Laplace transform with 2s(s^2+9) in denominator

[bmed90]

Homework Statement

Hi I would like to know how to expand 20/2s(s^2+9) in order to find the inverse Laplace Transform of the function K(s) to gt k(t). The (s^2+9) term in the denominator is throwing my calculations off for the constants because of the s^2 term.

Homework Equations

The Attempt at a Solution

I attempted to use partial fraction decomposition and expanded like so

c1/2 + c2/s + c3/(s^2+9)

from there I multiplied both sides by the common denominator, attempted to distribute and collect like terms to equate coefficients in order to find the proper system of equations to find the constants in order to take the inverse laplace of each term to find k(t) however I don't think I am expanding this problem right in the partial decomp and s^2+9 is not factorable any further. Please help!

 

[Chestermiller]

Replace c3 with c3s, and lose the first term.

Chet

 

[bmed90]

Replace c3 with c3s, and lose the first term.

Chet

Ok, so I expanded it like this

c1/2 + c2/s + (c3(s)+c4)/(s^2+9)

got my system of eqs got my constants

c1=0
c2=-20/7
c3=20/7
c4=0

so


-20/7*(1/s) --> L^-1[ -20/7*(1/s) ] = -20/7*Us(t) from table

but I am a little unsure how to find inverse laplace for the next term


L-1[ 20/7*(s) / (s^2+9) ] where 20/7 is c3 the const found c4 was 0 so its gone.

 

[bmed90]

Nevermind we got it. It's simply 20/7(s/s^2+3^2) which is 20/7cos(3t)...do you agree?

 

[Ray Vickson]

Homework Statement

Hi I would like to know how to expand 20/2s(s^2+9) in order to find the inverse Laplace Transform of the function K(s) to gt k(t). The (s^2+9) term in the denominator is throwing my calculations off for the constants because of the s^2 term.

Homework Equations

The Attempt at a Solution

I attempted to use partial fraction decomposition and expanded like so

c1/2 + c2/s + c3/(s^2+9)

from there I multiplied both sides by the common denominator, attempted to distribute and collect like terms to equate coefficients in order to find the proper system of equations to find the constants in order to take the inverse laplace of each term to find k(t) however I don't think I am expanding this problem right in the partial decomp and s^2+9 is not factorable any further. Please help!

You could also do it by integrating, using the well-known fact that if $g(s)$ is the LT of $f(t)$, then $g(s)/s$ is the LT of $\int_0^t f(u) \, du$. So, you can get the inverse transform of $1/(s^2+9)$, then integrate it.

 

[bmed90]

Thanks for all the help, let me ask you guys this,

If you had F(s)=1/((s)^2+16)^2 could you expand it like so to find your constants

[ (c1(s)+c2) / ((s)^2+16)] + [(c3(s)+c4) / ((s)^2+16)^2]

 

[Ray Vickson]

Thanks for all the help, let me ask you guys this,

If you had F(s)=1/((s)^2+16)^2 could you expand it like so to find your constants

[ (c1(s)+c2) / ((s)^2+16)] + [(c3(s)+c4) / ((s)^2+16)^2]

If you restrict yourself to real constants, the function 1/(s^2 + 16)^2 is already in 'partial fraction' form. The only way you can break it down further is to use complex numbers, writing s^2 + 16 = (s+4*i)*(s-4*i), where i = √(-1). You would then get:
\frac{1}{(s^2+16)^2} = \frac{i}{256}\frac{1}{s+4i} -\frac{i}{256}\frac{1}{s-4i}<br /> -\frac{1}{64(s-4i)^2}-\frac{1}{64(s+4i)^2}

 

[bmed90]

If you restrict yourself to real constants, the function 1/(s^2 + 16)^2 is already in 'partial fraction' form. The only way you can break it down further is to use complex numbers, writing s^2 + 16 = (s+4*i)*(s-4*i), where i = √(-1). You would then get:
\frac{1}{(s^2+16)^2} = \frac{i}{256}\frac{1}{s+4i} -\frac{i}{256}\frac{1}{s-4i}<br /> -\frac{1}{64(s-4i)^2}-\frac{1}{64(s+4i)^2}

Thanks for your reply Vickson, but to be honest I have no idea how to understand that as of right now...how would you recommend I tackle this problem to find the inverse laplace to get the function back to the time domain?

 

[bmed90]

Does anyone know how to find the constants If you had

F(s)= 1 / ((s)^2+16)^2 ? I need to find the inverse laplace transform of F(s)