Homework4_Solutions

# Homework4_Solutions - S06 Professor T-C Tsao 8.25 MA 107...

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MA 107 Modeling And Analysis of Dynamic Systems S06 Professor T-C. Tsao HW 4 Sol. 1. 3.4 (c) 2. 3.5 3. 8.1

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MA 107 Modeling And Analysis of Dynamic Systems S06 Professor T-C. Tsao 8.4 Free response and step response have the same transient response behavior. (a) x(0)=0 f(t) =1, Xs.s. = 10, pole is at -0.5 so the settling time, which is where the 98% of the steady state value is 4/0.5 = 8 ( sec.) (b) x(0)=5, f(t)=0, Xs.s. = 0, at t=8 sec. x(t) =(1-0.98)*5 If we considered x(0)=5 and f(t)=1, then Xs.s. = 10, and at t=8 sec. x(t)= 0.02*5+0.98*10=9.9. (c) x(0) =0, f(t) = 20, Xs.s. = 200, at t=8 sec. x(t)= 200*0.98. 8.9 From Figure 8.1.2, we have the math model as follows: For constant input flow rate 10, the solution is
MA 107 Modeling And Analysis of Dynamic Systems S06 Professor T-C. Tsao 8.24

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MA 107 Modeling And Analysis of Dynamic Systems

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Unformatted text preview: S06 Professor T-C. Tsao 8.25 MA 107 Modeling And Analysis of Dynamic Systems S06 Professor T-C. Tsao 8.48 Notice that there are three unknowns: m,c,k. From the step response plot, we can extract the steady state value and two pieces of information regarding the transient response, which could be from peak time, overshoot, oscillation, frequency, convergence envelop, initial slope (velocity), etc. Here we use the peak time and overshoot since they are easily identified from the plot. MA 107 Modeling And Analysis of Dynamic Systems S06 Professor T-C. Tsao 8.50 MA 107 Modeling And Analysis of Dynamic Systems S06 Professor T-C. Tsao The math model is ( ) ( ) dE t kE t dt = − With E(to) = 98.6-70=28.6 E(11-to)= 94.6-70=24.6 E(11.5-to)=93.4-70=23.4 Two data and two inknowns to and k: E(t-to) = 28.6e-k(t-to)...
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Homework4_Solutions - S06 Professor T-C Tsao 8.25 MA 107...

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