Question

Still chunking through Laplace-related problems. Posted below in a moment.

Answer 1

\[\mathcal{L}\left\{\vphantom{} \frac{e^{-2s}}{s^3}\right\}\]

Answer 2

My first thoughts looking at this are that a Heaviside function is going to be necessary in the end result, and the shift in that Heaviside is going to have to be equal to -2; \[a=-2.\]

Answer 3

This is the first problem where I've had to move backwards, in the other direction, when dealing with Heavisides in Laplace, so I'm a little bewildered by it; The next step I'm guessing would be some algebraic manipulation to get the s^3 term in the denominator to be in the appropriate form to have the same argument as the Heaviside function.

Answer 4

Whoops, supposed to be a Laplace Inverse! NOT a Laplace!

Answer 5

\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{e^{-2s}}{s^3}\right\}\]

Answer 6

yes your approach is correct\[L[u(t-a)]=\frac{ e^{-as} }{ s}\]

Answer 7

Alright, so I need an additional s^2, maybe I could break those up for the sake of looking at them separately:

Answer 8

\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{e^{-2s}}{s^3}\right\}=\mathcal{L^{-1}}\left\{\vphantom{}\frac{e^{-2s}}{s}\frac{1}{s^2}\right\}\]

Answer 9

\[L^{-1}\frac{ e^{-2s} }{ s^3 }=\int\limits_{0}^{t}\int\limits_{0}^{t}u(t-2)\\\]
does that look correct

Answer 10

The double integral is really odd to me, but I can see why that's the case. I just wouldn't be writing it out that way, but I think I understand.

Answer 11

now we can take take u(t-2) out of the integral and then integrate

Answer 12

How can we take U out of t is involved in the integral?

Answer 13

We're integrating with respect to t, yes?

Answer 14

No variable of integration.

Answer 15

because
\[u(t-2)=1 ~~for~~t>2 \]

Answer 16

Okay, so should the lower limits of integration be changed to 2, then?

Answer 17

0 to t-2 i think

Answer 18

its too much lagging :(

Answer 19

so u should get \[L^{-1}\frac{ e^{-2s} }{ s^3 }=\frac{ 1 }{ 2 }(t-2)^2.u(t-2)\]

Answer 20

seems ok?

Answer 21

have u done a problem involving integrating heaviside before?

Answer 22

if not i'm writing a example to clear things

Answer 23

suppose u have --\[\int\limits_{-\infty}^{\infty}u(\tau-2).u(t -\tau-4)d \tau\\step ~1:u(t-\tau)=1,for~(\tau-2)>0~or~\tau>2\\step 2::u(t-\tau-4)=1,for~t-\tau-4>0~or~\tau<t-4\\so ~the~\int\limits~becomes\\f(t)=\int\limits_{2}^{t-4}1.1 d \tau=(t-6)u(t-6)\]
like this

Answer 24

still confused?

Answer 25

I haven't done a problem integrating heaviside before, this is my first; I'm looking at the example now, thank you!

Answer 26

Isn't this t minus tau stuff convolution?

Answer 27

yes the idea comes from here

Answer 28

i can give u another method but i dont know y'll understand it or not , i'm attaching a photo-

Answer 29

wolfram gives the same result
http://www.wolframalpha.com/input/?i=inverse+laplace+%5Be%5E%7B-2s%7D%2Fs%5E3%5D

Answer 30

Hey siddhartha, I'm guessing the character to left-is that gamma or something? http://i.imgur.com/IVSZwaN.pngf

Answer 31

Oh, "r" for "ramp"?

Answer 32

I've never seen a ramp function before, either; my book altogether doesn't mention it. Thanks for putting so much effort into all of this stuff, man. It's very generous of you.

Answer 33

My final answer will be in the form:\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{e^{-2s}}{s^3}\right\}=f(t-2)\mathcal{U}(t-2),\]Where \[f(t-2)\]is the inverse laplace transform of \[\frac{1}{s^3}.\]This makes sense to me.

Answer 34

Correction, \[\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s^3}\right\}=f(t),\]\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s^3}\right\} \neq f(t-2).\]

Answer 35

\[\mathcal{L^{-1}}\left\{\vphantom{}\frac{1}{s^{3}}\right\}=\frac{1}{2}t^2\]

Answer 36

Accounting for the shift and multiplying by the unit step function, \[\frac{1}{2}(t-2)^2U(t-2)\]

This is the correct answer, but I took a wholly different approach than yours; is my approach consistently valid, or am I making my own patterns that in fact don't exist?