Finding the Inverse Laplace Transform of a Rational Function

[Jamin2112]

Homework Statement

...find the inverse transform of the given function.

F(s)= (2s+2)/(s²+4s+5)

Homework Equations

On another page, I see these 2:

f(t)=e^atsin(bt)--------->F(s)=b/[(s-a)²+b²)
f(t)=e^atcos(bt)--------->F(s)=(s-a)/[(s-a)²+b²)

The Attempt at a Solution

This is my first time doing anything with Laplace transforms.

I can change the denominator to (s+2)² + 1 and factor a -2 out of the numerator. Then it looks like the relevant equations.

-2((s-1/2)/((s+2)²+1))

But I don't understand where to go from here. The -2, of course, is just a constant that can chill out front. Judging from the answer in the back of the book, I use a combination of both relevant equations. Explain the algebra involved here.

 

[Dick]

Ok, so a=(-2) and b=1, right? The numerator of the first expression is 1 and the second is s+2. Looks like you want to solve c*1+d*(s+2)=(2s+2) for c and d. Then multiply your first f(t) by c and the second f(t) by d.

 

[Jamin2112]

Ok, so a=(-2) and b=1, right? The numerator of the first expression is 1 and the second is s+2. Looks like you want to solve c*1+d*(s+2)=(2s+2) for c and d. Then multiply your first f(t) by c and the second f(t) by d.

Let's slow down for a second.

I have (s-1/2)/((s+2)²+1).

That isn't in the form b/((s-a)²+b²) because the b's are inconsistent. If I make a=-2, and b=1, the denominator works out, but then the numerator doesn't.

It also isn't in the form (s-a)/((s-a)²+b²) because the a's are inconsistent. If I do the same, the numerator doesn't work out.

Explain this to me as if I was a 5-year-old.

http://blog.oregonlive.com/petoftheday/2008/04/large_thanks.bmp

 

[Dick]

Ok, five year old. You got the a and b figured out. Multiply each of your t expressions by a separate constant. So the laplace transform of c*e^(-2t)*sin(t) is c*1/((s+2)^2+1) and the laplace transform of d*e^(-2t)*cos(t) is d*(s+2)/((s+2)^2+1). If you add those you get that the laplace transform of e^(-2t)*(c*sin(t)+d*cos(t)) is (c*1+d*(s+2))/((s+2)^1+1). Now you just have to figure out c and d so that (c*1+d*(s+2))=2*s+2.

 

[Jamin2112]

Ok, five year old. You got the a and b figured out. Multiply each of your t expressions by a separate constant. So the laplace transform of c*e^(-2t)*sin(t) is c*1/((s+2)^2+1) and the laplace transform of d*e^(-2t)*cos(t) is d*(s+2)/((s+2)^2+1). If you add those you get that the laplace transform of e^(-2t)*(c*sin(t)+d*cos(t)) is (c*1+d*(s+2))/((s+2)^1+1). Now you just have to figure out c and d so that (c*1+d*(s+2))=2*s+2.

Got it!

Thanks, bro!