Question

laplace transform of (t-1)^4

Answer 1

an explanation would be nice not just the answer

Answer 2

Using the integral definition, or a table of transforms?

Answer 3

table is fine directions say determine laplace transform of the given fxn using table and properties of the transform

Answer 4

The table I'm using says
\[\mathcal{L}\left\{t^n\right\}=\frac{n!}{s^{n+1}}\]
I can't quite recall the transform of a function being shifted, though... I'm sure it can easily be derived from the definition.

Answer 5

\[\mathcal{L}\left\{f(t-c)\right\}=\int_0^\infty f(t-c)~e^{-st}~dt\]
Assume \(c>0\). Let \(u=t-c~\Rightarrow~du=dt\):
\[\large\begin{align*}\mathcal{L}\left\{f(t-c)\right\}&=\int_0^\infty f(t-c)~e^{-st}~dt\\
&=\int_{-c}^\infty f(u)e^{-s(u+c)}~du\\
&=e^{-sc}\int_{-c}^\infty f(u)e^{-su}~du\\
&=e^{-sc}\left(\int_{-c}^0 f(u)e^{-su}~du+\int_{0}^\infty f(u)e^{-su}~du\right)\\
&=e^{-sc}\left(\color{red}{\int_{-c}^0 f(u)e^{-su}~du}+\mathcal{L}\left\{f(u)\right\}\right)
\end{align*}\]
The final obstacle... We can probably compute the integral using the given function: http://www.wolframalpha.com/input/?i=Integrate%5B%28u%5E4%29*Exp%5B-s*u%5D%2C%7Bu%2C-c%2C0%7D%5D

I'm not sure if there's a general formula for the transform of a shifted function.

Answer 6

cant you just exapnd it and look for laplace of each term should be much easier o.O

Answer 7

The simplest route is often overlooked...

Answer 8

But yeah, you'll be using the first formula I mentioned

Answer 9

yes you would need to use only that formula

Answer 10

You certainly could expand it. :)
But then when you come across a similar problem like (t-1)^10 you run into a bit of problem. heh.

Answer 11

how would I expand it?

Answer 12

yes true that but since its in power of 4 guess its lot easier to just expand it

Answer 13

@rperez36, binomial expansion.
\[(t-1)^4=t^4-4t^3+6t^2-4t+1\]
Then transform term-by-term.

Answer 14

ok thank you I will try it from there