c24 - 18.03 Class 24, April 10, 2006 Unit impulse and step...

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18.03 Class 24, April 10, 2006 Unit impulse and step responses [1] In real life one often encounters a system with unknown system parameters. If it's a spring/mass/dashpot system you may not know the spring constant, or the mass, or the damping constant. But we can watch how it responds to various input signals. The simpler the input signal, the easier it will be to interpret the system response and get information about the system parameters, which will, in turn, allow us to predict the system response to other signals. For a start, we should be sure that the system is at rest before we do anything to it. So we'll start our experiment at t = 0 , and assume that before it the output signal, x(t) , is zero: x(t) = 0 for t < 0 . This is "rest initial conditions." Then we apply some input signal, and solve from this starting point. [2] Unit step response. Suppose we start loading a reactor at a constant rate. For a simple model, solve x' + kx = u(t) , rest initial conditions. (*) For t > 0 (*) is the same as x' + kx = 1 with x(0) = 0 . (**) The solution to (**) : xp = 1/k , xh = c e^{-kt} , so x = (1/k)(1 - e^{-kt})
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c24 - 18.03 Class 24, April 10, 2006 Unit impulse and step...

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