# Thread: Laplace Transform

1. ## Laplace Transform

What is the Laplace Transform of $\displaystyle f(x)=e^{bx}.sin ax$

2. Originally Posted by roshanhero
What is the Laplace Transform of $\displaystyle f(x)=e^{bx}.sin ax$
put it in formula for Laplace transformation and try to solve

Bilateral Laplace transformation :

$\displaystyle \displaystyle F(S)= \int _{-\infty} ^{+\infty} e^{-st}f_{(t)} \;dt$

3. I am trying but the things are getting messy though..........

4. Originally Posted by roshanhero
I am trying but the things are getting messy though..........
hmmmm.... as I recall from signal and systems, Laplace transformation doesn't exist on periodical signals like u have there ...
there is Laplace transformation of $\displaystyle \displaystyle sin (\omega_0 t) u(t)$ because with step function $\displaystyle u(t)$ u put to zero your signal for all t<0, But with like this one u have (periodical) it doesn't converge

P.S. perhaps only if u use unilateral Laplace transformation (but still I'm spinning a lot even that way) which limits go from (-0) to infinity... but as I said i think there's no point going there because it just start going in circles (at least I do)

Edit : without your "constants" (a) and (b) and using unilateral formula (don't know if you are bounded to use bilateral) u get this :
$\displaystyle \displaystyle \frac {1}{S^2-2S+2}$ that's for $\displaystyle ROC : Re(S) >1$ (region of convergence )

without (a), or should i say a=1 u'll get :

$\displaystyle \displaystyle \frac {1}{(b-S)^2+1}$ that's for $\displaystyle ROC :Re(b)< Re(S)$

5. Originally Posted by roshanhero
What is the Laplace Transform of $\displaystyle f(x)=e^{bx}.sin ax$
There is a standard operational formula that will handle this. I suggest you go to your table of Laplace transforms and find it. Then use it. If you have any further difficulties, please post your work and say where yuo are stuck.

6. yes, always u can use table of Laplace transformations, but on the exams i don't think anywhere would someone allow you to use table of Laplace transformations... or that's just me
anyway using unilateral u can get to the solution, but using bilateral u can't

solution should be something like:

$\displaystyle \displaystyle \frac {a}{(S-b)^2+a^2}$ with $\displaystyle ROC: Re(S)>Re(a)$