Find the inverse laplace transform of Y(s)=((82)/(s−6)−2s+2)/(s^2+6s+10)

Liam Keller

Liam Keller

Answered question

2022-09-07

Find the inverse laplace transform of   Y ( s ) = 82 s 6 2 s + 2 s 2 + 6 s + 10  
Let   L 1   be the inverse laplace operator.
Then,
y ( t ) = L 1 [ Y ( s ) ; t ] y ( t ) = L 1 [ 82 s 6 2 s + 2 s 2 + 6 s + 10 ]   = L 1 [ 2 s 2 + 12 s + 70 ( s 6 ) ( s 2 + 6 s + 10 ) ]
Now,
2 s 2 + 12 s + 70 ( s 6 ) ( s 2 + 6 s + 10 ) = A s 6 + B s + C s 2 + 6 s + 10   where A,B,C are unknown constants to be determined.

Answer & Explanation

Zayden Dorsey

Zayden Dorsey

Beginner2022-09-08Added 18 answers

You are on the right track, after you work out the constants A, B and C you should end up with
Y ( s ) = 1 s 6 3 s + 10 s 2 + 6 s + 10 = 1 s 6 3 ( s + 3 ) + 1 ( s + 3 ) 2 + 1 = 1 s 6 3 ( s + 3 ) ( s + 3 ) 2 + 1 1 ( s + 3 ) 2 + 1
Now use the fact that
L [ e a t sin b t ] = b ( s a ) 2 + b 2
and
L [ e a t cos b t ] = s a ( s a ) 2 + b 2
gemuntertjx

gemuntertjx

Beginner2022-09-09Added 1 answers

So far so good.
2 s 2 + 12 s + 70 ( s 6 ) ( s 2 + 6 s + 10 ) = A s 6 + B s + C s 2 + 6 s + 10
Use Heavy-side method with s=6 to find A = 70 82
Subtract 70 82 ( s 6 ) from both sides to find B and C.
Complete the square in s 2 + 6 s + 10 and use shifting formulas to finish the work.

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