View Full Version : Laplace transform
alamleh
Apr 25, 2008, 07:33 AM
Need to find laplace transform for
L{exp(2t)/t}
galactus
Apr 25, 2008, 08:57 AM
-ln(s-2)=ln(\frac{1}{s-2})
ebaines
Apr 25, 2008, 10:56 AM
Galactus - you seem to be talking about L(e^{2t}), but what about the t in the denominator? How do you do L(\frac {e^{2t}} t) ?
galactus
Apr 25, 2008, 11:59 AM
I included that.
L(\frac{1}{t})=ln(\frac{1}{s})=-ln(s)
L(e^{2t})=\frac{1}{s-2}
L(\frac{e^{2t}}{t})=ln(\frac{1}{s-2})
ebaines
Apr 25, 2008, 02:49 PM
I'm afraid I'm losing you at L(1/t) = ln (1/s). I tried calculating L(1/t) and I get infinity:
Starting with the definition L[f(t)]= F(s), and the fact that
L [ \frac {f(t)} t]) = \int _s ^ \infty F(s) ds \\
then with f(t) = 1:
L( \frac 1 t ) = \int _s ^ \infty (1/s) ds \ = \ ln(\infty ) - ln(s) = \infty
Where am I going wrong?
I must admit that I last struggled with Laplace transforms 25 years ago, so I am quite rusty on this stuff.
galactus
Apr 25, 2008, 03:32 PM
I must also admit that I looked these up in a LaPlace table or ran them through my TI-92, which has a Laplace program. I didn't do it from scratch.
The Laplace form is \int_{0}^{\infty}e^{-st}f(t)dt
This one would be \int_{0}^{\infty}\frac{e^{-st}e^{2t}}{t}dt
Which is a booger.
The inside cover of a DE book I have has a bunch of them.
I just used L(\frac{e^{at}}{t}) = -ln(s-a)
alamleh
Apr 26, 2008, 08:56 AM
thanks all for the help,
I think I can prove it now :
L{t . 1/t} = - dF/dS where F = L { 1/t }
the left hand side is L { 1 } = 1/S
therefore, dF/dS = - 1/S which means that F= - Ln(S)
i.e. L { 1/t } = - Ln (S)
and L { exp( a.t). /t } = - Ln(S-a)