Inverse Laplace transform

  • Thread starter Neoon
  • Start date
  • #1
25
0
Gents,

I have this problem:

find the inverse laplace transfor for

Y1(s) = exp(-s)/s^2

Y2(s) = {1/[4*(s+1)]}*exp(-2*s)

my solution is:

using the 2nd shifting theroem

y1(t) = (t-1) H(t-1)
y2(t) = (1/4)*exp(2-t)*H(t-2)


Is my solution correct?
 

Answers and Replies

  • #2
25
0
I want to clarify that H(t-1) and H(t-2) is the Hiviside function.
 
  • #3
809
0
I want to clarify that H(t-1) and H(t-2) is the Hiviside function.
What is the laplace transform of?
y1(t) = (t-1) H(t-1)
y2(t) = (1/4)*exp(2-t)*H(t-2)
 

Related Threads on Inverse Laplace transform

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
3K
Replies
3
Views
901
Replies
2
Views
423
Replies
2
Views
7K
Top