Question

laplace transformation

Answer 1

\[Y(t)= \frac{ e^{-\pi s} }{ (s+1)^2+1 }\]

Answer 2

Take the inverse

Answer 3

First, the left side should be Y(s).

Answer 4

So, this is a form of \[L \left\{ f(t) \right\} = e^{-as}F(s)\] Which would mean your f(t) has the form
\[U(t-a)f(t-a)\]
In this case, "a" is just pi.
As for your F(s) portion, it comes from an f(t) of the form\[e^{at}\sin(kt)\]
So looking at the F(s) portion\[\frac{ 1 }{ (s+1)^{2} + 1 }\] we see a is -1 and k is 1, meaning F(s) is\[e^{-t}sin(t)\]Therefore our inverse transform gives us
\[Y(t) =U(t-\pi )[e^{-(t- \pi )}\sin(t-\pi )]\]

Answer 5

I got it, but I though the answer is \[e^{-t} \sin(t) [u(t-\pi)]\]

Answer 6

how did you get e^-(t-pi) sin(t-pi)

Answer 7

It's just the form of that kind of inverse transform. I didn't really get it from anywhere other than saying that since our laplace transform is \[e^{-as}F(s)\] where F(s) is the transform of
\[f(t) = e^{at}\sin(kt)\]Taking the inverse transform of \[e^{-as}F(s) \] gives
\[U(t-a)f(t-a)\] So that required be to do the t-pi substitution into
\[f(t) = e^{at}\sin(kt)\] also, not just the U part.

Answer 8

My only question is how you got U(t-a) f(t-a)

Answer 9

is it because it is transformed?

Answer 10

Right. If I were to separate original functions f(t) and their transforms L{f(t)}, for this problem we have our f(t)'s equal to:
\[U(t-a)f(t-a)\]
\[e^{at}\sin(kt)\]
These are functions prior to taking their laplace transform. Our function was a laplace transform of these.
\[f(t) = U(t-a)f(t-a)\]
\[L \left\{ f(t) \right\} = L[U(t-a)f(t-a)] = e^{-as}F(s)\]
F(s) was the laplace transform of \[f(t) = e^{at}\sin(kt)\]So that gave us:
\[L \left\{ f(t) \right\}= L[e^{at}\sin(kt)] = \frac{ k }{ (s-a)^{2}+k^{2} }\]
So everything I have labeled as f(t) is the result after taking the inverse transform (or before taking the laplace transform if you prefer)