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Homework 6: Problem 6 Previous Problem Problem List Next Problempoint) Find the Laplace transform of 7/' + 2sinh(at):...

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Homework 6: Problem 6 Previous Problem Problem List Next Problempoint) Find the Laplace transform of 7/' + 2sinh(at):

Homework 6: Problem 6 Previous Problem Problem List Next Problem point) Find the Laplace transform of 7/' + 2sinh(at):



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Use the results of Problem 19 to find the inverse Laplace transform of the given function. $$ F(s)=\frac{2 s+1}{4 s^{2}+4 s+5} $$

Hello and Welcome to Probleble 21 of Chapter six. Section 3 you were asked to find the universal past transform of the group in function which is shown on the screen using what we learned in the previous problem, which is the inverse transform role. And then transform rule says that if s is offset by constant amounts an inverse transform that will have an E. To the minus offset term on that. So let's get into simply buying this uh nervous. We'll get to s plus one over uh to us plus one squared plus four. And from this we can say that the the results we'll have um and eat the minus T over two terms because negative one half is what will solve this top and bottom and get rid of this that offset. Oops. So from there we'll have um E to the minus T. Or two times the inverse laplace transform of two s over to S squared plus four is equal to. Well the universal pause transform of F F. S. And now we want to find whatever the universal post transform of to S over two X squared plus four is. And um let's drop by taking out factors so we can take out of four from the bottom and two from the top. Um This will get us one half E to the -2 over tea times the plus transformed of I believe S over S squared plus one as you go to the universal plus transform of FMS which is what we're trying to find. And simply this right here is a common universal post transform of co sign of teeth. The final answer is 1/2 E to the minus T. Or two. There's co sign of T. Mhm. Yeah.

Hello and Welcome to Probleble 31 of Chapter six, Section 3 you were asked to funnel plus transforming the given function F. F. T. S. T. We also know that this is a periodic function with a period of Capital T. Equals one. So we can use uh the equation we learned in chapter or question 27 I believe to a final pass transform of. Yes. And that equation we learned is that the low cost transform of a periodic function Is equal to the integral from 0 to capital T. E. To the minus S. T. F F T. D. T. Is all standard. All all over one minus E. To the minus S. Comes tap capital T. But if we notice that it's not periodically plug in the period is infinity, this will reduce to the normal plus transformation. Okay? Um in this case like you can plug in with F F. T. B. M. T. And Capital T Being one. So in our case this will turn into and drill from 0 to 1. Either minus S. T. T. T. T over one minus E to the minus s. Okay. And from here we can use integration by parts. To solve the top integral. We'll let you be equal to T. Do U. Is equal to D. T. We'll have a D. V. E. With my S. T. D. T. And V is equal to E to the minus S. T over negative s. Okay. And the next step is just to do extended integration by parts. So we'll have T. E. to the -4. Over -S evaluated at zero and 1 minus zero from 0 to 1 V. D. U. Well we have that plus in the outside so we'll turn this into a Plus one of us. E. to the -4. Gt. Okay I'll evaluate this again and I did leave out this bottom part but we'll bring that back later. Um So what happens when we evaluate this at one we'll have one times E. To the minus s over um minus S. So that will be even minus S over minus s minus weapons. Going to evaluate this zero? Well that's just zero so that will cancel out. Okay let's do the next part, add one of our s. And if we take the anti derivative will be left with. Um But we had up here so we can just take out another one of our negative s will be left with minus one over S squared times E. to the -4. Evaluated at 0 & one. Okay go down to the next line or have either minus s over minus s. One of us squared. What happens when we evaluate this at T. Equals one will get um either the minus S. And when we evaluate this T. Equals zero we'll get one. So this turns into E. To the minus s minus one. Great. And if we want to we can stick this all under one denominator S squared. And to this year we'll have um but the denominators squared will have um negative esque squared from this one term. So there will be a plus S squared. Uh huh, positive S squared. Um Then we'll have a a plot very minus S. E minus S and a minus E. To the minus s. Let's just do double check. We'll have this term there because multiply the top and bottom by S. Will have this term there because I will have an escort on the bottom and then we'll have yeah that term there is a plus one over S corp great, that should be right. And now we bring back this bottom term right there so I'll write this um on the bottom solution or capital F. Of S. Is equal to S squared minus S. E. To the minus s minus e minus s. All over. Um S squared times one minus E. To the minus s. One minus E. To the minus s. That concludes our problem.


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