HW9_soln

HW9_soln - ME 132 Solutions 9 1 Step Responses First note...

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Unformatted text preview: ME 132 Solutions # 9 1 Step Responses First note that each of the given systems is stable. The poles, zeros, steady-state gains, dampings, and natural frequencies in each case are tabulated below. poles steady state gain damping natural frequency (a) − 1 , ( − 1 ± j )5 √ 2 1 . 707 10 (b) − 1 1 n/a n/a (c) − 10 , − . 3 ± j . 9539 1 . 3 1 (d) − . 6 ± . 8 j 1 . 6 1 (a) The transfer function is 100 s 3 + ( 1 + 10 √ 2 ) s 2 + ( 100 + 10 √ 2 ) s + 100 The form of the unit-step response is y ( t ) = 1 + c 1 e- t + e- 5 √ 2 t bracketleftBig c 2 cos(5 √ 2 t ) + c 3 sin(5 √ 2 t ) bracketrightBig The constants above can be found by forcing the initial conditions to be zero. This yields c 1 = − 1 . 1513 ,c 2 = 0 . 1513 ,c 3 = − . 0115. (b) The transfer function is 1 s + 1 , and the unit-step response is easily found to be y ( t ) = 1 − e- t (c) The transfer function is 10 s 3 + 10 . 6 s 2 + 7 s + 10 and the unit-step response is y ( t ) = 1 + c 1 e- 10 t + e- . 3 t [ c 2 cos(0 . 9539 t ) + c 3 sin(0 . 9539 t )] Forcing the initial conditions yields c 1 = − . 0105 ,c 2 = − . 9895 ,c 3 = − . 4215. (d) The transfer function is 1 s 2 + 1 . 2 s + 1 and the unit-step response is y ( t ) = 1 + e- . 6 t [ c 1 cos(0 . 8 t ) + c 2 sin(0 . 8 t )] Forcing the initial conditions yields c 1 = − 1 ,c 2 = − . 75. These step responses are plotted below. 5 10 0.5 1 1.5 Time (sec) (a) (b) 5 10 0.5 1 1.5 Time (sec) (c) (d) 1 2 PI control with Anti-windup A general simulink block diagram (with blocks included for saturation and anti-windup) for this situation is shown in Figure 1. step r u TL u_desired w r x’ = Ax+Bu y = Cx+Du motor MATLAB Function antiwindup.m zout To Workspace3 uout To Workspace2 thetaout To Workspace Sum1 Sum Saturation Mux Mux1 Mux Mux s 1 Integrator Ki Gain2 Kp Gain1 Constant r - theta Figure 1: Block Diagram The responses θ ( t ) and input signals generated by the controller u desired ( t ) are plotted below. In part (b) with actuator saturation, the actuator is at the saturation limit for much of the time. We are not able to push the motor as hard as we could in part (a)–the input is limited. This is reflected in the response, which is slower to rise and has larger oscillations in part (b) than (a). Note that in the case with saturation, at some points where θ is already greater than the reference signal r the input is positive, that is the controller is still pushing the motor to make θ larger still–for example at time t = 7 . 5. This is due to integrator windup that happened during the relatively long time span before...
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This note was uploaded on 10/20/2010 for the course ME 132 taught by Professor Tomizuka during the Spring '08 term at Berkeley.

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HW9_soln - ME 132 Solutions 9 1 Step Responses First note...

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