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Laplace transform 1

Introduction to the Laplace Transform. Created by Sal Khan.

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  • blobby green style avatar for user Louis Balboa
    At of the video Sal begins integration. He starts with -1/s times e to the -st but it gets hairy for me because what happened to adding 1 to the exponent??
    (14 votes)
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  • piceratops seed style avatar for user Arimi
    My doubt is with the equation at , why we multiply the f(t) with e^-st why not with e^+st?
    (8 votes)
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    • blobby green style avatar for user e98cuenc
      You can also ask why do we do the integral from 0 to inf instead of 0 to -inf, or -inf to 0, or something completely different. There are, in fact, other different transformations (like the Fourier transformation), and each of these transformations is useful for something different.

      Why is exactly the integral from 0 to inf of e^-st * f(t) so useful that it has a proper name and that we want to study is something that has not yet been explained (it's explained later, in this video http://www.khanacademy.org/video/laplace-transform-to-solve-an-equation?playlist=Differential+Equations). Hint: it's very useful to solve differential equations.
      (18 votes)
  • leafers ultimate style avatar for user BSG
    I really didn't get how int. e^-st = -1/s e^-st..Please help!!
    (5 votes)
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  • blobby green style avatar for user vkgoku2012
    Why does e show up so much I can't seem to get away from it its used everywhere in math.
    (4 votes)
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    • blobby green style avatar for user pleitch
      e is very common in nature. If you look up the definition of e it arrises with any form of natural growth (think of a graph of compound interest). That's financial growth or even the growth (increase) of randomness around a norm (i.e. Gausian/normal distribution). It's everywhere so you see it in lots of places.
      (13 votes)
  • male robot hal style avatar for user abrule
    Would you do a video about the proof of the Laplace Transform definition ?
    (3 votes)
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  • blobby green style avatar for user Wajid Raza
    sir my question is that in laplace transform 1 ,we use definite integral can we not use indefinite integral?
    (2 votes)
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    • blobby green style avatar for user Grant Sherer
      the definition of the laplace transform is: the integral from 0 to infiniti of (e^(-st))*f(t)dt
      this is just a definition, the laplace transform is a specific operation you can perform on a function, and removing the limits would give you a different operation that may or may not be useful for solving differential equations
      (14 votes)
  • mr pants teal style avatar for user anushadrums
    at "" Sal mentions anti derivitive. what is that?
    (0 votes)
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  • Why can you assume that s > 0 at ?
    (4 votes)
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    • blobby green style avatar for user Gus.Gillen
      S is representing a frequency as in a "per second" or 1/second which is why you get 1/s for the Laplace of 1. From a physics perspective there is no such thing as a negative frequency because in reality it is the same thing as the frequency, saying something happens -3 times a second does not make a lot of sense. You can also tell that s cannot equal zero from the 1/s.
      (7 votes)
  • ohnoes default style avatar for user Nameless
    What if s < 0?

    Then this integral would diverge.
    (4 votes)
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    • leaf green style avatar for user alwick1234
      Yes it will diverge.

      Remember that a laplace transform is essentially telling you how close the function is to e^(st). If the integral diverges that just means the function gets farther and farther away from e^(st) as s gets closer to 0 (in this case).

      Really what it means is you need to pick an s much larger than 0.
      (7 votes)
  • male robot hal style avatar for user Oliver Turnbull
    why does this transform actually work in helping to solve differential equations?
    (4 votes)
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    • marcimus purple style avatar for user ABHISHEK
      It's a property of Laplace transform that solves differential equations without using integration,called"Laplace transform of derivatives".
      Laplace transform of derivatives: {f'(t)}= S* L{f(t)}-f(0).
      This property converts derivatives into just function of f(S),that can be seen from eq. above.
      Next inverse laplace transform converts again function F(S) into f(t).
      If my ans. looks confusing .Just observe am example of solving D.E. using laplace,i hope droughts will disappear.
      (0 votes)

Video transcript

I'll now introduce you to the concept of the Laplace Transform. And this is truly one of the most useful concepts that you'll learn, not just in differential equations, but really in mathematics. And especially if you're going to go into engineering, you'll find that the Laplace Transform, besides helping you solve differential equations, also helps you transform functions or waveforms from the time domain to this frequency domain, and study and understand a whole set of phenomena. But I won't get into all of that yet. Now I'll just teach you what it is. Laplace Transform. I'll teach you what it is, make you comfortable with the mathematics of it and then in a couple of videos from now, I'll actually show you how it is useful to use it to solve differential equations. We'll actually solve some of the differential equations we did before, using the previous methods. But we'll keep doing it, and we'll solve more and more difficult problems. So what is the Laplace Transform? Well, the Laplace Transform, the notation is the L like Laverne from Laverne and Shirley. That might be before many of your times, but I grew up on that. Actually, I think it was even reruns when I was a kid. So Laplace Transform of some function. And here, the convention, instead of saying f of x, people say f of t. And the reason is because in a lot of the differential equations or a lot of engineering you actually are converting from a function of time to a function of frequency. And don't worry about that right now. If it confuses you. But the Laplace Transform of a function of t. It transforms that function into some other function of s. and And does it do that? Well actually, let me just do some mathematical notation that probably won't mean much to you. So what does it transform? Well, the way I think of it is it's kind of a function of functions. A function will take you from one set of-- well, in what we've been dealing with-- one set of numbers to another set of numbers. A transform will take you from one set of functions to another set of functions. So let me just define this. The Laplace Transform for our purposes is defined as the improper integral. I know I haven't actually done improper integrals just yet, but I'll explain them in a few seconds. The improper integral from 0 to infinity of e to the minus st times f of t-- so whatever's between the Laplace Transform brackets-- dt. Now that might seem very daunting to you and very confusing, but I'll now do a couple of examples. So what is the Laplace Transform? Well let's say that f of t is equal to 1. So what is the Laplace Transform of 1? So if f of t is equal to 1-- it's just a constant function of time-- well actually, let me just substitute exactly the way I wrote it here. So that's the improper integral from 0 to infinity of e to the minus st times 1 here. I don't have to rewrite it here, but there's a times 1dt. And I know that infinity is probably bugging you right now, but we'll deal with that shortly. Actually, let's deal with that right now. This is the same thing as the limit. And let's say as A approaches infinity of the integral from 0 to Ae to the minus st. dt. Just so you feel a little bit more comfortable with it, you might have guessed that this is the same thing. Because obviously you can't evaluate infinity, but you could take the limit as something approaches infinity. So anyway, let's take the anti-derivative and evaluate this improper definite integral, or this improper integral. So what's anti-derivative of e to the minus st with respect to dt? Well it's equal to minus 1/s e to the minus st, right? If you don't believe me, take the derivative of this. You'd take minus s times that. That would all cancel out, and you'd just be left with e to the minus st. Fair enough. Let me delete this here, this equal sign. Because I could actually use some of that real estate. We are going to take the limit as A approaches infinity. You don't always have to do this, but this is the first time we're dealing with improper intergrals. So I figured I might as well remind you that we're taking a limit. Now we took the anti-derivative. Now we have to value it at A minus the anti-derivative valued at 0. And then take the limit of whatever that ends up being as A approaches infinity. So this is equal to the limit as A approaches infinity. OK. If we substitute A in here first, we get minus 1/s. Remember we're, dealing with t. We took the integral with respect to t. e to the minus sA, right? That's what happens when I put A in here. Now what happens when I put t equals 0 in here? So when t equals 0, it becomes e to the minus s times 0. This whole thing becomes 1. And I'm just left with minus 1/s. Fair enough. And then let me scroll down a little bit. I wrote a little bit bigger than I wanted to, but that's OK. So this is going to be the limit as A approaches infinity of minus 1/s e to the minus sA minus 1/s. So plus 1/s. So what's the limit as A approaches infinity? Well what's this term going to do? As A approaches infinity, if we assume that s is greater than 0-- and we'll make that assumption for now. Actually, let me write that down explicitly. Let's assume that s is greater than 0. So if we assume that s is greater than 0, then as A approaches infinity, what's going to happen? Well this term is going to go to 0, right? e to the minus-- a googol is a very, very small number. And an e to the minus googol is an even smaller number. So then this e to the minus infinity approaches 0, so this term approaches 0. This term isn't affected because it has no A in it, so we're just left with 1/s. So there you go. This is a significant to moment in your life. You have just been exposed to your first Laplace Transform. I'll show you in a few videos, there are whole tables of Laplace Transforms, and eventually we'll prove all of them. But for now, we'll just work through some of the more basic ones. But this can be our first entry in our Laplace Transform table. The Laplace Transform of f of t is equal to 1 is equal to 1/s. Notice we went from a function of t-- although obviously this one wasn't really dependent on t-- to a function of s. I have about 3 minutes left, but I don't think that's enough time to do another Laplace Transform. So I will save that for the next video. See you soon.