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8.5: Ideal Impulse Response of a Standard Stable First Order System

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    7674
  • From Equation 3.4.8, the problem statement for a standard stable 1st order system is

    \[\mathrm{ODE}+\mathrm{IC}: \quad \dot{x}+\left(1 / \tau_{1}\right) x=b u(t), \quad x(0)=x_{0}, \quad \text { find } x(t) \text { for } t>0\label{eqn:6.4}\]

    Let the input function be the ideal impulse, \(u(t)=I_{U} \delta(t)\). Although there are several methods for finding response to an ideal impulse, the conventional Laplace-transform approach is relatively simple and probably the most instructive, so we will use this method. With \(L[\delta(t)]=1\) from Equation 8.4.6, the steps of the solution are:

    \[L[\mathrm{ODE}]: \quad s X(s)-x_{0}+\left(1 / \tau_{1}\right) X(s)=b U(s)=b I_{U} \times 1\]

    \[\Rightarrow \quad X(s)=\frac{x_{0}+b I_{U}}{s+1 / \tau_{1}}\label{eqn:8.14}\]

    \[\Rightarrow \quad x(t)=\left(x_{0}+b I_{U}\right) e^{-t / \tau}\label{eqn:8.15}\]

    As a check, we evaluate the solution at \(t\) = 0:

    \[x(0)=\left(x_{0}+b I_{U}\right) e^{-0 / \tau_{1}}=x_{0}+b I_{U} \neq x_{0}\label{eqn:8.16}\]

    It appears that something is wrong with response solution Equation \(\ref{eqn:8.15}\), because Equation \(\ref{eqn:8.16}\) for \(x(0)\) contradicts the original IC, \(x(0)=x_{0}\). In order to explain this discrepancy, we must account mo e carefully for the nature of response to an ideal impulse, and for the distinction between the three different reference instants that are defined in the previous section:

    1. \(t=0^{-}\), the instant just before activity of the ideal impulse function;
    2. \(t=0\), the instant when \(\delta(t-0)\) acts; and
    3. \(t=0^{+}\), the instant just after activity of the ideal impulse function.

    The more detailed analysis follows.

    We begin by integrating the basic ODE, \(\dot{x}+\left(1 / \tau_{1}\right) x=b I_{U} \delta(t)\), across the ideal impulse function, just from \(t=0^-\) to \(t=0^+\):

    \[\int_{\tau=0^{-}}^{\tau=0^{+}} \frac{d x}{d \tau} d \tau+\frac{1}{\tau_{1}} \int_{\tau=0^{-}}^{\tau=0^{+}} x d \tau=b \int_{\tau=0^{-}}^{\tau=0^{+}} I_{U} \delta(\tau-0) d \tau\label{eqn:8.17}\]

    The fi left-hand-side term of Equation \(\ref{eqn:8.17}\) is identically equal to \(x\left(0^{+}\right)-x\left(0^{-}\right)\); this result introduces the new quantity \(x\left(0^{+}\right)\), which apparently is the post-impulse initial value at \(t=0^{+}\); the pre-impulse initial value, \(x\left(0^{-}\right) \equiv x_{0}\), is the original IC specified in Equation \(\ref{eqn:6.4}\), but now understood to exist at \(t=0^{-}\). The right-hand-side integral of Equation \(\ref{eqn:8.17}\) gives the finite area \(I_{U}\) under the infinite ideal impulse function. In contrast, the second left-hand-side term of Equation \(\ref{eqn:8.17}\) has a finite integrand; we can use the trapezoid rule to approximate the value of the integral and find that it is zero:

    \[\int_{\tau=0^{-}}^{\tau=0^{+}} x d \tau \approx \lim _{\Delta t \rightarrow 0}\left\{\frac{1}{2}\left[x\left(0^{+}\right)+x\left(0^{-}\right)\right] \Delta t\right\}=0\label{eqn:8.18}\]

    In Equation \(\ref{eqn:8.18}\), \(\Delta t \equiv 0^{+}-0^{-} \rightarrow 0\), in the spirit of the idealization that the ideal impulse function acts over an infinitesimal time interval. Therefore, Equation \(\ref{eqn:8.17}\) gives \(x\left(0^{+}\right)-x_{0}+1 / \tau_{1} \times 0=b \times I_{U}\), or

    \[x\left(0^{+}\right)=x_{0}+b I_{U}\label{eqn:8.19}\]

    Comparing Equations \(\ref{eqn:8.16}\) and \(\ref{eqn:8.19}\) shows that the term \(x(0)\) in the former equation really should be the post-impulse initial value, \(x\left(0^{+}\right)\).

    Equation \(\ref{eqn:8.19}\) can also be found directly from the Laplace transform \(X(s)\) by application of the initial-value theorem:

    \[\lim _{t \rightarrow 0^{+} \text {from } t>0} f(t) \equiv f\left(0^{+}\right)=\lim _{s \rightarrow \infty}[s F(s)]\label{eqn:8.20}\]

    We present only the relevant application of the initial-value theorem here; derivation of the theorem itself is postponed to Section 8.6. Applying Equation \(\ref{eqn:8.20}\) to Laplace transform solution Equation \(\ref{eqn:8.14}\) gives

    \[x\left(0^{+}\right)=\lim _{s \rightarrow \infty}[s X(s)]=\lim _{s \rightarrow \infty}\left(s \times \frac{x_{0}+b I_{U}}{s+1 / \tau_{1}}\right)=\lim _{s \rightarrow \infty}\left(\frac{s}{s} \times\left(x_{0}+b I_{U}\right)\right)=x_{0}+b I_{U}\]

    This consequence of the initial-value theorem is identical to Equation \(\ref{eqn:8.19}\), which was derived by a different method.

    Finally, we present an application of Equation \(\ref{eqn:8.19}\) to a mechanical system, and we show that the results conforms with a principle of mechanics. Consider a mass-damper system with mass \(m\) and viscous damping constant \(c\), as drawn at right. The mass is initially, moving with velocity \(v\left(0^{-}\right)=v_{0}\) when, at \(t=0^-\), the mass is disturbed by an ideal force impulse, \(f_{x}(t)=I_{F} \delta(t)\). The equation of motion is \(m \dot{v}+c v=f_{x}(t)\), or in terms of a standard stable 1st order system, \(\dot{v}+\left(1 / \tau_{1}\right) v=b u(t)\), in which \(\tau_{1}=m / c\), \(b=1 / m\), and \(u(t)=f_{x}(t)=I_{F} \delta(t)\), so that \(I_{U}=I_{F}\). The initial (pre-impulse) momentum of the mass is \(m v_{0}\). According to Equation \(\ref{eqn:8.19}\), the post-impulse velocity of the mass is \(v\left(0^{+}\right)=v_{0}+(1 / m) \times I_{F}=v_{0}+I_{F} / m\), so that the post-impulse momentum of the mass is \(m v\left(0^{+}\right)=m v_{0}+I_{F}\). In words, the momentum of the mass is increased exactly by the magnitude of the ideal impulse, in agreement with the impulse-momentum theorem, Equation 8.2.2. Force impulse \(I_{F} \delta(t)\) is a mathematically ideal impulse, not a physically realizable excitation, so the mathematical change in momentum occurs instantly, and the viscous dashpot has no influence on this instantaneous response. From Equation \(\ref{eqn:8.15}\), the post-impulse response of the mass is \(v(t)=\left(v_{0}+I_{F} / m\right) e^{-t / \tau_{1}}\), \(t>0\).