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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 4uc(t)
is a switch that is off until t = c
and then turns on and takes a value of 4.
Likewise, -7uc(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 one. 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.
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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, the first Heaviside function is now on while
the remaining two are still off. So,
in this case the function has the value.
In the third interval, the first two Heaviside functions are one
while the last remains off. Here the
function has the value.
In the last interval, all three Heaviside function are one
and the function has the value.
So, the function has the correct value in all the
intervals.
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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 wrote 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 buy 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(t-c)! 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 by f(t) = 1. Doing this gives us
Putting all of this together leads to the following two
formulas.
Let’ do some examples.
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Example 2 Find
the Laplace transform of each of the
following.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
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!
(a)
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 the following function is shifted,
and this has been shifted by the correct amount.
The third function uses,
which has also been shifted by the correct amount.
With these functions identified we can now take the
transform of the function.
[Return to Problems]
(b)
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
sifted 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 t-5 which will be shifted as we
expect. There is an easier way to do
this one however. From out 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 along 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
and . They are just numbers.
[Return to Problems]
(c)
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 t4
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.
[Return to Problems]
(d)
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,
[Return to Problems]
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