SECTION 10: FREQUENCY- RESPONSE DESIGN - College Of Engineering
SECTION 10: FREQUENCYRESPONSE DESIGNESE 499 – Feedback Control Systems
Introduction2 We have seen how to design feedback control systemsusing the root locusIn this section of the course, we’ll learn how to do thesame using the open-loop frequency responseObjectives:Determine static error constants from the open-loopfrequency response Determine closed-loop stability from the open-loopfrequency response Use the open-loop frequency response for compensatordesign to: Improve steady-state error Improve transient response K. WebbESE 499
3K. WebbSteady-State Error from Bode PlotsESE 499
Static Error Constants4 For unity-feedback systems, open-loop transferfunction gives static error constantsstatic error constants to calculate steady-stateerror Use𝐾𝐾𝑝𝑝 lim 𝐺𝐺 𝑠𝑠𝑠𝑠 0𝐾𝐾𝑣𝑣 lim 𝑠𝑠𝐺𝐺 𝑠𝑠𝑠𝑠 0 𝐾𝐾𝑎𝑎 lim 𝑠𝑠 2 𝐺𝐺 𝑠𝑠𝑠𝑠 0We can also determine static error constants from asystem’s open-loop Bode plotK. WebbESE 499
Static Error Constant – Type 05 For a type 0 system𝐾𝐾𝑝𝑝 lim 𝐺𝐺 𝑠𝑠𝑠𝑠 0At low frequency, i.e.below any open-looppoles or zeros𝐺𝐺 𝑠𝑠 𝐾𝐾𝑝𝑝Read 𝐾𝐾𝑝𝑝 directly fromthe open-loop Bode plot K. Webb𝐺𝐺 𝑠𝑠 100 𝑠𝑠 30𝑠𝑠 3 𝑠𝑠 200Low-frequency gainESE 499
Static Error Constant – Type 16 For a type 1 system𝐾𝐾𝑣𝑣 lim 𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠 0At low frequencies, i.e. below any other open-looppoles or zeros𝐺𝐺 𝑠𝑠 𝐾𝐾𝑣𝑣𝑠𝑠and𝐺𝐺 𝑗𝑗𝑗𝑗 𝐾𝐾𝑣𝑣𝜔𝜔A straight line with a slope of 20 g this low-frequency asymptote at 𝜔𝜔 1yields the velocity constant, 𝐾𝐾𝑣𝑣On the Bode plot, extend the low-frequency asymptoteto 𝜔𝜔 1 K. WebbGain of this line at 𝜔𝜔 1 is 𝐾𝐾𝑣𝑣ESE 499
Static Error Constant – Type 17𝐺𝐺 𝑠𝑠 K. Webb85 𝑠𝑠 0.1 𝑠𝑠 50𝑠𝑠 𝑠𝑠 2 10𝑠𝑠 125ESE 499
Static Error Constant – Type 28 For a type 2 system𝐾𝐾𝑎𝑎 lim 𝑠𝑠 2 𝐺𝐺 𝑠𝑠𝑠𝑠 0At low frequencies, i.e. below any other open-looppoles or zeros𝐺𝐺 𝑠𝑠 𝐾𝐾𝑎𝑎𝑠𝑠 2and𝐺𝐺 𝑗𝑗𝑗𝑗 𝐾𝐾𝑎𝑎𝜔𝜔2A straight line with a slope of 40 g this low-frequency asymptote at 𝜔𝜔 1yields the acceleration constant, 𝐾𝐾𝑎𝑎On the Bode plot, extend the low-frequency asymptoteto 𝜔𝜔 1 K. WebbGain of this line at 𝜔𝜔 1 is 𝐾𝐾𝑎𝑎ESE 499
Static Error Constant – Type 29𝐺𝐺 𝑠𝑠 K. Webb1600 𝑠𝑠 0.1 𝑠𝑠 5𝑠𝑠 2 𝑠𝑠 100ESE 499
10K. WebbStability from Open-Loop Bode PlotsESE 499
Stability11 Consider the following systemWe already have a couple of tools for assessingstability as a function of loop gain, 𝐾𝐾 RouthHurwitz Root locus Root locus: Stablefor some values of 𝐾𝐾 Unstable for othersK. WebbESE 499
Stability12 In this case gain is stablebelow some valueOther systems may bestable for gain abovesome valueMarginal stability point: Closed-looppoles on theimaginary axis at 𝑗𝑗𝜔𝜔1 For gain 𝐾𝐾 𝐾𝐾1K. WebbESE 499
Open-Loop Frequency Response & Stability13 Marginal stability point occurs when closed-looppoles are on the imaginary axis Anglecriterion satisfied at 𝑗𝑗𝜔𝜔1𝐾𝐾𝐾𝐾 𝑗𝑗𝜔𝜔1 Note 1 and 𝐾𝐾𝐾𝐾 𝑗𝑗𝜔𝜔1 180 that 180 180 𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 is the open-loop frequency responseMarginal stability occurs when: Open-loopgain is: 𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 0 𝑑𝑑𝑑𝑑 Open-loop phase is: 𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 180 K. WebbESE 499
Stability from Open-Loop Bode Plots14 Varying 𝐾𝐾 simply shifts gain response up or downHere, stable for smallergain values 0 𝑑𝑑𝑑𝑑 when𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 0 𝑑𝑑𝑑𝑑 when 𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 180 Often, stable for largergain values 𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 180 Root locus provides thisinformation K. WebbBode plot does notESE 499
Open-Loop Frequency Response & Stability15 Open-loop Bode plot can be used to assess stability Here, we’ll assume open-loop-stable systems But, we need to know if system is closed-loop stable for low gainor high gainClosed-loop stable for low gainOpen-loop Bode plot can tell us: Is a system closed-loop stable?If so, how stable? I.e. how close to marginal stabilityTwo stability metrics: K. WebbGain marginPhase marginESE 499
16K. WebbStability MarginsESE 499
Crossover Frequencies17 Two important frequencies when assessing stability:Gain crossoverfrequency, 𝜔𝜔𝑃𝑃𝑃𝑃 The frequency atwhich the open-loopgain crosses 0 𝑑𝑑𝑑𝑑Phase crossoverfrequency, 𝜔𝜔𝐺𝐺𝐺𝐺 K. WebbThe frequency atwhich the open-loopphase crosses 180 ESE 499
Gain Margin18 An open-loop-stable system will be closed-loop stableas long as its gain is less than unity at the phasecrossover frequencyGain margin, GM K. WebbThe change in openloop gain at thephase crossoverfrequency requiredto make the closedloop system unstableESE 499
Phase Margin19 An open-loop-stable system will be closed-loop stableas long as its phase has not fallen below 180 at thegain crossover frequencyPhase margin, PM K. WebbThe change in openloop phase at thegain crossoverfrequency requiredto make the closedloop system unstableESE 499
Gain and Phase Margins from Bode Plots20K. WebbESE 499
21Phase Margin and Damping Ratio, 𝜁𝜁 PM can be expressed as a function of damping ratio, 𝜁𝜁, as𝑃𝑃𝑃𝑃 tan 12𝜁𝜁 2𝜁𝜁 2 1 4𝜁𝜁 4For 𝑃𝑃𝑃𝑃 65 or so, we can approximate:𝑃𝑃𝑃𝑃 100𝜁𝜁 or 𝜁𝜁 K. Webb𝑃𝑃𝑃𝑃100ESE 499
22K. WebbFrequency Response Analysis in MATLABESE 499
bode.m23[mag,phase] bode(sys,w) If no outputs are specified, bode response is automaticallyplotted – preferable to plot yourselfFrequency vector input is optional sys: system model – state-space, transfer function, or otherw: optional frequency vector – in rad/secmag: system gain response vectorphase: system phase response vector – in degreesIf not specified, MATLAB will generate automaticallyMay need to do: squeeze(mag) and squeeze(phase)to eliminate singleton dimensions of output matricesK. WebbESE 499
margin.m24[GM,PM,wgm,wpm] margin(sys) sys: system model – state-space, transfer function, or otherGM: gain marginPM: phase margin – in degreeswgm: frequency at which GM is measured, the phase crossoverfrequency – in rad/secwpm: frequency at which PM is measured, the gain crossoverfrequencyIf no outputs are specified, a Bode plot with GM andPM indicated is automatically generatedK. WebbESE 499
25K. WebbFrequency-Response DesignESE 499
Frequency-Response Design26 In a previous section of notes, we saw how we can useroot-locus techniques to design compensatorsTwo primary objectives of compensation Improve steady-state errorProportional-integral (PI) compensation Lag compensation Improve dynamic responseProportional-derivative (PD) compensation Lead compensation Now, we’ll learn to design compensators using asystem’s open-loop frequency response K. WebbWe’ll focus on lag and lead compensationESE 499
27K. WebbImproving Steady-State ErrorESE 499
Improving Steady-State Error28 Consider the system above with a desired phase margin of𝑃𝑃𝑃𝑃 50 According to the Bode plot: K. Webb𝜙𝜙 130 at𝜔𝜔𝑃𝑃𝑃𝑃 3.46 ain is 𝐾𝐾𝑃𝑃𝑃𝑃 12.1 𝑑𝑑𝑑𝑑at 𝜔𝜔𝑃𝑃𝑃𝑃Set 𝐾𝐾 𝐾𝐾𝑃𝑃𝑃𝑃 12.1𝑑𝑑𝑑𝑑 4for desired phase marginESE 499
Improving Steady-State Error29 Can read the position constant directly from the Bodeplot: 𝐾𝐾𝑝𝑝 14.8 𝑑𝑑𝑑𝑑 5.5Note that 𝑃𝑃𝑃𝑃 50 , as desiredGain margin is𝐺𝐺𝐺𝐺 17.9 𝑑𝑑𝑑𝑑K. WebbESE 499
Improving Steady-State Error30 Steady-state error to a constant reference is𝑒𝑒𝑠𝑠𝑠𝑠K. Webb1 0.154 15.4%1 𝐾𝐾𝑝𝑝ESE 499
Improving Steady-State Error31 Let’s say we want to reduce steady-state error to 𝑒𝑒𝑠𝑠𝑠𝑠 5%Required positionconstant1 1 19𝐾𝐾𝑝𝑝 0.05Increase gain by 4xBode plot showsdesired positionconstant But, phase marginhas been degradedsignificantly K. WebbESE 499
Improving Steady-State Error32 Step response shows that error goal has been met But, reduced phase margin results in significant overshootand ringingError improvement cameat the cost of degradedphase marginWould like to be able toimprove steady-stateerror without affectingphase marginIntegral compensation Lag compensation K. WebbESE 499
33K. WebbIntegral CompensationESE 499
PI Compensation34 Proportional-integral (PI) compensator:𝐷𝐷 𝑠𝑠 Low-frequency gain increase Infinite at DCSystem type increaseFor 𝜔𝜔 1/𝑇𝑇𝐷𝐷 1 𝑇𝑇𝐷𝐷 𝑠𝑠 1𝑇𝑇𝐷𝐷𝑠𝑠Gain unaffectedPhase affected littlePM unaffectedSusceptible to integrator overflow K. WebbLag compensation is oftenpreferableESE 499
35K. WebbLag CompensationESE 499
Lag Compensation36 Lag compensator𝐷𝐷 𝑠𝑠 𝛼𝛼,𝛼𝛼 1Objective: add a gain of 𝛼𝛼 at low frequencies without affecting phasemarginLower-frequency pole: 𝑠𝑠 1/𝛼𝛼𝛼𝛼Higher-frequency zero: 𝑠𝑠 1/𝑇𝑇Pole/zero spacing determined by 𝛼𝛼For 𝜔𝜔 1/𝛼𝛼𝛼𝛼 𝑇𝑇𝑇𝑇 1𝛼𝛼𝛼𝛼𝛼𝛼 1Gain: 20 log 𝛼𝛼 𝑑𝑑𝑑𝑑Phase: 0 For 𝜔𝜔 1/𝑇𝑇 K. WebbGain: 0 𝑑𝑑𝑑𝑑Phase: 0 ESE 499
37Lag Compensation vs. 𝛼𝛼 Gain increased at lowfrequency onlyDependent on 𝛼𝛼 DC gain: 20log 𝛼𝛼 𝑑𝑑𝑑𝑑 Phase lag added betweencompensator pole andzero𝐷𝐷 𝑠𝑠 𝛼𝛼𝑇𝑇𝑇𝑇 1𝛼𝛼𝛼𝛼𝛼𝛼 10 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 90 Dependent on 𝛼𝛼 Lag pole/zero well belowcrossover frequency K. WebbPhase margin unaffectedESE 499
Lag Compensator Design Procedure38 Lag compensator adds gain at low frequencies withoutaffecting phase marginBasic design procedure:Adjust gain to achieve the desired phase margin Add compensation, increasing low-frequency gain toachieve desired error performance Same as adjusting gain to place poles at the desireddamping on the root locus, then adding compensationRoot locus is not changed Here, the frequency response near the crossover frequencyis not changed K. WebbESE 499
Lag Compensator Design Procedure391. Adjust gain, 𝐾𝐾, of the uncompensated system to provide thedesired phase margin plus 5 10 (to account for smallphase lag added by compensator)2. Use the open-loop Bode plot for the uncompensated systemwith the value of gain set in the previous step to determinethe static error constant3. Calculate 𝜶𝜶 as the low-frequency gain increase required toprovide the desired error performance4. Set the upper corner frequency (the zero) to be one decadebelow the crossover frequency: 1/𝑇𝑇 𝜔𝜔𝑃𝑃𝑃𝑃 /10 Minimizes the added phase lag at the crossover frequency5. Calculate the lag pole: 1/𝛼𝛼𝛼𝛼6. Simulate and iterate, if necessaryK. WebbESE 499
Lag Example – Step 140 Design a lag compensator for the above system to satisfy the followingrequirements 𝑒𝑒𝑠𝑠𝑠𝑠 2% for a step input%𝑂𝑂𝑂𝑂 12%First, determine the required phase margin to satisfy the overshootrequirement𝜁𝜁 ln 𝑂𝑂𝑂𝑂𝜋𝜋 2 ln2𝑂𝑂𝑂𝑂 0.559𝑃𝑃𝑃𝑃 100𝜁𝜁 55.9 Add 10 to account for compensator phase at 𝜔𝜔𝑃𝑃𝑃𝑃K. Webb𝑃𝑃𝑃𝑃 65.9 ESE 499
Lag Example – Step 141 Plot the open-loop Bode plot of the uncompensated systemfor 𝐾𝐾 1Locate frequency wherephase is 180 𝑃𝑃𝑃𝑃 114.1 Gain at 𝜔𝜔𝑃𝑃𝑃𝑃 is 𝐾𝐾𝑃𝑃𝑃𝑃 This is 𝜔𝜔𝑃𝑃𝑃𝑃 , the desiredcrossover frequency𝜔𝜔𝑃𝑃𝑃𝑃 2.5 ���𝐾𝑃𝑃𝑃𝑃 8.4 𝑑𝑑𝑑𝑑 0.38Increase the gain by1/𝐾𝐾𝑃𝑃𝑃𝑃 K. Webb𝐾𝐾 8.4 𝑑𝑑𝑑𝑑 2.63ESE 499
Lag Example – Step 242 Gain has now been set to yield the desired phase margin of𝑃𝑃𝑃𝑃 65.9 Use the new open-loopbode plot to determinethe static error constantPosition constant of theuncompensated systemgiven by the DC gain:𝐾𝐾𝑝𝑝𝑝𝑝 11.14 𝑑𝑑𝑑𝑑 3.6K. WebbESE 499
Lag Example – Step 343 Calculate 𝛼𝛼 to yield desired steady-state error improvementSteady-state error:𝑒𝑒𝑠𝑠𝑠𝑠1 0.021 𝐾𝐾𝑝𝑝The required positionconstant:1𝐾𝐾𝑝𝑝 1 49 𝐾𝐾𝑝𝑝 50𝑒𝑒𝑠𝑠𝑠𝑠Calculate 𝛼𝛼 as the requiredposition constantimprovement𝛼𝛼 K. Webb𝐾𝐾𝑝𝑝 13.9 𝛼𝛼 14𝐾𝐾𝑝𝑝𝑝𝑝ESE 499
Lag Example – Steps 4 & 544 Place the compensator zero one decade below the crossoverfrequency, 𝜔𝜔𝑃𝑃𝑃𝑃 2.5 /𝑇𝑇 0.25 ���𝑇 4 𝑠𝑠𝑠𝑠𝑠𝑠The compensator pole:1/𝛼𝛼𝑇𝑇 0.25141/𝛼𝛼𝛼𝛼 0.018 ag compensator transferfunction𝑇𝑇𝑠𝑠 1𝐷𝐷 𝑠𝑠 𝛼𝛼𝛼𝛼𝛼𝛼𝛼𝛼 1K. Webb𝐷𝐷 𝑠𝑠 144𝑠𝑠 156𝑠𝑠 1ESE 499
Lag Example – Step 645 Bode plot ofcompensatedsystem shows: 𝑃𝑃𝑃𝑃 60.5 𝐾𝐾𝑝𝑝 50.5K. WebbESE 499
Lag Example – Step 646 Lag compensatoradds gain at lowfrequencies onlyPhase near thecrossover frequencyis nearly unchangedK. WebbESE 499
Lag Example – Step 647 Steady-state errorrequirement hasbeen satisfiedOvershoot spec hasbeen met Thoughslow tailmakes overshootassessment unclearK. WebbESE 499
Lag Compensator – Summary48 𝑇𝑇𝑇𝑇 1𝐷𝐷 𝑠𝑠 𝛼𝛼𝛼𝛼𝛼𝛼𝛼𝛼 1Higher-frequency zero: 𝑠𝑠 1/𝑇𝑇 Lower-frequency pole: 𝑠𝑠 1/𝛼𝛼𝛼𝛼 Place one decade below crossover frequency, 𝜔𝜔𝑃𝑃𝑃𝑃𝛼𝛼 sets pole/zero spacingDC gain: 𝛼𝛼 20 log10 𝛼𝛼 𝑑𝑑𝑑𝑑Compensator adds low-frequency gainStatic error constant improvement Phase margin unchanged K. WebbESE 499
49K. WebbImproving Dynamic ResponseESE 499
Improving Dynamic Response50 We’ve already seen two types of compensators toimprove dynamic response Proportionalderivative (PD) compensation Lead compensation Unlike with the lag compensator we just looked at,here, the objective is to alter the open-loop phaseWe’ll look briefly at PD compensation, but will focuson lead compensationK. WebbESE 499
51K. WebbDerivative CompensationESE 499
PD Compensation52 Proportional-Derivative (PD) compensator:𝐷𝐷 𝑠𝑠 𝑇𝑇𝐷𝐷 𝑠𝑠 1Phase added near (andabove) the crossoverfrequency Increased phase marginStabilizing effectGain continues to rise athigh frequenciesSensor noise is amplified Lead compensation isusually preferable K. WebbESE 499
53K. WebbLead CompensationESE 499
Lead Compensation54 With lead compensation, we have three designparameters: Crossover frequency, 𝜔𝜔𝑃𝑃𝑃𝑃 Phase margin, PM Determines damping, 𝜁𝜁, and overshootLow-frequency gain Determines closed-loop bandwidth, 𝜔𝜔𝐵𝐵𝐵𝐵 ; risetime, 𝑡𝑡𝑟𝑟 ; peak time,𝑡𝑡𝑝𝑝 ; and settling time, 𝑡𝑡𝑠𝑠Determines steady-state error performanceWe’ll look at the design of lead compensators for twocommon scenarios, eitherDesigning for steady-state error and phase margin, or Designing for closed-loop bandwidth and phase margin K. WebbESE 499
Lead Compensation55 Lead compensator,𝛽𝛽 1Objectives: add phase lead near the crossover frequency and/oralter the crossover frequencyLower-frequency zero: 𝑠𝑠 1/𝑇𝑇Higher-frequency pole: 𝑠𝑠 1/𝛽𝛽𝛽𝛽Zero/pole spacing determined by 𝛽𝛽For 𝜔𝜔 1/𝑇𝑇 𝑇𝑇𝑇𝑇 1𝐷𝐷 𝑠𝑠 𝛽𝛽𝛽𝛽𝛽𝛽 1Gain: 0 𝑑𝑑𝑑𝑑Phase: 0 For 𝜔𝜔 1/𝛽𝛽𝛽𝛽 K. WebbGain: 20 log 1/𝛽𝛽 𝑑𝑑𝑑𝑑Phase: 0 ESE 499
56Lead Compensation vs. 𝛽𝛽 𝑇𝑇𝑇𝑇 1𝐷𝐷 𝑠𝑠 𝛽𝛽𝛽𝛽𝛽𝛽 1,𝛽𝛽 determines: Zero/pole𝛽𝛽 1spacing Maximumcompensator phaselead, 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 High-frequencycompensator gainK. WebbESE 499
57Lead Compensation – 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 𝛽𝛽, zero/pole spacing, determines maximum phase lead 1 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 sin1 𝛽𝛽1 𝛽𝛽Can use a desired 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 todetermine 𝛽𝛽1 sin 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚𝛽𝛽 1 sin ��𝑚𝑚𝑚 occurs at ��𝑚𝑚𝑚 𝑇𝑇 K. Webb1𝑇𝑇 𝛽𝛽1𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 𝛽𝛽ESE 499
Lead Compensation – Design Procedure581. Determine loop gain, 𝐾𝐾, to satisfy either steady-state errorrequirements or bandwidth requirements:a) Set 𝐾𝐾 to provide the required static error constant, orb) Set 𝐾𝐾 to place the crossover frequency an octave below the desiredclosed-loop bandwidth2. Evaluate the phase margin of the uncompensated system, using thevalue of 𝐾𝐾 just determined3. If necessary, determine the required PM from 𝜁𝜁 or overshootspecifications. Evaluate the PM of the uncompensated system anddetermine the required phase lead at the crossover frequency toachieve this PM. Add 10 additional phase – this is 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚4. Calculate 𝛽𝛽 from 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚5. Set 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 𝜔𝜔𝑃𝑃𝑃𝑃 . Calculate 𝑇𝑇 from 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 and 𝛽𝛽6. Simulate and iterate, if necessaryK. WebbESE 499
Closed-Loop Bandwidth and Transient Response59 Closed-loop bandwidth, 𝜔𝜔𝐵𝐵𝐵𝐵 , is one possible design criterion How is it related to transient response?For a second-order system (or approximate second-order system): K. WebbClosed-loop bandwidth and damping ratio and natural frequency, 𝜁𝜁 and 𝜔𝜔𝑛𝑛𝜔𝜔𝐵𝐵𝐵𝐵 𝜔𝜔𝑛𝑛1 2𝜁𝜁 2 4𝜁𝜁 4 4𝜁𝜁 2 24.6 𝑡𝑡𝑠𝑠 𝜁𝜁1 2𝜁𝜁 2 4𝜁𝜁 4 4𝜁𝜁 2 2Closed-loop bandwidth and 1% settling time, p bandwidth and peak time, 𝑡𝑡𝑝𝑝𝜔𝜔𝐵𝐵𝐵𝐵 4𝑡𝑡𝑝𝑝 1 𝜁𝜁 21 2𝜁𝜁 2 4𝜁𝜁 4 4𝜁𝜁 2 2ESE 499
Double-Lead Compensation60 A lead compensator can add, at most, 90 of phaseleadIf more phase is required, use a double-leadcompensator𝐷𝐷 𝑠𝑠 𝑇𝑇𝑇𝑇 1𝛽𝛽𝛽𝛽𝛽𝛽 12For phase lead over 60 70 , 1/𝛽𝛽 must be verylarge, so typically use double-lead compensationK. WebbESE 499
Lead Compensation – Example 161 Consider the following system Design a compensator to satisfy the following 0.1 for a ramp input %𝑂𝑂𝑂𝑂 15% 𝑒𝑒𝑠𝑠𝑠𝑠 Here, we’ll design a lead compensator tosimultaneously adjust low-frequency gain andphase marginK. WebbESE 499
Lead Example 1 – Steps 1 & 262 The velocity constant for the uncompensated system is𝐾𝐾𝑣𝑣 lim 𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠 0 𝐾𝐾 𝐾𝐾𝐾𝐾𝑣𝑣 lim𝑠𝑠 0 𝑠𝑠 1Steady-state error is1𝑒𝑒𝑠𝑠𝑠𝑠 0.1𝐾𝐾𝑣𝑣𝐾𝐾𝑣𝑣 𝐾𝐾 10Adding a bit of margin𝐾𝐾 12Bode plot shows the resultingphase margin is 𝑃𝑃𝑃𝑃 16.4 K. WebbESE 499
Lead Example 1 – Step 363 Approximate required phase margin for %𝑂𝑂𝑂𝑂 15% Design for 13%First calculate the required damping ratio𝜁𝜁 ln 𝑂𝑂𝑂𝑂𝜋𝜋 2 ln2𝑂𝑂𝑂𝑂 0.545Approximate corresponding PM, and add 10 correctionfactor𝑃𝑃𝑃𝑃 100𝜁𝜁 10 64.5 Calculate the required phase leadK. Webb𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 64.5 16.4 48 ESE 499
Lead Example 1 – Steps 4 & 564 Calculate 𝛽𝛽 from 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚1 sin 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚𝛽𝛽 0.1471 sin 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚Set 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 𝜔𝜔𝑃𝑃𝑃𝑃 , as determined from Bode plot, andcalculate 𝑇𝑇𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 𝜔𝜔𝑃𝑃𝑃𝑃 3.4 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠𝑠𝑠𝑠𝑠 𝑇𝑇 1𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 𝛽𝛽 13.4 0.147 0.766The resulting lead compensator transfer function isK. Webb𝑇𝑇𝑇𝑇 10.766𝑠𝑠 1𝐾𝐾𝐷𝐷 𝑠𝑠 𝐾𝐾 12𝛽𝛽𝛽𝛽𝛽𝛽 10.113𝑠𝑠 1ESE 499
Lead Example 1 – Step 665 0.766𝑠𝑠 1𝐾𝐾𝐷𝐷 𝑠𝑠 120.113𝑠𝑠 1The lead compensator Bode plotK. WebbESE 499
Lead Example 1 – Step 666 Lead-compensated system: 𝑃𝑃𝑃𝑃 48.5 𝜔𝜔𝑃𝑃𝑃𝑃 7.2 igh-frequency compensator gainincreased the crossover frequency Phase was added at theprevious crossover frequency PM is below targetMove lead zero/pole to higherfrequencies K. WebbReduce the crossoverfrequency increaseImprove phase marginESE 499
Lead Example 1 – Step 667 As predicted by theinsufficient phasemargin, overshootexceeds the target %𝑂𝑂𝑂𝑂 20.9% 15%Redesign compensatorfor higher 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚Improve phase margin Reduce overshoot K. WebbESE 499
Lead Example 1 – Step 668 The steady-state errorrequirement has beensatisfied 𝑒𝑒𝑠𝑠𝑠𝑠 0.08 0.1Will not change withcompensator redesign Low-frequencygainwill not be changedK. WebbESE 499
Lead Example 1 – Step 669 Iteration yields acceptable value for ��𝑚𝑚𝑚 5.5 rad/sec Maintain same zero/pole spacing, 𝛽𝛽, and, therefore, same𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 Recalculate zero/pole time constants:𝑇𝑇 1𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 𝛽𝛽 15.5 0.147 0.4742𝛽𝛽𝛽𝛽 0.147 0.4742 0.0697The updated lead compensator transfer function:K. Webb0.4742𝑠𝑠 1𝐷𝐷 𝑠𝑠 120.0697𝑠𝑠 1ESE 499
Lead Example 1 – Step 670 Crossover frequency hasbeen reduced Phase margin is close tothe target 𝜔𝜔𝑃𝑃𝑃𝑃 5.58 ���𝑃𝑃𝑃 58.2 Dip in phase is apparent,because 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 is nowplaced at point of loweropen-loop phaseK. WebbESE 499
Lead Example 1 – Step 671 Overshoot requirementnow satisfied %𝑂𝑂𝑂𝑂 14.7% 15%Low-frequency gain hasnot been changed, soerror requirement isstill satisfiedDesign is completeK. WebbESE 499
Lead Compensation – Example 272 Again, consider the same system Design a compensator to satisfy the following 1.2 𝑠𝑠𝑠𝑠𝑠𝑠 ( 1%) %𝑂𝑂𝑂𝑂 10% 𝑡𝑡𝑠𝑠 Now, we’ll design a lead compensator tosimultaneously adjust closed-loop bandwidth andphase marginK. WebbESE 499
Lead Example 2 – Step 173 The required damping ratio for 10% overshoot isln 𝑂𝑂𝑂𝑂𝜁𝜁 𝜋𝜋2 ln2 𝑂𝑂𝑂𝑂𝜔𝜔𝐵𝐵𝐵𝐵 4.6𝑡𝑡𝑠𝑠 𝜁𝜁 0.5912Given the required damping ratio, calculate the required closed-loopbandwidth to yield the desired settling time1 2𝜁𝜁 2 4𝜁𝜁 4 4𝜁𝜁 2 2𝜔𝜔𝐵𝐵𝐵𝐵 7.52 e’ll initially set the gain, 𝐾𝐾, to place the crossover frequency, 𝜔𝜔𝑃𝑃𝑃𝑃 ,one octave below the desired closed-loop bandwidth𝜔𝜔𝑃𝑃𝑃𝑃 𝜔𝜔𝐵𝐵𝐵𝐵 /2 3.8 . WebbESE 499
Lead Example 2 – Step 174 Plot the Bode plot for 𝐾𝐾 1 Determine the loop gain atthe desired crossoverfrequency𝐾𝐾𝑃𝑃𝑃𝑃 23.3 𝑑𝑑𝑑𝑑Adjust 𝐾𝐾 so that the loopgain at the desiredcrossover frequency is 0 𝑑𝑑𝑑𝑑1𝐾𝐾 23.3 𝑑𝑑𝑑𝑑 14.7𝐾𝐾𝑃𝑃𝑃𝑃K. WebbESE 499
Lead Example 2 – Steps 2 & 375 Generate a Bode plot using the gain value just determinedPhase margin for theuncompensated system:𝑃𝑃𝑀𝑀𝑢𝑢 14.9 Required phase margin to satisfyovershoot requirement:𝑃𝑃𝑃𝑃 100𝜁𝜁 59.1 Add 10 to account forcrossover frequency increase𝑃𝑃𝑃𝑃 69.1 Required phase lead from thecompensatorK. Webb𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 𝑃𝑃𝑃𝑃 𝑃𝑃𝑀𝑀𝑢𝑢 54.2 ESE 499
Lead Example 2 – Steps 4 & 576 Calculate zero/pole spacing, 𝛽𝛽, from required phase lead, 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚1 sin 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚𝛽𝛽 0.10401 sin 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚Calculate zero and pole time constants𝑇𝑇 1𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 𝛽𝛽 0.8228 𝑠𝑠𝑠𝑠𝑠𝑠𝛽𝛽𝛽𝛽 0.0855 𝑠𝑠𝑠𝑠𝑠𝑠The resulting lead compensator transferfunction:𝑇𝑇𝑇𝑇 1𝐾𝐾𝐾𝐾 𝑠𝑠 𝐾𝐾𝛽𝛽𝛽𝛽𝛽𝛽 1K. Webb0.8228𝑠𝑠 1𝐾𝐾𝐾𝐾 𝑠𝑠 14.70.0855𝑠𝑠 1ESE 499
Lead Example 2 – Step 677 Bode plot of thecompensated system 𝑃𝑃𝑃𝑃 49.8 Substantially belowtarget Crossover frequency iswell above the desiredvalue 𝜔𝜔𝑃𝑃𝑃𝑃 9.44 teration will likely berequiredK. WebbESE 499
Lead Example 2 – Step 678 Overshoot exceeds thespecified limit %𝑂𝑂𝑂𝑂 19.1 10%Settling time is fasterthan required 𝑡𝑡𝑠𝑠 0.98 𝑠𝑠𝑠𝑠𝑠𝑠 1.2 𝑠𝑠𝑠𝑠𝑠𝑠Iteration is required StartK. Webbby reducing thetarget 𝜔𝜔𝑃𝑃𝑃𝑃ESE 499
Lead Example 2 – Step 679 Must redesign the compensator to meet specificationsMust increase PM to reduce overshoot Can afford to reduce crossover, 𝜔𝜔𝑃𝑃𝑃𝑃 , to improve PM Try various combinations of the followingReduce crossover frequency, 𝜔𝜔𝑃𝑃𝑃𝑃 Increase compensator zero/pole frequencies, 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 Increase added phase lead, 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 , by reducing 𝛽𝛽 Iteration shows acceptable results for:𝜔𝜔𝑃𝑃𝑃𝑃 2.4 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠𝑠𝑠𝑠𝑠 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 3.4 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠𝑠𝑠𝑠𝑠 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 52 K. WebbESE 499
Lead Example 2 – Step 680 Redesigned lead compensator:0.8542𝑠𝑠 1𝐾𝐾𝐷𝐷 𝑠𝑠 6.270.1013𝑠𝑠 1Phase margin:𝑃𝑃𝑃𝑃 62 Crossover frequency:𝜔𝜔𝑃𝑃𝑃𝑃 4.84 . WebbESE 499
Lead Example 2 – Step 681 Dynamic response requirements are now satisfied Overshoot: %𝑂𝑂𝑂𝑂 8%Settling time:𝑡𝑡𝑠𝑠 1.09 𝑠𝑠𝑠𝑠𝑠𝑠K. WebbESE 499
Lead Compensation – Example 282 Lead compensatoradds gain at higherfrequenciesIncreased crossoverfrequency Faster response time Phase added near thecrossover frequencyImproved phasemargin Reduced overshoot K. WebbESE 499
Lead Compensation – Example 283 Step responseimprovements: Fastersettling time Faster risetime Significantly lessovershoot and ringingK. WebbESE 499
Lead-Lag Compensation84 If performance specifications require adjustment of:Bandwidth Phase margin Steady-state error Lead-lag compensation may be used𝐷𝐷 𝑠𝑠 𝛼𝛼𝑇𝑇𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠 1𝛼𝛼𝑇𝑇𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠 1𝑇𝑇𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠 1𝛽𝛽𝑇𝑇𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠 1Many possible design procedures – one possibility:1. Design lag compensation to satisfy steady-state error andphase margin2. Add lead compensation to increase bandwidth, whilemaintaining phase marginK. WebbESE 499
K. Webb ESE 499 15 Open-Loop Frequency Response & Stability Open-loop Bode plot can be used to assess stability But, we need to know if system is closed-loop stable for low gain or high gain Here, we'll assume open-loop-stable systems Closed-loop stable for low gain Open-loop Bode plot can tell us: Is a system closed-loop stable? If so, how stable?
Thursday, March 24, 2016 - Friday, March 25, 2016 4. The New Frequency Response Obligation 4.1. Frequency Response Standard The new frequency response standard will require each BA to achieve a Frequency Response Measure (FRM) that meets its FRO starting in the 2017 compliance period (i.e. December 2016 through November 2017).
response is easily obtained by means of the inverse Laplace Transform. Frequency-Response Method – Frequency response is the steady-state response of a system to a sinusoidal input. In frequency-response methods, we vary the frequency of the input signal over a certain
the 2012 Frequency Response Initiative Report because the actual frequency nadir was accurately captured. The frequency response analysis tool11 is being used by the NERC Bulk Power System Awareness group for frequency event tracking in supp
High-Frequency Response of CS Amp Take the following circuit and investigate its high-frequency response – First, redraw using a high-frequency small-signal model for the nMOS There are two ways to find the upper 3-dB frequency ω H – Use open-circuit time constant method
The simulation model in this circuit is designed for frequency response, and the functions not related to frequency response are not implemented. Table 4. BD9E200FP4_AC model pins used for frequency response Pin Name Description VIN Power supply input. EN Enable input. BOOT Pin for bootstrap. SW Switching node. FB Output voltage feedback pin.
2 Mirror Frequency Filter 2.1 Mirror Frequency In radio reception using heterodyning in the tuning process, the mirror frequency is an undesired input frequency that is capable of producing the same intermediate frequency (IF) that the desired input frequency produces. It is a potential source of interference to proper reception.
ORGANIZING QUALITATIVE DATA Section 2.2 . 5! Common Graphical Methods for Qualitative Data ! Frequency Tables ! Bar Charts ! Pie Charts Frequency Tables ! Standard frequency tables have three columns: - Categories for the variable - Frequency - Relative frequency . 6! Frequency Tables .
exponential, the forced response will also be of that form. The forced response is the steady state response and the natural response is the transient response. To find the complete response of a circuit, Find the initial conditions by examining the steady state before the disturbance at t 0. Calculate the forced response after the disturbance.File Size: 773KB
