Unilateral Laplace transform I tried to do the same unilateral Laplace transform in two ways, but I got different results. I have to calculate: $\mathcal{L}[r(t-1)]$, where $r(t)$ is the ramp function, that is $r(t)=t, t\ge0$.
$1^{st}$  way: $\mathcal{L}[r(t-1)]=\mathcal{L}[t-1]=\mathcal{L}[t]-\mathcal{L}[u(t)]=\frac{1}{s^2}-\frac{1}{s}$, where $u(t)$ is the unit step function.
$2^{nd}$ way (using Laplace transform properties): $\mathcal{L}[r(t-1)]=\mathcal{L}[t-1]=e^{-s}\frac{1}{s^2}=\frac{1}{e^s s^2}$
What's wrong with the second way? Is it right?
 A: The Laplace Transform of $r(t-1)$ is
$$\begin{align}
\mathscr{L}\{r(t-1)\}(s)&=\int_0^{\infty}r(t-1)e^{-st}dt\\\\
&=\int_0^{\infty}(t-1)u(t-1)e^{-st}dt\\\\
&=\int_1^{\infty}(t-1)e^{-st}dt \tag1\\\\
&=e^{-s}\int_0^{\infty}te^{-st}dt \tag 2\\\\
&=e^{-s}\mathscr{L}\{r(t)\}(s)\\\\
&=e^{-s}\frac1{s^2}
\end{align}$$
In going from $(1)$ to $(2)$, we used the substitution $t-1 \to t$.  Then, $e^{-st}\to e^{-s(t+1)}=e^{-s}e^{-st}$ and the new integration limits begin at $0$.
A: Both of your methods of computing $\mathcal{L}[r(t-1)u(t-1)]$ will work, but your 1st method needs fixing.


*

*1st way (Linearity): You tried to write $\mathcal{L}[r(t-1)]=\mathcal{L}[t-1]=\mathcal{L}[t]-\mathcal{L}[u(t)]=\frac{1}{s^2}-\frac{1}{s}$.
This is wrong because $r(t-1)u(t-1)\neq t - u(t)$.  


To fix this, use the correct expression $r(t-1)u(t-1) = t - g(t)$, where 
$$g(t) = \begin{cases} t,& 0\leq t\leq 1 \\ 1 ,& t\geq 1\end{cases}$$
(You can see that $g(t)$ is equal to $u(t)$, except for the part between $0\leq t\leq 1$.)
\begin{align}
\mathcal{L}[r(t-1)u(t-1)] 
&= \mathcal{L}[t] - \mathcal{L}[g(t)]\\
&= \frac{1}{s^2} - \frac{1}{s^2}(1 - e^{-s}) \\
&= e^{-s}\frac{1}{s^2}
\end{align}
This way (using Linearity) works, but is not as easy as using the Time Shifting Property, because the Laplace transform of $g(t)$ is more difficult to compute.


*

*2nd way (Time Shifting): Easier method.  You used the Time Shift Property correctly.
$$\mathcal{L}[r(t-1)u(t-1)] = e^{-s}\mathcal{L}[r(t)] = e^{-s}\frac{1}{s^2}$$

