Laplace question

  • Thread starter daf1
  • Start date
  • #1
1
0
I want to find the following inverse Laplace transform:

[5/(s+1)]exp(-2s)

I was thinking of using the second shift theorem to no avail.

I have tried the question and my answer is :
5(exp-t)u(t-2)

This does not seem to be a plausible answer since I modified the second
shift theorem.

{If only the question was to inverse [5/(s+2)]exp(-2s),
then my answer would have been 5(exp-2t)u(t-2) (which hopefully is right).
However the question is different.}
 

Answers and Replies

  • #2
107
0
I just want to make sure I understand the equation. Is it:

[tex]\frac{5e^{-2s}}{s+1}[/tex] ?

If it is, I think this is simpler than you're making it.

[tex]\frac{5e^{-2s}}{s+1} = \frac{K}{s+1}[/tex]

Multiply both sides by [tex]s+1[/tex] and evaluate at [tex]s = -1[/tex]:

[tex]5e^{-2(-1)} = K[/tex]

[tex]5e^{2} = K[/tex]

Now plug it back in:

[tex]\frac{5e^{2}}{s+1}[/tex]

Now the inverse laplace would be:

[tex]L^{-1} = 5e^{2}e^{-t} u(t)[/tex]

i.e.

[tex]L^{-1} = 5e^{2-t} u(t)[/tex]

I hope this helped.
 

Related Threads on Laplace question

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
696
  • Last Post
Replies
2
Views
24K
Replies
3
Views
3K
Replies
5
Views
2K
  • Last Post
Replies
0
Views
1K
Replies
8
Views
3K
Top