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- A lot of what we do with Laplace transforms, taking
- them and taking their inverse, it's a
- lot of pattern matching.
- And it shouldn't just be a mechanical thing, and that's
- why I've gone through the exercise of showing
- you why they work.
- But in order to just kind of make sure we don't get
- confused, I think it might be useful to review a little bit
- of everything that we've learned so far.
- So in the last video, we saw that the Laplace transform--
- well, let me just write something.
- The Laplace transform of f of t, let me just get some
- notation down, and we can write that as big capital F of
- s, and I've told you that before.
- And so given that, in the last video I showed you that if we
- have to deal with the unit step function, so if I said,
- look, the Laplace transform of the unit step function, it
- becomes 1 at some value c times some shifted function f
- of t minus c, in the last video, we saw that this is
- just equal to e to the minus cs times the Laplace transform
- of just this function right there, so times the F of s.
- And it's really important not to get this confused with
- another Laplace transform property or rule, or whatever
- you want to call it, that we figured out.
- I think it was one of the videos that I made last year,
- but if you're just following these in order I think it's
- three or four videos ago.
- And that one told us that the Laplace transform of e to the
- at, times f of t, that this is equal to-- and I want to make
- this distinction very clear.
- Here we shifted the f of t and we got just kind of a
- regular F of s.
- In this situation, when we multiply it times a e to the
- positive at, we end up shifting the actual transform.
- So this becomes F of s minus a.
- And these two rules, or properties, or whatever you
- want to call them, they're very easy to
- confuse with each other.
- So we're going to do a couple of examples that we're going
- to have to figure out which one of these two apply.
- Let's write all the other stuff that we learned as well.
- The very first thing we learned was that the Laplace
- transform of 1 was equal to 1/s.
- We know that that's a pretty straightforward one, easy to
- prove to yourself.
- And more generally, we learned that the Laplace transform of
- t to the n, where n is a positive integer, it equaled n
- factorial over s to the n plus 1.
- And then we had our trig functions
- that we've gone over.
- Let me do this in a different color.
- I'll do it right here.
- The Laplace transform of sine of at is equal to a over s
- squared, plus a squared.
- And the Laplace transform of the cosine of at is equal to s
- over s squared plus a squared.
- And you'll be amazed by how far we can go with just what
- I've written here.
- In future videos, we're going to broaden our toolkit even
- further, but just these right here, you can already do a
- whole set of Laplace transforms and inverse Laplace
- transforms. So let's try to do a few.
- So let's say I were to give you the Laplace transform.
- And you know, this is just the hard part.
- I think you know how to solve a differential equation, if
- you know how to take the Laplace transforms
- and go back and forth.
- The hard part is just recognizing or inverting your
- Laplace transforms. So let's say we had the Laplace
- transform of some function F of s.
- Let's say it's 3 factorial over s minus 2 to the fourth.
- Now, your pattern matching, or your pattern recognition part
- of your brain, should immediately say, look, I have
- a Laplace transform of something that has a factorial
- in it, and it's over an exponent.
- This must be something related to this
- thing right here, right?
- If I just had the Laplace transform-- let me write that
- down-- the Laplace transform of-- you see a 3 factorial and
- a fourth power, so it looks like n is equal to 3.
- So if you write the Laplace transform of t to the 3, this
- rule that we showed right here, this means that it would
- be equal to 3 factorial over s to the fourth.
- Now, this thing isn't exactly this thing.
- They're not quite the same thing.
- You know, I'm doing this to instruct you, but I find
- these, when I'm actually doing them on an exam-- I remember
- when I did them when I first learned this, I would actually
- go through this step because you definitely don't want to
- make a careless mistake and you definitely want to make
- sure you have a good handle on what you're doing.
- So you're like, OK, it's something related to this, but
- what's the difference between this expression right here and
- the expression that we're trying to take the inverse
- Laplace transform of, and this one here?
- Well, we've shifted our s.
- If we call this expression right here F of s, then what's
- this expression?
- This expression right here is F of s minus 2.
- So what are we dealing with here?
- So you see here, you have a shifted F of s.
- So in this case, a would be equal to 2.
- So this is the Laplace transform of e to the at times
- our f of t.
- So let me write this down.
- This is the Laplace transform of e to the-- and what's a?
- a is what we shifted by.
- It's what we shifted by minus a, so you have a positive a,
- so e to the 2t times the actual function.
- If this was just an F of s, what would f of t be?
- Well, we figured out, it's t the 3, t to the third power.
- So the Laplace transform of this is equal to that.
- Or we could write that the inverse Laplace transform of 3
- factorial over s minus 2 to the fourth is equal to e to
- the 2t times t to the third.
- Now, if that seemed confusing to you, you
- can kind of go forward.
- Let's go the other direction, and maybe this will make it a
- little bit clearer for you.
- So let's go from this direction.
- If I have to take the Laplace transform of this thing, I'd
- say, OK, well, the Laplace transform of t to
- the third is easy.
- I think the tool isn't working right there properly.
- Let me scroll up a little bit.
- So I could write it right here.
- So if I wanted to figure out the Laplace transform of e to
- the 2t times t to the third, I'll say, well, you know, this
- e to the 2t, I remember that it shifts something.
- So if I know that the Laplace transform of t the third, this
- is an easy one.
- It's equal to 3 factorial over s to the fourth.
- That's 3 plus 1.
- Then the Laplace transform of e to the 2t times t the third
- is going to be this shifted.
- This is equal to F of s.
- Then this is going to be f of s minus 2.
- So what's F of s minus 2?
- It's going to be equal to 3 factorial over s minus 2 to
- the fourth.
- I think you're already getting an appreciation that the
- hardest thing about these Laplace transform problems are
- really kind of all of these shifts and kind of recognizing
- the patterns and recognizing what's your a, and what's your
- c, and being very careful about it so you don't make a
- careless mistake.
- And I think doing a lot of examples probably helps a lot,
- so let's do a couple of more to kind of make sure things
- really get hammered home in your brain.
- So let's try this one right here.
- This looks a little bit more complicated.
- They give us that the Laplace transform of some function is
- equal to 2 times s minus 1 times e to the minus 2s, all
- of that over s squared minus 2s plus 2.
- Now this looks very daunting.
- How do you do this?
- I have an e here.
- I have something shifted here.
- I have this polynomial in the denominator here.
- What can I do with this?
- So the first thing, when I look at these polynomials in
- the denominator, I say can I factor it somehow?
- Can I factor it fairly simply?
- And actually, in the exams that you'll find in
- differential equation class, they'll never give you
- something that's factorable into these weird numbers.
- It tends to be integers.
- So you see, OK, what two numbers?
- They have to be positive.
- When you give their product, you get 2.
- And then when you add them, you get negative 2, or they
- could both be negative.
- But there's no two easy numbers, not 1 and 2.
- None of those work.
- So if you can't factor this outright, the next idea is
- maybe we could complete the square and maybe this will
- match one of the cosine or the sine formulas.
- So how can we complete the square in this denominator?
- Well, this can be rewritten as s squared minus 2s.
- And I'm going to put a plus 2 out here.
- And you can watch, I have a bunch of videos on the
- completing of the square, if all of this
- looks foreign to you.
- And to complete the square, we just want to turn this into a
- perfect square.
- So to turn this into a perfect square-- so something when I
- add it to itself twice becomes minus 2, and so that when I
- square it, when I add it to itself twice, it becomes minus
- 2, it's minus 1.
- And when I square it, it'll become plus 1.
- I can't just add plus 1 arbitrarily to some
- expression, I have to make it neutral.
- So let me subtract 1.
- I haven't changed this.
- I added 1 and I subtracted 1.
- A little bit of a primer on completing the square.
- But by doing this, I now can call this expression right
- here, I can now say that this thing is s minus 1 squared.
- And then this stuff out here, this out here is 2 minus 1.
- This is just plus 1.
- So I can rewrite my entire expression now as 2 times s
- minus 1 times e to the minus 2s-- make sure I'm not
- clipping off at the top-- e to the minus 2s, all of that over
- s minus 1 squared plus 1.
- So a couple of interesting things seem to
- be happening here.
- Let's just do a couple of test Laplace transforms. So if a
- Laplace transform of cosine of t, we know that this is equal
- to s over s squared plus 1, which this kind of looks like
- if this was an s and this was an s squared plus 1.
- If this was F of s, then what is this?
- Well, let's ignore this guy right here for a little bit.
- So what is it?
- We know, actually, from the last video.
- We saw, well, what if we took the Laplace transform of e to
- the-- I'll call it 1t.
- But let's say e to the-- yeah I'll just write it e to the 1t
- times cosine of t?
- Well, then this will just shift this Laplace
- transform by 1.
- It will shift it by 1 to the right.
- Wherever you see an s, you would put an s minus a 1.
- So this will be equal to s minus 1 over s minus 1
- squared plus 1.
- We're getting close.
- We now figured out this part right here.
- Now, in the previous video, I think it was two videos ago,
- or maybe it was the last video, I forget.
- Memory fails me.
- I showed you that if you have the Laplace transform of the
- unit step function of t times some f of t shifted by some
- value of c, then that this is equal to e to the minus cs
- times F of s.
- OK, And this can get very confusing.
- This can get very confusing, so I want to be
- very careful here.
- Let's ignore all of this.
- I called this F of s before, but now I'm going to backtrack
- a little bit.
- And let's just ignore this, because I'm going to redefine
- our F of s.
- So let's just ignore that for a second.
- Let's define our new f of t to be this.
- Let's say that that is f of t.
- Let's say f of t is equal to e to the t cosine of t.
- Then if you take the Laplace transform of that, that means
- that F of s is equal to s minus 1 over s minus 1
- squared plus 1.
- Nothing fancy there.
- I just defined our f of t as this, and then
- our F of s is that.
- Now, we have a situation here.
- Let's ignore the 2 here.
- The 2 is just kind of a scaling factor.
- This expression right here, we can rewrite as that expression
- is equal to-- this is our F of s.
- This expression right here is equal to 2 times our F of s
- times e to the minus 2s.
- Or let me just write it.
- Let me switch the order, just so we make it look right.
- 2 times e to the minus 2s times F of s.
- Well, that looks just like this if our 2 was
- equal to our c.
- So what does that tell us?
- That tells us that the inverse Laplace transform, if we take
- the inverse Laplace transform-- and
- let's ignore the 2.
- Let's do the inverse Laplace transform of the whole thing.
- The inverse Laplace transform of this thing is going to be
- equal to-- we can just write the 2 there as a scaling
- factor, 2 there times this thing times
- the unit step function.
- What's our c?
- You can just pattern match.
- You have a 2 here.
- You have a c, a minus c, a minus 2, so c is 2.
- The unit step function is zero until it gets to 2 times t, or
- of t, so, then it becomes 1 after t is equal to 2, times
- our function shifted by 2.
- So this is our inverse Laplace transform.
- Now, what was our function?
- Our function was this thing right here.
- So if our inverse Laplace transform of that thing that I
- had written is this thing, an f of t, f of t is equal to e
- to the t cosine of t.
- Then our inverse-- let me write all of this down.
- Let me write our big result.
- We established that the inverse Laplace transform of
- that big thing that I had written before, 2 times s
- minus 1 times e to the minus 2-- sorry, e to the minus 2s
- over s squared minus 2s plus 2 is equal to this thing where f
- of t is this.
- Or we could just rewrite this as 2 times the unit step
- function starting at 2, where that's when it becomes
- non-zero of t times f of t minus 2. f of t minus 2 is
- this with t being replaced by t minus 2.
- I'll do it in another color, just to ease the monotony.
- So it would be e to the t minus 2 cosine of t minus 2.
- Now, you might be thinking, Sal, you know, he must have
- taken all these baby steps with this problem, because
- he's trying to explain it to me.
- But I'm taking baby steps with this problem so that I myself
- don't get confused.
- And I think it's essential that you do
- take these baby steps.
- And let's just think about what baby steps we took.
- And I really want to review this.
- This is actually a surprisingly good problem.
- I didn't realize it when I first decided to do it.
- We solved this thing.
- We wanted to get this denominator into some form
- that is vaguely useful to us, so I completed the square
- there and then we rewrote our Laplace transform,
- our f of s like this.
- And then we used a little pattern recognition.
- We said, look, if I take the Laplace transform of cosine of
- t, I'd get s over s squared plus 1.
- But this isn't s over s squared plus 1.
- It's s minus 1 over s minus 1 squared plus 1.
- So we said, oh, well, that means that we're multiplying
- our original time domain function.
- We're multiplying our f of t times e to the 1t.
- And that's what we got there.
- So the Laplace transform of e to the t cosine of t became s
- minus 1 over s minus 1 squared plus 1.
- And then we had this e to the minus 2s this entire time.
- And that's where we said, hey, if we have e to the minus 2s
- in our Laplace transform, when you take the inverse Laplace
- transform, it must be the step function times the shifted
- version of that function.
- And that's why I was very careful.
- And you had this 2 hanging out the whole time, and I could
- have used that any time.
- But the simple constants just scale.
- A function is equal to two times the Laplace transform of
- that function and vice versa.
- So the 2's are very easy to deal with, so I kind of
- ignored that most of the time.
- But that's why I was very careful.
- I redefined f of t to be this, F of s to be this, and said,
- gee, if F of s is this, and if I'm multiplying it times e to
- the minus 2s, then what I'm essentially doing, I'm fitting
- this pattern right here.
- And so the answer to my problem is going to be the
- unit step function-- I just throw the 2 out there-- the 2
- times the unit step function times my f of t shifted by c.
- And we established this was our f of t, so we just
- shifted it by c.
- We shifted it by 2, and we got our final answer.
- So this is about as hard up to this point as you'll see an
- inverse Laplace transform problem.
- So, hopefully, you found that pretty interesting.

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