0.7 Laplace example

Find the step response for the system above, when $u(t)$ is the input and $y(t)$ is the output (i.e. find $y(t)$ for $u(t)=\mathrm{step}(t)$ ).

With our previous definition of $q(d-(t))y(t)=p(d-(t))u(t)$ we can define the Laplace domain equivalents of $q$ and $p$ as:

When we multiply $q(s)$ times $Y(s)$ , we have to remember to include terms relating to the initial conditions of $y(t)$ . We normally think of the Laplace transform of $\frac{d y(t)}{d t}$ as $sY(s)$ . However, in reality, the general transform is as follows:

Therefore, using the initial conditions stated above, we can find the Laplace transforms of the first three derivatives of $y(t)$ .

We can now get a complete $s$ -domain equation relating the output to the input by taking the Laplace transform of $q(d-(t))y(t)=p(d-(t))u(t)$ . The transform of the right-hand side of this equation is simple as the initial conditions of $y(t)$ do not come into play here. The result is just the product of $p(s)$ and the transform of the step function $(1-s)()$ .

The left-hand side is somewhat more complicated because we have to make certain that the initial conditions are accounted for. To accomplish this, we take a linear combination of Laplace transform of the third derivative of y(t) , Laplace transform of the second derivative of y(t) , and Laplace transform of the first derivative of y(t) according to the polynomial $q(s)$ . That is to say, we use the coefficients of the s terms in $q(s)$ to determine how to combine these three equations. We take 1 of Laplace transform of the third derivative of y(t) plus 2 of Laplace transform of the second derivative of y(t) plus 1 of Laplace transform of the first derivative of y(t) plus 2.

When we sum these components, collect the $Y(s)$ terms, and set it equal to the right-hand side, we have:

Rearranging, we can find the solution to $Y(s)$ :

This solution can be looked at in two parts. The first term on the right-hand side is the particular (or forced) solution. You can see how it depends on $p(s)$ and $u(s)$ . The second term is the homogeneous (or natural) solution. The numerator of this term describes how the initial conditions of the system affect the solution (recall that $s^2+1$ was the part of the result of the linear combination of Laplace transform of the third derivative of y(t) , Laplace transform of the second derivative of y(t) , Laplace transform of the first derivative of y(t) ). The denominator of the second term is the $q(s)$ polynomial; it serves to describe the system in general.