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CONTROL SYSTEMS

CONTROL SYSTEMS. B.Tech – II – ECE – II SEMESTER. M.MYSIAH Assistant Professor ACE ENGINEERING COLLEGE. 5. UNIT - II. Time. R esponse. An a l y sis. UNIT – II: Time R esponse Anal y sis. . Time Do m ain A n al y sis.  . T r ansi e n t and S t eady S t at e R esponse

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CONTROL SYSTEMS

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  1. CONTROL SYSTEMS B.Tech – II – ECE – II SEMESTER M.MYSIAH Assistant Professor ACE ENGINEERING COLLEGE M.MYSAIAH 5

  2. UNIT - II Time Response Analysis M.MYSAIAH

  3. UNIT – II: TimeResponseAnalysis  TimeDomainAnalysis   TransientandSteadyStateResponse StandardTestInputs:Step,Ramp,ParabolicandImpulse,Need,Significance and correspondingLaplaceRepresentation PolesandZeros: Definition,S-planerepresentation   First andSecondorderControl System FirstOrderControlSystem: AnalysisforstepInput,Conceptof Time Constant SecondOrderControlSystem:Analysisforstepinput,Concept,Definitionand effectofdamping  TimeResponse Specifications    TimeResponse Specifications( noderivations ) Tp,Ts,Tr,Td, Mp,ess– problems ontime responsespecifications Steady State Analysis – Type 0, 1, 2 system, steady state error constants, problems M.MYSAIAH 6

  4. SpecificObjectives: Appreciatetheimportanceofstandard apply them in analysis ofcontrolsystem. Differentiatebetweenpolesand zeros. inputs and Analyze input. Calculate systems. 1st& 2nd order control system for step timeresponse specificationsfor different M.MYSAIAH 12

  5. UNIT - II :TimeResponseAnalysis  TimeDomainAnalysis   Transientand SteadyStateResponse StandardTestInputs:Step,Ramp,ParabolicandImpulse,Need,Significance and correspondingLaplaceRepresentation PolesandZeros: Definition,S-planerepresentation   First andSecondorderControl System FirstOrderControlSystem: AnalysisforstepInput,Conceptof Time Constant SecondOrderControlSystem:Analysisforstepinput,Concept,Definitionand effectofdamping  TimeResponse Specifications    TimeResponse Specifications( noderivations ) Tp,Ts,Tr,Td, Mp,ess– problems ontime responsespecifications Steady State Analysis – Type 0, 1, 2 system, steady state error constants, problems M.MYSAIAH 13

  6. TimeResponse In time domain analysis, time is the independent variable. When a system is given an excitation, thereis a response (output). Definition:Theresponseofasystemtoan applied excitation is called “Time Response” and it is a function of c(t). M.MYSAIAH 14

  7. TimeResponse-Example Theresponseofmotor’sspeedwhena command is given to increase the speed is shown in figure, M.MYSAIAH 15

  8. Asseenfromfigure,themotors speedgraduallypicksup from 1000 rpm and moves towards 1500 rpm. It overshootsand again corrects itself and finally settles downat the last value M.MYSAIAH 16

  9. TimeResponse-Example The response of lift to a input to move up is shown in figure; M.MYSAIAH 17

  10. TimeResponse Generallyspeaking, theresponse of any system thus has two parts (i)TransientResponse (ii) Steady State Response M.MYSAIAH 18

  11. TransientResponse Thatpartofthetimeresponsethatgoestozeroas time becomes very large is called as “Transient Response” 0 L t c(t) i.e. As the name suggests that transient response remains only for some time from initial state to final state. M.MYSAIAH 19

  12. TransientResponse Fromthetransientresponsewecanknow;   Whensystembeginsto respondafteraninputisgiven. Howmuchtimeit takesto reachthe time. Whethertheoutputshootsbeyond &howmuch. Whetherthe outputoscillatesabout outputforthefirst  the desiredvalue   its finalvalue. Whendoesitsettleto thefinalvalue. M.MYSAIAH 20

  13. SteadyStateResponse That partof the response that remains after the transients have died out is called “Steady State Response”. Fromthesteadystatewe How longittook before canknow; steady statewasreached. Whether there is any error between the desired and actualvalues. Whetherthiserror totracktheinput. isconstant,zeroorinfinitei.e.unable M.MYSAIAH 21

  14. SteadyState Response M.MYSAIAH 22

  15. Steady State Response M.MYSAIAH 23

  16. TimeResponseAnalysis  TimeDomainAnalysis   TransientandSteadyStateResponse StandardTestInputs:Step,Ramp,ParabolicandImpulse,Need, and correspondingLaplaceRepresentation PolesandZeros: Definition,S-planerepresentation Significance   First andSecondorderControl System FirstOrderControlSystem: AnalysisforstepInput,Conceptof Time Constant SecondOrderControlSystem:Analysisforstepinput,Concept,Definitionand effectofdamping  TimeResponse Specifications    TimeResponse Specifications( noderivations ) Tp,Ts,Tr,Td, Mp,ess– problems ontime responsespecifications Steady State Analysis – Type 0, 1, 2 system, steady state error constants, problems M.MYSAIAH 24

  17. StandardTestSignal It is very interesting fact to know that most controlsystemsdonotknow aregoingtobe. whattheirinputs Thussystemdesigncannot bedonefrominput point of view as we are unable to know in advance the type input M.MYSAIAH 25

  18. Needof StandardTestSignal From example; Whenaradartracks anenemy planethenature random. oftheenemyplane’svariationis Theterrain,curvesonroadetc. drivesinanautomobilesystem. arerandom fora The loading on a shearing machine when and whichloadwill beappliedor thrownof. M.MYSAIAH 26

  19. Needof StandardTestSignal Thusfromsuchtypesofinputswe can expect a system in general to get aninputwhichmaybe; a) b) c) d) A A A A suddenchange momentaryshock constantvelocity constantacceleration Hencethesesignalsformstandardtestsignals. Theresponse to these signalsis analyzed.Theaboveinputsare calledas, a) b) c) d) Stepinput - Signifies asuddenchange Impulse input– Signifies momentaryshock Rampinput– Signifies aconstantvelocity Parabolicinput – Signifies constantacceleration M.MYSAIAH 27

  20. StandardTestSignal Step Input Mathematical Representations Graphical Representations r(t) =R. = 0 u(t) t>0 t<0 Thissignal signifies a sudden change in the reference input r(t) at time t=0 R s L{Ru(t)} LaplaceRepresentations M.MYSAIAH 28

  21. StandardTestSignal Unit StepInput Graphical Representations Mathematical Representations r(t) =1.u(t)= 1 = 0 t>0 t<0 Thissignalsignifiesa suddenchange in the reference input r(t) at time t=0 1 s L{u(t)} LaplaceRepresentations M.MYSAIAH 29

  22. StandardTestSignal Ramp Input Mathematical Representations Graphical Representations r(t) = = R.t 0 t>0 t<0 Signal have constantvelocity i.e.constantchange in it’svaluew.r.t. time R L{Rt} LaplaceRepresentations s2 M.MYSAIAH 30

  23. StandardTestSignal Unit Ramp Input Mathematical Representations GraphicalRepresentations r(t) =1.t = 0 t>0 t<0 If R=1it iscalled aunitramp input 1  L{1t} LaplaceRepresentations s2 M.MYSAIAH 31

  24. StandardTestSignal ParabolicInput Graphical Representations Mathematical Representations 2 Rt r(t) = t>0 2 = 0 t<0 R L{Rt} LaplaceRepresentations s3 M.MYSAIAH 32

  25. StandardTestSignal Impulse Input Graphical Representations Mathematical Representations (t) r(t) = =1 t>0 =0 t<0 Thefunctionhasaunit valueonly for t=0. Inpractical cases, a pulse whose time approacheszero is takenas animpulse function. L{ (t)}1 LaplaceRepresentations M.MYSAIAH 33

  26. Module II –TimeResponseAnalysis  TimeDomainAnalysis (4Marks)   TransientandSteadyStateResponse StandardTestInputs:Step,Ramp,ParabolicandImpulse,Need,Significance and correspondingLaplaceRepresentation Poles and Zeros :Definition,S-planerepresentation   First andSecondorderControl System (8 Marks) FirstOrderControlSystem: AnalysisforstepInput,Conceptof Time Constant SecondOrderControlSystem:Analysisforstepinput,Concept,Definitionand effectofdamping  TimeResponse Specifications (8Marks)    TimeResponse Specifications( noderivations ) Tp,Ts,Tr,Td, Mp,ess– problems ontime responsespecifications Steady State Analysis – Type 0, 1, 2 system, steady state error constants, problems M.MYSAIAH 34

  27. Poles&Zerosof TransferFunction Thetransferfunctionisgivenby, C(s) G(s)  R(s) polynomials Both C(s) and R(s)are ins m1 m bms bm1s .............bo G(s) n1 n s an1s ..................an K(sb1)(sb2)(sb3)............(sbm)  (sa1)(sa2)(sa3)............(san) Where, K= n= system Typeof gain system M.MYSAIAH 35

  28. Poles Thevaluesof‘s’,forwhichthetransferfunction magnitude |G(s)| becomes infinite after substitution in the denominator of the system arecalled as “Poles”of transfer function. M.MYSAIAH 11/21/2016 36

  29. Example1 Determine the polesofgiven transferfunction. s(s2)(s4) G(s) s(s3)(s4) be obtained Solution: Thepolescan withzero by equating denominator s(s3)(s4) s0 s30 s40  0 s3 s4 Thepoles are s=0,-3, -4 M.MYSAIAH 37

  30. S-plane Representation of Poles j 3j 2j j  0 -1 -2 -5 -4 -3 -j -2j -3j M.MYSAIAH 11/21/2016 38

  31. Zeros The values of ‘s’, for which the transfer function magnitude |G(s)| becomes zero after substitution in the numerator of the system are called as “Zeros” of transfer function. M.MYSAIAH 39

  32. Example2 Determinethe zeros ofgiven transferfunction. s(s2)(s4) G(s) s(s3)(s4) Solution:The zeros canbe obtained by equating numerator with zero s(s2)(s4) s0 s20 s40  0 s2 s4 Thepoles are s=0,-2, -4 M.MYSAIAH 40

  33. S-plane Representation of Zeros j 3j 2j j  0 -1 -2 -5 -4 -3 -j -2j -3j M.MYSAIAH 41

  34. Pole-ZeroPlot  Thediagramobtainedbylocatingallpolesandzerosof thetransferfunctioninthes-planeiscalledas“Pole-zero plot”.  The s-planehastwo axis real andimaginary. Since sj , theX-axis standsfor real axis  andshows avalueof . j  Similarly, imaginary Y-axis axis. stands for and represents the M.MYSAIAH 42

  35. Pole- Zero Plot for Example 1and 2 j 3j 2j j  0 -1 -2 -5 -4 -3 -j -2j -3j M.MYSAIAH 43

  36. CharacteristicsEquation Definition: The equation obtainedby equating transfer the denominator polynomial calledasthe of a functionto Equation” zerois “Characteristics n1 n2 n s an an2s ...............an 1s M.MYSAIAH 44

  37. Example3 For the given transfer function, K(s6) T.F. 2)(s5)(s27s12) s(s Find:(i)Poles (iii)Pole-zero Plot (ii)Zeros (iv)Characteristics Equation Solution:(i)Poles The poles can be obtainedby equating s(s2)(s5)(s27s12)0 denominator with zero s0 s20 s50 s2 s5 M.MYSAIAH 45

  38. Example 3 Cont…. s(s2)(s5)(s27s12) 0 (s27s12)  (s3)(s4) s30 s40 Thepoles are s=0, -2, -3, -4, (ii)Zeros: s3 s4 -5 Thezeroscan be obtainedby equating numerator withzero s60 s6 Thezerosare s=-6 M.MYSAIAH 46

  39. Example3 Cont…. (iii) Pole-zero plot: j 3j 2j j  0 -1 -2 -6 -5 -4 -3 -j -2j -3j M.MYSAIAH 47

  40. Example3 Cont…. (iv) Characteristics Equation: s(s2)(s5)(s27s12)0 s(s27s10)(s27s12)0 (s37s210s)(s2 7s12) 0 s57s412s37s449s384s210s370s2120s  0 s514s471s3154s2120s0 M.MYSAIAH 48

  41. Example4 For the given transfer function, (s2) C(s)  s(s22s2)(s27s12) R(s) Find:(i)Poles (iii)Pole-zero Plot (ii)Zeros (iv)Characteristics Equation Solution:(i)Poles The poles can be obtainedby equatingdenominator s(s22s2)(s27s12)0 with zero s0 s30 s40 s3 s4 M.MYSAIAH 49

  42. Example4 Cont…. s(s2 2s2)(s27s12)  0 b2 b 4ac roots 2a s1 s1 j j Thepoles are s=0, -3, (ii)Zeros: -4, -1+j,-1-j Thezeroscan be obtainedby equating numeratorwithzero s20 s2 Thezerosare s=-2 M.MYSAIAH 50

  43. Example4 Cont…. (iii) Pole-zero plot: j 3j 2j j  0 -2 -1 -6 -5 -4 -3 -j -2j -3j M.MYSAIAH 51

  44. Example4 Cont…. (iv)Characteristics Equation: s(s22s2)(s27s12)0 (s32s22s)(s27s12)0 s57s412s32s414s324s22s314s224s  0 s59s428s338s224s  0 M.MYSAIAH 52

  45. Example5 For the given transfer function, (s2) T .F. s(s4)(s26s25) Find:(i)Poles (iii)Pole-zero Plot (ii)Zeros (iv)Characteristics Equation Solution:(i)Poles The poles can be obtained by equating denominator with zero s(s4)(s26s25)0 s0 s40 s4 M.MYSAIAH 53

  46. Example5 Cont…. s(s4)(s26s25) 0 b2 b 4ac roots 2a s3j4 s3j4 Thepoles are s= 0, -4, (ii)Zeros: -3+j4, -3-j4 Thezeroscan be obtainedby equating numerator withzero s20 s2 Thezerosare s=-2 M.MYSAIAH 54

  47. Example 5 Cont…. j (iii) Pole-zero plot: 4j -4 - 3j 2j j -j -2j -3j  0 -2 -1 -6 -5 -4 -3 -4j M.MYSAIAH 55

  48. Example5 Cont…. (iv) Characteristics Equation: s(s4)(s26s25)0 (s24s)(s26s25)0 s46s325s24s324s2100s  0 s410s349s2100s  0 M.MYSAIAH 56

  49. Module II –TimeResponseAnalysis  TimeDomainAnalysis (4Marks)   TransientandSteadyStateResponse StandardTestInputs:Step,Ramp,ParabolicandImpulse,Need,Significance and correspondingLaplaceRepresentation PolesandZeros: Definition,S-planerepresentation   First andSecondorderControl System (8 Marks) FirstOrderControlSystem:AnalysisforstepInput,ConceptofTimeConstant SecondOrderControlSystem:Analysisforstepinput,Concept,Definitionand effectofdamping  TimeResponse Specifications (8Marks)    TimeResponse Specifications( noderivations ) Tp,Ts,Tr,Td, Mp,ess– problems ontime responsespecifications Steady State Analysis – Type 0, 1, 2 system, steady state error constants, problems M.MYSAIAH 57

  50. Analysisof firstordersystem forStep input Considera first order system as shown; 1 Ts + C(s) R(s) - 1 1 H(s)  G(s) Here and Ts 1 C(s) R(s) G 1 Ts     1GH 1Ts 1 1 Ts M.MYSAIAH 58

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