Question

Find the inverse transform by convolution: 4 (s-3)(s-2) C.27.

Answer 1

I suppose you're asking about the inverse Laplace transform of

`F(s)=\frac4{(s-3)(s-2)}`

by way of the convolution theorem.

The theorem itself says that

`\mathscr L^(-1)\{A(s)B(s)\}=(a*b)(t)`

where
`A(s)` and
`B(s)` are the Laplace transforms of
`a(t)` and
`b(t)`, respectively.

Let
`A(s)=\frac1{s-3}` and
`B(s)=\frac1{s-2}`. Then

`a(t)=\mathscr L^(-1)\left\{\frac1{s-3}\right\}=e^(3t)`

`b(t)=\mathscr L^(-1)\left\{\frac1{s-2}\right\}=e^(2t)`

So we have

`\mathscr L^(-1)\left\{\frac4{(s-3)(s-2)}\right\}=e^(3t)*e^(2t)`

Convolution is defined as

`(a*b)(t)=\displaystyle\int_0^t a(t-\tau)b(\tau)\,\mathrm d\tau`

The convolution of
`e^(3t)` and
`e^(2t)` is

`e^(3t)*e^(2t)=\displaystyle\int_0^t e^(3(t-\tau))e^(2\tau)\,\mathrm d\tau`

`=\displaystyle e^(3t) \int_0^t e^(-\tau)\,\mathrm d\tau`

`=e^(3t)(1-e^(-t))=e^(3t)-e^(2t)`

We end up with

`\mathscr L^(-1)\left\{\frac4{(s-3)(s-2)}\right\}=\boxed{4(e^(3t)-e^(2t))}`