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# CONTROL SYSTEM DESIGN MEEN 651

Texas A&M

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MEEN 651 Lectures 13 amp 14 Texas AampM University Department ofMechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lectures 13 amp 14 Stability Margins and Compensation Closed Loop Frequency Response and System Bandwidth Stability Margins Two quantities that measure the stability margin of a system are the gain margin and phase margin Gain margin GM is the factor by which the forward loop system gain ie the gain of KGjw is less than the neutral stability value For the typical case it can be read directly from the Bode plot by measuring the vertical distance between the lKGjw l curve and the lKGjw l 1 line at the frequencyWhere 4Gjw 180 Phase margin PM is de ned as the amount by which the phase of KGjw exceeds 180 When lKGQw l 1 The frequency at Which the gain is unity is usually referred to as the crossover frequency It is relatively easy to determine these stability margins from the Bode plot For example consider the open loop system given by G ss1s539 Bode Dwagvams 4 as ml 2 2351rad scmPm25deg mt 1 7271 radSec Magmlude mm P base deg Frequency rad39sec MEEN 651 Lectures 13 amp 14 Relation of the margins to other measures of performance Consider the open loop system G s 7 wz 7 ss2 a With unity feedback the closed loop system is Z T S s2 2 to swz The relation between the PM and the damping ratio of the system is given by 1 0 4 0 H 20quot It 40 50quot 60 70 xn Plum mm gm Figure 1 Damping ratio vs phase margin PM Note that the function is approximately a straight line up to about PM 60 The dashed line shows a straight line approximation to the function W ere 7PM 710039 2 It is clear that the above approximation only holds for phase margins below about 70 Furthermore equation 1 is only accurate for the second order system In spite of these MEEN 651 Lectures l3 amp 14 limitations equation 2 is often used as a rule of thumb for relating the damping ratio to PM The gain margin GM for this second order system is in nite since the phase curve does not cross 180 as the frequency increases Example 1 Consider the system de ned by 85s ls2 2s 4325 G s 32s22382sz23101 Determine the stability margins From the bode plot it can be seen that there are three crossings of the magnitude 1 line at 074 95 and 98 radsec The corresponding PM values are 37 43 and 71 respectively We choose the minimum PM as this is the most conservative assessment of the stability Bode Diagrams Gm20248 dB at 10343 radsec Pm36737 deg at 074365 radsec Phase deg Magnitude dB l l 1072 inquot m 10 W Frequency radsec MEEN 651 Lectures 13 amp 14 Closedloop Frequency response The bandwidth of the system is de ned as the frequency at Which the magnitude of the transfer function is equal to 0707 This is depicted in Figure 2 Decl h mem punk M Bundmmlh 1W 7 1 human Figure 2 De nitions of bandwidth and resonant peak The closed loop frequency response magnitude is approximated by K6050 N 1 a ltltwc 1KGU50 7 IKGI a gtgtwci Where we is the crossover frequency The above relation is shown in Figure 3 ITI MUlnn I M r I PM Ji 11mm I M 11 it 171 NY and My my o Luvdwnllll w w Hm m 1 MAIwt Figure 3 Closedloop frequency response MEEN 651 Lectures 13 amp 14 C ompensation Dynamic elements or compensation are added to feedback control systems to improve their stability and error characteristics The frequency response stability analysis to this point has considered the closed loop system to have the characteristic equation lKGs 0 With the introduction of compensation the closed loop characteristic equation becomes lKDsGs 0 and all the previous discussion pertaining to the frequency response of KGs applies directly to the compensated case if we analyze the frequency response of KDsGs PD Compensation The compensator transfer function is given by Ds KTDsl 3 The frequency response characteristics of equation 3 are shown in Figure 4 A stabilizing in uence is apparent in the increase in phase at frequencies above the break point lTD We use this compensation by locating lTD so that the increased phase occurs in the vicinity of crossover thus increasing the phase margin m n I m m 1 I I m Figure 4 Frequency response of PD control Note that the magnitude of the compensation continues to grow with increasing frequency This feature is undesirable since it ampli es the high frequency noise that is typically present in any real system Reading Assignment Read pages 403408 and 415418 from the textbook Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 8 Dynamic Response of Linear Systems Impact of Pole amp Zero Locations The objective of this lecture is to provide you with some background on the use of transfer function poles and zeros for determination of system dynamic response in the time domain Time Response Versus Pole Locations Once the transfer function of a dynamic system is calculated it can almost always be expressed as the ratio of two polynomials These polynomials can be used to compute the system poles and zeros which completely determine the system response to within a constant Poles and zeros can be used to determine the time history of a dynamic system and relate time histories to the location of the poles and zeros on the s plane The impulse response of a system is a time function that corresponds to the transfer function of a system As such it is usually called the natural response of the system Why First Order Systems For example for a rst order pole Hltsgt lt1 507 which corresponds to the following impulse response Mt 67011007 2 where 1t is the unit step function For 039 gt 0 the pole is located on the left half LHP of the complex plane Res lt 0 and the exponential decays resulting in a stable impulse 1 I Time see I T Figure 1 First order system response responsei On the contrary7 if a lt 0 the pole resides in the RHP and the impulse response is said to be unstable Further7 we de ne the time constant of the impulse response as 7 37 3 as shown in Figure 11 Examples of Transient Response Versus Real Pole Locations Compare the time response and pole locations of the system with transfer function 25 1 ii 4 52 35 2 lt gt The denominator can be written as s 1s 2 and the poles of the system are at 71 and Hs 721 The system has only one zero at 7 Performing a partial fraction expansion on the transfer function 4 results in 1 H 7 5 S s 1 s 2 lt Obtaining the inverse of the components results in the following impulse response function ht 7equot 362 t 2 0 6 2 1mm HQ 42 7 l l Rcs O Zero X Pole 1 Figure 2 Polezero locations of rst order systemi Equation 6 reveals that the shape of the impulse response depends on the pole locations7 as shown in Figure 2 ln fact7 this observation is more general The shape of the natural response of a system depends on the location of the poles of the transfer function This generalization is described in Figure 3 The role of the numerator zeros in the overall system response is that it in uences the size of the coef cients multiplying each component of the response Also7 a fast pole is one that decays faster compared to another pole7 which would be called a slow polei Second Order Systems A pole category that requires special attention is that of complex polesi These can be expressed in terms of their real and imaginary parts7 or s 70 ijwdi 7 Since there are always complex conjugate pairs of complex poles7 the denominator corre sponding to complex poles becomes 515 s a 7 jwds I jwd s 72 w i 8 1mm LHP Figure 3 Time functions associated with pole locationsi Figure 4 Second order system pole locations However7 it is common practice to write a second order transfer function as 2 w H 9 S s2 ZCwns 0 lt Comparing the denominator polynomial of equation 9 with equation 8 we conclude that a mm 10 and Mdwm1ng 11 where C is the damping ratio and Mn is the undamped natural frequency7 and Lad is the damped natural frequency The pole locations for a second order transfer function are shown in Figure 4 Equation 9 can be rewritten as 0J2 0J2 H 5 n n 12 w ltsltwngt2wilt17lt2gt ltsagt2w5 l which corresponds to an impulse response function of ht Lavalwwwiw 13 W 5 m Prove this to yourselves The time response of a second order system for different values of the damping ratio are shown in the top segment of Figure 5 One should note that as the damping ratio increases the actual frequency wd decreases slightly For very low values of damping C close to zero the response is oscillatory while for large values of damping C close to one the response shows no oscillations Three complex pairs of poles are shown in Figure 6 corresponding to different values of the damping ratio The imaginary part of the poles determines the damped natural frequency which in this case remains the same This despite changes in the undamped natural frequency and damping ratio The real part of the poles determines the decay rate of the exponential envelope For 039 lt 0 the response decays whereas for 039 gt 0 becomes unbounded or unstable For 039 0 the response neither decays nor grows Practically the system is considered unstable Timedomain Speci cations In the design of a mechanical system of any other type of system or even a control system we often specify requirements in terms of a systems time response These requirements shown in Figure 7 are de ned as follows 0 rise time tr the time it takes for the system to reach the vicinity of its new set point o settling time Its the time it takes the system transients to decay o overshoot Mp the maximum system overshoot divided by its nal value 0 peak time It the time it takes for a system to reach the maximum overshoot point For a second order system with no zeros the rise time can be approximately expressed as t g 14 For the overshoot more precise expressions can be obtained At the overshoot value the derivative of the response is zero The step response of a second order system in the time domain is given by yt 1 7 6quot 005wdt ismwdt 15 l0 08 06 0 4 ya 00 0A2 4 06 08 l 0 MI Figure 5 Second order system response for di erent damping a impulse responses b step responses lm3 1m 5 11115 75 Re 5 Re 5 R5 4 1 0707 g 05 539 03 Figure 6 Pole locations corresponding to di erent damping ratios Figure 7 Timeidomain system speci cations where wd wnx1 7 2 and 039 Cw Taking the derivative of expression 15 and setting it to zero we obtain 06quot coswdt isz39mudt 7 6quot 7wdsmwdt acoswdt 0 Wd 2 desmwdt wdsmwdt 0 16 6 The zero point occurs when wdtp 7T7 17 and we have the following formula for the peak time tp 18 For this value of time7 the overshoot is equal to yap 1 M 1 e Wme gm 1 WWW 19 So7 we have the following formula for the overshoot MpeWltM7 ogclt1 20 For a second order system7 a plot of the overshoot Mp versus the damping ratio is shown in Figure 8 Finally7 we would like an expression for the settling time We can de ne the settling time as that value of Its when the decaying exponential reaches 1 So7 We 0017 21 or 46 46 9 4 22 run 039 ln summary7 for a second order system with no zeros7 we have the following equations that determine its time domain characteristics run 2 23 C 2 CMP7 24 46 Mp Figure 8 Overshoot versus damping ratio for a second order system ms Imv Im Imx sin 39 W n m Re 3 Re Re 5 Re 5 C d Figure 9 Time domain speci cations in the s plane These inequalities can be graphed in the s plane and used for design7 as shown in Figure 9 The Effects of Zeros and Additional Poles So far we have only accounted for the time domain characteristics of a second order system with no zeros If the system has zeros or additional poles higher than second order system then the transient response of the system is more complex and it cannot be expressed in terms of simple equations7 as before Mathematically speaking7 the presence of zeros in a transfer function is to modify the coef cients of the exponential terms in the transient response For example7 consider two transfer functions with the same poles but different zeros7 as follows 2 s 15 2 2 e lt26 5 1 s 2 25 11 115 1s 2 2 01 09 lt gt 11 s 1 s 2 018 164 27 T s 1 s 239 The two transfer functions are normalized to have the same DC gain7 ie the transfer H15 H2lt8gt function value at s 0 Notice that the coef cient of the s 1 term has been reduced dramatically from 2 to 018 This reduction is the result of the zero at s 711 which is close to the pole at s 71 If the zero is place exactly where the pole is located7 then this term will disappear To account for the effect of a zero on the overall system transient response7 consider a transfer function with a zero and two complex poles7 as follows Sa wn 1 rML 2lt8wn 139 The zero for the transfer function is locate at s iag wn 7040 If 04 is large then the zero 1115 28 will be far away from the real part of the poles and its impact will be minimal If 04 1 then the zero will be close to the poles and its impact will be signi cant The step response of such a system with 05 damping ratio and for different values of 04 is shown in Figure 10 We see that the major impact of the zero is to increase the overshoot Mp7 with little impact on the settling time 16 I 14 a x g 12 a w a quot0 al a 08 u 2 N4 5 06 IOO 04 1 02 00 2 4 a s 10 a Figure 10 Impact of zero location on transient response Il Lu Figure 11 Impact of an additional pole on transient response When the value ofa is negative then there is a zero on the RHP also called a nonminimumi phase zero The transient response of the resulting system is quite different In fact the overshoot is suppressed to the point that the response rst starts in the Wrong direction and then changes sign In studying the effect of an additional pole let us consider the following transfer function 1 504an 1swn2 2swn1 Plots of the transient response of such a system with 05 damping ratio for different values H S 29 of a are shown in Figure 11 In this case the major effect is the increase in the rise time ie the system response slowsidown as the additional pole gets closer to the real part of the complex poles The Effects of PoleZero Patterns on Dynamic Response 7 A Summary 0 For a second order system with no zeros the transient response parameters can be 13 approximately obtained by 18 Rise Time t 2 w 30 Overswot Mp MPK7 see Figure 8 31 46 Settlmg Itsz t5 g 32 a Note This kind of system is also said to have second order dominance A zero in the LHP will increase the overshoot if the zero is within a factor of 4 of the real part of the complex poles7 otherwise its impact is minimal A zero in the RHP will depress the overshoot and will cause the response to start in the wrong direction non minimum phase response A additional pole in the LHP will increase the rise time significantly7 if the extra pole is within a factor of 4 of the real part of the complex poles7 otherwise its impact is minimal Reading Assignment Read pages 135 156 the textbook Read examples Handout E15 posted on the course web page Texas AampM University Department of Mechanical Engineering MEEN 651 Control System esign Dr AlexanderG Parlos Fall 2 03 Lecmre 513 Thermal and Fluidic Sysmms Basic Mechanisms m Heat Transfer Energy may be transfered aernss the bnandanes nf a system ether tn nr fem the system it neeurs nnly when there is a temp eatare difference between the system and the smnundmgs Energy is transfrre by some mammanan tamponwhrehmay nemr separately er in enmbrnatmn We start nur drseussmn by rntrndueng the base physreal meehamsms by wheh thermal enegy is transferred Cm39n lllctimi cnnduetrnn heat transfer neeurs nnly when there is physical enntaet between bndres systems at different temperatures lt ean alsn be de ned as the transfer nf enegy substanee Fnr nnerddmenslnnal heat ennduetrnn m the r dsreetrnn the rate nfheat nwls detemrned by the Fnuner equaann 1 amt ar ape 4 L nne degree in a unit dsstanee Fnr example nnerddmenslnnal steady state heat ennduetrnn is depretedrn Frgurel All Figure l Onerdlmenslnnal steady state heat ennduetrnn me equaann l the rate nfheat aw ean be wntten as dT Qhk kAE Integrating the above equation with respect to x we obtain 72 thkdx J 194511 0 T A 71 2 Q de If the thermal conductivity of the material k does not depend on temperature the rate of heat transfer can be expressed as Q T T2 Convection Convective heat transfer is de ned as the heat transfer between a uid and a solid when there is temperature difference between the solid and the uid It is usually associated with the signi cant motion of the uid around the solid The rate of heat transfer by convection between a solid and a uid owing around it is given by th lipKT Tf where he is the convective heat transfer coef cient A is the area of heat transfer and TS and Tf represent the solid and uid temperatures respectively Radiation This is the means by which heat is transferred for example from the sun to the earth through mostly empty space The rate of heat transfer by radiation between two separated bodies having temperatures T1 and T2 is determined by the StefanBoltzmann law Qhr O FEFAAYTA T24 9 where 6 5667 X 10398 Wmzk4 The StefanBoltzmann constant FE is the effective emissivity FA is the shape factor A is the heat transfer area The effective emissivity accounts for the deviation of the radiating systems from black bodies The values of the shape factor range from 0 to l and represent the fraction of the radiative energy emitted by one body that reaches the other body 2 Lump ed Models of Thermal Sysmms Mathn ucal models of thermal systmns are usually denved from lhe baslc energy a balance equauons lhat follow the gen ml form rate of energy heat aw heat ow rate of heat rate of Wm stared e rate 7 rate generated done m to system out afsystem wlthm system rpm system wthm me system Fluid Systel Elenwnts Fluid Cap aciwrs A uld capacltor is shown m Flgure 2 Flgure 2 Symbollc dlagram ofa uld capaclwr The pressure ln a uld capacltor must be refmed w a refermce pressure P when the when lhe ver V vacuum The volume ow rate QC ls gwm by dPlv Q 1 dt where crrs the uld capacltance The nerc ow mm the capacltor ls sLored and corresponds somewhat to lhe energy swred C I r Er 7a Fluidlnerwrs The symbollc dlagram ofa uld lna wr ls shown m Flgure 3 Flgure 3 A symbollc dlagram ofa Fluld rnenor The elernental equatlon for the lnertor ls a 1 dt39 where 1 ls the uld lnerlance For frlctlonless Incornpresslble ow ln a unlforrn passage haVlng cross sectlonal area A and length L the Inerlance ls I where p ls the rnass denslty ofthe uld The klnetlc energy stored ln an loleal lnerlor ls glven by 1 2 E quot 2 Fluid Resistors The syrnllollc dlagrarn ofz uld reslstor ls shown ln Flgure 4 l r L V 7 Flgure 4 A syrnllollc dlagrarn ofz uld reslstor The elemental equatlon ofan ldeal reslstor ls R R QR Fluid Sources The ldeal sources employed ln uld systern analysls are shown ln Flgure 5 An ldeal ressure so ce ls capable of dellverlng the lndlcated pressure regardless of the ow requlred by what lt ls olerlng whereas an ldeal ow source ls c p ble of dellverlng the V a a 1 Flgure 5 Ideal uld sources a pressure source and b ow source Interconnection Laws The two uid system interconnection laws are the law of continuity and compatibility The continuity law says that the sum of the ow rates at a junction must be 2 and the compatihi ity law laws are illustrated in Figure 5 7quot 5 39 333 L1 7 Figure 5 Interconnection laws The continuity law states that QnQaQc0 The Compatibility law states that RIRZPZV 0 3Rz ery Example 1 rMndeling nf a Heal Exchanger A heal orchanger 15 shown m Flgure 7 Sleam mlers lhe chamber through lhe oohlrollable valve al lhe lop and lhe cooler sleam leaves al lhe bollom There IS a oohslahl ow ofvvaler through lhe plpe lhal vvlhds through lhe mlddle othe chamber so al ll plcks up lhermal me y from lhe sleam Flhd lhe dlffermtlal equallohs lhal descrlbe lhe dynamlcs othe measured waler oulaovv lemperalure as a fuholloh of lhe area Flgure7 HeaLExchanger v d am The r heal lrahsfer from lhe sleam lo waler IS proporllohal lo lhe dlfferehoe h lhese lemp eralures 1h olhervvords lhe heal lrahsfer 15 through convectlon hence q MT 7 Tl The ow othamal energy hlo lhe chamber from lhe lhlel sleam depends on lhe sleam ow rale and lls lemp eralure accordlng lo 4 WA Ta Tgt where w KAr mass ow rale ofthe sleam A rea of e sleam lnlelvave Ka w ooemolehl ofthe lhlel valve c lflc heal ofthe sleam er Th mp e In ow sleam T lempaalure ofthe oul ovv sleam 5732 L L is lie 39 hot incoming sceam and lie liemial energy owlng to lie water This net ow detelmines clie rate onempemmie change omie steam according to QT AiK uTrThAT Ti 2 when cs mses is die liemial capacity omie steam in clie chamber whh mass ms Likewise clie differential equation describing clie Water temperature is CWTW wwcw TM rTwhAT 77quot 3 when mass ow rate ofthe water peeilie heat ofwater ei tux omie incoming mtquot mpeiamie omie out owing Water Example 2 7 Modeling d a Hydraulic Piston m 39 39 pinnieeliambei Figuie s Hydraulic piston actuator wiiu39ng clie Newton s second law ofmotion for clie piston we have Mx Ap e FD when A area of the piston P pressure in the chamber M mass of the piston X position of the piston In many cases of uid ow problems the flow is restricted either by a constriction in the path or by friction The general form of the effect of resistance is given by 1 V w 391 R P1 102 w mass flow rate p1 p2 pressures at ends of the path through which ow is occurring R 0c constants whose values depend on the type of restriction The constant at takes on values between 1 and 2 The most common value is approximately 2 for high flow rates through pipes or through short constrictions or nozzles Note that for this value the ow is proportional to the square root of the pressure difference and therefore will produce a nonlinear differential equation Reading Assignment Read pages 5667 from the textbook Read examples Handout E9 posted on the course web page Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 10C SteadyState System Tracking and System Stability The objective of this lecture is to discuss certain aspects of steady state feedback system tracking issues and look into the methods by which we can determine system stability given a transfer function or a state space model SteadyState Tracking and System Type In general we would like to investigate the steady state performance of a feedback system to disturbances andor reference inputs that are not only constants but can be expressed as arbitrary polynomials in time Assuming that the system under consideration is stablel we classify such systems into system types depending to the polynomial degree for which the tracking error is constant System types can be de ned with respect to the reference input or disturbance Let us consider the system with block diagram representation given by Figure 1 This block diagram can be simpli ed as shown in Figure 2 where we have ignored the disturbance signal So we will rst concentrate on the steady state tracking performance of the system to reference inputs rt Now lets consider the following reference input tk rlttgt E1097 lt1 with the corresponding Laplace transform 1Otherwise steadystate response does not even exist Commller Sens Figure 1 Typical singleiloop systemi Figure 2 Simpli ed block diagram of a typical singleloop systemi To compute the steady state errors we start from the following reference to output transfer function see Figure 2 Ys Tlt8gt DTDG5 1 HDyDGs7 3 01 From the nal value theorem 655 tlim 6t limJ 5Es limJ 51 7 TsRs 6 So7 the steady state error can be expressed as 7 1 1 7 T 5 1m 5 7 SS 9410 skfl 9410 5k The result of the above expression can be zero7 a non zero constant7 or in nity If it is a non zero constant7 the system is referred to as Type k The Special Case of Unity Feedback Assume now that the system has unity feedback7 with Ds 17 and that G0s DTDG5 8 Then7 from the closed loop transfer function we obtain that 1 1 7 T s 9 1 G0s The system error is given by 1 E s R s 10 ltgt1Qwltgt ltgt Using the Final Value Theorem7 we nd that the steady state value of the system error is 1 1 655 lim lim 11 swuaww swu mm Now7 depending on the system type and the nature of the reference signal7 we have differing steady state errors These are summarized in the following table 3 Step Ramp Parabola Type 0 HKF oo 00 Type I 0 Ki 00 Type ll 0 0 Kim The various coef cients mentioned in the table are de ned as Position 7 error Constant Kp lin C0s7 12 Velocity Constant Ky lin sC0s7 13 Acceleration Constant Ka lin s2C0s 14 System Type With Respect to Disturbance Similar to the ability of a system to track references7 it is possible to de ne system type with respect to disturbance rejection To do so7 we de ne the following transfer function Tws 15 This transfer function determines the system error from a disturbance input The system is Type 0 if a step disturbance results in a constant steady state output error The system is Type 1 if7 for a ramp input disturbance7 the steady state value of the output is a constant7 ie yss slirn0sTwsl constant 16 or equivalently7 the steady state output for a step input disturbance is7 1 0 17 yss slganTw In other words7 Tw0 07 18 or DC gain of zero System Stability There are varying de nitions for stability but we will choose to use the concept of bounded inputbounded output BIBO stability That is a system is said to be BlBO stable if for every bounded input ut the resulting system response yt is also bounded In terms of the system impulse response ht BlBO stability can be stated as follows a system is BIBO stable if and only if the integral f lh7 ld7 lt oo 19 In general if a system has any poles on the imaginary axis or on the RHP the integral of its impulse response will not be nite and it will not be a BlBO stable system Whereas if all of its poles are in the LHP then it will be BlBO stable So we can use the pole locations of a system to determine stability Equivalence of Transfer Function Poles and System Eigenvalues In the past we derived transfer function models for dynamic systems and de ned the poles of such transfer functions as the zeroes of the denominator polynomial Further we have developed state space models either directly from physical principles or from transfer function models We then de ned the system modes as the eigenvalues of the A matrix of the state space model For a given system transfer function model and its equivalent state space representation the poles of the system are identical to the eigenvalues of the A matrix of the system statespace representation As a result we can investigate system stability using system poles in the frequency domain or system eigenvalues in the time domain Stability Criterion Based on Poles or Eigenvalues Consider the transfer function Ys i bosm blsm l bm Rs T snals 1an 7 K m 7 i Hl18 2 m lt n H15 pi 7 T 7 where p are the system poles and z are the system zeros The denominator polynomial set Ts 20 to zero is also called the characteristic equation The system response in the time domain can now be expressed as V L W ZKiemt7 21 i1 where the exponents are the poles and the coef cients depend on the system zeroes and the initial conditions This system will be stable if and only if all terms on the right hand side Row n s 1 a2 a4 Row n 1 sn l a1 a3 a5 Row n 2 5 b1 b2 b3 Row n 3 sn g c1 c2 Cg Row 2 52 gtk gtk Row 1 51 gtk Row 0 so gtk of equation 21 go to zero as t a 00 This happens if all of the poles of the system are strictly on the LHP7 ie gke lt 0 22 This is called asymptotic internal stability Since the poles of a system are identical to the eigenvalues of its equivalent state space representation7 we can also state that a system will be stable if all of its eigenvalues are strictly on the LHP7 ie Eei lt 0 23 So7 we can determine the stability of a system by computing its poles or its eigenvalues and checking their real parts Computing the poles of a system from a general order denominator polynomial or computing the eigenvalues of a system from a general matrix requires tools such as MATLAB Routhls Stability Criterion A more indirect way of determining system stability without the use of MATLAB is the so called Routh criterion This criterion states that a necessary and su cient condition for stability is that all of the elements in the rst column of the Routh array be positive To construct the Routh array7 we consider the following characteristic equation as s als 1an1san 0 24 Then the Routh array is constructed as shown in the table above The elements b and c in the array are computed as follows 103 7 1102 1102 7 13 7 7 bl a1 a1 8 l 3 l 4 s5 4 2 4 0 4i 5 7 43712 7 41741 7 44710 s 5 i 5 4 0 i 5 4 i 4 2740 4744 53 275 2 71 52 g 2 0 2 2 20 E 3 2472 0 52 3 25 2 gt 4 2 51 7E 239778391 39 1 7 3 7 470 50 4 J 39 b 105 7 1104 i 1104 7 15 2 77 7 77 01 01 b 7 117 7 1106 i 1106 7 17 3 7 77 7 77 01 01 C Clin 7 bias 7 bias 7 Clin 1 7 7 b1 b1 C 1le 7 bias 51 7 1le 2 7 b1 b1 C 0154 7 5107 i 5107 7 0154 3 7 7 b1 b1 25 Example Routh7s Test Determine the stability of a system that has the following characteristic polynomial 1556455354253524540 26 The Routh array is given in the table above Since the elements of the rst column of the Routh array are not all positive7 the poly nomial has RHP roots ln fact7 there are two roots on the RHP7 since there are two sign changes on the rst column of the Routh array Reading Assignment Read pages 157 166 and 230 242 of the textbook Read the examples in Handout E22 and Handout A6 posted on the course web page Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Farlos Fall 2003 Lecture 12 Dynamic Compensation Using RootLocus In addition to Bode plots we can use the root locus method to design dynamic compen sators The simplest form of a dynamic compensator takes the structure 5 z Ds Hp 1 where if z lt p it is a lead compensator and if z gt p it is a lag compensator Lead Compensation To understand the stabilizing effect of a lead compensator we rst consider Ds 52 This is a PD controller and we apply it to the following second order system K KG5 2 The uncompensated and compensated root locus is shown in Figure 1 The effect of a zero is to move the locus towards the stable region of the s plane Whereas before compensation achieving wn of say 2 would have resulted in very low damping and high overshoot following compensation we can achieve the same wn with damping ratio of more than 05 The problem with the a pure lead compensator zero only is that its implementation requires use of a differentiator which is very sensitive to sensor noise Furthermore it is impossible to build a pure differentiator However the addition of a fast pole would not greatly reduce the effect of the zero So for example we could suggest the following lead compensator 2 s Dlt8gt s 2039 3 The effect of the pole on the compensation can be seen in Figure 2 Figure 1 Root locus without compensation solid line and with PD compensation dashed line Selecting exact values of z and p is usually done by trial and error Generally the zero is placed near the closediloop pole The choice of the compensator pole is a compromise between noise suppression and compensation effectiveness The process can be made more analytical in nature if the closediloop pole is selected rst Then we arbitrarily select one of the leadicompensator parameters and use the angle criterion to select the other Lag Compensation Once the desired transient response is obtained one might discover that the steadyistate response of the feedback loop is not ati factoi T r t in the t 39l t t errors can be made by placing a pole near the origin which is usually accompanied by a zero nearby so that the polezero pair does not signi cantly interfere with the overall dynamic system response shaped by the lead compensator 4 4 2 For the problem stud1ed prev1ously the lead compensator 520 could be followed by a lag compensator 5543011 The lag will not impact the faster dynamics However a root close to the imaginary axis will persist and it will slow down the overall response As a result it is important to place the lag poleizero at as high frequency as possible without impacting the 11113 Im 5 05 DS 2 s2 20 6 D H10 1 G 1 5 5 1 hm 7 l0 3 b Ims a 39 6 DZ i s2 DU 20 4 2 39 35 5 s10 Figure 2 Effect of pole on compensation Ims 1mm 05 y I a 1 A3 2 1 T Rem 7L0 08 fora o4 702 7 Ram Figure 3 E ects of lag compensation on root locusi dominant system dynamics The e ect of the lag compensation on the root locus is shown in Figure 3 Reading Assignment For material on dynamic compensation using rootilocus read pages 310328 of the text book Read the examples in Handout E28 posted on the course web page MEEN 364 Parasuram Lecture 28 Nov 24 2001 HANDOUT E28 EXAIVTPLE HANDOUT ON COlVTPENSATION USING ROOT LOCUS Example 1 Design a lead compensation for the system given by the transfer function Gs 1 that will provide a closedloop damping C gt 05 and natural frequency 3 s on gt 7 radsec S0 The general transfer function of a lead compensator is given as Let us design for the limiting condition of the damping ratio and natural frequency Therefore let us choose C 05 and Q1 7 radsec Hence the closed loop poles of the system are given by 0 ijwnl za 35r j6062 Let us choose 2 2 The angle subtended by all the poles and zeros of the feed forward transfer function with the closed loop pole at 35 6062 is 356062j U 2 1 0 gt X 9p ll24l l20 103898 This angle must be equal to 7180 degrees Hence the angle subtended by the compensator pole with the closed loop pole is 51488 degrees MEEN 364 Parasuram Lecture 28 Nov 24 2001 Therefore tan51488 2 x gt x 4824 p X 35 4824 35 8324 a 9 To calculate the gain K we have 1DsGs 0 gt DsGs 1 2 K s 2 1 s 9 ss 1 gt K a 60 1 573 55 052 Hence the lead compensator is given by 32 39 Ds 60 Plotting the root locus of the feed forward transfer function given by DsGs the specification of the location of the location of the closedloop pole can be veri ed Example 4 Consider the system whose feed forward transfer function is given by K s ss 2 system are located at s l i j and the steady state error to a unit ramp input is less than 02 G Design a lag compensator so that the dominant poles ofthe closed loop S0 The general transfer function for lag compensation is given by The forward transfer function is given as MEEN 364 Parasuram Lecture 28 Nov 24 2001 K X s 2 6mm ss2 sp39 For the specification that the steady state error of the system must not exceed 02 we have E0 wile ss2spKsz gt as lim SES lim MRUil H0 H0 ss2spKsz For a ramp input we have 655 w LZPlt02 H0 ss2spKsz s2 Kz Let 2 p 02 Kz Let us choose p 001 therefore we have Kz 01 We know that the closed loop poles lie in the root locus and hence 1DsGs 0 DsGs ss 2s 001 sz gtK S1 Solving for K and z we get K 188 and since Kz 01 we getz 00532 Therefore the lag compensator is given by s 00532 DU 2 s 001 39 Plotting the root locus of the feed forward transfer function given by DsGs the specification ofthe location ofthe closed loop pole can be verified MEEN 364 Parasuram Lecture 28 Nov 24 2001 Recommended Reading Feedback Control of Dynamic Systems 43911 Edition by Gene F Franklin etal 7 pp 310 Recommended Assignment Feedback Control of Dynamic Systems 43911 Edition by Gene F Franklin etal 7 problems 526 527 Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 1B Fourier Analysis and Fourier Transforms The objective of this lecture is to present to you some of the mathematics involved in the Fourier analysis of periodic functions and present the Fourier transform The latter is widely used in various signal acquisition and processing functions7 as you will witness in the lab part of the course Basic Continuous Time Signals Complex Exponential and Sinusoidal Signals The complex exponential signal is given by W O 6 1 where C and a are7 in general7 complex numbers If these are real numbers7 then t is called a real exponential and depending on the sign of a it can become either a growing exponential a gt 0 or a decaying exponential a lt 0 If a 0 then t is the constant C An equally important class of complex exponential signals is obtained when a is purely imaginary In this case7 we obtain t ejwot 2 An important property of this signal is that it is periodic A class of signals closely related to the complex exponential is the sinusoidal signal t Acosw0t g57 3 where it is common to write we 27Tf07 where f0 has the units of cycles per second or Hertz The fundamental period of the sinusoid is given by To 27139 4 7m 1 ut 0 Figure 1 Continuous time Unit Step Function The sinusoid of equation 3 can be expressed in terms of the complex exponential signal as Acosw0t Aggeea39wotltz7 where the right hand side denotes the real part of a complex number Unit Step and Unit Impulse Functions The unit step function is de ned as 0gtlt0 ut rtgt07 shown in Figure 1 Note that it is discontinuous at t 0 The unit impulse function 6t is related to the unit step function by the equation Therefore7 this suggests that 5 8t 0 t Figure 2 Unit Impulse Function A unit impulse is shown in Figure 2 Although the value at t 0 is in nite7 the height of the arrow used to depict the scaled impulse will be chosen to be representative of its area Continuous Time Fourier Series and the Continuous Fourier Transform Recall that a signal is periodic if for some positive7 nonzero value T7 W W T7 9 for all t The fundamental period To is the minimum value of the of T for which equation 9 is satis ed Further is called the fundamental frequency We also introduced the periodic complex exponential t ejwot 10 The signal 10 has fundamental frequency we and fundamental period To Now we can speak of the collection of harmonically related complex exponentials mt eWOE k 07i17i27 7 11 Each of these signals has a fundamental frequency that is a multiple of we Therefore7 a linear combination of harmonically related complex exponentials of the form 00 zt Z akejkwo 7 12 k7oo is also periodic with period To In this in nite sum the term for k 0 is a dc or a constant term The two terms for k 1 and k 71 both have fundamental period equal to To and are collectively referred to as the fundamental components or the rst harmonic components More generally7 the components for k N and k 7N are referred to as N h harmonic components The representation of equation 12 is called the Fourier Series representation Having stated the Fourier Series expansion7 it is now important to derive an expression for the expansion coef cients ak Multiplying both sides of equation 12 by e jWOt we obtain te 7W0t Z akemote 7WOt 13 k7oo lntegrating both sides from 0 to To 3 one fundamental period7 we have To To 1 00 te W0tdt Z ake kwotei wotdt 14 0 0 k7oo lnterchanging the order of integration and summation yields T 00 T O te quot 0tdt 2 aa 0 e7k 0tdt 15 0 167 0 The bracketed integral can be expressed as 1 To T k 77 57ltk gtwotdt 07 16 0 07 k 31 n Therefore7 the right hand side of equation 15 reduces to Teak If we denote integration over any interval of length To by T07 then we can summarize the Fourier series representation of a periodic signal as 00 t Z ake k w 7 17 k7oo and 1 t wotdt 18 at TOTozoe lt gt 1Prove this to yourselves as an exercise Equation 17 is referred to as the synthesis equation and equation 18 is referred to as the analysis equation The coef cients ak are often called the Fourier series coef cients or the spectral coef cients of These complex coef cients indicate the portion of the signal t that is at each harmonic of the fundamental component For k 0 the coef cient 10 is given by 10 ztdt7 19 which represents the average value of the signal t over one period The continuous time Fourier transform can be obtained by considering the behavior of the Fourier series as the fundamental period To a 00 or as the fundamental frequency we a 0 If we de ne the envelope Xw as Xw Tea7 20 then the Fourier series equations7 17 and 187 can be expressed as m immewm 21 and 00 Xw LOO xte dt 22 where for the limiting case of To a 007 the summations are replaced with integrals Equations 21 and 22 are referred to as the Fourier transform pair with equation 22 called the Fourier transform or the Fourier integral and equation 21 called the inverse Fourier transform The transform Xw of any signal t is also commonly referred to as the spectrum of xt7 because it provides us with information regarding how t is composed of sinusoidal signals at different frequencies Basic Discrete Time Signals Unit Step and Unit Impulse Sequences The unit step function in discrete time is de ned as 0 n lt 0 23 ulnl MEG lt gt shown in Figure 3 The unit impulse or unit sample sequence 6M is de ned as 0gn7 0 6M 1n0 24 5 un Figure 3 Discrete time Unit Step Sequence A unit impulse is shown in Figure 4 Complex Exponential and Sinusoidal Signals As in the continuous time an important signal in discrete time is the complex exponential signal or sequence de ned by Ca 25 where C and 04 are in general complex numbers If C and 04 are real we can have one of several types of behavior If lal gt 1 the signal grows exponentially with 71 while if lal lt 1 we have a decaying exponential Furthermore if 04 gt 0 all of the values of Ca are of the same sign but if 04 lt 0 then the sign of Ca alternates If 04 1 then is constant whereas if 04 7 1 then alternates in value between 0 and 70 Another important class of complex exponential signals is obtained if we let Oz i and 8n Figure 4 Unit Impulse Function let 6 be purely imaginary In this case7 we obtain 6790 26 A class of signals closely related to the complex exponential is the sinusoidal signal Acos90n i57 27 where if we take 71 to be dimensionless7 then both 90 and b have units of radians The sinusoid of equation 27 can be related to the complex exponential signal as follows A A Acos90n b 367790 36 56 790 28 There are some important differences between the complex exponential signals in con tinuous and discrete time regarding their range of de nition and periodicity Recall that in the continuous time case the larger the magnitude of we the higher the rate of oscillation in the signal Furthermore7 the function 67 is periodic for any value of we In the discrete time case the complex exponential is only de ned in the range 0 90lt27T70T 77T 20lt7T7 29 7 Table 1 Differences Between Continuous and Discrete Exponential Signals ewe ej en Distinct signals for ldentical signals for distinct values of we exponentials at frequencies separated by 27139 Periodic for any choice of we Periodic only if no 2quot for some integers N gt 0 and m Fundamental frequency Fundamental frequency to Fundamental period Fundamental period we 0 unde ned 9e 0 unde ned we 31 0 3 7 9e 31 0 and the signal 67 is not periodic for arbitrary values of Ge Rather for the signal to have period N gt 07 there must exist an integer m such that the following relation is true 90 m 30 27139 N Table 1 summarizes the differences between the signals 67 and 67 Read handout A1 for a presentation of the discrete time Fourier series and discrete Fourier transform Reading Assignment Read Handout Al Fourier Analysis and Fourier Transforms77 and the examples Hand out E2 posted on the course web page MEEN 364 Parasuram July 6 2001 HANDOUT A2 LAPLACE TRANSFORMS NOTE All the transformations have to be done using the analytical method outlined MATLAB has to be used only to verify the result obtained Introduction The Laplace transform is the mathematical tool that can be used for transforming differential equations into an easiertomanipulate algebraic form The advantages of this modern transform method for the analysis of lineartimeinvariant LTI systems are the following 1 It includes the boundary or initial conditions 2 The mathematics involved in the solution is simple algebra 3 The work is systematized 4 The use of table of transforms reduces the required labor 5 Discontinuous inputs can be treated 6 The transient and the steadystate components of the solution are obtained simultaneously The disadvantage of transform methods is that if they are used mechanically without the knowledge of the actual theory involved they sometimes yield erroneous results De nition of the Laplace transform The direct Laplace transformation of a function of time t is given by Lftl IN a dr no 1 where Lft is a shorthand notation for the Laplace integral Evaluation of the integral results in a function Fs that has s as the parameter This parameter s is a complex quantity of the form a bi Derivations of Laplace transforms for simple functions A number of examples are presented to show the derivation of the Laplace transform for several time functions A list of common transform pairs is given at the end of the handout MEEN 364 Parasuram July 6 2001 Example 1 Step Function The step function of size a is de ned as follows a OStltoo ut 0 0 oolttlt0 The Laplace transform of the above defined step function is obtained by substituting the function in equation 1 Lut fume wt Us 0 since ut has the value of a over the limits of integration w in Us Iaei dt 0 Decaying exponential function 6 quot The Laplace transform of the abovementioned exponential function is l w ersat Sda 0 Le m Ie meistdt Ie WWzdt 0 0 Sd39a Using MATLAB to calculate the Laplace transform To find the Laplace transforms of functions using MATLAB use the laplace command To know more about the command type the following command in the MATLAB command window help laplace help for symlaplacem LAPLACE Laplace transform L LAPLACEF is the Laplace transform of the scalar sym F with default independent variable t The default return is a function of s If F Fs then LAPLACE returns a function of t L Lt By definition Ls intFtexp stOinf where integration occurs with respect to t MEEN 364 Parasuram July 6 2001 L LAPLACEFt makes L a function of t instead of the default s LAPLACEFt ltgt Lt intFxeXp txOinf L LAPLACEFWZ makes L a function of Z instead of the default s integration with respect to w LAPLACEFWZ ltgt LZ intFwexp ZwOinf Examples syms a s t w X laplacetA5 returns 120sA6 laplaceexpas returns lt a laplacesinwxt returns wtA2wA2 laplacecosXwwt returns ttA2XA2 laplacexAsym32t returns 34piAl2tA52 laplacediffsym Ft returns laplaceFttss FO See also ILAPLACE FOURIER ZTRANS Note that there are also a few examples in the help le The laplace command is further explained with the help of the following examples Example 2 1 Determine the Laplace transform of the following step function using MATLAB 2 OStltoo ut 0 0 oolttlt0 The MATLAB code is as follows syms t f 2 tAO ans laplacef Note that the syms command in the first statement of the code implies that the variable t is to be considered as a symbol The laplace command works if and only if the argument is a function of time Since the function defined in the example is a constant it is converted to a function of time by multiplying the constant with t to the power zero The result of the above code is ans 2s The answer can be verified by following the procedure outlined in example 1 MEEN 364 Parasuram July 6 2001 2 Determine the Laplace transform of the following ramp function using MATLAB bt 0 S t lt oo 1 0 00 lt t lt 0 Analytical method Fs bjfie alt bjbteistdt w 7 w w gt bjie wiwie bLj1e di 0 S 0 S 0 w in w 0 Ie 5 dt e Li2 s 0 s s 0 s s s The integration by parts technique is used to obtain the above integral The MATLAB code to verify the result obtained is syms bt f bt ans laplacef Again note that the variable b also has to be defined as a symbolic variable The result is ans bSAZ Properties of Laplace transforms 1 Linearity If a is a constant or is independent of s and t and if t is transformable then Llaftl aLftl 61NS 2 Superposition If f1t and fzt are both Laplacetransformable the principle of superposition applies Llf1tir fztl Lf1tliLfztl 11 in S MEEN 364 Parasuram July 6 2001 3 Translation in time Ifthe Laplace transform offt is Fs and a is a positive e P O l real number then the Laplace transform of the translated function fta is Lft a e mFU Complex Di erentiation If the Laplace transform of t is F s then al thft1 FS als Multiplication by time in the time domain entails differentiation with respect to s in the sdomain Translation in the 3 domain If the Laplace transform of t is Fs and a is either real or complex then Lle ftl FS a Real Differentiation If the Laplace transform of t is Fs and if the first derivative of t with respect to time Dft is transformable then L1310 3FS f0 The term 0 is the value of the righthand limit of the function t as the origin t 0 is approached from the right side thus through positive value of time For simplicity the plus sign following the zero is omitted although its presence is implied The transform of the second derivative Dz t is LDZ0 SZFU Sf0 Df0 where Df0 is the value of the limit of the derivative of t as the origin t 0 is approached from the right side Final value Theorem If ft and D t are Laplacetransformable if the Laplace transform of t is F s and if lim f t exists then llIIOI sFs lim ft MEEN 364 Parasuram July 6 2001 Example Find the steady state value of the system corresponding to y3 ss 2 Solution From the nal value theorem the steady state value of the function is given by 3 3 11m ss2 H032 2 lim yt lim sYs lim s law SgtO Sgt0 Thus after the transients have decayed to zero yt will settle to a constant value of 15 00 Initial value Theorem If the function ft and its first derivative are Laplace transformable if the Laplace transform 0fft is F s and if lim sFs exists then lim sFs ft Applications of the Laplace transform to differential equation Let us consider a simple massspringdamper system whose equation of motion is given by mxcxkxft 2 The unknown quantity for which the equation is to be solved is the displacement of the mass Xt The above equation can be easily solved using the Laplace transform ft is the forcing function Now Lft FS Lm x ms2Xs sx0 x0 Lc x csXs x0 leC k XU The above results are obtained based on property 6 Substituting the above results in equation 2 we have MEEN 364 Parasuram July 6 2001 ms2Xs sx0 x0 csXs x0 kXs Fs gt ms2 cs kXs ms cx0 m x0 Fs From the abOVe equation we can see that X0 and x0 are the initial conditions for the displacement and the velocity If both the initial conditions are equal to zero then the above equation reduces to ms2 cs kXs Fs gt X s 37 ms cs k Based on the forcing function the Laplace transform of ft can be easily found as a result of which the value of Xs can be found In order to get the value of the displacement in the timedomain we need to determine the inverse Laplace transform of Xs to get Xt To get the inverse Laplace transform the above relation is expanded using partial fractions and then the inverse Laplace transform is obtained by looking in the table given in page 733 in Feedback control of dynamic systems Third Edition by Franklin etal To get the inverse Laplace transform using MATLAB use the ilaplace function in MATLAB By typing the following command in the command window yields the result help ilaplace help for symilaplacem ILAPLACE Inverse Laplace transform F ILAPLACEL is the inverse Laplace transform of the scalar sym L with default independent variable s The default return is a function of t If L Lt then ILAPLACE returns a function of X F FX By definition Ft intLseXpstsc iinfciinf where c is a real number selected so that all singularities of Ls are to the left of the line s c i sqrt l and the integration is taken with respect to s F ILAPLACELy makes F a function of y instead of the default t ILAPLACELy ltgt Fy intLyeXpsysc iinfciinf Here y is a scalar sym F ILAPLACELyX makes F a function of X instead of the default t ILAPLACELyX ltgt Fy intLyeXpXyyc iinfciinf integration is taken with respect to y Examples syms s t w X y ilaplacels l returns eXpt MEEN 364 Parasuram July 6 2001 ilaplaceltA2l returns sinX ilaplacetA sym52X returns 43piAl2XA32 ilaplaceyyA2 wA2yx returns coswx ilaplacesym laplaceFXXs sX returns FX Example 3 Determine the inverse transform of the function 2 X s 32 Analytical method By looking at the table of Laplace transforms on page 733 of the Feedback control of dynamic systems Third edition by Franklin etal it can be seen that the inverse Laplace transform of is 6 So the inverse Laplace transform of the above problem is s a 26722 The MATLAB code to verify the above result is syms s f 2s2 ans ilaplacef Note that the variable s is defined as a symbolic variable The result of the above code is ans 2exp 2t In other words the result is 26 Example 4 Solve the differential equation given by ytyt0 3 The initial conditions are y0 0 y0 0 MEEN 364 Parasuram July 6 2001 Solution First let us de ne the Laplace transform of each of the individual terms in the equation LJim SZYU Sy0 y0 LJim Ys Substituting the initial conditions in the above equations we have Lyt SZYS sot Substituting the Laplace transforms in equation 3 we have SZYS SaYS0 306 gtYsm After looking up in the transform tables the inverse laplace transform of Ys can be obtained as yt 06 cost It is always better to remember some of the inverse Laplace transform formulae Partial fraction expansion theorems The partial fractions technique is used when one needs to find the inverse Laplace transform For example consider the Laplace transform of a function to be Ps ansquot aHsH r r r als a0 Qs Sm bHsquot 1 gtb1sb0 The a s and b s are real constants and the coefficient of the highest power of s in the denominator has been made equal to unity The first step is to factor Qs into firstorder and quadratic factors with real coefficients Fs Fm Pm Pm Qs S SlS Szquot393 Skquot39S Sn The values s1 s2 sn that make the denominator equal to zero are called the roots of the denominator These values of s may be either real or complex To calculate the roots of Qs Qs is equated to zero ie MEEN 364 Parasuram July 6 2001 Qs 0 71 gts quot bHsquot gtblsb0 0 The transform F s can be expressed as a series of fractions Fs A1 A2 H Aquot FSQs s s1 s s2 39 s sn The procedure is to evaluate the constants A1 A2 A3 corresponding to the poles s1 sz s The coefficients A1 A2 An are termed as the residues of Fs at the corresponding poles There are four cases of problems depending on the denominator Qs 0 Case 1 Fs has first order real poles 0 Case 2 Fs has repeated first order real poles 0 Case 3 Fs has a pair of complex conjugate poles a quadratic factor in the denominator 0 Case 4 Fs has repeated pairs of complex conjugate poles a repeated quadratic factor in the denominator Case 1 First order real poles Consider the Laplace transform P3 135 40 A1 A2 4 Fs Qs ss s1s s2 s s s1 s s2 where s1 and s2 may be positive or negative or zero To evaluate a typical coefficient Ak multiply both sides of equation 4 by ss1 the result is i s sans s sags s sam lx z ampA0 AlAz ss s2 s s s2 MEEN 364 Parasuram July 6 2001 The multiplying factor ss1 on the left side of the equation and the same factor of Qs should be cancelled By letting s s1 each term on the right hand side of the equation is zero except A1 Thus a general rule for evaluating the constants for single order poles is Ps Ak k i S Qsim Example 5 For eXamPle 02 A0 A A2 FS 1 SS1S3 S sl s3 32 2 A0 sFSLo WSO g A1 S1FSrl SSs23 571 A2 s 3FSgts3 zisis21gti 573 Hence the partial transform expansion is Fltxgt21iLi 1 3s 2 s1 6 33 gt Fs 0396667 E 01667 S 31 33 MEEN 364 Parasuram July 6 2001 Case 2 Multiple order poles For the general transform with repeated real roots FSPS 133 Q0 3 SgrS Sl39 gt Fs AW 39 AQM w A A1 s sqquots sq 1 s sqs s1quot39 The coefficient Aqr can be obtained by following the procedure outlined in the previous case ie Ag 3 sq 133 QS To determine Aqr1 differentiate Aqr with respect to s and then substitute the value of s sq ie 1 mm A leis 0 Sq leim Repeating the differentiation gives the coefficient Aqrz as 161 2 r 133 AW l2 W l Sq leim In general 1 dk P s Aquila k S sq k ds Qs 55 Example 6 1 A13 L A12 L A11 L A2 s23s3 s23 39s22 39s2 39 s3 The constants are MEEN 364 Parasuram July 6 2001 3 3 1 An M2 Holy 0 2 S 2 3 3L 1 d s i 1 1 All Eb 2 19le d3 3 3 3 3 3 l 1 2 ii 1 l HZ dim 2 dsz s352 2 alss32 2 s33 A s 3Fsl3 1 s23 1612 A Eds2 Therefore 1 l l Fs3 Z s2 s2 s2 H3 Case 3 Complex conjugate poles The procedure is the same as that of case 1 Case 4 Multiple order complex poles The procedure is the same as that of case 2 Using MATLAB to calculate the partial fractions The residue function in MATLAB is used to obtain the coefficients and the poles of the transform The online help gives help residue RESIDUE Partial fraction expansion residues RPK RESIDUEBA finds the residues poles and direct term of a partial fraction expansion of the ratio of two polynomials BsAsL If there are no multiple roots Bs Rl R2 Rn Ks As s Pl s P2 s Pn Vectors B and A specify the coefficients of the numerator and denominator polynomials in descending powers of s The residues are returned in the column vector R the pole locations in column vector P and the direct terms in row vector K The number of MEEN 364 Parasuram Ju1y62001 poles is n lengthA l lengthR lengthP The direct term coefficient vector is empty if lengthB lt lengthA otherwise lengthK lengthB lengthAl If Pj Pjm l is a pole of multplicity m then the expansion includes terms of the form RU Rjl Rjm l sPJ SPUHAZ SPUHAI H BA RESIDUERPK with 3 input arguments and 2 output arguments converts the partial fraction expansion back to the polynomials with coefficients in B and A Warning Numerically the partial fraction expansion of a ratio of polynomials represents an ill posed problem If the denominator polynomial As is near a polynomial with multiple roots then small changes in the data including roundoff errors can make arbitrarily large changes in the resulting poles and residues Problem formulations making use of state space or zero pole representations are preferable The use of the command is explained with the help of an example Example 7 For the transform given calculate the poles and the coef cients of the partial fraction expansion s2 F 33 432 33 Note that in the residue command the coef cients of the numerator and the denominator should be given in the form of a matrix Analytical method 32 s2 A1 A2 i ss24s3 ss3s1 s 39s3gt 39 31 Fs s 2 3 A1 SFSL s3sls 0 3 0396667 l A2 s3FsS3 KHDLH 32 6 01667 H2 1 i A s1Fs Sm 3L 12 2 05 MEEN 364 Parasuram July 6 2001 Therefore 06667 01667 05 F 3 33 sl The MATLAB code verify the above result obtained is numerator l 2 denominator l 4 3 O rpk residuenumerator denominator The result of the above code is r Ol667 O5000 06667 p 3 l O k In the above result r denotes the coefficients Al A2 and A3 and p indicates the poles of the system MEEN 364 Parasuram July 6 2001 Assignment 1 Determine the Laplace transform of the following functions asinat bt3 c4t5 dte Verify the results using MATLAB 2 Determine the Inverse Laplace transform of the following functions 4 s5 l sa2 1 6 Verify the result using MATLAB 3 Solve the following differential equation ya 4yltrgt 0 Given the initial conditions y0 0 y0 Recommended Reading Feedback Control of Dynamic systems Fourth Edition by Gene F Franklin etal pp 96 7 115 Recommended Assignment Feedback Control of Dynamic systems Fourth Edition by Gene F Franklin etal pp 7182 32a 32c 33a 34b 34c 35a 35b 37b 37e pp 7 183 39 Texas A amp M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 6 Linearization and Scaling Operating Points and Impedance Matching The objective of this lecture is to give you an overview of the mathematical method involved in linearizing the dynamics of nonlinear systems in order to express them in stan dard state space form Linearization allows us to analyze complex dynamics using simple mathematics and analytical methods rather than computer simulations Linearization The differential equations of motion for most practically interesting systems are nonlinear For example most useful forms of damping contains nonlinear terms As mentioned before it is much easier to deal with linear models of a system than nonlinear ones Linearizatz39on is the process of nding a linear model of a system that approximates a nonlinear one Over 100 years ago Lyapunov proved that if a linearized model of a system is valid near an equilibrium point of the system and if this linearized model is stable then there is a region around this equilibrium point that contains the equilibrium within which the nonlinear system is also stable Basically this tells us that at least within a region of an equilibrium point we can investigate the behavior of a nonlinear system by analyzing the behavior of a linearized model of that system This form of linearization is also called small signal linearization Assume that the nonlinear equations of motion for a system model with one controlled input u are expressed in the form fx 1 In equation 1 the derivatives of the state are relate to the state andor the control through a nonlinear relation f In order to linearize this equation we must rst determine the equi librium values of the system The equilibrium values for the state7 x07 and control7 uo are such that the derivative of the state vector is zero That is7 we can compute the equilibrium values by solving x00fxo7u0 2 Equation 2 has two unknowns Therefore7 we must choose arbitrarily the value of uo and solve equation 2 for the equilibrium state7 X0 We now expand the nonlinear equation in terms of the perturbations from these equilib rium values that is7 we let Xt x0 5xt7 3 and ut uo Su 7 4 then we can write the following linear approximation to the nonlinear dynamics 1 x0 6Xt fx07u0 F6xt GSut7 5 where F and G are the best linear ts to the nonlinear function f at the point x07u0 Canceling the equilibrium from both sides of equation 5 results in 6Xt F6xt G ut7 6 which is a linear model approximating the nonlinear dynamics at the point x07u0 If an analytical expression for f is available7 then the best linear ts F and G can be obtained through differentiation7 as follows 3f F xuo7 7 6Xlo and 6f G MO 8 611 In case when an analytical expression is not available7 numerical differentiation is performed7 as shown in the following example Note For more details regarding linearization of nonlinear dynamics read the handout A5 Example Linearization of Motion in a Ball Levitator Figure 1 Laboratory scale magnetic ball levitator Figure 2 Model for a ball levitator Force fquot 10393N 0 x 1 I l 2 x1 3 4 5 6 Distance x mm Figure 3 Experimentally determined force curves Figure 1 shows a laboratory scale magnetic levitator where one electromagnet is used to levitate a ball bearing The physical arrangement of the levitator is depicted in Figure 2 The equation of motion for the ball is derived from Newton7s law as mid fmx 7 mg 9 where the force fmx is caused by the eld of the electromagnet Theoretically speaking the force is proportional to the inverse of the square distance from the magnet but the exact expression is dif cult to derive So we do not have an analytic expression for the force However the force can be measured and plotted Figure 3 shows the experimental curves for a ball with a 1 cm diameter mass and a mass of 84 X 10 3 kg From the experimental curves we infer that at the current value of 2392 600 mA and the displacement 11 the magnetic force fm just cancels the gravity force mg 82 X 10 3 N The mass is 84 X 10 3 Icy and the acceleration of gravity is 98 Therefore the point 1125 represents an equilibrium point We now want to nd the linearized equations of motion for this system First we expand the magnetic force in terms of deviations from the equilibrium point 11 2392 as follows fmx1 6xi2 6z39 z fmx1i2 Kz6xKi6zl 10 The linear gains Km and K can be computed as follows Km is the slope or derivative of the curve in Figure 3 for 2392 600 mA around the point 1 This is found to be about 14 K is the change of force with current at the value x 1 This is found as 122 10 3 i 42 10 3 N Kimm04 11 700 i 500 A So7 the linearized force expression becomes fm6z7 6239 82 gtlt 10 3 146z 04621 12 Considering that 5it7 the equation of motion 9 becomes 84 gtlt 10 36it 1469500 04611 13 621 16676zt 47662 t7 14 which is the linearized equations of motion about the equilibrium point We can select the state vector as Xt 5xt7 6it and the control ut 6ut This selection results in the following state matrices 0 1 0 7 G 15 1667 0 476 Amplitude Scaling Amplitude scaling is usually performed by simply picking units that make sense for the problem investigated In selecting the units we try to make the numbers of the problem comparable For example7 for the ball levitator expressing the displacement in millimeters and the current in milliamps would make the numbers easy to work with A method for accomplishing the best scaling for a complex system is rst to estimate the maximum values for each state and then scale the system so that each element varies between 71 and 1 The amplitude scaling is performed by de ning the scaled variables for each element lf xt S t7 16 then i t Smit i t Smit 17 We select the scaling Sm to accomplish our scaling objective outline above Time Scaling Variables involving time are usually measured in units of seconds Sometimes it is con venient to express time in other units We de ne a scaled time to be 739 wot 18 such that ift is measured in seconds and we 1000 then 739 will be measured in milliseconds The effect of time scaling is to change the differentiation as follows dx dx dx 19 x w dt dTw0 0dr7 and d2 d2 x 2 x z W wow 20 If the original system is in state variable or state space form Fx Gru7 21 then the time scaled system is expressed as Xe i 1 Fx 1 Gu 22 7 100 100 39 Read example 225 on page 76 of the text Note For more on operating points read the handout A5 For an exaple describing load or impedance matching read handout A4 Reading Assignment Read pages 68 76 the textbook Read examples Handout E12 and Handout A5 posted on the course web page MEEN 364 Parasuram August 5 2001 HANDOUT A4 MODELING EXAMPLES OF DYNAMIC SYSTEMS Introduction This handout consists of various modeling examples of dynamic systems It does not cover the entire subject It is always advisable for you to practice more such problems to become familiar with modeling of dynamic systems Example 1 DC Servomotor The dc motor is one example of an electromechanical system The circuit diagram of the armature controlled dc motor is given as follows Rm Lm Fixed eld TL 0 Description of the variables used im Armature current ea Applied voltage Rm Armature resistance Lm Armature inductance D Speed ofthe motor em Back emf Tm Torque developed by the motor Note that the time dependence of all the variables is ignored Unless specified explicitly all variables are time dependent The developed torque of a dc motor is proportional to the magnitude of the ux due to the field current if and the armature current im Therefore the developed torque can be expressed as Tm K3 im 1 For any given motor the only two adjustable quantities are the ux and the armature current There are two modes of operation of a servomotor For the armaturecontrolled MEEN 364 Parasuram August 5 2001 mode circuit diagram shown above the led current is held constant and an adjustable voltage is applied to the armature In the eld control mode the armature current is held constant and a voltage is applied to the eld circuit Since the above circuit is armature controlled the led current is held constant and therefore the equation 1 can be represented as Tm KTz39m 2 where KT is called the motor torque constant When the motor armature is rotating a voltage em is induced that is proportional to the product of the ux and the speed of the motor Since the polarity of this voltage opposes that of the applied voltage this voltage is called the back emf Since the ux is held constant the induced voltage is proportional to the speed of the motor mm This can be represented as emK1 wwamKb 3 where Kb is called the generator constant From the circuit diagram shown above the circuit loop equation can be written as 61139 ea Lm Rmim em 4 t d To write the mechanical equations of motion consider the diagram shown below T TL Motor Load Let J be the inertia of the entire system and B be the damping constant Then writing the torque balance equation we get 5129 d9 dtz 3 5 Tm TL J Ba Combining equations 2 and 5 we have 5129 de dtz 37K71m TL 6 J Similarly combining equation 4 and equation 3 we get MEEN 364 Parasuram August 5 2001 en Lm 61quotquot Rmim Kb 7 ch ch Equations 6 and 7 represents the governing equations of motion of the dc motor Let the individual states of the system be given as xl9 d9 x a 8 2 dt x3im From the rst two equations of equation 8 we get x1 x2 9 From equation 8 and equation 6 we have 2 d9 d9 Kz39 T J B dtz dt T m L gtJx2Bx2 KTx3 TL 10 gtxZ x2 x3 T J J L From Equation 8 and equation 7 we get 61 ea Lm Rmim Kb dt dt gtea Lm x3Rmx3Kbx2 11 R Kb m 1 gt X3 x2 x3 ea Lm L L Combining equations 9 10 and 11 and representing them in matrix format we get x1 1 K0 x1 0 0 x2 7 x2 0 ea 0 TL 12 J J 1 1 X3 K R x3 0 17 m L Lm Lm 1 mi The above equation is in the form of MEEN 364 Parasuram August 5 2001 XAXBuGw Where w is the disturbance to the system From equation 12 it can be concluded that the applied voltage ea is the input to the system The developed torque by the dc motor is the output to the system Representing equation 2 in matrix format we have 5 T KTim0 0 KT x m 13 N Equations 12 and 13 represent the state space model of an armature controlled dc motor Characteristics of a DC motor Combining equations 3 and 4 we get em ea Rmim m 14 dim dt gtwam ea Rmim Lm But we know that from equation 2 that Tm K7im Substituting the value of im in equation 14 we have T 51139 wamea Rm m Lm quot 15 K 7 dt Assuming a very low inductance the equation 15 can be rewritten as mm K1ea K2Tm 16 The torque speed curve for the armature controlled DC motor is as shown below A perating point Load torque curve constant torque Torque Input torque curve gtSpeed Torquespeed curve of a DC motor MEEN 364 Parasuram August 5 2001 Load matching The things that are usually regarded as important about an electric motor are its maximum speed and maximum power output Another is the motor s power and torque characteristic which are often overlooked but these need to be considered carefully because the torque and the power characteristics determine whether or not the motor can drive the attached load correctly To illustrate a motor driving a fan might require the same power output as a motor driving a conveyor belt However the torque and power characteristics required of the motor would be completely different To be able to successfully match a motor to a load we need to consider carefully the characteristics of the load In other words in the torquespeed curve of the motor the load torque and the input torque curves must intersect The point of intersection represents the condition at which the motor tends to operate There are different types of load each giving different characteristics and to select the correct motor the knowledge of the load profile is essential For example the most commonly found in the industry is the quadratic torque load In this case the torque varies as the square of the speed whereas the power varies as the cube of the speed This is the typical torque and speed characteristics of a fan or a pump Consider the following diagram Power Input Power output I MOIOI I Electrical Mechanlcal From the above figure it can be seen that the power input must be equal to the power output But since the Motor does not operate at its full efficiency the following relation is obtained P mzch PEZZL L714 m n But since the power output is equal to the product of the load torque and the speed of the motor Tm E12 L m Pm From the above relation it can be concluded that for a rated speed and voltage of the motor there is a fixed amount of load torque that the motor can drive MEEN 364 Parasuram August 5 2001 Example 2 A Liquid Level system The above gure represents a twotank liquid level system De nitions of the system parameters qi q1 qz Flow rates of uid hl hz Heights of the uid level is the tanks R1 R2 Flow resistance A1 A2 Crosssectional area of the tanks The basic linear relationship between the ow rate q change in the height h and the resistance to the ow r is given by q ow rate through ori ce h r The mass balance equation of the system can be written as Rate of uid storage in the tank tank input ow rate 7 tank output ow rate net ow rate A Applying the above two relations to tanks 1 and 2 we have dh h h A 1 1 2 1 dt q 91 9 R1 14 dh h h h A 2 2 2 ql qZ R1 R2 Let the individual states of the system be x1 hl x h2 15 The levels of the two tanks are the output of the system ie y1 h1 and y hz MEEN 364 Parasuram August 5 2001 From equations 14 and 15 the state space model of the system can be written as l l 1 M M M x1 A q 1 1 1 x 1 XZ 2 0 R1 A2 RIAZ R2 A2 y1 l 0 x1 y2 0 1 x2 Example 3 A Thermal System Consider the simple thermal system shown below Mi er Liquid in gt Temperature Ti Liquid out gt Temperature T Assume that the tank is thermally insulated and the liquid in the tank is kept at uniform temperature by perfect mixing with the help of a mixer Assume that the steady state temperature of the incoming uid is T and that of the out owing liquid is T The steady state thermal input rate from the heater is H and the liquid ow rate is assumed to be constant Let AH be a small increase in the thermal input rate from its steady state value This increase in thermal input will result in the increase in the thermal output ow rate by an amount AH1 and an increase in the thermal storage rate of the liquid in the tank by an amount AHz Consequently the temperature of the liquid in the tank and the out owing liquid rises by AT Since the insulation is perfect the increase in the thermal output ow rate is only due to the rise in temperature of the out owing liquid and is given by AH1 meAT 16 where m is the liquid ow rate and Cp is the specific heat of the liquid MEEN 364 Parasuram August 5 2001 Let us de ne the thermal resistance as l me I Therefore equation 16 reduces to 1 AH 17 R The rate of heat storage in the tank is given by AH MC MCM 18 2 F dt dt dAT where M is the mass of the liquid in the tank is the rate of rise of temperature in the tank and C which is equal to the product of M and Cp is called the thermal capacitance Therefore the heat ow balance equation or the energy balance equation can be stated as the thermal increase in the input must be equal to the sum of the thermal increase in the output and the thermal increase of the liquid stored in the tank For the above system the energy balance equation is given by AHAH1AH2 ATCdltATgt R dt AH 19 gtRAH ATRC The third equation of equation 19 describes the dynamics of the thermal system with the assumption that the temperature of the incoming uid is constant MEEN 354 Pauslxam August 5 2mm Exampk e A Hydnul39l syslzem spam x Tap Tasump me hgm Presslxe saute The shave gue shwws a ample hynkmhc annular m which we mmmn uf spam n memmmmehegepzemesmeeem s a yh nan 1 WexpmnThsauusadnff alpzesslx ans epmmwmhemesme er mum mta hen pushng39hzadm mtuf ta sump emlxs pusslxxzedby plxnpmdxsrecxrcu a a me a u m pmnmwes a awe y rm 5 Jaime paaummrespmsetathz splacanent x39 uf39hevalve spam rm 5 mm pasum Thar zxms anunhnear xelaumsh bemenme vuhxnemc ml aw me q39 mm me paw pm me exennal pzesme Ap39 acmssthz pmer mall values uf spam splacanent x39 The xelahmdmp beween q39 x39 and up maybewnuzn as Xv Av an Expmdnng me shave equahm m nym39s senes abvut me mama apauung pm qupwxn manegenmgm 39heterms uf secmd swing denwhves we gal 113 Xin Aviva 01 um A e Fax ms systzm me mama apenmng pm cmwpm s m a n AP n xquot meagre me equananal reducesta MEEN 364 Parasuram August 5 2001 q K1xK2Ap where Kz 1 a 22 160 AF K aq BAP 2 160 Apo Equation 22 gives a linearized relationship among q X and Ap Assuming leakage and compressibility ows to be negligible the rate of oil ow into the piston is proportional to the rate at which the piston moves ie dy A A 23 q dt y where A is the area of the piston The force on the piston is Ap which moves the load consisting of mass M and viscous friction with coef cient 7quot Writing the Newton s second law motion or the force balance equation we have AAp Myfy 24 From equations 22 and 23 we get A K1xqMyfy 25 K2 Substituting the value of q from equation 23 in equation 25 we have A K1xAy Myfy K2 AK A2 26 gt 1 xM KZ y f K2y Taking the Laplace transform of the second equation of equation 26 we get AK1 2 A2 XSMSYS sYs K2 f K2 AK 27 3 Yo K2 X Ms2 f 2 2s MEEN 364 Parasuram August 5 2001 The second equation of equation 27 represents the transfer function between the input X which is the displacement of the spool and the output y which is the displacement of the load attached to the piston The second equation of equations 26 represents the governing differential equation of the system To denote the equations in the statespace form let the states be defined as y x1 28 y x2 From the above relations it can be concluded that x1x2 29 Substituting the relations given by equation 28 in the second equation of equation 27 we get AK1 A2 xM K y f K2y 2 AK1 A2 gt xMx x 30 K2 2 f K2 2 2 AK gtxz ifA x2 1x M K2 MK2 Representing the equation 29 and the third equation of equations 30 in the matrix format we have 0 1 x 0 x1 1 A2 1 AK x 0 f x 1 x 31 2 M K 2 MK The output of the system is the displacement of the load y for the given input X Therefore representing the output equation in matrix format we get Yy gt Y 1 org 32 x 2 Equation 31 and the second equation of equation 32 represent the statespace model of the hydraulic model discussed above MEEN 364 Parasuram August 5 2001 Example 5 A Mechanical system X direction y direction The Disc with mass moment of inertia I rotates in the counterclockwise direction The block of mass m moves a distance of y units from the static equilibrium position in the positive ydirection From the above gure it can be seen that the system has two degrees of freedom One is the rotation of the disc and the other the linear displacement of the block Let the two degrees offreedom be represented as 9 and y respectively The next step is to determine the velocity and acceleration components of the block and disc The linear velocity and the linear acceleration of the block are given by y and y respectively Similarly the angular velocity and angular acceleration of the disc are given by 9 and 9 respectively This stage is called the kinematics stage The next step is to draw the free body diagram of the block and the disc This stage is called the kinetics stage Free body diagram of the block Fs my MEEN 364 Parasuram August 5 2001 Note that the gravity force is not considered in the free body diagram The reason for this is that y is considered from the static equilibrium position and hence the spring force at the equilibrium position is cancelled by the weight of the block Writing the Newton s second law of motion for the block which states that sum of all the forces acting on the block must be equal to the product of its mass and acceleration 2F ma gt FS m y 33 gt m y FS 0 Since the disc is rotating in the counter clockwise direction there is an elongation in the right hand spring by an amount of Re units The block is assumed to move down which in turn will again produce an elongation in the right hand spring by an amount of y units Therefore the total elongation of the right hand spring due to the movement of the disc and the block is R9 y units Therefore the spring force is given by F kR9 y Therefore the equation of motion of the block is given by my kR9 y 0 34 Free body diagram of the disc F51 F52 Taking moments about the center of the disc we get 19 F R F52R gt 19 kR9 R kR9 yR 35 gtIQ kR29 kRR9 y0 The third equation of equation 35 and equation 34 together represent the governing differential equation of motion for the system defined MEEN 364 Parasuram August 5 2001 To represent the abovederived differential equations in statespace form the states of the system have to be defined Let the states be given by y x1 y x2 36 9 x3 9 x4 From the above relations the following two state equations can be derived x x 1 2 37 x3 x4 Substituting the relations given by equation 36 in equation 34 we get mykR9 y0 gtmmkRx3x10 38 k kR gt 962 x1 x3 m m Similarly substituting the relations given by equation 36 in the third equation of equations 35 we have 19 kR29 kRR9 y0 gtIx4kR2x3 kRRx3 x10 39 CR 2kR2 2 x4 x1 x3 I I Representing the equations 37 the third equation of equations 3 8 and the third equation of equations 39 in the matrix format we have 0 l 0 0 x X1 1 1 0 kR 0 x m m 2 0 0 0 1 x3 40 XS kR 2kR2 0 0 x4 x4 I I MEEN 364 Parasuram August 5 2001 If the output of the system is the displacement of the block then representing the output relation in the matrix format we get 5 41 N lt II c o o o 88 w x4 Equation 40 and equation 41 represent the statespace representation of the above system MEEN 364 Parasuram August 5 2001 Assignment 1 An electromechanical actuator contains a solenoid which produces a magnetic force proportional to the current in the coil f K 1139 The coil ahs resistance and inductance a Write the differential equations of performance b Write the state equations Fll Z h J 797 x L M1 39i L e Solenoid quotQ R M2 B1 K2 B2 K1 2 Problems 21 23 28 220 222 in Feedback Control of Dynamic Systems 4111 Edition by Gene F Franklin etal Recommended reading Feedback Control of Dynamic Systems 4111 Edition by Gene F Franklin etal 7 pp 22 68 Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 2B Sampling and Aliasing Phenomena Sampling Rate Selection and Nyquist Frequency The objective of this lecture is to introduce and explain some of the most basic concepts that are involves in the measurement of continuous time signals The basic rules regard ing sampling are discussed and the consequences of incorrect sampling rate selection are mentioned De nition The instantaneous values of a signal at equally or unequally spaced values of time are called samples Sampling if done correctly can help us represent a continuous time signal very accurately The correct way of sampling a continuous time signal is described by the so called sampling theorem Sampling of ContinuousTime Signals In general if we were to sample a signal we would not expect to be able to guess the precise continuous time signal that generated the samples In fact there are in nite number of signals that pass through a nite number of points as shown in Figure 1 Strangely enough for most signals if we take the samples suf ciently close to each other we can uniquely identify or reconstruct the continuous time signal that generated the samples The above statement forms the basis for the accurate measurement of dynamic signals such as vibration level ImpulseTrain Sampling x3 x1 x2tt Figure 1 Three Continuous Time Signals with Identical Values at Integer Multiples of T W un hn n yml ll hnml Figure 2 Pulse Amplitude Modulation As A gt 07 the pt Approaches an Impulse In general7 a simple way to sample a continuous time signal is to measure it for a nite period of time using pulses7 as shown in Figure 2 As the duration of the pulses approaches zero7 we obtain the impulse train shown in Figure 37 for which the individual impulses have values corresponding to instantaneous samples of the continuous time signal at time instants spaced T seconds apart The problem is that if we arbitrarily space the impulses shown in Figure 37 then we will not be able to uniquely know the signal that generated the samples However7 if the impulses are appropriately spaced7 that is if we sample the continuous time signal with a predetermined sampling rate7 then we can uniquely determine its nature in the continuous time domain The determination of the sampling rate needed for a signal is computed from the sampling theorem using the formula ws gt ZWM 0 l T pm ll MM 0 t 0 r lt T gt x X Iquot 1 x xpm 1 i f x f in Figure 3 Pulse Amplitude Modulation with an Impulse Train t where MM is the maximum frequency contained in the signal and ws is the sampling rate Equation 1 says that for accurate representation in the discrete time domain7 a continuous time signal must be sampled at least twice the maximum frequency present in that signal That is7 the sampling frequency must be suf ciently high7 so that the spacing of the impulses shown in Figure 3 is close enough to give us an accurate picture of the continuous time signal being sampled The sampling frequency ws is called Nyquz st frequency The frequency ZwM which must be exceeded is called Nyquz st rate The Effect of Undersampling Aliasing In our previous discussion we argued that if in our measurements the sampling rate used is high enough7 we can identify or reconstruct a continuous time signal uniquely If for some reason our sampling rate is not high enough7 that is if we undersample a signal7 then we end up with an effect called aliasing This phenomenon results in obtaining very inaccurate and false information regarding the signal being sampled A simple explanation for the reasons behind this phenomenon is given in Figures 4 and 5 In this gure a sinusoidal signal with increasing frequency is sampled using impulse trains of constant spacing In part a of the gure the sampling rate is six times the frequency of the sinusoid7 ie six samples are collected during each complete cycle Since the sampling rate is more than twice the frequency of the sinusoid7 the signal can be reconstructed In part b the sampled signal is twice the frequency of the signal in part a7 whereas the sampling rate is kept the same As a result7 the sampling rate is three times the frequency of the Original signal W W l Reconstructed signal 4 m 3 Samples I 4 Figure 4 Effects of Aliasing on a Sinusoidal Signal sinusoid and7 again7 the signal can be uniquely reconstructedl In part c of the gure the sinusoidal signal frequency is increased to four times the frequency of the signal in part a By keeping the sampling rate the same as in part a we have created a serious problem The sampling rate is now 15 times the signal frequency less than the 2 required by the sampling theorem As a result we are not able to uniquely reconstruct the sinusoid shown in part c of the figureZ Any further increase in the frequency of the sinusoidal signal without appropriate increases in the sampling rate results in similar aliasing This is shown in part d of the gure also3 Reading Assignment Read the examples Handout E4 posted on the course web page 1Note that there are three samples collected per cycle7 including the sample at the zero crossing points 2Note that there are three samples every two cycles 3Note that there are six samples every ve cycles Figure 5 Effects of Aliasing on a Sinusoidal Signal continued Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Farlos Fall 2003 Lecture 4B Electrical and Electromagnetic System Components The objective of this lecture is to review the fundamental components of electric and electromagnetic circuits The laws governing electric circuits will be presented Basic Laws of Circuits ln electric circuits we talk about two terminal elements such as resistors capacitors etc as shown in Figure 1 The electric potential at each terminal is measured by its voltage with respect to the ground or some other local reference potential such as a machine frame or chassis The rate of ow of electrical charge through the element its current is measured in terms of amperes or A The fundamental equation relating these two quantities usually takes the form 5120i f1 At 07 MW f2512t 1 In addition to the two terminal elements there are two types of ideal sources used to drive circuits as shown in Figure 2 The ideal voltage source capable of delivering designated voltage level 65 regardless of the current drawn and the ideal current source capable of delivering the designated current is regardless of the voltage required to drive the load Two basic laws govern the operation of circuits These are known as Kirchoff7s voltage law and Kirchoff7s current law The voltage law says that the sum of the voltage drops around a loop must be zero The current law says that the sum of the currents at a node the junction of two or more elements must be zero These two laws are illustrated in Figures 3 and 4 Twoterminal 1 elemenl A 2 I1 Is 715 ell IS positive when 21g gt 923 Figure 1 Circuit diagram of a twoeterminal electrical elementi i10ad current gt o 1 652 2 25 is positive when F gt 525 ie when en 15 positive Figure 2 Circuit diagram of a voltage and current sources n l 2 3 4 aegle2clg00r b914 2523 34 ell 2392 4 22g Figure 3 Kircho 7s Voltage Lawr iAiBiCOor Quirk Figure 4 Kircho 7s Current Lawr if d qCd gt i g Figure 5 Circuit diagram of an ideal capacitor Capacitors A capacitor is used to store electric charge The equation describing the capacitor charge 100 06120 2 In terms of the current the rate of change of the charge the governing equation for a capacitor is 7 dead 20t 7 CT 3 The state variable of a capacitor is the its voltage e12t A capacitor like a mass m is an energy storage element and it is used to store electrical energy This energy is in the form of a static eld and it can be expressed as C 54 gem lt4 As in the arguments about a mass m attempts to suddenly change the voltage across a capacitor would require an in nite power source However one can suddenly change the capacitor current An ideal capacitor is shown in Figure 5 Inductors The variable governing the operation of an inductor is the ux linkage A12 It can be expressed in terms of the current owing through the inductor as A12t ma 5 4 a quot7 z A I 3913 Pig flgzl lz e21a dlZdrLdiLdr 3 2 Figure 6 Circuit diagram of an ideal inductor In terms of the voltage e12t Q and the governing equation for an inductor is expressed as diL t 612t 1 6 Note the similarity of an inductor to the ideal spring The state variable of an inductor is its current iLt The energy stored in an inductor is in the magnetic eld surrounding its conductors and it is known as magnetic eld energy The stored magnetic eld energy can be expressed as L 5amp0 52 7 As in the arguments about a spring 1c attempts to suddenly change the current owing though an inductor would require an in nite power source However one can suddenly change the voltage across an inductor An ideal inductor is shown in Figure 6 Transformers If two coils of wire are installed very close to each other so that they share the same core without ux leakage an electric transformer results as shown in Figure 7 A transformer is a fouriterminal element and two equations are needed to describe its operation The equations describing the operation of a transformer are e34t ne12t 8 where n is the ratio of the number of turns between 3 and 4 to the number of turns between 1 and 2 and 1 w We 9 5 1h gt 39 lt3 4 i SH 2 l n 2 ratio of turns between 3 and 4 0 turns between I and 2 I Figure 7 Circuit diagram of an ideal transformeri Figure 8 Circuit diagram of an ideal resistori Equation 8 and 9 indicate that transformers do not store energy7 rather are used to couple circuits dynamicallyi Resistors The equation governing the operation of an ideal resistor is Ohm7s law It can be ex pressed as 612t Riga 10 Note that the voltage across a resistor and the current through it are related instantai neously77 to each other This is because there is no energy storage7 rather dissipationi A circuit diagram of an ideal resistor is shown in Figure 8 Examples of Circuit Analysis Figure 9 An R7 L7 C circuit driven by a current source Example 1 Develop the inputioutput di erential equation for the circuit shown in Figure 9 We use the soicalled loop method to derive the equations describing the circuit operation As shown in Figure 9 there are two independent loops in this circuit We name the current owing through these loops as if and in It is obvious that from loop I we can immediately write it isti For Loop ll we can write Kircho 7s voltage law as follows d39 t 1 32mg Liz If 5 211tdt R12Ht 7 25m or We can relate the capacitor voltage 60 t with the capacitor current iHt as follows 7 de0t 211t 7 C dt i As a result7 equation 12 can be rewritten as deg t 1260 t 7 1561 R2CT LOW 60t 7 Rlzst or by rearranging we have a9 t d t l jglmerG9wamp w 7 11 12 13 14 15 Figure 10 Highigain OpAmp with capacitor feedback Example 2 The circuit shown in Figure 10 involves the use of a highigain operational ampli er OpAmp with feedback to achieve desired dynamic response in automatic controllers The gain km of the OpAmp is negative7 and its input current in is so small that it can be considered negligible The objective is to develop an inputioutput model for this circuit We use Kircholf7s current law at node This yields iRt iot because the current in is assumed negligiblei Equation 16 can be rewritten as EN 629t Cd 2gt 6290 R dt For the ampli er we have the following equation 6290 kaezg l 5290 E5390 Combining equations 17 and 19 results in d M egg ext 7 kiaeggm ROW 8 16 17 18 19 20 l F15 R7 flit 93g 2350 Figure 11 Simulation block diagram of an integrator with time constant Rearranging yields 17 71 de t 1 C 1 39lt 7 k 639t feiti 21 Considering that for an ampli er7 1cm is very large7 we have R LEG z 7 22 em egg 10gt 0i eilttgtdt mar lt23gt The use of a capacitor in the feedback with a resistor at the input results in an integrator with time constant RC as shown in the block diagram of Figure 11 Reading Assignment Read pages 4650 the textbook Read examples Handout E8 posted on the course web page Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 16 Use of Bode Plots in Dynamic Compensation The objective of this lecture is to complete the discussion on the use of Bode plots for dynamic compensation in control systems A Note of Steadystate Errors In previous lectures we have seen that the steady state error of a feedback system de creases as the gain of the open loop transfer function increases The low frequency open loop gain of a control system is given by KGjw K0jw 1 As a result the larger the value of the magnitude of the open loop DC gain the lower the steady state error will be for the closed loop system This inference is very useful in compensator design Summary of Compensation Characteristics In many control system design problems a mere change in the controller proportional gain cannot satisfy desired speci cations As a result controller dynamics are introduced to alter compensate the overall closed loop system dynamics There are two widely used techniques for dynamic compensation Lead compensation acts mainly to lower the rise time and decrease the overshoot of a feedback loop lt approximates the effects of derivative control Lag compensation acts mainly to lower steady state error and as such it approximates the effects of integral control In many instances both lead and lag compensation are used simultaneously resulting in lead lag compensation We can summarize the compensation characteristics as follows 1 1 PD Compensation It adds phase lead at frequencies above the break point If no changes are made to the low frequency gain7 PD compensation will increase the system bandwidth and speed of response As a result the system sensitivity to noise will increase 2 Lead Compensation It adds phase lead only in between the two break points Similarly to the PD compensation7 the system speed of response will increase Also7 unless adjustments to the low frequency gains are made the steady state error of the system will increase 3 PI Compensation It increases the magnitude of the frequency response below the break point7 resulting in lower steady state errors It also adds phase lag beyond the break point7 which degrades the overall system stability 4 Lag Compensation It increases the frequency response at frequencies below the two break points7 and decreases the steady state error It also contributes phase lag between the two break points7 which must be kept low enough as to not degrade the phase margin PM7 and as a result system stability Reading Assignment Read page 440 of the textbook Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 1A Introduction to Signals and Systems The objective of this lecture is to introduce you to the various concept regarding sig nals and systems Furthermore the widely used representation of signals and systems are introduced and the mathematical tools needed for their treatment are mentioned De nitions of Signals and Systems Signals and systems arise in many different areas of science and engineering such as aeronautics and astronautics electrical circuit design automobiles energy systems process control speech processing etc Though the physical nature of the underlying signals and systems might be different they have two very basic features in common The signals we are referring to are functions of one or more independent variable such as time and spacel and they typically contain or carry information about the underlying natural phenomena they represent For example an acoustic sound signal consists of pressure variations propagating through air that carry information about the music being played The systems on the other hand we are interested respond to particular signals the input signals by producing other signals output signals Voltages and currents as a function of time in an electrical circuit are examples of signals whereas the circuit itself is an example of a system Representations of Signals and Systems In general there are two frameworks for signal and system analysis one for phenomena that are described in continuous time and one for those described in discrete time Occa sionally phenomena in the continuous time are dealt with either in continuous or discrete time All physical phenomena can be described using continuous time or analog signals 1In this course we will primarily deal with signals that are only functions of time xm A xl11 x1 m 57 432 ll 110 1 61111012345 7891 n b 3 Figure l Graphical representation of a continuous time and b discrete time signals ln the case of continuous time signals the independent variable is continuous and thus these signals are defined for a continuum of values of the independent variable On the other hand discrete time signals are only defined at discrete times and consequently for these signals the independent variable takes on only a discrete set of values An example of a continuous and a discrete time signal is shown in Figure 1 Similarly to signals systems can also be classified as continuous time or discrete time A continuous time system transforms a continuous time input signal into a continuous time output signal Similarly a discrete time signal transforms a discrete time input signal into a discrete time output signal A schematic representation of continuous time and discrete time systems is shown in Figure 2 Use of Mathematical Transforms in Dealing with Signals and Systems ln dealing with the analysis of signals and systems we use various mathematical tools ln particular we use differential equations and their discrete counterparts difference equations extensively We will do so in this course especially when we model various dynamic systems Continuoustime x v l a Discretetime x n Vlnl b Figure 2 a continuous time systems and b discrete time system Table 1 Use of Transforms in Signals and Systems Continuous time Discrete time Steady state Behavior Continuous Fourier Transform Discrete Fourier Transform Transient Behavior Laplace Transform z Transform One particular class of mathematical tools that is very useful in the analysis of dynamic signals and systems is the various mathematical transforms Two particular transforms deserve our attention and we will spend the next a couple of lectures brie y talking about them these are the Fourier transform FT and the Laplace transform LT The former is primarily used for signals and systems when we are interested to analyze their steady state behavior The latter is a generalization of the former and it is used to analyze the transient behavior of signals and systems Both transforms have continuous time and discrete time equivalents For the FT we have a continuous and discrete time formulation ie continuous time FT and discrete time FT DFT The LT is used exclusively for continuous time whereas its discrete time equivalent is called the z Transform Table 1 summarizes the use of these transforms in the analysis of signals and systems Reading Assignment Read Chapter 1 of the course text by Franklin Powell and Emami Naeini and Handout El with examples on signals and systems posted on the course web page Furthermore make sure you download and read handouts Ml through M4 on the required math background MEEN 364 Parasuram Lecture 67 August 7 2001 HANDOUT E6 EXAIVTPLES ON MODELLING OF ROTATIONAL NIECHANICAL SYSTEMS Note that the time dependence of variables is ignored for all manipulations Example 1 A single DOF system Consider a simple pendulum shown below Kdir ction Kinematics stage There is only one rigid body Let the degree of freedom of the rigid body of mass m be de ned by the angle 9 moved by the body from the vertical position Therefore the angular velocity and angular acceleration of the body is given by 9 and 9 respectively Therefore the linear velocity and linear acceleration of the body is given by L9 and L9 respectively Kinetics stage Free body diagram of the body mLO mgsinG mgcosG mg T is the tension in the string Writing the Newton s force balance equation we get MEEN 364 Parasuram Lecture 67 August 7 2001 T mgcosO 0 mLO mgsin9 Therefore the governing differential equation of motion for the system is given by g 9 s1n 9 0 l L For small angles ie if 0 is very small then sin 9 cs 9 Therefore the equation of motion reduces to g 9 9 0 2 L Equation 2 represents the final linearized differential equation of motion for a simple pendulum The generalized second order differential equation is given by 9 0039 0 where on is the natural frequency of the system Comparing equation 2 with the abovegeneralized equation we get State space representation Let the states of the system be defined by 9 x1 3 9 x2 From the above relations we get 961 x2 4 MEEN 364 Parasuram Lecture 67 August 7 2001 Substituting the relations given by equation 3 in equation 2 we get gt 962 x1 5 Rewriting equations 4 and 5 in matrix format we have 0 l x X1 0 1 I 6 m L x2 If the output of the system is the angular velocity of the bob of the pendulum then expressing the output equation in a matrix format we have 7 gt y 0 XI x2 Equations 6 and 7 represent the statespace form of the above system Example 2 A single DOF system Consider the system shown below JLdirection Kinematics stage From the above gure it can be seen that there is only one rigid body and the degree of freedom is represented by 9 in the counter clockwise direction The angular velocity and the angular acceleration of the rod are given by 9 and 9 respectively This completes the kinematics stage MEEN 364 Parasuram Lecture 67 August 7 2001 Kinetics stage Free body diagram of the rod Ry If we assume that 0 is very small then point B essentially moves horizontally Therefore the distance moved by point B is the arc length L0 This is the amount by which the spring is compressed Therefore the spring force is given by F5 kLO Furthermore since 0 is small the spring is essentially horizontal Therefore the moment of the spring force about the pivot A is given by LkL9 Taking moments about the pivot A we have kL29 mgsin9 IA9 8 For small values of 0 we have sin 9 cs 9 Therefore equation 8 reduces to L 1A9 kL29 mg39 0 9 Equation 9 represents the governing equation of motion State space representation Let the states of the system be defined as 10 MEEN 364 Parasuram Lecture 67 August 7 2001 From the above relations it can be seen that x1x2 11 Substituting the relations given by equation 10 in equation 9 we have 2 L 1A9 kL 9 mg39 0 gt IA xZkL2 MgLx1 0 2 kLZ gtx2 2x1 12 IA Rewriting the equations 1 l and 12 in matrix format we have 0 1 x1 chZ mgL x1 13 x2 2 0 x2 IA If the output of the system is the angular displacement of the rod then expressing the output relation in matrix format we have s y 1 14 Equations 13 and 14 represent the statespace form of the abovede ned system MEEN 364 Parasuram Lecture 67 August 7 2001 Example 3 Two DOF system Consider the system shown in the gure below k direction Kinematics stage From the above gure it can be seen that there are two rigid bodies which are hinged and can only rotate about the xed point Since the two bodies are interconnected by a spring there are two degrees of freedom for the system Let the degrees of freedom be the rotation of the two discs in the clockwise direction Let the angle of rotation of the two discs be de ned as 91 and 92 respectively Then the angular velocities ofthe discs are 9 1 and 9 2 respectively and the angular acceleration of the two discs are given by 9 1 and 9 2 respectively This completes the kinematics stage Kinetics stage Note that since the discs are identical they have the same the moment of inertia 1 Assume 92 to be greater than 91 Free body diagram of the disc 1 kR9 2 9 1 kRQ 1 MEEN 364 Parasuram Lecture 67 August 7 2001 Writing the torque balance equation ie taking moments about the center of the disc we get 2M 19 gt kR91RkR9 2 9 1R 1 1 gt1912kR291 kR29 2 0 15 Free body diagram of the disc 2 kR9 2 9 1 Ft Taking moments about the center of the disc we have 19 2 kR9 2 9 1RFquot 19 2 kR291kR29 2Fr 16 Equations 15 and 16 represent the equations of motion of the system Rewriting the equations in matrix format we have I 0 91 2kR2 kR2 91 0 019 kR2kR292Fr39 2 State space representation Let the states of the system be de ned as 9 1x1 9 1x2 17 9 2 x3 9 2 x4 From the above relations the following two differential equations can be derived MEEN 364 Parasuram Lecture 67 August 7 2001 X1 x2 X3 x4 Substituting the relations given by equation 17 in equation 15 we get 19 12kR29 1 kR29 2 0 gt IXZ2kR2x1 lltR2x3 0 2 2 Zkf x1 kRx3 19 gtXZ Substituting the relations given by equation 17 in equation 16 we have 19 z kRZQ 1 kR29 2 Fr gt 1964 IltR2x1 ksz3 Fr kR2 kR2 x1 r gtx x F 20 4 I 3 I Rewriting equations 18 19 and 20 in matrix format we get o 1 o o 0 M 2kR2 kRZ x1 x o 0 x 0 2 0 0 6 1 2 0 F 21 x3 kRZ kRZ x3 r x 0 0 x4 4 L 1 1 1 1 If the output of the system is the angular displacements of both the discs then expressing the output equation in matrix format we get y191x1 y292x3 x1 10 0 0 x gty1 2 22 yz 0 010 x3 x 4 Equations 21 and 22 represent the statespace form of the system de ned above MEEN 364 Parasuram Lecture 67 August 7 2001 Example 4 Two DOF system Consider the system shown below Wtion Kinematics stage There are two rigid bodies and since they are attached by means of a spring the number of degrees of freedom of the system is two Let the two DOF s be represented by the amount of angular displacement of both the rods Therefore let the DOF s be 91 and 92 Then the angular velocities and angular accelerations will be 9 19 29 19 2 respectively This completes the kinematics stage Assume 91 and 92 to be small Kinetics stage Free body diagram of the rod 1 kae1 nen mlg Taking moments about the point A we have ZMQQQ1 L gt m1g71sin9 1 kL19 1 L29 2L1cos9 1 11A9 1 23 MEEN 364 Parasuram Lecture 67 August 7 2001 For very small angles sin9 1 29 1 cos9 1 a 1 Substituting the above relation in equation 23 we have m1 gL 1119 10ch 2 19 1 kL1L26 2 0 24 Free body diagram of the rod 2 kL91 L29 2 ng Taking moments about the point B we get 2M 13 9 2 B L 25 gt m2g2sin9 2 kL19 1 L29 2L2 cos9 2 I 9 2 Assuming small angles the above equation reduces to 2 quot12ng 1239 2kLZT9 z kLlLZQ 1 0 26 Equations 24 and 26 represent the governing differential equations of motion State space representation Let the states of the system be de ned as 9 1 x1 27 9 1 x2 MEEN 364 Parasuram Lecture 67 August 7 2001 9 2 x3 28 9 2 x4 From the relations given by equations 27 and 28 we have X1 x2 29 X3 x4 Substituting the relations given by equations 27 and 28 in equation 24 we have m1 gL 1A9 1ka 2 19 1 kL1L29 2 0 L gt 11A x2 ka x1 kL1L2x3 0 mlng kLL 2 xl xy ka x2 30 11A 1A Similarly substituting the relations given by equations 27 and 28 in equation 26 we get L 1239 2kL22 9 2 kLlL26 1 0 L gt 28 x4kL quot122g 2 x3 kL1L2x1 0 L kLL M M 2 x4 x1 I 2 x3 31 28 28 Rewriting equations 29 30 and 31 in matrix format we have 0 L 1 0 0 x Z g 0 1amp3 L 1 x 32 x4 kLle 125 x4 I 123 MEEN 364 Parasuram Lecture 67 August 7 2001 If the output if the system is the angular displacement of rod 1 then the output expression can be expressed in matrix format as y91 x1 x gty10 0 0 2 33 x3 x4 Equations 32 and 33 represents the statespace form of the system de ned MEEN 364 Parasuram Lecture 67 August 7 2001 Assignment 1 For the system shown below derive the governing differential equation 121112 The above system consists of two point masses m1 and ml each suspended by strings of length L1 and L2 2 Derive the governing differential equations of motion for the system shown below Mass m Wag NIP Recommended Reading Feedback Control of Dynamic Systems 43911 Edition by Gene F Franklin etal 7 pp 24 45 Recommended Assignment Feedback Control of Dynamic Systems 43911 Edition by Gene F Franklin etal 7 problems 23 Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 10B The Classical Threeterm PID Controller The objective of this lecture is to present the elements of the most widely used control algorithm7 that is the proportional P7 integral l and derivative D controller ProportionalIntegral Derivative Feedback Controllers Proportional P Feedback Control When the control signal is made proportional to the error in the measured output7 we call the control law proportional feedback or P control Speci cally7 if ut Kezt7 1 where K is the P control gain The controller transfer function Ds is Ds K 2 As we saw in the case of the DC motor control7 unless the control gain is large7 there will be steady state error in the measured output However7 as we shall see later7 as the control gain is made larger7 the feedback loop stability becomes a serious issue So7 a P controller has some fundamental limitations regarding its largest allowed gain in order to obtain well damped response Of course7 such a gain limit might result in unacceptable steady state error ProportionalIntegral PI Feedback Control The main reason for adding integral action is to eliminate any steady state errors that 1 might appear in the system response Integral feedback has the form of K t W T eltngtdm lt3 I to where TI is called the integral reset time The l control transfer function Ds is K D s 4 lt gt T15 lt gt To see the effect of l control on the steady state error7 take the derivative of equation 3 This results in dut K t 5 dt T1elt gt lt gt which means that the control effort7 ut7 of an l controller will stop varying7 and take a non zero value7 only if the error 6t becomes zero So7 using l control we can drive the steady state error of 6t to zero The combination of P and l control7 so called Pl control can be expressed as 1 t W Kltelttgt T comm lt6 I to where the proportional gain K and the integral reset time T1 are tunable parameters In general7 l control improves the steady state response of a control systern7 but it also slows down its response That is7 unless we are willing to increase the overshoot in an attempt to improve the systems response speed Derivative D Feedback Control Derivative or D feedback has the form det t KT 7 ult gt D dt lt gt with the corresponding transfer function being Ds KTDS 8 where TD is called the derivative time This form of control is used to increase damping and improve system stability It is usually used in conjunction with P andor l control because D control by itself may not be effective if the control error7 6t7 is constant D control is anticipatory in nature7 leading a P only control by TD seconds D control has additional limitations such as implementation problems and problems associated with the existence of sensor noise ProportionalIntegral Derivative PID Control For more effective control over the steady state and transient errors we can combine all three elements discussed before to obtain proportional integralderivative PID control Some form of PID control including P or Pl or PD only is used in probably over 90 or 95 of the industrial control loops The resulting PID control law is 1 t d6 t W Kltelttgt mm TD l lt9 TI to dt with the corresponding transfer function given by 1 Ds K1 T135 10 T18 The design of PID controllers deals with nding the parameters K TI and TD such that design speci cations are met This controller parameter adjustment process is called con troller tuning Example Impact of PID Gains We can see the effects of PID controller elements by considering the DC motor control problem we had studied before First look at the disturbance rejection impact see Figure 1a P control only shows a large impact of the output as a result of a step disturbance input As l control is added the Pl controller results in zero steady state error for a step input in disturbance Further addition of the D element resulting in PID control results in a better behaved response A similar behavior is observed for a step input change in the reference see Figure 1b P control only results in some steady state error Addition of the l control removes the steady state error whereas addition of D control results in a better damped better behaved response Reading Assignment W J1LL 5 i 04 02 l 0 L02 004 006 008 010 TlnlE sec 0 uidsec 16 L4 7 10 08 0 6 04 02 004 Time sec 006 b Figure 1 PID Control of DC Motor a step disturbance input b step reference input Read pages 215 226 of the textbook Read the examples in Handout E21 posted on the course web page MEEN 364 Parasur Lecture 19 20 August 25 2001 HANDOUT E19 EXAIWIPLES ON FEEDBACK CONTROL SYSTEMS Examplel Consider the system shown below ss1 The open loop transfer function is given by K G J rr 1 The closed loop transfer function is K Yr7 G0 7 srl 7 K 7 K RU 1Gr 1 K rrlK J2JKI rr1 ExampleZ Consider the system shown a Determine the transfer function from r to y b Determine the transfer function from w to y To obtain the transfer function between r to y assume w to be equal to zero Therefore the block diagram reduces to Lecture 19 20 August 25 2001 Hence the open loop transfer inction is G M i M x xr1 20 rr2 s 20 The closed loop transfer function is given by 10K sz m Gr m2 r 20 10K sz Rs 1Gs 110Ki Kzs 5 3 x2010K210K39 rr2 s 20 To obtain the transfer inction between W and y assurne r to be equal to zero Hence the block diagram reduces to For the above block diagram the open loop transfer inction is Gm L 10 ss 1 20 and the feedback transfer function is given by H K K S Therefore the transfer inction from W to y is given by MEEN 364 Parasuram Lecture 19 20 August 25 2001 10 Y3 G3 32 320 103 WU 1GSHS 1 10 39K1Kzs 333232010K210K139 32 3 20 3 Example 3 A unity feedback system has the open loop transfer function 31 GS 3 3 3 2 Find the error constants KP KV and Ka for the system The closed loop transfer function is given as 31 YS GU 3332 31 R3 1G3 1 31 33323l39 3332 The transfer function for the error signal is given by 31 3332 E3R3 Y3R3 S3 R3 R3 3231 333231 Therefore the steady state error for the system can be obtained by applying the final value theorem as egg lim 3E3 Sigt0 Therefore 33 32 es 11m3 R3 S Srgt 33 32 3l For a unit step input we have R3 therefore the steady state error is 33 32 l 33 32 l eSSlim3gt 0 1Kp f zlimf S gt 3 3 3l 3 S gt 3 3 3l Therefore the value of Kp is given as Lecture 19 20 August 25 2001 K P gt m For a ramp input We have Rr J2 therefore the steady state error is 5 1 x2x 1 es limxf zlimf 0 HE x r x1x Hus x s1 K Hence the value of Kv is given as K gt m For a parabolic input We have Rr J3 therefore the steady state error is r 1 r1 e limxf 3limf H x r x1x Hus r s1 Therefore Kn 1 Example 4 The block diagram shown below shows a control system in Which the output member of the system is subject to a disturbance In the absence ofa disturbance the output is equal to the reference Investigate the response ofthis system to a unit step disturbance Since We are interested in the response of the system to a unit step disturbance assume the reference input to be equal to zero Therefore the block diagram reduces to MZEEN 364 Parasuram Lecture 19 20 August 25 2001 W5 W5 The open loop transfer functroh IS glyeh by 65 and the feedback transfer functloh ls glyen as s Hs1lt Ys Gs 7 E 1 ms lssHs H Js1lt s For a umt step dlsturbahce the output reduces to 125 1 Js1lt s sltJsKgt The steady l h p be obtalhed pp as y llmSYS llmS hm 7L Hquot w sltJs K Hquot Js K K Therefore the system ls lncapable of rejecung the dlsturbahce completely as there ls a steady state offset Example 5 ll ldl whose opehl p re 7 1 65 7 so Ts where m the feedback loop transfer functloh IS glyeh as MEEN 364 Parasuram Lecture 19 20 August 25 2001 H s h with h gt 0 Determine the system type with respect to the reference input The closed loop transfer function is given as 1 Ys Gs s1Ts 1 Rs 1GSHs 1 h Ts2sh39 s1Ts The transfer function for the error signal is given by 1 T32 s h 1 E WSW 1 mim WW The steady state error is obtained as T32 sh 1 Rs Ts2sh egg 11msgtEs 111113 Sigt0 Sigt0 For a step reference input the error is Ts2sh 1 1 eSS11msgt 2 Srgt Ts sh s M T32 sh 1 h l 36 Ts2sh h I If the system is not unity feedback then h 72 1 therefore the system is of Type 0 If the system is unity feedback ie h 1 then the system is of Type 1 as the steady state error for the step reference input is zero MEEN 364 Parasuram Lecture 19 20 August 25 2001 Assignment 1 Consider a unity feedback system whose open loop transfer function is given by s l ss2 23 5 39 Determine the closed loop transfer function of the system Gs 2 Problem 432a from the textbook Feedback control of dynamic systems by Franklin etal Recommended Reading Feedback Control of Dynamic Systems 4111 Edition by Gene F Franklin etal 7 pp 201 7 215 230 7 242 Recommended Assignment Feedback Control of Dynamic Systems 43911 Edition by Gene F Franklin etal 7 problems 422 428 433 435 437 Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 10A Basic Properties of Feedback A Case Study of Speed Control The objective of this lecture is to present some of the most fundamental elements of control loops7 including open loop and closed loop or feedback control There are two basis structures for control of dynamic systems a open loop control and b feedback or closed loop control These two control structures are shown in Figures 1 and 2 Case Study of Motor Speed Control Let us write the equations of motion for a DC motor coupled to an inertial load The electrical dynamics can be expressed as dim t Ke m it La dt l Raina 1097 whereas the mechanical dynamics can be expressed as Jm6mt b9mt Km n 2 Let us de ne the output to the motor speed yt and lets name the motor load as the disturbance wt Tl Taking the Laplace transforms of equations 1 and 2 and eliminating the current Ias we obtain an equation of the form1 JmLa 2 JmRabLa bRaKth b12aKth 1mg Kt l 3 ma MKS 5 be mew 1NOTE Please7 do the algebra to convince yourselves of this result Conuml inp Process 0 Output or plant Figure 1 Openeloop Control Systeml Reference input Reference input Figure 2 Closediloop or Feedback Control Systeml Equation 3 can be simpli ed to the form 718 172s 1Ys AVas BWs7 4 where the time constants 71 and 72 as well as the constants A and B are expressed in terms of the DC motor variables Equation 4 can be written in a familiar transfer function form7 as follows Ys yas A B W 5 715 1725 1 715 1725 1 At steadyestate7 when both wt and vat are constant7 we have the steadyistate motor response as y Ava 3w 6 Figures 3 and 4 show the DC motor in openeloop and closediloop speed control con gu7 ration We now compare the various properties of the two control systems we considered7 iiel open and closeloop control systems Reference 1 speed r Figure 3 Openiloop DC Motor Speed Control Systemr Reference speed r Figure 4 Closediloop Feedback DC Motor Speed Control Systemr Disturbance Rejection One of the important characteristics of a speed control system is how well it maintains regulates speed at steady state7 in the presence of disturbances or torques For the open loop controller the control input the ampli er voltage is 1a KT7 where the controller gain is chosen such that the motor speed is equal to the desired speed 7 when the torque w is zero That is 1 K 8 A lt gt The DC motor speed with open loop control and zero load torque will be given by 1 yss Ava AZ r 9 For the case of the feedback controller the ampli er voltage will be vaKr7y 10 In view of this control law7 the closed loop transfer function in this case will be AK B WE f WW8 11gt Ys At steady state 5 a 0 with zero load torque we have AK 95 7 y 1AK7 where if the gain K is selected such that AK gtgt 17 then yss E r So7 for both control systems we obtain the desired speed if there is no load torque What happens when the load torque is not zero For the open loop system we have that the steady state speed is yss AKr Bu7 13 or with the controller K i we have yss r Bun 14 and if we de ne the speed variation caused by the load torque as 6y yss 7 r then 6y Bw 15 4 So the speed error is proportional to the load torque where the constant of proportionality is B and x for a given problem For the closed loop system we have the steady state speed given by AK B 16 r w 1 1 AK 1 AK If the controller is designed such that AK gtgt 1 and AK gt B then there will be no signi cant error in the motor speed despite the presence of any amount of load torque Advantage of Feedback Reduce Impact of Disturbances Output errors can be made less sensitive to disturbances with feedback than Without feedback by an amount of 1 AK Sensitivity to Gain Changes Another comparison that can be made between the two controllers is that of changing the controller gain value Assume that temperature effects have resulted in the motor gain to shift from A to 1 A 6A In the open loop case the controller gain K will still be and the new overall Z7 system gain would be 1 6A Toz6TozKA6A ZA6A 1j7 17 where To is the open loop torque So the gain error is In terms of percent changes de ned as we have 6Tol 7 6A 18 To 7 A 7 which means that a 10 error in A would translate in a 10 in T01 The ratio of STT to SAA is called the sensitivity of the gain from r to y with respect to A For the open loop case this ratio is 1 Applying the same change in A to the feedback case yields A 6AK Tc 6Tc 1 1 1A6AK 19 One can compute the sensitivity using equation 19 and differential calculus However an easier approach is to rst de ne the sensitivity as A de To dA 7 55d 2 20 5 and then apply it to the close loop transfer function AK To 21 l 1AK l The result is 1 STCl 22 A 1AK7 l which reveals another major advantage of feedback control Advantage of Feedback Reduce Impact of Uncertainties When using feedback control output errors can be made less sensitive to variations in the plant gain A by a factor of 1 AK as compared to openloop control Dynamic Tracking So far we have looked at steady state system responses under constant disturbances and references However most control systems must track time varying inputs The differences between open loop and closed loop control with respect to dynamic tracking can best be seen through an example 1 1 7 7 E m A i 10 and B i 50 and With the open loop controller gain set at 01 Determine the output for a step change in the load Consider a servomechanism with T1 7392 torque of 701 N m For the open loop case assuming that r 0 the output is expressed as 50 Ys 23 6 105 1 s 1 The disturbance is given by 01 Ws 7 24 So 75 55 25 6 105 16 l 08 1s39 The system response is shown in Figure 5 Now lets assume that we wish to use feedback control in order to improve the systems ability to reject steady state disturbances by a factor of 100 with respect to the open loop case The output is given by 50 lt26 Y8 y ladsec 6 0 002 004 006 0 08 010 012 014 016 018 020 Timesec Figure 5 Openiloop transient response to a disturbance inputi Since we want the error reduction to be 100 we must have 1 AK 100i So7 K 99 Considering that Ws 7051 from the nal value theorem we have that 7005 27 7 5 7 955 1AK The transfer function from the reference to the output is given by Y 10K NS i 28 Rs 717252 71 72s 1 AK The system response to the given disturbance is shown in Figure 6a7 whereas the system response to a step refernce input shown in Figure 6b Notice the differences in the responses of the two control systems Reading Assignment Read pages 200215 of the textbook Read the examples in Handout E19 posted on the course web page 31 K 0 0 0010 0030 0040 0050 0 0010 0020 0030 0040 0050 Time sec Time see a b Figure 6 Closed 100p transient responses to a disturbance input and b reference input MEEN 651 Lecture 15 Texas AampM University Department ofMechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 15 Compensation Lead Compensation In order to alleviate the h39 h 9 q quot 39 of PD 1 a rst order pole is added in the denominator at frequencies higher than the breakpoint of the PD compensator Thus the phase increase still occurs but the ampli cation at higher frequencies is limited The resulting lead compensator has the transfer function of DsK 1 a Ts 1 Where 0 lt1 Figure 1 shows the frequency response of this lead compensator A lead compensator is generally used Whenever a substantial improvement in system damping is required 04 n Figure 1 Lead compensation frequency response With 11 10 The phase contributed by the lead compensator in equation 1 is given by tanquotTw 7tanquota Ta The frequency Where the phase is maximum is given by MEEN 651 Lecture 15 The maximum phase contribution corresponds to sin max 1 a 10 l sin max l sin lm Usually a single lead compensator can contribute a maximum of 60 to the phase If a greater phase lead is required then a double lead compensation is used where 2 DSK a Ts l In leadnetwork designs there are three primary design parameters 1 The crossover frequency which determines the bandwidth rise time and settling time 2 The phase margin PM which determines the damping ratio and the percent overshoot 3 The low frequency gain which determines the steadystate error characteristics Design procedure for Lead Compensation 1 Determine the loop gain to satisfy the specified error constant 2 Using the value of K determine the PM of the uncompensated system Determine the phase lead required using the relation E 1 5 1 8 where Is specified phase margin 11 phase margin of the uncompensated system and s margin of safety approximately 5 to 12 4 Let Im 11 and determine the value of at using the relation 1 sin l sin lm 5 For the calculated value of 0c obtain the frequency at which the magnitude of the system is lOlogl0c dB This frequency is the new crossover frequency and is equal to mm 0 Compute the two comer frequencies using the relation MEEN 651 Lecture 15 l a Z a T J Draw the compensated frequency response check the PM and iterate the design if necessary gt1 PI compensation In many problems it is important to keep the bandwidth low and also to reduce the steadystate error For this purpose a proportionalintegral Pl compensator is useful A PI controller has the transfer function of the form Ds 5 s L s n The frequency response characteristic of the Pl compensator is shown Figure 2 The desirable aspect of this compensator is the in nite gain at zero frequency which reduces the steady state error This is accomplished at the cost of a phase decrease below the break point 124 u m 4 m u I w m ml Figure 2 Frequency response of a PI controller Lag Compensation Lag compensation approximates Pl control The transfer function of a lag compensator is given by DSa Tsl 2 aTs1 MEEN 651 Lecture 15 where or gt1 Although equation 2 is similar to equation 1 the fact that or gt1 causes the pole to have a lower breakpoint frequency than the zero This relationship produces the lowfrequency increase in amplitude and phase decrease apparent in the frequency response plot shown in Figure 3 The primary objective of the lag compensator is to provide an additional gain of 2010gx in the lowfrequency range and to leave the system with suf cient phase margin Thus the lag compensator increases the open loop DC gain thereby improving the steady state response characteristics without changing the transient response characteristics signi cantly Design procedure for Lag Compensation 1 Determine the openloop gain for the speci ed error constant 2 Find the frequency meg where the uncompensated system makes a phase margin contribution oz y 5 where d speci ed phase margin and a margin of safety approximately 5 to 12 3 Measure the gain of the uncompensated system at meg and equate it to the required high frequency attenuation 2010g0 4 Choose the upper comer frequency of the network one octave or one decade below the crossover frequency Deg ie l a z 22 to oz T 2 10 5 With 1 and T determined draw the frequency response to check the resulting phase margin nquot 411 In u 760 7 no In Figure 3 Frequency response of lag compensation with x 10 PID Compensation For problems that need phase margin improvement at the crossover frequency and low frequency gain improvement it is effective to use both lead and lag compensation By MEEN 651 Lecture 15 combining the derivative and integral feedback we obtain PID control With transfer function given by Ds TDS ls and With frequency response characteristics shown in Figure 4 This compensation is roughly equivalent to combining lead and lag compensators in the same design and so is sometimes referred to as a leadlag compensator Hence it can provide simultaneous improvement in transient and steady state responses Ulml Um u r m 7 m39 m v2 I 2 m mu m u 3 In Mr 1 ml J Figure 4 Frequency response of PID compensation With T1TD 20 Reading Assignment Read pages 418440 from the textbook Read the examples in Handout E24 posted on the course web page MEEN 364 Parasuram Lecture 897 August 8 2001 HANDOUT E8 EXAIVTPLES ON MODELLING OF ELECTRICAL ELECTRONTECHANICAL SYSTEMS Note that the time dependence of variables is ignored for all manipulations Example 1 An electrical circuit Consider the circuit shown below R L 1 T Writing the Loop equation for the above circuit we have Va Ri L i idt0 1 dt C Weknowthat i dt Therefore substituting the above relation in equation 1 we have R 2 dt Equation 2 represents the governing differential equation of the circuit shown above State space representation Let the states of the system be defined as q x19 3 q x2 From the above relations we get x1x2 4 MEEN 364 Parasuram Lecture 897 August 8 2001 Substituting the relations given by equation 3 in equation 2 we get dq q 2 Va Ld qR dtz dt C 1 V szRx2 x1 C 1 R 1 gtx x x V 5 2 LCILZL 0 Rewriting equations 4 and 5 in matrix format we have M l ll gl l l 1L5 1 6 X2 LC L 9 L If the output of the system is the voltage drop across the capacitor then the output equation can be written as y x19 1 i C C 1 x1 gt y 5 0L 7 Equations 6 and 7 represent the statespace form of the circuit Example 2 An electrical circuit Note that dql dqz 1 1 8 1 dt 2 dt MEEN 364 Parasuram Lecture 897 August 8 2001 Writing the loop closure equation for the loop ABFGHA we have 1 Vm EJ11 lz dt R111 039 9 Substituting equation 8 in equation 9 we have Knq192Rl 0 10 C dt Writing the loop closure equation for the loop BCDEFB we get 51139 1 1 R212LEEIzzdtEIOZ zldt 0 11 d2 d 2 gtL 2 R2 12 i 0 I d2 dt C C Equations 10 and 11 represent the governing differential equations for the circuit shown State space representation Let the states of the system be de ned as ql x1 92 x2 12 qz x3 From the above relations the following equation can be deduced x2 x3 13 Substituting the relations given by equation 12 in equation 10 we have Vm 311612 R10 C dt gtVm x1 x2 R1x1 0 gtx1 xlx2 Vm 14 RIC RIC R1 MEEN 364 Parasuram Lecture 897 August 8 2001 Similarly substituting the relations given by equation 12 in equation 1 l we get 612 dtz dqz 2ampi dt C C L R 0 2 2 l gtLX3R2x3 Ex2 Ex1 0 2 sz gtx x 3 LC2L3 1 15 LCX1 Rewriting equations l3 l4 and 15 in matrix format we have V l6 i LC LC From the circuit it can be seen that the output is given by l Vm EjzzdtRlzp l dq Vouz Eq2 R161 1 d 1 from equation 10 1n the above relatlon we get I Substituting the value of R1 1 q1 q2 V V out qu m C 2 l 0142 Ex2 Ex1 Vm9 0 x2 1114 17 Therefore equations 16 and 17 represent the statespace form of the above circuit MEEN 364 Parasuram Lecture 897 August 8 2001 Example 3 An electromechanical system Consider the following electromechanical system shown X 4Lx The electromechanical system shown above represents a simpli ed model of a capacitor microphone The system consists a parallel plate capacitor connected into an electric circuit Capacitor plate a is rigidly fastened to the microphone frame Sound waves pass through the mouthpiece and exert a force fst on plate b which has mass M and is connected to the frame by a set of springs and dampers The capacitance C is a function of the distance X between the plates The electric field in turn produces a force fe on the movable plate that opposes its motion Kinematics stage Let the movable plate b move a distance X units Then the velocity and the acceleration of the plate is given by x x respectively It can be seen from the figure that the current i ows through the electric circuit in the counter clockwise direction Kinetics stage First let us consider the electric circuit MEEN 364 Parasuram Lecture 897 August 8 2001 Writing the loop closure equation for the above circuit we get all39 1 L R39 39dtV 18 d l Cll Since i dt equation 18 reduces to dig altz dqq L R V 19 d C Free body diagram of the movable plate of massM Jam lax p gt cx fe gt Writing the Newton s second law of motion we have 2F ma gt kxcxf2 fSt mx gtmxcxkxf2 fst 20 Equations 19 and 20 represent the governing differential equations of motion for the electromechanical system considered The force fe is de ned as Z q f2 28 A where A is the surface area of the plates and 8 is the dielectric constant of the material between the plates MEEN 364 Parasuram Lecture 897 August 8 2001 Since fe is nonlinear in nature equation 20 is a nonlinear equation The linearization of a nonlinear equation is explained in the handout on linearization Once the equation is linearized then it can be represented in the statespace form This section is dealt with in detail in a later handout Assignment 1 Find the differential equations for the circuit shown below and put them in state variable form R1 R2 R3 2 Consider the schematics of an electromechanical shaker as shown below This system consists of a table of mass M and a coil whose mass is m A permanent magnet rigidly attached to the ground provides a steady magnetic eld ie the motion of the coil through the magnetic eld induces a voltage in the coil that is proportional to its velocity The passage of current through the coil causes it to experience a magnetic force proportional to the current Derive the equations governing the dynamics of this system 3 Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fall 2003 Lecture 2A Laplace Transforms The objective of this lecture is to introduce the concepts and mathematics involved in the Laplace transform LT and its use in dynamic systems and controls LTs will be instrumental in the development of transfer functions and they will be used throughout this course in the analysis of dynamic systems and in the design of control systems De nition of the Laplace Transform and the Inverse Laplace Transform The LT allows us to transform functions of time into functions of a complex variable 5 and it is a generalization of the Fourier Transform we discussed last week It is de ned as M mo 0 rowan lt1 where s is a complex variable 5 039 jw and ft is a continuous function of timel For the integral in equation 1 to be computable there must exist some real numbers A and b such that lftl lt A61 Most functions encountered in engineering are of this form The most common use of LTs is to solve ODEs By using LTs we can transform ODEs form the t domain into algebraic equations in the 5 domain The algebraic equations are typically easier to solve that the equivalent ODEs The solutions of the algebraic equations can then be transformed back to the time domain using the inverse LT de ned by the integral 1 cjoo flttgt rims Feeds lt2 2717 07700 Typically the de nition of the inverse LT is not used in practice Rather the method of partial fraction expansion is used to obtain the inverse LT as it will be shown in the examples Table 1 below summarizes the LTs of the most commonly used time functions 1Furthermore the function is assumed zero for t lt 0 and the resulting transform is called onesided LT Table 1 Laplace Transforms of Most Common Functions of Time Continuous Function Laplace Transform Impulse Step t t2 9 67 sle w 7 at W 5271 wt m 005 wt W Basic Properties of the Laplace Transform The LT has a number of useful properties that are handy in manipulating transfer func tions of dynamic systems The most important properties of the LT are listed below 1 Linearity a1f1t a2f2t 01F18 a2F2lts lt3 2 Integration 0 mew lt4 3 Differentiation ff ms lt5 4 Shifting in the time domain ft 0 6 SF8 6 5 Shifting in the complex domain ft6 t F8 a 7 Fw m J X0 0000 Uh Figure 1 Schematic Diagram of Mass Spring Dashpot System Additionally7 the initial and nal value theorems associated with the LT nd very useful applications The initial value theorem can be expressed as imagma w The nal value theorem allows calculation of the steady state value of a function as follows tlim ft limJ 5Fs 9 Partial Fraction Expansion Method See Handout A2 An Example Consider the mass spring dashpot system shown in Figure 1 Use the LT to nd the system transfer function from the forcing function Ft to the displacement The equations of motion for this system are ma Ft 7 bvt 7 kxt7 10 and d t mw7 m Taking the LT of equations 10 and 11 results in msVs 7 110 Fs 7 bVs 7 kXs7 12 and 5Xs 7 z0 Vs7 13 where z0 and 00 are the initial conditions for displacement and velocity of mass m just before the force is applied Now7 eliminating Vs from equations 12 and 13 results in Xs 77152 b5 k Fs x0 ms b Vii0 14 Assuming zero initial conditions7 the system transfer function is found to be E 1 15 Reading Assignment Read Handout A2 Laplace Transforms77 and the examples Handout E3 posted on the course web page Texas A 85 M University Department of Mechanical Engineering MEEN 651 Control System Design Dr Alexander G Parlos Fh 2003 Lecture 9 Frequency Response Analysis Bode Plots of First and Second Order Systems The objective of this lecture is to provide you with some background on the use of transfer functions to perform the so called frequency response analysis of dynamic systems7 also known as analysis based on Bode plots This method will be later used in the design of control systems Frequency Response Consider a system with transfer function Y s 65 lt1 where the input is ut Uosmhut 2 with a Laplace transform of we 82 lt3 Considering zero le7 the output of the system is Uow 52 LUZ 55 Ge 4 Assuming that Cs has distinct poles we can perform a partial fraction expansion and take the inverse Laplace transform The result of the partial fraction expansion can be written as r 041 042 Otn 040 040 i 7 seal 8 02 57a sjw sijw 5 020 MS I gt E 005 l O 0 l 005 39 39 39 39 41100123456789 Time sec Figure 1 Linear System Response to Sinusoidal lnput where a1 a2 an are the poles of Cs Taking the inverse Laplace transform of equation 5 results in yt ale a t 0426 anew Zlaolsmwt b 6 Where I lt gt 7 71 m 0 0 lt1 7 tan R6010 7 If the system is stable the exponential will all die out and the response of the system will be yt Zlaolsmwt 8 This equation indicates that if a linear system is excited by a sinusoidal input then its response is also sinudoidal with the same frequency but with possibly di erent amplitude and phase This is shown in Figure 1 From equation 4 we can see that the response of the system after the exponential terms have decayed can also be expressed as yt UoAsinth b 9 Decibels Resonant peak M Amplitude ratio 3 l l Bandwidth m w gt w radsec Figure 2 De nitions of Bandwidth and Resonant Peak where A lGUW GS S w 10 and I Glt39 gt1 7 m 7 tan1 ltGwl 11 Belem lt gt lt gt Equations 10 and 11 allow us to compute the magnitude A and phase 1 of the system 05 A very important speci cation in system performance is that of bandwidth This is the frequency above which the system output does not track the system input7 if the latter is a sinusoidl Assume a feedback look with unity feedbackl Then a typical closediloop transfer function Bode plot is shown in Figure 27 which also depicts the concept of the BW and resonant peak7 MW Bode Plot Techniques The idea of Bode plotting is to express the log of the system magnitude and its phase as a function of the input signal frequency Following some simple rules we can plot the frequency response of some fairly complex transfer functions The principle of the Bode plots resides in the fact that a transfer function 05 for s jw can be written in polar form as follows So7 T T 1Gjwl 7 lt13 or 10g10 lGUWl loglo Ti 10g10 72 10g10 Ts 10g10 T4 10g10 75 14 Also7 lt Gm 6162 703 704 765 15 Bode plots are usually expressed in terms of log 6 versus logw and b versus logw A particularly important unit to remember is that of decibels7 or st7 de ned as lGldB 20log1O ln general7 we can express a transfer function in terms of its poles and zeros7 in the following form Kg5 16 5 P15 102 The transfer function of equation 16 can be transformed to the so called Bode form7 as follows jm39l 1jw7 21 17 ija 1ju777 1 7 where K0 and the 77s are all related to the K7 the poles and zeros of the transfer function 16 For example7 suppose that KGjw K0 KGjw 18 Then7 loglKGjwl loglKOl logljw7391 11 7logljwzl 7logljw739a 117 19 or in st7 lKGjwldB 20log lKOl 20logljw7391 117 20log 1jwzl 7 20log lija 11 20 So7 it is suf cient to analyze the most commonly encountered transfer function terms These terms can be classi ed as follows 0 Class 1 Singularities at the origin7 K0ju 7 4 Slope is 1 or 20 dBdecade l 000 Wmquot B 01 l 10 00 1000 a radsec Figure 3 Magnitude of u 0 Class 2 First order terms7 jUJT Dil 0 Class 3 Second order terms7 1i1 Class 1 Since logK0ljw l logKonlogljwl7 21 the magnitude plot of the Class 1 term is a straight line with slope no A value of n 1 denotes 20 dBdec7 so the slope is in multiples of this value Examples of di erent Class 1 terms are depicted in Figure 3 The phase of QM is n gtlt 90 deg this is a horizontal line Class 2 The sketch of this term7s magnitude is obtained by looking at its asymptotes at low and 5 Break point 100 100 m T lt2 E g 18 3 01 00 01 10 10 100 1000 mradsec Figure 4 Magnitude of jw 10 1 for 7 10 high frequencies In particular oforo ltlt 1jw71 1 o foro gtgt 1 jw71 jUJT If we call the frequency value of La the break point then we observe that for frequencies below the break point the magnitude is approximately constant whereas for frequencies above the break point the magnitude behaves like a Class 1 term K jw An example is shown in Figure 4 The slope at high frequencies is 20 5135166 The magnitude plot of the term jo 1 1 can be easily obtained as the mirror image of the magnitude of term jo 1 with respect to the horizontal axis The phase curve can be obtained by using similar low and high frequency asymptotes In particular 0 foro ltlt 1 lt1 Odeg o for MT gtgt 1 ltjw7 90deg o for MT 1 lt 1107 1 45deg The overall phase varies from Odeg to 90 deg An example of a phase plot is shown in Figure 5 The phase plot of the term jUJT 1 1 can be easily obtained as the mirror image Asymplute 4ij0 n 001 002 H 2 04 l w radsec Figure 5 Phase of jw 10 1 for 7 10 of the phase of the term jUJT 1 with respect to the horizontal axis The overall phase varies from Odeg to 790 deg Class 3 This term behaves like the Class 2 term but there are some di erences in the details In particular 0 the break point is now at La Lu 0 the magnitude changes with slope of 2 or 40 dBdec when the term is in the numerator and 72 or 740 dBdec when the term is in the denominator o the phase changes from 0 deg to 180 deg when the term is in the numerator and 0deg or 7180 deg when the term is in the denominator o the values of the magnitude and phase near the break point depend on the value of C A rough sketch of the magnitude when the term appears in the denominator can be obtained by observing that in addition to the above rules l lGwl 2 4 at La Law 22 7 No such easy rule exists for the phase plot however the phase also depends on C In the limiting cases when C 0 the phase plot is a step function from 0 deg to i180 deg at w w whereas when C 1 it can be treated like two rst order terms Class 2 terms For other values of C the phase is in between these two limits An example of a magnitude and phase plot for Class 3 systems is shown in Figure 6 The magnitude and phase plots ofthe Class 3term in the numerator can be obtained from the equivalent plots in the denominator by considering the mirror image of the magnitude and phase plots with respect to the horizontal axis The Composite Curve When a dynamic system is composed of many poles and zeros plotting the Bode plot requires that we plot the individual component Bode plots and then combine them into a composite plot The composite curve is the sum of the individual curves and a sketch can be easily obtained by hand More accurate Bode plots are obtained using Matlab NOTE Please read carefully the Summary of Bode Plot Rules on page 379 ofthe text Examples of how to sketch the composite Bode plot of transfer functions will be provided during the problem solving session Nonminimum phase Systems A system with one or more zeros in the RHP behaves very differently that its counterpart with zeros in the LHP For example consider the transfer functions 51 G 10 23 15 510 lt gt 571 The magnitude curves for these two transfer functions is identical The phase curve however is different Both are shown in Figure 7 The phase for G2s starts at 180 degrees instead of 0 and this makes it nonminimum phase Reading Assignment Read pages 364 386 of the textbook Read the examples in Handout E17 posted on the course web page Magnitude Phase 02 0 l 008 006 004 730 90 A 20 150quot 180 01 0 6 0x 1 2 4 llwl 04 b Figure 6 Magnitude and phase of second order term Decibels Illilliiiii illlilllllili llllil lllil Illi h IIIIII IIIlllllllllllllllllllmll 120 quot39quotquotquotquotquotquotquotquotFiiiiiliii IlullllIiii39llillllll lllllllli 9D g lGIJWGZJw A m 9 V IllquotquotquotIiI w IquotquotillIliw39lllmllllllm uquot IIquotquot on u Illlllll IIIIIIII 10 00I 01 1 0 0 100 1030 ndSEC a b Figure 7 Bode plot for minimum and nonminimum phase systems a magnitude b phase MEEN 364 Parasuram Lecture 57 August 7 2001 HANDOUT E5 EXAIVIPLES ON MODELLING OF TRANSLATIONAL NIECHANICAL SYSTEMS A generalized procedure has been followed in all the examples in this handout to derive the governing differential equations of motion Note that the time dependence of all variables is ignored for all manipulations Example 1 One DOF system Consider the system shown below Kinematics stage In this stage the position velocity and the acceleration of all the rigid bodies in the system are defined In the above system there is only one rigid body Let the displacement moved by the body from the static equilibrium position be equal to y units Then the velocity and the acceleration of the body is defined as y and yunits respectively This completes the kinematics stage Kinetics stage In this stage the Newton s second law of motion which states that the sum of all the forces acting on the body is equal to the product of its mass and its acceleration is applied to obtain the final governing differential equation of motion Free body diagram of the body of mass m Jry m ky Cy MEEN 364 Parasuram Lecture 57 August 7 2001 Note that the gravity force is not considered in the free body diagram as the displacement of the body is considered from the static equilibrium position Hence the spring force due to the initial compression of the spring balances the gravity force Writing the Newton s second law of motion we have ZFy ma gt ky c y m y mycyky0 l The block does not move in the Xdirection Equation 1 represents the governing differential equation of motion of the above defined system Equation 1 can be rewritten as c k y y y0 2 m m The generalized second order differential equation is given by y 2 a y a 0 3 where C is the damping ratio and Q1 is the natural frequency of the system Comparing equations 2 and 3 we have 2 a 2 n c m k a m State space representation Let the states of the system be defined as y x1 9 4 yx2 MEEN 364 Parasuram Lecture 57 August 7 2001 From the above relations it can be concluded that x1 x2 5 Substituting the relations given by equation 4 in equation 2 we have c k y y y0 m m M x1 x2 6 m Rewriting equations 5 and 6 in matrix format we get 0 1 X1 x1 7 952 m m x2 If the output of the system is the displacement of the block then the output equation can be written in the matrix form as follows Yyxp Y 1 org 8 x2 Equations 7 and 8 represent the statespace representation of the system de ned MEEN 364 Parasuram Lecture 57 August 7 2001 Example 2 Two DOF system Consider the system given below X20 y X10 X Note The time dependence is ignored for all future manipulations Kinematics stage In this stage the position velocity and the acceleration of all the rigid bodies are de ned From the above gure it can be seen that there are two rigid bodies The total number of degrees of freedom of the system is two The degrees of freedom of the system are defined as the horizontal displacement of the two bodies of mass ml and mz Let the displacement ofthe bodies of mass ml and mz be equal to X1 and X2 respectively The velocity and acceleration of the body with mass ml is given by 361 and 361 respectively Similarly the velocity and the acceleration of the body with mass mz is M and 362 respectively This completes the kinematics stage Kinetics stage In this stage the Newton s second law of motion is used to obtain the final governing differential equation of motion To write the force or the torque balance equations we need to draw the free body diagram of each rigid body Assume X2 to be greater than X1 Free body diagram of body of mass mI m1g klxl kzX2X1 1111 N1 Writing the Newton s second law of motion which states that the sum of the forces acting on the body must be equal to the product of its mass and acceleration MEEN 364 Parasuram Lecture 57 August 7 2001 ZFX ma gtk2x2 x1 k1x1 m1x1 m1 x1k1 k2x1 k2x2 0 9 Similarly ZFy ma gtN1 m1g 0 where N1 is the reaction force of the ground on the block Note that the block does not move in the ydirection and hence does not have any acceleration in that direction Free body diagram of body of mass mz ng k2X2X1 k3X2 1112 N2 Writing the Newton s second law of motion for this body we have ZFX ma gt k2x2 x1 k3x2 m2 962 m2 Cz klek2 k3x2 0 10 ZFy ma gt N2 ng 0 where N2 is the reaction force of the ground on the block of mass mz Even in this case the block does not move in the ydirection and has no acceleration in that direction Equations 9 and 10 represent the governing equation of motion for the system de ned Rewriting the equations in the form of a matrix we have MEEN 364 Parasuram Lecture 57 August 7 2001 m1 0 x1k1 k2 k2 x10 11 0 m2 x2 k2 k2 k3 x2 0 The above equation is of the form MX KX 0 where M is the mass matrix K is the stiffness matrix and X is the vector containing the displacements of the blocks To calculate the Eigen values and Eigen vectors Let the constants in the above system be de ned as follows m1 3 Kg m2 l5 Kg k1 2000 Nm k2 1000 Nm k3 3000 Nm Therefore the mass and the stiffness matrices are 3 0 M 0 15 3000 4000 1000 4000 To obtain the Eigen values and Eigen vectors of the above system use the following MATLAB code W this code calculates the eigen values and eigen vectors associated with the defined system w m 3 OO 15 k 3000 10001000 4000 vd eigkm wnat sqrtd The result of the above code is MEEN 364 Parasuram Lecture 57 August 7 2001 v 09372 O183O 03489 09831 d lOeOO3 08759 0 0 27908 wnat 295957 0 0 528276 The diagonal elements in the matrix d represents the Eigen values of the system and the corresponding column vector in the matrix v represents the Eigen vector associated with that particular Eigen value Also note that the natural frequency of the system is de ned as the square root of the Eigen values The vector wnat gives the natural frequency of the system The physical significance of the Eigen values and Eigen vectors is explained in detail in the handout on Eigen values and Eigen vectors State space representation Let the states of the system be defined as E Yv M Xz 12 x2 X3 xzin From the above relations the following equations can be derived X3 4Yp 13 KB X3 Substituting the relations given by equation 12 in equation 9 we get MEEN 364 Parasuram Lecture 57 August 7 2001 m1 x1k1 k2x1 k2x2 0 gtm1X2k1 k2X1 k2X3 0 X2 mX1k 2X3 14 m m 1 1 Similarly substituting the relations given by equation 12 in equation 10 we get m2 xz kle k2 k3x2 0 gtm2X4 k2X1 k2 k3X3 0 k k k X4 m 2X1 XY 15 2 2 Rewriting equations l3 l4 and 15 in matrix format we get 0 0 1 0 X1 k1k2 0 k2 0 X1 X2 m1 m1 X2 16 X 0 0 0 X3 3 k2 k2k3 X 0 0 X4 4 m 2 2 If the output of the system is the displacement of the block of mass mz then the output equation can be written in the matrix format as 17 Equations 16 and 17 represent the statespace form of the abovede ned system MEEN 364 Parasuram Lecture 57 August 7 2001 Example 3 Base excitation problem Consider the system shown below in which the base of the system moves 37 X Notice that the base of the system moves by y units independent of the mass But in this case we however know how the base moves as given by the harmonic function So in effect this is just a one degree of freedom problem That is in other words there are two degrees of freedom for this system but out of which one degree of freedom is known Kinematics stage The position velocity and the acceleration of the mass are x x x respectively This completes the kinematics stage Kinetics stage Free body diagram of the mass kxy Cx y Note that the displacements X and y in the system are from the static equilibrium position Hence the spring force due to the initial compression of the spring will balance the gravity force Writing the Newton s force balance equation we have MEEN 364 Parasuram Lecture 57 August 7 2001 2Fma gtkxycxymx gtmxcxkxcyky 18 Since yFe gtyFjw 6 Substituting the above relations in equation 18 we get mxcxkxcyky gt mxcxkx CFja ejmkFem gtmxcxkxFkjca em 19 Equation 19 represents the governing differential equation of motion for the system in which the base is excited with a known amplitude and phase This equation is similar to that of a forced one degree of freedom system State space representation Let the states of the system be x x1 20 xm From the above relations it cab be concluded that x1x2 21 Substituting the relations given by equation 20 in equation 19 we have you mxcxkxFkjca e gtmxzcx2 qu FI where m F1 Fkjca e MEEN 364 Parasuram Lecture 57 August 7 2001 gtx2 5 5x1 ix2 22 m m m Combining equations 21 and 22 in a matrix format we get tdzi quotil 39 0 If the output of the system is the velocity of the mass then the output equation can be written in the matrix format as Yxx 2Yb if 09 Equations 23 and 24 represent the statespace form of the system de ned Example 4 The qualter car model Consider the system shown below Road surface r2 Inertial reference MEEN 364 Parasuram Lecture 57 August 7 2001 The system shown below is an approximation of a suspension model for one wheel of an automobile The displacements of the masses X and y are from their equilibrium positions Kinematics stage The velocity and acceleration of the two masses are given as x x y y respectively Kinetics stage Free body diagram of the body of mass m1 k1yx c yx k2 x r Note that the gravity force is neglected as the spring forces due to the initial compression of the springs balance the gravity force Writing the Newton s second law of motion we get 2F ma gt k1y xcy x kzx V quot11 96 gtm1xk1k2x k1y cy xk2r 25 Free body diagram of body of mass m 2 k1 y x Cy x MEEN 364 Parasuram Lecture 57 August 7 2001 Writing the Newton s force balance equation we have M y k1 y x Cy x gtm2yk1y xcy x0 26 Equations 25 and 26 represent the equations of motion for the quarter car model State space representation Let the states of the system be de ned as x x1 x 2 27 y x3 y x4 The following two equations can be derived based on the above relations x x 1 2 28 x3 x4 Substituting the relations given by equation 27 in equation 25 we get m1 xk1 k2xk1y cyx kzr gt m1 xZ k1 k2 x1 k1x3 Cx4 x2 kzr k k k k gtxZ 2r Mx1 ix2 1x3ix4 29 m1 m1 m1 m1 m1 Similarly substituting the relations given by equation 27 in equation 26 we get quot12 y k1 y x Cy x 0 gt m2 czzk1x3 x1cx4 x2 0 k k 9 1x1ix2 1x3 ix4 30 m2 m2 m m2 Rewriting equations 28 29 and 30 in matrix format we have MEEN 364 Parasuram Lecture 57 August 7 2001 0 1 0 0 0 x1 k1 k2 L E i x1 k2 x x2 311 311 r31 quot1 x2 m r 31 63 L L k1 L x3 0 x4 m2 m2 m2 m2 4 0 If the output of the system is the displacement of mass m2 then the output equation can be expressed in the matrix format as Yzyzxs 5 N Y0 0 1 0 32 88 u 5 Equations 31 and 32 represent the statespace form of the system de ned MEEN 364 Parasuram Lecture 57 August 7 2001 Assignment 1 Derive the differential equation of motion for the system shown below F 2k 4 M m2 k 2 For the system shown below derive the governing differential equation of motion Y3 Recommended Reading Feedback Control of Dynamic Systems 43911 Edition by Gene F Franklin etal 7 pp 24 45 Recommended Assignment Feedback Control of Dynamic Systems 43911 Edition by Gene F Franklin etal 7 problems 21 28

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