Lesson 1 - Laplace Transform Definition (Engineering Math)

hello I'm Jason welcome to the Laplace transform tutor and here what we're going to do in this lesson and in this whole set of lessons is really build your skills so that you can master how to use the Laplace transform now I expect you to know absolutely nothing about the Laplace transform going into this course but at the end of it you'll not only know what the transform is but also how to use it and how to use a table of Laplace transforms and also the Laplace transform properties to take Laplace transforms and also inverse Laplace transforms so it's a very practical skill that we use in science and engineering classes you know in order to really solve some more complicated problems and you'll find that lots and lots of different classes that you'll take as you go on in your engineering sequence or in your science sequence we'll use the Laplace transform so if you ever get to a class in electrical engineering or mechanics or pure mathematics or differential equations where Laplace transforms are used then what you'll be able to do is come into here and get that whole sequence of information on how to use this very powerful tool along with lots of problems to practice your skills so I'm going to obviously write some math on the board in a second but in the general scheme of things what the heck is this thing called a Laplace transform well first of all it is it is calculus so you have to know some calculus in order to be able to calculate the transform so I expect you up into this point to know a little bit of calculus really not much more than calculus one to be honest with you will enable you to be able to do the math here but you will have to be proficient with it also obviously your algebra skills because I know that you all know how to do basic algebra but when you get into taking some of these transforms it can it can quickly explode into a bunch of things to simplify none of its hard it's just a lot of little terms to keep track of and so we'll be doing that here and I'll show you as I go but it's basically calculus and what you're basically doing is you're using a technique to take a function so here we're going to be talking about functions of time but really it's it's any function you know f of X you normally study in algebra but you know when you get an engine marrying and studying real systems you're usually talking about functions of time things that change with time for instance an electrical engineering you might have an input signal you know like a sine you know wave or some kind of random voice recording of a voice by giving an amplitude as a function of time that would be a function of time right and so the Laplace transform takes a function of time and transforms it to a new function that's a new new way of representing that original function and you might say well why would you ever care about doing that and the truth is that the reason why it works is beyond the scope of what I'm talking about right now but basically Laplace transforms lets you solve a lot of different types of differential equations in an easier way than doing them by hand so in the original way so if you think back to my differential equations tutor or any class that you may have taken in differential equations you should agree with me that it's kind of a nightmare to solve a lot of differential equations there's a lot of math involved a lot of theory and differential equations really have a lot of different techniques so you may do method of undetermined coefficient new coefficients and differential equations you might do integrating factors and differential equations you might use exact differential equations there's lots of different ways to solve differential equations and a lot of it boils down to recognizing what the equation is figuring out the method that works and then applying it and so it seems to be disconnected though because there's lots of different techniques for different styles of differential equations Laplace transform is a unified method that really allows you to take a differential equation and apply the transform to it and change it into a new form that's easier to solve and then when you get the answer in the Laplace domain or the S domain we'll talk about in a minute then you can transform the answer back into the time domain that's why it's called a transform we take the problem or the function or whatever it is we transform it into the Laplace domain which we're going to call the S domain and a lot of times doing the solution in that domain it just becomes algebra or some other easier simplest solution method M differential equations first method so we solve it over here in the Laplace domain or the S domain we get an answer and then we can take that in inverse transform it back into a function of time so that's kind of without getting into the details that's basically why it's useful and as you know differential equations are used in all branches of science and engineering we use them in electric circuits we use them in mechanical systems use them in control systems theory use them in chemical engineering aerospace engineering differential equations are everywhere and because of that and because Laplace transform is kind of a general method to to make these types of problems easier you're going to find that Laplace transforms pop up in all kinds of different situations and so that's why I want to put this tutorial in its own little course because you can use it if you're mechanical engineer electrical engineer physicists you know anything so let's just dive into it first of all and talk about it a little bit so we have what we call the Laplace transform okay and I'm not going to beat around the bush too much we're just going to write it on the board we'll talk about it so the Laplace transform gives you a function in the Laplace domain which is the S domain that's why you're getting a function capital f of X of s that's what you're arriving at when you perform this thing called the Laplace transform and it's the integral from 0 up to infinity of e to the minus s T times f of T DT so this ladies and gentlemen is the famous Laplace transform all right so what you're doing all right is if I give you a function of time think about any function of time could be a square wave could be a sine of T could be sawtooth could be any function of time you want it could be x squared or some function of x squared you stick that function of time in f of T then you multiply by this exponential e to the minus st right and so you're getting something inside of the integral sign you integrate it over time right and then you apply the limits of integration to the answer that you get which are obviously going into your into your time spots because you're integrating over time so the limits of integration is from 0 to infinity in terms of time right then when you do all of that you're going to get just a function of s back because if you integrate over time and then you plug in limits of integration over time there's no T anymore left in what you've done there's just the other aspects of it here so there's no T anymore here there's no T anymore here because you plugged in limits of integration only s remains so when you do a Laplace transform properly you get purely a function of s and so that's the notation here you put F of T little F of T is your input function and out of this computation comes another function we call it capital F we use the same letter capital F and a little F to imply that little F and capital F are related to one another we put a little F into the Laplace transform and we get a capital F out which is a function of s all right so we can also write we can also write as the following F capital F of S which is the Laplace transform is curly L know how she would say that write f of T so this is the shorthand way of saying hey I'm going to take the Laplace transform of some function of time or some function and when I do that I mean it's going to get a pure function of s so these two things are exactly the same thing this is the shorthand way of writing what you're doing you're taking the Laplace transform of a function of time this is one of the details this is what you actually do you have to multiply this you have to integrate it you have to plug in the limits of integration so what we're doing is we're transforming a function from the time domain to the S domain that's why I'm telling you that because the answers that you get are just functions of what we call s and so we get a new function as a result of that and so I know that you've seen this this kind of action before where you have 0 and infinity up on an integral you probably saw that in calculus but just to refresh your memory that's called an improper integral right so when you have an improper integral I know you've studied this once probably a long time ago but basically when you have any kind of 0 to infinity e to the minus st f of T DT what you really are doing is you do the integration but what you do is you take the limit so what I'm going to do is I'll say let me write it in a different color what you're really doing here is you're taking the limit and you'll see what I mean in a minute as H goes to infinity of the integral from 0 to H e to the minus s T F of T DT I'll just mostly explain what you're doing with the calculus when you see an integral like this with one of the limits of integration is an infinity then what you're really doing is you're taking the integral and you pretend that this top number is just some variable you calculate the integral and then you take the limit as H goes to infinity that's how you would write it mathematically but in practice what you really do is you enter evaluate the integral you plug infinity in one limit and zero in the other limit and then you know how to evaluate infinities you know if they're on the denominator if they're on the numerator they might drive your answer a different way and so that's basically what you're doing but mathematically rigorously what you're doing is you're integrating the guy and then you're taking the limit as this limit goes to infinity all right so what I would like to do at this point is calculate a Laplace transform basically what you have is we have the definition that was on the board a minute ago that's really the bread and butter of it if you understood how to how to apply that to all situations and you wouldn't need me you would just do the Laplace transform all the time but the truth is it as you do some of these things there's a little bit of tricks along the way to help make it comprehensible to you and I'm going to show you those here what you're going to find pretty quickly in any book that deals with the Laplace transform is they're going to give you a table of transforms in other words there are some pretty basic functions that are pretty easy to apply this definition to that you get the answer and that answer is very useful going forward because some functions pop up in nature all the time you know Exponential's pop up all the time for instance so we want to learn how to take the Laplace transform of things like Exponential's and things like cosines and sines because those things pop up all the time so what we're going to do in this lesson in the next few lessons is we're going to apply that definition of the Laplace transform this this full-blown definition of the Laplace transform to a few core functions and then we're going to assemble our own table of Laplace transforms which in many cases you'll just find them listed in a book but I'm going to derive how they get how we get there so you'll understand how we're applying this once you have a basic table of Laplace transforms of common functions than what you typically do is you use that table to explore more complicated functions so we're kind of getting our footing we first learn what the real definition of Laplace transform is then we're going to apply it to some simple functions we're going to assemble our table of common functions that we're useful to know the Laplace transform and then we'll apply that to many problems going forward so the most important function that you could probably know how to or want to know how to take the Laplace transform of would be the Laplace transform of the function e to the lambda T e to the lambda T now this is just an exponential it's e to the T it's a function of time right but there's a constant in front we're calling it lambda in your book it might say e to the a T or it might say e to the be T or might say e to the Alpha T e to the beta T it doesn't matter but there is some number in front of T T it could be 1 it could be to its left open-ended because a lot of times Exponential's pop up in solutions to differential equations they pop up in lots of physical systems right so it's an exponential if it's e to the T the actual value of lambda just changes the slight shape of what it looks like so we're leaving that as a constant left open interpret we want to find out in general what would this Laplace transform be so the way you do it is you just apply it directly 0 to infinity and you apply that Laplace transform which if you remember was e to the minus s T F of T DT this was the general equation for the Laplace transform so then what we do is we say zero to infinity e to the minus s T then we put our function of time in here which is e to the lambda T D T and this is what we want to integrate and we suspect and we claim and I'm telling you that once you do this properly all you get is a function of s and we say that that Laplace transform function of s is inexorably tied to the function of time through this thing called the transform and you have to trust me on faith that once you know how to do these transforms and once you get proficient at them that they help you solve real problems you know I have to take that part on faith we're going to get to that part a little bit later all right so what we're going to do is begin to evaluate this integral I could just give you the answer of course I could do that but I want to walk you through it so that you can really feel like you understand what's really going on so notice this is two Exponential's so we can simply combine the exponents at the top that's something we can do because these Exponential's are multiplied together right so we can say that we'll have negative s minus lambda DT make sure you agree that there's nothing different here if you distribute the negative and you get negative s which is what we have here it would have a t here as well right negative s T and then here this would be positive lambda T positive lambda T so basically we are adding these guys together in the exponent we're basically adding negative s T plus lambda T we're adding that together but I pull the negative sign out so but it still becomes negative s T plus a lambda T in the exponential there all right so what you need to do is integrate this right integrate this this is an exponential integrals of Exponential's are relatively simple the only thing is notice we're integrating over time here is your time variable lambda is just a constant s is going to end up being a variable that we you're going to have an hour transform in the end but since we're not integrating over s we're integrating over T you basically treat s is a constant and that's really important for you to understand you know anytime you take derivatives or integrals you have to look at what variable you're taking it with respect to if I give you an expression I say take derivative with respect to X if you're talking about derivatives then you pretend everything else besides X is a constant in that in that function that's how you take partial derivatives right well in intervals it's the same thing if we're integrating over time then this is the variable we care about every other letter or symbol in there you pretend as a constant so for the purposes of this integral you pretend that s is a constant and lambda is also a constant which means that everything in front of the T is really just a constant so it's like a giant number here so this integral looks intimidating but really since it's all just a constant it's not that intimidating so many of you can look at this and write the answer to the integral down but a lot of times we have to make a substitution to make it absolutely clear and so what you would do is you would say U is equal to minus s minus lambda right T so we want to make a substitution because we want to make it an easy integral to solve so when I do this off to the side I'm doing a substitution I'll say D u DT is equal to now notice if I'm taking a derivative with respect to time this whole thing is a constant so the derivative is just going to be minus s minus a lambda because it's almost like this is just a number three or in the number five or the number seven out in front of the time the teachers disappears for a first derivative now since we're you're going to end up substituting it back here we want us we want to solve for DT and so what we're going to get is negative one over s minus lambda D u all we've done here is move the DT over here and then we've taken all of this stuff and moved it over here so we could solve for DT getting 1 over s minus lambda D U alright so now that we have that we want to take this and stick it back in and substitute into our integral so we'll change colors again and for now we're going to leave the limits of integration 0 to infinity we're not change them right now because we're going to you know we'll see how to handle that in a minute it becomes e ^ you because this is exactly coming from that and DT just becomes what we have found here so it's going to be negative 1 over s minus lambda D you right so that substitution just goes in now remember this is an integral over D you now right so everything in here is again just a constant it's just a constant so what we're going to have here we pull the whole thing out negative 1 over s minus lambda comes out we integrate from 0 to infinity e to the you D you now we did this whole thing so that we could get an integral into a form that we know how to solve very easily right and so what we're going to do then is say we have negative 1 over s months lambda the integral of e to the u still remains e to the U and again I'm leaving my limits of integration 0 to infinity for now because what I'm going to do before I apply the limits of integration is I'm going to substitute back in for you so what I'm going to have is negative 1 over s minus lambda and then U is going to come right back substitute it in as it was before e to the minus s minus lambda T now it goes from zero to infinity so you see I realize that as you have the limits of integration here this really if this were a true statement you would have to transform the limits of integration into limits of U and this would have to be limits of you but I'm not going to substitute the limits in until after I've plugged in for you so I end up having a function of time here's a limits in terms of T and so then I can substitute everything in as exactly as I want to and then what I'm going to get is negative 1 over s minus lambda right and then we have to apply this n so what this basically becomes if you think about it this is negative s minus lambda times T and I'm putting infinity in for T so really it becomes e to the negative negative infinity it doesn't matter what s and T are if I put infinity here it's going to negative infinity and then you do a subtraction because you're evaluating the limits eetu the zero because zero goes in there now this becomes very tractable because we know that e to the negative infinity it's like taking that limit what you're going to have either negative infinity is like 1 over e to the infinity so what you're going to have this guy is going to be zero and this guy anything to the power of zero is just a 1 so you have a negative 1 here and this whole thing evaluates to negative 1 which makes a positive 1 when you multiply this whole thing here so you get 1 over s minus lambda 1 over s minus lambda alright so what I'm going to do then is show you that through this whole situation all we do is we found the Laplace transform of this guy and said that it was equal to that so let me go to the next board and summarize that because it's a very important result so what we found that is that the Laplace transform of e to the lambda times T which is just e to the power of some number times T is just equal to 1 over s minus lambda so for instance if it was e to the 2t then it would be 1 over s minus 2 if it were we to the 7t it would be 1 over s minus 7 if it were e to the 4t it would be 1 over s minus 4 so you see what we give as a result is just a function of s lambda is just a constant it's going to be locked down by whatever we start with right so the answer that we get F of s capital F of s remember we said let me go back we said that when you do this Laplace transform you should get just a function of s and that's what we call it capital f of s and that's exactly what we got we got a single function pretty simple looking function just a function of s this the guy is just a constant there now since it's in the denominator and we don't want any zeros to be in our denominator because then you get undefined you know poor parts of the Laplace transform or in fin 80 parts then what you can also say this is valid for s greater than lambda that just ensures that the denominator is not going to be zero and ensures the denominator is going to be positive this is something that mathematically you write down just to lock it down you don't have any zeros in the bottom but realistically you don't really use this back to all that terribly much when you're solving essential basic problems with the Laplace transform now this is an important result and it's important enough that I'm going to circle it for you right let's circle the whole thing in fact and it's also important enough that we want to draw your attention to something else let me change the blue here in the special case let's say lambda is equal to zero so let's say that lambda is just equal to zero then you would have e to the 0 times T right which would be e to the 0 that would be my function and of time if lambda were zero be either the 0 t and e to the 0 is just 1 because anything to the power of 0 is just 1 so because of this we can draw kind of another conclusion that's already here but we can kind of write it ourselves we can say that the Laplace transform of the number 1 right would be 1 over s minus 0 right because the way you would come up with that is you would say which would be 1 over s right the way you would come up with that is you would say all right I can take the special case when lambda is 0 in that case this exponential just becomes a 1 so it's the same as taking the Laplace transform of 1 and then we'd be putting lambda equals 0 in here getting an answer of 1 over s so the Laplace transform of the number 1 is just 1 over s and that's important enough well I will also circle it and of course it's for s greater than 0 because you don't want any denominator driving the whole thing to infinity so this is the first important conclusion of what we have have done here I'll go quickly through a brief history of what we've done we basically said there's this thing called well PLAs transform it's an integral notice it's not a double integral or a triple integral or spherical coordinates or anything crazy like you get into calculus three territory it's really a calculus one maybe a calculus two type of thing but the implications of how you use it is really why it's interesting and we're going to get to that later but basically you just stick a function of time in here you evaluate the integral evaluate the limits of integration and if you do it correctly you should get just a function of s and we label it capital f of s because capital F and little F are related to one another by this thing we call a transform we say that cap that little F yields capital F of s and later on we'll find out that you can go in the reverse direction and start with a Laplace transform function of s and get the corresponding function of time so these things are kind of linked by an invisible chain and that's the chain which being is the transform that's why we can take a problem which has which has functions of time transform it into the Laplace domain which is function of s solve it usually algebraically so you don't have to deal with differential equations you do with algebra and then you get a function of s and then you we're going to learn how to transform that back to the time domain and you get your answer in terms of time and if you do it right it should be simpler we also say that the Laplace transform F of S is squigly l operating on the function of time and we just talked about improper integral saying that it's basically like taking the limit but really what you're doing is plugging in the limits of integration there and then we do our real problem so this is how a lot of these problems are going to go you take your function of time you put it in and you simplify and then you realize the critical step here is that since you're integrating over T everything here is a constant right so I've done the details here we did the substitution of you putting it all in there and pulling this junk out because it's a constant giving us an integral that everyone watching is should know how to solve right so we do that we plug the limits of integration in the limits of integration greatly simplify what's happening because this becomes a zero this becomes a one because it's negative infinity remember e to the negative infinity is like 1 over e to the positive infinity which means 1 over infinity giving you zero so all this stuff drops away giving you negative 1 giving you positive 1 over s minus lambda so the conclusion block here is the Laplace transform of e to the lambda t is just a function of s this lambda is locked down with whatever your specific function is alright and then as a special case of that we say hey what happens if lambda is 0 then what you should get if lambda 0 is 1 over s right and if lambda is 0 then it's the whole thing just goes to 1 so what we're saying is the Laplace transform of the number 1 is 1 over s if you remember back to calculus 1 however many years ago that was for you first thing we talked about besides the limits but for as far as derivatives the first thing we talked about was how to take simple derivatives how to take the derivative of a constant remember you had to learn that at one point in the past how to take the derivative of X how to take the derivative of x squared then you learn how to take derivatives of polynomials then you learn how to do it when there's a giant fraction then you learn about sines and cosines you build those skills learning how to take those derivatives we're doing the same thing with Laplace transforms we're taking very simple Laplace transforms first and we're recording these answers which are useful then as we go on we're going to take more and more complicated Laplace transforms to the point where you can actually get pretty proficient then I'm going to show you how to use it to solve real problem so right now you look at this and you're like what's it for you know well I can't get into that until you know a little bit so just trust me follow me on to the next lesson we'll build our skills deriving these essential transforms and then we'll get a repertoire going so that you have some skills that we can apply to real problems and whenever you see how how much simpler it makes solving certain kinds of differential equations you'll understand that it's worth its weight in gold just for that application but also since differential equations are used in all branches of science and engineering they really lends the Laplace transform to lots and lots of different situations in real math science and engineering so follow me on to the next lesson in mastering the Laplace transform here in the Laplace transform tutor

Channel: Math and Science

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Keywords: laplace transform, laplace transform definition, engineering math, engineering, definition of laplace transform, definition of the laplace transform, laplace transform concept, laplace transforms, laplace transform engineering mathematics, laplace, lesson on laplace transform definition, laplace transform tutor, help with laplace transform, laplace transform assistance, laplace transform math tutor, transform ODEs, transform of functions, laplace transform review

Id: 8oE1shAX96U

Length: 28min 54sec (1734 seconds)

Published: Thu Feb 04 2016