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

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 $$v(t)=\left(v_{0}+I_{F} / m\right) e^{-t / \tau_{1}}$$, $$t>0$$.