Question

laplace transform of t^2 u(t-1). How do I make t^2 in terms of t-1?

Answer 1

(t-1)^2+2(t-1)+1...?not quite sure what u r asking...

Answer 2

Question asks for Laplace transform of \[f(t) = t ^{2} u(t-1)\] Hint given rewrite in terms of t-1 first

Answer 3

dont think too complicatedly....just stretch it out x=t-1
f(t)=x^2u(t-1)+2xu(t-1)+u(t-1)

Answer 4

f(t)=x^2u(x)+2xu(x)+u(x) sorry...

Answer 5

\[\large\mathcal{L}\{t^2u_c(t-1)\}=\int_0^\infty t^2u_c(t-1)e^{-st}~dt\]
Missing the subscript on the Heaviside step function, assuming that's what you're referring to.

Answer 6

Ah I am still so lost in this equation :(

Answer 7

Sorry, my mistake! I wasn't familiar with your notation of the Heaviside function. I know it as \(u_c(t)\), where \(c\) is the shift. Your notation is actually more obvious (a shift of 1 unit to the right), so \(u(t-1)=u_1(t)\). Sorry for the confusion.


The Laplace transform of this function is
\[\mathscr{L}\{t^2u(t-1)\}=\int_0^\infty t^2u(t-1)e^{-st}~dt\]
Since \(t^2u(t-1)=0\) for \(0\le t<1\) and \(t^2u(t-1)=t^2\) for \(t\ge1\), you can write the integral as
\[\int_1^\infty t^2e^{-st}~dt\]
This shouldn't be too difficult for you. Integrate by parts twice, or use the formula for the Laplace transform of functions of the form \(t^n\).

As for the hint, I'm not really sure what "in terms of t-1" could mean.