Laplace Transform

• February 24th 2010, 04:42 PM
Aryth
Laplace Transform
I have only had a one hour lecture on these, so it is hard to think about these.

Show that $\cal{L}$ $[f'(t)] = sF(s) - f(0)$
• February 24th 2010, 06:33 PM
chisigma
Integrating by parts we obtain...

$\mathcal {L} \{f^{'}(t)\}= \int_{0}^{\infty} e^{-st}\cdot f^{'}(t)\cdot dt =$

$= |e^{-st}\cdot f(t)|_{0}^{\infty} + s\cdot \int_{0}^{\infty} e^{-st}\cdot f(t)\cdot dt =$

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

... where we use the fact that if $\mathcal{L} \{f(t)\}$ exists, then is...

$\lim_{p \rightarrow \infty} e^{-sp} f(p) =0$ (2)

Kind regards

$\chi$ $\sigma$
• February 24th 2010, 06:37 PM
Aryth
I appreciate the help. Thanks.
• February 24th 2010, 08:28 PM
Aryth
I have one question with inverse transforms.

$\cal{L}$ $\left[\frac{s-1}{(s-3)(s+1)}\right]$
• February 24th 2010, 09:06 PM
chisigma
Applying the residue theorem we find...

$F(s)=\frac{s-1}{(s-3)\cdot (s+1)} = \frac{r(3)}{s-3} + \frac{r(-1)}{s+1}$ (1)

... where...

$r(3)= \lim_{s \rightarrow 3} F(s)\cdot (s-3)= \frac{1}{2}$

$r(-1)= \lim_{s \rightarrow -1} F(s)\cdot (s+1)= \frac{1}{2}$ (2)

... so that is...

$\mathcal {L}^{-1} \{\frac{s-1}{(s-3)\cdot (s+1)} \}= \frac{1}{2}\cdot e^{-t} + \frac{1}{2}\cdot e^{3t}$ (3)

Kind regards

$\chi$ $\sigma$
• February 24th 2010, 09:08 PM
oblixps
for the inverse transform use partial fractions to split the expression up. you can check if i did it correctly but i got (1/2)/(s-3) + (1/2)/(s+1). so you should be able to recognize these as laplace transforms of the exponential function. so you should get (1/2)e^(3t) + (1/2)e^(-t).
• February 25th 2010, 03:34 PM
oblixps
the laplace transform of te^(-t) should be 1/(s+1)^2, not 1.
• February 25th 2010, 03:46 PM
Aryth
Quote:

Originally Posted by oblixps
the laplace transform of te^(-t) should be 1/(s+1)^2, not 1.

I figured out that my solution was riddled with errors. Thanks for the help though. I appreciate it.