Question

Laplace

Answer 1

\[L ^{-1}(\frac{ 1 }{ (s-4)^{2}}\]

Answer 2

Do you know the inverse Laplace transform of 1/s^2 alone? So that we might use the shifting theorem for s to s-4.

Answer 3

would it be t

Answer 4

Yup, that sounds right. Then do you know the shifting theorem for moving s to s - a?

Answer 5

not quite

Answer 6

It is essentially that, a factor of e^(at) correlates to the shift \( s \to s - a\) in the function:
\( \mathcal{L} \left[ t \right] = \dfrac{1}{s^2} \)
\( \mathcal{L} \left( t \times e^{4t} \right] = \dfrac{1}{\left(s - 4\right)^2} \)
But if you're not familiar with that, I gotta rethink my strategy a bit.

Answer 7

i have studied the shift theorem but i can't see how it is being applied to the above

Answer 8

The denominator doesn't fit the description of any of the trig functions ( s^2 + a^2 ). It also cannot be decomposed for exponentials.
It is close to 1/s^2, except the s is shifted over to s-4. We know the inverse of 1/s^2 to be t, so if we use the shifting theorem to deal with the s-4, we would have t times the exponential factor that is used to shift it.

Answer 9

doesn't it matter that it's not s^2-4 but s^2-8s+16

Answer 10

\[\text{ If } F(s)=\frac{1}{s^2} \text{ then } F(s-4)=\frac{1}{(s-4)^2}\]
So we can use the shifting theorem access mentioned

Answer 11

http://mathfaculty.fullerton.edu/mathews/c2003/LaplaceShiftingMod.html

Answer 12

Yes. I mean, this is NOT in the form of trig function s^2 - a^2. So we cannot invert it into one.
It also does not decompose into anything simpler because of the repeating factor in the denominator. So we must use the fact that 1/s^2 has inverse Laplace transform of t, along with the shifting theorem for the difference between 1/s^2 and 1/(s-4)^2, , to determine the entire inverse Laplace transform.

[tried to make simpler example, it just looked more confusing >.<]

Answer 13

\[L (t ^{n} e ^{at})=F(s-a)=\frac{ n! }{ (s-a)^{n+1}}\]this is the first time i've read this identity

Answer 14

it makes everything u were saying come to light

Answer 15

our n=1 a=4 sub into t^n e^(at) and bob's your uncle!

Answer 16

thanks a million

Answer 17

thx @myininaya

Answer 18

mm, glad to help!

similar behaviors happen for all functions, so i didn't think about the formula but rather the general concept that:
\( \mathcal{L} \left[ f(t) \right] (s)= F(s) \)
\( \mathcal{L} \left[ e^{at} f(t) \right](s) = F(s - a) \)
but i suppose whichever works for you best? :)