How do you evaluate the inverse transform below using convolution ? ccL^(−1)[(s)/((s^2+a^2)^2)]

Zackary Diaz 2022-11-19 Answered
How do you evaluate the inverse transform below using convolution ? L 1 [ s ( s 2 + a 2 ) 2 ]
I tried
L 1 [ s ( s 2 + a 2 ) 2 ] ( t ) = 0 t sin t cos ( a t a τ ) d τ = sin t 0 t cos ( a t a τ ) d τ = sin t [ 1 a sin ( a t a τ ) ] 0 t = sin t [ 0 ( 1 a sin ( a t ) ) ] = 1 a sin 2 ( a t )
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Answers (2)

siriceboynu1
Answered 2022-11-20 Author has 12 answers
Instead of
L 1 ( s 2 ( s 2 + a 2 ) 2 ) ( t ) = 0 t sin t cos ( a t a τ ) d τ
you should have
L 1 ( s 2 ( s 2 + a 2 ) 2 ) ( t ) = 1 a 0 t sin ( a τ ) cos ( a t a τ ) d τ
because the convolution of functions f,g is defined as
( f g ) ( t ) = R g ( τ ) f ( t τ ) d τ
where f ( t ) := cos ( a t ) 1 [ 0 , ) ( t ), g ( t ) := 1 a sin ( a t ) 1 [ 0 , ) ( t )
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Keshawn Moran
Answered 2022-11-21 Author has 2 answers
Assume someone tells you to derivative 1 ( s 2 + a 2 ) respect to s wherein a is a constant. You certainly will reply
2 s ( s 2 + a 2 ) 2
( 1 ( s 2 + a 2 ) ) = 2 s ( s 2 + a 2 ) 2
or
( 1 2 ( s 2 + a 2 ) ) = s ( s 2 + a 2 ) 2
or
1 2 × ( 1 ) 1 ( 1 ( s 2 + a 2 ) ) = s ( s 2 + a 2 ) 2
But surely know that
L ( t 1 f ( t ) ) = ( 1 ) 1 F ( s )
Now can you find the proper f(t) form tow last equalities when you know that
L ( 1 a sin ( a t ) ) = 1 s 2 + a 2
This another approach besides to yours and first answer.
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