Laplacian Translation

  • #1
Given:
F(s) = (s-1)/(s+1)^3

Find:
f(t)

Solution:

Using the equation that when F(s) = n!/(s-a)^(n=1), L^(-1){F(s)} = t^n*e^(at)

So far I find that f(t) = e^(-t)*(-t^2+__)

The book says that f(t) = e^(-t)*(t-t^2)
How did they get the t?
 

Answers and Replies

  • #2
10
0
You should rewrite F(s)
[tex]
F(s) \equiv \frac{s-1}{(s+1)^3} \equiv \frac{1}{(s+1)^2} - \frac{2}{(s+1)^3}
[/tex]
and apply the inverse laplace transform
when F(s) = n!/(s-a)^(n=1), L^(-1){F(s)} = t^n*e^(at)
for each term.
 

Related Threads on Laplacian Translation

  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
4
Views
497
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
803
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
3
Views
493
Top