CENG 4480 Lecture 06: PID Controlbyu/CENG4480/2016Fall/L06_pid.pdf · 2016-11-01 · Discuss what will happen if dead -band is changed, say (a) 0.5 or (b) 2. Example of a dead band

CENG4480Lecture06:PIDControl

BeiYu

CENG4480: feedback control V5a 1

Objectives

1) StudyDCmotors

2) Studyopen-loopandclosed-loop control

3) Controlmethods– I)Proportional feedbackcontrol

– II)PID(proportional-integral-derivative) control


1)DCmotors

Forrobots


Motors

• DCmotors:Directcurrentmotor,easytocontrolanduse.Formakingwheeledrobots

• Servomotorsformakingrobotlegshttp://www.lynxmotion.com/


SmallDirectCurrentD.C.motors

• Speed(~1200-2000rpm).• Operatesona3~5Volt,Canusegearbox(e.g.ratio58:1)toincreasetorque

• UseH-bridgecircuittoboostupcurrentfromtheTLLleveltomotordrivinglevel.


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http://item.taobao.com/item.htm?id=1606576457&tracelog=newcardfavirate

Motorcontrolchip

• H-bridgeChips

• L293D:H-bridgecircuit,up2A

• LDIR:leftmotordirection;RDIR:rightmotordirection

• LEN:leftmotorenable;REN:rightmotorenable


LEN

LDIR

REN

RDIR

2 (1A)1Y(3)

1(EN1/2)

7(2A) (2Y)6

10(3A)(3Y)11

9(EN3/4)

15(4A)(4Y)14

Left-motor

Right-motor

2)open-loopandclosed-loopcontrol

FeedbackcontrolPIDtheoryandimplementation


Open-loopmotorcontrolanditsproblems

• Changemotorsupplypowerchangespeed• Problem:Howmuchpowerisright?

– Ans:don’tknow,dependsoninternal/externalfrictionsofindividualmotors.

• Problem:Howtomaketherobotmovestraight?• Howtocontrolpower(Ton)byISR&anMCU?

– Solution:Usefeedbackcontroltoreadactualwheel:– Slower,increasepower(+Ton)– Faster,reducepower(- Ton)


PWMSignal• PulseWidthModulation• Analogresultswithdigitalmeans• asquaresignalswitchedbetweenonandoff• changingtheportionthesignalon


Exercise

• Whenusingtheopen-loop controlmethodwithaconstantPWMsignalforbothwheels,explainwhytherobotwouldslowdownwhenclimbinguphill.– Ans:Whenclimbinguphill,energyisusedtoactagainstgravity,hencetherobotisrunningslower.


Feedbackcontrol

• Therealsolutiontorealspeedcontrolisfeedbackcontrol

• Requirespeedencodertoreadbacktherealspeedofthewheelatrealtime.


Firstyouneedtohavespeedencoders

• Readwheelspeed.• Usephotointerrupter• Usereflectivedisktosavespace• Basedoninterrupts


Wheelencoder


Our motor and speed encoder

Each wheel rotation=88 on/off changes

IR receiver

IR light source

Darkenedpart blocks light


Newspeedencoder

•


Demomovie

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http://item.taobao.com/item.htm?id=1606576457&tracelog=newcardfavirate

3)Controlmethods

I)ProportionalfeedbackcontrolII)PID(proportional-integral-derivative)control


I)Proportionalclosed-loopfeedbackcontrolsystem

• Showtheleftmotorcontrolonly


Required speed=leftRPMset

leftRPM

+

-

leftErr Motor

Alter PWM for driver L293

leftPWM

if (leftErr >deadband)leftPWM increase by (Pgain * leftErr)

II)PID(proportional-integral-derivative)control

AmoreformalandprecisemethodUsedinmostmodernmachines


HistoryofPID

• By NicolasMinorskyin1922• observationsofa helmsman• steeredtheshipbasedon

– thecurrentcourseerror– pasterror– thecurrentrateofchange


Introduction• Controlforbetterperformance• UsePID,choosewhateverresponseyouwant


Good performanceCriteria depends on users and applications

Too much overshoot/undershoot, not stable

Response too slow

time

Motor speed (w)

required

ValuestoevaluateacontrolsystemExercise2:Describethetermsnthefollowing

diagrams•


Typicallyvalue=10%Depends on application

Rise time Settling time0 time

Target value

overshoot

Steady state error

undershoot

PIDControl


PIDControl


UseofPIDcontroltermsareintertwined

http://en.wikipedia.org/wiki/PID_controller

• Kp(Pgain):ProportionalGain - LargerKptypicallymeansfasterresponse sincethelargertheerror,thelargertheProportional termcompensation.Anexcessivelylargeproportionalgainwillleadtoprocess instabilityandoscillation.

• Ki(Igain):IntegralGain - LargerKiimpliessteadystateerrorsareeliminatedquicker.Thetrade-off islargerovershoot:anynegativeerrorintegratedduringtransientresponsemustbeintegratedawaybypositiveerrorbeforewereachsteadystate.

• Kd(Dgain):DerivativeGain - LargerKddecreasesovershoot,butslowsdowntransientresponse andmayleadtoinstabilityduetosignalnoiseamplificationinthedifferentiationoftheerror.


Effectsofincreasingparametershttp://en.wikipedia.org/wiki/PID_controller

•

Parameter Rise Time Overshoot Settling Time

Steady state error

Kp (Pgain) Decreasestep1

Increase Small Change

Decrease

Ki (Igain) Decrease Increase Increase Eliminatestep3

Kd (Dgain) Small Change

Decreasestep2

Decrease Small Change


Software

PIDrobotdemo.c(atcoursewebpage)

http://www.cse.cuhk.edu.hk/~khwong/www2/ceng2400/PIDRobotDemo093.c


controlmethod:PID(proportional-integral-derivative)control

•


Required speed=

leftRPMset

IR wheelSpeedencoder

LeftRPM

+

-

leftRPM

Motor

generates PWM for driver L293

leftPWM

integral control

Igain* leftErr dt

Proportional control

Pgain*leftErr

Derivative control

Dgain*[d(leftErr)/dt]

sumleftErr

leftPWM

PIDcontrolalgorithmusinginterrupt


IR receiver Speed Encodersensor

interruptstime

Main( ){Setup( );::}

_IRQ( )//1000Hz{:read wheel speedPID:}

1000 interrupts per second

Overview• //////////////main//////////////////////////////////////• Main()• {setup()• forward(Lstep,Rstep,Lspeed,Rspeed)…..• }• ////////////subroutine //////////////////////////////////////////• forward(Lstep,Rstep,Lspeed,Rspeed)• {lcount=0• if(lcount>=Lstep)• Stopleftmotor• …sameforrightmotor…• .}• //////////timertriggered,interruptserviceroutine,1000Hz///////• __irqexception//Interrupt()runningat1000HZ• {lcount++• readwheelspeeds• PIDcontroltoachieveL/Rspeed• ….sameforrightmotor….• }


_IRQ see next slide

Interrupt rate 1000Hz

Speedencoderinterfacing(showleftmotoronly)

• Blockdiagram


IR transmitter

IR receiver(speed encoder sensor)

GPIOParallel interface

/int0

CPU

1000Hz timer interrupt

ARM7-microcontrollerLPC2131

In our robot88 pattern changesin one rotation

Exercise3_irqinterruptprogrammingmethodforthemainPIDloop,usingacounter(intcount)

• __irq()exception//timer interrupt1000HZ,• {intcount++;//intcount• //increasesat1000times/sec• //readmotorspeed:• //updatethemotorspeedinevery1/4seconds• if(intcount==250)//happensatevery¼seconds• {• readwheelspeed• ControlthePWMusingPID• intcount=0;• }• }• Question:If“if(intcount==250)“ischangedto“if(intcount==500)“• Whathappens? CENG4480: feedback control V5a 32

PID CORE

Theinterruptserviceroutineenablesthelooptorun4timesinasecond

•


Loop once in ¼ seconds

Speedcontrolinterruptcore_irq()part1:Findmotorspeed,leftRPM

• void__irqIRQ_Exception()• {ms++; intcount++;• //getthecurrentwheelsensorvalues• lcur=IO0PIN&LWheelSen;rcur=IO0PIN&RWheelSen;• if(lcur!=lold) { lcount++;lold=lcur;lefttimeval++;}• :• //updatemotor speedbyPIDfeedbackcontrolinevery¼Seconds• if(intcount==250) {//calculatethespeedofleftmotor inRPM• leftRPM=lefttimeval;• :

• intcount=0;• lefttimeval=0;• }• }


PID core : See following slides

Speed encodersensor

timeInterrupts 1000 Hz

lefttimeval

•lcount :steps of wheel •lcur : sensor read 1 or 0•lefttimeval: time lapsed since sensor last change of state•leftRPM : left wheel RPM

Part 2PID control

Part1Read Wheelspeeds

Part2:AlgorithmforPIDcore

• Forevery¼seconds– Finderror=(desired value- measuredvalue)– {If(error>deadband)

• {find– Error,– Accumulatederror(addupallpreviouserrors)– Derivativeerror(currenterror– previouserror)– PWM+=Kp*Error+– Ka*(Accumulatederror)+– Kd*Derivativeerror;

• }• Else

– Error<=dead_band,errortoosmalldonothing


In our experiment leftPWM=276000 at the beginning and 192800 at steady state

Set_point

part2:PIDcore1) if(intcount==250){//forevery¼seconds2) //caculatethespeedofleftmotorinRPM3) leftRPM=lefttimeval;4) leftErr=leftRPMset- leftRPM; //caculatelefterror5) if((leftErr<DeadBand*(-1))||(leftErr>DeadBand))//seenextslide6) { //if|lefterror|>deadband7) leftP=Pgain*leftErr; //calculatePProportionalterm8) leftI=Igain*leftaccErr; //calculateIIntegralterm9) leftD=Dgain*(leftErr- leftlastErr); //calculateDDerivativeterm10) leftPWM+=(leftP+leftI+leftD);//update leftmotorPWMusingPID11) if(leftPWM>PWM_FREQ)12) leftPWM=PWM_FREQ;//preventoverrange(max.=PWM_FREQ)13) if(leftPWM<0)leftPWM=0;//(min.=0)14) leftaccErr+=leftErr; //accumulateerror15) leftlastErr=leftErr;//update leftlasterror16) :17) ://handlerightmotorsimilarly…….18) current_leftRPM=leftRPM*240/88;19) current_leftPWM=leftPWM;//20) lefttimeval=0;21) }


PID core

P=Proportional

I Integral

D Derivative

Pgain, Igain, Dgain are constants found by a trial and error method, here we havePgain = 8000; Igain = 6000; Dgain = 5000;

because each rotation has 88 counts, the ISR loop is in ¼ seconds, each minutes 60 seconds. see line 18


•


Required speed=

leftRPMset


LeftRPM

+

-

leftRPM

Motor


leftPWM

integral control

Igain* leftErr dt


Pgain*leftErr

Derivative control


sumleftErr

leftPWM

InsidethePIDcore,wewillstudytheselines

• 5)if((leftErr<DeadBand*(-1))||(leftErr>DeadBand))• :• :• 7)leftP=Pgain*leftErr;//calculatePProportionalterm• 8)leftI=Igain*leftaccErr;//calculateIIntegralterm• 9)leftD=Dgain*(leftErr- leftlastErr);//calculateDDerivativeterm• 10)leftPWM+=(leftP+leftI+leftD);//updatemotorPWMbyPID• :• :• 14)leftaccErr+=leftErr; //accumulateerror


Deadbandline5)if((leftErr<DeadBand*(-1))||(leftErr>DeadBand))

• Dead-band :ADead-band (sometimescalledaneutralzone)isanareaofasignal range orbandwherenoactionoccurs :

• onlyenablemotorwhenleftErr>asmallvalue(deadband,ie=1inourrobot)

• OtherwisemayoscillatewhenleftErrissmall


Dead-band

leftErr = leftRPM - leftRPMset; //calculate left errorif(leftErr>DeadBand ){ activate motor}

Exercise4:Discusswhatwillhappenifdead-bandischanged,say(a)0.5or(b)2.

Exampleofadeadband;donothing“if10-1<leftPRM<10+1

•




overshoot

Steady state error

undershoot

Dead band+/- 1

Target speed=leftRPMset =10


When leftRMPset=10, the real RPM is RPM*240/88=27.3., because each rotation has 88 counts, the ISR loop is in ¼ seconds, each minutes 60 seconds.see line 18

Parametersforevaluatingacontrolsystem

•




Target value

overshoot

Steady state error

undershoot

near steady stateSee next slide

(line7)EffectsofincreasingKp(P)http://en.wikipedia.org/wiki/PID_controller

•


Steady state error

(1) Kp (Pgain)

Decreasestep1


Decrease



Decreasestep2



Example:t0àt1,Theproportionaltermwhenthemeasurementisbelowthesetpoint(leftRPMset),

proportionalPtermis+ve•


leftP = Pgain * leftErr;leftP= 8000* (10-6)= 8000*4 is positiveThis term pushing up “leftPWM” (more energy delivered to the wheel).

leftErr=leftRPMset –leftPRM=10-6=4

Rise time0 time

overshoot

undershoot

t1 t3t2 t4

6

P is +ve


Usage of the P proportional termEach ¼ seconds a new left RPM (speed of wheel) is measured



P is +ve

e.g.Pgain = 8000; Igain = 6000; Dgain = 5000; In our experiment leftPWM=276000 at the beginning and 192800 at steady state

8

(leftErr = leftRPMset-leftRPM )

Exercise5:Fillin?__:t0àt1,Theproportionaltermwhenthemeasurementisbelowthesetpoint(leftRPMset),





Rise time0 time

overshoot

undershoot

t1 t3t2 t4

6

P is +ve




leftP = Pgain * leftErr;leftP=?___________ ,is +ve or -ve?“leftPWM” is increasedor decreased?__more/less energy delivered to wheel?__

P is +ve

e.g.Pgain = 8000; Igain = 6000; Dgain = 5000; In our experiment leftPWM=276000 at the beginning and 192800 at steady state

8

Example:t1àt3,Theproportionaltermwhenthemeasurementisbelowthesetpoint(leftRPMset),the



leftP = Pgain * leftErr;leftP= 8000* (10-13)= 8000*(-3) is negativeThis term lowering down “leftPWM” (less energy delivered to the wheel).

leftErr=leftRPMset –leftPRM=10-13= -3

Rise time0 time

overshoot

undershoot

t1 t3t2 t4

13

P is +ve



P is -ve

e.g.Pgain = 8000; Igain = 6000; Dgain = 5000;


UnderstandingPID–alittlesummaryforthePproportionalterm

• Whenthemeasurementisbelowthesetpoint(leftRPMset)– Themotoriscurrentlytooslow– ThePtermcalculatedis+ve,needtopushthespeedhigher

• Whenthemeasurementisabovethesetpoint(leftRPMset)– Themotorisrunningtoofast– ThePproportionaltermis-ve,loweringdownthespeed.

• IncreaseinPgain(Kp)willdecreaserisetime(meaningfastertoreachsetpoint),decreasesteadystateerror(studyitlater)– Italsoincreasesovershoot


(line9)EffectsofincreasingKd(D)http://en.wikipedia.org/wiki/PID_controller

•


Steady state error

Kp (Pgain)Igain

Decreasestep1


Decrease

Ki (Igain)gainI

Decrease Increase Increase Eliminatestep3


Decreasestep2



Derivativeterm

• Derivativecontrol– Dgain*[d(leftErr)/dt]

• d(leftErr)/dt =– =Derivativeterm– =current_Err- last_Err– =leftErr– leftlastErr;inourprogram


Example:time0à t1,theDerivativeterm(=current-previous)Whenthemeasurementisrising,theDerivativetermis-ve

•


leftD = Dgain * (leftErr – leftlastErr); leftD= 5000* (3-7)= 5000*( -4) is negativeTo do:This term lowering down “leftPWM” (less energy delivered to the wheel).

leftlastErr=leftRPMset –eftlastPRM=10-3=7


Rise time0 time

Target Value=leftRPMset=10

overshoot

undershoot

¼ seconds

t1 t3t2 t4

3

7

D is -ve

leftRPMset =10

Usage of the D Derivative termEach ¼ seconds a new left RPM (speed of wheel) ismeasured



Example:timet1àt2,theDerivativetermWhenthemeasurementisrising,theDerivativetermis-ve

•


Rise time0 time

overshoot

undershoot

leftD = Dgain * (leftErr – leftlastErr); leftD= 5000* (-5- (-2))= 5000*(-3) is negativeThis term lowering down “leftPWM” (less energy delivered to the wheel).

leftlastErr=leftRPMset -leftlastPRM=10-12= -2

¼ secondsleftErr=leftRPMset -leftPRM=10-15= -5

t1 t3t2 t4

leftRPMset =10

D is -ve

1512

leftPRM

D is -ve




LittleSummary:NegativeDDerivativeterm• Whenthemeasurement(leftRPM)isrising,thechange

(change=current-previous)oferror(error=setpoint-leftRPM)is-ve=d(Err/dt)=-ve

• DDerivativeterm=Kd*d(Err/dt)is–VE• Decreaseenergytomotorà decreaseovershoot

CENG4480: feedback control V5a 51Rise time0 time

undershoot

t1 t3t2 t4

D is -ve

D is -ve

Setpoint��

overshoot

Example:timet2àt3,theDerivativetermWhenthemeasurementisfalling,theDerivativetermis+ve

•


Rise time0 time

overshoot

undershoot

leftD = Dgain * (leftErr – leftlastErr); leftD= 5000* (-1- (-4))= 5000*3 is positiveThis term pushing up “leftPWM” (more energy delivered to the wheel).


leftlastErr=leftRPMset -leftlastPRM=10-14= -4

¼ seconds

leftErr=leftRPMset -leftPRM=10-11= -1

t1 t3t2 t4

leftRPMset =10

D is -ve

1411

leftPRM

D is +ve

D is -ve



Exercise6:Fillin?_timet2àt3,theDerivativetermWhenthemeasurementisfalling,theDerivativetermis+ve

•

CENG4480: feedback control V5a53

Rise time0 time

overshoot

undershoot

leftD = Dgain * (leftErr – leftlastErr); leftD= ?___________, which is +ve or –ve?____“leftPWM” increased or decreased?____More/less energy delivered to the wheel?___


leftlastErr=leftRPMset -leftlastPRM=10-9= 1

¼ seconds

leftErr=leftRPMset -leftPRM=10-7= 3

t1 t3t2 t4

leftRPMset =10

D is -ve

97

leftPRM

D is +ve

D is -ve

D is +ve



LittleSummary:PositiveDDerivativeterm• Whenthemeasurement(leftRPM)isfalling,thechange

(change=current-previous)oferror(error=setpoint-leftRPM)is+ve

• DDerivativeterm=Kd*d(Err/dt)is+VE• Increaseenergytomotorà decreaseundershoot

CENG4480: feedback control V5a 54Rise time0 timet1 t3t2 t4

D is +ve

D is +ve

Setpoint��

overshoot

undershoot

UnderstandingPID–alittlesummaryfortheDderivativeterm

• Whenthemeasurement(leftRPM) isrising,– Themotor isgainingspeed– TheDderivativetermis–ve,solowering themotorspeedà decrease

overshoot

• Whenthemeasurement(leftRPM) isfalling,– Themotor isreducingspeed– TheDerivativetermis+ve,sopushing themotorspeedhigherà decrease

undershoot– Inconclusion,thegradientoftheerror(Err)determinesthe

adjustment.Dependsonwhetherd(Err)/dtis+veor–ve.• IncreaseinDgain(Kd)willdecreaseovershoot/undershootand

settlingtime(systemmorestable)


(line8)EffectsofincreasingKi(I)http://en.wikipedia.org/wiki/PID_controller

•


Steady state error

Kp (Pgain) Decreasestep1


Decrease



Decreasestep2




•


Required speed=

leftRPMset


LeftRPM

+

-

leftRPM

Motor


leftPWM

integral control

Igain* leftErr dt


Pgain*leftErr

Derivative control


sumleftErr

leftPWM

At steady state, leftErrà0, soSo, Pgain*leftErr=0

also Dgain*[d(leftErr)/dt] =0

And if no Integral { Igain* leftErr dt} term, left PWM à0

It is a problem.

Timeà nearsteadystateleftRPMset=leftRPM

henceleftErr=0,leftlastErr=0à leftP=0,leftD=0

• leftErr=leftRPMset– leftRPM=0• SoasleftlastErr=0• Therefore• Steps• 7)leftP=Pgain*leftErr//=0 //calculatePProportional term• 8)leftI=Igain*leftaccErr; //calculateIIntegralterm• 9)leftD=Dgain*(leftErr - leftlastErr)//=0 //calculateDDerivativeterm• 10)leftPWM+=(leftP+leftI+leftD);//update leftmotorPWMusingPID

Theonlyvalidtermistheintegralterm“leftI”• leftPWM+=leftI• Themainideaistocreateasmallterm(leftI)tomaintaintheleftPWMvalue




near steady state : leftP=0 leftD=0

Theintegraltermisfoundbyaddingallpreviouserrors

sameasleftaccErr=sum{leftErr(t=0)+leftErr(t=1)+.. +leftErr(t=now)}• 14)leftaccErr+=leftErr;//LeftaccErristhesummationofallpreviouserrors• 8)leftI=Igain*leftaccErr//integralterm,• :• 10)leftPWM+=leftI//nearsteadystate,onlyleftIisvalid• :

• Don’tworryitwillnot becomeinfinitive,onemeasuretosafeguardthisis:• 11)if(leftPWM>PWM_FREQ);// PWM_FREQ=maximumPWMallowed• 12)leftPWM=PWM_FREQ;//preventoverrange• (max.=PWM_FREQ)

• Also,becausenearsteadystate,leftaccErrwilladjustitselfautomatically,seenextslide




Howtheintegraltermadjustsitselfautomaticallynearsteadystate

• Nearsteadystate– 8)leftI=Igain*leftacceErr– 10)leftPWM+=leftI

• Ifmeasuredspeed(leftRPM)>setpoint– leftErris-ve– leftaccErr(Acculumation)decreases,

hencereducingleftaccErratasuitablevaluetomaintainleftPWM

• Ifmeasuredspeed(leftRPM)<setpoint– leftErris+ve– leftaccErr(Acculumation)increases,

henceincreasingleftaccErratasuitablevaluetomaintainleftPWM


Pgain = 8000; Igain = 6000; Dgain = 5000;


UnderstandingPID–alittlesummaryfortheIIntegralterm

• Whenthemeasurementisbelowthesetpoint(leftRPMset)– Themotorisslow– TheIIntergraltermis+ve,pushingthespeedhigher

• Whenthemeasurementisabovethesetpoint(leftRPMset)– Themotorisrunningtoofast– TheIintergraltermis-ve,loweringdownthespeed.

• IncreaseinIgain(Ki)willresultin– ItissimilartothePterm(seeabovetwopoints)– Soincreaseovershoot, settlingtime– anddecreaserisetime– Note:themainfunction ofIntegralcontrolistoreducesteadystateerror


PIDTuning(usuallydonebytrailanderror)

• Tune(adjustmanually)– step1)Pgainproportional_gain(Kp),– step2)Dgainderivative_gain(Kd),– step3)Igainintegral_gain(Ki)

• SetconstantspeedV1forawhile(5seconds)andreducedto(V1)/2atT1

– RecordthespeedbythecomputerafterT1andseeiftheperformanceisokornot• Yes(acceptKp,Ki,Kd)• No(tuneKp,Ki,Kd again)


Motor speed

V1

(V1)/2

Accepted performance

unstable

done

T1 time

Summary

• StudiesPIDcontroltheoryandimplementation


Documents

CENG 4480 Lecture 06: PID Controlbyu/CENG4480/2016Fall/L06_pid.pdf · 2016-11-01 · Discuss what will happen if dead -band is changed, say (a) 0.5 or (b) 2. Example of a dead band