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- 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.

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