Find inverse laplace for: f ( s ) = 1 ( s − 2 ) ( s + 2 ) 2

Name: Find inverse laplace for: f ( s ) = 1 ( s − 2 ) ( s + 2 ) 2
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find inverse laplace for: f ( s ) = 1 ( s − 2 ) ( s + 2 ) 2

The inverse Laplace Transform is given by f ( t ) = 1 2 π i ∫ 2 + − i ∞ 2 + + i ∞ 1 ( s − 2 ) ( s + 2 ) 2 e s t d sFor t&gt;0, we close the contour in the left-half plane s &lt; 2 + . Note that F ( s ) e s t has a simple pole at s=2 and a second-order pole at s=−2.The associated residues of F ( s ) e s t are e 2 t 16 at s = 2 and − 1 4 t e − 2 t − 1 16 e − 2 t at s = − 2Hence, we find that for t&gt;0 f ( t ) = 1 16 e 2 t − 1 16 e − 2 t − 1 4 t e − 2 t For t&lt;0, we close the contour in the right-half plane. Since F ( s ) e s t is analytic in that plane, f(t)=0 for t&lt;0.

"Find inverse laplace for: f ( s )..." was answered by students like you

College
Calculus and Analysis
Differential Equations
Laplace transform

jatericrimson8b

Answered question

2022-09-12

Find inverse laplace for:

f (s) = \frac{1}{(s - 2) (s + 2)^{2}}

Answer & Explanation

Baluttor7

Beginner2022-09-13Added 17 answers

The inverse Laplace Transform is given by

f (t) = \frac{1}{2 π i} \int_{2^{+} - i \infty}^{2^{+} + i \infty} \frac{1}{(s - 2) (s + 2)^{2}} e^{s t} d s

For t>0, we close the contour in the left-half plane

s < 2^{+}

. Note that

F (s) e^{s t}

has a simple pole at s=2 and a second-order pole at s=−2.
The associated residues of

F (s) e^{s t}

are

\frac{e^{2 t}}{16}

s = 2

and

- \frac{1}{4} t e^{- 2 t} - \frac{1}{16} e^{- 2 t}

s = - 2

Hence, we find that for t>0

f (t) = \frac{1}{16} e^{2 t} - \frac{1}{16} e^{- 2 t} - \frac{1}{4} t e^{- 2 t}

For t<0, we close the contour in the right-half plane. Since

F (s) e^{s t}

is analytic in that plane, f(t)=0 for t<0.

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