Computing Inverse Laplace Transforms for F(s) = (-2s-2)/((s-1)^2+9)

[jumbogala]

Homework Statement

F(s)=(−2s−2)/((s−1)²+9)

Find f(x) by computing L^-1[F(s)] = f(x).

Homework Equations

The Attempt at a Solution

I made put (s-1) in place of s in the numerator, then compensated for that by subtracting 2:
F(s) = (-2(s-1) - 2 -2 ) / ((s-1)² + 9)

Then s-1 becomes s...

F(s) = (-2s - 4) / (s² + 3²)

Split this into 2 parts:

-2s / s² + 3² and (-4/3)(3 / s² + 3²)

This gives -2cos(3x) - (4/3)sin(3x) as f(x). But it's wrong... what should I have done differently?

 

[rock.freak667]

$$\frac{-2s-2}{(s-1)^2+9} =-2 \frac{s+1}{(s-1)^2+9}$$


Now use a similar principle of (a+b)/c = a/c + b/c


(shift formula will help you here)

 

[jumbogala]

Okay so I split it up:

-2 [ s / ((s-1)² + 9) + 1 / (s-1)² + 9)

I'll work with s / (s-1)² + 9 first. Rewrite the denominator as s - 1 + 1. Then split it up into two, so I have:

s-1 / (s-1)² + 9 and 1 / (s-1)² + 9)

Now working with just the first part of the above. If the shift formula was used then a = 1, and the other part was cos(3t). So this part of f(t) must have been e^tcos(3t).

Now for 1 / (s-1)² + 9. if I multiply the whole thing by 1/3 then I can put a 3 in the denominator, so this part of f(t) is 1/3sin(3t). Since I have two of those (one from the second split) I get 2/3sin(3t).

Now I multiply everything by -2.

Final answer: -2e^tcos(3t) -4/3sin(3t).

Is that correct?

 

[jumbogala]

I think there's still a problem with the sin(3t). Because I have 1/ (s-1)^2 +9, not just s in the denominator. Did I need to use the shift formula there too? I'm not sure how as I have no s in the numerator to work with!

 

[rock.freak667]

I think there's still a problem with the sin(3t). Because I have 1/ (s-1)^2 +9, not just s in the denominator. Did I need to use the shift formula there too? I'm not sure how as I have no s in the numerator to work with!

yes you have to shift that as well

$$L[sinkt]=\frac{k}{s^2+k^2}$$

 

[jumbogala]

How do you shift it though? I can't put 1-s+s in the numerator, that doesn't help...

 

[rock.freak667]

How do you shift it though? I can't put 1-s+s in the numerator, that doesn't help...

you don't need to, the second term (befor multiplying by -2) is


$$\frac{2}{(s-1)^2+9} = \frac{2}{3}\frac{3}{(s-1)^2+(3^2)}$$


So you just shift it the same way with a=1

 

[jumbogala]

So -2[e^tcos(3t) - 0.67e^tsin(3t)]?

 

[rock.freak667]

So -2[e^tcos(3t) - 0.67e^tsin(3t)]?

-2[e^tcos(3t) + 0.67e^tsin(3t)]

 

[jumbogala]

Oops yeah, silly mistake. I think I finally understand, yay! Thanks very much for your help!