# Find the inverse Laplace

## Homework Statement

I have some problem finding the inverse laplace transformation of the function: $$\frac{s}{s^2+2s+5}$$

## Homework Equations

http://math.fullerton.edu/mathews/c2003/laplacetransform/LaplaceTransformMod/Images/Table.12.2.jpg [Broken]

## The Attempt at a Solution

I tried to factorise the denominator: $$\frac{s}{(s+1)^2+2^2}$$

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Chestermiller
Mentor
This is a good start. Now write the s in the numerator as (s+1)-1, and then separate the expression into two fractions.

Chet

1 person
You mean like this?
$$\frac{s+1}{(s+1)^2+2^2} - \frac{1}{(s+1)^2+2^2}$$

The inverse laplace of the first term would be I think then: $$e^{-t}*cos(2t)$$

But I'm not so sure what to do then since I don't recognize the term in the table, would is be something like this?
$$- \frac{0s+1}{(s+1)^2+2^2}$$

So the second term doesn't have an inverse? Oh and thanks for responding!

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gneill
Mentor
You mean like this?
$$\frac{s+1}{(s+1)^2+2^2} - \frac{1}{(s+1)^2+2^2}$$

The inverse laplace of the first term would be I think then: $$e^{-t}*cos(2t)$$
Looks good.
But I'm not so sure what to do then since I don't recognize the term in the table, would is be something like this?
$$- \frac{0s+1}{(s+1)^2+2^2}$$

So the second term doesn't have an inverse? Oh and thanks for responding!
Would you recognize it if it were written:
$$-\frac{1}{2} \frac{2}{(s + 1)^2 + 2^2}$$
?

1 person
Yes now I see it, thanks!!!

rude man
Homework Helper
Gold Member
Or, you could have factored the denominator s^2 + 2s + 5 = (s + a)(s + b), then done a partial fraction expansion into
F = s/(s^2 + 2s + 5) = A/(s+a) + B/(s+b).

We all know 1/(s+a) transforms to exp(-at).

The discriminant is a negative number so it cannot be factorised. At least to my knowledge.

rude man
Homework Helper
Gold Member
The discriminant is a negative number so it cannot be factorised. At least to my knowledge.

F(s) = s/(s2 + 2s + 5)

The factors for the denominator are a,b = 1 +/- j2.
So we get A/(s + 1 + j2) + B/(s + 1 - j2) → Aexp(-1 + j2)t + Bexp(-1 - j2)t

with A = (-1 + j2)/j4 and B = -(-1 - j2)/j4

and by using Euler's formula plus some algebra you can reduce this to the real answer
f(t) = (1/2)exp(-t){2cos(2t) - sin(t)}.

It's a bit messy but how did you work with 2/[(s+1)2 + 22] without looking it up?

1 person
I did thought of doing that way but I felt that it would take more time doing that method plus my book and my teacher never showed doing that way. But thanks for showing that it's possible to work with imaginary numbers!