Inverse Laplace Transform of $\frac{s^2+2s+2}{s+1}$

Question

I want detailed steps of this if anyone can help.

Try doing polynomial division to simplify the fraction. – John Habert Mar 20 '14 at 17:43 — John Habert, Mar 20 '14 at 17:43
@user136837 Is this formatting ok? – imranfat Mar 20 '14 at 17:43 — imranfat, Mar 20 '14 at 17:43

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

1

$$\frac{s^2+2s+2}{s+1}=\frac{(s+1)^2+1}{s+1}=s+1+\frac1{s+1}$$

Now, can you set the proper value of $a$ in $$L\{e^{at}\}=\frac1{s-a}$$

Set $c=0$ in $$L\{\delta(t-c)\}=e^{-cs}$$

edited Apr 13 '17 at 12:21

Community

answered Mar 20 '14 at 17:45

lab bhattacharjee

Is that the whole answer?? – imranfat Mar 20 '14 at 17:47
@imranfat, I've just completed . – lab bhattacharjee Mar 20 '14 at 17:49
Your link is "closed" and I though when doing an inverse laplace we are supposed to get t-terms back? – imranfat Mar 20 '14 at 17:56
@imranfat, but there is one valid answer, I believe – lab bhattacharjee Mar 20 '14 at 18:02
makes sense, just not used to inverses laplace of plain polynomial terms since that is kind of...rare? – imranfat Mar 20 '14 at 18:10

score 0 · Answer 2 · edited Nov 25 '20 at 13:42

0

Hint: $$\frac{s^2 + 2s + 2}{s + 1} = \frac{(s + 1)^2 + 1}{s + 1} = s + 1 + \frac{1}{s + 1}$$

edited Nov 25 '20 at 13:42

Sam Wong

answered Mar 20 '14 at 17:43

DeepSea

2 Answers2