## UnkleRhaukus 3 years ago $\mathcal L\big\{f(x-a)h(x-a)\big\}$

1. UnkleRhaukus

\begin{align*} \mathcal L\big\{f(x-a)h(x-a)\big\}&=\int\limits_0^\infty f(x-a)h(x-a)e^{-px}\text dx\\ \text{let } t=x-a\\ \text dt=\text dx\\ x=0\rightarrow t=-a\\ x=\infty\rightarrow t=\infty\\ &=\int\limits_{-a}^\infty f(t)h(t)e^{-p(t+a)}\text dt\\ &= \\ \end{align*}

2. UnkleRhaukus

\begin{align*} &=\int\limits_{-a}^\infty f(t)h(t)e^{-p(t+a)}\text dt\\ &=e^{-pa}\int\limits_{0}^\infty f(t)e^{-pt}\text dt\\ &=e^{-pa}F(p)\ \\ \end{align*} is this right?

3. UnkleRhaukus

i dont get it

4. ChmE

Thats the same thing I show in my book. I just learned this last week. I don't get it either

5. ChmE

The h(x-a) or in my book u(t-a) as I understand it is just an on off switch for the discontinuous function

6. UnkleRhaukus

yeah , h, u , or $$\theta$$ are used for the heaviside unit step function

7. UnkleRhaukus

$\mathcal L\big\{f(x-a)h(x-a)\big\}=e^{-pa}F(p)$ $f(x-a)h(x-a)=\mathcal L^{-1}\big\{e^{-pa}F(p)\big\}$

8. UnkleRhaukus

@gohangoku58