Mechatronics Exercise: Modeling, Analysis, & Control of an Electrohydraulic Valve-Controlled Servomechanism
|
|
- Roderick Mathews
- 6 years ago
- Views:
Transcription
1 Mechatronics Exercise: Modeling, Analysis, & Control of an Electrohydraulic Valve-Controlled Servomechanism 11
2 Introduction Although a wide variety of detailed hydraulic control schemes are in use, a useful classification is that of pump-controlled versus valve-controlled systems. Pump-controlled systems are usually relatively high power (above 1 or 2 hp) applications, where efficiency is economically significant and response speed requirements are modest (less than 1 Hz frequency response). 111
3 Valve-controlled systems are faster but are generally quite inefficient. For a low-power system, inefficiency has little economic impact. For fast high-power systems where speed specifications can only be met by valve control, the economic cost of low efficiency must be accepted. In pump-controlled systems, the fluid power supply must be included in the system model, while the analysis of valve-controlled systems can proceed without consideration of power supply details if one assumes the existence of a power source of constant supply pressure, p s, irrespective of flow demand. Power supplies that approximate this behavior are available in several different forms that trade off complexity, cost, efficiency, and static/dynamic pressure regulation accuracy. 112
4 For example, the spring-loaded relief valve is completely shut until pressure reaches the low end of the regulating range, whereupon it opens sufficiently to bypass any pump flow not required by the servovalve. The fluid power of the bypassed flow is completely lost and converted to heat. pump flow 113
5 When the servo system requires no flow, all the pumpgenerated power is converted to heat, giving zero efficiency. Supply pressure p s varies by about ± 3% over the regulating range for steady conditions and response to transient flow requirements is quite fast. Pump size must be chosen to match the largest anticipated servo-system demand. Thus the standard assumption of constant p s used in servo-system analysis is usually reasonable. 114
6 The figure below shows a flight simulator, a sophisticated application of valve-controlled servos where the motions of six actuators are coordinated to provide roll, pitch, and yaw rotary motions plus x, y, z translation. 115
7 Physical System: Valve-Controlled Servo 116
8 Real valves always exhibit either underlap (x u > ) or overlap (x u < ) behavior. Underlap / Overlap effects are usually very small. spool valve 117
9 Physical Modeling Assumptions supply pressure is constant at 1 psig reservoir pressure is constant at psig valve is zero lap actuator pressures p cl and p cr each come to p s /2 at the servo rest condition neglect inertia of the fluid cylinder and piston are rigid sensor dynamics are negligible parameters are constant 118
10 compressibility effects are neglected in the orifice flow equations, but not in the cylinder equations as pressures can be high during acceleration and deceleration periods and oil compressibility can have a destabilizing effect both flow orifices are identical, i.e., the flow and pressure coefficients are identical for both disturbance to the mass is zero 119
11 f U p r Physical Model Parameters x u inches, zero lap condition p s 1 psig (constant), supply pressure C d.6, orifice discharge coefficient w.5 in, valve port width ρ 7.8E-5 lbf-s 2 /in 4, fluid density A p 2. in 2, piston area β 1, psi, bulk modulus of fluid M.3 lbf-s 2 /in, mass K pl.1 in 3 /s-psi, piston leakage coefficient B 1 lbf-s/in, viscous damping coefficient V l 4. in 3, volume at operating point of left cylinder V r 4. in 3, volume at operating point of right cylinder p cl 5 psi, initial pressure of left cylinder p cr 5 psi, initial pressure of right cylinder x C in, initial displacement of mass ẋ C in/sec, initial velocity of mass lbf, disturbance psig, return pressure 12
12 Nonlinear Mathematical Model Equations for the orifice volume flow rates Q cl and Q cr for the left and right ends of the cylinder Equations for conservation of mass (continuity equation) for the left and right ends of the cylinder Newton s 2 nd Law applied to the moving mass Kinematic relation representing the mechanical feedback 121
13 The Variable Orifice The variable orifice is at the heart of most fluid power systems and is the most popular device for controlling fluid flow. It is the fluid equivalent of the electrical resistor and like the resistor its use leads to energy dissipation. Overriding advantages include simplicity, reliability, and ease of manufacture. The orifice can be used in analog (infinite number of positions) or discontinuous (fully open or fully closed) modes, depending on the application. 122
14 Knowledge of the orifice equations for incompressible and compressible flow is essential. Here we consider incompressible flow. The orifice equation for the volume flow rate Q of an incompressible fluid, assuming that the upstream area is much greater than the orifice area A, is: Q = CA d ( P ) 2P u ρ d C d = orifice discharge coefficient In a given system the dominating variables are usually the pressure drop and the orifice area. 123
15 The predominant nonlinearity is the square root term, but C d depends on the Reynolds number and cavitation. Cavitation refers to the formation and collapse of cavities, containing air or gas, in the liquid. If the pressure is reduced far enough hydraulic oil will vaporize and vapor cavities will form. The pressure at which vaporization commences is called the vapor pressure of the liquid and is very dependent on the temperature of the liquid. As the temperature increases, the vapor pressure increases. 124
16 The phenomenon of cavitation damage in hydraulic machinery, turbines, pumps, and propellers is well known. It has been shown both analytically and experimentally that when cavities collapse as a result of increased hydraulic pressure, very large pressures can be developed. However, there is still controversy about the exact mechanism of the damaging process. 125
17 Orifice Flow-Rate Equations x u positive: valve underlap x u negative: valve overlap x u zero: valve zero lap x v is displacement of valve spool 2(ps p cl) Qcl = Cw(x d u + x) v valid when (x u+ x v) is >. ρ This is flow into the left cylinder. 2(pcl p) r Qcl = Cw(x d u x) v valid when (xu x) v is >. ρ This is flow out of the left cylinder. 2(pcr p) r Qcr = Cw(x d u + x) v valid when (xu + x) v is >. ρ This is flow out of the right cylinder. 2(ps p cr) Qcr = Cw(x d u x) v valid when (xu x) v is >. ρ This is flow into the right cylinder. 126
18 Conservation of Mass Equations Conservation of Mass = ρ dv+ ρvda i t CV CS =ρ V + V ρ +ρq CV CV CV CV net V ρ CV = VCV + ρ+ Qnet Here we assume that all of the densities of the system (inlet flow, outlet flow, and control volume) are the same and equal to ρ. 127
19 This assumption is justified for incompressible fluids and is quite accurate for compressible fluids if pressure variations are not too large and the temperature of flow into the control volume is almost equal to the temperature of the flow out of the control volume. The isothermal bulk modulus is given by: Therefore: ρ dp ρ= = Pdt ρ β P P β=ρ = ρ P,T Conservation of Mass can be written as: V = V + P CV + Q β net P ρ/ ρ P,T 128
20 Evaluating terms: Left cylinder Q = Q + K (p p ) net cl pl cl cr C V= Ap dt V P β CV = dx (V + Ax ) dp β dt l p C cl The resulting equations for the left and right cylinders are: (V Ax ) dp dx Q K (p p ) A β dt dt r p C cr C cr + pl cl cr = p (V + Ax ) dp dx Q K (p p ) A β dt dt l p C cl C cl pl cl cr = p 129
21 V l and V r are the volumes at the operating point of the left and right cylinders, respectively. β is the bulk modulus (isothermal) of the fluid defined by the expression: dp β= V dv (V l + A p x C ) and (V r - A p x C ) represent the compressed volumes of the left and right sides of the cylinder, respectively, which include the lines from the valve to the actuator plus the ends of the cylinder. 13
22 Newton s Second Law & Feedback Equations 2 dxc dxc cl cr p + U = 2 (p p )A B f M dt dt xv = xv xc 131
23 Simulink Block Diagrams Xv Step Command ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM input To Workspace6 Xv Command Xc Xv - Xc Positive Xv - Xc Negative Controller Xv - Xc Positive Xv - Xc Negative Pcl Xv - Xc Pos Xv - Xc Neg Pcr Flow Qcl Qcr Qcl To Workspace2 Qcr To Workspace3 Qcl Qcr Xc To Workspace1 Pcl To Workspace4 Pcl Xc Pcr Xc dot Mass Xc Xc dot Xc Qcl Qcr Xc dot Cylinder Pcl Pcr Pcl Pcr Pcr To Workspace5 Clock time To Workspace 132
24 CONTROLLER SUBSYSTEM 1 Xv Command Sum Xv - Xc Switch1 1 Xv - Xc Positive Zero 2 Xc Switch 2 Xv - Xc Negative 1 MASS SUBSYSTEM Pcl 2 Pcr Sum3 Ap Gain2 Disturbance fu Sum 1/M Gain B Gain5 Xc dot 1/s 1/s Integrator Integrator1 1 Xc 2 Xc dot 133
25 Ps Supply P FLOW SUBSYSTEM 1 Pcl Sum2 2/rho Gain4 sqrt(u) Fcn Cd*w Gain Product 1 2/rho sqrt(u) Cd*w Sum Qcl 2 Sum3 Gain5 Fcn1 Gain1 Product1 Xv - Xc Pos 3 Xv - Xc Neg Sum4 2/rho Gain6 sqrt(u) Fcn2 -Cd*w Gain2 Product2 Sum1 2 Qcr 4 Pcr Sum5 Pr Reservoir P 2/rho Gain7 sqrt(u) Fcn3 -Cd*w Gain3 Product3 134
26 Vlo Constant1 1/MB 1/u 1 Xc Ap Gain Sum Gain1 Fcn CYLINDER SUBSYSTEM 1/MB 1/u Vro Constant2 Sum1 Gain2 Fcn1 1 Pcl 2 Qcl Kpl Sum2 Product 1/s Integrator Pcl Gain3 Sum4 3 Qcr Sum3 Product1 1/s Integrator1 Pcr 4 Xc dot Ap Gain4 2 Pcr 135
27 MatLab File of Constants and Expressions M=.3; B=1; Ap=2.; Kpl=.1; Vlo=4.; Vro=4.; MB=1; Pr=; Ps=1; rho=7.8e-5; Cd=.6; w=.5; Pclo=5; Pcro=5; Xcdoto=; Xco=; Vo=4.; Cx= ; Cp=; A=[-(Cp+Kpl)*MB/Vo Kpl* MB /Vo -Ap* MB/Vo; Kpl*MB/Vo -(Cp+Kpl)* MB/Vo Ap* MB/Vo; 1;Ap/M -Ap/M -B/M]; B1=[MB*Cx/Vo ;- MB*Cx/Vo ; ; 1/M]; C=[1 ; 1 ; 1 ; 1]; D=[ ; ; ; ]; K=(2*Cx*Ap)/(2*Ap^2+B*(Cp+2*Kpl)); omegan=sqrt((mb*(2*ap^2+b*(cp+2*kpl)))/(m*vo)); zeta=(b+((2*mb*m)/vo)*kpl+((mb*m)/vo)*cp)/(2*sqrt(((mb*m)/vo) *(2*Ap^2+B*(Cp+2*Kpl)))); XcXvNum=K*omegan^2; XcXvDen=[1 2*zeta*omegan omegan^2 ]; %[XcXvNumCL,XcXvDenCL]=cloop(XcXvNum,XcXvDen,-1); Kc=1; XcXvNumCL=Kc*K*omegan^2; XcXvDenCL=[1 2*zeta*omegan omegan^2 Kc*K*omegan^2]; 136
28 Linear System Analysis Restrict the analysis to small perturbations around a chosen operating point. A linearized approximate model may be obtained that provides many useful results. Valve flow equations can be thought of as relations between a dependent variable (flow rate) and two independent variables (spool motion and cylinder pressure) and thus can be linearized about any desired operating point. 137
29 Q Q Q Q + x + p v v v v, v,p c,p xv p operating point c operating point flow gain = C pressure coefficient = C x p Q = x v Q = p v operating point v c operating point Qv Qv, + Cx x v,p Cp p c,p 138
30 Assume that Q v, = and that the numerical values of C x and C p are equal for the Q cl and Q cr equations (correct assumptions for commonly used operating points). Qcl Cx x v,p Cp p cl,p Qcr Cx x v,p Cp p cr,p Take the volumes (V l + A p x C ) and (V r - A p x C ) to be constant at V l = V r = V, a good approximation for small changes in x C. 139
31 Linearized Set of Equations: V dp Cx Cp K p p = A β dt ( ) cl,p ( ) x v,p p cl,p pl cl,p cr,p p dx dt C,p V dp Cx Cp + K p p = A β dt ( ) cr,p ( ) x v,p p cr,p pl cl,p cr,p p dx dt C,p 2 dxc,p dxc,p cl,p cr,p p + U,p = 2 ( ) p p A B f M dt dt 14
32 Simulink Block Diagram of the Linear System Pcl_l Cp Sum5 To Workspace1 MB/Vo 1/s Pcl Gain6 Ap Sum Sum3 Gain2 MB/Vo Gain3 Integrator2 1/s Pcr Integrator3 Sum2 Cp Gain7 Gain4 Kpl Gain5 5 Constant1 Sum7 Qcl_l To Workspace4 Pcr_l Clock time_l To Workspace Ap Gain9 Sum6 To Workspace2 Sum8 input_l To Workspace6 Sum4 Cx Gain8 Disturbance fu Sum1 1/M Gain 1/s Integrator B Gain1 1/s Xc Integrator1 Qcr_l To Workspace5 Xc_l To Workspace3 Xv Step Command ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR) 141
33 If we take the Laplace Transform of these equations, we can derive six useful transfer functions relating the two inputs, x v and f U, to the three outputs, p cl, p cr, and x C. ( ) Vs +β Kpl + Cp Kpl As p Cxβ Cx Cx Vs β K + C ( ) p x cl v Kpl pl p As p p cr = x v Cx Cxβ C x x 2 C fu Ap Ap Ms + Bs 142
34 One of these transfer functions is: x ( s) K = s C 2 x v s 2s ζ 2 ωn ωn where K = ω = n ζ= x p ( ) 2 p p pl 2C A 2A + BC + 2K ( ) β 2A + BC + 2K MV 2 p p pl 2βM βm B+ Kpl C V + V βm 2 2 2Ap + BCp + 2Kpl V p ( ) 143
35 The state variables for the linearized equations are: The state-variable equations are: q q 1 p p cl 2 cr = q3 xc q x 4 C ( ) β Cp + Kpl Kplβ Apβ βcx V V V q 1 q 1 V q Kpl ( Cp K β β + pl ) Apβ 2 q 2 βcx = V V V + x q q V q 4 q 4 Ap Ap B M M M [ v ] 144
36 Simulink Block Diagram: Linear System in State-Variable Form Cx Cp Clock time_lss To Workspace1 input_lss To Workspace6 Gain Gain1 Sum1 Qcl_lss To Workspace4 Sum3 Pcl_lss To Workspace2 5 Xv Step Command Sum Mux Mux x' = Ax+Bu y = Cx+Du System Demux Demux Constant Pcr_lss Distrbance fu Xc_lss To Workspace Sum4 To Workspace3 Cx Gain2 Sum2 Cp Gain3 Qcr_lss To Workspace5 ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR - STATE SPACE) 145
37 Simulation Results: Step Command x V =.2 in. applied at t =.3 sec x V + x v x C - CONTROLLER K c PLANT G(s) x KG(s) KKω x 1 KG(s) s 2 s s KK 2 C c c n = = V + c + ζω n +ω n + c ωn 146
38 Nonlinear and Linear Simulation Results: x C vs. time x C (in) solid: nonlinear dashed: linear time (sec) 147
39 Nonlinear and Linear Simulation Results: Q cl vs. time 25 2 Q cl (in 3 /sec) 15 solid: nonlinear dashed: linear time (sec) 148
40 Nonlinear and Linear Simulation Results: Q cr vs. time -5 Q cr (in 3 /sec) -1 solid: nonlinear dashed: linear time (sec) 149
41 Nonlinear and Linear Simulation Results: p cl time 8 75 p cl (psig) 7 65 solid: nonlinear dashed: linear time (sec) 15
42 Nonlinear and Linear Simulation Results: p cr vs. time p cr (psig) 4 35 solid: nonlinear dashed: linear time (sec) 151
43 Open-Loop Frequency Response Plots with K c = 1 Bode Diagrams Phase (deg); Magnitude (db) Frequency (rad/sec) 152
44 GM = 16.2 db = 6.46 PM = 74.8 Bode Diagrams Gm=16.2 db (at rad/sec), Pm= deg. (at rad/sec) 2 Phase (deg); Magnitude (db) Frequency (rad/sec) 153
45 Closed-Loop Frequency Response Plots with K c = 1 Closed-Loop Bandwidth = 123 Hz = 774 rad/sec Bode Diagrams At 774 rad/sec: Mag =.77 Phase = Phase (deg); Magnitude (db) Frequency (rad/sec) 154
46 Simulink Block Diagram: Nonlinear Control with Time Delay input_l_nl Cx Cp To Workspace6 Gain Gain1 Xv Step Command Sum1 Qcl_l_nl Sum3 Pcl_l_nl To Workspace2 Delay To Workspace4.1 Sum Sign Gain4 Mux Mux x' = Ax+Bu y = Cx+Du System Demux Demux 5 Constant Pcr_l_nl Distrbance fu Xc_l_nl To Workspace Sum4 To Workspace3 Clock time_l_nl To Workspace1 Cx Gain2 Sum2 Cp Gain3 Qcr_l_nl To Workspace5 ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR) with Nonlinear On-Off Controller and Time Delay 155
47 Curve A: Gain =.1, Delay =.1 sec Curve B: Gain =.5, Delay =.1 sec Curve C: Gain =.5, Delay = sec x c (in).15 A C.1 B time (sec) 156
48 Simulink Block Diagram: Linear System with Proportional Control Cx Cp time_l_p Gain Gain1 Clock To Workspace1 Sum1 Pcl_l_p input_l_p Qcl_l_p Sum3 To Workspace2 To Workspace6 To Workspace4 5 Xv Step Command Sum Kc Controller Mux Mux x' = Ax+Bu y = Cx+Du System Demux Demux Constant Pcr_l_p Disturbance fu Xc_l_p To Workspace Sum4 To Workspace3 Cx Gain2 Sum2 Cp Gain3 Qcr_l_p To Workspace5 ELECTROHYDRAULIC VALVE-CONTROLLED SERVOMECHANISM (LINEAR) With Proportional Control 157
49 Root Locus Plot Pt. #1: K c = 6.46 Pt. #2: K c = 3.7 Pt. #3: K c = 1.36 Pt. #4: K c = Imag Axis Real Axis 158
50 Open-Loop Poles:, ± 1993i K c =6.46 Closed-Loop Poles: -3384, ± 2614i K c = 3.7 Closed-loop Poles: -2677, -353 ± 2195i K c = 1.36 Closed-Loop Poles: -1133, ± 1737i K c = 1. Closed-Loop Poles: -732, ± 1771i 159
51 Closed-Loop Time Response (Step) Plots.3.25 K c = x C (in) K c = 1.36 K c = time (sec) 16
52 Closed-Loop Frequency Response Plots Bode Diagrams K c = K c =1. K c =1.36 Phase (deg); Magnitude (db) K c =1. K c =3.7 K c = Frequency (rad/sec) 161
53 Nyquist Diagram: K c = 1. Nyquist Diagrams Imaginary Axis Real Axis 162
54 Nyquist Diagram: K c = 6.46 Nyquist Diagrams Imaginary Axis Real Axis 163
55 Nyquist Diagram: K c = 1. Nyquist Diagrams Imaginary Axis Real Axis 164
56 Stability Considerations Closed-Loop Transfer Function x KG(s) KKω x 1 KG(s) s 2 s s KK 2 C c c n = = V + c + ζω n +ω n + c ωn K n = ω = ζ= x p ( ) 2 p p pl 2C A 2A + BC + 2K ( ) β 2A + BC + 2K MV 2 p p pl 2βM βm B+ Kpl C V + V βm 2 2 2Ap + BCp + 2Kpl V ( ) 165 p
57 Neglect leakage (K pl = ) and consider the load as mainly inertia (B =, friction is ignored). The closed-loop transfer function becomes: x x C V = KC c x Ap VM MC KC s + s + s+ 2βA 2A A 3 p 2 c x 2 2 p p p Since the bulk modulus β of the fluid is defined as: P β= V/V The combined stiffness k of the two columns of fluid is: k = 2βA V 2 p 166
58 The valve stiffness k v is defined as: C C x p Q = x v Q = p v operating point v c operating point k v = C x 2A p C p The closed-loop transfer function can now be written as: KC c x x x C V = Ap M CM KC s + s + s+ k Ak A 3 x 2 c x p v p 167
59 Applying the Routh Stability Criterion to the characteristic equation of the closed-loop transfer function gives the relationship for stability as: k > k v In other words, the stiffness of the oil column must be greater than the effective valve stiffness if stability is to be satisfactory. 168
Model-Based Design, Analysis, & Control: Valve-Controlled Hydraulic System K. Craig 1
Model-Based Design, Analysis, & Control: K. Craig 1 K. Craig K. Craig 3 K. Craig 4 K. Craig 5 Mission: It s All About Process Dynamic System Investigation K. Craig 6 K. Craig 7 K. Craig 8 K. Craig 9 K.
More informationModelling the Dynamics of Flight Control Surfaces Under Actuation Compliances and Losses
Modelling the Dynamics of Flight Control Surfaces Under Actuation Compliances and Losses Ashok Joshi Department of Aerospace Engineering Indian Institute of Technology, Bombay Powai, Mumbai, 4 76, India
More informationAppendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2)
Appendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2) For all calculations in this book, you can use the MathCad software or any other mathematical software that you are familiar
More informationD(s) G(s) A control system design definition
R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure
More informationRESEARCH ON AIRBORNE INTELLIGENT HYDRAULIC PUMP SYSTEM
8 TH INTERNATIONAL CONGRESS OF THE AERONAUTICAL SCIENCES RESEARCH ON AIRBORNE INTELLIGENT HYDRAULIC PUMP SYSTEM Jungong Ma, Xiaoye Qi, Juan Chen BeiHang University,Beijing,China jgma@buaa.edu.cn;qixiaoye@buaa.edu.cn;sunchenjuan@hotmail.com
More informationLecture 5. Labs this week: Please review ME3281 Systems materials! Viscosity and pressure drop analysis Fluid Bulk modulus Fluid Inertance
Labs this week: Lab 10: Sequencing circuit Lecture 5 Lab 11/12: Asynchronous/Synchronous and Parallel/Tandem Operations Please review ME3281 Systems materials! 132 Viscosity and pressure drop analysis
More informationLecture Note 8-1 Hydraulic Systems. System Analysis Spring
Lecture Note 8-1 Hydraulic Systems 1 Vehicle Model - Brake Model Brake Model Font Wheel Brake Pedal Vacuum Booster Master Cylinder Proportionnig Valve Vacuum Booster Rear Wheel Master Cylinder Proportioning
More informationA m. Q m P. piston or diaphragm
Massachusetts Institute of echnology Department of Mechanical Engineering 2.141 Modeling and Simulation of Dynamic Systems 2.141 Assignment #3: GAS-CHARGED ACCUMULAOR he figure below (after Pourmovahed
More informationLecture 5. Labs this week:
Labs this week: Lab 10: Bleed-off Circuit Lecture 5 Lab 11/12: Asynchronous/Synchronous and Parallel/Tandem Operations Systems Review Homework (due 10/11) Participation is research lab Hydraulic Hybrid
More informationOutline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
More informationDynamic Redesign of a Flow Control Servo-valve using a Pressure Control Pilot
Dynamic Redesign of a Flow Control Servo-valve using a Pressure Control Pilot Perry Y. Li Department of Mechanical Engineering University of Minnesota Church St. SE, Minneapolis, Minnesota 55455 Email:
More informationHYDRAULIC CONTROL SYSTEMS
HYDRAULIC CONTROL SYSTEMS Noah D. Manring Mechanical and Aerospace Engineering Department University of Missouri-Columbia WILEY John Wiley & Sons, Inc. vii Preface Introduction xiii XV FUNDAMENTALS 1 Fluid
More informationIndex. Index. More information. in this web service Cambridge University Press
A-type elements, 4 7, 18, 31, 168, 198, 202, 219, 220, 222, 225 A-type variables. See Across variable ac current, 172, 251 ac induction motor, 251 Acceleration rotational, 30 translational, 16 Accumulator,
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3.. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid -
More informationChapter Four fluid flow mass, energy, Bernoulli and momentum
4-1Conservation of Mass Principle Consider a control volume of arbitrary shape, as shown in Fig (4-1). Figure (4-1): the differential control volume and differential control volume (Total mass entering
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationLab-Report Control Engineering. Proportional Control of a Liquid Level System
Lab-Report Control Engineering Proportional Control of a Liquid Level System Name: Dirk Becker Course: BEng 2 Group: A Student No.: 9801351 Date: 10/April/1999 1. Contents 1. CONTENTS... 2 2. INTRODUCTION...
More informationReceived 21 April 2008; accepted 6 January 2009
Indian Journal of Engineering & Materials Sciences Vol. 16, February 2009, pp. 7-13 Inestigation on the characteristics of a new high frequency three-way proportional pressure reducing ale in ariable ale
More informationHYDRAULIC LINEAR ACTUATOR VELOCITY CONTROL USING A FEEDFORWARD-PLUS-PID CONTROL
HYDRAULIC LINEAR ACTUATOR VELOCITY CONTROL UING A FEEDFORWARD-PLU-PID CONTROL Qin Zhang Department of Agricultural Engineering University of Illinois at Urbana-Champaign, Urbana, IL 68 ABTRACT: A practical
More informationHydraulic (Fluid) Systems
Hydraulic (Fluid) Systems Basic Modeling Elements Resistance apacitance Inertance Pressure and Flow Sources Interconnection Relationships ompatibility Law ontinuity Law Derive Input/Output Models ME375
More informationCDS 101/110a: Lecture 8-1 Frequency Domain Design
CDS 11/11a: Lecture 8-1 Frequency Domain Design Richard M. Murray 17 November 28 Goals: Describe canonical control design problem and standard performance measures Show how to use loop shaping to achieve
More informationHARDWARE-IN-THE-LOOP SIMULATION EXPERIMENTS WITH A HYDRAULIC MANIPULATOR MODEL
HARDWARE-IN-THE-LOOP SIMULATION EXPERIMENTS WITH A HYDRAULIC MANIPULATOR MODEL Jorge A. Ferreira, André F. Quintã, Carlos M. Cabral Departament of Mechanical Engineering University of Aveiro, Portugal
More informationSome tools and methods for determination of dynamics of hydraulic systems
Some tools and methods for determination of dynamics of hydraulic systems A warm welcome to the course in Hydraulic servo-techniques! The purpose of the exercises given in this material is to make you
More informationMathematical Modeling and response analysis of mechanical systems are the subjects of this chapter.
Chapter 3 Mechanical Systems A. Bazoune 3.1 INRODUCION Mathematical Modeling and response analysis of mechanical systems are the subjects of this chapter. 3. MECHANICAL ELEMENS Any mechanical system consists
More informationTopic # Feedback Control Systems
Topic #19 16.31 Feedback Control Systems Stengel Chapter 6 Question: how well do the large gain and phase margins discussed for LQR map over to DOFB using LQR and LQE (called LQG)? Fall 2010 16.30/31 19
More informationLoad Prediction-based Energy-efficient Hydraulic Actuation. of a Robotic Arm. 1 Introduction
oad rediction-based Energy-efficient Hydraulic ctuation of a Robotic rm Miss Can Du, rof ndrew lummer and Dr Nigel Johnston fixed displacement pump. This can reduce the weight of plant compared with the
More informationCHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System
CHAPTER 1 Basic Concepts of Control System 1. What is open loop control systems and closed loop control systems? Compare open loop control system with closed loop control system. Write down major advantages
More informationModeling, Control and Experimental Validation of a Device for Seismic Events Simulation
Modeling, Control and Experimental Validation of a Device for Seismic Events Simulation Paolo Righettini, Roberto Strada, Vittorio Lorenzi, Alberto Oldani, Mattia Rossetti Abstract Single and multi-axis
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? Open-Loop vs Closed-Loop (Feedback) Why Use Feedback Control? Closed-Loop Control System Structure Elements of a Feedback Control System
More informationControl Systems. Design of State Feedback Control.
Control Systems Design of State Feedback Control chibum@seoultech.ac.kr Outline Design of State feedback control Dominant pole design Symmetric root locus (linear quadratic regulation) 2 Selection of closed-loop
More informationToday (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10
Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:
More informationECSE 4962 Control Systems Design. A Brief Tutorial on Control Design
ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got
More informationMechatronics Engineering. Li Wen
Mechatronics Engineering Li Wen Bio-inspired robot-dc motor drive Unstable system Mirko Kovac,EPFL Modeling and simulation of the control system Problems 1. Why we establish mathematical model of the control
More informationCHAPTER 3 QUARTER AIRCRAFT MODELING
30 CHAPTER 3 QUARTER AIRCRAFT MODELING 3.1 GENERAL In this chapter, the quarter aircraft model is developed and the dynamic equations are derived. The quarter aircraft model is two degrees of freedom model
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More informationTopic # Feedback Control. State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback
Topic #17 16.31 Feedback Control State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall
More informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN MECHANICAL ENGINEERING SEMESTER 1EXAMINATION 2017/2018
ENG00 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN MECHANICAL ENGINEERING SEMESTER EXAMINATION 07/08 ADVANCED THERMOFLUIDS & CONTROL SYSTEMS MODULE NO: AME6005 Date: 8 January 08 Time: 0.00.00
More informationMechatronics Assignment # 1
Problem # 1 Consider a closed-loop, rotary, speed-control system with a proportional controller K p, as shown below. The inertia of the rotor is J. The damping coefficient B in mechanical systems is usually
More informationObjectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation
Objectives Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation Conservation of Mass Conservation of Mass Mass, like energy, is a conserved
More informationPositioning Servo Design Example
Positioning Servo Design Example 1 Goal. The goal in this design example is to design a control system that will be used in a pick-and-place robot to move the link of a robot between two positions. Usually
More information6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0.
6. Sketch the z-domain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)
More informationModeling and Simulation Revision III D R. T A R E K A. T U T U N J I P H I L A D E L P H I A U N I V E R S I T Y, J O R D A N
Modeling and Simulation Revision III D R. T A R E K A. T U T U N J I P H I L A D E L P H I A U N I V E R S I T Y, J O R D A N 0 1 4 Block Diagrams Block diagram models consist of two fundamental objects:
More informationMech 6091 Flight Control System Course Project. Team Member: Bai, Jing Cui, Yi Wang, Xiaoli
Mech 6091 Flight Control System Course Project Team Member: Bai, Jing Cui, Yi Wang, Xiaoli Outline 1. Linearization of Nonlinear F-16 Model 2. Longitudinal SAS and Autopilot Design 3. Lateral SAS and Autopilot
More informationWind Turbine Control
Wind Turbine Control W. E. Leithead University of Strathclyde, Glasgow Supergen Student Workshop 1 Outline 1. Introduction 2. Control Basics 3. General Control Objectives 4. Constant Speed Pitch Regulated
More informationUniversity of Alberta ENGM 541: Modeling and Simulation of Engineering Systems Laboratory #7. M.G. Lipsett & M. Mashkournia 2011
ENG M 54 Laboratory #7 University of Alberta ENGM 54: Modeling and Simulation of Engineering Systems Laboratory #7 M.G. Lipsett & M. Mashkournia 2 Mixed Systems Modeling with MATLAB & SIMULINK Mixed systems
More informationFEEDBACK CONTROL SYSTEMS
FEEDBAC CONTROL SYSTEMS. Control System Design. Open and Closed-Loop Control Systems 3. Why Closed-Loop Control? 4. Case Study --- Speed Control of a DC Motor 5. Steady-State Errors in Unity Feedback Control
More informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationMass of fluid leaving per unit time
5 ENERGY EQUATION OF FLUID MOTION 5.1 Eulerian Approach & Control Volume In order to develop the equations that describe a flow, it is assumed that fluids are subject to certain fundamental laws of physics.
More informationLecture 4. Lab this week: Cartridge valves Flow divider Properties of Hydraulic Fluids. Lab 8 Sequencing circuit Lab 9 Flow divider
91 Lecture 4 Lab this week: Lab 8 Sequencing circuit Lab 9 Flow divider Cartridge valves Flow divider Properties of Hydraulic Fluids Viscosity friction and leakage Bulk modulus Inertance Cartridge Valves
More informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS BRANCH : ECE YEAR : II SEMESTER: IV 1. What is control system? 2. Define open
More informationTitre. AMESim and Common Rail type Injection Systems. Technical Bulletin n 110
Titre AMESim and Common Rail type Injection Systems Technical Bulletin n 110 How to contact IMAGINE: North America Europe Asia imagine-us@amesim.com imagine@amesim.com imagine-japan@amesim.com Visit www.amesim.com
More informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Online Lecture Material at (www.brunel.ac.uk/~emstaam)» C. W. De Silva, Modelling and Control of Engineering Systems, CRC Press, Francis & Taylor, 2009.» M. P. Groover,
More informationStability of CL System
Stability of CL System Consider an open loop stable system that becomes unstable with large gain: At the point of instability, K( j) G( j) = 1 0dB K( j) G( j) K( j) G( j) K( j) G( j) =± 180 o 180 o Closed
More informationAcceleration Feedback
Acceleration Feedback Mechanical Engineer Modeling & Simulation Electro- Mechanics Electrical- Electronics Engineer Sensors Actuators Computer Systems Engineer Embedded Control Controls Engineer Mechatronic
More informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationImproving the Control System for Pumped Storage Hydro Plant
011 International Conference on Computer Communication and Management Proc.of CSIT vol.5 (011) (011) IACSIT Press, Singapore Improving the Control System for Pumped Storage Hydro Plant 1 Sa ad. P. Mansoor
More information1 x(k +1)=(Φ LH) x(k) = T 1 x 2 (k) x1 (0) 1 T x 2(0) T x 1 (0) x 2 (0) x(1) = x(2) = x(3) =
567 This is often referred to as Þnite settling time or deadbeat design because the dynamics will settle in a Þnite number of sample periods. This estimator always drives the error to zero in time 2T or
More informationManufacturing Equipment Control
QUESTION 1 An electric drive spindle has the following parameters: J m = 2 1 3 kg m 2, R a = 8 Ω, K t =.5 N m/a, K v =.5 V/(rad/s), K a = 2, J s = 4 1 2 kg m 2, and K s =.3. Ignore electrical dynamics
More informationControl of Electromechanical Systems
Control of Electromechanical Systems November 3, 27 Exercise Consider the feedback control scheme of the motor speed ω in Fig., where the torque actuation includes a time constant τ A =. s and a disturbance
More informationCOMPARISON OF TWO METHODS TO SOLVE PRESSURES IN SMALL VOLUMES IN REAL-TIME SIMULATION OF A MOBILE DIRECTIONAL CONTROL VALVE
COMPARISON OF TWO METHODS TO SOLVE PRESSURES IN SMALL VOLUMES IN REAL-TIME SIMULATION OF A MOBILE DIRECTIONAL CONTROL VALVE Rafael ÅMAN*, Heikki HANDROOS*, Pasi KORKEALAAKSO** and Asko ROUVINEN** * Laboratory
More informationTherefore, the control volume in this case can be treated as a solid body, with a net force or thrust of. bm # V
When the mass m of the control volume remains nearly constant, the first term of the Eq. 6 8 simply becomes mass times acceleration since 39 CHAPTER 6 d(mv ) CV m dv CV CV (ma ) CV Therefore, the control
More informationDynamic Redesign of a Flow Control Servovalve Using a Pressure Control Pilot 1
Perry Y. Li Department of Mechanical Engineering, University of Minnesota, 111 Church St. SE, Minneapolis, MN 55455 e-mail: pli@me.umn.edu Dynamic Redesign of a Flow Control Servovalve Using a Pressure
More informationMAE 142 Homework #5 Due Friday, March 13, 2009
MAE 142 Homework #5 Due Friday, March 13, 2009 Please read through the entire homework set before beginning. Also, please label clearly your answers and summarize your findings as concisely as possible.
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 9. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid -
More informationMechanical Systems Part A: State-Space Systems Lecture AL12
AL: 436-433 Mechanical Systems Part A: State-Space Systems Lecture AL Case study Case study AL: Design of a satellite attitude control system see Franklin, Powell & Emami-Naeini, Ch. 9. Requirements: accurate
More informationECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]
ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative
More informationModeling and Simulation Revision IV D R. T A R E K A. T U T U N J I P H I L A D E L P H I A U N I V E R S I T Y, J O R D A N
Modeling and Simulation Revision IV D R. T A R E K A. T U T U N J I P H I L A D E L P H I A U N I V E R S I T Y, J O R D A N 2 0 1 7 Modeling Modeling is the process of representing the behavior of a real
More informationCDS 101/110 Homework #7 Solution
Amplitude Amplitude CDS / Homework #7 Solution Problem (CDS, CDS ): (5 points) From (.), k i = a = a( a)2 P (a) Note that the above equation is unbounded, so it does not make sense to talk about maximum
More informationSimulation Study on Pressure Control using Nonlinear Input/Output Linearization Method and Classical PID Approach
Simulation Study on Pressure Control using Nonlinear Input/Output Linearization Method and Classical PID Approach Ufuk Bakirdogen*, Matthias Liermann** *Institute for Fluid Power Drives and Controls (IFAS),
More informationControl for. Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e
Control for Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e Motion Systems m F Introduction Timedomain tuning Frequency domain & stability Filters Feedforward Servo-oriented
More informationLab 1: Dynamic Simulation Using Simulink and Matlab
Lab 1: Dynamic Simulation Using Simulink and Matlab Objectives In this lab you will learn how to use a program called Simulink to simulate dynamic systems. Simulink runs under Matlab and uses block diagrams
More informationME 4232: FLUID POWER CONTROLS LAB. Class #5 Valve Modeling
ME 4232: FLUID POWER CONTROLS LAB Class #5 Valve Modeling Notes No Office Hours Today Upcoming Labs: Lab 9: Flow Divider Lab 10: Sequencing Circuits 2 Agenda Wrap-up: Leakage Calculations Fluid Compressibility
More informationREPETITIVE LEARNING OF BACKSTEPPING CONTROLLED NONLINEAR ELECTROHYDRAULIC MATERIAL TESTING SYSTEM 1. Seunghyeokk James Lee 2, Tsu-Chin Tsao
REPETITIVE LEARNING OF BACKSTEPPING CONTROLLED NONLINEAR ELECTROHYDRAULIC MATERIAL TESTING SYSTEM Seunghyeokk James Lee, Tsu-Chin Tsao Mechanical and Aerospace Engineering Department University of California
More informationDynamic Modeling of Fluid Power Transmissions for Wind Turbines
Dynamic Modeling of Fluid Power Transmissions for Wind Turbines EWEA OFFSHORE 211 N.F.B. Diepeveen, A. Jarquin Laguna n.f.b.diepeveen@tudelft.nl, a.jarquinlaguna@tudelft.nl Offshore Wind Group, TU Delft,
More information100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =
1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot
More informationTHE INFLUENCE OF THERMODYNAMIC STATE OF MINERAL HYDRAULIC OIL ON FLOW RATE THROUGH RADIAL CLEARANCE AT ZERO OVERLAP INSIDE THE HYDRAULIC COMPONENTS
Knežević, D. M., et al.: The Influence of Thermodynamic State of Mineral S1461 THE INFLUENCE OF THERMODYNAMIC STATE OF MINERAL HYDRAULIC OIL ON FLOW RATE THROUGH RADIAL CLEARANCE AT ZERO OVERLAP INSIDE
More informationR10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1
Code No: R06 R0 SET - II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry
More informationThe First Law of Thermodynamics. By: Yidnekachew Messele
The First Law of Thermodynamics By: Yidnekachew Messele It is the law that relates the various forms of energies for system of different types. It is simply the expression of the conservation of energy
More informationHYDRAULIC EFFICIENCY OF A HYDROSTATIC TRANSMISSION WITH A VARIABLE DISPLACEMENT PUMP AND MOTOR. A Thesis presented to the Faculty
HYDRAULIC EFFICIENCY OF A HYDROSTATIC TRANSMISSION WITH A VARIABLE DISPLACEMENT PUMP AND MOTOR A Thesis presented to the Faculty of the Graduate School at the University of Missouri-Columbia In Partial
More informationModeling and Analysis of Dynamic Systems
Modeling and Analysis of Dynamic Systems Dr. Guillaume Ducard Fall 2017 Institute for Dynamic Systems and Control ETH Zurich, Switzerland G. Ducard c 1 / 34 Outline 1 Lecture 7: Recall on Thermodynamics
More informationYTÜ Mechanical Engineering Department
YTÜ Mechanical Engineering Department Lecture of Special Laboratory of Machine Theory, System Dynamics and Control Division Coupled Tank 1 Level Control with using Feedforward PI Controller Lab Date: Lab
More informationControls Problems for Qualifying Exam - Spring 2014
Controls Problems for Qualifying Exam - Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function
More information5 ENERGY EQUATION OF FLUID MOTION
5 ENERGY EQUATION OF FLUID MOTION 5.1 Introduction In order to develop the equations that describe a flow, it is assumed that fluids are subject to certain fundamental laws of physics. The pertinent laws
More informationThe Bernoulli Equation
The Bernoulli Equation The most used and the most abused equation in fluid mechanics. Newton s Second Law: F = ma In general, most real flows are 3-D, unsteady (x, y, z, t; r,θ, z, t; etc) Let consider
More informationCHAPTER 5 QUASI-STATIC TESTING OF LARGE-SCALE MR DAMPERS. To investigate the fundamental behavior of the 20-ton large-scale MR damper, a
CHAPTER 5 QUASI-STATIC TESTING OF LARGE-SCALE MR DAMPERS To investigate the fundamental behavior of the 2-ton large-scale MR damper, a series of quasi-static experiments were conducted at the Structural
More informationθ α W Description of aero.m
Description of aero.m Determination of the aerodynamic forces, moments and power by means of the blade element method; for known mean wind speed, induction factor etc. Simplifications: uniform flow (i.e.
More informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies.
SET - 1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies..
More informationProgrammable Valves: a Solution to Bypass Deadband Problem of Electro-Hydraulic Systems
Programmable Valves: a Solution to Bypass Deadband Problem of Electro-Hydraulic Systems Song Liu and Bin Yao Abstract The closed-center PDC/servo valves have overlapped spools to prevent internal leakage
More information(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:
1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.
More informationChapter 5. Mass and Energy Analysis of Control Volumes
Chapter 5 Mass and Energy Analysis of Control Volumes Conservation Principles for Control volumes The conservation of mass and the conservation of energy principles for open systems (or control volumes)
More informationRobust Control Design for a Wheel Loader Using Mixed Sensitivity H-infinity and Feedback Linearization Based Methods
25 American Control Conference June 8-, 25. Portland, OR, USA FrB2.5 Robust Control Design for a Wheel Loader Using Mixed Sensitivity H-infinity and Feedback Linearization Based Methods Roger Fales and
More informationDSCC PASSIVE CONTROL OF A HYDRAULIC HUMAN POWER AMPLIFIER USING A HYDRAULIC TRANSFORMER
Proceedings of the ASME 25 Dynamic Systems and Control Conference DSCC25 October 28-3, 25, Columbus, Ohio, USA DSCC25-9734 PASSIVE CONTROL OF A HYDRAULIC HUMAN POWER AMPLIFIER USING A HYDRAULIC TRANSFORMER
More informationDesign and Modeling of Fluid Power Systems ME 597/ABE Lecture 7
Systems ME 597/ABE 591 - Lecture 7 Dr. Monika Ivantysynova MAHA Professor Fluid Power Systems MAHA Fluid Power Research Center Purdue University Content of 6th lecture The lubricating gap as a basic design
More informationCourse roadmap. Step response for 2nd-order system. Step response for 2nd-order system
ME45: Control Systems Lecture Time response of nd-order systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More informationTable of Laplacetransform
Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e- at, an exponential function s + a sin wt, a sine fun
More informationME 475/591 Control Systems Final Exam Fall '99
ME 475/591 Control Systems Final Exam Fall '99 Closed book closed notes portion of exam. Answer 5 of the 6 questions below (20 points total) 1) What is a phase margin? Under ideal circumstances, what does
More informationTopic # Feedback Control Systems
Topic #1 16.31 Feedback Control Systems Motivation Basic Linear System Response Fall 2007 16.31 1 1 16.31: Introduction r(t) e(t) d(t) y(t) G c (s) G(s) u(t) Goal: Design a controller G c (s) so that the
More informationCHAPTER INTRODUCTION
CHAPTER 3 DYNAMIC RESPONSE OF 2 DOF QUARTER CAR PASSIVE SUSPENSION SYSTEM (QC-PSS) AND 2 DOF QUARTER CAR ELECTROHYDRAULIC ACTIVE SUSPENSION SYSTEM (QC-EH-ASS) 3.1 INTRODUCTION In this chapter, the dynamic
More informationModeling Mechanical Systems
Modeling Mechanical Systems Mechanical systems can be either translational or rotational. Although the fundamental relationships for both types are derived from Newton s law, they are different enough
More informationSAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015
FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequency-domain analysis and control design (15 pt) Given is a
More information