Finding the Laplace transform of a step function

Question

I am trying to find the Laplace transform of the function $f(t)=(7-t)(u(t-1)-u(t-4))$ for $s \neq 0$. As far as I know this is a function of the form $f(t-c)u_c(t)$ where $u_c(t)=(t-c)$. As such I tried to solve it by performing the laplace on each term. $$\mathcal{L}[7-t]=\frac{7}{s}-\frac{1}{s^2}=\frac{7s-1}{s^2}\ ,$$ $$\mathcal{L}[u(t-1)]=\frac{e^{-s}}{s}\ ,$$ $$\mathcal{L}[u(t-4)]=\frac{e^{-4s}}{s}$$ With each of these Laplace functions found, I then put it all together back in the original equation, and simplified. $$\frac{7s-1}{s^2}*\frac{e^{-s}}{s}-\frac{7s-1}{s^2}*\frac{e^{-4s}}{s}\ ,$$ $$\frac{7s-1}{s^3}(e^{-s}-e^{-4s})$$ This became my final answer, but it seems like I am incorrect. I reworked the problem a few times, so I am fairly confident I didn't make an arithmetic mistake, but I am unsure where else I went wrong. Did I approach the problem in the wrong way or take a laplace incorrectly? Any guidance on where the issue arises would be appreciated.

Answer 1

$$f(t)=(7-t)(u(t-1)-u(t-4))$$ You need to have functions of the form $f(t-c)u_c(t)$. You can rewrite $f(t)$ as: $$f(t)=-(t-1)u(t-1)+6u(t-1)+(t-4)u(t-4))-3u(t-4)$$ Then apply Laplace Transform: $$F(s)=-\dfrac {e^{- s}}{s^2}+6\dfrac {e^{-s}}s+\dfrac {e^{- 4s}} {s^2}-3\dfrac {e^{- 4s}} {s}$$ Don't take Laplace Transform on $(7-t)$ alone. You need to change it first since you have the step function with it. $$(7-t)u(t-1)=-(t-1)u(t-1)+6u(t-1)$$