DYNAMIC SYST AND CONTROL
DYNAMIC SYST AND CONTROL MEEN 364
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MEEN 364 Parasuram Lecture 12 August 28 2001 HANDOUT E12 EXANIPLES ON LINEARIZATION Example 1 Consider the system shown below m The governing differential equations of motion for the above system is given by mrkr klO mr9 2 mgcos9 0 l mr9 2mr9 mgsin9 0 2 where 10 is the initial length of the spring and k is the stiffness constant of the spring Note that the above differential equations are nonlinear in nature First to find the equilibrium point equate all the derivative terms to zero Therefore equation 2 reduces to mgsinO 0 gtsin9 0 gt9 mr There 90 0 is one equilibrium point for the above system Following the same procedure for equation 1 we get kr kl0 mg cos9 0 gtkr kl0 mg 0 mg kl0 k gtr V0 3 Therefore r r0 is the equilibrium value for the variable r MEEN 364 Parasuram Lecture 12 August 28 2001 Expanding each term in equation 1 by Taylor s series about the equilibrium point and neglecting the higher order terms we have mrkr kl0 mr9 2 mgcos9 0 gtmrkr kl0 mr9 2 7 3mr9 2 7 gtr r0 imr9 2 gt9 0 Ef e 3 111 89 ne 2 mgcos9 mn mgcos9 9 9 o0 9 39 51 gt mrkr kl0 mg 0 gtmlr r0H kr r00 4 Following the same procedure for equation 2 we get mr9 2mr9 mgsin9 0 gtmr9 3mr9 gtr r0imr9 9 0 mm 3r mu m 9 9D 9 0 39 9 3 E E 2mr9 7 2mr9 rr 0 2mr9 9 0 Egg 2 I Wm B 9 rru 9 0 9 0 3 mgs1n9 mn mgs1n9 9 9 o0 9 9u 39 536 gt mro 9 mgO 0 5 Equations 4 and 5 represents the linearized differential equation of motion for the above system MEEN 354 Puaslxxm Leann 12 AugustZX 2mm Exampk 2 Cmaderthz deckamagmhc mpmsmsymm daman glxe An deckamagwlxs laca39zdntthz uppupm fth Exp1xan systzm ULLhnnthz deckamagwhc fmcef we dame m suspend m um ball um am this ample deckamagmhc suspmsm system is essenna y mwmkahle Hence mamk annual xs mtiypensahle As a gap smsax a mmmumpmbe uf eddycmemtypexsplaced belwwthz bu The deckmnagm has an inductance L39 and a mum 39 Us sppmxxmmanfm the ammath rum Th2 amen 15 x on apexahngpmnlmd x39xs39hzvmahle Th 55 uf39hzbalh 39 x whzxe x mm szuhngpmnlmd x39 mm unable Th2 elemmnagwhcfmce xs gven e a mm ms 9 Wm In 5 a X z mum ufmmmn by f 4 Wu tkv 5 mm Dam a mama gume mam cp m anm the gure Wmng gnvemmg exznhal equananfm the mm urcuL we gal L lt6 mm anba mungmumm39s secandlnw afmmmn wehm MEEN 364 Parasuram Lecture 12 August 28 2001 gtmxki 1 mg 7 x Equations 6 and 7 represent the governing differential equations of motion for the electromagnetic suspension system Note that equation 7 is nonlinear in nature In order to linearize the equations the operating point has to be calculated Equating all the derivative terms in equation 6 and 7 to zero we get RI0 v Z I mg 0 where 10 and X0 are the operating points Therefore equation 6 reduces to di L R 1 139 v dt 0 gt Li Rz39 0 8 dt Using the Taylor series expansion for equation 7 we get 2 2 2 mxkI 1 k 1 gti1 IOi ki gtxg X0 mg X0 Bil xg 7 Bx x 7 1171 g g 11Iu g XU xg XU i Zki 39i10 1 1 39x X0 115U xg xg 2 111 g H Xu gtmx 2k 2k 2 X 021 0 x 9 X03 0 Equations 8 and 9 represent the linearized differential equation for the system defined MEEN 364 Parasuram Lecture 12 August 28 2001 Assignment Derive the equations of motion for the inverter pendulum problem and linearize the equations about the equilibrium point Do not assume small angle approximation m MEEN 364 Parasuram Lecture 21 August 21 2001 HANDOUT E22 EXAIVIPLES ON STABILITY ANALYSIS Example 1 Determine the stability ofthe system whose characteristics equation given by as s6 435 3s4 2s3 s2 4s4 The above polynomial satis es the necessary condition for stability since all the coefficients are positive and nonzero Writing the Routh array we have s6 1 3 1 4 s5 4 2 4 0 S4 543 12 41 41 44 10 2 4 4 4 32 40 24 44 S3 22 22 0 2 5 2 2 2 20 Egg 24 30 Sz 2 5 2 2 2 12 SI E3 5 42 15 3 4 0 0 4 S E 15 We conclude that the system has roots in the right half plane since the elements of the rst column are not all positive In fact there are two roots in the right half plane since there are two sign changes In other words two closed loop poles of the system lie in the right half plane and hence the system is unstable MEEN 364 Parasuram Lecture 21 August 21 2001 Example 2 Determine the stability of the following polynomial 613 SS 53quot 1133 2332 283 12 Writing the Routh array we have Since the entire row is zero we construct an auxiliary equation by taking the coefficients of the previous row ie 6113 332 12 Differentiating the above equation with respect to s we get 61013 ds 6s 1 So the Routh array is continued by taking the coefficients of equation 1 Since there are no sign changes in the first column of the Routh array there are no roots in the right half plane However since one entire row in the Routh a1ray was zero there are roots in the imaginary axis The roots in the imaginary axis can be obtained by solving the auxiliary equation Therefore 3szl20 gtsz40 gtsij2 MEEN 364 Parasuram Lecture 21 August 21 2001 Example 3 Consider the system shown below The stability properties of the system are a function of the proportional feedback gain k Determine the range of k over which the system is asymptotically stable sl ss ls6 39 The characteristics equation for the system is given by kS 10 ss ls6 33 532 k 6sk0 Therefore the corresponding Routh array is For the system to be stable it is necessary that all the elements in the first column of the Routh array must be positive Therefore gt0 and kgt0 gtkgt75 and kgt0 gtkgt75 MEEN 364 Parasuram Lecture 21 August 21 2001 Example 4 The statespace representation of a system is given as twat y 1 21x For the system to be stable the poles of the system should lie in the left half plane In other words all the real poles should be negative or the real parts of complex poles must be negative The poles of the system are nothing but the eigenvalues of the A matrix of the system The MATLAB code is shown below a 7 12 l O vd eiga The result is O970l 09487 02425 O3l62 The diagonal elements of the matrix d are eigenvalues of the system and columns of the matrix v represent the corresponding eigenvectors Note that since all the eigenvalues are negative the system is stable MEEN 364 Parasuram Lecture 21 August 21 2001 Assignment 1 Determine the stability of the system whose characteristics equation is given by as SS 334 233 632 6s 9 2 Find the range of the controller gains k k1 so that the PI feedback system is stable 9 kr V s s1s2 39 3 If the A matrix of the system is given by 0 4 0 0 1 4 0 0 A 5 7 1 15 0 0 3 3 Determine the stability of the system Recommended Reading Feedback Control of Dynamic Systems Fourth Edition by Gene F Franklin etal 7 pp 157 166 Recommended Assignment Feedback Control of Dynamic Systems Fourth Edition by Gene F Franklin etal 7 problems 339 340 341 342 MEEN 364 Parasuram July 13 2001 HANDOUT M2 DIFFERENTIATION AND INTEGRATION Section 1 Differentiation De nition of derivative A derivative f x of a function fX depicts how the function x is changing at the point X It is necessary for the function to be continuous at the point X for the derivative to exist A function that has a derivative is said to be differentiable In general the derivative of the function y x also denoted dydX can be defined as Q dfx lim meme fx dx dx AHO Ax This means that as AX gets very small the difference between the value of the function at X and the value of the function at X AX divided by AX is defined as the derivative Derivatives of some common functions xl d Et22t 0652 logx d a ear 2 agar sin x cos x cos x sin x General rules of differentiation 1 The derivative of a constant is equal to zero If y c dy d 0 dx 51x where c is any arbitrary constant MEEN 364 Parasuram July 13 2001 2 The derivative of the product of a constant and a function is equal to the constant times the derivative of the function If y c x dy d df dx dxcfx 6 dx Examplel Ify8Xthen dy d d 8 8 81 8 dx dx x deC 3 The derivative of the sum or difference of two functions is equal to the sum or difference of the derivatives of the functions If y x i gX d d d d 61 006 i gx 006 iagx Example 2 Ify 8XXz then dy d 2 d d 2 8 8 8 2 dx dxx x dxx deC x 4 Product rule The derivative of the product of two functions is equal to the first function times the derivative of the second plus the second function times the derivative of the first If y fXgX i i 1 dx dx f xgx f x dx gx gx dx f 96 Example 3 If y xex then dy d d d xe x ex e x xe e l xe e dx dx dx dx Example4 If y XZSlnX then MEEN 364 0 gt1 Parasuram July 13 2001 dy d 2 2 d d 2 dx x s1nx x dx s1n x s1n x dx x x2 c0sxsinx 2x x2 c0sx2xsinx Division rule If y fXgX then d d i x dx dx gx gov2 Example 5 If y ex X then i x x i d yii M xe e 1 xe e dx dx x x2 x2 x2 Chajn rule Ify fu and u gX that is if y is a function of a function then dy dx du dx If y ft and X gt that is if y and X are related parametrically then d ydt dx dx dt Example6 If y t3 and X 2t2 then 61 d 3 2 er oh ziza dx dxdt iatz 4t 4 Higher derivatives The operation of differentiation of y fX produces a new function y f x called the first derivative If we again differentiate y f x we produce another new function y y f 39x called the second derivative If we continue this process we have MEEN 364 Parasuram July 13 2001 y f x y f x imx y f x d zmx dx y f x d 3fx dx n n 61quot yfx nfx dx Example7 Ifyx5 then 5x4 dx 2 d yi i5x420x3 abc2 dx dx dx dsy d dzy d 3 2 20x 60x de dx dxz dx d4y d d3y d 2 60x 120x abc4 dxdx3 dx dsy d d4y d 120x 120 dxs dxdx4 dx dsy d dsy d 120 0 abc6 dxdx5 dx MEEN 364 Parasuram July 13 2001 Partial differentiation A partial derivative is the derivative with respect to one variable of a multivariable function assuming all other variables to be constants For example if y Xy is a function depending on two variables X and y then the partial derivative of f with respect to X is obtained by assuming the variable y to be a constant and taking the 3 der1vat1ve of f w1th respect to X Thls 1s represented as a f x y x Example 8 If y xsint then xsint sintx sint In this example since the partial derivative with respect to the variable X is required the variable t is assumed to be a constant and the derivative with respect to X is obtained by following the general rules of differentiation Example 9 If 2 X2y3 then 32 a 2 3 2 a 3 2 2 2 2 3 3 By ayOCy x ayy xy xy General rules of partial differentiation 0 If the function 2 is dependent on two variables X and y ie if z xy then 322 3 32 axz a 322 3 32 ay2 5 322 3 32 3x3y g g 322 3 32 3y3x 3 0 If 2 xy and X ht y gt then the total derivative of 2 with respect to t is given by MEEN 364 Parasuram July 13 2001 dz 1 Eli ifiy dt dtfx y Bx dt By dt Example 10 Ifz xy and X cost y sint then 21 Z zai i 3xyicostiwisint dt dt Bx dt By dt Bx dt By dt y sint xcos t ysint xcost MEEN 364 Parasuram July 13 2001 Section 2 Integration Introduction The basic principle of integration is to reverse differentiation An integral is sometimes referred to as antiderivative De nition Any function F is said to be an antiderivative of another function f if and only if it satisfies the following relation F f where F 39 derivative of F Note that the definition does not say the antiderivative it says an antiderivative This is because for any given function f if there are any antiderivatives then there are infinitely many antiderivatives This is further explained with the help of the following example Example 11 For the function x 3x2 the functions F x x3 Gx x3 15 and Hx x3 38 are all antiderivatives off In fact in order to be an antiderivative off all that is required is that the function be of the form Kx x3 C where C is any real number This is so because the derivative of a constant function is always zero so the differentiation process eliminates the C F39x3x2 fx G39x3x2 fx H39OC 3x2 fx Although integration has been introduced as an antiderivative the symbol for integration is l So to integrate a function x you write I f xdx It is very essential to include the dx as this tells someone the variable of integration De nition The expression l x dx Fx C where C is any real number means that F x is an antiderivative of x This expression represents the inde nite integral of fx MEEN 364 Parasuram July 13 2001 The C in the above de nition is called the constant of integration and is absolutely vital to inde nite integration If it is omitted then you only nd one of the infinitely many antiderivatives Some properties of inde nite integrals mm gx dx Ifmdxi goodx 0 Icfxdxcjfxdx where c is any real constant Some of the common integration formulae Jde c 2 dex x c 2 3 Ixzdx x c 3 n 1 x Ix39 x c n1 l I dx lnxc x Isin xdx cosxc Icos xdx sinxc The constant c can be found if some more information about the antiderivative is given This is explained with the aid of the following example Example 12 Find the function whose derivative is f x 3x2 1 and which passes through the point 25 In other words the function f has to be integrated with respect to X So Fx jfxdx 3 ldx j3x2dx jidx gtFxx3 xc It is given that the function FX passes through the point 25 ie when X 2 F2 5 MEEN 364 Parasuram July 13 2001 FxF2232c10c gt510c gtc 5 Therefore the function FX is given by Fxx3 x 5 De nite integration This is very much similar to the inde nite integration except that the limits of integration are speci ed Since the limits are speci ed there is no need to put the constant of integration In other words ZF Hm j mwFo Example 13 Evaluate the following integral dex 1 2 2 2 2 2 J xdxzi ampQ2li 1 2 1 2 2 Example 14 5 5 5 5 5 I I2x 3dx I2xdx j3dx x20 3x0 0 0 0 gt 1 52 02 35 30 gt125 015 0 gt140 MEEN 364 Parasuram July 13 2001 Integration by parts Formulae If u x and V gX and if f and g are continuous then Iudvuv Ivdu 1 For de nite integrals jfxg39xdx wagon goo make The following example illustrates how the integrationbyparts formula works Example 15 Evaluate the integral I Ixe dx Let u X and dV ede 2 So dulandvex 3 Since v Idv Ie dx e Thus substituting equations 2 and 3 in equation 1 we have Ixe dx xe Ie ldx c gtIxe e c MEEN 364 Parasuram July 13 2001 Example 16 Find Ie cos xdx u e dvcosxdx x du e v sin x Substituting the above relation in equation 1 we get Ie cosxabcex sinx Iex sinxdxc 4 Consider again the integral Ie sin xdx Let u e dvsinxdx due v cosx Thenwe have Ie sin xdx e cos x Ie cos xdx c 5 gt Ie s1n xdx e cos x Ie cos xdx 0 Substituting the second equation of equation 5 in equation 4 we get Ie cos xdx 6x sin x ex cos x Ie cos xdx 0 gt 218x cos xdx e sinx cos x c l gt Ie cosxdx Ee s1nxcosxc MEEN 364 Parasuram July 13 2001 Leibniz rule The fundamental theorem of integral calculus states that if f is a continuous function on the closed interval a b then d t E fxdx f0 foraStSb The Leibniz rule can be derived from the above fundamental theorem of calculus If 172 10 Ifxtdx 6 a0 The final relation which outlines the Leibniz rule is given by 11t ibffx odx Mfg x rdxibr fbr r iarfar r 7 dt dtam imat dt dt This rule is useful when one needs to find the derivative of an integral without actually evaluating the integral The rule is further explained with the aid of the following example Example 17 Given 2 1t Icostx2 dx 72 d Evaluate I t dt The Leibniz rule can be applied to evaluate the derivative without actually evaluating the integral Comparing the above integral with equation 6 we have fx t costx2 at t bt t2 fx t costx2 x2 sintx2 MEEN 364 Parasuram July 13 2001 Substituting the above relations in equation 7 we get 1t x2 sintx2 dx 2tcostt2 2 lcost t2 22 gt 00 ij sintx2 dx 2tcost5 cost3 L Hospitals rule if the numerator and f x gx denominator both tend to zero that is lim f x lim gx 0 The rule states that if the f x x L Hospitals rule gives a way to evaluate limits of the form lim is numerator and the denominator both tend to zero then the limit of the quotient f g In other words x 96 equal to the limit of the quotient of the derivatives mamp lim ax H gx H g x Example 18 sin x Evaluate 11m 16 x It can be seen that both the numerator and the denominator tend to zero as X tends to zero Therefore applying the L Hospitals rule we have sin x d Sin x cos x 1 lim limdx lim 1 xgt0 x xgt0 d 1 1 dx MEEN 364 Parasuram July 13 2001 Assignment 1 Calculate the derivatives of the following functions with respect to X a y2x3 4 x5 b y 2x3 c yx2 sinx 2 Calculate the partial derivatives g y and for the following functions x a y x2 sin2 I b y x3t6 tanxsintcost 3 Calculate the following indefinite integrals analytically and verify the result Using MATLAB a Icosz xdx b 1x2 sin xdx c Ixtan xdx 4 Evaluate the following definite integrals and verify the result using MATLAB Z a x 7x3 5x2 6x 9dx 1 quotA b Isin xdx 0 5 Evaluate the derivative with respect to time of the following integral by means of Leibniz rule a 1t J sinaxz dx 0 MEEN 364 Parasuram July 13 2001 6 Evaluate the following limits x2 2x a 11m H0 s1nx x b x2 1 sinx3 0 lim H0 sinx2 NOTE This handout is not a comprehensive tutorial for differentiation and integration This just deals with the very basics of differentiation and integration It is advisable always to go through some MATH book for various other techniques of performing differentiation and integration MEEN 364 Parasuram Lecture 1 August 8 2001 HANDOUT E1 EXAIVIPLES ON SIGNALS AND SYSTEMS Signals A signal is de ned as any physical quantity that varies with time space or any other independent variable or variables Mathematically we describe a signal as a function of one or more independent variables Examples of signals 1 Current from a current sensor 2 The variation in pressure in the cylinder of an IC engine during operation 3 The variation in the number of shoes sold in a certain shop during the month 4 The variation in temperature at noon on successive days at a holiday resort Although all these signals have time as their independent variable the signals in examples 1 and 2 are basically different from those in examples 3 and 4 In examples 1 and 2 the signals representing the variables are present at all instants of time These types of signals are known as continuous time signals In examples 3 and 4 the signals are only available at discrete time intervals These type of signals are called discretetime signals It should be noted that the processes producing the variables in examples 3 and 4 are continuous processes the resort temperature has a value at all times as does the number of shoes sold However because the measurements are taken at discrete intervals they give rise to discretetime signals The discretetime signals arise due to the sampling of a continuous signal Selecting values of an analog signal at discretetime instants is called sampling A11 measuring instruments that take measurements at a regular interval of time provide discretetime signals For example counting the number of cars using a street every hour or recording the value of gold every day results in discretetime signals The continuous time signals can also be de ned as those signals that can take any value at any given instant of time Whereas the discretetime signals are de ned only for discrete instants of time That is the time variable of these signals can take only certain values The difference between a continuous and a discretetime signal is explained further with the help of the following gure MEEN 364 Parasuram Lecture 1 August 8 2001 Continuous Vs Discrete Signal Amplitude 6 7 8 9 10 0 1 2 3 4 5 Time The continuous signal in the above figure is represented by xt sin t To obtain the discretetime signal values of the continuous time signals are taken at equally spaced time intervals of 04 seconds In other words the continuous time signal is sampled at the rate of amp samples per second or 25 samples per second So the discrete time signal is represented as xn sin04n Therefore in general if the continuous time signal is represented as xt f t then the discretetime signal which is obtained by sampling the continuous time signal every T seconds is represented as xn fnT Quantization of the discrete signals gives rise to digital signals or discrete valued signals A quantized signal assumes only discrete amplitude values In other words in these signals both the amplitude and time variable can take only certain values A digital signal is shown in the figure given below MEEN 364 Parasuram Lecture 1 August 8 2001 Digital Signal Amplitude 4 Sample Number Systems A system may be de ned as a physical device that performs an operation on a signal It can also be regarded as a process that transforms one signal into another The input to a system and the output from a system are signals A system can also be de ned as a group of physical components assembled to perform a specific function It may be electrical mechanical hydraulic pneumatic thermal or a combination of any of these systems In general a system may be represented as Input signal Output signal gt System All existing systems change with time and when the rates of change are significant the systems are referred to as dynamic systems A car riding on a road can be considered to be a dynamic system especially on a crooked or a bumpy road The input variables could be the throttle position position of the steering wheel or the road conditions like the slope and roughness The choice of the output variables is arbitrary and is determined by the objective of the analysis The position velocity or acceleration of the car or the average fuel rate or the engine temperature can be selected as the outputs MEEN 364 Parasuram Lecture 1 August 8 2001 Another example of a system is an ampli er which modi es the amplitude of the input signal by a gain factor at every time instant This system is shown below X0 Yt Ampli er Gain a In general the relationship between the input signal to the system Xt and the output signal from the system Yt is given by Yt aXt where a is a constant System modeling and types of models of a system operating under speci ed A To A J and control complex systems one must therefore establish an adequate mathematical model of the system The term model means a set of differential equations that describe the dynamic behavior of the system The main objective of system analysis is to predict relevant performance characteristics I 39 39 System models can be classi ed into various types Some of the types are considered below Continuous time systems discrete time systems and digital systems Systems where the input and the output signals are continuous in nature are continuous systems Example of a continuous time system or an analog system is a sensor Continuous systems can be expressed in the form of differential equations Analog inputI Analog Analo g Butput Signal System Signal Systems where both the input and the output signals are discrete in nature are called discretetime systems These systems are expressed in the form of timedifference equations Discrete input Discrete output Discrete Time Signal System Signal MEEN 364 Parasuram Lecture 1 August 8 2001 If both the input and output of a system are digital in nature then the system is called a digital system Example of a digital system is a computer Digital input Digital output Signal Digital Signal System Linear and Non linear systems The concept of linear system can be illustrated by an example For example one is listening to some instrumental music from a tape via an ampli er One instrument plays the tune this instrument stops and another instrument plays the same tune then both instruments play the tune together When the instruments play together one would expect the sound produced to be the combination of the sounds of the individual instruments and if the system is linear this will be the case But we all know that this is not the case Hence the basis of linear system is that if the inputs are superimposed then the responses to these individual inputs are also superimposed In other words If an arbitrary input x1t produces the output y1t and an arbitrary input xzt produces an output y2t then if the system is linear input x1txzt will produce an output y1tyzt This principle is called the principle of superposition All linear systems satisfy the principle of superposition One example of a linear system is the simpli ed model of the suspension system of an automobile springmassdamper system Example of a nonlinear system is a pendulum of a clock Nonlinear systems do not satisfy the principle of superposition Time domain and frequency domain models For any simple system the mathematical model is obtained by applying the laws of physics The result is a differential equation This equation is nothing but the time domain model of the physical system By taking the Laplace transform of the differential equation we get the frequency domain model of the same physical system So time domain and frequency domain models represent the same physical system in two different domains For example consider the suspension system of an automobile which can be simplified into a massspringdamper system The governing differential equation of motion for the system is given by m x c x kx f f is the input to the system This equation represents the time domain model of the system under consideration Taking the Laplace transform of the above equation we get MEEN 364 Parasuram Lecture 1 August 8 2001 mszXs chx kXs Fs 1 ampGUZ 3 ms csk F Gs represents the frequency domain model of the same physical system considered MEEN 364 Parasuram Lecture 897 August 8 2001 HANDOUT E8 EXAIVTPLES ON MODELLING OF ELECTRICAL ELECTRONTECHANICAL SYSTEMS Note that the time dependence of variables is ignored for all manipulations Example 1 An electrical circuit Consider the circuit shown below R L 1 T Writing the Loop equation for the above circuit we have Va Ri L i idt0 1 dt C Weknowthat i dt Therefore substituting the above relation in equation 1 we have R 2 dt Equation 2 represents the governing differential equation of the circuit shown above State space representation Let the states of the system be defined as q x19 3 q x2 From the above relations we get x1x2 4 MEEN 364 Parasuram Lecture 897 August 8 2001 Substituting the relations given by equation 3 in equation 2 we get dq q 2 Va Ld qR dtz dt C 1 V szRx2 x1 C 1 R 1 gtx x x V 5 2 LCILZL 0 Rewriting equations 4 and 5 in matrix format we have M l ll gl l l 1L5 1 6 X2 LC L 9 L If the output of the system is the voltage drop across the capacitor then the output equation can be written as y x19 1 i C C 1 x1 gt y 5 0L 7 Equations 6 and 7 represent the statespace form of the circuit Example 2 An electrical circuit Note that dql dqz 1 1 8 1 dt 2 dt MEEN 364 Parasuram Lecture 897 August 8 2001 Writing the loop closure equation for the loop ABFGHA we have 1 Vm EJ11 lz dt R111 039 9 Substituting equation 8 in equation 9 we have Knq192Rl 0 10 C dt Writing the loop closure equation for the loop BCDEFB we get 51139 1 1 R212LEEIzzdtEIOZ zldt 0 11 d2 d 2 gtL 2 R2 12 i 0 I d2 dt C C Equations 10 and 11 represent the governing differential equations for the circuit shown State space representation Let the states of the system be de ned as ql x1 92 x2 12 qz x3 From the above relations the following equation can be deduced x2 x3 13 Substituting the relations given by equation 12 in equation 10 we have Vm 311612 R10 C dt gtVm x1 x2 R1x1 0 gtx1 xlx2 Vm 14 RIC RIC R1 MEEN 364 Parasuram Lecture 897 August 8 2001 Similarly substituting the relations given by equation 12 in equation 1 l we get 612 dtz dqz 2ampi dt C C L R 0 2 2 l gtLX3R2x3 Ex2 Ex1 0 2 sz gtx x 3 LC2L3 1 15 LCX1 Rewriting equations l3 l4 and 15 in matrix format we have V l6 i LC LC From the circuit it can be seen that the output is given by l Vm EjzzdtRlzp l dq Vouz Eq2 R161 1 d 1 from equation 10 1n the above relatlon we get I Substituting the value of R1 1 q1 q2 V V out qu m C 2 l 0142 Ex2 Ex1 Vm9 0 x2 1114 17 Therefore equations 16 and 17 represent the statespace form of the above circuit MEEN 364 Parasuram Lecture 897 August 8 2001 Example 3 An electromechanical system Consider the following electromechanical system shown X 4Lx The electromechanical system shown above represents a simpli ed model of a capacitor microphone The system consists a parallel plate capacitor connected into an electric circuit Capacitor plate a is rigidly fastened to the microphone frame Sound waves pass through the mouthpiece and exert a force fst on plate b which has mass M and is connected to the frame by a set of springs and dampers The capacitance C is a function of the distance X between the plates The electric field in turn produces a force fe on the movable plate that opposes its motion Kinematics stage Let the movable plate b move a distance X units Then the velocity and the acceleration of the plate is given by x x respectively It can be seen from the figure that the current i ows through the electric circuit in the counter clockwise direction Kinetics stage First let us consider the electric circuit MEEN 364 Parasuram Lecture 897 August 8 2001 Writing the loop closure equation for the above circuit we get all39 1 L R39 39dtV 18 d l Cll Since i dt equation 18 reduces to dig altz dqq L R V 19 d C Free body diagram of the movable plate of massM Jam lax p gt cx fe gt Writing the Newton s second law of motion we have 2F ma gt kxcxf2 fSt mx gtmxcxkxf2 fst 20 Equations 19 and 20 represent the governing differential equations of motion for the electromechanical system considered The force fe is de ned as Z q f2 28 A where A is the surface area of the plates and 8 is the dielectric constant of the material between the plates MEEN 364 Parasuram Lecture 897 August 8 2001 Since fe is nonlinear in nature equation 20 is a nonlinear equation The linearization of a nonlinear equation is explained in the handout on linearization Once the equation is linearized then it can be represented in the statespace form This section is dealt with in detail in a later handout Assignment 1 Find the differential equations for the circuit shown below and put them in state variable form R1 R2 R3 2 Consider the schematics of an electromechanical shaker as shown below This system consists of a table of mass M and a coil whose mass is m A permanent magnet rigidly attached to the ground provides a steady magnetic eld ie the motion of the coil through the magnetic eld induces a voltage in the coil that is proportional to its velocity The passage of current through the coil causes it to experience a magnetic force proportional to the current Derive the equations governing the dynamics of this system 3
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