Laplace Transform Practice

hello welcome to this lesson of the Laplace transform tutor what we're going to do now is use this nice table of transforms that we have derived and have on the board to solve some real problems we're going to start with the easy problems and gradually Ratchet it up I think you're going to find that now that we've gotten over that hump of understanding what the transform is what the integral is and how to have you use it how to build this table that these problems here are quite simple now that we know how to do that so let's say that we want to find the Laplace transform of the function of time T to the fourth so that's f of T is equal to T to the fourth so the way you would do that on your paper is you would just say Laplace transform of F of T is equal to the Laplace transform of T to the fourth right so how do you do that you go over to your table of Laplace transforms and say hey is there anything that looks like that ah T to the power of n it's going to be n factorial over s to the power of n plus one in this case n is equal to four all right so you have 4 factorial over s to the 4 plus 1 all right and then for factorial you should know is 4 times 3 times 2 times 1 over s to the fifth power and so what you're going to get is capital F which is a function of s which represents the Laplace transform of the function of F transformed into the s domain 4 times 3 is 12 times 2 is 24 over s to the fifth all right and that's what you would circle on your test so you have transformed from T to the fourth on into the S domain which gives you 24 over s to the fifth and now you have a function of s so that's why we say function of time function of s the f goes with the big f that just lets you know that these are linked by this transform we use that sort of symbology there but really it's its own independent function of s which is representing the function but in the Laplace domain or in the s domain all right now the next problem switch colors what if we have f of T is sine of 2 times T and you want to find the Laplace transform of this so you say Laplace transform of sine 2t and you go over here and say is there anything that looks like that and of course we have the sine function right here in this case beta would be 2 sine 2t beta would be 2 it's going to be beta over s squared plus beta squared here beta is equal to 2 so you can just write refresh my memory at beta over s squared plus beta squared where beta is equal to 2 so it's going to be 2 over s squared plus 2 squared and so the write the full answer you can say f is a function of s which implies that you've already transformed it it's 2 over s squared plus 4 and that's what you would circle again on your exam so notice you have a function of s there's nothing else here except SS alright we've locked down this value of beta when I was given the specific problem at hand all right so let me go ahead and erase the board here and work on our next couple problems which are really no harder but we're just going to work our way through our skills all right now the next problem says let's take the Laplace transform of 5 times e to the 2t minus T cubed and furthermore let's try to write it out in integral form now because we want to do all the calculus but just to see how it would look because that can definitely give you some experience too all right so what you would have if you were doing it by the definition of the transform which we're just doing here for clarity just to kind of show you in a roll e to the minus s T and you open up parentheses and dump your function and now your function is a linear combination of two things it's got an exponential and it's going to minus sign here actually and it's got T cubed D T so if I didn't know anything about this table of transforms I would dump this in and I would try to integrate it and furthermore I can rewrite this let me go to the left a little bit I can rewrite this as integral 0 to infinity e to the minus s T 5 e to the 2t DT plus another integral from 0 to infinity e to the minus s t- t cubed d t make sure you'll understand this all I'm doing is I'm saying look I can break this integral up into two pieces and I can say because what I can do is I can take this exponential and multiply it in times this that's going to be one integral I can also multiply it in by this this negative I'm keeping track of here and so that becomes a positive and so basically I have two guys right there and actually I can if I want to since it's just a constant I can take that negative and I can pull it out it's the same exact thing so what I've done here is I've shown you from calculus because it's just an integral that because you have a linear combination of two things it's basically applying the Laplace transform separately once to this guy and then secondly to this guy any pluses or - just basically link the two guys together so another way to write this is Laplace transform of 5 e to the 2t - Laplace transform T cubed so this is kind of what your first instinct would have been anyway if I given you this problem and I said what's the plas transform you apply it to the first term you apply it to the second term whatever is here plus minus just links them together I'm just going through the integral here to kind of show you that by calculus that that basically it all connects together and at work so that's linearity that's the property we discussed now that we got down to this level we can apply the transform separately let's go ahead and do it right so does anything look like this e to the 2t well we see right here e to the 2t fits this 1 over s minus 2 is what it would be there but we have this 5 out front but don't forget we actually said when we talked about linearity that any constants in the function can be pulled out because just as you see here it's just an integral so the constants can come out here there's really only a one that comes out so there's nothing to show there so what it becomes then is it becomes five times the Laplace transform of this which we already said is 1 over s minus 2 that's the Laplace transform of e to the 2t and we subtract off we say what's the Laplace transform of t cubed we go here we say that it's going to be three factorial s to the n plus 1 which would be s to the fourth power so we say just to show it we say three factorial s to the 3 plus one I don't like to do too many things in one step so I like to show everything and so what I would have here is five over s minus 2 minus 3 times 2 is 6 times 1 is still 6 over s to the fourth power so I get 5 over s minus 2 minus 6 over s to the fourth this is f capital of s because it's a function of it's a pure function of s so notice that what's happening every time we transform these guys is I go from a function of T to a pure function of s it gets confusing whenever you first learn about the transform because you see betas and lambdas and stuff running around but those are just constants those are locked down by your problem you should end up with a pure function of s that's what the Laplace transform is all right so let's go on to the final problem in this lesson which is going to be what if I have the function f of T is T squared minus 7 plus cosine of 2t how do we take the Laplace transform of this so we say Laplace transform of F of T is equal to well first of all we have three terms here we have linked by a minus here we have linked by a plus but from the linearity you should now know that we just applied the Laplace transform to each little piece separately just like we do an integration the minuses and pluses will just link the answer so we go and look at T squared T squared fits this mold so it'll be 2 factorial over s to the 2 plus 1 so what we're going to write here is 2 factorial over s to the 2 plus 1 that's going to be the first term and then we carry the minus sign which comes from here and now we have to take the Laplace transform of seven now we don't have a Laplace transform of seven ok obviously we don't have that but we have the Laplace transform of one so notice that seven right is just a constant you can write 7 as 7 times one right so you can think of it as pulling the 7 outside four lack of a better word and now it's the Laplace transform of 1 which is 1 over s so 7 times 7 times 1 7 is pulled out because it's a constant Laplace transform of 1 is 1 over s that's how you get this guy and then the plus sign links it here and now you have cosine 2t so here's cosine so you have s over s squared plus beta squared so that's going to be a 2 squared so it's going to be s over s squared plus 2 squared and so the final answer is f of s here 2 factorial is just 2's cubes what you'll have on the bottom minus 7 over s for the middle term plus s over s squared plus 4 this is the final answer 2 over s cubed minus 7 over s plus s over s squared plus 4 notice again this is a pure function of time we have transformed into a pure function of s if you end up with the Laplace transform that has anything other than just SS and numbers you've done something wrong so that's just something you think about what we've done in this lesson is we're just inching our way forward I think we're taking what we've done here and we're applying it to some common problems using some of the properties of the transform such as linearity pulling out constants of things and these kinds of skills are really important because what we're going to do here in a little while we get a little more practice is we'll start applying it to differential equations and so then you'll basically take the Laplace transform of both sides of that differential equation and basically get a new equation that can be solved simpler than the original differential equation could could ever have been and then you'll be able to get the answer an inverse transform back that's the basic roadmap so follow me on to the next lesson we'll get some more practice with taking transforms and also inverse transforms before we get to doing those differential equations solution methods

Channel: Math and Science

Views: 380,685

Keywords: laplace transform, fourier transform, calculus, differential equations, inverse transform, transform, fourier series, engineering math, control systems, impulse response, engineering, math, physics, math laplace transform, calculus laplace transform, laplace transform rules, laplace transform problems, laplace transform applications, uses of the laplace transform, what is the laplace transform, laplace transform engineering

Id: RDtITuZDZi4

Length: 10min 54sec (654 seconds)

Published: Mon Sep 30 2013