PROBLEM 3.1 KNOWN:K = 2 V/kg F(t) = constant = A Possible range of A: 1 kg to 10 kg FIND:y(t) SOLUTION We will model the input as a static value and interpret the static output that results. To do this, this system is modeled as a zero order equation. y(t) = KF(t) where F(t) is constant for all time; so y(t) is constant At the low end of the range, F(t) = A = 1 kg, then y = KF(t) = (2 V/kg)(1 kg) = 2 V At the high end of the range, F(t) = A = 10 kg, then y = KF(t) = (2 V/kg)(10 kg) = 20 V Hence, the output will range from 2 V to 20 V depending on the applied static input value. Clearly, the model shows that if K were to be increased, the static output y would be increased. Here K is a constant, meaning that the relationship between the applied input and the resulting output is constant. The calibration curve must be a linear one. Notice how K, through its value, takes care of the transfer in the units between input F and output y. COMMENT Because we have modeled this system as a zero order responding system, we have eliminated any accommodation for a transient response in the system model solution. The forcing function (i.e., input signal) is constant (i.e., static) for all time. So in the transient sense, this solution for y is valid only under static conditions. However, it is correct in its prediction in the steady output value following any change in input value.
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PROBLEM 3.2 KNOWN:System model FIND:75%, 90% and 95% response times ASSUMPTIONS:Unless noted otherwise, all initial conditions are zero. SOLUTION We seek the rise time to 75%, 90% and 95% response. For a first order system, the percent response time is found from the time response of the system to a step change in input. The error fraction for such an input is given by C(t) = e-t/τfrom which the percent response at time t is found by % response = (1 - Γ(t)) x 100 = (1 - e-t/τ) x 100 from which t is computed directly. Alternatively, Figure 3.7 could be used. For a second order system, the system response depends on the damping ratio and natural frequency of the system and can be established from either (3.15) or Figure 3.14. By direct comparison to the general model of a first order system, τ= 0.4 s. Hence, 75% = (1 - e-t/0.4) x 100 or t75= 0.55 s 90% = (1 - e-t/0.4) x 100 or t90= 0.92 s 95% = (1 - e-t/0.4) x 100 or t95= 1.2 s Alternatively, Figure 3.7 could be used. )(44.0)tUTTa=+•