You are about to erase your work on this activity. Are you sure you want to do this?
Updated Version Available
There is an updated version of this activity. If you update to the most recent version of this activity, then your current progress on this activity will be erased. Regardless, your record of completion will remain. How would you like to proceed?
We introduce the unit step function and some of its applications.
The Unit Step Function
In the next section we’ll consider initial value problems
where , , and are constants and is piecewise continuous. In this section we’ll
develop procedures for using the table of Laplace transforms to find Laplace
transforms of piecewise continuous functions, and to find the piecewise continuous
inverses of Laplace transforms.
Use the table of Laplace transforms to find the Laplace transform of
Since the formula for changes at , we write
To relate the first term to a Laplace transform, we add and subtract
in (eq:8.4.2) to obtain
To relate the last integral to a Laplace transform, we make the change of variable and rewrite
the integral as
Since the symbol used for the variable of integration has no effect on the
value of a definite integral, we can now replace by the more standard and
This and (eq:8.4.3) imply that
Now we can use the table of Laplace transforms to find that
Laplace Transforms of Piecewise Continuous Functions
We’ll now develop the method of Example example:8.4.1 into a systematic way to find the Laplace
transform of a piecewise continuous function. It is convenient to introduce the unit
step function, defined as
Thus, “steps” from the constant value to the constant value at . If we replace by
in (eq:8.4.4), then
that is, the step now occurs at (See figure below)
The step function enables us to represent piecewise continuous functions conveniently.
For example, consider the function
where we assume that and are defined on , even though they equal only on the
indicated intervals. This assumption enables us to rewrite (eq:8.4.5) as
To verify this, note that if then and (eq:8.4.6) becomes
If then and (eq:8.4.6) becomes
We need the next theorem to show how (eq:8.4.6) can be used to find .
Let be defined on Suppose and exists for Then exists for , and
From this and the definition of ,
The first integral on the right equals zero. Introducing the new variable of
integration in the second integral yields
Changing the name of the variable of integration in the last integral from to
Second Shifting Theorem] If and exists for then exists for and
Recall that the First Shifting Theorem (Theorem thmtype:8.1.3) states that multiplying a function
by corresponds to shifting the argument of its transform by units. Theorem thmtype:8.4.2 states
that multiplying a Laplace transform by the exponential corresponds to shifting the
argument of the inverse transform by units.