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Dynamic Responses of Systems The most important function of a model devised for measurement or control systems is to be able to predict what the output will be for a particular input. We are not only interested with a static situation but wanted to see how the output will change with time when there is a change of input or when the input changes with time. To do so, we need to form equations which will indicate how the system output will vary with time when the input is varying with time. This can be done by the use of differential equations. 1 Examples of Dynamic Systems • An example of a first order system is water flowing out of a tank. q1 is the forcing input. • Another example, is a thermometer being placed in a hot liquid at some temperature. The rate at which the reading of the thermometer changes with time. TL is the forcing input. dh RA pgh q1 dt dT RC T TL dt 2 Electromechanical System Dynamics, energy Conversion, and Electromechanical Analogies Modeling of Dynamic Systems Modeling of dynamic systems may be done in several ways: Use the standard equation of motion (Newton’s Law) for mechanical systems Use circuits theorems (Ohm’s law and Kirchhoff’s laws: KCL and KVL) Another approach utilizes the notation of energy to model the dynamic system (Lagrange model). 3 Mathematical Modeling and System Dynamics Newtonian Mechanics: Translational Motion • The equations of motion of mechanical systems can be found using Newton’s second law of motion. F is the vector sum of all forces applied to the body; a is the vector of acceleration of the body with respect to an inertial reference frame; and m is the mass of the body. • To apply Newton’s law, the free-body diagram in the coordinate system used should be studied. F ma 4 Force : Fcoulomb Translational Motion in Electromechanical Systems • Consideration of friction is essential for understanding the operation of electromechanical systems. • Friction is a very complex nonlinear phenomenon and is very difficult to model friction. • The classical Coulomb friction is a retarding frictional force (for translational motion) or torque (for rotational motion) that changes its sign with the reversal of the direction of motion, and the amplitude of the frictional force or torque are constant. • Viscous friction is a retarding force or torque that is a linear function of linear or angular velocity. 5 Newtonian Mechanics: Translational Motion • For one-dimensional rotational systems, Newton’s second law of motion is expressed as the following equation. M is the sum of all moments about the center of mass of a body (Nm); J is the moment of inertial about its center of mass (kg/m2); and is the angular acceleration of the body (rad/s2). M j 6 The Lagrange Equations of Motion • • • • • Although Newton’s laws of motion form the fundamental foundation for the study of mechanical systems, they can be straightforwardly used to derive the dynamics of electromechanical motion devices because electromagnetic and circuitry transients behavior must be considered. This means, the circuit dynamics must be incorporated to find augmented models. This can be performed by integrating torsional-mechanical dynamics and sensor/actuator circuitry equations, which can be derived using Kirchhoff’s laws. Lagrange concept allows one to integrate the dynamics of mechanical and electrical components. It employs the scalar concept rather the vector concept used in Newton’s law of motion to analyze much wider range of systems than F =ma. With Lagrange dynamics, focus is on the entire system rather than individual components. , D, are the total kinetic, dissipation, and potential energies of the system. qi and Qi are the generalized coordinates and the generalized applied forces (input). d d dt . d qi d dD d . Qi dqi dqi d q i 7 Electrical and Mechanical Counterparts Energy Mechanical Electrical Kinetic Mass / Inertia 0.5 mv2 / 0.5 j2 Inductor 0.5 Li2 Potential Gravity: mgh Spring: 0.5 kx2 Capacitor 0.5 Cv2 Dissipative Damper / Friction 0.5 Bv2 Resistor Ri2 8 Mathematical Model for a Simple Pendulum 1 2 1 The kinetic energy of the pendulum bob is : mv m l 2 2 The potential energy is : mgh mgl 1 cos . 2 y Ta, 0 l mg x x y mg cos 9 Electrical Conversion Input Electrical Energy Output Mechanical Energy Coupling Irreversible Electromagnetic Energy Conversion Field Energy Losses Energy Transfer in Electromechanical Systems For rotational motion, the electromag netic torque, as a function dWc (i, ) of current and angular displaceme nt, is : Te (i, ) d Where Wc di; where is the flux. i 10 Electromechanical Analogies • From Newton’s law or using Lagrange equations of motions, the secondorder differential equations of translational-dynamics and torsionaldynamics are found as m j d 2x dt 2 d 2 dt 2 Bv dx k s x Fa (t ) (Translational dynamics) dt Bm d k s Ta (t ) (Torsional dynamics) dt 11 For a series RLC circuit, find the characteristic equation and define the analytical relationships between the characteristic roots and circuitry parameters. d 2i R di 1 1 dva i 2 L dt LC L dt dt R 1 2 s s 0 L LC The characteristic roots are 2 s1 R 1 R 2L LC 2L 2 s2 R 1 R 2L LC 2L 12 Translational Damper, Bv (N-sec) Appied force Fa (t ) in Newton Linear vel ocity v(t ) (m/sec) Linear position x(t ) (m) Fa (t ) Bm v(t ) 1 v(t ) Fa (t ) Bm dx(t ) Fa (t ) Bm v(t ) Bm dt 1 t x(t ) Fa (t )dt Bv t0 x(t) Fa(t) Bm 13 Translational Spring, k (N) Appied force Fa (t ) in Newton Linear vel ocity v(t ) (m/sec) Linear position x(t ) (m) x(t) Fa (t ) k s x(t ) 1 x(t ) Fa (t ) ks Fa(t) dx(t ) 1 dFa (t ) v(t ) dt k s dt t Fa (t ) k s v(t )dt t0 14 Rotational Damper, Bm (N-m-sec/rad) Appied torque Ta (t ) (N - m) Angular velocity (t ) (rad/sec) Angular displacement (t ) (rad) (t) (t) Ta (t ) Bm (t ) 1 (t ) Ta (t ) Bm Fa(t) Bm d (t ) Ta (t ) Bm (t ) Bm dt 1 t (t ) Ta (t )dt Bm t 0 15 Rotational Spring, ks (N-m-sec/rad) Appied torque Ta (t ) (N - m) Angular velocity (t ) (rad/sec) Angular displacement (t ) (rad) (t) (t) Ta (t ) Bm (t ) 1 (t ) Ta (t ) ks d (t ) 1 dTa (t ) (t ) dt k s dt Fa(t) ks t Ta (t ) k s (t )dt t0 16 Mass Grounded, m (kg) Appied torque Ta (t ) (N - m) Linear vel ocity v(t ) (m/sec) x (t) v (t) Linear position x(t ) (m) dv d 2 x(t ) Fa (t ) m m dt dt 2 Fa(t) m 1 t v(t ) Fa (t )dt m t0 17 Mass Grounded, m (kg) Appied torque Ta (t ) (N - m) Angular velocity (t ) (rad/sec) Angular displaceme nt (t ) (rad) d d 2 (t ) Ta (t ) J J dt dt 2 1 t (t ) Ta (t )dt J t0 (t) (t) Fa(t) m 18 Steady-State Analysis • State: The state of a dynamic system is the smallest set of variables (called state variables) so that the knowledge of these variables at t = t0, together with the knowledge of the input for t t0, determines the behavior of the system for any time t t0. • State Variables: The state variables of a dynamic system are the variables making up the smallest set of variables that determine the state of the dynamic system. • State Vector: If n state variables are needed to describe the behavior of a given system, then the n state variables can be considered the n components of a vector x. Such vector is called a state vector. • State Space: The n-dimensional space whose coordinates axes consist of the x1 axis, x2 axis, .., xn axis, where x1, x2, .., xn are state variables, is called a state space. • State-Space Equations: In state-space analysis we are concerned with three types of variables that are involved in the modeling of dynamic system: input variables, output variables, and state variables. 19 State Variables of a Dynamic System x(0) initial condition y(t) Output u(t) Input Dynamic System State x(t) The state variables describe the future response of a system, given the present state, the excitation inputs, and the equations describing the dynamics 20 Electrical Example: An RLC Circuit x1 vC (t ); x2 iL (t ) 1 / 2Li L2 1 / 2Cvc2 L iL x1 (t 0 ) and x2 (t 0 ) is the total initial energy of the network USE KCL at the junction dv ic C c u (t ) iL dt iC u(t) vC C R 21 The State Differential Equation . x1 a11 x1 a12 x2 ... a1n xn b11u1 ... b1m u m . x 2 a 21 x1 a 22 x2 ... a 2 n xn b21u1 ... b2 m u m . x n a n1 x1 a n 2 x2 ... a nn xn bn1u1 ... bnmu m State Vector x1 a11 a12 a1n d x2 a 21 a 22 a 2 n . . . dt . xn a n1 a n 2 a nn x1 b11....b1m u1 x2 .......... .. . . bn1....bnm u m xn . x Ax Bu (State Differenti al Equation) A : State matrix; B : input matrix C : Output matrix; D : direct transmission matrix y Cx Du (Output Equation) 22 The Output Equation y1 h11 x1 h12 x2 ... h1n xn y 2 h21x1 h22 x2 ... h2 n xn yb ab1 x1 ab 2 x2 ... abn xn . y1 h11 h12 h1n y 2 h21 h22 h2 n y . . . . yb hb1 hb 2 hbn x1 x2 Hx . xn x Ax Bu (State Differenti al Equation) A : State matrix; B : input matrix H : Output matrix; D : direct transmission matrix y Hx Du (Output Equation) 23 Example 1: Consider the given series RLC circuit. Derive the differential equations that map the circuitry dynamics. dvc C i dt di L vc Ri v(t ) dt dvc 1 i dt C di 1 vc Ri v (t ) dt L R i(t) L V(t) C 24 Example 2: Using the state-space concept, find the state-space model and analyze the transient dynamics of the series RLC circuit. dv(t ) 1 i dt C di 1 (vc Ri v(t )) dt L x1 (t ) vc (t ); x2 (t ) i (t ) (These are the states) v(t ) is the control dx1 dvc 0 dx dt dt dt dx2 di 1 dt dt L 1 0 C v c 1 va Ax Bu R i L L 25 Continue with Values.. • • • Assume R = 2 ohm, L = 0.1 H, and C = 0.5 F, find the following coefficients. The initial conditions are assumed to be vc(t0)=vc0=15 V; and I (t0) = i0 = 5 A. Let the voltage across the capacitor be the output; y(t)= vc(t). The output equation will be • The expanded output equation in y • The circuit response depends on the value of v (t) 2 0 0 A and B - 10 - 20 10 x10 15 x0 x20 5 vc y 1 0 Hx; H 1 0 i v c y 1 0 0va Hx Du i 26