Constant coefficient linear systems describe LTI
systems and form the beginnings of the study of state-space
systems. In general, an

This is a linear differential equation with real-valued constant coefficients

If we go ahead and say that

Looking at differential equations in terms of polynomials such as

### Example 1

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.

RLC circuit: 2nd order (Impedances Labeled) |
---|

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:

The results from this example encourage us to apply Laplace techniques to earlier equations in this section. By taking the Laplace transform of Equation 4, we find that the transfer function of this system is simply the ratio of the two polynomials

**Non-zero Initial Conditions**

By saying above that the Laplace transform of

Note:

#### Example 2

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

By combining the

If we say that

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,

### Example 3

Let's say that we know the homogeneous solution,

Goal: Using this solution, we want to try and figure out the system's

Solution:

From above, we know that for a homogeneous solution

We can clear the denominator by moving the

The solution can quickly be determined by inspection because we know that the derivative of

Again, we will do our work in the Laplace domain. By equating the Laplace transform of our homogeneous solution with the ratio of

Directly, we can see the solution for

### Example 4

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

Solution:

Again, we take the Laplace transform of

Next, we equate the denominators of the last two fractions to find

Recalling the start of this module, multiplying