in response toaunit stepinputsiytal. K h w — k :_ KG _ s(s+2) _ k _ w] w” n ' R 1+KG H k s2+25+k sl+2gwns+wf s(s+2) 21an = 2 bviously. W" Z J1? I-C _ 1 £1 _ _ 0 Formp —e <15Aweneedg> lln(0.15)l :05? ,/12+|1n(0.15)| 1 1 2 —>0.52:>k>(—) 23.9 J? 052 For the system to have comple poles, we also needo < 4 <1 <=> 0 < < 1 :> k > I J? Combining (l) and (2), we have that ‘k>3.9' is sufficient to guarantee the requirement ' fiv 7 w 1000 _i l (s2 +s+l)40000 40 s2 +s+l Note that the time-domain specification. c m t [x ofG are the same that of p, r, 1 G: 2 s +s+l because oflinean'lty 2 m l _ w” z _ 2 wehave W" = land 24W” =1 hence s +s+l s2+2§wns+wn 1 ‘ If; _,lHl>Z C=§m =e “‘ =e 2 :0.163,z, = 4’6 =fl=92 " CW" 1 2 K L j Y _ KG _ S(S+2) _ K (R2,l,K),(Rl,2.0),{R0,K,0} R 1+KG H K SZ+ZS+K S(S+ 2) By Routh-Hurtwiz Stability Criteria, the system is stable ifan donly ifK>0. 1 1 R(s) = L(r(t)) = - + — 3b) 2 hence l l l l 5 s Ets)=——1(;+s—2) E s = R 3 1+ K— () 1+KG () s(s+2) By final value theorem . . l 1 1 2 e(c:o)=l1mx_,o sE(s) = limHo s K (— +~2—) = I < 0.001 1 + __ S S s(s + 2) 3c) Note that the reference r(t) is oftypezfor arbitrary 90. The system is oftype 1 thus it is not possibleto specinysothat the steady state erroris finite no matter how small Gis [Q I cribe the integal—diffemttial equation ol‘a PlD ' de(l) controlleru(t) = kPe + k, £e(t)dl + k0 7 lb) Describe the transfer function ofaPlD controller = k + .11]. + sz E(s) P 5 lo) Describe the procedure ofthe Ultimate Gain Method to design a PI controller Consider the system show. Let r(t) = 1(t), Let the distyurbance signal be 2 where ‘w‘ is an unknownoonslant. Letthecontrollerbe C(s =l+l. Let 6(5): 1 . 3+2 @Showthaty(s)= R+ DwhereKDaretheLaplacelransfonnsofKt),d(t) 1+CG 1+CG respectively. E=R—YandY=G(CE+D) R :>Y=G[C(R-Y)+D]:>(1+CG)Y=CGR+GD:>Y= CG + GD 1+CG 1+CG 3b) Show that the steady state error will be 0 for any unknown constant w l W _ G_ E=R—Y: R .— GD 25=+— 5] 1+CG 1+CG 1+1F “510 A A e(m)=11mm‘,s£=1ith0 1 —lim,floi=lim1a04—limmnfl: _ = l+.t_+lG l+§+_lG s+(s+l)G S+(S+l)G .Y S I 3c) Desigt a controller C such that the steady state error will be 0 for disturbance signal n d([) = w] (Ii—l wheren is a given natural number andw;, i=l, ,.,, n are unknown constants. I=| n—l The reference Sign] is type 0, the disturbance contains component ofthe form t 10 achieve zero steady state, the system should be type n or 5&0 S "CG qt 0 . For simplicity, we choose I l l - - . _ ' '1 _. ' _ . A n orous proof tsas follows. — —n 311th0 5 CG — 11mmo 3 s s+2_“2_ E: R _ GD mmR=l,C=i_&D=:(i-1)1W 1+CG 1+CG s s" ,=. s‘ ‘ 1 G: (i :1)! w. H G:(i_ WNW; E: S _ l-l 3 = 5 _ M 1+‘1_ “LG s"+G s"+G 3.6 x. ” GZ(i—1)!x"“" . . S . = W4 '=l 2... n e00=lt 35-1 11 " —0 0=of°' I" y y ( ) mtso HIE—.0ch mm “+0 4 _ ‘ - A 'w Deten'nine PlD controller parameters according to the ultimate gain method ‘-. l0, Pu=0.25. PlD: K=0.6Ku=6. Tz=(l/2)Pu=0. ls. Td=(l/8)Pu=0.0255 '6 (s)_ (s+lOO)(s+500) C(S)_ 100(500) _ 12.5 V _ (s2 +s+ l)(s+4000) " 4000(52 +s+ 1) Due to the linearility. the parameters for this is the same as sz+s+l z 2 _ _ _ _ GO): 2 l :Tifi w” —1,w,, —1,2§’wn —l,{—0.5 s +s+l s +24’w‘,s+wu _ n; _ n(o.5) M =e W =e lbw-5’2 =o.163, =fl=fl=923 p I Cw" 0.5 I N . '% ow that the feedback system IS stable K 1 F“): Y(s)_ K(s)G(s) : s(s+1) _ 2 k ,GQ): l R(s) l+K(s)G(s) 1+K l ' s +s+K s(s+l) s(s+l) l k l 0 lSZ,l,k),(Rl,l,o),(R0,)c) x = —— = k inordertobestable—>k>0 l b) R(t)=(l+t)(l(t), R(s)=L[r(t)\=(l/s)+(l/s"2) KGR R Es =Rs -Y5,Ys =—,ES :— () () ()() HKG ()1+KG . R . l l l l e(°°)= llmtsoSHKG =11m1s05(;+s—2)(1+—K)=I s(s+1) lfe <0.00l-> l/k 0001-) K>l000 {[3] or the feedback system. Let G ( S) = 1 ,which is type (n-l). Thus, @R-Y -> Y=R-E -> Y=G(CE+D)=R-E -> R=E+CGE+DG=E(HCG)+DG~> E=R/(I+CG)-GD/(1+CG) s(s + 2) 2a) Describe the conditions on the controller gain K so that the closed loop system is stable k 5(5 »+ 2) k Let 52+25+k=0 => two roots I + k _ 51 + 25+ k 2(s + 2) Y(s)_ k s _—2:J4—4k R(s) s2+2s+k’ "2 2 We want the real part ofs, and $1 to be negative. We consider three cases Casel: kSO,Case11:0<kSI_Case1n:k>l InCasel, v4—4k > x/Z = 2 husoneofthepolesis _2+___ V4_4k> 0 2 C0 the closed loop system is unstable InCNIL—2+V4“4k <0and —2-—\j4—4k <0 2 2 I$ the closed loop system is stable lnCasel'l], k >12, .[4 _ 4k isanimaginarynumber =3 all the poles in open lefi halfpltute b) Detemtine controller gain K so that the output has an overshoot ofno more than l0% in response to a unite step ' SEE P324 c) Let ,- (l) = (I + ,)1 (I) . lsitpogibletospecinyso thatthesteadystateerrorisless than I? Find the‘controller gain K orjustify whin is not possible. R(s)=L(r(r))=%+;lz—E(5)_ 1 R_ 1 LL —1+KG ‘1 1 (5 s1) s+2 e(oo)=lim _ Ho 52 + 25+k (s + : %Thus < i... 2 d) Let : +t . Is it possible to specify Kso that the steady state error is less than I? Find the controller gain K orjustify why it is not possible 13(5) = L(r(t)) = L((1+ (51(1)) = 1+ _23_ S S ~ I 2 e(°°>=hm,-. ‘ H-) l+k l 2 5’ 5(s+2) . 5+2 2 ‘l'mmom(s+;)—°° °) r(t) = (21(t),e(oo) = limHo SKG = limHo SK - This indicats the system is type l, Type l systems cannot track type 2 signal with finite error. show where K D are laplace .. 4b) show stability Y=R-E=R-Rl( l +CG)+GD/(l +CG)=CGR/( l +CG)9-GD/( l+CG)=(CGR+GD)/(1+CG) C(S)=K/S, G(S)l/(S+|) (R2 1, KHRl. l, 0) (RD. K) Wl-HSN K>0 FEEDBACK SYS STABLE _ CGR + CD _ 3(s + l)(CSR + CD) ‘HE‘L’ Si+S+K 3 5+1 AC) Show steady state error will be 0 R(S)=A/S, D(S)=W/S, E(S)=R/(HCG)-GD/(l +CG) l’ a w l 5— 51*] e(w)=l1ml_msE = lth,, ~ l1m,__‘, l+—— l+—— 35+] sx+l . as s+l . sw —lth_,0 Z( )—lth_,0 2 -0—0—0 3 +s+ s +s+k It is clear that e() is not dependent on a and w. e()=0 for any constant a and any unknown constant w. a, Let G be a linear stable system with frequency response shown by Figure 1. Let the input be r(t) = [3 sin(t) + 2005(10t) + sin(0.1)]1(t). Determine the steady state response y(t). 3a sin(t + (9) - 3 = 2010;;10 x :> x = 10"3/20’ Thus: y(t) = 3(0.7079)sin(t — 45° ) + 2(0.1)cos(10t — 85“)+ l(1)sin(0.1t — 5°) . _...‘_#/—*\W Mflwu. SO4#2 Consider the feedback system show by Fig2. Derive the sensitivity function with respect to the variation of plant model G. What is the significance of studying the sensitivity function? SUM : Cast 6:81 mm W! unc : = = — = — \x _/ 1+CG 1+( 1 ME) H k _ s + a s s2 + as “raw”... . . .o _ , .. ..._,. a- .._,!t’s.the faster trimming.“ .SXEtem error ' 5 +1 55 +s“ +53 +232 +7s+K parameter K so that the system is stable? Justify your answer or determine the range ofK that guarantees stability. _-'-'-----—_. _.. __... --m—-——.....m.____..-_. __. 3‘\‘_II-r wrSO4il9l3 Consider a system with transfer function G“) _ where K is a parameter. [s it possible to find 1 1 7 la cl II 1| |17| |10| l 2 K X b d l 2 _ l unstable because negative!!! x __ l k _ (k 7) x = 1 0 _ 0,etc, etc -1 7-k 0 b l 1 ’ l 9-k K 0 0 0 Note: if no negatives, you would find K that satisfies all equations. faw_ '.'\—'|.lk_"-C\I ---.-- -—-A--lm¢--"-=--- --=- “MI-W—rnuw—umuu“h,flw_ “I” SO4#4 Consider a feedback system shown in Fig2. The transfer funct of the plant is 0(3) 2 1 where p E [~1,1] is an uncertain s + p parameter. The transfer funct of the controller is as) = fl+ k . Determine the conditions for Ki and Kp so that the feedback is stable [I S for ANY value of the uncertain parameter p 6 [—1,1] k, 1 + k )- T(S)_ CG _ S p S+P _ ki+kps _ ki+kpS Row2 I] A2 _1+CG_ k, 1 _s(s+p)+k.+ks—s2+(k +p)s+k. Rm“ A‘ 0 (~+kP)—- I P P I Row0 A2 0 l + . S S k p + p > 0 k p > 1 <-(needs to be positive) A l>0, a2>0 kl. > 0 k, > 0 W‘s-("1&8 Consider a feedback system shown in Fig2. The transfer funct of the plant is 0(5) = A) Suppose that the reference 32 + 25 + 1 input signal is r(t)=l(t) and that C(s)=K is a proportional controller with gain K. Can the steady state error be arbitrarily decreased by increasing the controller gain K? B) Suppose that the reference input signal is r(t)=tl(t) and that C(s)=K/s is an integral controller with gain K. What is the smallest error can be achieved by increasing gain K? 1 1 1 E(s)=—R=——— _. 1 1 a) 1 + 5 S 9(00) — llms——>0 *5— 1+k2— 1+1. . .s S +23+1 s2+23+1 k 5 .7?___.._ 2 Tm: CG 2 s +2s+1 : 2 5k e(oo)=1imHO 1 zlimHO 7s +2s+1 z 1 1+CG 1+k__ 5 s +2s+5k+1 1+k 5 s‘+2s+5k+1 5k+1 sz+25+l s2+25+1 £__ 5 ___ 2 b) T(s)= : SS +2S+1 = 2 = 3 1+CG 1+7]: 5 s(s +2s+1)+5k s +2s +s+5k ssz+23+l ,I -(5/2)k>0 —> 0<k<(2/5) 5k>0 a -.._— lm‘fl‘w “nu-n... nan ks; fin-{I‘M " Steady-state error: 8(00) = lim_H0 SE(S) yrrwk . r? 0 4 Tm-r‘v’ 1 ' _ er ' I 1 1 1 15= K—{wr-u) H?" ‘ E(s)= =———, v = (Made?) my: = e ,3, 1+CG 1+£7M 5 S' v: Ga+<e[/~“"W"V\3 Hc’ T] 6 L.;+§L-:g"""" s 52 +2s+l 493739.93"? _ - — r d” ’ "" 7 33163534959: . 1 1 . 1 1 . 1 . , s2+2s+1 1 , e(oo)=l1ms———;=11m——=llm——=11m%=——)O>lh 1+5 5 s 1+£_7V_5 S S+k 5 s(s +2S+1)+5k 5k ssz+25+1 ss2+25+l s2+2s+1 _h_i_l__M._._.«.n__1'-c .....-...~n~.-.—.... .1 . _‘___7_ J I \ I.“ . tuna :memnuw- _ _ _ __ l where a is an uncertain parameter N303?“ ConSIdei‘ a le’éflfiacfi system shown in Fig2. The transfer function of the plant is 6(5) = s+a varying in (0, 00) . The transfer function of the controller is as) = Determine the condition for K so that the feedback system is s stable for ANY value of uncertain parameter a. 1 k k _ia 0‘ ~(0—ak) ak Us): CG ___ sz+as : k xl=_=——————=—= , I+CG 1+7k 7 52+as+k a a a sz+as u _ -..-~. ‘ ‘ _u_ _u “a _._-mu“... J. _ __I L" F ‘ .v. "'"'" —-_.‘ _._ ._- :- SO3#5 Electronic Circuit with 4 op—amps R(T)-)SUM-)AMP-)INTEGRATOR-)y(t) / (-—-_—switch —----AMP ------------- -- I No feedback if switch is switched. _ _ ,_--’ .r' ‘ Fa “' ._.r,. ...
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Controls Formula Sheet Test 2b - Desigt a PE controller...

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