Systems Coursework

8/22/2019 Systems Coursework

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1.Question1

AnalyticalSolution:

TheLaplaceformoftheequationcanbewrittenas:

! ! !!+ 3! + 2 =

1

!

! ! =1

2!+

1

2(! + 2)

1

! + 1

! ! =1

2+1

2!!!!

!!!

SimulinkSolution:


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2.Question2

AnalyticalSolution:

!! + !!"! + !" = ! !

! ! !!!+ !!"! + ! = ! !

SubstitutingM=10,b12=0.5andk=1

! ! =!(!)

!(!)=

0.1

!! + 0.05! + 0.1

!!= 0.1 , ! = 0.05

2 0.1

!! = 1+ exp !"

1 !!= 1.779 1.8

SimulinkSolution:


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3.Question3

(a) Theplotofpart(a)isdepictedbythebluecurve.(b) Aswecanseefromthegraphabove,fora10stepinput,aswe

decreasetheparameterJ,thetransientresponseissignificantly

changed.Theovershoot,risetimeandsettlingtimedecreasesaswe

reduceJ.Wecanconcludethatasthemomentofinertiaisdecreased,

theeffectofdamping(dampingratio)increasesandhencethe

dampedfrequencywillalsodecrease,causingtheresponsetosettle


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faster.Withthatbeingsaid,thechangeinparameterJ,hasnoeffecton

thesteady-stateresponseofthesystem.

4.Question4

(a)AnalyticalSolution:

Objectiveistofindsensitivityofsystemtoparametera.

! ! =!(!)

!(!)=

2

! ! + 2

Whena=1,foraunitstepresponse:

! ! = 21

!

1

! + 1

!! = 2 2!

!!

!!!= lim

!!

! ! = 2

! 4 = 2 2!!!= 1.9633

! 4

! 100% = 98.17% > 98%

Thereforeitcanbeverifiedthatat4secondstheresponseiswithin2%ofthe

steadystateresponse.

(b)

Whena=0.5,

!(!)

!(!)=

2

! + 1.5

Weseethatthesystemwillbehaveasafirst-ordersystemsinceithas

onlyonepole.Solvingfory(t),wehave:

! ! =4

3(1 !!!.!!)


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AsshownontheplotusingSimulink,thesystembehavesasasimplefirst-

ordersystemwithatimeconstantof1.5sandamplitudeof4/3.Thiswas

expectedsincethepolesofthetransferfunctionliesontheleftsideofthe

imaginaryaxisandhencethesystemisstable.

Whenweincreasetheparameteratoa=2,!(!)

!(!)=

2

!

! ! = 2!

Weseethatthesystemisnolongerstable.Thepoleofthetransfer

functionliesontheimaginaryaxis.Thiscanbeviewedasthecriticalpoint

ofinstability,where! < 2inorderforthesystemtobestable.Fromthe

graphitcanbeseenthattheresponseincreaseslinearlywithtimeand

doesnotconvergetoasteady-statefinitevalue.


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Whenwefurtherincreaseparameteratoa=5,!(!)

!(!)=

2

! + 1.5

! ! =2

3(1+ !!!)

Weseethattheresponsetendstoinfinityexponentially,withtime.Thisis

becausethepoleofthetransferfunctionliesontherighthandsideofthe

imaginaryaxisandhencethesystemisunstable.Therefore,wecanconcludethattheparameterahastobekeptlessthan2forthesystemto

remainstable.

5.Question5

(a)Theopen-loopresponseforaunit-stepdisturbanceis,! ! =

1

5[

5

! !! + 0.9! + 5

]

! ! =!

! (11

.

02089!!!.!"!

sin 2.

190! + 1.

3682 )


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(b)Theclosedloopresponsetoaunitstepdisturbancecanbewrittenas:! ! =

1

55[

55

! !! + 0.9! + 55

]

! ! =1

55(1 1.001846!!!.!"! sin 7.4025! + 1.51 )


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(c)

Anopen-loopsystemoperateswithoutfeedbackanddirectly

generatestheoutputinresponsetoaninputsignal.Whereasaclosed-

loopsystemusesameasurementoftheoutputsignalanda

comparisonwiththedesiredoutputtogenerateanerrorsignalthatis

usedbythecontrollertoadjusttheactuator,inthiscasethe

disturbance.Hencewecanseethattheclosedloopgreatlyimproves

thedisturbancerejection,minimisingthesteadystatedisturbance

from1/5intheopen-loopto1/55intheclosedloop.

6.Question6

(a)Usingaproportionalcontroller,withGc(s)=2,wehave

!(!)

!(!)=

20

! + 3

! ! = (120

! + 3)!(!)

!!!= lim

!!

!"(!

)=

1

3,

!"#! ! = 1

!(b)UsingaPIcontroller,withGc(s)=2+20/s,wehave


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!(!)

!(!)=

20

! + 20

! ! = (120

! + 20)!(!)

!!!= lim

!!

!"(!

)= 0

,

!"#! ! = 1

!Plottingthegraphfor(a)and(b):

(c)Thesteadystatetrackingerrorwasreducedfrom1/3,whenaproportionalcontrollerwasusedto0,whenaproportionalplus

integralcontrollerwasused.HoweverthePIcontollerisfarmore

complexthanasimpleproportionalcontroller.Designingsucha

controllerwouldincurmorecost.Thereforetheengineerwouldhave

toweighthemarginalbenefitsofreducingerrorbeforeimplementing

suchcomplexcontrollers.Furthermore,ifaPIcontrollerwaschosen

incorrectly,itmayleadtoinstabiltyofthesytem.


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7.Question7(a)

(b)

Therampresponseswereplottedforbothpart(a)and(b).Fromthegraphwe

canseehowtheactualresponsedeviationfromthedesiredrampresponse.The

errorforagiventimecanbeevaluatedfromtheverticaldisplacementawayfrom

thedesiredattitude.Foraunityfeedbackloop:

! ! = !! ! !(!)

! ! = !! !!(!)


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HoweverfromSimulink,wecanplottheerroragainsttime.Thisgivesus

abetterrepresentationoftheerrorsignalofthesystem.

(a)Attitudeerrorafter10s=0.3(Usingproportionalcontroller)

(b)Thesteadystateerrorforthesystemwhenaproportionalcontrollerwasusedwas0.3,andthesteadystateerrorforthe

systemusingaPIcontrollerwaszero.Thereforeitcanbe

concludedthattheproportionalplusintegralcontrollergreatly

reducedtheerrorofthesystemresultingintheactualresponse

tobeveryclosetothedesiredresponse.

8.Question8! ! =

!(!)

!!(!)=

10!!+ 510! + 500

(! + 0.83)(!! + 11.6! + 600.44)

Usinganalyticalformulasforsecond-ordersystem,wehave:

!!= 600.44 = 24.5, ! = 11.6

2 600.44= 0.2367

!! = 1+ exp !"

1 !!= 1.465

!!=

4

!!!

= 0.69!

!! =!

!!

1 !!= 0.132!


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UsingSimulinksolution:

Fromthegraphabove,

!! = 1.25

!!= 2.55!

!! = 0.13!

Fromtheresults,itcanbeseenthat!!and!!wasclosetothepredictedresults

ofthesecond-ordersystem,andtheactual!!wasverymuchhigherthanthatofthepredicted!

!.Thiscanbeexplainedbyobservingthepolesofthetransfer

function.Thesecond-orderpoleshavearealvalueof-5.58whereasthethird

polehasarealvalueof-0.83.Sincethethirdpoleliesclosertotheimaginary

axis,itisthemoredominantpole,sowecannotassumethesystemtobe

approximatedasasecond-ordersystem.

Furthermorewecanevaluatethetimeconstantofthethirdpole,! = !!.!"

= 1.2.

Anditrequiresthat!

! 10!!

!forathird-ordersystemtobeapproximatedby

thedominantrootsofthesecond-ordersystem.Thus,whatweobserveisthat

transientresponseinitiallyfollowsasecond-ordersystem,andafter0.7s,of

whichisthesettlingtimeofthesecond-ordersystem,theresponsefollows


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closelytothatofafirst-ordersystem.Thisexplainspredictionsfor!!and!!

wereclosetotheactualresults,whereas!!wasverydifferentfromthepredicted

result.

9.Question9

Fromthegraph,

!! = 0.9784

!"#$%!"#$%,!" = 0.4995!"#$"%&'("!"#$%!!" =!! !"

!"100% = 95.87%

!"##$%&'!"#$,!!(!"#!"2%!"!") = 19.4!

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