Laplace Transforms (Part 5. Step Functions)
Posted by Suhaivi Hamdan on 8:17 PM with No comments
Step Functions
Before proceeding into solving differential equations we
should take a look at one more function.
Without Laplace transforms it would be
much more difficult to solve differential equations that involve this function
in g(t).
The function is the Heaviside function and is defined as,
Here is a graph of the Heaviside function.
Heaviside functions are often called step functions. Here is some alternate notation for Heaviside
functions.
We can think of the Heaviside function as a switch that is
off until t = c at which point it
turns on and takes a value of 1. So what
if we want a switch that will turn on and takes some other value, say 4, or 7?
Heaviside functions can only take values of 0 or 1, but we
can use them to get other kinds of switches.
For instance 4u_{c}(t)
is a switch that is off until t = c
and then turns on and takes a value of 4.
Likewise, 7u_{c}(t)
will be a switch that will take a value of 7 when it turns on.
Now, suppose that we want a switch that is on (with a value
of 1) and then turns off at t = c. We can use Heaviside functions to represent
this as well. The following function
will exhibit this kind of behavior.
Prior to t = c the
Heaviside is off and so has a value of zero.
The function as whole then for t
< c has a value of 1. When we hit
t = c the Heaviside function will
turn on and the function will now take a value of 0.
We can also modify this so that it has values other than 1
when it is on. For instance,
will be a switch that has a value of 3 until it turns off at
t = c.
We can also use Heaviside functions to represent much more
complicated switches.
Example 1 Write
the following function (or switch) in terms of Heaviside functions.
Solution
There are three sudden shifts in this function and so
(hopefully) it’s clear that we’re going to need three Heaviside functions
here, one for each shift in the function.
Here’s the function in terms of Heaviside functions.
It’s fairly easy to verify this.
In the first interval, t
< 6 all three Heaviside functions are off and the function has the
value
Notice that when we know
that Heaviside functions are on or off we tend to not write them at all as we
did in this case.
In the next interval,
In the third interval,
In the last interval,
So, the function has the correct value in all the
intervals.

All of this is fine, but if we continue the idea of using
Heaviside function to represent switches, we really need to acknowledge that
most switches will not turn on and take constant values. Most switches will turn on and vary
continually with the value of t.
So, let’s consider the following function.
We would like a switch that is off until t = c and then turns on and takes the
values above. By this we mean that when t = c we want the switch to turn on and
take the value of f(0) and when t = c + 4 we want the switch to turn on
and take the value of f(4), etc.
In other words, we want the switch to look like the following,
Notice that in order to take the values that we want the
switch to take it needs to turn on and take the values of ! We can use Heaviside functions to help us
represent this switch as well. Using
Heaviside functions this switch can be written as
Okay, we’ve talked a lot about Heaviside functions to this
point, but we haven’t even touched on Laplace
transforms yet. So, let’s start thinking
about that. Let’s determine the Laplace transform of (1). This is actually easy enough to derive so
let’s do that. Plugging (1)
into the definition of the Laplace transform gives,
Notice that we took advantage of the fact that the Heaviside
function will be zero if t < c and
1 otherwise. This means that we can drop
the Heaviside function and start the integral at c instead of 0. Now use the
substitution u = t c
and the integral becomes,
The second exponential has no u’s in it and so it can be factored out of the integral. Note as well that in the substitution process
the lower limit of integration went back to 0.
Now, the integral left is nothing more than the integral
that we would need to compute if we were going to find the Laplace transform of
f(t).
Therefore, we get the following formula
In order to use (2) the
function f(t) must be shifted by c, the same value that is used in the
Heaviside function. Also note that we
only take the transform of f(t) and
not f(tc)! We can also turn this around to get a useful
formula for inverse Laplace transforms.
We can use (2) to get the Laplace transform of a Heaviside function by itself. To do this we will consider the function in (2)
to be f(t) = 1. Doing this gives us
Putting all of this together leads to the following two
formulas.
Let’s do some examples.
Example 2 Find
the Laplace transform of each of the
following.
Solution
In all of these problems remember that the function MUST be in the form
before we start taking transforms. If it isn’t in that form we will have to
put it into that form!
So there are three terms in this function. The first is simply a Heaviside function
and so we can use (4)
on this term. The second and third
terms however have functions with them and we need to identify the functions
that are shifted for each of these. In
the second term it appears that we are using the following function,
and this has been shifted by the correct amount.
The third term uses the following function,
which has also been shifted by the correct amount.
With these functions identified we can now take the
transform of the function.
This part is going to cause some problems. There are two terms and neither has been
shifted by the proper amount. The
first term needs to be shifted by 3 and the second needs to be shifted by
5. So, since they haven’t been
shifted, we will need to force the issue.
We will need to add in the shifts, and then take them back out of
course. Here they are.
Now we still have some potential problems here. The first function is still not really
shifted correctly, so we’ll need to use
to get this shifted correctly.
The second term can be dealt with in one of two ways. The first would be to use the formula
to break it up into cosines and sines with arguments of t5 which will be shifted as we
expect. There is an easier way to do
this one however. From our table of Laplace transforms we have #16 and using that we can see that if
This will make our life a little easier so we’ll do it
this way.
Now, breaking up the first term and leaving the second
term alone gives us,
Okay, so it looks like the two functions that have been
shifted here are
Taking the transform then gives,
It’s messy, especially the second term, but there it
is. Also, do not get excited about the
This one isn’t as bad as it might look on the
surface. The first thing that we need
to do is write it in terms of Heaviside functions.
Since the t^{4}
is in both terms there isn’t anything to do when we add in the Heaviside
function. The only thing that gets
added in is the sine term. Notice as
well that the sine has been shifted by the proper amount.
All we need to do now is to take the transform.
Again, the first thing that we need to do is write the
function in terms of Heaviside functions.
We had to add in a “8” in the second term since that
appears in the second part and we also had to subtract a t in the second term since the t in the first portion is no longer there. This subtraction of the t adds a problem because the second
function is no longer correctly shifted.
This is easier to fix than the previous example however.
Here is the corrected function.
So, in the second term it looks like we are shifting
The transform is then,

Without the Heaviside function taking Laplace
transforms is not a terribly difficult process provided we have our trusty table of transforms. However, with the advent of Heaviside
functions, taking transforms can become a fairly messy process on occasion.
So, let’s do some inverse Laplace
transforms to see how they are done.
Example 3 Find
the inverse Laplace transform of each of the
following.
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.
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.
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 the 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,
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,
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,

So, as this example has shown, these can be a somewhat
messy. However, the mess is really only
that of notation and amount of work. The
actual partial fraction work was identical to the previous sections work. The main difference in this section is we had
to do more of it. As far as the inverse
transform process goes. Again, the vast
majority of that was identical to the previous section as well.
So, don’t let the apparent messiness of these problems get
you to decide that you can’t do them.
Generally they aren’t as bad as they seem initially.
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