Laplace Transforms Using the Definition

Today we’re going to be talking about how to perform a Laplace transform using the definition
of Laplace transforms. And in this particular problem, we’ve been asked to perform the
Laplace transform of sin 3t. And I’ve got the formula here for Laplace transform. It’s
denoted to this F of s and it’s equal to the integral from zero to infinity of e to
the negative st where s is a constant multiplied by our original function F of t.
So to perform our Laplace transform here, we’ll get F of s equals zero to infinity
and again this is all part of the formula. e to the negative st, we’re going to multiply
this e to the negative st by our original function. And our function is sine of 3t so
we go ahead and plug that in there. So now that we’ve got our function plugged into
our formula, all we need to do is evaluate the integral and then plug in our limits of
integration, evaluate the definite integral. In order to solve this integral, we’re going
to need to use integration by parts so we’ll go ahead and set u equal to sin 3t and we’ll
say that dv is going to be equal to e to the negative st dt. And remember, s is a constant
here. So then we’ll find the derivative of u. We’ll get du. The derivative of sin
is cos. Remember, according to chain rule, we’re going to have to multiply by the derivative
of 3t as well. So we’ll get 3 cos 3t dt and then we’re going to take the integral
of dv to find v. So when we do that, we get 1 over negative s e to the negative st. So
that will be v. So now we go ahead and plug in here for our
integration by parts formula which remember when we have the integral of u dv is equal
to u times v times the integral of vdu. So we’ll have u times v. We’ve got here,
negative one over s e to the negative st is our v times u sin of 3t. And actually before
we do that, we’re going to have to evaluate this on the range zero to infinity, then we’re
going to subtract the integral from zero to infinity of vdu. So now we’re onto this
part of the formula here, vdu. So when we do that, we’ll get negative 3 over s e to
the negative st times cos 3t dt. And remember that s is a constant. So this negative 3 over
s is a constant that can come out in front of our integral. So since we have two negatives
here, we’re going to go and make this a positive 3 over s. So we’ll bring that out
in front and we’ll get this out of our integral to make it as simple as possible.
Alright, so now we have e to the negative st cos 3t. And we haven’t made a lot of
progress. We’re actually going to need to use integration by parts again to evaluate.
So to do that, we’ll do something very similar to what we did before but this time we’ll
set u equal to cos of 3t, we’ll take the derivative to get negative 3 sin 3t because
the derivative of cos is negative sin. So we’ll have negative 3 sin of 3t and remember,
this 3 here comes from the derivative of 3t. According to chain rule, we have to multiply
by that. So we’ll have dt. Then dv is the other half of our original integral which
is e to the negative st dt. When we integrate that to get v, again we get the same thing
we got before, negative one over s e to the negative st. Alright so we will use our formula
again. We’ll get F of s equals negative one over s e to the negative st sin 3t and
of course, we have to evaluate that on zero to infinity, plus 3 over s. Because it’s
the start of our integral, this is where the start of our second integration by parts formula
is going to come into play. So again, u times v, so we’ll get negative one over s e to
the negative st times cos 3t is u times v. Then we’ll subtract the integral from zero
to infinity just like before I want to make sure that I say here that we’re going to
evaluate on zero to infinity. So then minus the integral from zero to infinity of vdu.
So again, because we have this negative one over s and we also have this negative 3, that’ll
turn into a positive 3 over s so we’ll have positive 3 over s e to the negative st sin
3t dt. Okay. So now we have a lot of simplification
to do. We will start by evaluating this first part here and maybe we can chunk this into
colors so we don’t get lost here. So we’ll start by evaluating this first part and we’ll
say, for F of s, you could evaluate on the range zero to b. When we’re dealing with
improper integrals, remember we could evaluate from zero to b and then we plug in infinity
for b, which is fine. I like to plug in infinity because I don’t like to deal with the additional
step. So remember, essentially, we’re plugging in for t not for s because again, s is a constant.
t is our variable. So when we plug in infinity, our upper limit of integration, for t, this
negative s of t here, we’re going to get negative s times infinity. In other words,
negative infinity. e to the negative infinity is going to become zero. So essentially, here,
we’ll end up with zero for our first term when we plug in for infinity because this
whole thing will be multiplied by zero. But then we’re going to subtract what we get
when we plug in zero or the lower limit of integration. So we’ll have a minus negative
1 over s so that’s going to be a plus 1/s e to the zero. Negative s times zero is going
to give us zero. So in fact, let’s just erase that because e to the zero is going
to be 1. Remember, anything raised to the first power is 1. So we get 1/s times sin
3t. Well, sine of 3 times zero is going to be sine of zero and we know that sine of zero
is zero. So that whole thing is just going to give us zero. We’ll go ahead and just
cancel both of these. We’ve now taken care of that entire term boxed there in orange.
Now, let’s go ahead and simplify. We’ll do this separately from the next part here
in color because there’s a lot going on here. So remember that we have to distribute
this 3s here to both pieces. So we’ve got 3/s times a negative 1/s so we’re going
to have minus 3/s squared times e to the negative st cos 3t and we’re going to be evaluating
this on zero to infinity. Then the second part here, we’re going to distribute that
3s again so we’ll have minus this 3s here and this 3s and that’s going to pull out
in front of the integral as well because it’s a constant. So we’ll end up with minus nine
over s squared times the integral from zero to infinity of e to the negative st sin 3t
dt. So the cool thing about this, now that we’ve kind of broken this apart, is we should
realize a couple of things. We’ll be evaluating but this integral from zero to infinity of
e to the negative st sin 3t dt is our original function. It’s the same as this up here
F of s. So we ended up right back in the same place where we started. We’re going to go
ahead and substitute F of s for this whole thing. So we’ll get there in a second but
for now, let’s go ahead and continue to evaluate here.
We’ll get F of s equals, remember that this whole part’s going to go away, and we’re
going to evaluate on the range zero to infinity. So remember, we’ll plug in infinity first
for t. When we plug it in here for t, we’ll get e to the negative infinity which is going
to end toward zero. so again, when we plug in infinity, we’re just going to get zero.
then we’re going to subtract what we get when we plug in zero. So when we plug in zero,
we’ll get plus because we get a minus negative 3s squared so we get plus 3s squared. e to
the zero is just 1 and then cosine of 3 times zero is going to give us cosine of zero. We
know cosine of zero to be equal to 1. So 3/s squared times 1 just gives us 3/s squared.
And that’s the whole middle section here. Now, the last part, we’ll get minus 9/s
squared times F of s. We said that we’re going to substitute F of s for that integral
because we ended up with the same integral that we’ve started out with, F of s equals.
We already defined it. So now what we’re going to want to do is solve for F of s. So
let’s go ahead and add 9/s squared times F of s to both sides to move it to the left-hand
side. So we’ll get F of s, so plus 9/s squared times F of s is equal to 3/s squared. now,
let’s go ahead and factor out the F of s. So when we do that, we get F of s times 1
plus 9/s squared equals 3/s squared. We’ll divide both sides by the quantity 1 plus 9/s
squared to solve for F of s which is what we wanted to do because remember that F of
s denotes the Laplace transform. So we’ll get 3/s squared all divided by 1 plus 9/s
squared. If we find a common denominator, and let’s actually go ahead and move up here, we’ll say F of
s equals 3/s squared in the denominator here we’ll multiply that 1 by s squared over
s squared so that essentially we get s squared plus 9 over s squared. And now instead of
dividing by a fraction, we’ll multiply by its reciprocal so we’ll get 3/s squared
and instead of dividing by s squared plus 9/s squared, we’ll multiply by what we get
when we flip it upside down, which is s squared divided by s squared plus 9. So as you can
see, we’ll get s squared and s squared to cancel and we’ll be left with F of s equals
3/s squared plus 9. And that’s it. That’s our final answer.
This F of s here is the Laplace transform of sine of 3t. and if you are looking at a
table of Laplace transforms, you’ll see that the formula for the Laplace transform
for sin 3t would be sin at where a is a constant. We already know that a is equal to 3 for us.
So it will be a/s squared plus a squared which matches what we get here for our answer when
we use the definition of the Laplace transform. So that’s it. I hope this video helped you,
guys and I will see you in the next one.