How do I compute the following transform? (s−1)/(2s^2+s+6)

Jamarcus Lindsey

Jamarcus Lindsey

Answered question

2022-10-10

How do I compute the following transform?
s 1 2 s 2 + s + 6
I've gotten this far:
1 2 s 1 ( s + 1 4 ) 2 + 47 16

Answer & Explanation

Jeremy Mayo

Jeremy Mayo

Beginner2022-10-11Added 8 answers

You can now use:
L s ( e λ t e i t ω ) = 0 e s t e λ t e i t ω d t = 1 s + λ i ω
valid as long as s + λ > 0, and ω R . From here, taking real and imaginary parts you conclude:
L s ( e λ t cos ( ω t ) ) = s + λ ( s + λ ) 2 + ω 2 , L s ( e λ t sin ( ω t ) ) = ω ( s + λ ) 2 + ω 2
Now you the decomposition you obtained and read off the coefficients, keeping in mind that the inverse Laplace transform has the form e λ t ( α cos ( ω t ) + β sin ( ω t ) ) for some λ , ω , α and β
Parker Pitts

Parker Pitts

Beginner2022-10-12Added 1 answers

F ( s ) = s 1 2 s 2 + s + 6 = s 1 2 ( s 2 + s 2 + 3 )
a x 2 + b x = a [ ( x + b 2 a ) 2 ( b 2 a ) 2 ]
So:
2 ( s 2 + s 2 + 3 ) = 2 [ ( s + 1 4 ) 2 ( 1 4 ) 2 + 3 ] = 2 [ ( s + 1 4 ) 2 + ( 47 15 ) ]
therefore:
F ( s ) = ( 1 2 ) s 1 + 5 4 5 4 ( s + 1 4 ) 2 + ( 47 15 ) = ( 1 2 ) [ s + 1 4 ( s + 1 4 ) 2 + ( 47 15 ) 5 4 ( s + 1 4 ) 2 + ( 47 15 ) ]
We know:
L [ e a t c o s ( b t ) ] = s + a ( s + a ) 2 + b 2
L [ e a t s i n ( b t ) ] = b ( s + a ) 2 + b 2
therefore:
f ( t ) = ( 1 2 ) e t 4 [ c o s ( 47 4 t ) 5 47 s i n ( 47 4 t ) ]

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