Math Exam- Laplace Transform

Jul 2010
4
0
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
 
Jul 2010
456
138
\(\displaystyle F(S) = \displaystyle \int _{-0} ^{\infty} e^{at} e^{-st} dt = \frac {1}{S-a} \) with \(\displaystyle ROC : Re(S)>-Re(a) \)
 

chisigma

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Mar 2009
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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

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
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