Inverse Laplace Transform

I need to inverse Laplace transform this:
2s/((s^2+4)(s+2))

I know the answer is cos(t)^2+sin(t)*cos(t)-1/2-1/2*exp(-2*t).
I can inverse Laplace transform stuff, but I'm a little unsure with this specific type of problem. Can you explain how to do it, please?

I think your answer is wrong. I get:
f(t)=.5(sin(2t)+cos(2t))-.5e^(-2t)

Start with partial fraction expansion for complex roots:
F(s)=(As+B)/(s^2+4) + C/(s+2)
2s=(A+C)s^2 + (2A+B)s + 2B+4C

set coefficients equal
0=A+C
2=2A+B
0=2B+4C

I get A=.5, B=1, C=-.5

The -.5/(s+2) section is straightforward

Manipulate the complex section to get in a recognizable form
(Mw+N(s+a))/((s+a)^2+w^2)=(.5s+1)/(s^2+4)
where w=2 and a=0 as determined by the roots of (s^2+4)
solve for the amplitudes M and N by setting coefficients equal
2M+Ns=.5s+1
M=.5, N=.5

F(s)= (.5*2+.5(s+0))/((s+0)^2+2^2)-.5/(s+2)
Use tables for ILT
f(t)=.5(sin(2t)+cos(2t))-.5e^-2t

I skipped some of the details, but these links should fill them in:
**broken link removed**
https://mathworld.wolfram.com/LaplaceTransform.html

Well personally I have no idea about the whole transform part, but have you tried splitting the fraction, "partial fractions" style, in order to make your life easier?

2s/[(s^2+4)(s+2)]


A/(s+2) + (Bs+C)/(s^2+4) = 2s/[(s^2+4)(s+2)]

A(s^2+4) + (Bs+C)(s+2) = 2s
let s = -2
8A = -4
A = -1/2

-1/2*(s^2+4) + (Bs+C)(s+2) = 2s
let s=0
-2+2C=0
C = 1

-1/2*(s^2+4) + (Bs+1)(s+2) = 2s
let s=1
-5/2+3(B+1)=2
(B+1)=3/2
B=1/2

Sooo, after all that;
2s/[(s^2+4)(s+2)] = -1/[2(s+2)] + (1/2s+1)/(s^2+4)
Tidy it up a bit
=(s+2)/(2s^2+8) - 1/(2s+4)

I hope that helps you cause, sorry I am TeX n00b, i'll make sure I learn it at some stage. As I said, I'll leave you to the transform part...

EDIT: jeg beat me to it typing it out, but he atleast finished it for you... lol, also learned something new from him... excellent lol.

At least we got the same answers for the coefficients. Thats a good sign

jeg223 said:

I think your answer is wrong. I get:
f(t)=.5(sin(2t)+cos(2t))-.5e^(-2t)

Start with partial fraction expansion for complex roots:
F(s)=(As+B)/(s^2+4) + C/(s+2)
2s=(A+C)s^2 + (2A+B)s + 2B+4C

set coefficients equal
0=A+C
2=2A+B
0=2B+4C

I get A=.5, B=1, C=-.5

Click to expand...

Perfect solution I just want to point out that it's exactly what the OP expected.

sin(2x) = 2sin(x)cos(x)
cos(2x) = 2cos²(x)-1

Hence, f(t) = .5(sin(2t)+cos(2t))-.5e^(-2t) = cos(t)^2+sin(t)*cos(t)-1/2-1/2*exp(-2*t)

I never could remember those trig identities

jeg223 said:

Manipulate the complex section to get in a recognizable form
(Mw+N(s+a))/((s+a)^2+w^2)=(.5s+1)/(s^2+4)
where w=2 and a=0 as determined by the roots of (s^2+4)
solve for the amplitudes M and N by setting coefficients equal
2M+Ns=.5s+1
M=.5, N=.5

Click to expand...

Thanks!
Where is the table that you found that transform on?

It's actually the sum of the sin(wt) and cos(wt) transforms with a scalar multiplier and a frequency shift.
The mathworld link above has a table, but if you look you should be able to find a more extensive table with the (Mw+N(s+a))/((s+a)^2+w^2)=e^-(at)(Msin(wt)+Ncos(wt)) transform.

I am glad you guys are still transforming Laplace...... If I remember my IT89 will do all of this ........will almost all of it... getting the equation into the correct recognizable form is the last hard step........... yes I am happy now! I should practice.

Oh, we don't do them anymore. That was three years ago. Ha ha just kidding... Actually, I haven't done one in a while and I should probably refresh my skills.

(sin(2t)+cos(2t)-e^(-2t))/2

is the right answer.

That's correct and it's equivalent to all the answers that have already been posted.