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Time domain response specifications Defined based on unit step response with i.c. = 0 Defined for closed-loop system

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  • Time domain response specificationsDefined based on unit step response with i.c. = 0Defined for closed-loop system

  • Prototype 2nd order system:target

  • Prototype 2nd order system:

  • Settling time:Remember:

  • +-Example:When given unit step input, the output looks like:Q: estimate k and .

  • Effects of additional zerosSuppose we originally have:i.e. step responseNow introduce a zero at s = -zThe new step response:

  • Effects:Increased speed,Larger overshoot,Might increase ts

  • When 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.

  • L.P.F. has smoothing effect, oraveraging effectEffects:Slower,Reduced overshoot,May increase or decrease ts

  • StabilityBIBO-stable:Def: A system is BIBO-stable if any bounded input produces bounded output. Otherwise its not BIBO-stable.

  • Asymptotically StableA system is asymptotically stable if for any arbitrary initial conditions, all variables in the system converge to 0 as t when input=0.A system is marginally stable if for all initial conditions, all variables in the system remain finite, but for some initial conditions, some variable does not converge to 0 as t.A system is unstable if there are initial conditions that can cause some variables in the system to diverge to infinity.

    A.S., M.S. and unstable are mutually exclusive.

  • Asymptotically Stable

  • Asymptotically Stable vs BIBO-stableThm: If a system is A.S., then it is BIBO-stable

    If a system is not BIBO-stable, then it cannot be A.S., it has to be either M.S. or unstable.

    But BIBO-stable does not guarantee A.S. in general.

    If there is no pole/zero cancellation, then BIBO-stable Asymp Stable

  • Characteristic polynomialsThree types of models:Assume no p/z cancellationSystem characteristic polynomial is:

  • A polynomial

    is said to be Hurwitz or stable if all of its roots are in O.L.H.P

    A system is stable if its char. polynomial is HurwitzA nxn matrix is called Hurwitz or stableif its char. poly det(sI-A) is Hurwitz, orif all eigenvalues have real parts

  • Routh-Hurwitz MethodFrom now on, when we say stability we mean A.S. / M.S. or unstable.

    We assume no pole/zero cancellation,A.S.BIBO stableM.S./unstable not BIBO stable

    Since stability is determined by denominator, so just work with d(s)

  • Routh Table

  • Repeat the process until s0 row

    Stability criterion:d(s) is A.S. iff 1st col have same signthe # of sign changes in 1st col = # of roots in right half plane

    Note: if highest coeff in d(s) is 1,A.S. 1st col >0

    If all roots of d(s) are

  • Example:has roots:3,2,-1

  • (1*3-2*5)/1=-7(1*10-2*0)/1=10(-7*5-1*10)/-7

  • Remember this

  • Remember this

  • e.g.

  • Routh CriteriaRegular case: (1) A.S. 1st col. all same sign(2)#sign changes in 1st col. =#roots with Re(.)>0

    Special case 1: one whole row=0Solution: 1) use prev. row to form aux. eq. A(s)=0 2) get:

    3) use coeff of to replace 0-row 4) continue as usual

  • Examplewhole row=0

  • Replace by e

  • Useful case: parameter in d(s)How to use: 1) form table as usual2) set 1st col. >03) solve for parameter range for A.S.2) set one in 1st col=03) solve for parameter that leads to M.S. or leads to sustained oscillation

  • Examples+3s(s+2)(s+1) Kp+

  • Q: find region of stability in K-a plane. aK

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