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ME_446_-_Lec_10-Signal_Simulation

ME_446_-_Lec_10-Signal_Simulation - System Simulation...

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Unformatted text preview: System Simulation 446-10 446- Prof. Neil A. Duffie University of Wisconsin-Madison Wisconsin- 10 Neil A. Duffie, 1996 All rights reserved 1 Utility of System Simulation Understanding system behavior - tests on actual system may be difficult - system may not yet exist in hardware Aid in design - answer "what if" questions - iterate to final system design - verify design before implementation Choice of simulation method 10 Correctness of models used 2 Spring-Mass-Damper System Spring- Massdamper c [N-s/m] [N2 mass M [kg] y(t) [m] spring k [N/m] x(t) [m] M d y(t) dy(t) +c + ky(t) = kx(t) 2 dt dt 1 d 2 y(t) 2 dy(t) + + y(t) = x(t) 2 n dt n dt 2 where n = 10 k M = c 2 kM 3 System Damping > 1: overdamped (unforced response not oscillatory) < 1: underdamped (unforced response is oscillatory) = 1: critically damped (unforced response not oscillatory) Two examples with n = 1.414 rad/s: = 0.3536 (underdamped) 10 = 1.0610 (slightly overdamped) 4 Obtaining Input-Ouput Response Input Solution of system equations with given input function for resulting output function: - yields explicit, exact output function - different solution each input function - difficult for all but simple input functions Time-based simulation of system: Time- formulation same for all input functions 10 - approximate solution based on time steps and numerical integration 5 Spring-Mass-Damper Solution Spring- Mass1 2 n (s2Y(s) - sy(0 - ) - y(0 - )) + 2 sY(s) - y(0 - ) + Y(s) = x(s) n ( ) For x(t) = 0 and initial conditions: mass position = 0.15 [m] mass velocity = 0 [m/s] 2 1 2 s Y(s) - 0.15s + (sY(s) - 0.15) + Y(s) = 0 2 n n 10 6 ( ) Spring-Mass-Damper Solution ( < 1) Spring- Mass( 2 s + 2 0.15 0.15(s + 2 n ) n n Y(s) = = 2 s 2 2s (s + n )2 + 1- 2 n 2 + + 1 n n ( ( ) ) Using table of Laplace transforms: y(t) = 0.15 1 1- 2 e - nt sin n 1- 2 t - ( ) 7 where = cos -1 10 Response Plot: x(t) = 0 [m], n = 1.414 [rad/s], = 0.3536 0.15 y(t) Displacement [m] 0.1 0.05 0 -0.05 0 1 2 3 4 5 6 7 8 9 10 10 Time [sec] 8 Spring-Mass-Damper Solution ( > 1) Spring- Mass( 2 s + 2 0.15 n n 0.15(s + 2 n ) Y(s) = = s 2 2s (s + a)(s + b) + + 1 2 n n where a,b = n n 2 - 1 Using table of Laplace transforms: 1 y(t) = 0.15 ( - a)e -at - ( - b)e -bt (b - a) ( ) 9 10 where = 2 n Response Plot: x(t) = 0 [m], n = 1.414 [rad/s], = 1.0610 0.14 y(t) Displacement [m] 0.12 0.1 0.08 0.06 0.04 0.02 0 0 1 2 3 4 5 6 7 8 9 10 10 Time [sec] 10 Spring-Mass Damper Simulation SpringState equations: 2 dy (t) 2 y(t) - y(t) = n x(t) - dt n dy(t) = y(t) dt Example: Euler integration with time step t 2 2 y (t + t) y (t) + n x(t) - y(t) - y(t) t n y(t + t) y(t) + y(t)t 10 11 Choice of Time Step t Choose time step so that t is shorter than the (significant) characteristic times of the system to be simulated. Characteristic times: - process and controller time constants - process delays - closed-loop system time constants closed- damped natural frequencies 1 1 c = = d n 1- 2 10 12 Spring-Mass-Damper Time Step Spring- Mass n = 1.414 [rad / s], = 0.3536 : d = n 1- 2 = 1.323 [rad / s] t << 1 = 0.756 [s] d n = 1.414 [rad / s], = 1.0610 : a,b = n n 2 - 1 = 2,1 1 1 1 1 t << min , = min , = 0.5 [s] a b 2 1 10 Use t = 0.01 [s]. Compare to t = 0.001 [s]. 13 Response Plot: x(t) = 0 [m], n = 1.414 [rad/s], = 0.3536 0.15 y(t) Displacement [m] 0.1 0.05 0 -0.05 0 1 2 3 4 5 6 7 8 9 10 10 Time [sec] 14 Response Plot: x(t) = 0 [m], n = 1.414 [rad/s], = 1.0610 0.14 y(t) Displacement [m] 0.12 0.1 0.08 0.06 0.04 0.02 0 0 1 2 3 4 5 6 7 8 9 10 10 Time [sec] 15 Position Control Simulation y(t) [m] x(t) [m] damper spring mass Actuator M [kg] c [N-s/m] k [N/m] [N t Integral control: x(t) = k c 0 (r(t) - y(t))dt x(t + t) x(t) + k c [r(t) - y(t)]t 2 2 y(t) - y(t) t y (t + t) y (t) + n x(t) - n y(t + t) y(t) + y(t)t 10 16 Underdamped System Step Response (overdamped process) 0.2 y(t) Displacement [m] 0.15 0.1 0.05 r(t) y(t) 0 0 1 2 3 4 5 6 7 8 9 10 10 Time [sec] 17 ...
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