Question

Find the Laplace transform of f(t) when:

f(t)= 9 , 0 = or < t < 2

f(t)= (t-5)^2 , 2= or < t < 5

f(t)= 2te^6t , t > or = 5

Answer 1

`f(t)=\begin{cases}9&\text{for }0\le t<2\\(t-5)^2&\text{for }2\le t<5\\2te^(6t)&\text{for }t\ge5\end{cases}`

and presumably 0 for
`t<0`. We can express
`f(t)` in terms of the unit step function,

`u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}`

`f(t)=9(u(t)-u(t-2))+(t-5)^2(u(t-2)-u(t-5))+2te^(6t)u(t-5)`

Quick explanation:
`9u(t)=9` for
`t\ge0`, and
`9u(t-2)=9` for
`t\ge2`. So subtracting these will cancel the value of 9 for all
`t\ge2` and leave us with the value of 9 over the interval we want,
`0\le t<2`. The same reasoning applies for the other 3 terms.

Recall the time displacement theorem:

`\mathcal L_s\{f(t-c)u(t-c)\}=e^(-sc)\mathcal L_s\{f(t)\}`

By this property, we have

`\mathcal L_s\{9u(t)\}=\mathcal L_s\{9\}=\frac9s`

`\mathcal L_s\{9u(t-2)\}=e^(-2s)\mathcal L_s\{9\}=\frac{9e^(-2s)}s`

`\mathcal L_s\{(t-5)^2u(t-2)\}=\mathcal L_s\{((t-2)-3)^2u(t-2)\}`

`=e^(-2s)\mathcal L_s\{(t-3)^2\}=\left(\frac2{s^3}-\frac6{s^2}+\frac9s\right)e^(-2s)`

`\mathcal L_s\{(t-5)^2u(t-5)\}=e^(-5s)\mathcal L_s\{t^2\}=(2e^(-5s))/(s^3)`

`\mathcal L_s\{2te^(6t)u(t-5)\}=\mathcal L_s\{2e^(30)(t-5)e^(6(t-5))+10e^(30)e^(6(t-5))\}`

`=2e^(30-5s)\mathcal L_s\{te^(6t)+5e^(6t)\}=2e^(30-5s)\left(\frac1{(s-6)^2}+\frac5{s-6}\right)`

Putting everything together, we end up with

`\boxed{\mathcal L_s\{f(t)\}=((2-6s)e^(-2s)-2e^(-5s))/(s^3)+\frac9s-(2e^(30-5s)(29-5s))/((s-6)^2)}`