Laplace Transforms of $(t-2)u(t-2)$ and $t u(t-2)-2u(t-2)$?


Your solution is right. Note that the Laplace transform of $tu(t-2)$ is not $\dfrac{e^{-2s}}{s^2}$. Rather $$\mathcal{L}(t u(t-2)) = \int_{2}^{\infty} t e^{-st}dt = \frac{2s+1}{s^2}e^{-2s}. \tag{1}$$ Using $(1)$, you can easily verify that $$\mathcal{L}((t-2)u(t-2)) = \mathcal{L}(tu(t-2)-2u(t-2))=\frac{e^{-2s}}{s^2}.\tag{2}$$


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Updated on August 01, 2022


  • most venerable sir
    most venerable sir over 1 year

    Let $u(t)$ be the unit-step function; i.e., it is equal to $0$ for $t<0$, and $1$ for $t>0$. My question is why are not the Laplace transforms of $(t-2)u(t-2)$ and $t u(t-2) - 2u(t-2)$ equal to each other?

    I first calculated the Laplace transform of $(t-2)u(t-2)$. I used the formula for the Laplace transform of $f(t-a)u(t-a)$ is equal to $e^{-as}F(s)$. Therefore, for $a=2$, and $f(t)=tu(t)$, whose $F(s)=\frac{1}{s^2}$, I get the Laplace transform of $(t-2)u(t-2)$ to be $\frac{e^{-2s}}{s^2}$.

    But my professor split it into $tu(t-2)-2u(t-2)$ and got its Laplace tranform to be $e^{-2s}(\frac{1}{s^2}-\frac{2}{s})$.

    I don't know who is wrong.