DYNAMIC SYST AND CONTROL
DYNAMIC SYST AND CONTROL MEEN 364
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364 Recitation Notes 952007 EXAMPLE 1 Derive the Fourier Series of the following function Ft V H First Identify the function Ft l for XT S t S TXT X 2l0l2 Then plug this equation into the Fourier transform equation 1 k 1 T t k c j Fte wa dt j 1 e wquot k T T T 0 T Integrating over one period Solving for ca 1T tjkwao1Tt 00 IOl e dt J0l dt 1 t2 1 c T quot T 2T 2 Continue solving for the series coefficients T T T l l l l l ck j1 e W j e JWuk dr j e 1de T 0 T T 0 T 0 T Solving the first integral 1 T 1 1 r 1 1 W Ie JWade I e Mk I 6 J ng1 T 0 T w0k 0 T w0k To solve the second integral us integrate by parts lJT39ie 39Wak dt T 0 T 739 k ut dve 1W ejwakr du 1 v e jw0k 1 eijwakt T T effwok Iudvuv Jvdu 2 t Idt T Jw0k 0 0 onk 1 Te jwakT 1 e fWak 1 jwosz jw0k Adding this result to the result of the rst integral 1 1 7M H 1 7 39 kT 1 7 39 kt c a1 waa 61 1 k T ijk I jwokT2 i jw0k ck l 1 2 emkf 1 T jw0k jwokT jw0k l l c k Tjka T 2wozk2 eW 1 27139 s1nce w0 ck becomes 1 l ck 2 j27l39k 27zk This result is then plugged into the Fourier series 7 2 2 1 177E161 l w 177E161 Z cke Z cke 6700 2 kl eijwakt EXAMPLE 2 Find the Fourier coefficients for the following discrete function Xn First identify the function Xn0 l 2 2 01 From this it can be seen that the signal has periodicity N4 Using the discrete coefficient expression 27 27 27 27 1 N 1 ij kn 1 ij n ij 2n ij 3n akz Zxne N e 4 2e 4 2e 4 N M 4 This result is then plugged into the Fourier series 27r N l kn 3 j27rkn xn Zake N Zake 4 k0 k0 MEEN 364 Recitation 3 Fall 07 September 19 2007 Recitation 3 1quot Derive the Equations of motion L4 3L4 3k 6 2k b 2m 1X F Asinwt 1 Count masses 7 2 2 Each mass can move independently of the other so there are 2 degrees of freedom 3 The coordinate systems are given as X being positive downward and theta is counter clockwise positive 4 Motion of the bar in the X direction where the mass is attached is L4sin6 Up and down motion of the bar at its end is 3L4sin6 The spring force of the 2k spring will be dependent on the change in vertical distance between the mass and the point where it is attached to the bar This change in distance is XL4sin9 The force from the dampers will be dependent on the rate of change of this distance d L L gtE xZs1n6 xzcos06 The spr1ng force of the 3k spr1ng w1ll depend on the change in vertical distance between the end of the bar and the spring s anchor point This change in distance is Lsin6 5 Draw FBDs MEEN 364 Recitation 3 Fall 07 September 19 2007 Bar FBD L4cos6 3L4cos6 Sum moments ZM 19 cos6Fsz 13 L cos61 1 E1 3kL sin6 FS2 2kLx sin6 E bx cos69 cos 92kLx sinw bx cos69 L cos63kL sin6 Mass FBD F52 1b F Asinwti Sum forces 2F 2m5c39 2kLxsin0 bxcos69 Asinwt MEEN 364 Recitation 3 Fall 07 September 19 2007 2 Derive equations of motion and put in state space form X1 X3 Hm X2 2m F0 3m 3k k 2k 3k m Output taken at X1 1 Count masses 7 3 2 Each mass can move independently of the others so there are 3 degrees of freedom 3 The coordinate systems are given as X1 X2 X3 being positive to the right 4 Draw FBDs FBD s X13k X3X14k X2X1k X3X14k X33k X3X22k X3X22k X2X1k x13kx2 x1kx3 x14k mzjc392 x2 x1k x3 x2 2kF quot13563 3kx3 x3 x14k x3 x2 2k MEEN 364 Recitation 3 Fall 07 September 19 2007 x12q15x1 12 x2 2 13 x2 6145353 2 15 x3 2 16 11 12 W q3 14 15 16 0 1 0 0 0 0 8k 0 i 0 0 0 2m 2m 2m 0 0 0 0 1 0 0 X k 7k 2k X 0 FI 0 0 0 1 m m m 0 0 0 0 0 0 1 0 0 2k 0 ik 0 3m m m y1 0 0 0 0 0X0FI pr0b2m function ydprob2ty ml k11000 k21lOO blO Ll w25 Al MEEN 364 Recitation 3 Fall 07 September 19 2007 xldylt2gt xlddEXPRESSIONyl y2 x2dylt4gt x2ddEXPRESSIONyl y2 yd xld xlddx2d x2dd hw3pr0b2m tspan O 15 y0OOOO tyode45 39probZ39 tspany0 p10ttyl figure p10tty3 Texas A amp M University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr Alexander G Parlos Lecture 2 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 process ing functions 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 9515 C 6 l Where C and a are in general complex numbers If these are real numbers then 9515 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 9515 is the constant C An equally important class of complex exponential signals is obtained when a is purely imaginary In this case we obtain 9515 6M 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 9515 Acosw0t q 3 Where it is common to write we 27Tf0 Where f0 has the units of cycles per quotD Figure 1 Continuous time Unit Step Function second or Hertz The fundamental period of the sinusoid is given by 27F 4 To 7 lwol The sinusoid of equation 3 can be expressed in terms of the complex expo nential signal as Acosw0t q A R6ejwot 7 5 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 0t lt 0 t 6 M gt M gt 0 lt gt shown in Figure 1 Note that it is discontinuous at t 0 The unit impulse function 605 is related to the unit step function by the 8t o t Figure 2 Unit Impulse Function equation t ut L 6mm 7 Therefore this suggests that 60 digit lt8 A unit impulse is shown in Figure 2 Although the value 7 at t 0 is in nite 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 positive nonzero value T 9515 xtT 9 for all t The fundamental period To is the minimum value of the of T for which equation 9 is satis ed Further 2T1 is called the fundamental frequency 0 We also introduced the periodic complex exponential 95t ejwot 10 The signal 10 has fundamental frequency we and fundamental period To 37 Now we can speak of the collection of harmonically related complex exponentials Mt e wot k0i1i2 11 Each of these signals has a fundamental frequency that is a multiple of we Therefore a linear combination of harmonically related complex exponentials of the form 9515 Z ake mh 12 k7oo is also periodic with period To In this in nite sum the term for k 0 is a do or a constant term The two terms for k 1 and k 1 both have fundamental period equal to To and are collectively referred to as the fundamental components or the rst harmonic components More generally the components for k N and k N are referred to as N th harmonic components The representation of equation 12 is called the Fourier Series representation Having stated the Fourier Series expansion it is now important to derive an expression for the expansion coef cients ak Multiplying both sides of equation 12 by e jm0t we obtain 00 xte m t Z akejkwoteimwot 13 16700 lntegrating both sides from 0 to To 37 one fundamental period we have T T 00 0 0 xte tdt 0 0 Z akejkwoteijmotdt l4 16700 lnterchanging the order of integration and summation yields To 00 To xte 7 ldt Z akl e k tdt 15 0 k 0 700 The bracketed integral can be expressed as1 T T k n 61k tdt 0 16 0 0 k y 71 Therefore the right hand side of equation 15 reduces to Teak If we denote integration over any interval of length To by T0 then we can summarize the Fourier series representation of a periodic signal as 00 k 9505 Z dice W7 17 16700 and 1 at TOTO xte k tdt 18 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 9515 that is at each harmonic of the fundamental component For k 0 the coef cient a0 is given by do TieToxtdt 19 which represents the average value of the signal 9515 over one period The continuous time Fourier transform can be obtained by considering the behavior of the Fourier series as the fundamental period To gt 00 or as the fundamental frequency we gt 0 If we de ne the continuous frequency as w E km 20 and the envelope X w as Xw E Teak 21 then the Fourier series equations 17 and 18 can be expressed as 1 00 i 7 at m i 2 L00 Xwe dw 22 and Xw LTOOxte jmdt 23 1Prove this to yourselves as an exercise un Figure 3 Discrete time Unit Step Sequence where for the limiting case of To gt 00 the summations are replaced with integrals Equations 22 and 23 are referred to as the Fourier transform pair with equation 23 called the Fourier transform or the Fourier integral and equation 22 called the inverse Fourier transform The transform X w of any signal 9515 is also commonly referred to as the spectrum of 0505 because it provides us with information regarding how 9515 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 Hi 0nlt0 24 um T ln207 shown in Figure 3 5m Figure 4 Unit Impulse Function The unit impulse or unit sample sequence 6 is de ned as 071 y 0 6 25 n 1 n 0 lt gt 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 26 where C and oz are in general complex numbers If C and oz are real we can have one of several types of behavior If a gt 1 the signal grows exponentially with 11 while if a lt 1 we have a decaying exponential Furthermore if oz gt 0 all of the values of Ca are of the same sign but if oz lt 0 then the sign of Ca alternates If oz 1 then 7 is constant whereas if oz 1 then alternates in value between 0 and C Another important class of complex exponential signals is obtained if we let oz 6B and let 5 be purely imaginary In this case we obtain 619 27 A class of signals closely related to the complex exponential is the sinusoidal signal A005Qon q 28 where if we take 71 to be dimensionless then both 90 and q have units of radi ans The sinusoid of equation 28 can be related to the complex exponential signal as follows Acoslt90n gememon ge be jnon 29 There are some important differences between the complex exponential signals in continuous 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 Furthermore the function em is periodic for any value of we In the discrete time case the complex exponential is only de ned in the range OSQOlt27T0r 7T QOlt7T 30 and the signal 790 is not periodic for arbitrary values of 90 Rather for the signal to have period N gt 0 there must exist an integer m such that the following relation is true 90 i m ow m Table 1 summarizes the differences between the signals ejwot and 6790 Read handout Al for a presentation of the discrete time Fourier series and discrete Fourier transform Reading Assignment Table 1 Differences Between Continuous and Discrete Exponential Signals ejwot ej on Distinct signals for Identical signals for distinct values of we exponentials at frequencies separated by 27f Periodic for any choice of we Periodic only if no 27r39m for some integers N gt 0 and mi Fundamental frequency Fundamental frequency 00 we W Fundamental period Fundamental period we 0 unde ned 90 0 unde ned 2 2 we 0i i 90 f 0 Read Handout Al Fourier Analysis and Fourier Transforms and the examples Handout E2 posted on the course web page Texas A amp M University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr Alexander G Parlos Lecture 5 Modeling of Translational Mechanical Systems The objective of this lecture is to review the basic building blocks of lumped parameter translational mechanical systems and to build the foundations that will enable you to model more complex dynamic systems Translational Inertia Elements or Masses Analysis of mechanical systems is based on Newton s laws of motion Usu ally ideal or point masses are considered in the analysis as shown in the Figure 1 Motion is considered with respect to a non accelerating reference frame usually a xed point on earth or another non accelerating object The equation of motion for a mass mt is based on Newton s second law which expresses the conservation of linear momentum as follows ammmmn ggggiigam n where if we assume that the mass is constant m we can rewrite this equation as dv1gt i n i gtigamimm7 m where U190 is the velocity of the mass relative to the ground reference g m and Fmt is the net force acting on the mass Because of the non accelerating nature of the reference frame dflg i d1 dt E resulting in the following equation of motion for the ideal mass m M 3 Enos lt4 Furthermore the action of the applied force represents work being done on the mass as it accelerates increasing its kinetic energy The rate at which energy is stored in the system is equal to the rate at which work is expended on it Using the rst law of thermodynamics or the law of conservation of energy we have 0 mama lt5 To nd the energy of the system we need to integrate over a period 015 as follows Ska t t dv19t v1gt km dSK 0 melgaw 0 mv19t d dtm1g0 U19 tdv19 6 or 6m 8140 211706 lt7 This is the well know formula for the kinetic energy of a point mass Equation 4 indicates that because of the integrations involved it takes some time for the moving object to build up velocity and displacement As such it would not be realistic to attempt to apply a step change in velocity or displacement of the mass This would require an in nite amount of force and an in nite source of powerl Translational Stiffness Elements or Springs An ideal translational spring stores potential energy as it is de ected along its axis This is depicted in Figure 2 The gure shows a spring in its relaxed state Fk 0 and with the force Fk acting at both ends in free body diagram fashion Because an ideal spring has no mass the force transmitted by it is 2 Figure l Freebody diagram of an ideal mass undiminished during acceleration Therefore the forces acting on its ends must be equal and opposite Newton s third law of motion The elemental equation for such a spring derives from Hooke s law namely Fklt 952105 95210l7 8 where x210 is the free length of the spring Because x2105 x210 x205 0510 we can write the equation for a spring as FM WW 96105 9 where x205 05105 is the de ection of the spring from its initial free length In this development we have assumed that the spring has a constant stiff ness k If this is not the case we can write the general form of equation 9 as FNLSO fNLfL 2t 951057 10 where N L stands for a non linear spring Simpli cation of this equation via linearization to a spring with an equivalent linear stiffness near an operating point will be discussed in future lectures 3 39l 39g Fk lt J30 n Force acting Figure 2 Freebody diagram of an ideal spring Similar to the case of an ideal mass we can investigate the energy point of view of an ideal spring The conservation of energy for an ideal spring can be expressed as d8 t CZ kalga 11 To nd the energy of the system we need to integrate as follows 8130 t 1 t dFkt 1 Fk 5130 dsp 0 Fktv19tdt g 0 am dt dt g AW Fmde 12 or 1 2 51205 l 13 This is the formula for the potential energy of an ideal spring As in the case of the point mass it would not be realistic to attempt to apply a step change in force to a spring Such a force would have to move in nitely fast to de ect the spring suddenly which would require an in nite 4 Figure 3 Freebody diagram of an ideal damper source of powerl Translational Damping Elements or Dampers An ideal damper is shown in Figure 3 Because an ideal damper contains no mass the force transmitted through it is undiminished during acceleration Therefore the forces acting at its ends must always be equal and opposite The basic equation of an ideal damper is FM b01290 111905 b11210 14 With a damper there is no storage of retrievable mechanical work as the work being done by an applied force becomes dissipated as thermal internal energy The relationship between the force and velocity is instantaneous Thus it is realistic to apply step changes of either force or velocity to such an element An ideal damper arises from viscous friction between well lubricated mov ing mechanical parts of a system This is the only form of damping that is linear Non ideal forms of damping are very common in practice However non ideal damping is characterized by non linearities such as dry Coulomb friction aerodynamic damping structural damping These forms of damp ing can be linearized for simpli cation when no discontinuities exist in the force velocity characteristics of the damper at the expense of accuracy of the analysis Non ideal damping is found for example in poorly lubricated metal metal contact surfaces In general for a nonlinear damper we can write FNLth fNLfU12t7 15 where N L stands for a non linear damper Simpli cation of this equation via linearization to a damper with equivalent linear damping near an operating point will be discussed in future lectures Example A Massspringdamper System in a Gravity Field The system to be analyzed is shown in Figure 4 We apply Newton s second law to the mass m resulting in d t mt mg Fm mt lt16gt The equations for the damper and the spring can be expressed as FM 57119037 17 and Fkt kx1t A1 959 18 Equations 16 through 18 can be combined to obtain the following equation of motion 2 d x t dx t m 61 l W C l kltxllttgt xggt Em 19gt where the kA1 term was eliminated using the mg term With my 0 we obtain the following second order equation d21t d1t ll dt2 dt M mmnEm am This second order system also called a 1 degree of freedom or 1D OF system requires two variables to be completely described velocity and position We 6 I Relaxed pas I wnhoul gm Al I Rclnxtd pnsiu with gravity m CHlsl 0 hi Figure 4 Diagram of a massspring damper system in a gravity eld call these variables the states of the system This system also has two energy storage devices the mass and the spring exchanging kinetic and potential energy The damper is an energy dissipation element Reading Assignment Read pages 24 28 of the textbook Read the examples Handout E5 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 amp M University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr Alexander G Parlos Lecture 19 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 loops 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 con trol 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 diat dt whereas the mechanical dynamics can be expressed as 697105 La Raia vat 1 Jm ma b mot Km n 2 Let us de ne the output to the motor speed yt and let s name the motor load as the disturbance wt Tg Taking the Laplace transforms of equations 1 and 2 and eliminating the current Ias we obtain an equation of the form1 JmLa 2 JmRa bLa 1 y bRa Kth bRa KtKes gt 5 Kt l 1NOTE Please do the algebra to convince yourselves of this result 1 bRa glows bRa mom3 Figure l Openloop Control Systemi Figure 2 Closedloop or Feedback Control Systemi Equation 3 can be simpli ed to the form 715 1Tgs 1Ys AVas BIA5 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 form as follows B Y W 5 8 715 1Tgs 1 8 At steady state when both wt and vat are constant we have the steady A 715 1Tgs 1Va8 l state motor response as ysS Ava Bw 6 Figures 3 and 4 show the DC motor in open loop and closed loop speed control con guration We now compare the various properties of the two control systems we considered ie open and close loop control systems Figure 3 Openloop DC Motor Speed Control Systemr Reference speed r I Figure 4 Closedloop 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 state in the presence of disturbances or torques For the open loop controller the control input the ampli er voltage is va K r 7 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 7 8 A lt gt The DC motor speed with open loop control and zero load torque will be given by 1 SS A a A7 9 y 1 A 7 For the case of the feedback controller the ampli er voltage will be vaKr y 10 In view of this control law the closed loop transfer function in this case will be AK B Ylt8gt 715 l725 1 AKRlt8gt l 715 l725 1 AK At steady state 5 gt 0 with zero load torque we have AK 355 m7 where if the gain K is selected such that AK gtgt 1 then ysS E 7 So for Ws 11 12 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 333 AK Bu 13 or with the controller K i we have yssrBw7 and if we de ne the speed variation caused by the load torque as 63 yss r then 63 Bw 15 So the speed error is proportional to the load torque where the constant of proportionality is B and x for a given probleml For the closed loop system we have the steady state speed given by AK B W If the controller is designed such that AK gtgt 1 and AK gt B 7 then there 235 16 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 A 6A In the open loop case the controller gain K will still be i and the new overall system gain would be T016T0KA6A A6A 1 17 where T0 is the open loop torque So the gain error is In terms of percent changes de ned as 7 we have 6T0 7 6A T01 I7 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 333 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 YbT nF1h4amp K39 One can compute the sensitivity using equation 19 and differential calculus However an easier approach is to rst de ne the sensitivity as A dT hzgi C 20 A TC dA7 gt and then apply it to the close loop transfer function AK TC 21 l 1AK The result is 1 E4447 22 A 1AK which reveals another major advantage of feedback control Advantage of Feedback Reduce Impact of Uncertainties When us ing 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 and with the open loop controller gain set at 01 Determine the output for Consider a servomechanism with 71 72 a step change in the load torque of 01 N m Figure 5 Openloop transient response to a disturbance inputl For the open loop case assuming that r 0 the output is expressed as 50 Y s W s 23 The disturbance is given by 01 W5 24 So Yltsgt 5 25gt 5 1X61 8 Us The system response is shown in Figure 5 Now let s assume that we wish to use feedback control in order to improve the system s ability to reject steady state disturbances by a factor of 100 with respect to the open loop case The output is given by Ylt8gt 2 50 71725 71 73928 1 AK Since we want the error reduction to be 100 we must have 1 AK 100 So K 99 Considering that Ws from the nal value theorem we W5 26 7 Figure 6 Closedloop transient responses to a disturbance input and b reference inputi have that 5 38 005 27 y 1 AK gt The transfer function from the reference to the output is given by Y 10K ms 5 28gt R5 717252 71 73928 1 AK The system response to the given disturbance is shown in Figure 6a 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 200 2 15 of the textbook Read the examples in Handout E19 posted on the course web page MEEN 364 Parasuram Lecture 1415 August 22 2001 HANDOUT E 15 EXAIVIPLES ON TRANSIENT RESPONSE OF FIRST AND SECOND ORDER SYSTEMS SYSTEM DAIVIPING AND NATURAL FREQUENCY Example 1 In the system shown below Xt is the input displacement and 9t is the output angular displacement Assume all masses involved are negligibly small and that all motions are restricted to be small Obtain the response of the system for a unit step input Assume zero initial conditions No Friction Writing the force balance equation for the above system we get bx L6 kLG L6 L6 x 1 Equation 1 represents the governing differential equation of motion Taking the Laplace transforms of equation 1 we get Ls Ls SXS 83 1 s XsLs39 2 MEEN 364 Parasuram Lecture 1415 August 22 2001 Equation 2 represents the transfer function of the system shown above For a unit step input X s A therefore the output s becomes 1 l s 4 L s ltgt Taking the inverse Laplace transform of equation 4 we get 9 t equot 5 Equation 5 represents the response of the system for a given step input The MATLAB sequence to obtain the step response for a given L k and b is given below L k b 2 100 20 lL l kb num den sys tfnumden stepsys Xlabel Time ylabel Angular Displacement Theta Title Step response of a first order system The response of the system is as shown below Step response of a first order system our 005 e Angular Displacement Theta ml 003 MEEN 364 Lecture 14 15 Example 2 What is the unit step response of the system shown below Parasuram August 22 2001 10 s The closed loop transfer function is Cs 10s 10 Rs s2 10s1039 For a unit step input Rs 13 therefore 10s 10 1 Co 3 103 10 3 10310 gt Cs s5JEs5 JES CS 4 JE 1 4JE 1 1 6 35 shE 3 5 5 5 s The inverse Laplace transform of the above equation yields Ct 4 E ahE 4 JG 7675 1 3 15 3 JG gt Ct 11455e 8 87 01455e 1 13 1 7 Equation 7 represents the response of the system to a unit step input The MATLAB code for obtaining the unit step response of the above second order system is given below num 10 10 den l 10 10 sys tfnumden stepsys MEEN 364 Lecture 1415 Xlabel Time ylabel Output Title step response of a second order system The response is shown below Note that there is an overshoot Step response of a second order system Output 06 Time Parasuram August 22 2001 MEEN 364 Parasuram Lecture 1415 August 22 2001 Example 3 When the system as shown in Figure a is subjected to a unit step input the system output responds as shown in Figure b Determine the values of K and T from the response curve Na K 08 39 s2quots 1 a Step Response 0254 Amplitude P TIme sec The maximum overshoot from the response curve is 254 Therefore M P 0254 1 gte W0254 gt 204 MEEN 364 Parasuram Lecture 1415 August 22 2001 From the response curve we have 53 7 7t 7 P 3 an com Z cowl 042 60 t l14 From the block diagram the closed loop transfer function is Cs K srs2sK39 Hence OFF 2 wtL T T Therefore the values of K and T can be determined as T 10 25 a n 2X04Xll4 K wa 1142 Xl09l42 3 Example 4 Figure a shows a mechanical vibratory system When a 2 N force step input is applied to the system the mass oscillates as shown in Figure b Determine m b and k of the system from the response curve The displacement X is measured from the equilibrium position P2 N Force a MEEN 364 Parasuram Lecture 1415 August 22 2001 Step Response 1 1 1 1 00095 01 Amplitude xt in m o 8 Tim sec The transfer function of the system is Xs 1 Ps ms2 bski Since for step inputon lb Ps we obtain 2 Xs sms2 bs k From the response curve the steady state value of X is 01 hence from the nal value theorem we have xoo1i113sXs1jm i 2 0L rgt0ms2 bsk N k20 From the response curve the maximum overshoot is 00095 hence applying the formula for the maximum overshoot we get MP 2 00095 J gte W00095 gt 206 MEEN 364 Parasuram Lecture 1415 August 22 2001 The peak time tp is given by 727 7T 7T 2 a a a ll 2 a ll 06V rad 2m n 196 Age F Since 2 m 52 a 1962 Then b is determined as zgwn m gtb122 NSe Example 5 Consider the second order system whose transfer function is given as Cs 0 Rs s2 2 0 quotHm For a unit step input Rs 13 therefore the output is given by ml Cs quot ss22 a nswj Expressing the above equation in partial fraction format we have s a n g a n Cori s s an2aj s an2aj39 Taking the inverse Laplace transform of the above equation and rearranging the terms we get MEEN 364 Parasuram Lecture 1415 August 22 2001 7501 l 2 8 sin a a1ttan 1 1 g 8 111 Equation 8 represents the generalized solution of a second order system The following plot shows the step response of a second order system for various values of co 1 MEEN 364 Parasuram Lecture 1415 August 22 2001 Example 6 Derive the governing differential equation of motion of a swinging bar supported at its ends by a cord Solve the derived differential equation and plot the initial response of the system for the following initial conditions a Arbitrary initial condition Choose 11 1m 12 1m 1112 5Kg The differential equations of motion for the above system when represented in a matrix form is 2 m l cos9 mzll 22 W2 Sin m212 9 21119 milll2 cos9 mill2 9 2 L 2 3 J wzl2 sin9 milll2 q sin9 2 Initial response The second order differential equation has to be converted into a first order differential equation Let y1 y2 9 y3 9 y4 Substituting the above relations in the original nonlinear differential equation we get the following nonlinear first order differential equation which when represented in matrix form is MEEN 364 Parasuram Lecture 1415 August 22 2001 0 0 1 y2 mzll 0 quot1212 COSy3 y1 y W2 Sin y 1 mzlzyz 4siny3 yl 2 y2 0 1 0 2 ya y4 0 W212 sin ylt3gt m21112y22siny3 y1 2 3 lly4J 2 MATLAB Code In this type of a problem where the inertia matrix mass matrix is a function of the states or the variables a separate Mflle has to be written which incorporates a switchcase programming with a ag case of mass For example if the differential equation is of the form M t Y Y tF t Y then the right hand side of the above equation has to be stored in a separate mflle called Fm Similarly the inertia matrix mass matrix should also be stored in a separate mflle named Mm So when the ag is set to the default the function Fm is called and later when the ag is set to mass the function Mm is called The code with the switch case is given below Note that it is a function file and should be saved as indmotiodem in the current directory function varargoutindmotodetyflag switch flag case no input flag varargoutlFFty case mass flag of mass calls MMm varargoutlMMty otherwise error unknown flag flag end To store the right hand side of the original matrix form of differential equation a separate function file must be written as shown below Note that the name of the function is FF so this file must be saved as FFm the following function contains the right hand side of the differential equation of the form Mtyy Fty ie it contains Ftyit is also stored in a separate file named FFm function ypFFty 111 121 MEEN 364 Parasuram Lecture 1415 August 22 2001 m25 g981 w2m2g ypzeros4l yply2 2 w2sinyl m2122y4 Y4i w212siny32m2ll122y2 yp A2siny3yli yp3 yp4 A 2siny3yli Similarly to store the mass matrix a separate function le is written which is stored as MMm the following function contains the mass matrix it is separately stored in a file named MMm function n MMtY n20 m2ll O m2122cosy3 yl n30 O l O n40 m2ll122cosy3yl 0 m212123 nnln2n3n4 To plot the response the main le should call the function indmotiodem which has the switchcase programming which in turn calls the corresponding functions depending on the value of the ag For the main file to recognize the inertia matrix the MATLAB command ODESET is used to set the mass to M t y tspan O 30 y0052330l04670 Arbitrary Initial condition optionsodeset mass Mty tyodell3 indmotode tspany0options subplot2ll Xlabel Time ylabel phiW subplot2l2 plotty3 grid Xlabel Time ylabel Theta The above code plots the values of theta and phi with respect to time for the arbitrary initial condition case MEEN 364 Parasuram Lecture 1415 August 22 2001 Notice the command subplot subplotmnp breaks the Figure window into an mbyn matrix of small axes and selects the pth axes for the current plot The axes are counted along the top row of the Figure window then the second row etc For example subplot2l l plotincome subplot2l 2 plotoutgo plots income on the top half of the window and outgo on the bottom half Arbitrary initial condition phi Time Theta Time MEEN 364 Pamsuram Lecture 1415 August 22 2001 Assignment 1 Determine the values of K and k ofthe closed loop system shown below so that the maximum overshoot in unit step response is 25 and the peak time is 2 sec Assume that J 1 Kgm2 o 010 2 Consider the closed loop system given by C S 60 Rs 52 2 a Mac Determine the values of Q and a so that the system responds to a step input With approximately 5 overshoot and With a settling time of 2 sec Use the 2 criterion 3 Use MATLAB to plot the unit step response of the following system Cs 7 10 25 10 39 output ct respectively Where Rs and Cs are the Laplace transforms of the input rt and 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 AampM University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr AlexanderG Parlos Lecture 10 Thermal and Fluidic Systems Basic Mechanisms mquot Heat Transfer Energy rnay be transferred acmss the bnandanes nf a systern erther tn nr fern the systern it neeurs nni when therers aternneratare drffereneebetween the system and the sarrnundrngs momma an Imam whehrnay art nur drseussrnn by rntrnduerng the base n ysreai rneehanrsrns by wheh therrnai energy rs transferred Cmiductimi cnnduetrnn heat transfer neeurs nniy when there rs nhysreai enntaet between bndres systerns at drfferent temperatures it ran aisn be de ned as the transfer nf energy substanee Fnr nnerdimermnnal heat ennduetrnn m the r drreetrnn the rate nfheat nwis deterrnrned by the Fnuner equaann 1 nnrt ar ayer 4 L nne degree in a unrt drstanee Fnr example nnerdimensmnal steady state heat ennduetrnn rs depretedrn Frgurei Frgure i Onerddmmsnnal steady state heat ennduetrnn me equaann i 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 Vvacuum The volume ow rate QC ls grvm by We Q 1 dz where Cnsthe uld capacltance The nec ow lnLo the capacltor ls sLored and corresponds somewhat to ma energy swred C I Er 7P Fluid Inerwrs The symbollc dlagram ofa uld lna wr ls shown m Flgure 3 Flgure 3 A symbollc dlagram ofa Fluld rnenor The elernental equatton for the tnertor IS where 1 IS the utd Inerlance For fncttonless Incornpresslble ow tn a dz untforrn passage havtng cross secuonal area A and length L the Inerlance IS I zpjf z 521 where p 15 the rnass denstty ofthe utd The ktnettc energy stored tn an toleal lnerlor ts gtven by 1 2 E quot 2 Fluid Resistors The syrnboltc dtagrarn ofz utd reststor 15 shown tn thure 4 thure 4 A syrnboltc dtagrarn ofz utd reststor The elemental equatton ofan tdeal reststor IS R RQR Fluid Sources The tdeal sources employed tn utd systern analysts are shown tn thure 5 An tdeal ressure so ce IS capable of deltvertng the tndtcated pressure regardless of the ow n e tdeal ow source 15 c p ble of deltvertng the A l requtred by what tt IS drlvl g wher as an thure 5 Ideal utd 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 amp M University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr Alexander G Parlos Lecture 21 The Classical Threeterm PID Controller The objective of this lecture is to present the elements of the most widely used control algorithm that is the proportional P integral 1 and derivative D controller Prop ortionalIntegralDerivative Feedback Controllers Proportional P Feedback Control When the control signal is made proportional to the error in the mea sured output we call the control law proportional feedback or P control Speci cally if W K6007 1 where K is the P control gain The controller transfer function D5 is 195 K 2 As we saw in the case of the DC motor control unless the control gain is large there will be steady state error in the measured output However as we shall see later as the control gain is made larger the feedback loop stability becomes a serious issue So a P controller has some fundamental limitations regarding its largest allowed gain in order to obtain well damped response Of course 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 might appear in the system response Integral feedback has the form of K t ww dmm 3 where T1 is called the integral reset time The l control transfer function D5 is K 7 4 ES To see the effect of l control on the steady state error take the derivative D5 of equation This results in dut K d7EWL a which means that the control effort ut of an l controller will stop varying and take a non zero value only if the error 615 becomes zero So using l control we can drive the steady state error of 615 to zero The combination of P and I control so called Pl control can be expressed as umKMwifwwm o T to where the proportional gain K and the integral reset time T1 are tunable parameters In general l control improves the steady state response of a control system but it also slows down its response That is unless we are willing to increase the overshoot in an attempt to improve the system s response speed Derivative D Feedback Control Derivative or D feedback has the form det t KT 7 7 M D ltgt with the corresponding transfer function being MKamp 2 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 conjunc tion with P and or l control because D control by itself may not be effective if the control error 615 is constant D control is anticipatory in nature 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 Prop ortional IntegralDerivative PID Control For more effective control over the steady state and transient errors we can combine all three elements discussed before to obtain proportional integral derivative PID control Some form of PlD control including P or Pl or PD only is used in probably over 90 or 95 of the industrial control loops The resulting PlD control law is l t det not a Kltelttgt E 0 mm TD lt9 with the corresponding transfer function given by 1 Ds K17TD5 10 T18 The design of PlD controllers deals with nding the parameters K T1 and TD such that design speci cations are met This controller parameter adjustment process is called controller tuning Example Impact of PID Gains We can see the effects of PlD 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 Figure 1 PH Control of DC Motor a step disturbance input b step reference input the Pl oontroller results in zero steady state error for a step input in distur bance Further addition of the D element resulting in PlD oontrol results in a better behaved response A similar behavior is observed for a step input change in the reference see Figure 1b P oontrol only results in some steady state error Addition of the l oontrol removes the steady state error Whereas addition of D oontrol results in a better damped better behaved response Reading Assignment Read pages 215 226 of the textbook Read the examples in Handout E21 posted on the course web page Texas A amp M University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr Alexander G Parlos Lecture 26 Introduction to RootLocus The objective of this lecture is to introduce you to another method used in the analysis and design of control systems namely the root locus tech nique The basic idea exploited in root locus design is that one can track the movement of the closed loop poles as parameters of interests such as a controller gain are varied This gives a designer the map of the closed loop pole movement on the s plane and proper design choices can be made Root Locus of a Basic Feedback System Assume the following closed loop transfer function Y8 KAKGa8 1 R8 1 KAKGa87 Where KA and KG is the controller and system gain respectively The closed loop poles are the roots of l KAKGG8 0 2 Let us assume that the transfer function Cs can be expressed in terms of its poles and zeros as b 7 3 Gltsgt L5 15 2 lt3 08 1118 P2 and that K KAKg 4 We now express the characteristic equation in various root locus forms as follows 1KG5 0 5 1 1K 3 0 Q a5 Kb5 Cs 07 W l 7 8 gt K ltgt The root locus can be thought of as a method for inferring the location of the closed loop poles from examining the open loop transfer function K Cs Example 1 Root Locus with Respect to Controller Gain A normalized transfer function of a DC motor is 6m5 l K G 9 mg a 8 881 ltgt The characteristic equation for this system is 1 1 K 0 10 88 1 7 gt or sK0 n The solution of equation 11 gives the closed loop pole locations as follows 1 M1 4K mm igjfi m Plotting these two roots as K is allowed to vary results in the following root locus shown in Figure 1 Several simple observations can be made from Figure 1 There are two roots and thus two branches to the locus At K 0 these branches begin at the open loop poles because for K 0 the system is open loop As K is increased the closed loop poles move towards each other and they meet at 5 At that point they break away from the real axis Fiom the breakaway point the poles move towards in nity while their sum is equal to 1 We have now characterized the path the closed loop poles transverse as the gain K is allowed to increase Example 2 Root Locus with Respect to Plant Parameters 2 Figure 1 Root Locus for Gs 451 as a function of the controller gain Ki Now consider the following transfer function Glt8gt 5510 We want to nd the root locus with respect to 0 Here we assume that K 1 13 The corresponding closed loop characteristic equation is 1 CS 0 14 or 52cs10 15 We can express this as 5 1 7 0 16 c82 1 7 placing it in the standard root locus form The roots of equation 16 can be expressed as c 02 4 T1772 E l Plotting these two roots as c is allowed to vary results in the following root locus shown in Figure 2 Note that when 0 0 the closed loop poles are also the open loop poles The closed loop poles are daInped as c grows At 5 1 the two segments 3 Figure 2 Root Locus for Gs as a function of the damping factor 0 1 ssc of the root locus abruptly change direction and move away from each other one towards the origin and the other towards in nity This point of change in direction is called the breakin point Guidelines for Sketching a Root Locus The root locus of a transfer function can be accurately drawn using for example MATLAB However one can obtain sketches of the root locus by following the following steps We will demonstrate the use of these steps on the following transfer function 1 G 5 55 4 16 STEP 1 Place the open loop poles and zeros of Cs on the s plane See 18 Figure 3 STEP 2 Find the segments of the real axis belonging to the root locus The acceptable segments must have an odd number of poles plus zeros to their right See Figure 4 STEP 3 Draw the asymptotes for large values of K As K gt 00 the Figure 3 Step 1 of the Root Locus Figure 4 Step 2 of the Root Locus Figure 5 Step 3 of the Root Locust root locus equation 1 Gltsgt 77 19gt can only be satis ed if CS 0 This can occur in two ways rst at the zeros of C5 and second when some asyrnptotes approach in nity If the system has n poles and m zeros then there will be 71 m asymptotes These asyrnptotes forrn angles with the real axis given by ZWll27n m7 20 n m and they intercept the real axis at 219239 22239 n m oz 21 See Figure 5 STEP 4 Compute the departure angles from the open loop poles and the arrival angles to the open loop zeros If a pole appears q times then the departure angle is given by q dep 2 Zq i 1800 36001 22 6 Figure 6 Step 4 of the Root Locus where 23 is the sum of the angles of the remaining poles and 211 is the sum of the angles of all the zeros Also 1 takes q values and it is selected such that 610 in the range 1800 1800 Similarly for arrival angles we have the following formula IQJaw l 1800 36001 See Figure 6 STEP 5 Estimate the points where the root locus crosses the imaginary axis This can be done using Routh s criterion For the third order example we are using the characteristic equation is K 1 55 42 16 07 24 or equivalently 53 852 325 K 0 25 The Routh array is then given by Figure 7 Step 5 of the Root Locust 1 8 x 32 K 8 50 K 26 For 0 lt K lt 256 there are no positive roots of the characteristic polynomial If we substitute K 256 and 5 jw into equation 25 we obtain the non trival solution w l566 Note that the asymptote crosses the imaginary axis at 462 See Figure 7 STEP 6 Estimate the location of multiple roots especially on the real axis breakaway points The location 50 of the breakaway point is computed by solving the following equation d 1 330 27 STEP 7 Complete the root locus sketch by combining the information obtained from Steps 1 through 5 and 6 if necessary See Figure 8 Uses and Gain Selection from the Root Locus Figure 8 Step 7 of the Root Locus The root locus technique is a method by which the closed loop pole lo cations of a system can be tracked as a parameter is allowed to vary from zero to in nity So far we only mentioned the use of the control gain as a parameter to vary However any system parameter can be allowed to vary and the same procedures of root locus can be used in mapping the location of the closed loop poles The key is to express the characteristic equation in the form 1 KG5 0 28 where K can be any parameter to be varied In doing so one would collect the terms that do not multiply the parameter to be varied and name it a5 whereas the terms that do multiple the parameter as bs These two polynomials de ne the poles and zeros of the system respectively The rest of the procedure for plotting the root locus is as described before Further one could generate a root locus when two parameters must be allowed to vary This is done by allowing one parameter to vary while xing the other 9 One more use of the root locus is that it allows us to compute the gain required to place the closed loop poles on certain locations on the root locus Every point on the root locus satis es the condition 1 KG5 0 29 This is a complex relation and the magnitude part of equation 29 corre sponds to 1 K 7 30 G8 So for any point on the root locus so the required gain K0 can be obtained from 1 K0 31 lG5l8So Reading Assignment For material on basic procedures for plotting and using root locus read pages 272 310 of the textbook Read the examples in Handout E26 posted on the course web page MEEN 364 Parasuram Lecture 67 August 7 2001 HANDOUT E7 EXAIVTPLES ON MODELLING OF lVlECHANICAL SYSTEMS MIXED ROTATIONAL AND TRANSLATIONAL Note that the time dependence of variables is ignored for all manipulations Example 1 One DOF system Consider the system shown below y The disc is of radius R and has a moment of inertia I There is friction between the disc and the block of mass m Kinematics stage From the above gure it can be seen that there are two rigid bodies The coordinates representing their independent movement is given by X and 9 respectively But since there exists a relation between these two coordinates which is given by x R0 the number of degrees of freedom of the system is one Let the degree of freedom be X The velocity and the acceleration are given by x x 9 9 respectively This completes the kinematics stage Kinetics stage Free body diagram of block F kx cx I Ff Note that the gravity force is not considered as the reaction force of the disc on the block balances this force Writing the Newton s second law of motion we get MEEN 364 Parasuram Lecture 67 August 7 2001 2Fma gtF kx cx Ff mx gtmxcxkxFf F 1 Free body diagram of the disc gtFf Note that the gravity force is ignored as the reaction force due to the block on the disc balances the gravity force Taking moments about the center of the disc we have 2M 19 gtFfR 19 gtFf 9 Since x R0 we get Ff x Substituting the value of Ff in equation 1 we get mxcxkxFf F gtmxcxkx xF gtmRI 2xcxkxF 2 Equation 2 represents the governing equation of motion for the system de ned MEEN 364 Parasuram Lecture 67 August 7 2001 State space representation Let the states of the system be x x1 3 x From the above relations we get x1 x2 4 Substituting the relations given by equation 3 in equation 2 we have I mFxcxkx F I gt mR 2XZCC2 kx1 F F k xZ x1 x2 5 I I I W quotHR 2 quotHF Rewriting the equations 4 and 5 in matrix format we get k c 1 x1 x1 F 6 I I x 1 x2 mR 2 mF 2 mR 2 If the output of the system is the displacement of the block then the output relation can be expressed in matrix format as follows y 1 org 7 x2 Equations 6 and 7 represent the statespace form of the system defined MEEN 364 Parasuram Lecture 67 August 7 2001 Example 2 One DOF system PKG X Consider the system shown above Assume no friction Kinematics stage There is one rigid body The number of degrees of freedom of the system is one Let the degree of freedom be represented as X The relation between the rotation of the disc and the linear displacement moved by the disc is given by x R0 Assume rolling without slipping The velocity and the acceleration of the disc are x x respectively Kinetics stage Free body diagram of the disc 10 10 Writing the Newton s second law of motion we get 2F ma gt 2kx m x m x 2kx 0 8 Substituting the above relation in equation 8 we get me 2kR9 0 9 MEEN 364 Parasuram Lecture 67 August 7 2001 Equation 9 represents the governing differential equation of motion State space representation Let the states of the system be de ned as 9 x1 10 9 x2 From the above relations the following equation can be written x1 x2 1 1 Substituting the relations given by equation 10 in equation 9 we have mR 9 2kR9 0 2 m XZ kal 0 2k gtxz x1 12 m Rewriting equations 1 l and 12 in matrix format we have x 0 1 x1 1 0 x 13 m m 2 If the output of the system is the linear displacement of the disc then the output relation can be expressed in the matrix format as yxR9 Rx1 y R of x 2 14 Equations 13 and 14 represent the statespace form of the system defined MEEN 364 Parasuram Lecture 67 August 7 2001 Example 3 Two DOF system 37 direction X m Kinematics stage There are two rigid bodies and the number of degrees of freedom of the system is two Let the two degrees of freedom be represented by the angular displacement of the bob 9 and the linear displacement of the block X Therefore the linear velocity and the linear acceleration of the block are x x respectively Similarly the angular velocity and the angular acceleration of the bob are 9 9 respectively Assume x gt 0 Kinetics stage Free body diagram of the bob kx L9 The linear acceleration of the bob is towards the right and is equal to L9 Therefore writing the Newton s law of motion we get ZFX ma gt kx L9 Tsin9 mLO 15 MEEN 364 Parasuram Lecture 67 August 7 2001 Similarly ZFy ma gt T cos9 mg Substituting the value of tension T in equation 15 we have kx L9 mgtan9 mL9 16 For small angles we have tan0 z 0 Therefore equation 16 reduces to kx L9 mg9 mL9 17 Free body diagram of block kx L9 F 1 Note that the gravity force is neglected as the normal reaction force of the ground balances the gravity force Writing the Force balance equation we get 2F ma gtF kx L9 Mx gtMxkx L9 F 18 Equations 17 and 18 represent the governing differential equation State space representation Let the states of the system be defined as 9 x1 9 x 19 xx3 xx MEEN 364 Parasuram Lecture 67 August 7 2001 From the above relations the following two equations can be derived X1 X2 20 X3 X4 Substituting the relations given by equation 19 in equation 17 we get kx L9 mg9 mLO gt kx3 Lx1 mgx1 mLxZ kLmgx1 k m gtx x 21 2 mL L 3 Similarly substituting the relations given by equation 19 in equation 18 we get Mxkx L9 F gtMx4kx3 Lx1F F gtX4 HHxl Hx3 22 Rewriting equations 20 21 and 22 in matrix format we have 0 l 0 0 x 0 x1 kL mg k 1 x L 0 7 0 x 0 2 m m 2 0 0 0 1 x 0 F 23 X3 kL k 3 l 0 0 x x4 M M 4 M If the output of the system is the linear displacement of the block then the output relation can be expressed in the matrix format as yxx39 x1 x2 y0 0 1 0 x 24 3 X Equations 23 and 24 represent the statespace form of the abovedefined system MEEN 364 Parasuram Lecture 67 August 7 2001 Example 4 Two DOF system Consider the system shown below 37 J zmn X Moment of inertia of the drum is equal to I Kinematics stage There are two rigid bodies and they can move independent of each other Therefore the number of degrees of freedom of the system is two Let the degrees of freedom be chosen as the linear displacement of the block y and the angular displacement of the drum 9 Therefore the linear velocity and the linear acceleration of the block is given by y y respectively Similarly the angular velocity and the angular acceleration of the drum is given by 9 9 respectively Assume y gt9 Kinetics stage Free body diagram of the block R CyRQ 16939 MEEN 364 Parasuram Lecture 67 August 7 2001 Note that the displacements chosen are from the static equilibrium position Hence spring force due to the initial elongation of the spring balances the gravity force Writing the Newton s second law of motion we have ZFy ma gtcy R9 ky 9 my R mycy R9 ky 39 0 25 Free body diagram of the drum R CyRQ My 39 Taking moments about the center of the drum we have 2M 1 9 gt Tcy R9 Rky 9 g kRQ R c59 g 19 2 Z 2 19 cR2d 9 cRykR2k9 k yT 26 Equations 25 and 26 represent the governing differential equations of motion State space representation yx19 yx29 27 9 x3 9 x MEEN 364 Parasuram Lecture 67 August 7 2001 From the above relations the following equations can be deduced X1 x2 28 X3 x4 Substituting the relations given by equation 27 in equation 25 we get R mycy R9 ky 39 0 R gt mxzcx2 Rx4kx1 3x3 0 gtxZ x1 x2 Ex3 x4 29 m m 2m m Similarly substituting the relations given by equation 27 in equation 26 we get cR2 kR2 R 19 cR2 9 cRykR2 9 k yT 4 4 2 2 2 gt 1964 SCR x4 cRx2 Skf x3 k x1 T gtx Zk R x 5kR2 x SCR2 x 30 4121x112 413 41 Rewriting equations 28 29 and 30 in matrix format we have 0 l 0 0 0 M i LR 2 x1 0 X2 m 2m 2m x2 0 0 0 1 x3 0 T 31 M E g 5kR2 5ch x 1 x4 21 1 41 41 4 I If the output of the system is the displacement moved by the block then the output equation can be expressed in the matrix format as Yzyle 5 N Y10 0 0 32 88 u 4 Equations 3 l and 32 represent the statespace form of the system defined MEEN 364 Parasuram Lecture 67 August 7 2001 Assignment 1 Derive the governing differential equation of motion for the following system Mass m I L2 39 I L2 2k k Hinged support The bar can rotate about This point 2 Consider the system shown below O m The above gure consists of a drum to the center of which a simple pendulum is attached The bob of the pendulum is displaced slightly to the right Derive the governing differential equation of motion Assume small angle motion Recommended Reading Feedback Control of Dynamic Systems 4th Edition by Gene F Franklin etal 7 pp 24 45 Texas A amp M University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr Alexander G Parlos Lecture 28 Dynamic Compensation Using RootLocus In addition to Bode plots we can use the root locus method to design dynamic compensators The simplest form of a dynamic compensator takes the structure 5 z 05 8 lt1 P 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 D5 5 z This is a PD controller and we apply it to the following second order system KG5 2 55 l The uncompensated and compensated root locus is shown in Figure l The effect of a zero is to move the locus towards the stable region of the s plane Whereas before compensation achieving can of say 2 would have resulted in very low damping and high overshoot following compensation we can achieve the same can with damping ratio of more than 05 The problem with the a pure lead compensator zero only is that its im plementation 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 1 Figure 1 Root locus without compensation solid line and with PD compensation dashed line The effect of the pole on the compensation can be seen in Figure 2 Selecting exact values of z and p is usually done by trial and error Gen erally the zero is placed near the closed loop pole The choice of the com pensator pole is a compromise between noise suppression and compensation effectiveness The process can be made more analytical in nature if the closed loop pole is selected rst Then we arbitrarily select one of the lead compensator parameters and use the angle criterion to select the other Lag Compensation Once the desired transient response is obtained one might discover that the steady state response of the feedback loop is not satisfactory Improvements in the steady state errors can be made by placing a pole near the origin which is usually accompanied by a zero nearby so that the pole zero pair does not signi cantly interfere with the overall dynamic system response shaped by the lead compensator For the problem studied previously the lead compensator 8320 could be 1m 5 05 2 Dm nzo 6 31 4 1 03 56 4 55 2 x 710 8 6 4 2 I Ru lO 8 a Imr I Dx2 39l Z Dm xfZO lt l l l2 8 6 4 Re Figure 2 Effect of pole on compensation Figure 3 Effects of lag compensation on root locus s0l 300l 39 dynamics However a root close to the imaginary axis will persist and it will followed by a lag compensator The lag will not impact the faster slow down the overall response As a result it is important to place the lag pole zero at as high frequency as possible without impacting the dominant system dynamics The effect of the lag compensation on the root locus is shown in Figure 3 Reading Assignment For material on dynamic compensation using root locus read pages 310 328 of the textbook Read the examples in Handout E28 posted on the course web page MEEN 364 Parasuram June 29 2001 HANDOUT M3 EIGEN VALUES AND EIGEN VECTORS De nition Let A be an n x n matrix Then a real number 7 is called an eigenvalue of the matrix A if and only if there is a ndimensional nonzero vector V for which Avillvi z39ln 1 Any such vector V is called an eigenvector of the matrix A associated with the eigenvalue 11 Example 1 Consider the 2 x 2 matrix given by 7 l A 6 2 Then since Eigenvector l2 lill ifi ifll lelil Eigenvalue we see that 7 4 is the 39 39 ofA 39 A with the 39 V ll 1 3 Also since 2 liHlilelil We see that 7 5 is also an eigenvalue of A associated with the eigenvector wt MEEN 364 Parasuram June 29 2001 From the above example note that the vector V1 is not the only eigenvector associated with matrix A Any nonzero scalar multiple of vector the V1 is also an eigenvector of A associated with the eigenvalue 7 4 In other words v 2 which is twice the vector V1 is also an 39 of A 39 A with 39 39 7 4 A similar statement holds for the eigenvector V2 Note 1 Eigenvalues and eigenvectors are de ned only for square matrices ie the number of rows must be equal to the number of columns in the matrix 2 In the above example the number of rows and number of columns is equal to 2 So the size of the matrix is represented as 2 x 2 The number of eigenvalues associated with the matrix is equal to the size of the matrix ie in this case there are two eigenvalues associated with the de ned matrix In general if the size of the matrix is n x n then there are n eigenvalues associated with the matrix and each one has an associated eigenvector with it Procedure for calculating eigenvalues and eigenvectors analytically If A is any square matrix of size n x n and 7 is an associated eigenvalue rewrite equation 1 as AVi l IVi where I is the identity matrix of size n x n Note The size of the identity matrix has to be the same as that of the matrix for which the eigenvalues and eigenvectors has to be calculated The above equation reduces to A 1 IVi 0 2 which is nothing more than a homogenous system of n equations in n variables In order to nd the value of 7 the above system of linear equations has to be solved V 0 is a trivial solution but from the de nition of eigenvectors V has to be a nonzero vector MEEN 364 Parasuram June 29 2001 The eigenvalues of A are those real numbers 7 for which the homogenous system de ned by equation 2 has a nontrivial or nonzero solution and the eigenvectors of A associated with 7 are the nonzero solutions of this system Equation 2 has a nonzero solution if and only if its coef cient matrix is noninvertible The coef cient matrix is noninvertible if and only if its determinant is equal to zero ie A a 10 3 Equation 3 is called the characteristic equation of matrix A The above procedure is further explained with the help of an example Example 2 Let A a 3 x 3 matrix be de ned as 0 5 7 A 2 7 7 l l 4 The size ofA is 3 x 3 So the identity matrix I has to be 1 0 0 I 0 l 0 0 0 l The characteristic equation is de ned as l 5 7 A M 2 7 1 7 0 1 1 4 1 gt 1 2l 4l 5 0 The eigenvalues of A are the solutions to this equation namely 1 2 1 4 and 1 5 1 Z 3 To nd the eigenvectors substituting the value of 7 5 in the coef cient matrix of equation 2 we have MEEN 364 Parasuram June 29 2001 5 5 7 v1 A 51v 2 2 7 v2 0 1 1 1 v 3 Rewriting the above matrix in the form of homogenous equations we get 5v1 5v2 7v3 0 2v1 2v2 7v3 0 4 v1 v2 v3 0 By multiplying the third equation of equation 4 by 2 and add the result to the second equation of equation 4 we get v3 0 Now by substituting this result in the third equation of equation 4 we get 1 V2 Let v1 s where s is any arbitrary nonzero constant Therefore the solution to this system is given by v1 3 1 v2 3 31 v 0 0 From the above solution it can be concluded that the vector 1 l is an eigenvector of the system associated with the eigenvalue 7 5 0 Usually the eigenvectors are normalized and the resulting vector represents the normalized eigenvector of the de ned system To normalize a vector determine its magnitude by taking the square root of the square of its components and divide each element by the magnitude The resulting vector represents the normalized vector For example the magnitude of the abovede ned eigenvector is given by MEEN 364 Parasuram June 29 2001 mag 12 12 02 J5 Therefore the normalized eigenvector of the matrix A is given by 1 1 1 mg J5 07071 n11 1 1 i07071 mag mag J5 0 0 0 0 W Lm ag 1 Also note that any nonzero scalar multiple of the normalized eigenvector is also an eigenvector of the system Similar procedure must be followed to obtain the normalized eigenvector associated with the eigenvalue 7 0 Using MATLAB to calculate eigenvalues and eigenvectors The eigenvalues and eigenvectors can be calculated in MATLAB using the eig command By typing the command help eig at the prompt in the MATLAB command window displays the following help message help eig EIG Eigenvalues and eigenvectors E EIGX is a vector containing the eigenvalues of a square matrix X VD EIGX produces a diagonal matrix D of eigenvalues and a full matrix V whose columns are the corresponding eigenvectors so that XV VD VD EIGX nobalance performs the computation with balancing disabled which sometimes gives more accurate results for certain problems with unusual scaling E EIGAB is a vector containing the generalized eigenvalues of square matrices A and B VD EIGAB produces a diagonal matrix D of generalized eigenvalues and a full matrix V whose columns are the corresponding eigenvectors so that AV BVD MEEN 364 Parasuram June 29 2001 Example 3 Consider the following 3 x 3 matrix A 2 3 4 A 2 3 0 0 0 5 The program to calculate the eigenvalues and eigenvectors of the de ned matrix is given below 0 1 This program calculates the eigenvalues and eigenvectors A2 3 42 3 00 0 5 vdeig1 Note that the first line of the program starts with a sign This represents that any character following the symbol is a comment and will not be included in the program while compiling Running the above program gives the following result in the MATLAB command window v 08321 0707l 07071 05547 0707l 07071 0 0 00000 d 0 0 0 0 5 0 0 0 5 d is a diagonal matrix whose diagonal elements represent the eigenvalues of the system From the above result it can be seen that for the eigenvalue 7 5 the corresponding eigenvectors are given by the second and the third columns of the v matrix The reason for the existence of two eigenvectors is that the eigenvalue 7 5 is a double root Properties of eigenvalues and eigenvectors l Eigenvalues and eigenvectors are defined only for square matrices 2 According to the definition the zero vector cannot be an eigenvector However the real number 0 can be an eigenvalue of a matrix MEEN 364 Parasuram June 29 2001 3 Every eigenvalue has an in nite number of eigenvectors associated with it as any nonzero scalar multiple of an eigenvector is also an eigenvector 4 A matrix is invertible if and only if none of its eigenvalues is equal to zero Importance of eigenvalues and eigenvectors I I There are various reasons for and This handout outlines one such reason Vibrations point of View Consider the system given below It consists of two masses interconnected by a spring and forced to move on a horizontal plane X E X1 I f1 f2 K1 K2 The governing differential equation of motion for the abovedefined system is Mxr lem f M is the mass matrix K is the stiffness matrix and x is the time dependent vector whose elements represent the displacement of the two masses from the fixed support and f is the constant force matrix The eigenvalues of the system represent the square of the natural frequencies with which the system will vibrate and the eigenvectors represent the mode shapes In other words the eigenvectors denote how the masses vibrate with respect to each other This is explained in detail with the help of an example Example 5 For the above defined system the mass matrix M and the stiffness matrix K are given by MEEN 364 Parasuram June 29 2001 Eigenvalues or in this case the natural frequencies and the eigenvectors are properties of the unforced system In other words the force f matrix is not used in the calculation of the eigenvalues and eigenvectors To calculate the eigenvalues and eigenvectors the eig command in MATLAB is used The code for the procedure is shown below m1 O O 2 k2 11 1 vdeigkm wnat sqrtd Note that the way the eig command is used is different from that used in the previous cases Always make sure that the first argument in the eig command is the stiffness matrix and then the mass matrix The diagonal elements of d matrix represent the eigenvalues and the columns of the v matrix represent the corresponding eigenvectors The natural frequency of the system is the square root of the eigenvalues The result of the above code is v 09628 04896 02703 08719 d 22808 0 0 02192 wnat 15102 0 0 04682 From the above result it can concluded that the natural frequencies of the system are 15102 and 04682 rads The eigenvectors associated with the natural frequencies are 09628 04896 V1 and V2 respectively The e1genvectors represent how 02703 08719 the two masses vibrate with respect to each other For example if the system is vibrating at 15102 rads which is one of the natural frequencies then from the eigenvector associated with this frequency it can be concluded that the masses vibrate in the opposite direction ie they are out of phase In other words if mass M1 moves a distance of 09628 units in one direction then the mass M2 moves a distance of 02703 units in the opposite direction MEEN 364 Parasuram June 29 2001 Similarly if the system Vibrates at 04682 rads then both the masses Vibrate in phase with each other ie in the same direction as can be seen from the same sign of the elements of the eigenvector Note The eigenvalues are also used to determine the stability of the system This is discussed in detail in a later handout MEEN 364 Parasuram June 29 2001 Assignment 1 For the following matrices determine the eigenvalues and eigenvectors analytically and also verify the result using MATLAB Al 3 4 3 l B 2 0 2 3 3 0 2 Consider the following system Derive the governing equation of motion and obtain the eigenvalues and eigenvectors Discuss in detail the mode shapes of the system ie explain in detail how the masses move with respect to each other when the system is excited with one of the natural frequencies Explain the same for all the natural frequencies Note The surface is frictionless f1 f2 f3 k1 k2 M1 M2 M3 Given M1 lKg M2 2Kg M3 3Kg k1 k2 lONm The force matrix is defined as f1 1 Ff21N f3 0 MEEN 364 Parasuram June 29 2001 Recommended reading Feedback Control of Dynamic Systems Third Edition by Gene F Franklin eta1 7 pp 752 MEEN 364 Parasuram Lecture 3 August 28 2001 HANDOUT E3 EXANIPLES ON LAPLACE TRANSFORMS Example 1 Evaluate the Laplace transform of the following functions a f t 6 0c is a positive real number The Laplace transform of a function t is given by mm Imam Therefore Lft je me s39dt je wm 0 0 ersat w 1 S060 30 b ftcosa I Here 0 is a positive real number Lft jcosw t 6de 0 Expressing cosa tin exponential form gives jwteqwt 2 e coswt Then MEEN 364 Parasuram Lecture 3 August 28 2001 Lcosa 112J39emesedtjemzeszdtj 0 0 1 11er is dtbje mimdt 2 0 0 1 eowenz ehvw l w 39f f 2mm S 10 Sj0 l 1 1 s 2 Uw n mr wr c ftte m Using the theorem which states that d Ltft EFS we get Llre mi Le i 1 ds sa sa239 d ft 6 quot cosa t Using the theorem which states that Le ft FS a we get sa Lequot cosa t 2 maYw Example 2 a Determine the inverse Laplace transform of the following function 3 Fm s2 3s 10 Simplifying F s using partial fractions we have MEEN 364 Parasuram Lecture 3 August 28 2001 3 A B Fs 2 3 s 3s 10 s5s 2 s5 3 2 3 7 A s 5Fs 575 3 52 7 3 P5 s 2 3 BS 2FSSZ m Therefore F gt 0 s5 3 2 Therefore 3 3 L 1Fs L 1 g A s 3 2 zip 1 251 1 7 s5 7 3 2 Ee 5 26 7 7 b Determine the inverse Laplace transform of the function Fs s 1s 2 We write the partial fraction as B C s1 s2 322 173 Then 33 571 322 571 A s1Fs MEEN 364 Parasuram Lecture 3 August 28 2001 B is22Fs 7 10 2 ds 5 2 ds s 1 572 C s22Fs 7 53 1 H s 1 H Therefore LL 1 31 s2 02 gt ft 26 26 te 2 Assignment 1 Determine the Laplace transform of the following functions a f0 10e539 b ht L t2 c gt 6te cost 2 Determine the inverse Laplace transform of the following functions 6 Fm 3s 2s 1 2s 1 b Gm s2 6s5 1 C Recommended Reading Feedback Control of Dynamic Systems Fourth Edition by Gene F Franklin etal 7 pp 96 115 Recommended Assignment Feedback Control of Dynamic Systems Fourth Edition by Gene F Franklin etal 7 problems 32a 32c 33b 35b 37 ad MEEN 364 Parasuram Lecture 23 24 April 1 2003 HANDOUT E24 EXAIVIPLE HANDOUT ON STABILITY MARGINS AND CONTPENSATION Example 1 Consider the system whose openloop transfer function is given by K Gm s02s 1005s139 Determine the value of K for a phase margin of 40 For the value of K computed determine the gain margin S0 The phase margin relation is given by PM 4Gja 180 3 40 90 tan391 02w tan391 03905w 180 3 tan391 02a tan391 005m 50 3 tan391 50 1 001a22 3a 4 radsec At this frequency the magnitude must be equal to 1 Hence Gjw imzzt 21 3 2 2 1 a 1 004a 1 1 00025a 1 W4 3 K 52 To determine the gain margin first compute the frequency where the phase is equal to 180 degrees Therefore we get 90 tan391 02a tan391 005m 180 3 tan391 02a tan391 005m 90 MEEN 364 Parasuram Lecture 23 24 April 1 2003 tan 1 90 1 001a 025m oo 1 001m2 Therefore the denominator must be equal to zero Hence 1 001a2 0 3 m2 001 3 a 10 Substituting this value of frequency in the magnitude we get IGUw I 0208 K 52 a 004m2 1l00025m2 1 1043XM Therefore the gain margin of the system is GM 0 2010g0208 1364 dB Example 2 Consider a type I unity feedback system with Gs K ss139 Design a Lead compensator so that kV 12 sec391 and Pm gt 40 Use MATLAB to verify that your design meets the specifications S0 The transfer function of the Lead Compensator is given by DsK1 wherealt l a Tsl For the design consider K1 always with the plant such that the plant transfer function reduces to KK1 ss 139 Gs MEEN 364 Parasuram Lecture 23 24 April 1 2003 Step 1 For the given steady state error constant obtain the value of the gain KK kV limsGs lim 1 KK1 12 sgt0 sgt0 S Therefore the plant transfer function reduces to 12 ss139 1 G0 Step 2 Plot the Bode plot of the transfer function given by equation 1 and compute the phase margin of the uncompensated system The Bode plot of the above system with KK1 12 is given by following the sequence of MATLAB code given below stf39s39 sys 12ss1 grid on margin sys The plot is given below Bode Diagrams Gm Inf Pm l62123 deg at 33927 radsec 1 5c 4r 7 Ed E J 7 r T C g E 61 a E 3 m 7 5 z D lamp 7 7 141 7 r 3r 7 r T 1 wquot n H Frequency radsec MEEN 364 Parasuram Lecture 23 24 April 1 2003 Step 3 It can be seen that the phase margin of the uncompensated system is 164 degrees The required phase margin should be greater than 40 degrees Let us design for 42 degrees Therefore the phase increase that is to be provided by the lead compensator is 1 42 l647326m33 Note that 7 is used as a margin of safety Put 1 max and using the relation a M obtain the value of 0c 1 s1n max Therefore a 02948 l s1n3 3 Step 4 Using the value of 0c calculated compute the frequency where the magnitude is equal to 71010gl0c Gja 1010g 12 1 32010 1010 gm sz 1 glows 3 a z 623 Step 5 Choose the above frequency as the maximum frequency given by the relation 1 wmax 39 NE Therefore T l mmx 3 T 1 623J02948 3 T 02956 Therefore the Lead compensator is given by the transfer function Ts 1 029563 1 Ds a Ts1 00871s1 MEEN 364 Parasuram Lecture 23 24 April 1 2003 Note that K1 is not considered in this transfer function as it had already been accounted for in the plant transfer function Plotting the Bode plot of the compensated system using the following sequence we get stf39s39 sys 12 ssl comp O2956s10087lsl grid on margin syscomp From the bode plot it can be seen that the phase margin is 4447 degrees Hence the design specifications have been satisfied Bode Dlagrams Gm In1 Pm4447 deg at 41734 radsec l x l of 4L J g E C V a A C 3 t C L39 U to E E a 1i a C n l 7 7 my 7 W g 7 i i i 10 0 we to if Frequency radsec Example 3 For the system with openloop transfer function 1 design a lag compensator such that the phase margin is approximately equal to 40 and the steady state velocity error constant is 10 sec39l G0 MEEN 364 Parasuram Lecture 23 24 April 1 2003 S0 The transfer function of the Lag compensator is given by Tsl Dsa aTsl For the case of design the gain at is always considered with the plant transfer function so that the plant transfer function reduces to Gs Step 1 For the given steady state error constant obtain the value of Koc K kV 1imsGslim K0 10 40 40 S S l l 14 3 Therefore the plant transfer function reduces to G0 Step 2 Since the specified phase margin is 40 degrees let us design the compensator such that the nal phase margin of the system is 45 degrees assuming a margin of safety of 5 degrees Compute the frequency at which the Bode plot of the system gives a phase margin of 45 degrees The bode plot is given below MEEN 364 Parasuram Lecture 23 24 April 1 2003 Bode Diagrams Gm 0 dB Pm 0 unstable closed loop Phase deg Magnitude dB 1U 10 w vo m Frequency radsec From the Bode plot the frequency 0302 at Which the phase margin is 45 degrees is 0807 radsec ie a 52 0807 radsec Step 3 Using the frequency computed compute the magnitude of the system at that frequency and equate it to 2010g0c Therefore Gja lm07012010ga 3a 1035 Step 4 Choose 1 a 52 0807 200807 T 10 10 3T1239 Therefore the transfer function of the Lag compensator is given by 1239s 1 D 128253s 139 MEEN 364 Parasuram Lecture 23 24 April 1 2003 The Bode plot of the compensated system is given below Bode Diagrams 211112478 dB at 1 9696 ladsec Pm3966 deg a1081093 ladsec l l l l Phase deg Magnitude dB mom w 7200 7 7 Frequency radsec From the above Bode plot it can be seen that the phase margin of the above system is approximately equal to 3966 degrees Hence the design specifications are satisfied Recommended Reading Feedback Control of Dynamic Systems Fourth Edition by Gene F Franklin etal pp 417 442 Recommended Assignment Feedback Control of Dynamic Systems Fourth Edition by Gene F Franklin etal problems 645 648 MEEN 364 Parasuram Lecture 4 April 2 2003 EXAIVIPLES ON SAIVIPLING AND ALIASING PHENOIVIENA Example 1 Consider the following two analog signals x10 cos 27239 10t x2 I cos 27239 50t which are sampled at a rate of FS 40 Hz The corresponding discrete time signals are 10 7239 x n cos27239 ncos n 1 40 2 50 57239 x n cos27239 ncos n 2 40 2 However 57239 7139 7 7239 cos n cos27239 n cos n 2 2 2 Hence x1n x201 Thus the signals are identical and consequently indistinguishable If we are given the sampled values generated by cosgn there is some ambiguity as to whether these sampled values correspond to x1t or xzt Since xzt yields exactly the same values as x1t when the two are sampled at Fs 40 samples per second we say that the frequency F2 50 Hz is an alias of the frequency F1 10 Hz at the sampling rate of 40 samples per second Example 2 Consider the analog signal xa t 3cos1007239 t a Determine the minimum sampling rate required to avoid aliasing b Suppose that the signal is sampled at the rate of Fs 200 Hz what is the discrete time signal obtained after sampling Suppose that the signal is sampled at the rate of Fs 75 Hz what is the discrete time signal obtained after sampling d What is the frequency of a sinusoid that yields samples identical to those obtained in part c O V MEEN 364 Parasuram Lecture 4 April 2 2003 a The frequency of the analog signal can be calculated as 27239 F1 1007239 3 F1 50 Hence the minimum sampling rate required to avoid aliasing is Fs 100 Hz b If the signal is sampled at FS 200 Hz then the discretetime signal is 1007239 x n 3cos 200 n 3cosn c Ifthe signal is sampled at FS 75 Hz then the discretetime signal is xn 3 cos10 47239 n3cos n 5 3 3cos27239 2 n 3 27239 3cos n 3 d For the sampling rate of Fs 75 Hz we have F2 fFS 75f The frequency ofthe sinusoid in part c is f Hence F2 75f 25 Clearly the sinusoidal signal yt 3cos 27239 Flt 3cos 507239 t sampled at FS 75 samples per second yields identical samples Hence F1 50 Hz is an alias of F2 25 Hz for the sampling rate Fs 75 Hz MEEN 364 Parasuram Lecture 4 April 2 2003 Example 3 Consider the analog signal xa t 3cos 507239 t10sin3007r t cos 1007239 I What is the Nyquist rate for this signal The frequencies present in the signal above are F1 25 F 150 F3 50 Thus Fmax 150 Hz and according to the sampling theorem F5 gt 2an 300 Hz The Nyquist rate is FN 2me Hence FN 300 Hz Discussion It should be observed that the signal component lOsin3007239 t sampled at the Nyquist rate FN 300 Hz results in samples lOsinn39 n which are identically zero In other words we are sampling the analog signal at its zerocrossing points and hence we miss the signal component completely This situation would not occur if the sinusoid is offset in phase by some amount 9 In such a case we have lOsin3007r tt9 sampled at the Nyquist rate F N 300 samples per second which yields the samples lOsin7239 n6 10sin7239 ncost9 cos7239 nsint9 10sint9 cosn39 n 1quot10sint9 Thus if 9 0 or 11 the samples of the sinusoid taken at Nyquist rate are not all zero However we still cannot obtain the correct amplitude from the samples when the phase 9 is unknown A simple remedy that avoids this potentially troublesome situation is to sample the analog signal at a rate higher than the Nyquist rate MEEN 364 Parasuram Lecture 4 April 2 2003 Example 4 Consider the analog signal xa t 3cos 20007239 t 5sin 60007239 t 10 cos 120007239 I a What is the Nyquist rate for this signal b Suppose that the signal is sampled at the rate of Fs 5000 samples per second what is the discretetime signal obtained after sampling c What is the analog signal we can reconstruct from the samples a The frequencies existing in the above signal are HI 1000 Hz F 3000 Hz F3 6000 Hz Thus Fmax 6000 Hz and according to the sampling theorem F5 gt 2an 12000 Hz The Nyquist rate is FN 2Fmax 12000 Hz b The discretetime signal ofthe signal sampled at 5000 samples per second is n 10 coxwm 00 5000 x01 3c 2000 6500 n5sin 5000 l 3 6 3cos27239 gn5s1n27r n10cos27r EM 1 2 l 3cos27239 gn5s1n27r l n10cos27r 1gn l 2 l 3cos27239 gn5s1n27r n10cos27r gn Finally we obtain 1 2 xn l3cos 27239 n 5s1n 27239 n Since FS 5000 Hz the folding frequency is Fs2 2500 Hz This is the maximum frequency that can be represented uniquely by the sampled signal MEEN 364 Parasuram Lecture 4 April 2 2003 c Since only the frequency components at 1000 Hz and 2000 Hz are present in the sampled signal the analog signal that can be recovered is yt l3cos 27239 1000t 5sin 27239 2000t which is obviously different from the original signal xt This distortion of the original analog signal was caused by the aliasing effect due to the low sampling rate used Assignment 1 Consider the analog signal xt 3 cos 6007239 t 2 cos18007239 t a Determine the minimum sampling rate required to avoid aliasing b What is the Nyquist rate for the signal xt c Suppose that the signal is sampled at FS 2000 Hz what is the discretetime signal obtained after the sampling What are the frequencies in the resulting discretetime signal d If the sampling rate is F s 600 Hz then what is the maximum frequency that can be recovered from the discretetime signal 2 An analog signal contains frequencies up to 10 kHz a What range of sampling frequencies allows exact reconstruction of the signal from its samples b What is the Nyquist rate for the signal c If the sampling rate Fs is chosen as 10 kHz then what is the maximum frequency that can be represented uniquely by the sampled signal MEENZ A Puaslxam Leann m n u 192nm HANDOOUT E 9 e EXAMPLES ON FLUID THERMAL AND MIXED SYSTEMS Enmpk 1 Adlumzlsyslzm The fa vwmg fxglxe shm a ample maaex uf an meme furnace A pauhng uf temp131 T1 5 bang heaudm The furnace by an deems hum Weng at me u p a me mcexsT1Lhz wa s are stump131 T and AT Lhzmmlcspacxtmcesuf epachng meme am 1 Q t The m emuem me ambmltzmpen xe xsT xmce mam xnacew n Deme equanansfm The man asmmmg me atxstxansfenedbycmvechmady wah The eameneve but transfer caef cxemshd gamma ha mum walls make madewa samhzntm wwww Mm m a WW The me afhznt kmsfer Q between a snhdwall and a nd awmg We ms gvm by gm44777 a whzxe h 15 Lb cmvechve heat transfer caef cxem A 15 the area ufhzat transfer and TV mamepxeeemm mu m mampenme eepemvexy Usng39hz shave xelahmsfm39hepachng we have C Ewe arm 771 7 M d n7 a MEEN 364 Parasuram Lecture 10 ll 7 August 19 2001 Similarly applying the relation for the furnace we have dT dT Q2 mzcz 7 Chz 7 Q10 hclA 1T2 T1 hczAz T2 T3 3 Applying the relation given by equation 1 to the walls we get dT dT Q3 m3637ch37h52 42 T2 T3h3A3T3 Ta 4 Equations 2 3 and 4 represent the governing differential equations of motion for the abovedefined system State space representation Let the states of the system be defined as T 1 x1 T2 x2 5 x3 T 3 Substituting the above relation in equation 2 we have dT Cm 1 11511410 T1 dt 6 h h 5 x1 LA1XI LA1XT Chl Chl Similarly substituting the relations given by equation 5 in equation 3 we have 611T2 Chz dt Q1t 11511410 T1 hczAz T2 T3 9 h A h A h A h A 1 7 2x2 1 1C1 1 1 2 Z x2 2 2x3 Chi Chi Chi Chi Chi Substituting the relations given by equation 5 in equation 4 we get dT CM 7 1152142 T2 T3 1153143 T3 Ta 9 8 h A h A h A h A 2x3 2 2 x2 2 2 3 3x3 53 3TH ChS ChS ChS ChS NIEEN 364 Parasuram Lecture 10 11 7 August 19 2001 Rewriting equations 6 7 and 8 in matrix format we have cl 1 cl 1 0 x1 Chl Chl x1 0 0 x2 2 wigs12 hczAz x2 L 0 gm x3 Chz Chz Chz Chz x3 Chz Ta 0 hczAz h52A2 lidl3 0 115313 L C113 C113 C113 J C113 Example 2 A mixed system A simplified sketch of a computer tape drive is shown below Write the equations of motion in terms of the parameters listed below N0 friction IF Vacuum column Free body diagram of the takeup capstan T JhBi MEEN 364 Parasuram Lecture 10 ll 7 August 19 2001 where J1 is the inertial of the motor B1 is the motor damping constant Tm is the torque developed by the motor T is the tension in the string Writing the torque balance equation we have J1 dwtlBlw1 Tr1T 9 But from the gure it can be concluded that TBx2 x1kx2 x1 10 We also know that the torque developed by the motor is proportional to the armature current Hence T ktz39a l l m Substituting the equations 10 and l l in equation 9 we have J1 d 1 310 1 Tr1 Tm 0 dt 12 gtJ1Blw 1 Br1xz x1 kr1x2 x1 kz39 0 Free body diagram of the idler wheel T2 032 J2 132 F Writing the Torque balance equation we get J2 d2 3200 2 T2r2 Fr2 0 13 But again T2 Bx2 xlkx2 x1 14 MEEN 364 Lecture 10 11 7 Substituting equation 14 in equation 13 we get J2 d 2 3200 2 T2r2 Fr2 0 611 610 2 gtJ2 dt 3200 2 Br2xz xlkr2x2 x1 Fr2 0 From the gure the following relation can be concluded X1 Vla 1 X2 r2a 2 Parasuram August 19 2001 15 16 Equations 12 15 and 16 represent the governing differential equation of motion State space representation Let the states of the system be defined as x1 XI a 1 X X 2 3 2 x a 2 X 4 Substituting the above relation in equation 16 we have X1 V10 1 Z X 1 VIXZ X2 V20 2 Z X 3 FZXA Substituting the relation given by equation 17 in equation 12 we have 6101 1 5h J 13101 13r1xz xl kr1x2 x1 ktia0 J1XZB1X2 Br1r2X4 r1X2kr1X3 X1 ktia 0 Bl Brl2 kr Br1 r k J J k gtX2 iX1 X2 1X3 2X4 ia J 1 J J 1 1 1 Substituting the relation given by equation 17 in equation 15 we have 17 18 19 20 MEEN 364 Parasuram Lecture 10 11 7 August 19 2001 610 2 J27Bzw 2 Br2Xz xlkr2x2 x1 Fr2 0 gtJZX4BZX4 Br2r2X4 r1X2kr2X3 X1 Fr2 0 B BZ BVZZX4 r 2F 21 2 J2 J2 k k gtX4 iX1 X2 iX3 J2 J J2 Rewriting equations 18 19 20 and 21 in matrix format we have 0 r 0 1 0 0 X1 kr1 Bl Br12 kr1 Bar2 X1 k 0 X2 J1 J1 J1 J1 X 2 J1 in 22 X3 0 0 0 r2 X3 0 0 F X4 Ci Brlrl BzBrzz X4 0 r 2 L J2 J2 J2 J2 J J2 Example 3 A uid system Introduction Fluid capacitor A uid capacitor is shown in the following gure Qc gt Pr Cf The pressure in a uid capacitor must be referred to a reference pressure Pr The volume ow rate Qc is given by Q C f d where Cf 1s the u1d capac1tance t Fluid inertor The symbolic diagram of a uid inertor is shown in the following figure Q1 PIE P2 lVlEEN 364 Parasuram Lecture 10 ll 7 August 19 2001 The elemental equation for the inertor is 1in d where I is the uid inertance For frictionless incompressible ow in a l 12 uniform passage haVing cross sectional area A and length L the inertance I p AL where p is the mass density of the uid Fluid resistor The symbolic diagram of a uid resistor is shown below QR gt wwwgg 0 P1 WQW Q Q i 4 P2 Rf The elemental equation of an ideal resistor is P12 RfQR39 Problem Develop the inputoutput differential equation relating the output pressure to the input pressure for the uid system shown below Figure 99 A simple uid R I C system For the uid resistor we have P12 2 RfQR39 23 MEEN 364 Parasuram Lecture 10 ll 7 August 19 2001 For the inertor we get P Q 23 d 24 t For the uid capacitor 1P 25 C QR f dt Writing the pressure balance equation we have PS Plr P12 P23 P37 d gtPS RfQR 1 3 P3 51133 dip C 1 3 P dt f dtz 3 d213 dP gtCf1 a 1fo 61 P3PS 26 gtPS Rfo Equation 26 represents the governing differential equation of motion for the uid system shown State space representation Let the states of the system be de ned as P3r x1 a 52123 27 x2 dt From the above relation we get ya x2 28 Substituting the relation given by equation 27 in equation 26 we have d213 dP Cf a Rfo d P3rPS gtCfIxzRfox2 x1 PS MEEN 364 Parasuram Lecture 10 ll 7 August 19 2001 l Rf 1 gt 362 x1 x2 PS 29 Cf 1 Cf Rewriting equations 28 and 29 in matrix format we get 0 l 0 M Rrx1 LP 30 m C f1 I x2 C f1 Assignment 1 The sewage system leading to a treatment plant is shown The variables qA and qB are input ow rates into tanks 1 and 2 respectively Pipes 1 2 and 3 have resistances as shown Derive the state equations 1A 1B LI gt All A2 Tank 1 h1 hz Tank 2 R l Pipe 2 R1 qz R3 11 gt q3 gt Pipe 1 Pipe 3 2 The temperatures of the side surfaces of the composite slab shown below areT1 and T2 The other surfaces are perfectly insulated The cross sectional areas of the two parts of the slab are A1 and A2 and their conductivities are k1 and k2 respectively The length of the slab is L a Find the equivalent thermal resistance of the slab and express it in terms of the thermal resistances of the two parts A1 k1 T1ltI gtTz k2 MEEN 364 Parasuram Lecture 10 11 7 August 19 2001 3 Write the equations of otion for the h nging crane shown below Assume that the driving force on the hanging crane is provided by the motor mounted on the cab with one of the support wheels connected directly to the armature shaft The motor constants are K6 K and the circuit driving the motor has a resistance Ra and no inductance The wheel has a radius r Texas A amp M University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr Alexander G Parlos Lecture 9 Mixed Systems Electromechanical System Operation The objective of this lecture is to give you some background on modeling systems with components in mixed energy domains such as electromechanical systems and discuss the principles of their operations In many engineering applications we must combine elements from several different energy domains such as electrical and mechanical domains In order to combine these domains we need coupling devices that convert one kind of energy or signal to another ie electrical to mechanical The general term transducer will be used for ideal coupling devices In cases where signi cant amounts of energy or power is involved these will be called energy converting transducers Otherwise they will be referred to as signal converting transducers EnergyConverting Transducers We start our discussion with ideal transducers where no energy is stored or dissipated in the elements Two types of such transducers are discussed TranslationaltoRotational Mechanical Transuders A diagram of such a transducer is shown in Figure l where n is the coupling coe icient relating output motion to input motion The equations for this transducer are 93 71111 vv 1 Figure l A symbolic diagram of an ideal translationalto rotational mechanical transducer Why Hint Power in must equal power out Examples of such transducers include pulley and cable systems level and shaft mechanisms and rack and gear mechanisms Electromechanical Energy Converters Again we have two types of electromechanical transducers as shown in Figure 2 Translational mechanical electromechanical transducers such as solenoids and linear motors and rotational mechanical electromechanical transducers such as rotaty electric motors or electric generators The same rotational device can operate as a motor and a generator depending on the direction of the power ow or the sign of the input power Also given in Figure 2 are the equations governing the operation of such devices Although these equations are intended primarily for use with direct current DC devices they could be applied to alternative current AC de vices as long as the root mean square rms values of the voltage and current are used Electric current and magnetic elds interact in many ways however three laws associated with these interactions are very important in understanding the operation of most electromechanical systems Figure 2 A symbolic diagram of an ideal t 39 t a 39 39 and b rotational mechanical Figure 3 Toroid wrapped with N turns of wire Magnetic Field Proportional to Current The rst relation states that an electric current establishes a magnetic eld The strength of the magnetic eld at a given point depends on the intensity of the current the materials and the geometry involved A typical case is shown in Figure 3 Here a toroid of radius R and of material with permeability u is wrapped with N turns of wire carrying t amperes The eld strength at the center of the toroid is given by 7 I B i 27TRN2 3 Law of Motors Force is Proportional to Current and Magnetic Field The second important law states that a charge q moving with a velocity v in a magnetic eld with intensity B experiences a force F given by F qv x B If the moving charge is composed of a current i amperes in a conductor of lenght l meters arranged in at right angle to a eld of strength B tesla then the force is at right angles to the plane of i and B 7 and has the magnitude F Bli Newtons 4 This equation forms the basis for converting electrical energy to mechanical energy and it is called the law of motors Law of Generators Voltage is Proportional to Velocity and Magnetic Field The last important law relates conversion of mechanical energy to electrical energy and it is the opposite of the previous law It states that if a charge is moving in a magnetic eld and is forced along a conductor an electric voltage is established between the ends of the conductor If a conductor of length l meters is moved at a velocity 1 meters per second through a constant eld of B tesla at right angles to the direction of the eld then the voltage between the ends of the conductor is given by e Blv Volts 5 This equation forms the basis for converting mechanical energy to electrical energy and it is called the law of generators DC motor actuators A common actuator used in control systems is the DC motor It provides rotary motion and it can also provide translational motion A sketch of a DC motor is given in Figure 4 The components of the motor are 1 housings and bearings 2 stator magnets permanent or electromagnets 3 rotor 4 commutator and 5 brushes The brushes force current through the wire wound around the rotor The rotating commutator causes current always to be sent through the armature a collection of conductor windings so that it will produce the maximum torque in the desired direction If the armature current direction is reversed 4 Figure 4 DC motor sketch the torque direction is reversed also The operation of a DC motor is usually expressed by the equation relating the torque developed in the rotor in terms of the armature current ta and the voltage generated called back electro motive force or emf as a result of rotation in terms of the shafts rotational velocity 6m Thus T Ktia7 A m V e Kedm A 1 V In consistent units Kt Kg AC motor actuators Another device used for electromechanical energy conversion is the AC induction motor Suppose we construct a rotor of conductors with no commutator or brushes Now suppose that the stator magnetic eld can be made to rotate at high speed say 1800 rpm The mov ing eld will induce currents in the rotor which will in turn be subject to forces according to the motor law These forces will cause the rotor to turn so that it follows the rotating magnetic eld This is a simpli ed principle of operation of AC induction motors Figure 5 Steadystate torquespeed curve for squirrelcage AC induction motor operating at constant supply frequency and voltage AC induction motor of the squirrel cage type operate well only at speeds close to their no load speed A typical torque versus speed curve of such a motor operating a constant supply voltage is shown in Figure 5 This curve accounts for friction effects and other losses Thus it is necessary to know the motor ef ciency in order to determine the motor current Given the motor ef ciency of a single phase AC induction motor 77m the motor current rms value is given by 1 T03 39m 7 8 SignalConverting Transducers These devices could be energy converting transducers with negligible load or specially designed devices that convert one type of signal to another Ex amples of such devices include tachometer generator linear velocity sensor pieZO electric force sensor etc Examples of Mixed Systems Example 1 Modeling of DC Motors Figure 6 DC motor a electric circuit of the armature b freebody diagram of the rotor Find a set of equations for the DC motor with the armature driven by the electric circuit shown in Figure 6a Assume the rotor has inertia Jm and friction coef cient b In the electrical part of the system we consider the back emf of the electrical circuit In the mechanical part of the system we need to include the motor torque developed The free body diagram of the rotor is shown in Figure 6b If we apply the law of conservation of angular momentum we have Jm ma Tmt Twat 9 The motor torque is given by Tmt Ktz39a f 10 and the load in this case is the rotor inertia The only torque generated by the load is the friction or damping torque expressed as Twas b9mt 11 Applying Kirchoff s voltage law on the electrical system we have d39a t La 1d l Raw vat memos 12 If we de ne the state vector as X05 l0mt9mtiatlT7 13 and the input as ut vat we obtain the following state space represen tation of a DC motor model Axt But 14 where 0 1 0 A 0 JET 15 K8 Ra O E E and 0 B 0 16 L La Note in most instances the electrical part of the system will be much faster than the mechanical component As a result an applied voltage will result in an almost instantaneous current flow Therefore it is often feasible to neglect the inductance La in equation 12 The resulting equation is now of second order h f w an Equation 17 indicates that the mechanical friction and the back emf are indistinguishable They both act to damp the motion of the DC motor Reading Assignment Read pages 51 56 the textbook Read examples Handout E9 posted on the course web page Texas A amp M University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr Alexander G Parlos Lecture 12 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 standard 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 sys tems 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 Linearization 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 X05 fX7u 1 In equation 1 the derivatives of the state are relate to the state and or the control through a nonlinear relation f In order to linearize this equation we must rst determine the equilibrium values of the system The equilibrium values for the state x0 and control uo are such that the derivative of the state vector is zero That is we can compute the equilibrium values by solving X0 0 fX0 uo Equation 2 has two unknowns Therefore we must choose arbitrarily the value of uo and solve equation 2 for the equilibrium state x0 We now expand the nonlinear equation in terms of the perturbations from these equilibrium values that is we let xlttgt x0 6x0 lt3 and not uo 6ut7 lt4 then we can write the following linear approximation to the nonlinear dy namics 1 x0 6Xt m fx0 uo F6xt G6ut 5 where F and G are the best linear ts to the nonlinear function f at the point x0 uo Canceling the equilibrium from both sides of equation 5 results in 6x05 F6xt G6ut 6 which is a linear model approximating the nonlinear dynamics at the point X0 Me If an analytical expression for f is available then the best linear ts F and G can be obtained through differentiation as follows 8f F 7 7 8XlX07u07 2 Figure 1 Laboratory scale magnetic ball levitatori and 8f G IX07u0 In case when an analytical expression is not available numerical differentia tion is performed 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 shows a laboratory scale magnetic levitator where one electro magnet 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 Newton s law as WW mm may 9 Figure 2 Model for a ball levitator where the force fmx is caused by the eld of the electromagnet Theoret ically 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 Fiom the experimental curves we infer that at the current value of 1392 600 mA and the displacement 951 the magnetic force fm just cancels the gravity force mg 82 x 10 3 N The mass is 84 x 10 3 kg and the m 8662 acceleration of gravity is 98 Therefore the point 9511392 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 equilib rium point 0512392 as follows fmx1 605 1392 6139 m fmx1 2392 K1605 K2621 10 The linear gains KI and K can be computed as follows K1 is the slope or 4 Figure 3 Experimentally determined force curvesi derivative of the curve in Figure 3 for 1392 600 mA around the point 951 This is found to be about 14 K2 is the change of force with current at the value x 951 This is found as N 122 x 103 42 x 103 N N Ki N W N 0394 Z39 11 So the linearized force expression becomes fm6x6z39 82 x 103 146 04622 12 Considering that Sight the equation of motion 9 becomes 84 x 10 3693t 146xt0461 t 13 or 65130 166769505 41mm 14 which is the linearized equations of motion about the equilibrium point We can select the state vector as Xt 6xt6jct and the control ut 6ut This selection results in the following state matrices 0 1 0 7 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 example 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 1 and 1 The amplitude scaling is performed by de ning the scaled variables for each element If xt 8359505 16 then dct 31330 a39 t Sz t 17 We select the scaling SE to accomplish our scaling objective outline above Time Scaling Variables involving time are usually measured in units of seconds Some times it is convenient to express time in other units We de ne a scaled time to be 739 wot 18 such that if t 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 E Clowe do 19 9395 and dgx 2 dgx w 7 0d72 If the original system is in state variable or state space form Fx Gu 21 then the time scaled system is expressed as 39t 1F 1G 22 x 7 x 7 u do do Read example 225 on page 76 of the text Note For more on operating points read the handout A5 For an exaple describ ing load or impedance matching read handout A4 Reading Assignment Read pages 68 76 the textbook Read examples Handout BIZ and Hand out A5 posted on the course web page 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