Response to Exponential Inputs

Exponential

1 Response to Exponential Input

We start with an equation of form

$$\frac{dy}{dt} = ay + e^{st} \tag{1}$$

The output, $y(t)$, is called the “exponential response.”

We have a starting deposit (initial condition) $y = y(0)$ at time $t = 0$, and additional deposits equalling $e^{st}$.

With Equation 1, the solutions will be a multiple of $e^{st}$, written

$$y_p = Ye^{st} \tag{2}$$

($y_p$ is a particular solution of y.)

So we can substitute Equation 2 into Equation 1 and solve for $Y$.

$$\begin{gathered} Yse^{st} = aYe^{st} + e^{st} \\ (s-a)Y = 1 Y = \frac{1}{s-a} \end{gathered}$$

We’ve found a particular solution, but to find the complete set of solutions, we need to add all null solutions, where the input is zero.

$$\begin{gathered} \frac{dy}{dt} = ay \\ y = e^at \end{gathered}$$

So the complete set of solutions $y_p + y_n$ is:

$$y(t) = \frac{e^{st}}{s-a} + Ce^{at}$$

and

$$y(0) = \frac{1}{s-a} + C$$

The complete solution that satisfies the initial conditions is:

$$\begin{aligned} y(t) &= \frac{e^{st}}{s-a} + \left[ y(0) - \frac{1}{s-a} \right]e^{at} \\ &= \frac{e^{st}-e^{at}}{s-a} + y(0)e^{at} \end{aligned}$$

$y(0)e^{at}$ is the term growing out of the initial deposit (to use the money-in-the-bank scenario again), and $\frac{e^{st}-e^{at}}{s-a}$ is the term growing out of the additional deposits.

$\frac{e^{st}-e^{at}}{s-a}$ is the “very particular solution”, since it equals $0$ at $t=0$, giving us the form $y_{vp} + y_n$.

1.1 Resonance

There is one problem: what if $s=a$?

Then you have resonance: you are putting money in with the same exponential as the natural growth of the money, and our formula has to change. The very particular solution will then be $\frac{0}{0}$: the formula has broken down.

So IF $s=a$,

$$y_{vp} + y_n = te^{at} + y(0)e^{at} \tag{3}$$

How do we get the resonant answer? We use l’Hôpital’s rule.

$$\begin{aligned} \lim_{s\to a} \frac{e^{st}-e^{at}}{s-a} &= \frac{\frac{d}{ds}\left(e^{st} - e^{at} \right)}{\frac{d}{ds}(s-a)} \\ &= \frac{te^{st}}{1} \\ &= te^{at} \end{aligned} \tag{4}$$

¹

¹ Clearly, Equation 4 works whether you take the derivative with respect to $s$ or with respect to $a$.

1.2 Sum

So, to recap:

1. Find a particular solution $y_p$
2. Find the nullspace, all solutions where the input (in this case $e^st$ equals zero)
3. Combine the particular and null solutions, then find an equation in terms of $y(0)$
4. Check for resonance