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$t^{2} u(t-2)$

$\frac{e^{-2 s}\left(4 s^{2}+4 s+2\right)}{s^{3}}$

Algebra

Chapter 7

Laplace Transforms

Section 6

Transforms of Discontinuous Functions

Polynomials

Oregon State University

Idaho State University

Lectures

01:32

In mathematics, the absolu…

01:11

03:15

$(t-1)^{2} u(t-1)$

02:17

$$t^{2}$$

00:49

$u(t-1)-u(t-4)$

01:35

$t u(t-1)$

03:24

$$e^{-t} t \sin 2 t$$

01:15

\begin{equation}\begin{arr…

01:21

$$t e^{3 t}$$

03:16

$$e^{-t} \sin 2 t$$

02:04

$$\begin{array}{l}{\te…

02:35

$f(t)=e^{-t^{2}}$

So in this problem were asked to find them Applause trains for this function. What we want to do first is we want to equate that to f T minus a multiplied by you t minus A and doing that, we can compare this with this. We can see that a in this case is equal to two. Now what we want to do is take T Square, Call that GFT and equate that to half of T minus a. So when we do that, we get g t equals F T minus A. But we do want to get to capital after Bess. So we have to add a on both sides. So we have g t plus A is equal to f t. Now we know we could get to the applause transform so g a T plus a turns out to be. It's a g of t plus A is equal to t plus A is equal to two in this case, too. And that's squared. So we wanted to take a look. Loss transform. We should foil this out. So when we do that, we get t squared plus 40 plus four. Now it's in a easier for him to take a pill plus transform up so we could set that in our brackets. So we have t squared plus 40 plus four, and that's multiplied by you to the T minus two. And when we take a look, loss transform, first thing I like to do is evaluate this first and that takes a form of E to the minus A s a being too. So we have e to the minus two s. Now we're going to multiply that by the applause, transform each of these terms. So when we do t squared, we end up with applause. Transform off that equally two over s to the 3/3 we're going to add our constant will stay on top in the numerator and the LaPlace transform of tea is over s square. So 14 with the Austrians were that is for over s square. And for this last term here, just a constant that turns out to be four over s. So there we have our laplace transform. And to simplify it, what will get e to the two s That's all multiplied by two plus four s times one plus s and all of that is over s race to the third. So there we have There's our answer

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