Today’s goalsGain and phase marginsExample 1Example 2Transient from closed-loop frequency response /1Transient from closed-loop frequency response /2Transient from closed-loop frequency response /3Transient from open-loop phase diagramsExampleExample: Proportional control in the frequency domainGain adjustment for phase margin specificationSteady-state errors from the frequency responseExampleLecture 32 – Monday, Nov. 262.004 Fall ’07 Today’s goals• Last week– Frequency response=G(jω)– Bode plots• Today– Using Bode plots to determine stability• Gain margin•Phase margin– Using frequency response to determine transient characteristics• damping ratio / percent overshoot• bandwidth / response speed• steady-state error– Gain adjustment in the frequency domainLecture 32 – Monday, Nov. 262.004 Fall ’07 Gain and phase marginsFigure 10.37Gain margin:the difference (in dB) between0dB and the system gain,computed at the frequencywhere the phase is 180°Phase margin:the difference (in °) betweenthe system phase and 180°,computed at the frequencywhere the gain is 1 (i.e., 0dB)A system is stableif the gain and phase margins are both positiveM (dB)Gain plot0 dB}log ωGMPhase plotPhase (degrees)ΦM180olog ωω}ΦωMGMFigure by MIT OpenCourseWare.Lecture 32 – Monday, Nov. 262.004 Fall ’07 Example 1Open—Loop Transfer FunctionKG(s)H(s)=20K(s +1)(s +2)(s +5).DC gain = 20 = 20lo g1020(dB) ≈ 6dBBreak (cut—off) frequencies: 1, 2, 5rad/sec.Final gain slope: − 60 dB/dec.Total phase change: − 270◦.PhaseMarginGainMarginBreak frequenciescurrentclosed-loopdominantpoles(K=1)Increasing the closed-loop gain by an amount equal to the G.M.(i.e., setting K=+G.M. dB or more) will destabilize the systemLecture 32 – Monday, Nov. 262.004 Fall ’07 Example 2Open—Loop Transfer FunctionKG(s)H(s)=K(s − 10)(s − 100)(s +1)(s +100).PhaseMargincurrentclosed-loopdominantpoles(K=1)GainMarginReducing the closed-loop gain by an amount equal to the G.M.(i.e., setting K=-G.M. dB or less) will stabilize the systemnegative Gain Margin→ closed-loop system is unstableLecture 32 – Monday, Nov. 262.004 Fall ’07 Transient from closed-loop frequency response /1Ks(s +1)Open-Loop systemKs(s +1)+−Closed-Loop systemKs2+ s + KEquivalent block diagramfor the Closed-Loop systemOpen-Loop systemω2ns(s +2ζωn)ω2ns(s +2ζωn)+−Closed-Loop systemω2ns2+2ζωns + ω2nEquivalent block diagramfor the Closed-Loop systemConsider a 1st—order system with ideal integral control:More generally, with the definition K ≡ ω2n:R(s)R(s)C(s)C(s)R(s)R(s)C(s)C(s)Lecture 32 – Monday, Nov. 262.004 Fall ’07 Transient from closed-loop frequency response /2Closed—Lo op Transfer Function:C(s)R(s)≡ T (s)=ω2ns2+2ζωns + ω2nFrequency Response magnitude: M (ω)=|T (s)| =ω2nn(ω2n− ω2)2+4ζ2ω2nω2o1/2Frequency response magnitude peaks at frequency ωp= ωnp1 − 2ζ2.Frequency response peak magnitude is Mp=12ζp1 − ζ2.Bandwidth:The frequency where the magnitude drops by 3dBbelow the DC magnitudeωBW=q(1 − 2ζ2)+p4ζ4− 4ζ2+2Image removed due to copyright restrictions. Please see: Fig. 10.39 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.Lecture 32 – Monday, Nov. 262.004 Fall ’07 Transient from closed-loop frequency response /3Images removed due to copyright restrictions. Please see: Fig. 10.40 and 10.41 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.Lecture 32 – Monday, Nov. 262.004 Fall ’07 Transient from open-loop phase diagramsRelationship between phase margin ΦMand damping ratio:ΦM=tan−12ζq−2ζ2+p1+4ζ2.Open—Loop gainvs Open—Loop phaseat frequency ω = ωBW(i.e., when Closed—Loop gainis 3dB below the Closed—Loop DC gain.)Images removed due to copyright restrictions. Please see: Fig. 10.48 and 10.49 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.Lecture 32 – Monday, Nov. 262.004 Fall ’07 ExampleBandwidth from frequency response:find where M = −6 ∼−7.5dB while Φ = −135◦∼−225◦⇒ ωBW≈ 3.5rad/sec.Damping ratio from phase margin:Find phase margin (≈ 35◦)and substitute into plot (ζ ≈ 0.32).Images removed due to copyright restrictions. Please see: Fig. 10.50 and 10.48 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.Lecture 32 – Monday, Nov. 262.004 Fall ’07 Example: Proportional control in the frequency domainSpecification: 9.5% overshoot.For 9.5% overshoot, the required damping ratio is ζ =0.6.Using the damping ratio—phase margin relationship, we findΦM=tan−12ζq−2ζ2+p1+4ζ4⇒ ΦM=59.2◦.Before compensation, the phase margin was ≈ 85◦(see the Bode plot on the right.)We must reduce the phase margin to 59.2◦,i.e. the Bode magnitude must be 0dBwhen the Bode phase is −180◦+59.2◦= −120.8◦.This occurs when ω ≈ 15◦andwe can see that the required gain adjustment is ≈ 44dB.What is the total gain for the compensator?In our uncompensated Bode plot, M =1whenω =0.1 ⇒the uncompensated gain is K ≈ 3.6.After compensation, the gain (in dB) should be20log3.6+44≈ 11 + 44 = 55 ⇒ K ≈ 570.K100PreamplifierDesiredpositionR(s)Power amplifierMotor andloadShaftvelocityShaftposition1sC(s)(s + 100)1(s + 36)+-Frequency (rad/s)1001010.1-80-70-60-50-40-30-20-10020 log MFrequency (rad/s)1001010.1-220-200-180-160-140-120-100-80Phase (degrees){Magnitude aftergain adjustmentMagnitude beforegain adjustmentFigures by MIT OpenCourseWare.Lecture 32 – Monday, Nov. 262.004 Fall ’07 Gain adjustment for phase margin specificationImage removed due to copyright restrictions. Please see: Fig. 11.1 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.Lecture 32 – Monday, Nov. 262.004 Fall ’07 Steady-state errors from the frequency responseType 1 system (one free integrator)G(s)=KΠ (s + zk)sΠ (s + pk)Steady—state velocity errore∞=1Kv, whereKv≡ KΠ zkΠ pk= ω—axis intercept.Kvlog ω20logM…Type 2 system (two free integrators)G(s)=KΠ (s + zk)s2Π (s + pk)Steady—state acceleration errore∞=1Ka, whereKa≡ KΠ zkΠ pk=(ω—axis intercept)2.(Ka)1/2log ω20logM…20log Kplog ω20logM…Type 0 system (no free integrators)G(s)=KΠ (s + zk)Π (s + pk)Steady—state position errore∞=11+Kp, whereKp≡ KΠ zkΠ pk=DCgain.Lecture 32 – Monday, Nov. 262.004 Fall ’07 ExampleType 0; steady—state position error20log
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