A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. CHAPTER 5 TRANSFER FUNCTION AND RESPONSE COMPONENTS 5.1TRANSFER FUNCTION: DEFINITIONS OneofthemostpowerfultoolsofControlSystem istheTransferFunctionrepresentation, which is a generalization of the impedance concept in the electrical and mechanical networks. TransferFunction,T(s), istheratioof theLaplace Transformoftheoutputvariable tothe Laplace Transform of the input variable, with all initial conditions assumed or taken to be zero.The TransferFunctionofaSystemorElementrepresentstherelationshipdescribingthedynamicsof the system under consideration. ATransferFunctionmayonlybedefinedforalinear,stationary(orconstantparameter) system. A Non-Stationary System, often called Time-Varying System, has one or more time-varying parameters but Laplace Transformation may be utilized. Furthermore,aTransferFunctionisaninput-outputdescriptionofthebehaviorofthe system.Thus,theTransferFunctiondescriptiondoesnotincludeanyinformationconcerningthe internal structure of the system and its behavior. The Transfer Function concept and approach is very important since it provides the analyst and designer with a useful mathematical model of system elements. 5.2SOLVING FOR THE TRANSFER FUNCTION/S, T(s) 5.2.1SINGLE-INPUT, SINGLE-OUTPUT SYSTEMS zero are conditions initial all ) s ( R) s ( Y) s ( T = where:y(t) = Output SignalY(s) = Laplace Transformed Output Signal r(t) = Input SignalR(s) = Laplace Transformed Input Signal Note: When solving for Transfer Function, even if the value of the input or input function is given, the input should revert back to simply as an input variable in the function of time. Example:SolvefortheTransferFunctionofthegivensystemexpressedintermsofits differential equation. r 5dtdry 8dtdy6dty d22+ = + + A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. Solution: Take the Laplace Transform of the given equation. All initial conditions are zero. LetL{ } ) s ( Y ) t ( y = and L{ } ) s ( R ) t ( r = . | | | | ) s ( R 5 s ) s ( R 5 ) s ( sR ) s ( Y 8 s 6 s ) s ( Y 8 ) s ( sY 6 ) s ( Y s2 2+ = + = + + = + +8 s 6 s5 s) s ( R) s ( Y) s ( T2+ + + = = Example: Given the Translational Mechanical System shown. Let M=5kg,B=20kg/sec,andK=20kg/sec2,solveforthe Transfer Function. Solution: Using FBD. ) t ( x 20dt) t ( dx20dt) t ( x d5 ) t ( f ) t ( f ) t ( f ) t ( f22K B M+ + = + + = Take the Laplace Transform of the equation. All initial conditions are zero. LetL{ } ) s ( F ) t ( f = and L{ } ) s ( X ) t ( x = . | | ) s ( X 20 s 20 s 5 ) s ( X 20 ) s ( sX 20 ) s ( X s 5 ) s ( F2 2+ + = + + =20 s 20 s 51) s ( F) s ( X) s ( T2+ += = 5.2.2MULTIPLE-INPUT, MULTIPLE-OUTPUT SYSTEMS zero are conditions initial all ) s ( Rj) s ( Yi) s ( Tij = where:i = Output Numberj = Input Number y(t) = Output Y(s) = Laplace Transformed Output Signalr(t) = Input SignalR(s) = Laplace Transformed Input Signal ToobtaintheTransferFunctionforeachinput-outputcombination,usethe SuperpositionPrincipletoisolateoneinputandoneoutputatatime.Ifthesystemhas multiple outputs, all outputs except the one being considered are ignored. If the system has multiple inputs, all inputs except the one being considered are set to zero. The number of Transfer Functions to be obtained from multiple-input, multiple output system is determined by multiplying the number of inputs to the number of outputs of the system. Note: When solving for Transfer Functions, even if the value of the inputs or input functions are given, the inputs should revert back to simply as input variables in the function of time. Example:SolvefortheTransferFunctionsofthegivenmultiple-input,multiple-output system expressed in its differential equations. A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. 2 1 11r 5 r y 2dtdy+ = +221 22r 4dtdrr 6 y 2dtdy+ + = + Solution: Take the Laplace Transform of the given equations. All initial conditions are zero.There are two inputs and two outputs.There are four Transfer Functions to be obtained. Let L{ } ) s ( Y ) t ( y1 1= , L{ } ) s ( Y ) t ( y2 2= , L{ } ) s ( R ) t ( r1 1= , and L{ } ) s ( R ) t ( r2 2= . ) s ( R 5 ) s ( R ) s ( Y 2 ) s ( sY2 1 1 1+ = + ) s ( R 4 ) s ( sR ) s ( R 6 ) s ( Y 2 ) s ( sY2 2 1 2 2+ + = +| | ) s ( R 5 ) s ( R ) s ( Y 2 s2 1 1+ = + | | | | ) s ( R 4 s ) s ( R 6 ) s ( Y 2 s2 1 2+ + = + Ignore) s ( Y2 and set0 ) s ( R2= Ignore) s ( Y2 and set0 ) s ( R1=2 s1) s ( R) s ( Y) s ( T1111+= =2 s5) s ( R) s ( Y) s ( T2112+= = Ignore) s ( Y1 and set0 ) s ( R2= Ignore) s ( Y1 and set0 ) s ( R1=2 s6) s ( R) s ( Y) s ( T1221+= =2 s4 s) s ( R) s ( Y) s ( T2222++= = Example: Given the circuit shown. E is constant.Solve for the Transfer Functions, wherein, ) s ( E) s ( I) s ( T111=and ) s ( E) s ( I) s ( T221= . Solution: Using Nodal Method ) t ( i ) t ( i ) t ( i2 1 T+ =) 0 ( v dt ) t ( iC1) t ( i R ) t ( idtdL ) t ( i R ) t ( v ) t ( v ) t ( v ) t ( v ) t ( et0C 2 R 2 L 1 R 1 C 2 R L 1 R+ + = + = + = + = ) t ( i ) t ( i ) t ( iL 1 R 1= = ) t ( i ) t ( i ) t ( iC 2 R 2= =) t ( idtdL ) t ( i R ) t ( e1 1 1+ = ) 0 ( v dt ) t ( iC1) t ( i R ) t ( et02 2 2+ + = Take the Laplace Transform of the Integra-Differential Equations.When solving for Transfer Function, all initial conditions must be zero. LetL{ } ) s ( I ) t ( i1 1=, L{ } ) s ( I ) t ( i2 2=, andL{ } ) s ( E ) t ( e =| | ) 0 ( i ) s ( sI L ) s ( I R ) s ( E1 1 1 + = s) 0 ( vs) s ( IC1) s ( I R ) s ( E22 2+ + =| | ) s ( I Ls R ) s ( E1 1 + = ) s ( ICs1 Cs R) s ( E22((

+=Ls R1) s ( E) s ( I) s ( T1111+= =1 Cs RCs) s ( E) s ( I) s ( T2221+= = A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. Example:GiventheRotationalMechanicalSystemshown, solve for the Transfer Functions, wherein ) s () s () s ( T111=and ) s () s () s ( T221= . Solution: Using AEE At node 1(t): | | ) t ( ) t ( 8dt) t ( d6dt) t ( d8 ) t ( ) t ( ) t ( ) t (2 112128 K 6 B 8 J + + = + + == = = equation 1 At node 2(t):) t ( ) t ( ) t (8 B 12 J 8 K = = =+ = | |dt) t ( d8dt) t ( d12 ) t ( ) t ( 822222 1 + = equation 2 Take the Laplace Transform of the equations.When solving for Transfer Function, all initial conditions are zero. LetL{ } ) s ( ) t (1 1 = , L{ } ) s ( ) t (2 2 =, andL{ } ) s ( ) t ( =| | ) s ( 8 ) s ( 8 s 6 s 8 ) s ( 8 ) s ( 8 ) s ( s 6 ) s ( s 8 ) s (2 122 1 1 12 + + = + + =equation 1a ) s ( s 8 ) s ( s 12 ) s ( 8 ) s ( 82 222 1 + = ( ) ) s ( 8 s 8 s 12 ) s ( 8 ) s ( s 8 ) s ( s 12 ) s ( 8222 2 221 + + = + + =equation 2a 8 s 8 s 12) s ( 8) s (212+ += equation 2b Substitute equation 2b in equation 1a. | |( ) ( )) s (8 s 8 s 1264 8 s 8 s 12 8 s 6 s 88 s 8 s 12) s ( 88 ) s ( 8 s 6 s 8 ) s (1 22 22112 )`+ + + + + +=((

+ + + + =( )( ) ( ) 64 8 s 8 s 12 8 s 6 s 8) s ( 8 s 8 s 12) s (2 221 + + + ++ +=( )( ) ( ) 64 8 s 8 s 12 8 s 6 s 88 s 8 s 12) s () s () s ( T2 22111 + + + ++ += = Substitute) s (1in equation 2b. ( ) ( ) ( ) ( ) 64 8 s 8 s 12 8 s 6 s 8) s ( 864 8 s 8 s 12 8 s 6 s 8) s ( ) 8 s 8 s 12 (8 s 8 s 128) s (2 2 2 222 2 + + + +=((

+ + + ++ +((

+ += ( ) ( ) 64 8 s 8 s 12 8 s 6 s 88) s () s () s ( T2 2221 + + + += = A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. 5.2.3PROBLEM SETS 5.2.3.1TRANSFER FUNCTION: GIVEN THE DIFFERENTIAL EQUATIONS 1. r 6 y 7dtdy3dty d22= + + 2. dtdr8dtr d5 y 4dtdy2dty d6dty d222233+ = + + +3. 2222122r 2dtr d6 r 4 y 8dtdy4dty d + = + + 4. dtdr4 y 6dtdy6r 3dtdry 2dtdy6dty d2211212= + = + + 5. dtdr4 ydtdy6dty d7dty dr r 3dtr dydtdy6dty d7dty d1222223232 121211212313= + + ++ + = + + + 5.2.3.2TRANSFER FUNCTION: GIVEN ELECTRICAL AND MECHANICAL NETWORKS (TRANSLATIONAL AND ROTATIONAL) 1. Solve for ) s ( E) s ( I) s ( T111=and ) s ( E) s ( I) s ( T221= . 2. Solve for ) s ( F) s ( X) s ( T111=and ) s ( F) s ( X) s ( T221= . 3. Solve for ) s () s () s ( T111=and ) s () s () s ( T221= . A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. 5.3RESPONSE AND RESPONSE COMPONENTS: DEFINITIONS The Response is the actual system output; obtained from a Feedback Control System. ThesystemoutputwhenalltheinitialconditionsarezeroistermedtheZero-State ResponseComponent.ItsLaplaceTransformisgivensimplybytheproductoftheTransfer Function and the input transform. Ifthesysteminitialconditionsarenotzero,thereareadditionaloutputcomponents present. These are the Zero-Input Response Component. The Zero-Input Response Component has the same denominator polynomial as the Transfer Function, T(s). TheTransferFunctionDenominatorPolynomialistheSystemsCharacteristicPolynomial and the roots are known as the systems Characteristic Roots or Poles. AnalternativetotheZero-Input/Zero-Statedecompositionofthesystemresponseisto separate the response into Natural Response Component and Forced Response Component. TheNaturalResponseComponentconsistsofallcharacteristicroottermsinthepartial fraction expansion for the response, which is inherent of the system. TheForcedResponseComponentistheremainderoftheresponseandiscomposedof terms associated with the Laplace Transform of the input variable. 5.3.1SINGLE-INPUT, SINGLE-OUTPUT SYSTEMS ) s ( R ). s ( T ) s ( Y = where:Y(s) = Laplace Transformed Output Signal T(s) = Transfer Function of the System R(s) = Laplace Transformed Input Signal Example:SolvefortheZero-StateResponsegiventheTransferFunctionand inputsignal.Classify,usingthepartialfractionexpansion,whichistheForcedandNaturalResponse Components. 3 s4) s ( T+= ) t ( u ) t ( r = Solution:SolvefortheLaplaceoftheoutputusing) s ( R ). s ( T ) s ( Y = .Then,proceedto partial fraction expansion, classify which are the Forced and Natural Response Components, and perform Inverse Laplace to obtain the Zero-State Response Component. L { }s1) s ( R ) t ( r = = ( ) 3 s34s343 sBsA3 s s4s13 s4) s ( Y+ =++ =+= +=The Zero-State Response: t 3e3434) t ( y = . Wherein s34 is the Forced Response Component and 3 s34+is the Natural Response Component. A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. Example:SolveforthecompleteresponsegiventheTransferFunction,inputsignal,and initialcondition.Fromthepartialfractionexpansion,identifywhicharetheForcedand NaturalResponseComponents.Fromtheresponse,identifywhicharetheZero-Inputand Zero-Output Response Components. 4 s10) s ( T+= t 8 ) t ( r = 4 ) 0 ( y = Solution: From the Transfer Function given, revert back to the form: ) s ( R ). s ( G ) s ( Y ). s ( F = . L { }2s8) s ( R ) t ( r = = ) s ( R) s ( Y4 s10) s ( T =+= ( ) ) s ( R 10 ) s ( Y 4 s = + Expandtheequationtoincludethe terms containing the initial conditions. | | ) s ( R 10 ) s ( Y 4 ) 0 ( y ) s ( Y s = + Apply the given initial condition. | ||.|

\| = + 2s810 ) s ( Y 4 4 ) s ( Y s ( ) 4s80) s ( Y 4 s2 + = + ( ) ( )4 s44 s s80) s ( Y2+++= Dontcombinethetermsofthe) s ( Y intoa singlefractionbyLCM.This istomakesure that the Zero-Input Response Component will be indentified later. Taking: ( ) 4 sCsBsA4 s s802 2++ + =+ 4 s44 s5s20s5) s ( Y2++++ +=The Complete Response: t 4 t 4e 4 e 5 t 20 5 ) t ( y + + + =Wherein the 2s20s5+ are the Forced Response Components and the 4 s44 s5+++ are the Natural Response Components. The Zero-Input Response Component is t 4e 4. The Zero-State Response Components are t 4e 5 t 20 5+ + .

Example: Solve for the complete response of the given system with the indicated input and initial condition.From the partial fraction expansion, classify which are Forced and Natural ResponseComponents.FromtheResponse,classifywhichareZero-StateandZero-Input Response Components.E, R, and L are constant. At t=0, i=Io, also a constant. Solution: Obtain the equation of the given circuit. Take the Laplace of the equation. LetL{ } ) s ( I ) t ( i = . dt) t ( i dL ) t ( i R ) t ( V ) t ( V EL R+ = + = | | LIo ) s ( I Ls ) s ( I R ) 0 ( i ) s ( sI L ) s ( I RsE + = + =| | ) s ( I Ls R ) s ( I Ls ) s ( I R LIosE+ = + = +A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. | | | | Ls RLIoLs R sE) s ( I+++=

Taking:| | | | | | Ls RRELsRELs RBsALs R sE+ =++ =+ | | | |((

++((

+ =((

++((

+ =+++ =LRsIoLRsREsRELRs LLIoLRs LRELsRELs RLIoLs RRELsRE) s ( I The Complete Response: tLRtLRIoe eRERE) t ( i + =Amp Wherein sRE is the Forced Response Component and ((

++((

+LRsIoLRsRE are the Natural Response Components.tLReRERE) t ( i =are the Zero-State Response Components while tLRIoe ) t ( i=is the Zero-Input Response Component. 5.3.2MULTIPLE-INPUT, MULTIPLE-OUTPUT SYSTEMS ... ) s ( R ). s ( T ) s ( R ). s ( T ) s ( R ). s ( T ) s ( Y3 13 2 12 1 11 1+ + + = where:j = Input Numberi = Output NumberYi(s) = Laplace Transformed Output Signal Tij(s) = Transfer Function Rj(s) = Laplace Transformed Input Signal Example: Solve for the Complete Response of the given multiple-input, multiple-output system expressed in its differential equations. 2 s1) s ( R) s ( Y) s ( T1111+= =2 s6) s ( R) s ( Y) s ( T1221+= = 10 ) t ( r1= 2 s5) s ( R) s ( Y) s ( T2112+= =2 s4 s) s ( R) s ( Y) s ( T2222++= = t 10 ) t ( r2= Solution: Take the Laplace Transform of the inputs. L{ }s10) s ( R ) t ( r1 1= = L{ }22 2s10) s ( R ) t ( r = = Solving for) t ( y1: |.|

\| |.|

\|++ |.|

\| |.|

\|+= + =22 12 1 11 1s102 s5s102 s1) s ( R ). s ( T ) s ( R ). s ( T ) s ( YA COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. ( ) 2 s215s25s2152 sCsBsA2 s s50 s 10) s ( Y2 2 21++ +=++ + =++=The Zero-State Response: t 21e215t 25215) t ( y+ +=Wherein 2s25s215+are the Forced Response Components while 2 s215+ is the Natural Response Component. Solving for ) t ( y2: |.|

\| |.|

\|+++ |.|

\| |.|

\|+= + =22 22 1 21 2s102 s4 ss102 s6) s ( R ). s ( T ) s ( R ). s ( T ) s ( Y( )( ) 2 s25s20s252 sCsBsA2 s s4 s 10 s 60) s ( Y2 2 22+ + =++ + =+ + +=The Zero-State Response: t 22e215t 25215) t ( y+ +=Wherein 2s20s25+ are the Forced Response Components while 2 s25+ is the Natural Response Component. 5.3.3PROBLEM SET Instruction: Given the Transfer Function/s, the input signal, and (in certain items) theinitialconditions.Fromthepartialfractionexpansion,classifywhicharethe ForcedandNaturalResponseComponents.Fromtheobtainedresponse,classify which are the Zero-State and Zero-Input Response Components. 1. 4 s10) s ( R) s ( Y) s ( T+= = t 4e 3 ) t ( r= 2. 8 s 4 s3 s) s ( R) s ( Y) s ( T2+ + += =t 3e 10 ) t ( r= 3. 2 s 3 s5 s) s ( R) s ( Y) s ( T2+ + = = ) t ( u ) t ( r = 3 ) 0 ( y = 4 ) 0 ( ' y = 4. 3 s3) s ( R) s ( Y) s ( T+= =) t ( u 6 ) t ( r = 10 ) 0 ( y = 5. 12 s 4 s6 s) s ( R) s ( Y) s ( T2111 += =2 s5) s ( R) s ( Y) s ( T212= = t 41e 6 ) t ( r = ) t ( u 10 ) t ( r2= A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. 6. ( ) ( ) 3 s 1 ss) s ( R) s ( Y) s ( T1111+ += =( ) ( ) 3 s 1 s3) s ( R) s ( Y) s ( T2112+ += =) t ( ) t ( r1 = ) t ( u ) t ( r2= 5.4TIME RESPONSE ThetypeofTimeFunctioncorrespondingtoeachpartialfractionexpansiontermfora Laplace-Transformed signal depends upon the terms root location in the complex plane and upon whether or not the root is repeated. Root LocationTime FunctionRoot LocationTime Function Origin Constant K f(t) = Origin and Repeated Equation of a Line 2 1K t K f(t) + = Negative Real Decaying Exponential atKe ) t ( f= Positive Real Expanding Exponential atKe ) t ( f = Negative Real and Repeated Decaying Exponential with Line ( )-at2 1e K t K f(t) + = Positive Real and Repeated Expanding Exponential with Line ( )at2 1e K t K f(t) + = Imaginary Sinusoidal ( ) + = bt cos A ) t ( f Imaginary and Repeated Line with Sinusoid ( ) ( ) + + = bt cos K t K f(t)2 1 A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 5 CATADMAN, LIZETTE IVY G. Complex with Negative Real Part Decaying Exponential with Sinusoid ( ) + = bt cos e f(t)-at Complex with Positive Real Part Expanding Exponential with Sinusoid ( ) + = bt cos e f(t)at 5.4.1PROBLEM SETS 5.4.1.1ROOT AND ROOT LOCATIONS Instruction:Solvefortherootsofthegivencharacteristicpolynomialfrom the Transfer Function given and plot them in scale on the Argand Diagram. 1. 2 s 3 s5 s) s ( R) s ( Y) s ( T2+ + = = 3. 13 s 6 s2 s) s ( R) s ( Y) s ( T2+ + += = 2. ( ) ( ) 20 s 4 s 5 s2 s) s ( R) s ( Y) s ( T2+ + ++= = 5.4.1.2TIME FUNCTIONS Instruction:Plotthefollowingtimefunctionsinscale.TaketheLaplace Transform of the given function and solve for the roots.Then, plot the root in scale on the Argand Diagram. 1. te 6 ) t ( f= 2.2 t 4 ) t ( f + = 3.t2sin 10 ) t ( f=4.( ) t4cos 4 t 2 ) t ( f+ = 5.( )t 4e 2 t ) t ( f+ = SOURCES / REFERENCES DAzzo,J.&C.Houpis.FeedbackControlSystemAnalysis&Synthesis,2ndEdition.New York: McGraw-Hill Publishing Company, 1966. Dorf,RichardC.andRobertH.Bishop.ModernControlSystems,7thEdition.Philippines: Addison-Wesley Publishing Company, Inc, 1995. Hostetter, Gene H., Clement J. Savant Jr., and Raymond T. Stefani. Design of Feedback Control Systems, 2nd Edition. USA: Saunders College Publishing, 1989.