First Order Unit Step Response

First order Unit Step Response 1. Unit Step Response Consider the initial value problem . x + kx = ru(t), x (0− ) = 0, k, r constants. This would model, for example, the amount of uranium in a nuclear reactor where we add uranium at the constant rate of r kg/year starting at time t = 0 and where k is the decay rate of the uranium. As in the previous note, adding an infinitesimal amount (r dt) at a time leads to a continuous response. We have x (t) = 0 for t < 0; and for t > 0 we must solve . x + kx = r, x (0) = 0. The general solution is x (t) = (r/k) + ce−kt . To find c, we use x (0) = 0: 0 = x (0) = r r +c ⇒ c = − . k k Thus, in both cases and u-format � 0 for t < 0 x (t) = r −kt ) ( 1 − e for t > 0 k = r (1 − e−kt )u(t). k (1) With r = 1, this is the unit step response, sometimes written v(t). To be more precise, we could write v(t) = u(t)(1/k)(1 − e−kt ). The claim that we get a continuous response is true, but may feel a bit unjustified. Let’s redo the above example very carefully without making this assumption. Naturally, we will get the same answer. The equation is . � x + kx = 0 r for t < 0 for t > 0, Solving the two pieces we get � c1 e−kt x (t) = r −kt k + c2 e x (0− ) = 0. (2) for t < 0 for t > 0. This gives x (0− ) = c1 and x (0+ ) = r/k + c2 . If these two are different there is a jump at t = 0 of magnitude x (0+ ) − x (0− ) = r/k + c2 − c1 . 18.03SC The initial condition x (0− ) = 0 implies c1 = 0, so our solution looks like � 0 for t < 0 x (t) = r −kt + c e for t > 0. 2 k To find c2 we substitute this into our differential equation (2). (We must use the generalized derivative if there is a jump at t = 0.) After substitution the left side of (2) becomes � . 0 for t < 0 x + kx = (r/k + c2 )δ(t) + −kc2 e−kt + r + kc2 e−kt for t > 0 � 0 for t < 0 = (r/k + c2 )δ(t) + r for t > 0. Comparing this with the right side of (2) we see that r/k + c2 = 0, or c2 = −r/k. This gives exactly the same solution (1) we had before. Figure 1 shows the graph of the unit step response (r = 1). Notice that it starts at 0 and goes asymptotically up to 1/k. 1/k v(t) t . Figure 1. Unit step is the response of the system x + kx = f (t) when f (t) = u(t). The Meaning of the Phrase ’Unit Step Response’ In this note looked at the system with equation . x + kx = f (t) and we considered f (t) to be the input. As we have noted previously, it sometimes makes more sense to consider something else to be the input. For example, in Newton’s law of cooling . T + kT = kTe it makes physical sense to call Te , the temperature of the environment, the input. In this case the unit step response of the system means the response to the input Te (t) = u(t), i.e. the solution to . T + kT = ku(t).