Question

Laplace transforms

@zepdrix
Can you please help me with the Laplace transform of the function
f(t) = 3t+12

using the integral method thing

Answer 1

\[\int\limits\limits_{0}^{\infty} e^{-st}(3t + 12) \text{ dt}\]

Answer 2

Tal, do you have to use the definition? Or are you allowed to use a few memorized transforms?
Cause we can break this function down really nicely.

\[\large \mathscr{L}\left[3t+12\right] \qquad = \qquad 3\mathscr{L}\left[t\right]+12\mathscr{L}\left[1\right]\]

Answer 3

yip we have to:(

Answer 4

Understand how to start by parts on this one?

\[\large \int\limits_{0}^{\infty} \color{orangered}{e^{-st}}\color{royalblue}{(3t + 12)}\;\color{orangered}{dt}\]

\[\large \color{royalblue}{u=3t+12} \qquad \qquad \qquad \color{orangered}{dv=e^{-st}\;dt}\]

Answer 5

Okay here is what I got:
\[u = 3t+12;\text{ }du = 3 dt\\dv=e^{-st};\text{ }v=\frac{1}{-s}e^{-st}\\\\uv-\]\[u = 3t+12;\text{ }du = 3 dt\\dv=e^{-st};\text{ }v=\frac{1}{-s}e^{-st}\\\\uv-\int\limits_{0}^{\infty}v \text{ du}\\\\(3t+12)(\frac{1}{-s}e^{-st})_{0}^{\infty}-(\frac{1}{-(s)^2})\int\limits_{0}^{\infty} \frac{-(s)^2}{-s}e^{-st}3\text{ dt}\]

Answer 6

Your \(\large uv\) looks good.
I'm a little confused what's going on with the \(\large \int v\;du\).

Answer 7

I already multiplied the part that was needed to do the e integration

Answer 8

and then I got\[\left[ (3t+12)\frac{1}{-s}e^{-st}-\frac{3}{-s^2}e^{-st} \right]_{0}^{?}\]
but now what?

Answer 9

that ? = infinity

Answer 10

Ummmm ok let's think about what's happening.

Answer 11

At the upper limit, as we approach infinity, what happens to our exponential terms?
(Assuming s is positive).

Answer 12

\[\huge e^{-t}\]Think about it from this standpoint if you're getting confused.
What happens as t approaches infinity?

Answer 13

the e part will become 1

but what about the "normal" parts, such as (3t+12)?

Answer 14

No not for the upper limit :O
As t gets bigger and bigger, the exponential is getting closer to 0.

Answer 15

\[\large e^{-999999} \qquad = \qquad \frac{1}{e^{999999}} \qquad \approx \qquad 0\]

Answer 16

Maybe you were thinking of the lower boundary though.

Answer 17

oh I missed the -

Answer 18

So we don't need to worry about the upper limit of integration, they're producing a zero in both terms.

So we just need to evaluate the lower limits.
Careful though, remember you're suppose to `subtract` the lower limit, so we need a negative sign in front.

\[\large -\left[ (3\cdot0+12)\frac{1}{-s}e^{-s0}-\frac{3}{-s^2}e^{-s0} \right]\]

Answer 19

Oh woops I think you missed some multiplication earlier.

Integrating the e^{-st} gave us a -1/s
When you integrated again, it should have given you another -1/s

So that coefficient in front of the second term should be
\[\large -\frac{3}{s^2}\]
Without the negative in the denominator.

Answer 20

Ahh okay I understand now!
I struggled with the t-> infinity part at 3.t I thought the term would go to infinity. It will, but I missed the part where it gets multiplied by 0 so it doesn't matter then

Answer 21

Oh good call.. if it's being multiplied by 0 we have the indeterminate form \(\large 0\cdot \infty\).

It's going to still give us 0, because the exponential is approaching 0 much much much faster than 3t is approaching infinity.
But maybe it's a good idea to justify that on the side.

It requires L'Hop Rule. Remember that one? :3

Answer 22

yip, lim of the two derivatives

Answer 23

So here's some side work, just in case your teacher wants to see this justified.

\[\large \lim_{t \rightarrow \infty} 3t e^{-st} \qquad =\qquad 3\lim_{t \rightarrow \infty}\frac{t}{e^{st}}\]

This limit is of the indeterminate form \(\large \dfrac{\infty}{\infty}\), which is one of the forms from which we're allowed to apply L'Hop Rule.

Taking the derivatives gives us,\[\large 3\lim_{t \rightarrow \infty}\frac{t}{e^{st}} \qquad = \qquad 3\lim_{t \rightarrow \infty} \frac{1}{s e^{st}} \qquad = \qquad 0\]

Answer 24

I don't mean to do all the work for ya. Sorry if I'm taking the fun away lol.
I just know that those steps can be a little tricky if you haven't done them in a while.

Answer 25

Nope, I like that you're doing it, then I know it's right. Thanks!

Answer 26

If you've just started working on Laplace Transforms recently, you wanna make a note by your answer (for s>0).

It's not a big deal, it's kind of just.. understood when you're working with these.
But just something to keep in mind. :3

No probs.

Answer 27

Thanks so much!
I'm writing exam tomorrow, so I really appreciate the help!