Inverse Laplace transformation (s2+s)/(s2+1)(s2+2s+2)

Name: Inverse Laplace transformation (s2+s)/(s2+1)(s2+2s+2)
Uploaded: 2022-02-23T17:22:57+00:00
Description: Inverse Laplace transformation (s2+s)/(s2+1)(s2+2s+2)

Question

Inverse Laplace transformation  (s2+s)/(s2+1)(s2+2s+2)

Ian Adams · Accepted Answer

First, we need to use partial fraction decomposition to simplify the expression:

$\frac{s^{2} + s}{(s^{2} + 1) (s^{2} + 2 s + 2)} = \frac{A}{s^{2} + 1} + \frac{B s + C}{s^{2} + 2 s + 2}$

To find A, we multiply both sides by $(s^{2} + 1)$ and then set $s = i$ , where i is the imaginary unit:

$(s^{2} + s) = A (s^{2} + 1) + (B s + C) (s^{2} + 1)$

$i^{2} + i = A (i^{2} + 1) + (B i + C) (i^{2} + 1)$

$- 1 + i = A (- 1 + 1) + (B i + C) (- 1 + 1)$

$- 1 + i = 2 B i$

$B = \frac{- 1 + i}{2 i} = \frac{1 - i}{2}$

To find C, we multiply both sides by $(s^{2} + 2 s + 2)$ and then set $s = - 1 + i$ :

$(s^{2} + s) = A (s^{2} + 1) + (B s + C) (s^{2} + 2 s + 2)$

$(- 1 + i)^{2} + (- 1 + i) = A (- 1 + 1) + (B (- 1 + i) + C) (- 1 + 2 i + 2)$

$0 = A (2) + (- 1 + i + B + C) (1 + 2 i)$

$0 = 2 A + (2 - i + C - i) (1 + 2 i)$

$0 = 2 A + (2 C - 2) (1 + 2 i)$

$2 C - 2 = 0$

$C = 1$

Now we can rewrite the expression using the partial fractions:

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

Substituting the values of A and B, we get:

$\frac{s^{2} + s}{(s^{2} + 1) (s^{2} + 2 s + 2)} = - \frac{i}{2} \cdot [\frac{1}{s^{2} + 1} - \frac{1 - i}{s^{2} + 2 s + 2}]$

Now we can use the Laplace transform table to find the inverse Laplace transform of each term:

$L^{- 1} {\frac{1}{s^{2} + 1}} = \sin (t) L^{- 1} {\frac{1}{s^{2} + 2 s + 2}} = (\frac{1}{2}) e^{- t} \sin (t)$

Therefore, the inverse Laplace transform of the original expression is:

$L^{- 1} {\frac{s^{2} + s}{(s^{2} + 1) (s^{2} + 2 s + 2)}} = - \frac{i}{2} \cdot [\sin (t) - (1 - i) e^{- t} \sin (t)]$

Inverse Laplace transformation of (s^2 + s)/(s^2 +1)(s^2 + 2s + 2)

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