Introduction to the Convolution

⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.

Introduction to the Convolution

相關課程

0 / 750

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 discreet 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.

載入中...

Introduction to the Convolution

Introduction to the Convolution