Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Prototype 2nd order system: 2222nnnsssH22,11, :PolesndndjpStep response: ssssYnnn12222   tutetutysdts212σ1tansin1)(   tutetesdt212 σ1tansin-1Step Response Specs & Parameters2dn= = 1ptp pww z-21σ max1 1tpy e epzz---= + = +21Overshoot : pM epzz--=21percentage : 100%epzz--= �1.5 2 1.8Rise time : rn ntw w-= �()24.6 ln 1stzs- -�Settling time:= (3 or 4 or 5)/In the complex plane :cosnsj zw= =Any pole on the ray havethe same Mp < … or  > … corresponds to a conic region about the neg real axisConstant σ : vertical lines“σ > # ” is the half plane to the leftConstant ωd : horizontal lineωd < · · · is a bandωd > · · · is the plane excluding bandConstant ωn : circlesωn < · · · inside of a circleωn > · · · outside of a circleConstant ζ : φ = cos-1ζ constantConstant ζ = ray from the origin ζ > · · · is the cone ζ < · · · is the other partIf more than one requirement, get the common (overlapped) areae.g. ζ > 0.5, σ > 2, ωn > 3 givesSometimesmeeting twowill also meetthe third, butnot always.Example: Given 0.6, 5, estimate specs.nz w= =sol: 0.6 5 3ns z w= = � =21 4d nw z w= - =peak time: 0.8sec.pdtpw= �210.96 10%pM ezz--= = �settling time for 1% :�5 51.67sec.3sts= = =42% : 1.33sec.3stol t� = �1.8rise time: 0.36sec.rntw= =Try to remember these:3 4 5 or or 5% 2% 1%sts s s=� � �5% or 10% or 16% or 25%0.7 0.6 0.5 0.4pMz==1.8rntw�( )1ks st ++-Example:When given unit step input, the output looks like:i.e. 25%pM �3sec.pt �Q: estimate k and τ.sol: from 25% 0.4pM z޻�3sec.3p dpttp pw� � = =21nz w-2 21.141 3 1 0.4dnwpwz\ = = �- -From block diagram:( )221k kH ss s ks s ktttt= =+ ++ +22 2Match this against 2nn ns swzw w+ +2 21.1412 2 0.4 1.14nnk t wzwt�= =��= = � ���1.091.42kt =��=�Solve for and , get:ktEffects of additional zerosSuppose we originally have:( )0H s( )0y ti.e. step responseNow introduce a zero at s = -z( ) ( )01sH s H sz� �= +� �� �The new step response:( ) ( ) ( )01 11sY s H s H ss z s� �= = +� �� �( )01sY sz� �= +� �� �( ) ( )011dy t y tz dt� �\ = +� �� �)(tus     tytyzty001 beginning.at rising is Typically,0ty 01,0 If)10 tyzz    overshoot.larger has and , before overshoots ,n faster tha rising is 00tytyty 01,0 If)20 tyzz    initially 0 i.e. direction in wrong offstart may ? overshoot smaller n slower tha rises 0tytyty:0 e.g. zEffects:•Increased speed,•Larger overshoot,•Might increase tspdrttt ,, i.e.pMWhen z < 0, the zero s = -z is > 0,is in the right half plane.Such a zero is called a nonminimum phase zero.A system with nonminimum phase zeros is called a nonminimum phase system.Nonminimum phase zero should beavoided in design.i.e. Do not introduce such a zero in your controller.Effects of additional poleSuppose, instead of a zero, we introducea pole at s = -p, i.e.     0110 psHsHps     ssHssHsYps11110 sYpsp0filter pass loworder first a is psp.at frequency corner with pL.P.F. has smoothing effect, oraveraging effectEffects:•Slower,•Reduced overshoot,•May increase or decrease tspdrttt ,,