Modeling and Analysis of Dynamic Systems (2nd Edition)
Ramin S. Esfandiari, Bei Lu
Language: English
Pages: 558
ISBN: B004VG2DRW
Format: PDF / Kindle (mobi) / ePub
Modeling and Analysis of Dynamic Systems, Second Edition introduces MATLAB(r), Simulink(r), and Simscape and then uses them throughout the text to perform symbolic, graphical, numerical, and simulation tasks. Written for junior or senior level courses, the textbook meticulously covers techniques for modeling dynamic systems, methods of response analysis, and provides an introduction to vibration and control systems. These features combine to provide students with a thorough knowledge of the mathematical modeling and analysis of dynamic systems.
See What s New in the Second Edition:
• Coverage of modeling and analysis of dynamic systems ranging from mechanical to thermal using Simscape Utilization of Simulink for linearization as well as simulation of nonlinear dynamic systems Integration of Simscape into Simulink for control system analysis and design
• Each topic covered includes at least one example, giving students better comprehension of the subject matter. More complex topics are accompanied by multiple, painstakingly worked-out examples. Each section of each chapter is followed by several exercises so that students can immediately apply the ideas just learned. End-of-chapter review exercises help in learning how a combination of different ideas can be used to analyze a problem.
• This second edition of a bestselling textbook fully integrates the MATLAB Simscape Toolbox and covers the usage of Simulink for new purposes. It gives students better insight into the involvement of actual physical components rather than their mathematical representations."
0 0 0 1− λ = (1− λ )3 − (1− λ ) = λ(λ − 1)(λ − 2) = 0 λ1,2,3 = 0,1, 2 For λ1 = 0, we solve (A − λ1I)v1 = 0 Av1 = 0 Elementary row operations 1 0 1 0 1 0 1 0 1 a 0 b = 0 c 0 1 0 0 0 1 0 1 0 0 a 0 b = 0 c 0 The second row gives b = 0. The first row yields a + c = 0. The last row of zeros suggests a free variable exists, which can be either a or c. Choosing a = 1 results in c = −1 and 1
Systems............................................................................266 6.4.1 Elemental Relations of Electromechanical Systems.........................266 6.4.2 Armature-Controlled Motors............................................................ 268 6.4.3 Field-Controlled Motors.................................................................... 272 6.5 Impedance Methods....................................................................................... 275 6.5.1
as the parts move relative to each other. If v2 > v1 > 0, then the right end of the damper moves to the right with respect to the left end. The force applied to the right end is dependent on the relative velocity vrel = v2 − v1. The force has a magnitude of f = b(v2 − v1) (5.17) and points to the right. Assume that the damper is massless, or of negligible mass. Then the forces at the two ends of the damper are equal in magnitude but opposite in direction. For a torsional damper as shown in
approach (as opposed to the energy approach). 169 Mechanical Systems x k m b f (a) x fk = kx m fb = bx f (b) FIGURE 5.29 A mass–spring–damper system: (a) physical system and (b) free-body diagram. Let us consider a simple system consisting of a block of mass m, a spring of stiffness k, and a viscous damper of viscous damping coefficient b. Figure 5.29 shows the physical mass–spring–damper system and the free-body diagram drawn for the mass. The motion of the system can be described
following example shows that the gravity term does not enter into the governing differential equation if the displacement is measured from the static equilibrium. Consider the mass–spring system shown in Figure 5.33, where the mass is assumed to move only in the vertical direction. The free length of the spring is y0. Due to gravity, the spring is stretched by δst when the mass is in static equilibrium and mg = kδst. Imagine the mass to be displaced downward 172 Modeling and Analysis of