Laplace Transform: Finding Q(t) - Confused!

[justinis123]

Homework Statement

I already got Q(s)=150/(s(s^2+20s+200)), then i complete the square on the quadratic.
I got Q(s)=150/(s((s+10)^2+10^2))). But then i can't find the Q(t) because the equation (s+10)^2+10^2=0 dosent have roots. Or i have to use complex numbers ? So I am confused.

Homework Equations

The Attempt at a Solution

 

[vela]

Did you use partial fractions to get separate terms first?

To get the inverse, look at the Laplace transforms of sine and cosine.

 

[justinis123]

Did you use partial fractions to get separate terms first?

To get the inverse, look at the Laplace transforms of sine and cosine.

U mean use partial fractions on s(s^2+20s+200)? but how?

 

[justinis123]

I can get 150/s * 1/((s+10)^2+10^2), but this dosent seems to fit either sin or cos

 

[vela]

You need to use partial fractions to separate it into terms that appear in the tables.

\frac{150}{s(s^2+20s+200)} = \frac{A}{s} + \frac{Bs+C}{s^2+20s+200}

Solve for A, B, and C.

 

[justinis123]

You need to use partial fractions to separate it into terms that appear in the tables.

\frac{150}{s(s^2+20s+200)} = \frac{A}{s} + \frac{Bs+C}{s^2+20s+200}

Solve for A, B, and C.

thanks for reply, after i solve partial fraction what should i do?
I solved: which = 3/40s + ((-3s/40-3/2))/(s^2+20s+200)

 

[vela]

You seem to have made some errors solving for the coefficients. You should go back and check your calculations. It looks like you have an extra factor of 10 somewhere.

After you get the coefficients, you can rearrange stuff slightly to get it into the form

\frac{150}{s(s^2+20s+200)} = \frac{A}{s} + \frac{B'(s+10)}{(s+10)^2+10^2} + \frac{C'}{(s+10)^2+10^2}

Then use the Laplace transform tables to convert from the s-domain to the time domain. Note that the second and third terms look like the Laplace transforms for sin 10t and cos 10t except with s replaced by s+10. You should be able to find a property of the Laplace transform (it's probably in a table somewhere) that tells you what this means.