Question

\[\mathcal L\big\{f(x-a)h(x-a)\big\}\]

Answer 1

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

Answer 2

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

is this right?

Answer 3

i dont get it

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

@gohangoku58