# Math Exam- Laplace Transform

#### andymarra

Hello, Im currently revising for my university maths exam and have found the following question:
F(s) denotes the laplace transform of the causal signal f(t), with the region of convergence Re(s) >σc. Prove the first shift theorem, using the basic definition of the laplace transform, by showing that the transform of e^at f(t) is given by L{f(t)}=F(s-a), with the region of conversion Re(s)>σc +Re(a).

Im not really sure what its asking. I know that the first shift theorem is:
L{e^at f(t)}=F(s-a), where F(s)=L{f(t)}

and i know that the definition of a laplace transform is:
F(s)=L{f(t)}=¦(meaning integral from 0- infinity in this case)of e^-st f(t) dt.

How do i use these to come up with the answer that is expected? In the question, this is worth 5 marks out of a total 20.
All help appreciated, Thank you.
Andy

#### yeKciM

$$\displaystyle F(S) = \displaystyle \int _{-0} ^{\infty} e^{at} e^{-st} dt = \frac {1}{S-a}$$ with $$\displaystyle ROC : Re(S)>-Re(a)$$

#### chisigma

MHF Hall of Honor
From definition of the LT is...

$$\displaystyle \displaystyle F(s)= \int_{0}^{\infty} f(t)\ e^{- s t} dt$$ (1)

... and the integral (1) converges for $$\displaystyle \Re (s) > \sigma$$. From (1) it follows that...

$$\displaystyle \displaystyle \int_{0}^{\infty} f(t)\ e^{a t} e^{- s t} dt = \int_{0}^{\infty} f(t)\ e^{- (s-a) t} dt = F(s-a)$$ (2)

... and the integral (2) converges for $$\displaystyle \Re (s-a) > \sigma$$...

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

andymarra