Unilateral Laplace transform

Question

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?

Answer 1

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 $0$.

Answer 2

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. $$\mathcal{L}[r(t-1)u(t-1)] = e^{-s}\mathcal{L}[r(t)] = e^{-s}\frac{1}{s^2}$$