I have only had a one hour lecture on these, so it is hard to think about these.

Show that $\displaystyle \cal{L} $ $\displaystyle [f'(t)] = sF(s) - f(0)$

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- Feb 24th 2010, 04:42 PMArythLaplace Transform
I have only had a one hour lecture on these, so it is hard to think about these.

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Feb 24th 2010, 06:37 PMAryth
I appreciate the help. Thanks.

- Feb 24th 2010, 08:28 PMAryth
I have one question with inverse transforms.

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

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

... where...

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

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

... so that is...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Feb 24th 2010, 09:08 PMoblixps
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).

- Feb 25th 2010, 03:34 PMoblixps
the laplace transform of te^(-t) should be 1/(s+1)^2, not 1.

- Feb 25th 2010, 03:46 PMAryth