Example 3 Find
the inverse Laplace transform of each of the
following.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
All of these will use (3)
somewhere in the process. Notice that
in order to use this formula the exponential doesn’t really enter into the
mix until the very end. The vast
majority of the process is finding the inverse transform of the stuff without
the exponential.
In these problems we are not going to go into detail on
many of the inverse transforms. If you
need a refresher on some of the basics of inverse transforms go back and take
a look at the previous section.
(a)
In light of the comments above let’s first rewrite the
transform in the following way.
Now, this problem really comes down to needing f(t).
So, let’s do that. We’ll need
to partial fraction F(s) up. Here’s the partial fraction decomposition.
Setting numerators equal gives,
We’ll find the constants here by selecting values of s.
Doing this gives,
So, the partial fraction decomposition becomes,
Notice that we factored a 3 out of the denominator in
order to actually do the inverse transform.
The inverse transform of this is then,
Now, let’s go back and do the actual problem. The original transform was,
Note that we didn’t bother to plug in F(s). There really isn’t a
reason to plug it back in. Let’s just
use (3)
to write down the inverse transform in terms of symbols. The inverse transform is,
where, f(t) is,
This is all the farther that we’ll go with the
answer. There really isn’t any reason
to plug in f(t) at this point. It would make the function longer and
definitely messier. We will give
almost all of our answers to these types of inverse transforms in this form.
[Return to Problems]
(b)
This problem is not as difficult as it might at first
appear to be. Because there are two
exponentials we will need to deal with them separately eventually. Now, this might lead us to conclude that
the best way to deal with this function is to split it up as follows,
Notice that we factored out the exponential, as we did in
the last example, since we would need to do that eventually anyway. This is where a fairly common complication
arises. Many people will call the
first function F(s) and the second
function H(s) and then partial
fraction both of them.
However, if instead of just factoring out the exponential
we would also factor out the coefficient we would get,
Upon doing this we can see that the two functions are in
fact the same function. The only
difference is the constant that was in the numerator. So, the way that we’ll do these problems is
to first notice that both of the exponentials have only constants as
coefficients. Instead of breaking
things up then, we will simply factor out the whole numerator and get,
and now we will just partial fraction F(s).
Here is the partial fraction decomposition.
Setting numerators equal and combining gives us,
Setting coefficient equal and solving gives,
Substituting back into the transform gives and fixing up
the numerators as needed gives,
As we did in the previous section we factored out the
common denominator to make our work a little simpler. Taking the inverse transform then gives,
At this point we can go back and start thinking about the
original problem.
We’ll also need to distribute the F(s) through as well in order to get the correct inverse
transform. Recall that in order to use
(3)
to take the inverse transform you must have a single exponential times a
single transform. This means that we
must multiply the F(s) through the
parenthesis. We can now take the
inverse transform,
where,
[Return to Problems]
(c)
In this case, unlike the previous part, we will need to
break up the transform since one term has a constant in it and the other has
an s. Note as well that we don’t consider the
exponential in this, only its coefficient.
Breaking up the transform gives,
We will need to partial fraction both of these terms
up. We’ll start with G(s).
Setting numerators equal gives,
Now, pick values of s
to find the constants.
So G(s) and its
inverse transform is,
Now, repeat the process for H(s).
Setting numerators equal gives,
Now, pick values of s
to find the constants.
So H(s) and its
inverse transform is,
Putting all of this together gives the following,
where,
[Return to Problems]
(d)
This one looks messier than it actually is. Let’s first rearrange the numerator a
little.
In this form it looks like we can break this up into two
pieces that will require partial fractions.
When we break these up we should always try and break things up into
as few pieces as possible for the partial fractioning. Doing this can save you a great deal of
unnecessary work. Breaking up the
transform as suggested above gives,
Note that we canceled an s in F(s). You should always simplify as much a
possible before doing the partial fractions.
Let’s partial fraction up F(s) first.
Setting numerators equal gives,
Now, pick values of s
to find the constants.
So F(s) and its
inverse transform is,
Now partial fraction H(s).
Setting numerators equal gives,
Pick values of s
to find the constants.
So H(s) and its
inverse transform is,
Now, let’s go back to the original problem, remembering to
multiply the transform through the parenthesis.
Taking the inverse transform gives,
[Return to Problems]
