(Solution Manual) Modeling of Dynamic Systems with Engineering Applications

Silva, 2

Engineering

Solution Manual

2nd Edition

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Solution Manual for Modeling of Dynamic Systems with Engineering Applications, The downloadable file consists of 302 pages, and the total size of the file is 4.34 MB. Purchase now only for highly discounted price.

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Solutions Manual for Modeling of Dynamic Systems with Engineering Applications, 2e by Clarence Silva (All Chapters) Chapter 1 Introduction to Modeling Solution 1.1 Integrated: Concurrent, all aspects (physical domains such as mechanical, electrical, fluid, and thermal ) are considered simultaneously . Unified: Uses analogous (similar) procedures t o model and analyze various domains . Unique: Leads to a single result (leads to the “best” result, usually through optimization). Systematic: Uses clear ly identified sequence of steps (no confusion in the procedure). _______________________________________________________________________ Solution 1 .2 (a) Muscle contraction, body movements , body temperature, heart rate , waste (b) Decisions, profits, finished products (c) Electric power, profit, customer satisfaction, pollution rate (d) Front wheel turn, direction of heading , noise level, accident, pollution level , ride quality (vehicle acceleration, jerk, etc.) (e) Motions (position, velocity, acceleration) of the joints and the end effector (hand) of the robot , trajectory error, collision , speech (f) Bridge oscillation, fault (due to material, aging, poor design, workmanship or maintenance), overload, noise . Other inputs: (a) Food, forces (impact, etc.), heat (b) Demand for product, economic environment (c) Outside temperature (d) Road disturbances, engine forces/torques, fuel rate, air flow rate (e) Loading on t he end effector, voice commands (f) Aerodynamic forces MODELING OF DYNAMIC SYSTEMS 2 System Typical Input Typical Outputs Human body Neuroelectric pulses (due to external stimuli), desired stimulus; distractions Muscle contraction, body movements, cursing, unlawful activity Company Information; stock ; price increase of raw material, drop in product selling price Decisions, finished products, wastage, product quality deterioration Power plant Fuel rate, deterioration in environmental conditions, increase of operating costs Electric power, pollution rat e, drop in production, noise Automobile Steering wheel movement, road disturbances, accidents, weather deterioration Front wheel turn, direction of heading, ride quality deterioration, noise, wrong motion Introduction to Modeling 3 Note : In this table, the undesirable quantities are given in Italics. __________________________________________________________________ _____ Solution 1 .3 (a) This system has sub -processes in the physical domains of thermal, electrical, mechanical, and fluid, working together. Hence, it is a multi -physics system. (b) The component s of the system include the sun, solar panel, inverter, and home (e.g., appliances such as refrigerator, microwave oven , hair dyer, and vacuum cleaner, light bulb s, fans, heat panels ). Some of the sub -processes in the system are listed below. Energy transfer from the sun to the solar panel: Thermal (radiation) Solar panel: Thermo -electric energy conversion Inverter: Electrical to electrical energy conversion Light bulb: Electrical energy conversion into light and thermal energy Fan: Electrical energy conversion into mechanical energy , noise, heat Refrigerator: Electrical energy conversion into thermal energy , mechanical energy (e.g., compressor), and light (bulbs in the refrigerator) Micro wave oven: Electrical en ergy conversion into thermal energy, mechanical energy, light (bulb). Many of the processes are nonlinear and rather complex. Some subsystems are distributed (e.g., solar panel). There are time -varying parameters (e.g ., fluid flow and drying in the refrigerator and cooking of food in the microwave oven). (c) An energy flow diagram for thi s system is shown in Figure S1.3 . Note the energy conversion among different physic al domains, in this multi -physics dynamic system. Robot Voltage to joint motor, collisions, noise in input signal Joint motions, effector motion, wrong heading (response error) MODELING OF DYNAMIC SYSTEMS 4 Figure S1.3 : Energy flow diagram of the system. _____ __________________________________ _______________________________ Solution 1.4 A system is an entity of interest that is demarcated by a system boundary (a physical or virtu al). Typically it has inputs, outputs and other quantities , which may vary with time. It may be represented by a set of algebraic equations or differential equations A system in which the rates of changes of the system/ response variables (outputs, or generally state varia bles) are not negligible, is a dynamic system. Also, in a dynamic system, the present output of the dynamic system depends not only on the present input but also on the past input. It is represented by a set of differential equations. __________ ______________________________________________________________ Solution 1.5 System analysis , simulation and e valuation; system design and modification; fault detection and diagnosis; system control. Thermal/Photo Radiation Electrical Solar Panel Inverter Electrical Light Bulb Fan Refrigerator Microwave Oven Photo, Thermal Mechanical , Thermal Thermal , Light Thermal, Mechanical , Light Introduction to Modeling 5 Advantages of experimental modeling : A real physical s ystem is used; realistic pr actical inputs (excitations) can be used; feasible /practical system modifications are known and can be incorporated into the “identified” model . Disadvantages of experimental modeling : A physical system or its prototype is needed ; test instrumentation and proced ures can be costly; testing is slow ; test /system modification s can be difficult and are costly . ________________________________________________________________________ Solution 1.6 (a) 1. Identify the system (incl uding its purpose, system boundary, key variables, and the boundary condition s) 2. Perform a functional analysis (study the function of each component or element of the system) , determine their interactions, and draw a structural diagram for the system where the “structure” of the system (e.g., are two components connected in series or in parallel) is clearly indicated 3. Represent each functional aspect of each component by one or more basic elements , properly interconnected, and consistent with the system struc ture, as identified in Step 2 4. Write element physical (constitutive) equations 5. Write continuity (node) and compatibility (loop) equations 6. Eliminate unknown variables. (b) Parameter values of the model; initial conditions; excita tion (input) signals. Note : Boundary conditions are not needed (implicit) for lumped -parameter model. Boundary conditions have to be specified/included in the model itself, in a distributed -parameter model . (c) Interfac e details (i.e., complete details of how the subsystems are interconnected). ________________________________________________________________________ Solution 1.7 Advantages of the MR approach: • Simple • Fast • Going from B to C (i.e., MR) is sensitive to changes in parameters of the physical system (and of the accurate model). Advantages of the DM+MR approach: • Much insight of the system is gained during DM. • The final degree of approximation can be chosen/changed as needed. • The MR process is more systematic (easier) and analytical th an the direct SM process. MODELING OF DYNAMIC SYSTEMS 6 • Easier to correlate the parameters of the real system to those in model B. ________________________________________________________________________ Solution 1.8 1. Errors in the model itself (model errors) including the model structure (how the components are interconnected) and the model parameters 2. Inaccurate or incomplete information on the in puts to the model (incompletely or inaccurately known inputs; presence of input noise and disturbances , which are stochastic —random and not completely known ) 3. Incompletel y-known (e.g., un -measurable), inaccurately processed, or inaccurately interpreted outputs. ________________________________________________________________________ Solution 1.9 What aspects of the system should be included in its model depends on the purpose of the model , allowable model complexity, and knowing that other (un -modeled) aspects have no signif icant influence on the modeled aspects. The shown two models may be used to analyze the lateral vibration dynamics in 2 -D or 1 -D of the HDD. Other aspects that may be included in an HDD model : 1. Dynamics in the remaining degrees of freedom (dof) of the unit (e.g., the third rectilinear dof —vertical and the three rotatory degrees of freedo m 2. Rotational dynamics, inertia and flexibility of the read-write swing arm and sensing head Electrical, electronic dynamics electro -magnetic -mechanical aspects of actuation and controller 3. Thermal aspects (heat transfer, cooling). ________________________________________________________________________ Solution 1.10 The shown model may be used to analyze the r igid-body dynamics of the vehicle, in the heave, lateral, roll, pitch and yaw modes. If the purpose of the model is the design or control of the overall elevated guideway -vehicle system, the following aspects of the system may be in cluded as well in a more detailed model: 1. Flexibility and damping of the vehicle 2. Other degrees of freedom (and the distribute d aspects) of the vehicle 3. Inertia, flexibility, and damping (and the distributed aspects) of t he elevated guideway 4. The nature and effect of multiple vehicles moving on the gu ideway 5. Dynamic interactions between the vehicle and the guideway 6. All the actuators , including the e lectro -magnetic -mechanical aspects of the actuators Introduction to Modeling 7 7. Controller . ________________________________________________________________________ Solution 1.1 1 (a) The model includes inputs, outputs, state variables, system boundary, and the environment. The initial conditions and boundary conditions of the system are missing, and are needed for complete analysis or simulation. (b) The analytical model of a dynami c system is represented by differential equations of the output/state variables with respect to time. (c) Causality dictates what ar e proper inputs and what are proper outputs (and states) of the system. ________________________________________________________________________ Solution 1.12 A universal model will be too complex, costly and difficult to handle (analyze, simulate, use, etc.). It may overshadow some key characteristics of the system while hi ghlighting relatively unimportant characteristics. The best model is the least complex model that meets the specific purpose. The best model depends on practical constraints (e.g., whether physics is known; how quickly we need a model; purpose of the model —e.g., control, design; cost). _______________________________________________________________________ 1 Chapter 2 Basic Model Elements Solution 2.1 (a) (i) Capacitor Element The constitutive equation is Cdv dti= (i) Since power is given by iv, the energy in a capacitor may be expressed as , Energy (Electrostatic) E Cv=1 22 (ii) Also, vtvCidtt () ()=+− −01 0 (iii) Observation 1: Energy of the capacitor may be expressed by v alone and not i alone (see equation (ii). Observation 2: If v is input and i is output, then a finite step input will cause an infinite output (see equation (i)), which is not physically realizable. This is not the case if i is input and v is output . Observation 3 : The voltage v at any future time t can be completely determined f rom the initial voltage (at 0− ) and the applied current from 0− to t (see equation (iii)). This is known to be a way to define state variable. All three of these properties justify that v, not i, is the natural output variable for a capacitor , and it properly represents the “dynamic state” of a capacitor. (ii) Inductor Element The constitutive equation is Ldi dtv= (i) Energy (Electromagnetic) E Li=1 22 (ii) Also, itiLvdtt ()()=+− −01 0 (iii) Observation 1: Energy of the inductor may be expressed by i alone and not v alone (see equation (ii). Observation 2: If i is input and v is output, then a finite step input will cause an infinite output (see equation (i)), which is not physically realizable. This is not the case if v is input and i is output . Observation 3 : The current i at any future time t can be completel y determined from the initial current (at 0− ) and the applied voltage from 0− to t (see equation (iii)). This is known to be a way to define state variable. All three of these properties justify that i, not v, is the natural output variable for an inductor, and it properly represents the “dynamic state” of a n induc tor. MODELING OF DYNAMIC SYSTEMS (b) (i) This choice of state variables will generate a “unique” state space model . (ii) This choice will provide a unified approach for modeling multi -physics (i.e., multi -domain, consisting of the physical domains mechanical, electrical, etc.) systems . In a “unified” approach we can exploit the analytical similarities (analogies) of different physical d omains , when developing the underlying procedures. (iii) With this choice, the s tate variable s will have physical meaning s. (c) (i) Required number of state variables (ii) Number of independent energy storage elements (iii) Required number of initial conditions (iv) The order of the input -output differential equation (v) Order of the characteristic polynomial (denominator of the transfer function, in linear time -invariant systems). ________________________________________________________________________ Solution 2.2 (i) Mass, spring, damper, force source, velocity source (ii) Capacitor , inductor, resistor, current source, voltage source. (a) lumped , (b) distributed , (c) lumped , (d) distributed , (e) distributed , (f) distributed. ________________________________________________________________________ Solution 2.3 (a) • We can determine similarities and differences in the dynamic behavior of the system . • By studying one type of system we can learn about other analogous systems (this is the basis for the “unified” approach in Mechatronics) . • By exploiting the analytical similarities (analogies) of different physical domains in a multi -physics (i.e., multi -domain or mixed) system, w e can develop a unified approach to analyze and d esign a multi -physics systems, w hich is particularly useful in electro -mechanical systems and M echatronics. (b) Inertia element (c) Yes. No, it is analogous to a stiffness (strictly, flexibility —the inverse of stiffness ) element. ________________________ ________________________________________________ Solution 2.4 (a) Force, current, fluid flow rate, heat transfer rate. Basic Model Elements (b) Velocity, voltage, pressure, temperature (c)-(f) No. These are state variables (or, output variables of the elements). To change them instantaneously, infinite inputs would be needed, which is unrealistic. (g) No. Thermal systems have only one type of energy (thermal energy). Two different types of energy would be needed (and back and forth energy exchange between them) in order to execute free (natural) oscillations. ________________________________________________________________________ Solution 2.5 2(1 ) 2rrPEb− =  Divide by the corresponding time increment t , and take the limit as 0t→ . We have , 2(1 ) 2dr r dP dt Eb dt−= (i) Also, 34 3Vr= Differentiate wrt t. We get, 224343dV dr drQ r rdt dt dt = =  = (ii) Substitute (ii) in (i): 2 2 (1 )42r dPQrEb dt−= or, 42 (1 )r dPQEb dt−= → Fluid capacitance 42 (1 ) frCEb−= ________________________________________________________________________ Solution 2.6 (a) 2 dV dhQrdt dt== (i) P gh= (ii) Hence, dP dhgdt dt= Substitute (i): 2dP Qgdt r= Henc e, 2r dPQg dt = We have, 2 frCg = MODELING OF DYNAMIC SYSTEMS Now substitute tanhrR=− We get, 2()tanfhCRg =− (b) For a cylindrical tank, 2= . Hence tan→ and the second term in the parentheses. We have , 2 fRACgg == with 2AR= ________________________________________________________________________ Solution 2.7 (a) () V V P= () dV dV P dPQdt dP dt= =  Compare with fdPCQdt= Hence () fdV PCdP= (b) Consider an elemental water slice of height h (See Figure S2.7 ). R h h r Elemental Water Slice Figure S2.7: Elemental water s lice. Elemental volume 2V r h  = with tanhrR=− Hence volume of water is , Basic Model Elements 2 2 3 0 00tan( ) ( ) ( )tan 3 tanh hhhhV r dh R dh R  = = − = − − or 33 tan()3 tanhV R R= − − (i) Now , dV dV dh dP dh dP= (ii) with 1 dhP ghdP g=  = (iii) Differentiate (i) w.r.t. h : 22 1tan ( 3)( ) ( ) ( )3 tan tan tandV h hRRdh  = − − − = − Substitute in (ii) with (iii): 2()tandV hRdP g =− This is the same re sult that we obtained in Problem 2.6 . ________________________________________________________________________ Solution 2.8 Suppose that the pressure in the cylinder increases by P in time period t . From the formulas for the corresponding radius increase r and the length increase L , we have, 2(1 / 2) r r P t Eb t  − = and (1/ 2 ) L rL P t Eb t  − = In the limit, as 0t→ , we get: 2(1 / 2) dr r dP dt Eb dt−= (i) and (1/ 2 ) dL rL dP dt Eb dt−= (ii) Internal volume of the cylinder, 324 3V r r L=+ (iii) The increase of the internal volume of the cylinder will accommodate additional fluid flow into the cylinder (we neglect the liquid compressibility here). The rate of this volume increase equals the fluid volume flow ra te Q into the cylinder. Hence, by differentiating (iii) we get, 2242dV dr dr dLQ r rL rdt dt dt dt  = = + + (iv) Substitute (i) and (ii) into (iv). We get , MODELING OF DYNAMIC SYSTEMS 2 2 12 (2 ) (1 ) ( )22r dP rL dPQ r r L rEb dt Eb dt  = + − + − With fdPCQdt= we have, 2 2 312 (2 ) (1 ) ( )22 12(2 )(1 ) ( )22fr rLC r r L rEb Eb rr L LEb   = + − + − = + − + − ________________________________________________________________________ Solution 2.9 In thermal systems there is only one form of energy, the heat transfer energy. Unlike in mechanical, electrical, and fluid systems, the rate of this energy represents the through - variable in a thermal system . The across -variable is temperature. In this manner , in a thermal system , the “thermal capacitance” relating the heat transfer rate to the temp erature rate is defined. This is an A -type element whose state variable is the temperature. But since there is no other form of energy, an element that would store such energy does not exist. In other words, an element whose constitutive equation relating the rate of the “heat transfer rate” to the temperature does not exist. Consequently, a T -type element does not exist in a thermal system. As a consequence of the absence of two types of energy storage elements, no oscillations in the free natural response are found in a purely thermal system (because, free oscillations result from the back -and-forth transfer of energy between two elements) . ________________________________________________________________________ Solution 2.10 Consider a cylindrical surface of length l at radius r. Its area is 2rl . From the standard law of 1 -D conduction, t he heat transfer rate out to the exterior, in the radial direction is , 2dTQ rlkdr=− Note : If the temperature gradient (in the increasing direction of r) is positive, the heat transfer will be into the interior. This is the reason for using t he –ve sign in this equation (as in the standard equation of 1 -D conduction), which considers the heat transfer out to the exterior . On dividing by r, we have , 2Q dTlkr dr=− Integrate wrt r from di to do (i.e., in the increasing direction of r), noting that Q is independent of r. We get [ln / 2 ln / 2] 2 [ ]o i o i Q d d lk T T  − =− − , or, Basic Model Elements ln 2 [ ]o oi idQ lk T Td=− − Manipulate: ln 2oi ioddT T Qlk−= Hence, the conductive resistance, ln 2oi kddRlk= ________________________________________________________________________ Solution 2.11 (a) Mass of the bar Al= Rate of “increase ” of thermal energy in the bar b pdTAlcdt= Note : 0bdT dt in this example , since the temperature of the bar decreases . Rate of heat transfer “into” the bar from the atmosphere at the free end ()c a bh A T T=− Note : This i s also -ve in this example , since the thermal energy leaves the bar . Energy balance gives , ()b p c a bdTc Al h A T Tdt =− Now , with , hpC c Al== thermal capacitance 1 c cRhA== convective resistance we have , b c h a bdTR C T Tdt=− Tb = state variable; Ta = input variable . (b) Biot number , /k c c cR h A lhBiR kA l k= = = Substitute numerical values: MODELING OF DYNAMIC SYSTEMS 0.2 100.00.16 0.1125.0Bi= =  It follows that the present state -space model (1st order) is not adequate. Note : smaller Bi means smaller Rk, which corresponds to faster conductive heat transfer. This means heat is transferred quicker over a given length, making the temperature more uniform, which determines the accuracy of the model. This may be interpreted as follows: When Rk is smaller for a given length of heat transfer, the time of heat transfer is shorter. This means, for a given time, heat can transf er over a longer distance , making the temperature in that region uniform . Hence, when Bi is smaller, the size of the subdivided lumped elements of the heat transfer object can b e made larger without sacrificing the model accuracy . In summary, the smaller t he Bi the more accurate the lumped -parameter model. ________________________________________________________________________ Solution 2.12 (c) Four types of fluid capacitance: 1. Due to the compressibility (bulk modulus) of the contained fluid, if the fluid is a liquid 2. Due to the compressibility of the contained gas (e.g., under isothermal or adiabatic conditions ), if the fluid is a gas 3. Due to the flexibility of the container that contains the fluid 4. Due to the gravity head of the liquid in the tank. Case 1: When the liquid in a container is subjected to pressure, the liq uid volume decreases. This allows more liquid to enter the container (thereby generating “capacitance”) . Case 2: When the gas in a container is subjected to pressure, the gas volume decreases. This allows more gas to enter the container. B oth isothermal (i.e., constant temperature) and adiabatic ( i.e., no heat transfer) conditions can be analyzed to determine expressions for the corresponding fluid capacitance. Case 3: A flexible contain er carrying fluid will expand due to its flexibility, when the fluid pressure increases. The extra volume that is created in the container, as a result, will allow more liquid to enter the container. Case 4: The height of a liquid column in a container inc reases when more liquid enters the container. Then, the pressure at the base of the liquid column increases as the column height increases. This process obeys the law of fluid capacitance. (d) Suppose that a small volume V of water e nters the tank. Then, the water level rises by h . Tank are a at the height h is ( tan )A a h b  = −  Then, the volume increment ( tan ) V A h a h b h   = = −  The volume flow rate into the tank is equal to the rate of change of V. Hence, Basic Model Elements 0lim ( tan ) tdV V dh dhQ A a h bdt t dt dt→= = = = −  (i) The water pressure at the bottom of the tank is, P gh= (ii) Differentiate (ii) wrt t: ( tan )dP dh ggQdt dt a h b==− {after substituting (i) Rearrange: ( tan )b a h dPQg dt −= We have, ( tan ) fb a hCg −= ________________________________________________________________________ Chapter 3 Analytical Modeling Solution 3.1 (a) Lumped ; (b) Distributed (distributed element mainly for the guideway) ; (c) Lumped ; (d) Distributed ; (e) Distributed (distributed element mainly for the aircraft body) ; (f) Distributed. _______________________________________________________________________ Solution 3.2 1. The end conditions of the actual spring and the lumped model have to be consistent 2. The assumed velocity distribution (e.g., linear) for the distributed -mass spring w ill limit the validity only to the particular mode shape that is compatible with that of the assumed velocity distribution (e.g., linear velocity distribution and first mode approximation ) 3. The frequency of operation should be limit ed to the considered mode of operation of the actual spring (e.g., for low -frequency operation, an operating frequency limit below the first natural frequency of vibration) . ________________________________________________________________________ Solution 3.3 First we will obtain a two -degree -of freedom lumped model for the spring alone and then add the end mass M. The exact first two natural frequencies of the heavy spring are : 12sk m= and 23 2sk m= A half of the spring has double the stiffness. Hence, we have the two -degree -of-freedom lumped -parameter model (for the spring alone) shown in Figure S3.3(a). m 2k m 2k 1y 2y Figure S3.3: (a) A two -degree -of-freedom model. It can be shown (from Newton's 2nd law) that the equations of motion (for free vibration) of the system are , 11 220 4 2 0 0 2 2 0yy m k k yym k k−         +=        −          Its characteristic equation (the condition for non -trivial sinusoidal solution for the system equations) is , MODELING OF DYNAMIC SYSTEMS ()()2 2 2 2 242det 0 4 2 4 0 22m k km k m k k k m k   −=  − − − = −  Roots for ω of this equation provide the natural frequencies. Define (for convenience of analysis) the normalizing parameter 02k m= Note : Physically, this is the natural frequency of a one -degree -of-freedom oscillator of mass m and stiffness 2 k. We have: ()()2 2 2 2 4 0 0 020    − − − = Now (for convenience) define the non -dimensional frequency parameter 2 0p =  We have: ()() ()22 1 1 0 2 1 1 0p p p p  − − − =  − + + = Roots: ()22 1 2 1 4 2p   + + −= 22 1 4 1 2 ++= or, 22 122 1 4 1 2 1 4 1; 22pp    + − + + + +== For natural frequency equivalence, equate these two roots to the exact natural frequencies of the heavy spring, given by (in the normalized form with respect to 0 ): 2 2 2 2 1 2 0 2 2 2 2 2 2 04 2 8 8 9 9 9 4 2 8 8ss ssk m m m k m k m m m k m       =  = = =  = = where sm m= (1) We have , 2 22 1 4 14  + − + = (i) 2 2 92 1 4 14  + + + = (ii) (i)+ (ii): 25214  += (iii) (ii) - (i): 2241  += (iv) (iii)  (iv): 22 1 5 4 41 += + 236 64 9 0 − + = Roots:  = 1.624, 0.154 (2) Analytical Modeling From (iii): 221 5 4 +=   = 0.212 or 0.688 (from (2)) (3) Hence, from (2) and (3), we have two p ossible solutions for the equivalent lumped model: 0.212s mm= and 0.344s mm= or, 0.688s mm= and 0.106s mm= Note : We can easily verify that both models have the same two natural frequencies. The two possible lumped -parameter models are (with the end object) shown in Figure S3.3 (b). 0.212sm 2k 2k 0.344smM+ 0.688sm 2k 2k 0.106smM+ Figure S3.3 : (b) Two equivalent lumped models. ________________________________________________________________________ Solution 3.4 (a) F; (b) T; (c) F; (d) T; (e) T. ________________________________________________________________________ Solution 3.5 (i) Required number of state variables (ii) Number of independent energy storage elements (iii) Required number of initial conditions (iv) The order of the input -output differential equat ion (v) The number of eigenvalues/poles (vi) The order of the characteristic polynomial (denominator of the transfer function, in the linear time -invariant case ) (a) One; (b) Two; (c) Two; (d) Two. _______________________________________________________________ _________ Solution 3.6 MODELING OF DYNAMIC SYSTEMS 1. Under steady operation at an operating point with only small variations about that point 2. When nonlinearities are weak 3. When a nonlinear model is too complex, and only a crude study needs to be conducted 4. When the model can be validated through physical experiments on the real system. (i) (a), (b), (d) ; (ii) (c); (iii) (a) ________________________________________________________________________ Solution 3.7 All six models are linear, 2nd order and time -invariant. All models have just one output (y). Other dynamic characteristics are identified below. (a) This is a completely c oupled system, and has two inputs and one output. The two state equations cannot be solved separately (i.e., the equations are coupled) . The response of one state is dependent on the other state response. There is no feed - forward characteristic in this sys tem. (b) This is a completely coupled system , and has one input and one output. Even though the 2nd state equation does not have an input term, the response of the 2nd state depends on the input (because it depends on the first state, which in turn depends on the input). There is no feed -forward characteristic in this system. (c) This is a completely coupled system. This system does not have an input. Hence the given model provides the “natural” or “free” or “unforced” response of the system. The output depends on both states. (d) The system has one input. The first state is completely uncoupled from the second state. Furthermore, the output depends on the first sate only. Hence, we are able to treat this equation as representing a 1st order system with state variable x1 and input u and solve it separately and completely. The 2nd state equation, however, is not uncoupled from the 1st state equation. Specifically , the driver of the 2nd state equation is the first state variable x1 (which depends on the input u). In summary, x1 and y are independent of the 2nd state equation while x2 is dependent on x1. There is no feed -forward characteristic in this system. (e) This is a completely coupled system, and has two inputs. Furthermore, the system has a “feed -forward” characteristic. Particularly, the output is directly and instantaneously fed by the input. The matrices A, B, C and D of the six models are given in Table S3. 8. Analytical Modeling Table S3.7 : Model matrices.     2 3 1 0( ) 1 2 0 01 -1 0 -1 2 3 1( ) 1 2 01 -1 0 23( ) 1 01 -1 2 0 1( ) 1 0 01 -1 0 2 3 1 0( ) 1 2 3 01 -1 0 -1a b c d e                                             System 00A B C D ________________________________________________________________________ Solution 3.8 (a) A =   − 0121 ; B =   20 ; C = 1 2− (b) Two. ________________________________________________________________________ Solution 3.9 The free -body diagrams are shown in Figure S3.9 Figure S3.9(a): FBD of mass. Figure S3.9 (b): FBD of spring. Spring k v Mass m Velocity v Applied Force f(t) Spring Force fk 0 fk fk MODELING OF DYNAMIC SYSTEMS (a) Method 1: Pick 1 2k x f x v= Constitutive equations : kf kv= ()k mv f t f=− State Model: 11 2200 1 0 1xx kuxx mm      =+      −       11 221/ 0 01yx k yx   =       (b) The order = t wo (c) The same state equations are still valid , but now the output equation is , 1 210xyx= (d) Two. Method 2 (This is not our preferred method) : (a) Pick 1 2x x x v= The state equations may be written as: xv= () mv f t kx=− State -space Model: um xx mk xx  +    −=  10 01 0 21 21     =  21 21 1001 xx yy (b) The order = t wo Analytical Modeling (c) The same state equations are valid here, but now the output equation is ,  = 210xxky (d) Two. ________________________________________________________________________ Solution 3.10 In this system there are two inertia elements and three spring elements —a total of five energy -storage elements. However, the three springs are not independent. The states (or motion s) of any two springs completely determines the state (or motion ) of the third spring . This indica tes that the system (model) is fourth -order . The state vector is chosen as  TTT2 1 2 1=x To develop the state -space model, we first formulate the state -space shell by writing the constitutive equations as fo llows: Newton’s 2nd law gives , 2 1 11 TT I +−= 2 2 2 3I T T=− − (i) Hooke’s law gives , ) (1 2 2 21 1  −== k Tk T  (ii) The first, third, and fourth equations above are already in the final state -equation form. Only the second equation has to be further manipulated to eliminate the auxiliary variable T3 (which is not a state variable ). To accomplish this we write the compatibility equation — the disp lacement compatibility relation at inertia I2 (this essentially is also the velocity compatibility or the traditional loop equation): 33 22 11 kT kT kT=+ (iii) Note : In this example, the continuity equations (torque balance) are already sa tisfied, as seen in Figure 3.10 (b). Hence they need not be explicitly written. Next, torque T3 in (i) is substituted in terms of T1 and T2, using (iii). This gives the remaining state equation (modified second equation of (i)): 33 2 2 1 2 121kkI T Tkk   =− − +        The system matrix of the resulting state model is , MODELING OF DYNAMIC SYSTEMS 11 33 2 1 2 2 1 221100 110 0 1 0 0 0 00II kk I k I k k kk−      − − + =            −A (iv) The output (displacement) vector is , T kT kT kT  + = 22 11 11, y (v) This corresponds to the following output -gain matrix:     = 2 11 1 1000100 k kkC (vi) Note : There is no B matrix in this example bec ause it concerns free (unforced) motion and there is no input. ________________________________________________________________________ Solution 3.11 (a) The mechanical circuit diagram of the vehicle model, with the free-bodie s of the elements separated, is shown in Figure S3.11 . The n odes and the loops are indicated in the figure. Note : The inertia forces 12 and mmff have to be marked in the direction opposite to the acceleration (or velocity) of the mass, to be consistent with D’Alemberts’s equation, as shown in the figure. Also, in the figure, the mass elements are shown as “nodes,” for the sake of brevity (which is what is normally done) . Instead, next to each mass, a node may be placed showing all the forces applied to the node whose net value is applied to the inertia and balanced by its inertia force. This force (through -variable) would go into the mass (inertia) eleme nt at the action point and come out at the reference point of the mass (inertial reference). (b) Loop Equations: Five loops are identified in the figure, of which only for (any four of these) are independent (e.g., loop 1 is obtained by combining loop 2 and loop 3). In f act, some more loops can be identified (e.g., by combining loop 2 and loop 4; or by combining loop 3 and loop 5; or by combining loop 4 and loop 5), but they will not be independent of the first four loops . The “increases ” of velocity across the el ements are indicated by arrows in Figure S3.11 . This is traditional. However, the positive direction of a loop (indicated by its loop arrow) is traditionally, the counter -clockwise (ccw) direction, which is not followed in this figure. Analytical Modeling The loop equations (going i n the clockwise —cw direction) are given below. Loop 1: 1 2 1 2 ( ) ( ( )) ( ) ( ) 0v t v v t v v v+ − + − + − = Loop 2: 11 ( ) ( ( )) ( ) 0v t v v t v+ − + − = Loop 3: 1 2 1 2( ) ( ) 0v v v v+ − + − = Loop 4: 11( ( )) ( ( )) 0v v t v v t− − − = Loop 5: 2 1 2 1( ) ( ) 0v v v v− − − = It is seen that all the loop equations are a utomatically satisfied. This happens because we did not use separate variables to represent the change in velocity (across -variables) for different elements in the system. Instead, we have used the “absolute” velocities at the joints of the elements. Also, the velocities of the masses and the source element (a velocity source) are measured with respect to a common inertial reference having zero velocity. Hence, they are also “absolute” velocities. Node Equations: Ten nodes are ident ified in Figure S3.11 . Take the forces acting to the right to be positive. The node equations for the isolated 10 nodes are given below. Node 1: 11 0bv f f f+ − = Node 2: 11 0cff−= Node 3: 11 0dbff−= Node 4 (mass 1m): 1 1 1 1 1 0c d p q mf f f f f+ − − − = Node 5: 21 0p ff−= Node 6: 21 0bqff−= Node 7: 22 0cff−= Node 8: 22 0dbff−= Node 9 (mass 2m ): 2 2 2 0c d mf f f+ − = Node 10 (Inertial reference): 12 0v m mf f f− + + = Note 1: There are only 9 independent node equations. The equation of node 10 can be obtained by combining the previous 9 node equations, and it serves as a check. Note 2: 1mf and 2mf are the inertial forces of the masses 1m and 2m , respectively. These forces are transmitted to the ground reference in some way (not directly, and he nce shown by dotted lines) . Similarly, the actuator (source) force fv is transmitted to the ground reference. Note 3: Forces are vectors. Hence their directions have to be properly taken into account (particularly in the node equations). Constitutive Equations: Mass 1m : 1 1 1 m m v f= Mass 2m : 2 2 2 m m v f= Spring 1k : 1 1 1 ( ( ) ) f k v t v=−