6.302 Feedback Systems Recitation 6: Steady-State Errors Prof. Joel L. Dawson S -

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1 6302 Feedback ytem Recitation 6: teadytate Error Prof Joel L Dawon A valid performance metric for any control ytem center around the final error when the ytem reache teadytate That i, after all initial tranient have died away, how cloe i the output to the deired output? Look at the error in a feedback ytem r(t) + e(t) c(t) G() r(t) e(t) c(t) G() Ue Black Formula: E() = + G() For teadytate, we re intereted in t e(t) Uing the final value theorem, we can evaluate in the frequency domain uing 0 F() Now, what kind of input might interet u when dicuing teady tate error? Let look back at our poition control ytem from recitation 2 (We ve added damping o that there no ocillating) A F m x x x β Page Cite a: Joel Dawon, coure material for 6302 Feedback ytem, pring 2007 MIT OpenCoureWare ( Maachuett Intitute of Technology Downloaded on [DD Month YYYY]

2 6302 Feedback ytem Recitation 6: teadytate Error Prof Joel L Dawon If we wanted to command the ma to be moved, ay, from the origin to x, our command might look 0 omething like: x 0 x (t) = x u(t) X CMD () = CMD 0 t x 0 uppoe, intead, we wanted the block to be moved at a contant velocity If we didn t have a velocity enor handy, our command would take the form of a ramp: d dt =v t x = v tu(t) CMD 0 X CMD () = v 20 Or, we could be intereted in moving with a contant acceleration Lacking a good accelerometer, we could achieve thi with our preent control trategy uing a parabolic input: ½ a 0 t 2 t = ½ a 0 t 2 u(t) X CMD () = a 0 3 Page 2 Cite a: Joel Dawon, coure material for 6302 Feedback ytem, pring 2007 MIT OpenCoureWare ( Maachuett Intitute of Technology Downloaded on [DD Month YYYY]

3 6302 Feedback ytem Recitation 6: teadytate Error Prof Joel L Dawon It not diffcult to imagine more complicated input to a control ytem Conider a rotating camera to track a running athelete: vt d Θ Figure by MIT OpenCoureWare We have: TAN Θ = vt d Θ (t) = TAN ( vt d ) Now we don t have a Laplace Tranform for TAN However, we can write a Taylor erie for TAN : Θ(t) = a + a t + a t 2 + a t Θ() = a 0 + a + 2a 2 + 6a DC teadytate Error X() Returning to our original ytem: G() Y() E = + G() By DC teadytate Error, we mean the final error in repone to a tep Page 3 Cite a: Joel Dawon, coure material for 6302 Feedback ytem, pring 2007 MIT OpenCoureWare ( Maachuett Intitute of Technology Downloaded on [DD Month YYYY]

4 6302 Feedback ytem Recitation 6: teadytate Error Prof Joel L Dawon Mathematically: CLA EXERCIE: t e(t) = 0 + G() Find a function G() for which the error in repone to a tep i zero It i intructive to work thing out for different type of ytem: X() k τ + Y() In repone to a tep: 0 k = + + k τ + (The more we turn up the gain, the better we do) In repone to a ramp: 0 = 2 k + τ + (Error grow without bound) Page 4 Cite a: Joel Dawon, coure material for 6302 Feedback ytem, pring 2007 MIT OpenCoureWare ( Maachuett Intitute of Technology Downloaded on [DD Month YYYY]

5 6302 Feedback ytem Recitation 6: teadytate Error Prof Joel L Dawon What can we do about that ramp? Try: X() k() 2 Y() Where k() ha no ingularitie at the origin In other word, k() =k =0 0 Now the error for a ramp i: k() k() = 0 o for zero error in repone to a tep ( ), we needed one integrator For zero error in repone to a ramp ( ), we needed two integrator Can we dicern a pattern? 2 Let input be X() = M generate expreion:, and let forward path be k() N, where k(0) = k 0 apply final value theorem to 0 M + k() N = 0 N M N + k 0 Page Cite a: Joel Dawon, coure material for 6302 Feedback ytem, pring 2007 MIT OpenCoureWare ( Maachuett Intitute of Technology Downloaded on [DD Month YYYY]

6 6302 Feedback ytem Recitation 6: teadytate Error Prof Joel L Dawon Nm+? 0 = 0 N + k 0 In order to be zero, we mut have N M + N M o in order to get zero teadytate error, we require: Input tep Ramp Parabolic # of origin 2 3 Now, to reemphaize a couple of point from lecture: ) Ocillation condition G H Break loop here + e jω 0 k ee what come back inject e jω 0 k Page 6 Cite a: Joel Dawon, coure material for 6302 Feedback ytem, pring 2007 MIT OpenCoureWare ( Maachuett Intitute of Technology Downloaded on [DD Month YYYY]

7 6302 Feedback ytem Recitation 6: teadytate Error Prof Joel L Dawon If what come back i jut e jw 0 t, that i, L(jω ) =, and L(jω ) = We have an ocillation condition, or a pole in the cloedloop tranfer function: G() +L() if L(jω ) =, and L(jω ) = denominator i zero WARNING: BEWARE OF FALE INTUITION! L(jω ) = 80 AND L(jω ) > 0 0 DOE NOT necearily imply intability/ocillation Why? Becaue in thi cae, + L(jω ) i not 0 zero, o jω i not a pole of the cloedloop ytem It i very common to get thi wrong Don t let it 0 happen to you!! Finally, a word about root locu It anwer the quetion: How do the cloedloop pole of the ytem move a I vary the gain k? k G() H() More on thi next week Page Cite a: Joel Dawon, coure material for 6302 Feedback ytem, pring 2007 MIT OpenCoureWare ( Maachuett Intitute of Technology Downloaded on [DD Month YYYY]