Find the inverse Laplace transform of F(s)=1(s+a)2(s+b) for t≥0 a) sin⁡(at)+e−bt b) te−bt(a−b)2+sin⁡(at) c) sin⁡(at)(b−a)2+e−bt(a−b)2 d) e−bt(a−b)2+te−atb−a−e−at(b−a)2 e)sin⁡(at)+te−bt(b−a)2−te−at(a−b)2

Name: Find the inverse Laplace transform of F(s)=1(s+a)2(s+b) for t≥0 a) sin⁡(at)+e−bt b) te−bt(a−b)2+sin⁡(at) c) sin⁡(at)(b−a)2+e−bt(a−b)2 d) e−bt(a−b)2+te−atb−a−e−at(b−a)2 e)sin⁡(at)+te−bt(b−a)2−te−at(a−b)2
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find the inverse Laplace transform of F(s)=1(s+a)2(s+b) for t≥0 a) sin⁡(at)+e−bt b) te−bt(a−b)2+sin⁡(at) c) sin⁡(at)(b−a)2+e−bt(a−b)2 d) e−bt(a−b)2+te−atb−a−e−at(b−a)2 e)sin⁡(at)+te−bt(b−a)2−te−at(a−b)2

Step 1 Here, F(s)=1(s+a)2(s+b) step 2 Rewrite the function. F(s)=1a−b×(s+a)−(s+b)(s+a)2(s+b) =1a−b×[(s+a)(s+a)2(s+b)−(s+b)(s+a)2(s+b)] =1a−b×[1(s+a)(s+b)−1(s+a)2] =1a−b×[1a−b×(s+a)−(s+b)(s+a)(s+b)−1!(s+a)1+1] =1(a−b)2[1s+b−1s+a]−1a−b×1!(s+a)1+1 step 3 Apply inverse laplace transform formula. f(t)=1(a−b)2[e−bt−e−at]−1a−b×t1e−at =e−bt(a−b)2+te−atb−a−e−at(b−a)2 Step 4 Answer: Option D is correct.

Expert Answer to "Find the inverse Laplace transform of F(s)=1(s+a)2(s+b) ..."

geduiwelh

Answered question

2021-09-10

Find the inverse Laplace transform of

F (s) = \frac{1}{{(s + a)}^{2} (s + b)} for t \geq 0

\sin (a t) + e^{- b t}

\frac{t e^{- b t}}{{(a - b)}^{2}} + \sin (a t)

\frac{\sin (a t)}{{(b - a)}^{2}} + \frac{e^{- b t}}{{(a - b)}^{2}}

\frac{e^{- b t}}{{(a - b)}^{2}} + \frac{t e^{- a t}}{b - a} - \frac{e^{- a t}}{{(b - a)}^{2}}

\sin (a t) + \frac{t e^{- b t}}{{(b - a)}^{2}} - \frac{t e^{- a t}}{{(a - b)}^{2}}

Answer & Explanation

Pohanginah

Skilled2021-09-11Added 96 answers

Step 1
Here,

F (s) = \frac{1}{{(s + a)}^{2} (s + b)}

step 2
Rewrite the function.

F (s) = \frac{1}{a - b} \times \frac{(s + a) - (s + b)}{{(s + a)}^{2} (s + b)}

= \frac{1}{a - b} \times [\frac{(s + a)}{{(s + a)}^{2} (s + b)} - \frac{(s + b)}{{(s + a)}^{2} (s + b)}]

= \frac{1}{a - b} \times [\frac{1}{(s + a) (s + b)} - \frac{1}{{(s + a)}^{2}}]

= \frac{1}{a - b} \times [\frac{1}{a - b} \times \frac{(s + a) - (s + b)}{(s + a) (s + b)} - \frac{1!}{{(s + a)}^{1 + 1}}]

= \frac{1}{{(a - b)}^{2}} [\frac{1}{s + b} - \frac{1}{s + a}] - \frac{1}{a - b} \times \frac{1!}{{(s + a)}^{1 + 1}}

step 3
Apply inverse laplace transform formula.

f (t) = \frac{1}{{(a - b)}^{2}} [e^{- b t} - e^{- a t}] - \frac{1}{a - b} \times t^{1} e^{- a t}

= \frac{e^{- b t}}{{(a - b)}^{2}} + \frac{t e^{- a t}}{b - a} - \frac{e^{- a t}}{{(b - a)}^{2}}

Step 4
Answer: Option D is correct.

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