Transform: F(s)=(2s^2+(a−6b)s+a^2−4ab)/((s^2−a^2)(s−2b))

Alberanteb4T

Alberanteb4T

Answered question

2022-11-24

inverse laplace transform - with symbolic variables:
F ( s ) = 2 s 2 + ( a 6 b ) s + a 2 4 a b ( s 2 a 2 ) ( s 2 b )
My steps:
F ( s ) = 2 s 2 + ( a 6 b ) s + a 2 4 a b ( s + a ) ( s a ) ( s 2 b )
= A s + a + B s a + C s 2 b + K
K = 0
A = F ( s ) ( s + a )

Answer & Explanation

inventorspotDkt

inventorspotDkt

Beginner2022-11-25Added 5 answers

The routine way is to use the way we get in partial fractions and you correctly noted that. If we set:
2 s 2 + ( a 6 b ) s + a 2 4 a b ( s 2 a 2 ) ( s 2 b ) = F = A s + a + B s a + C s 2 b
then by doing handy calculations we can find A,B and C. We have then:
A s a + 2 A a b + A s 2 2 A s b + B s a 2 B a b + B s 2 2 B s b + C s 2 C a 2 ( s + 2 b ) ( s 2 + a 2 )
Now if we put s=−a in the numerators, we get A = a + b a + 2 b and this is what you already got. With the similar approach s = + a for B and s = 2 b for C, we get:
B = 2 a 5 b a 2 b ,     C = 4 b 2 + 2 a b a 2 a 2 4 b 2

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