Find the inverse Laplace transform of F(s)=\frac{1}{(s+a)^2(s+b)} for t \geq 0

geduiwelh

geduiwelh

Answered question

2021-09-10

Find the inverse Laplace transform of F(s)=1(s+a)2(s+b)  for  t0
a) sin(at)+ebt
b) tebt(ab)2+sin(at)
c) sin(at)(ba)2+ebt(ab)2
d) ebt(ab)2+teatbaeat(ba)2
e)sin(at)+tebt(ba)2teat(ab)2

Answer & Explanation

Pohanginah

Pohanginah

Skilled2021-09-11Added 96 answers

Step 1
Here,
F(s)=1(s+a)2(s+b)
step 2
Rewrite the function.
F(s)=1ab×(s+a)(s+b)(s+a)2(s+b)
=1ab×[(s+a)(s+a)2(s+b)(s+b)(s+a)2(s+b)]
=1ab×[1(s+a)(s+b)1(s+a)2]
=1ab×[1ab×(s+a)(s+b)(s+a)(s+b)1!(s+a)1+1]
=1(ab)2[1s+b1s+a]1ab×1!(s+a)1+1
step 3
Apply inverse laplace transform formula.
f(t)=1(ab)2[ebteat]1ab×t1eat
=ebt(ab)2+teatbaeat(ba)2
Step 4
Answer: Option D is correct.

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