Laplace Transforms (Part 4. Inverse Laplace Transforms)
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Inverse Laplace Transforms
Finding the Laplace
transform of a function is not terribly difficult if we’ve got a table of
transforms in front of us to use as we saw in the last section.
What we would like to do now is go the other way.
We are going to be given a transform, F(s), and ask what function (or functions) did we have
originally. As you will see this can be
a more complicated and lengthy process than taking transforms. In these cases we say that we are finding the
Inverse Laplace Transform of F(s) and use the following notation.
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As with Laplace transforms,
we’ve got the following fact to help us take the inverse transform.
Fact
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Given the two Laplace
transforms F(s) and G(s) then
 
for any constants a
and b.
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So, we take the inverse transform of the individual
transforms, put any constants back in and then add or subtract the results back
up.
Let’s take a look at a couple of fairly simple inverse transforms.
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Example 1 Find
the inverse transform of each of the following.
Solution
I’ve always felt that the key to doing inverse transforms
is to look at the denominator and try to identify what you’ve got based on
that. If there is only one entry in
the table that has that particular denominator, the next step is to make sure
the numerator is correctly set up for the inverse transform process. If it isn’t, correct it (this is always
easy to do) and then take the inverse transform.
If there is more than one entry in the table has a
particular denominator, then the numerators of each will be different, so go
up to the numerator and see which one you’ve got. If you need to correct the numerator to get
it into the correct form and then take the inverse transform.
So, with this advice in mind let’s see if we can take some
inverse transforms.
From the denominator of the first term it looks like the
first term is just a constant. The
correct numerator for this term is a “1” so we’ll just factor the 6 out
before taking the inverse transform.
The second term appears to be an exponential with a = 8 and the numerator is exactly what it needs to be. The third term also appears to be an
exponential, only this time
 
So, with a little more detail than we’ll usually put into
these,
 
The first term in this case looks like an exponential with
 
The second term almost looks like an exponential, except
that it’s got a 3s instead of just
an s in the denominator. It is an exponential, but in this case
we’ll need to factor a 3 out of the denominator before taking the inverse
transform.
The denominator of the third term appears to be #3 in the table with
 
So, let’s first rewrite the transform.
 
So, what did we do here?
We factored the 19 out of the first term. We factored the 3 out of the denominator of
the second term since it can’t be there for the inverse transform and in the
third term we factored everything out of the numerator except the 4! since
that is the portion that we need in the numerator for the inverse transform
process.
Let’s now take the inverse transform.
 
In this part we’ve got the same denominator in both terms
and our table tells us that we’ve either got #7
or #8.
The numerators will tell us which we’ve actually got. The first one has an s in the numerator and so this means that the first term must be #8
and we’ll need to factor the 6 out of the numerator in this case. The second term has only a constant in the
numerator and so this term must be #7, however, in order for this to be
exactly #7 we’ll need to multiply/divide a 5 in the numerator to get it correct
for the table.
The transform becomes,
 
Taking the inverse transform gives,
 
In this case the first term will be a sine once we factor
a 3 out of the denominator, while the second term appears to be a hyperbolic
sine (#17). Again, be careful with the difference
between these two. Both of the terms
will also need to have their numerators fixed up. Here is the transform once we’re done
rewriting it.
 
Notice that in the first term we took advantage of the
fact that we could get the 2 in the numerator that we needed by factoring the
8. The inverse transform is then,
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So, probably the best way to identify the transform is by
looking at the denominator. If there is
more than one possibility use the numerator to identify the correct one. Fix up the numerator if needed to get it
into the form needed for the inverse transform process. Finally, take the inverse transform.
Let’s do some slightly harder problems. These are a little more involved than the
first set.
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Example 2 Find
the inverse transform of each of the following.
Solution
From the denominator of this one it appears that it is
either a sine or a cosine. However,
the numerator doesn’t match up to either of these in the table. A cosine wants just an s in the numerator with at most a multiplicative constant, while
a sine wants only a constant and no s
in the numerator.
We’ve got both in the numerator. This is easy to fix however. We will just split up the transform into
two terms and then do inverse transforms.
 
Do not get too used to always getting the perfect squares
in sines and cosines that we saw in the first set of examples. More often than not (at least in my class)
they won’t be perfect squares!
In this case there are no denominators in our table that
look like this. We can however make
the denominator look like one of the denominators in the table by completing
the square on the denominator. So,
let’s do that first.
 
Recall that in completing the square you take half the
coefficient of the s, square this,
and then add and subtract the result to the polynomial. After doing this the first three terms
should factor as a perfect square.
So, the transform can be written as the following.
 
Okay, with this rewrite it looks like we’ve got #19 and/or #20’s from our table of transforms. However, note that in order for it to be a #19 we want just a constant in the numerator
and in order to be a #20 we need an s
In correcting the numerator always get the s
So, we will leave the transform as a single term and
correct it as follows,
 
We needed an s + 4
in the numerator, so we put that in.
We just needed to make sure and take the 4 back out by subtracting it
back out. Also, because of the 3
multiplying the s we needed to do
all this inside a set of parenthesis.
Then we partially multiplied the 3 through the second term and
combined the constants. With the
transform in this form, we can break it up into two transforms each of which
are in the tables and so we can do inverse transforms on them,
 
This one is similar to the last one. We just need to be careful with the
completing the square however. The
first thing that we should do is factor a 2 out of the denominator, then
complete the square. Remember that
when completing the square a coefficient of 1 on the s2 term is needed!
So, here’s the work for this transform.
 
So, it looks like we’ve got #21 and #22
with a corrected numerator. Here’s the
work for that and the inverse transform.
 
In correcting the numerator of the second term, notice
that I only put in the square root since we already had the “over 2” part of
the fraction that we needed in the numerator.
This one appears to be similar to the previous two, but it
actually isn’t. The denominators in
the previous two couldn’t be easily factored.
In this case the denominator does factor and so we need to deal with
it differently. Here is the transform
with the factored denominator.
 
The denominator of this transform seems to suggest that
we’ve got a couple of exponentials, however in order to be exponentials there
can only be a single term in the denominator and no s’s in the numerator.
To fix this we will need to do partial fractions on this
transform. In this case the partial
fraction decomposition will be
 
Don’t remember how to do partial fractions? In this example we’ll show you one way of getting
the values of the constants and after this example we’ll review how to get
the correct form of the partial fraction decomposition.
Okay, so let’s get the constants. There is a method for finding the constants
that will always work, however it can lead to more work than is sometimes
required. Eventually, we will need
that method, however in this case there is an easier way to find the
constants.
Regardless of the method used, the first step is to
actually add the two terms back up.
This gives the following.
 
Now, this needs to be true for any s that we should choose to put in. So, since the denominators are the same we
just need to get the numerators equal.
Therefore, set the numerators equal.
 
Again, this must be true for ANY value of s that we want to put in. So, let’s take advantage of that. If it must be true for any value of s then it must be true for
   
We found A by
appropriately picking s. We can B
in the same way if we chose
   
This will not always work, but when it does it will
usually simplify the work considerably.
So, with these constants the transform becomes,
 
We can now easily do the inverse transform to get,
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The last part of this example needed partial fractions to
get the inverse transform. When we
finally get back to differential equations and we start using Laplace
transforms to solve them, you will quickly come to understand that partial
fractions are a fact of life in these problems.
Almost every problem will require partial fractions to one degree or
another.
Note that we could have done the last part of this example
as we had done the previous two parts.
If we had we would have gotten hyperbolic functions. However, recalling the definition of the hyperbolic functions we
could have written the result in the form we got from the way we worked our
problem. However, most students have a
better feel for exponentials than they do for hyperbolic functions and so it’s
usually best to just use partial fractions and get the answer in terms of
exponentials. It may be a little more
work, but it will give a nicer (and easier to work with) form of the answer.
Be warned that in my class I’ve got a rule that if the
denominator can be factored with integer coefficients then it must be.
So, let’s remind you how to get the correct partial fraction
decomposition. The first step is to
factor the denominator as much as possible.
Then for each term in the denominator we will use the following table to
get a term or terms for our partial fraction decomposition.
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Factor in
denominator
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Term in partial
fraction
decomposition
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Notice that the first and third cases are really special
cases of the second and fourth cases respectively.
So, let’s do a couple more examples to remind you how to do
partial fractions.
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Example 3 Find
the inverse transform of each of the following.
Solution
Here’s the partial fraction decomposition for this part.
 
Now, this time we won’t go into quite the detail as we did
in the last example. We are after the
numerator of the partial fraction decomposition and this is usually easy
enough to do in our heads. Therefore,
we will go straight to setting numerators equal.
 
As with the last example, we can easily get the constants
by correctly picking values of s.
 
So, the partial fraction decomposition for this transform
is,
 
Now, in order to actually take the inverse transform we
will need to factor a 5 out of the denominator of the last term. The corrected transform as well as its
inverse transform is.
 
So, for the first time we’ve got a quadratic in the
denominator. Here’s the decomposition
for this part.
 
Setting numerators equal gives,
 
Okay, in this case we could use
 
The “long” way is to completely multiply out the right
side and collect like terms.
 
In order for these two to be equal the coefficients of the
s2, s and the constants must all be equal. So, setting coefficients equal gives the
following system of equations that can be solved.
 
Notice that I used s0
to denote the constants. This is habit
on my part and isn’t really required, it’s just what I’m used to doing. Also, the coefficients are fairly messy
fractions in this case. Get used to
that. They will often be like this
when we get back into solving differential equations.
There is a way to make our life a little easier as well
with this. Since all of the fractions
have a denominator of 47 we’ll factor that out as we plug them back into the
decomposition. This will make dealing
with them much easier. The partial
fraction decomposition is then,
 
The inverse transform is then.
 
With this last part do not get excited about the s3. We can think of this term as
 
and it becomes a linear term to a power. So, the partial fraction decomposition is
 
Setting numerators equal and multiplying out gives.
 
Setting coefficients equal gives the following system.
 
This system looks messy, but it’s easier to solve than it
might look. First we get C for free from the last
equation. We can then use the fourth
equation to find B. The third equation will then give A, etc.
When plugging into the decomposition we’ll get everything
with a denominator of 5, then factor that out as we did in the previous part
in order to make things easier to deal with.
 
Note that we also factored a minus sign out of the last
two terms. To complete this part we’ll
need to complete the square on the later term and fix up a couple of
numerators. Here’s that work.
 
The inverse transform is then.
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So, one final time.
Partial fractions are a fact of life when using Laplace
transforms to solve differential equations.
Make sure that you can deal with them.
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