Introduction to the convolution - Mathematics Video Lecture - Engineering Mathematics

In this video, I'm going to
introduce you to the concept
of the convolution, one of the
first times a mathematician's
actually named something
similar to what
it's actually doing.
You're actually convoluting
the functions.
And in this video, I'm not
going to dive into the
intuition of the convolution,
because there's a lot of
different ways you
can look at it.
It has a lot of different
applications, and if you
become an engineer really of any
kind, you're going to see
the convolution in kind of a
discrete form and a continuous
form, and a bunch of
different ways.
But in this video I just want
to make you comfortable with
the idea of a convolution,
especially in the context of
taking Laplace transforms.
So the convolution theorem--
well, actually, before I even
go to the convolution theorem,
let me define what a
convolution is.
So let's say that I have
some function f of t.
So if I convolute f with g-- so
this means that I'm going
to take the convolution of f and
g, and this is going to be
a function of t.
And so far, nothing I've written
should make any sense
to you, because I haven't
defined what this means.
This is like those SAT problems
where they say, like,
you know, a triangle b means a
plus b over 3, while you're
standing on one leg or
something like that.
So I need to define this
in some similar way.
So let me undo this silliness
that I just wrote there.
And the definition of a
convolution, we're going to do
it over a-- well, there's
several definitions you'll
see, but the definition we're
going to use in this, context
there's actually one other
definition you'll see in the
continuous case, is the integral
from 0 to t of f of t
minus tau, times g of t-- let me
just write it-- sorry, it's
times g of tau d tau.
Now, this might seem like a very
bizarro thing to do, and
you're like, Sal, how
do I even compute
one of these things?
And to kind of give you that
comfort, let's actually
compute a convolution.
Actually, it was hard to find
some functions that are very
easy to analytically compute,
and you're going to find that
we're going to go into a lot of
trig identities to actually
compute this.
But if I say that f of t, if I
define f of t to be equal to
the sine of t, and I define
cosine of t-- let me do it in
orange-- or I define g of t to
be equal to the cosine of t.
Now let's convolute
the two functions.
So the convolution of f with g,
and this is going to be a
function of t, it equals this.
I'm just going to show you how
to apply this integral.
So it equals the integral--
I'll do it in purple-- the
integral from 0 to t of
f of t minus tau.
This is my f of t.
So it's is going to be sine of
t minus tau times g of tau.
Well, this is my g of t, so
g of tau is cosine of tau,
cosine of tau d tau.
So that's the integral, and
now to evaluate it, we're
going to have to break out
some trigonometry.
So let's do that.
This almost is just a very
good trigonometry and
integration review.
So let's evaluate this.
But I wanted to evaluate this
in this video because I want
to show you that this isn't some
abstract thing, that you
can actually evaluate
these functions.
So the first thing I want to
do-- I mean, I don't know what
the antiderivative of this is.
It's tempting, you see a sine
and a cosine, maybe they're
the derivatives of each
other, but this is the
sine of t minus tau.
So let me rewrite that sine of
t minus tau, and we'll just
use the trig identity, that
the sine of t minus tau is
just equal to the sine of t
times the cosine of tau minus
the sine of tau times
the cosine of t.
And actually, I just made a
video where I go through all
of these trig identities really
just to review them for
myself and actually to make a
video in better quality on
them as well.
So if we make this subsitution,
this you'll find
on the inside cover of any
trigonometry or calculus book,
you get the convolution of f and
g is equal to-- I'll just
write that f-star g; I'll just
write it with that-- is equal
to the integral from 0 to t of,
instead of sine of t minus
tau, I'm going to write this
thing right there.
So I'm going to write the sine
of t times the cosine of tau
minus the sine of tau times the
cosine of t, and then all
of that's times the
cosine of tau.
I have to be careful with my
taus and t's, and let's see, t
and tau, tau and t.
Everything's working so far.
So let's see, so
then that's dt.
Oh, sorry, d tau.
Let me be very careful here.
Now let's distribute this
cosine of tau out,
and what do we get?
We get this is equal to-- so f
convoluted with g, I guess we
call it f-star g, is equal to
the integral from 0 to t of
sine of t times cosine of
tau times cosine of tau.
I'm just distributing
this cosine of tau.
So it's cosine squared of tau,
and then minus-- let's rewrite
the cosine of t first, and I'm
doing that because we're
integrating with
respect to tau.
So I'm just going to write my
cosine of t first. So cosine
of t times sine of tau times
the cosine of tau d tau.
And now, since we're taking
the integral of really two
things subtracting from each
other, let's just turn this
into two separate integrals.
So this is equal to the integral
from 0 to t, of sine
of t, times the cosine squared
of tau d tau minus the
integral from 0 to t of cosine
of t times sine of tau cosine
of tau d tau.
Now, what can we do?
Well, to simplify it more,
remember, we're integrating
with respect to-- let
me be careful here.
We're integrating with
respect to tau.
I wrote a t there.
We're integrating with
respect to tau.
So all of these, this
cosine of t right
here, that's a constant.
The sine of t is a constant.
For all I know, t could
be equal to 5.
It doesn't matter that one
of the boundaries of our
integration is also a t.
That t would be a 5,
in which case these
are all just constants.
We're integrating only with
respect to the tau, so if
cosine of 5, that's a constant,
we can take it out
of the integral.
So this is equal to sine of t
times the integral from 0 to t
of cosine squared of tau d tau
and then minus cosine of t--
that's just a constant; I'm
bringing it out-- times the
integral from 0 to t of sine
of tau cosine of tau d tau.
Now, this antiderivative is
pretty straightforward.
You could do u substitution.
Let me do it here, instead
of doing it in our heads.
This is a complicated
problem, so we don't
want to skip steps.
If we said u is equal to sine of
tau, then du d tau is equal
to the cosine of tau, just
the derivative of sine.
Or we could write that
du is equal to the
cosine of tau d tau.
We'll undo the substitution
before we evaluate the
endpoints here.
But this was a little bit
more of a conundrum.
I don't know how to take the
antiderivative of cosine
squared of tau.
It's not obvious what that is.
So to do this, we're going
to break out some more
trigonometric identities.
And in a video I just recorded,
it might not be the
last video in the playlist, I
showed that the cosine squared
of tau-- I'm just using tau as
an example-- is equal to 1/2
times 1 plus the cosine
of 2 tau.
And once again, this is just a
trig identity that you'll find
really in the inside cover of
probably your calculus book.
So we can make this substitution
here, make this
substitution right there, and
then let's see what our
integrals become.
So the first one over here,
let me just write it here.
We get sine of t times the
integral from 0 to t of this
thing here.
Let me just take the 1/2 out,
to keep things simple.
So I'll put the 1/2 out here.
That's this 1/2.
So 1 plus cosine of 2 tau and
all of that is d tau.
That's this integral
right there.
And then we have this integral
right here, minus cosine of t
times the integral from--
let me be very clear.
This is tau is equal to 0
to tau is equal to t.
And then this thing right
here, I did some u
subsitution.
If u is equal to sine of
t, then this becomes u.
And we showed that du is equal
to cosine-- sorry, u is equal
to sine of tau.
And then we showed that du is
equal to cosine tau d tau, so
this thing right here
is equal to du.
So it's u du, and let's see if
we can do anything useful now.
So this integral right here, the
antiderivative of this is
pretty straightforward, so
what are we going to get?
Let me write this
outside part.
So we have 1/2 times
the sine of t.
And now let me take the
antiderivative of this.
This is going to be tau plus
the antiderivative of this.
It's going to be 1/2
sine of 2 tau.
I mean, we could have done
the u substitution.
we could have said u is equal to
2 tau and all of that, but
I think you could do that from
recognition, and if you don't
believe me, you just have to
take the derivative of this.
1/2 sine of 2 tau is the
derivative of this.
You multiply, you take the
derivative of the inside, so
that's 2, so the 2 and the 1/2
cancel out, and the derivative
of the outside, so
cosine of 2 tau.
And you're going to evaluate
that from 0 to t.
And then we have minus
cosine of t.
When we take the antiderivative
of this-- let
me do this on the side.
So the integral of u du,
that's trivially easy.
That's 1/2 u squared.
Now, that's 1/2 u squared, but
what was u to begin with?
It was sine of tau.
So the antiderivative of this
thing right here is 1/2 u
squared, but u is sine of tau.
So it's going to be 1/2u, which
is sine of tau squared.
And we're going to evaluate
that from 0 to t.
And we didn't even have to do
all this u substitution.
The way I often do it in my
head, I see the sine of tau,
cosine of tau.
if I have a function and I have
its derivative, I can
treat that function just like
as if I had an x there, so
it'd be sine squared of tau over
2, which is exactly what
we have there.
So it looks like we're
in the home stretch.
We're taking the convolution of
sine of t with cosine of t.
And so we get 1/2 sine of t.
Now, if I evaluate this thing
at t, what do I get?
I get t plus 1/2 sine
of 2t, that's when I
evaluated it at t.
And then from that I need to
subtract it evaluated at 0, so
minus 0 minus 1/2 sine of
2 times 0, which is
just sine of 0.
So this part right here, this
whole thing right there, what
does that simplify to?
Well this is 0, sine of 0
is 0, so this is all 0.
So this first integral right
there simplifies to 1/2 sine
of t times t plus
1/2 sine of 2t.
All right, now what does this
one simplify to over here?
Well, this one over here, you
have minus cosine of t.
And we're going to evaluate this
whole thing at t, so you
get 1/2 sine squared of t
minus 1/2 the sine of 0
squared, which is just 0,
so that's just minus 0.
So far, everything that we have
written simplifies to--
let me multiply it all out.
So I have 1/2-- let me just pick
a good color-- 1/2t sine
of t-- I'm just multiplying
those out-- plus 1/4 sine of t
sine of 2t.
And then over here I have minus
1/2 sine squared t times
cosine of t.
I just took the minus cosine t
and multiplied it through here
and I got that.
Now, this is a valid answer,
but I suspect that we can
simplify this more, maybe using
some more trigonometric
identities.
And this guy right there
looks ripe to simplify.
And we know that the sine of
2t-- another trig identity
you'll find in the inside cover
of any of your books--
is 2 times the sine of t
times the cosine of t.
So if you substitute that there,
what does our whole
expression equal?
You get this first term.
Let me scroll down
a little bit.
You get 1/2t times the sine of
t plus 1/4 sine of t times
this thing in here, so times
2 sine of t cosine of t.
Just a trig identity, nothing
more than that.
And then finally I have this
minus 1/2 sine squared t
cosine of t.
No one ever said this was
going to be easy, but
hopefully it's instructive
on some level.
At least it shows you that you
didn't memorize your trig
identities for nothing.
So let me rewrite the whole
thing, or let me just
rewrite this part.
So this is equal to 1/4.
Now, I have-- well let
me see, 1/4 times 2.
1/4 times 2 is 1/2.
And then sine squared
of t, right?
This sine times this sine is
sine squared of t cosine of t.
And then this one over here is
minus 1/2 sine squared of t
cosine of t.
And luckily for us, or lucky
for us, these cancel out.
And, of course, we had this
guy out in the front.
We had this 1/2t sine
t out in front.
Now, this guy cancels with this
guy, and all we're left
with, through this whole hairy
problem, and this is pretty
satisfying, is 1/2t sine of t.
So we just showed you that the
convolution-- if I define--
let me write our result.
I feel like writing this
in stone because
this was so much work.
But if we write that f of t is
equal to sine of t, and g of t
is equal to cosine of t, I
just showed you that the
convolution of f with g, which
is a function of t, which is
defined as the integral from 0
to t of f of t minus tau times
g of tau d tau, which was equal
to-- and I'll switch
colors here-- which was equal to
the integral from 0 to t of
sine of t minus tau times g of
tau d tau, that all of this
mess, all of this convolution,
it all equals-- and this is
pretty satisfying-- it all
equals 1/2t sine of t.
And the whole reason why I went
through all of this mess
and kind of bringing out the
neurons that had the trig
identities memorized or having
to reproof them or whatever
else is to just show you that
this convolution, it is
convoluted and it seems a little
bit bizarre, but you
really can take the convolutions
of actual
functions and get an
actual answer.
So the convolution of sine
of t with cosine of t is
1/2t sine of t.
So, hopefully, you have a little
of intuition of-- well,
not intuition, but you at least
have a little bit of
hands-on understanding of how
the convolution can be
calculated.