Inverse Laplace Transform

[drolex]

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?

 

[jeg223]

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

 

[erosennin]

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.

 

[jeg223]

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

 

[eng1]

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

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)

 

[jeg223]

I never could remember those trig identities

 

[drolex]

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

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

 

[jeg223]

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.

 

[RWD7980]

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.

 

[drolex]

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.

 

[MrAl]

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

is the right answer.

 

[drolex]

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