8/9/2019 88324 Control System - Solution Assignment 6 - 2010

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88-324Assignment6

E5.5(a)

Giventhat: Where istheclosedloopTF,wemayderivetheTFasthefollowing:

100 100 100

lim

Ifr(t)andy(t)arethetimedomainresponseandtherampinputtothesystem,tofindsteadystate

error,weareactuallymeasuring lim Since isastablesystemwithpolesonthelefthalfplane,wemayapplyfinalvaluetheorem

So, 1 lim ,

(b)Forzeroovershoot,weneedtomakesurethat theclosedloopTFiscriticallydamped.

Since:

2

Wehave: 50 butforcriticallydampedresponse 1 , 50 10 0.2Polesarelocatedat 1 50 101 25

arerealandthesystemisoverdampedandtherewillbenoovershoot.So,forK>0.2,thepoles

For0

8/9/2019 88324 Control System - Solution Assignment 6 - 2010

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E5.10

6 5 1 3 10 6 5ForstepinputwithstepmagnitudeofA:

Bythesamemethodinthelastquestion,

lim 1 lim 1 3 10 1 3 10 6 5 0ForrampinputwithstepmagnitudeofA:

lim 1 lim 1 3 10 1 3 10 6 5

5.1(a)

1P

, 1 1

Inputcanbeconsideredastepinputwithamagnitudeof25o/s

So,

lim 25 1 lim 1 1 25 251 (b) 1/

251

1 24

(c)WenoticethatthepoleoftheclosedloopTFhasarootat

Sincethisisafirstordersystem,whichhastheloopperformancecompletelycharacterizedbythe

efollowingmaybeused.associatedtimeconstant.Thesettlingtimeexpressionsuchasth

0.02 . 1=51.2

8/9/2019 88324 Control System - Solution Assignment 6 - 2010

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P5.3 Given ,

lim 1 1 lim 1 1Sincewerequirethat 0.5 , 2P5.5(a)

1 , 1

.. 100 51 0.05 ,

0.05 0.05 0.69 2 , 1.38

1.38

, 1.9(b)ForstepinputwithmagnitudeA:

lim 1 lim

0ForrampinputwithmagnitudeA:

lim

1 lim

0

(c)(seepg.308andpg.312ontextbook)

aracteristicequationis 1.4 0Forarampinput,theoptimumITAEcharacteristicequationis 3.2 0Thus, 3.2. So, 3.2 10.24

Forstepinput,theoptimumITAEch

8/9/2019 88324 Control System - Solution Assignment 6 - 2010

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P5.19(a)

10 12

lim 1 1201 1201 lim 10 12 1 10 12 =0

1 120 120

(b)For ess

1 120 AP5.1(a)

lim 1 1 lim 1 96 3 8 8 36 1 963836 0(b) 4 , 6 , 0.67Sincethezerowouldaffec theresponse,wecannotusethe2ndorderTFaloneforthecalculation.Here,

weneedtorefertothe

t

vsP.O.graphonp.289ofthetextbook.(aisthenegativeofthezero)

So,wehave 0.75

Fromthegraph,wefoundthatP.O.=45%

. 1

(c)WecanplotthiswithMATLABtogetasimilargraphasbelow.Actual P.O=33% and T =0.94secs

8/9/2019 88324 Control System - Solution Assignment 6 - 2010

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AP5.5(a)

For =0,[sameTFforR(s)orTd(s)]

50 2

3 4 50 2

lim 1 1 lim 3 4 3 4 50 2 4 312100 0.11

For 0,[forR(s)]

50 2 3 4 50 2

lim 1 1 lim 3 4 3 4 50 2 0

50 2 3 4 50 2[forTd(s)]

lim 1 1 lim 1 50 2 3 4 50 2 1)(b