PROFESSOR: Welcome back.
So in this session we're going to look at Laplace transform.
And we'll start with asking you to recall the definition that
you saw in class, then to use the definition to compute
the Laplace transform of the function 1, exponential a*t,
and the delta function.
For each one of these, give the domain of definition,
or convergence of the integral.
For the last question, you're asked
to use the results of question 2 to give
the Laplace transform of this linear combination
of functions.
In the last part, you're asked to compute Laplace transform
of cosine and sine.
So why don't you pause the video, take a few minutes,
and work through that.
Welcome back.
So let's start with the definition.
Laplace transform of the function s was defined
as the integral from 0 minus to infinity of the function f of t
exponential minus s*t dt.
So note here, the interval of integration
is 0 minus to infinity.
So using this definition, we can go ahead
and compute our first Laplace transform, L of 1.
So I'm just going to substitute 1 in that integral,
which gives me basically exponential minus s*t dt,
which is just the integral of the exponential minus s*t over
minus s from 0 minus to infinity.
And if I expand this, basically, I end up with 1/s,
the minus reverses the order of integration, so I start with 0,
which is 1, minus the limit when T goes to infinity,
of exponential minus s*T.
So here the sign of s becomes important.
If s was positive, then this term
would go to 0 as t goes to infinity.
If s is negative, then this term diverges,
and so we're not anymore in the domain of convergence
of the Laplace integral.
But really, s could be also complex.
So what we're interested in is really
the sign of the real part of s.
So if the real part of s is positive,
this term is goes to 0, and the Laplace transform of 1
is just 1/s.
And if the real part of s is negative,
then the Laplace diverges.
So the domain of convergence in which you want to be on
is the one where the real part of s is positive.
OK.
So let's move to the second one.
The second one is a Laplace of exponential of a*t.
So we can move a bit faster now, and just
merge the two exponentials.
Exponential minus 0 to infinity-- 0 to infinity
of this exponential.
Clearly this is just, again, a case of exponential integration
between the two bounds.
And here again we're going to hint a problem
with the domain of convergence where we need-- so
let me just write these again.
So we're going to have here a minus--
so we have our a minus s.
So we have the limit again when T
goes to infinity of exponential minus s plus a capital
T minus 1.
And here, again, we need to impose the condition
that the real part of a minus s be
negative to have the domain of convergence of the integral.
And then we're left with the Laplace integral
being 1 over s minus a.
If the real part is positive, then we have divergence.
So the domain of convergence of this Laplace
is the one defined by the real part of a less
than the real part of s.
OK.
So let's do the last one.
The last one is the Laplace transform of the delta function
that we saw before.
That's 0 minus to infinity delta exponential minus s*t dt.
So from the previous recitations,
we saw that on this domain, from 0 minus to infinity,
the delta is 0 everywhere except at 0, where it basically
assigned a value of this function at t equal to 0.
So basically we're just left with exponential of 0,
which is 1.
And this computation was really easy,
due to the properties of the delta function.
So that ends roughly this first part, except that you can also
notice here that the domain of convergence
for the Laplace for delta is all s.
There's no condition.
So the last part, next question, asked
us to compute the Laplace transform
of a linear combination of functions.
So this one is 7 plus 8 exponential 2t plus 9
exponential minus 3t.
So here, as you saw the Laplace is an integral,
and so the Laplace transform of this linear combination
of functions is the linear combination of the Laplace
transform of the functions individually.
And so we can just rewrite this as 7 Laplace of 1
plus 8 Laplace of exponential 2t plus 9 Laplace
of exponential minus 3t.
And here we can see how we can recycle the results from
the previous part, as we computed the Laplace transform
of 1, and we computed the Laplace transform exponential
a*t, which we're going to be able to apply in these two
cases.
So we can write the results directly here.
And I'm just going to not rewrite everything, just
include it.
So the Laplace of 1, we found it earlier to be 1/s.
The Laplace of exponential 2t we just found here,
and it would be s minus 2.
The Laplace of exponential minus 3t would be s minus minus 3,
so it's s plus 3 with the 9.
And we're done.
So for the last part, you're asked
to compute the Laplace transform of cosine and sine,
and you should know these by heart.
But just a trick to remember it--
I just want to remind you, again,
of the linearity and the fact that you could also
use the Euler formula.
Given what we just derived, you could also
write this, again due to the linearity of the integral
as a linear combination of Laplace of cosine
and sine here.
And we know that Laplace of the exponential a*t is just what we
computed here.
So that would be s minus i*omega,
which you can just rewrite like this.
And then identify just the real part
with the real part, the imaginary part
with the imaginary part, which finishes our problem.
And this is a quick way of checking
that you have that right.
To give you the Laplace transforms of cosine and sine.
So that ends the problem for today.
The key point was just remembering the definition
of the Laplace transform, and then
learning how to use it for different cases,
and identify the domains of convergence.