# Laplace Transform Within a Domain

1. Jan 20, 2014

### GreenPrint

1. The problem statement, all variables and given/known data

Find the Laplace transform of

f(t) = t $\forall$ 0≤t≤T, 0 otherwise

2. Relevant equations

3. The attempt at a solution
I write the function as

$tu(t)-t*u(t-T)$

That is turn on the function $t$ at $t=0$ and turn the function $t$ off at $t=T$. It seems to be right to me.

But now I struggle with trying to take the Laplace transform of this. I know that $L(u(t)) = \frac{1}{s}$. I know that $L(f(t-T)) = e^{-sT}F(s)$. So I know that $L(u(t-T)) = e^{-sT}\frac{1}{s}$, but I'm not sure how to evaluate $L(t*u(t-T))$ because of the extra $t$ term, hence I'm stuck.

Thanks for any help.

I seem to be confused because $F(s) = ∫_{0^{-}}^{∞}f(t)e^{-st}dt$. So I don't see how you can take the Laplace transform over a domain other than over $0^{-}≤t≤∞$, which this question seems to be asking me to do.

Last edited: Jan 20, 2014
2. Jan 20, 2014

### LCKurtz

You have a problem of the form $\mathcal L(f(t)u(t-T))$.$$\mathcal L(f(t)u(t-T))=\int_0^\infty e^{-st}f(t)u(t-T)~dt =\int_T^\infty e^{-st}f(t)\cdot 1~dt$$Now make the substitution $u = t - T$:$$=\int_0^\infty e^{-s(u+T)}f(u+T)~du=e^{-Ts}\int_0^\infty e^{-su}f(u+T)~du =e^{-Ts}\mathcal Lf(t+T)$$So for your problem$$\mathcal L(tu(t-T)) = e^{-Ts}\mathcal L(t+T) =~?$$

3. Jan 21, 2014

### GreenPrint

So this problem becomes

$L(t*u(t) - t*u(t-T)) = \frac{1}{s^{2}} - L(t*u(t-T)) = \frac{1}{s^{2}} - \int^{∞}_{0^{-1}}e^{-st}t*u(t-T)dt = \frac{1}{s^{2}} - \int^{∞}_{0^{-}}e^{-st}tdt$

Now I do $u$ substitution to evaluate the integral. $u(t) =t, \frac{du}{dt} = 1$. $\int dv = \int e^{-st}dt = -\frac{e^{-st}}{s}$

Now I continue solving

$\frac{1}{s^{2}} - \int^{∞}_{0^{-}}e^{-st}tdt = \frac{1}{s^{2}} -(-\frac{te^{-st}}{s} - \int^{∞}_{0^{-}}e^{-st}dt) = \frac{1}{s^{2}} + \frac{te^{-st}}{s} - \frac{e^{-st}}{s}$

The solutions has
$F(s) = \frac{1}{s^{2}} - \frac{1}{s^{2}}e^{-sT} - \frac{T}{s}e^{-sT}$. So I must be doing something but I don't see what.

4. Jan 21, 2014

### LCKurtz

You don't have to work out any integrals and re-do all the work. Just use the formula I gave you above. All you have to do is use that formula with $f(t)=t$.

5. Jan 22, 2014

### GreenPrint

Thanks, I was able to figure it out