ME 365: SYSTEMS, MEASUREMENTS, AND CONTROL SMAC) I Dynamicresponseof 2 nd ordersystem Prof.SongZhangMEG088)
Solutions to ODEs Forann@thorderLTIsystem a n yn) + a n 1 y n 1) ++ a 1 "y + a 0 y = b m u m) + b m 1 u m 1) ++ b 1 "u + b 0 u SdomainsoluFon: Y Free s) = Y Forced s) = ~ + ~ ++ s p 1 s p 2 ~ s p 1 + ~ s p 2 ++ yt ) = y ) t + y ) t Free Forced ~ s p k )s p k+1 ) + ~ s p k+2 )s p k+3 ) + ~ s p k )s p k+1 ) + ~ ++ F[Us)] s p k+2 )s p k+3 ) Fornonerepeatedpoles,systemresponsetypically canbedecomposedintothecombinafonof - Zeroordersystemresponse - 1stordersystemresponse - 2 nd $order$system$response$
Typical inputs Unitimpulseresponse δt)dt =1 ut)=δt) Us)=1 Unitstepresponse 0 t < 0 ut ) = 1 t 0 ut)=1 Us)= 1 s ut) ut) Sinusoidalresponse:Frequency$response$ ω ut ) = sin ωt)' ' Us ) = 2 2 s + ω 2 0 0 Timet" Timet"
2nd order system: Free response Fora2 nd ordersystemwithinifalcondifons y + d 1 y + d 0 y = b 1 u + b 0 u y0) = y 0 y0) = y 0 - CharacterisFcequaFon: - Poles Freeresponseins@domain
2nd order system: Free response FreeresponseinFmedomain - RealanddisFnctpoles - RealandidenFcalpoles - Complexpoles
2nd order system: Free response TwodisFnctrealpoles y Free t) = Ae p 1 t + Be p 2 t y Free t) = Ae p 1 t + Be p 2 t p 1 < 0 & p 2 < 0 Img. p 1 < 0 & p 2 = 0 Img. y Free t) = Ae p 1 t + Be p 2 t p 1 < 0 & p 2 > 0 Img. Real Real Real y H" t) y H" t) y H" t) Timet) Timet) Timet)
2nd order system: Free response TwoidenFcalpoles y Free t) = At + B)e p t y Free t) = At + B)e p t y Free t) = At + B)e p t p 1 = p 2 = p < 0 p 1 = p 2 = p = 0 p 1 = p 2 = p > 0 Img. Img. Img. Real Real Real y H" t) y H" t) y H" t) Timet) Timet) Timet)
2nd order system: Free response Twocomplexpoles y Free t) = Ce σ t sinωt +φ) y Free t) = C sinωt +φ) p 1,2 =σ ± jω &σ < 0 p 1,2 = ± jω &σ = 0 Img. Img. y Free t) = Ce σ t sinωt +φ) p 1,2 =σ ± jω &σ > 0 Img. Real Real Real y H" t) y H" t) y H" t) Timet) Timet) Timet)
2nd order system: Forced response Stable2 nd ordersystem y + d 1 y + d 0 y = bu y + 2ζω n y + ω n 2 y = Kω n 2 u ω n >0:NaturalFrequency[rad/s] ζ>0:dampingrafo K:StaFcSteadyState,DC)Gain Poles s2 +2ζω n s +ω n 2 = 0 s = ζω n ±ω n ζ 2 1) - ζ>1:twodisfnctrealroots/poles Img. ω n" - ζ=1:twoidenfcalrealroots/poles - ζ<1:twocomplexconjugateroots/poles Real )ω n"
2nd order system: Unit step response Under@dampedstable2 nd ordersystem y + 2ζω n y + ω 2 n y = Kω 2 n u CharacterisFcequaFon s2 +2ζω n s +ω n 2 = 0 s = ζω n ±ω n ζ 2 1) Unitstepresponse
TimedomainsoluFon 1 yt) = K 1 1 ζ 2 e ζω nt sin ω d t +φ) whereφ = tan 1 1 ζ 2 andω ζ d = ω n 1 ζ 2
Pole location and time response 1 yt) = K 1 1 ζ 2 e ζω nt sin ω d t +φ) whereφ = tan 1 1 ζ 2 andω ζ d = ω n 1 ζ 2 Im ω n" Re )ω n"
2nd order system: Unit step response yt) = K 1 1.6K 1.4K 1 1 ζ 2 e ζω nt sin ω d t +φ) whereφ = tan 1 1 ζ 2 andω ζ d = ω n 1 ζ 2 UnitStepResponse 1.2K K 0.8K 0.6K 0.4K 0.2K 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Time[sec]
2nd order system: Unit step response Peak$4me$forunderdampedsystem:t p yt) = K 1 e ζω n t cosω d t)+ ζ sinω 1 ζ 2 d t) PeakFme: π π t P = = ωd ω 1 ζ 2 n
2nd order system: Unit step response Overshoot$forunderdampedsystem:OS OS = y MAX y SS = K e πζ 1 ζ 2 Overshoot$percent$OS%) %OS = OS y SS y0) 100% = 100% e πζ 1 ζ 2
2nd order system: Unit step response Se9ng$4me$t s - Timerequiredfortheresponsetobewithina specificpercentofthefinalsteady@state)value. SometypicalspecificaFonsforseelingFmeare:5%, 2%and1%. %$$ 36.8% 13.5% 5% 2% 1% SeelingTimet S ) 1 2 3 3.9 4.6 ζω n ζω n ζω n ζω n ζω n
Under-damped 2 nd order system time response Damped$natural$frequency$ ω $ d = ω n 1 ζ 2, T d = 2π ω d Peak$4met P ","0< <1) t P = π ω d = π ω n 1 ζ 2 Overshoot$OS,"0< <1) OS = K e πζ 1 ζ 2 = K e πζ ω n ω d UnitStepResponse 1.6K y MAX 1.4K 1.2K K 0.8K 0.6K 0.4K 0.2K t P" OS" T d" 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 X%$se@ling$4met S )" Time[sec] t S = 1 ln 100 ζω n X,assumingζ issmall t S" Percent$overshoot$%OS)$ %OS = 100 e πζ 1 ζ 2 = 100 e πζ ω n ω d %$$ 5% 2% 1% Seeling Timet S ) 3 3.9 4.6 ζω n ζω n ζω n 18
Example 1 Onthecomplexs)planeidenFfytheregionwhere thepolesofasecondordersystemwillmeetthe followingspecificafons: - 5%seelingFme,t S 5%),lessthan1second - %OS<10%
Example 2 Fortheass@spring@dampersystem,whatisthestaFcsteady@ state)gainofthesystem?andhowwouldthephysical parametersm,b,k)affecttheresponseofthesystem? Mx+Bx+kx = ft) k" B" M" x" ft)
Transient and steady-state response ForaLTIsystemsubjecttoaninputut) a n yn) + a n 1 y n 1) ++ a 1 "y + a 0 y = b m u m) + b m 1 u m 1) ++ b 1 "u + b 0 u Totalresponsecanalsobedecomposedintotwoparts: yt) = y T t) + y t) SS Transient Response SteadyState Response Transientresponse:y T t) - decaytozeroataratethatisdeterminedbythecharacterisfc rootspoles)ofthesystem Steady@stateresponse:y SS t) - takethesameformastheforcinginputut) - Asinusoidalinput,thesteadystateresponsewillbeasinusoidal signalwiththesamefrequencyastheinputbutwithdifferent magnitudeandphasewill$be$discussed$later)
Steady-state response FinalvaluetheoremFVT) - Givenasignal slaplacetransformfs),if$the$poles$of$ sfs)$all$lie$in$the$lhp$stable$region),thenft) convergestoaconstantvaluef ).f )canbe obtainedwithoutknowingft)byusingthefvt f ) = lim ft) = lims Fs) t s 0 LTIsystemtransferfuncFon Gs) b m sm + b m 1 s m 1 ++ b 1 s+ b 0 = b s z )s z )s z ) m 1 2 m a n s n + a n 1 s n 1 ++ a 1 s+ a 0 a n s p 1 )s p 2 )s p n ) Steady@statevalueofthefreeresponse Fs,IC's) Y Free s) a n s p 1 )s p 2 )s p n ) y ) = lim y t) = lim s Y s) Free Free Free t s 0
Steady-state response Steady@statevalueofunitimpulseresponse Ys) = Gs) U s) = Steady@statevalueoftheunitstepresponse Ys) = Gs) U s) =
Forced response of harmonic input Steady@stateresponseofinputut)=sinωt) - 1stordersystem: - 2 nd ordersystem: - Forcedresponse: b0 Gs ) = as+ a Gs ) = ω Ys) = Gs) Us) = Gs) s 2 +ω 2 1 0 bs+ b as as a 1 0 2 2 + 1 + 0
Example 3 Forthefollowingsystem,ifainputut)=5is appliedatfmet=0 y + 4 y +12y = 4 u + 3u - Isyt)convergingtoaconstantvalue? - Ifyes,whatwillbethesteady@stateoutput?