0.8 Linear systems with constant coefficients

By using Laplace domain tools to examine the RLC circuit seen before, the differential equation describing the behavior of this system is easy to deduce.

where

and

This yields

By multiplying both sides by the denominator of the fraction and taking the inverse Laplace transform, the final differential equation describing the system is determined:

Let's look at an example to see how these non-zero initial conditions come into play. A first order system is described by the following differential equation:

When we take the Laplace transform of the derivative of $y(t)$ , we must remember to include a term that represents the initial values of the system output.

By combining the $Y(s)$ terms we get

If we say that $q(s)=s+{}_0$ and $p(s)=1$ , and define $r(s)$ as $y(0^{-})$ , we can rearrange terms to get an expression relating $Y(s)$ to $U(s)$ that takes the initial conditions into account:

What we have here is the Laplace domain solution to a differential equation describing a dynamical system. There are two terms in this solution: one that relies on the input and one that does not. These parts correspond to the particular and homogeneous solutions, respectively. Taking the inverse Laplace transform, we can write (14) as:

Here, $y_{\mathrm{part}}(t)$ corresponds to $\frac{p(s)}{q(s)}U(s)$ and $y_{\mathrm{homo}}(t)$ corresponds to $\frac{r(s)}{q(s)}$ . This makes a lot of sense. The particular solution (forced response) depends on a combination of $q(s)$ , which describes how the system behaves independently, and $p(s)U(s)$ , which describes how the system reacts to the input $U$ . The homogeneous solution (natural response) depends on a combination of $q(s)$ and $r(s)$ , the latter of which contains information about the initial conditions of the system.

Let's say that we know the homogeneous solution, $y(t)$ , to a differential equation describing a system.

Goal: Using this solution, we want to try and figure out the system's $q(\frac{}{(t)})$ function given zero initial conditions.

Solution:

From above, we know that for a homogeneous solution

We can clear the denominator by moving the $q(\frac{}{(t)})$ to the left-hand side. And since we have zero initial conditions, $r(\frac{}{(t)})$ goes to 0:

The solution can quickly be determined by inspection because we know that the derivative of $e^t$ is $e^t$ . Therefore a solution of $q(\frac{}{(t)})=\frac{}{(t)}-1$ would work. However, a more systematic approach will be necessary for more difficult situations. We will investigate this approach here.

Again, we will do our work in the Laplace domain. By equating the Laplace transform of our homogeneous solution with the ratio of $r(s)$ and $q(s)$ as discussed above, we have:

Directly, we can see the solution for $q(s)$ : by simply setting the denominators equal to each other, $q(s)=s-1$ . This, of course, is the Laplace transform of the solution of $q(\frac{}{(t)})$ that we found by inspection above.

Now that we have the basics down, we'll look at a more complicated example. We are given

Goal: We would like to find the differential equation whose homogeneous solution is $y(t)$ .

Solution:

Again, we take the Laplace transform of $y(t)$ , and then combine the two resultant fractions into one ratio of polynomials:

Next, we equate the denominators of the last two fractions to find $q(s)$ :

Recalling the start of this module, multiplying $q(s)$ by $Y(s)$ and taking the inverse Laplace transform will yield the differential equation whose homogeneous solution is $y(t)$ :