Translation and Inverse Laplace Transforms

[@juhisaxena3594]

The entire series on Laplace transform is just exceptional

[@mili3212]

Thank you so much for a clear and precise video!! I’m so lucky to have just discovered your channel and I hope you gain a lot more recognition soon!!

[@orueom7720]

Thanks so much for this really explanatory video! I'm learning the theory behind the Laplace transforms so that I can use it modelling situations on SIMULINK. As an ChemEng student this is a life saver as it really clears things up!

[@DrTrefor]

Glad it was helpful!

[@caosspearbr7268]

You're a life saver!! Thanks from Brazil!

[@PrachiVerma-h2n]

thank you very much for the clear explanation. you are the best teacher I have ever seen 💜💜

[@ricardobautista-garcia8492]

Thank you very much for your videos. I recall watching your integration ones 1 year ago when the Pandemic began and in the middle of everything I needed to submit an assignment.

[@ricardobautista-garcia8492]

This is where I discovered your channel for the first time.

[@DrTrefor]

Glad they’ve been helping over this last year!

[@scottboyer7774]

So far I have a 100 in differential equations because of these videos :D

[@ricardoraymond9037]

I am enjoying these tutorials👍

[@DrTrefor]

Glad to hear!

[@mileslegend]

Thank you 😊😊 because of your explanation was able to win a quick solve contest and I got a prize 🏆 😊🥰🔥🔥in Laplace transforms

[@chuxiaoguo6721]

What a legend, thank you!

[@tonywang7933]

Thank you so much exactly what I needed!!

[@ClvrFarmer]

saved my ass. 24 hrs for the HW deadline. i have zero clue what laplase is and what is it for. 3 hrs of concentration and finished the HW in 2 hrs. amazing

[@VndNvwYvvSvv]

At 12:00, this is division by 2 not by ½, or multiplication by ½

[@yazmrt]

Thanks for the video and explanations

[@asahbernard429]

i have a question. why di he put (s-1)+1 in the numerator at minute 8:53?

[@hanminoru_]

integral definition formula

[@ManojKumar-cj7oj]

at 11:35 I think you forget to include translation part e^t in second inverse transform isn't it Im sorry I didn't see that caption ,sorry again :)

[@user-wu8yq1rb9t]

Thank you so much ❤️💗

[@the_eternal_student]

You just threw the idea of t^2 becoming a cube in the denominator without deriving it, so I had no idea what you were talking about. I wish you had said more about how the shift applies when you have two functions being multiplied together as in e^t sin(kt).

[@carultch]

Here's how you derive those two standard Laplace transforms. For polynomial functions t^n in general: L{t^n} = integral t^n * e^(-s*t) dt, from 0 to inf Construct integration by parts table, with t^n differentiated, and e^(-s*t) integrated. Title the columns k for the row number, S for signs, D for differentiate, and I for integrate. Construct several rows until you see the pattern. Then construct the kth row, the nth row, and the (n+1)th row (which will be the final row once we differentiate to zero). k _ _ S _ _ _ _ _ _ _ D _ _ _ _ _ _ _ _ _ _ _ _ _I 0 _ _ + _ _ _ _ _ _ _ t^n _ _ _ _ _ _ _ _ _ _ _ e^(-s*t) 1 _ _ - _ _ _ _ _ _ _ n*t^(n - 1) _ _ _ _ _ _ _ _-1/s*e^(-s*t) 2 _ _ + _ _ _ _ _ _ _ n*(n-1)*t^(n-2) _ _ _ _ _ +1/s^2*e^(-s*t) k _ _ _(-1)^k _ _ _ _ n!/(n - k)! * t^(n - k) _ _ (-1)^k/s^k*e^(-s*t) n _ _ _(-1)^n _ _ _ _ n! _ _ _ _ _ _ _ _ _ _ _ _ _ (-1)^n/s^n*e^(-s*t) n+1 _ _(-1)^(n+1) _ 0 _ _ _ _ _ _ _ _ _ _ _ _ _ (-1)^(n+1)/s^(n+1)*e^(-s*t) All terms at infinity evaluate at zero, due to e^(-s*t). All terms containing any t^(nonzero power) will evaluate to zero at t=0. Thus, the only row that governs the definite integral, is the nth row. Connect the signs with the D-column, and then to the I-column from the next row down. n!*(-1)^n * (-1)^(n + 1) * 1/s^(n + 1) * e^(-s*t) The two alternators multiply to -1: -n!/s^(n + 1) * e^(-s*t) Evaluate at t=inf, evaluate at t=0, and subtract: -n!/s^(n + 1) * [e^(-s*inf) - e^(0)] -n!/s^(n + 1) * [0 - 1] Result: L{t^n} = n!/s^(n + 1) Plug in n = 2: L{t^2} = 2!/s^3

[@carultch]

For deriving L{sin(k*t)}: L{sin(k*t)} = integral sin(k*t)*e^(-s*t) dt Construct IBP table with sin(k*t) in the D-column, and e^(-s*t) in the I-column: S _ _ _ D _ _ _ _ _ _ I + _ _ sin(k*t) _ _ _ e^(-s*t) - _ _ k*cos(k*t) _ _ -1/s*e^(-s*t) + _ _ -k^2*sin(k*t) _ _ 1/s^2*e^(-s*t) Observe that we have a constant multiple of the original integral across the bottom row. Call it I, and construct the result: I = -1/s*sin(k*t)*e^(-s*t) - k/s^2*cos(k*t)*e^(-s*t) - k^2/s^2 * I Shuffle I-terms to the left: I*(1 + k^2/s^2) = -1/s*sin(k*t)*e^(-s*t) - k/s^2*cos(k*t)*e^(-s*t) Multiply thru by s^2 to clear fractions: I*(s^2 + k^2) = -s*sin(k*t)*e^(-s*t) - k*cos(k*t)*e^(-s*t) Isolate I: I = 1/(s^2 + k^2) * [-k*cos(k*t) - s*sin(k*t)]*e^(-s*t) Evaluate at t=inf, and t=0, and subtract: At t=inf, entire integral goes to zero At t=0, sin(k*t) is zero, and cos(k*t) = 1. Lower bound evaluates to -k/(s^2 + k) Thus, when subtracting from upper bound evaluation of zero, we get: L{sin(k*t)} = k/(s^2 + k)

[@ritvikkiragi8110]

You must watch the videos earlier in series

[@Yuen891]

Thank you so much.If teaching with laplace transform table simultaneously is way better . I love your video

[@wunboonail]

well done. You have planned this presentation so well.

[@DrTrefor]

Thank you very much!

[@continnum_radhe-radhe]

🔥🔥🔥

[@dread2much]

Damn these videos are a godsend. My professor sucks... at least the one I have in person ;)

[@yoyochan8615]

You explained very well, it was very clear and easy to understand, I understand how inverse Lapplace Transform works. But are we going to be asked to show the detailed steps of the process of inverse Laplace Transform? Because you didn't show them in the video, and doing the normal Laplace Transform can show the steps of the integration

[@DrTrefor]

It’s a bit like the relationship between differentiation and anti derivatives. If you already know the one and have put the work in to get it, you get the other “for free”.

[@yoyochan8615]

@@DrTrefor so you are saying normal laplace transform and the inverse often show up the same time? And you mean inverse transform requires less detailed steps? Or doesn't require detsiled steps at all?

[@carultch]

@@yoyochan8615 To do Laplace transforms from the t-domain to the s-domain, you can either use the integral definition as the first principals to derive it, or you can use a library of standard Laplace transforms to construct it (which is a much easier method, most of the time). To do inverse Laplace transforms, to get back to the t-domain from the s-domain, you need to manipulate the given expression with algebra and with properties of the Laplace transform to resemble Laplace transforms from the library of standard transforms. There is an integral that can invert a Laplace transform, but it is complicated, and it is a lot easier to use algebra to manipulate the terms to match the standard transforms in the library of transforms.

[@gatesfather007]

Hi, Trevor, what tablet do you use to write and share the screen with?

[@angwe-asencollins3987]

Laplace lecture is good

[@terramyr2074]

this might be a stupid question but why do we add a e^t in front of both cos(2t) and 1/2 sin(2t)

[@VndNvwYvvSvv]

That is the translation in s, from s -> (s-1) where we add e^(at) for (s-a).

[@josephhajj1570]

Nice one professor plz can you solve laplace-1(ln(1+s))

[@carultch]

Given: L^-1{ln(1 + s)} Use the s-derivative property of the Laplace Transform, where given F(s) = L{f(t)}: -d/ds L{F(s)} = t*f(t) Let G(s) = ln(1 + s), and g(t) be the solution. This means: -d/ds ln(1 + s) = t*g(t) Carry out derivative: d/ds ln(1 + s) = 1/(s + 1) Thus: L{t*g(t)} = -1/(s + 1) Invert: L^-1{-1/(s + 1)} = -e^(-t) = t*g(t) Solve for g(t), and we have our solution: g(t) = -1/t * e^(-t)

[@alyashour5861]

The GOAT

[@benetsonichulangula7361]

The final answer i thought it could have e^t as a coefficient of cos 2t +(1/2)sin 2t...i need help on this🙏🏾

[@xxxSoulreaperxxx99]

you are correct, note that at the end of the video OP left a note stating that the sin term also needs e^t as a coefficient.

[@ninasmith-pd8fv]

none of your ks look like ks

[@MrPatrickbuit]

This is insane

[@Knottieyarn]

Laplace makes me sleepy 😢

[@wesleytsao4685]

[@eatemadfanaee5954]

brilliant explanation + awful handwriting

[@eatemadfanaee5954]

= worth to watch any way

[@aayushpatel8913]

Why just basic things are covered? More complex Problems expected

[@VndNvwYvvSvv]

It's an adequate explanation of the theorem. If you want more problems, search for that