what is the laplace transform of a^t where "a "
is constant

Question

anonymous · Answer

bleh not sure on this one. always hated laplace transforms. there are some really good videos on this on youtube tho. http://www.youtube.com/watch?v=zvbdoSeGAgI

anonymous · Answer

http://www.youtube.com/watch?v=hqOboV2jgVo&feature=results_video&playnext=1&list=PL71E66F7E1E13C838

nikvist · Answer

$$f(t)=a^t\quad,a>0\quad a\neq 1$$$$F(s)=\int\limits_{0}^{+\infty}f(t)e^{-st}dt=\int\limits_{0}^{+\infty}a^te^{-st}dt=\int\limits_{0}^{+\infty}(ae^{-s})^tdt=$$$$=\frac{1}{\ln{(ae^{-s})}}(ae^{-s})^t|_0^{+\infty}=\frac{0-1}{\ln{a}+\ln{e^{-s}}}=\frac{1}{s-\ln{a}}$$

jamesj · Answer

Very nice nikvist.  You can see this all makes sense as in the limit when a = 1, then the Laplace transform is 1/s, which is what we'd expect for the constant function/unit step.  So actually you can relax your condition a not equal to 1 and let it be 1 as well.

For a < 1, it's clear the domain we should consider for the Laplace transform is Re(s) > ln(a).  The question is what to say about the domain if a > 1; I think we should impose the same condition.