Question

F(s)=5s^2+34s+53/(s+3)^2(s+1)
determine inverse laplace transform
will format in question!

Answer 1

\[F(s)=5s^2+34s+53/(s+3)^2(s+1)\]

Answer 2

I did...
\[A/(s+3)+Bs+C/(s+3)^2+D/(s+1)\]

Answer 3

Just spread out 53/(s+3)2(s+1) and then use the inverse rules to do the rest. This one is pretty straight forward.

Answer 4

but then got stuck at
\[A(s+3)^2(s+1)+Bs+C(s+3)(s+1)+D(s+3)^2(s+3)\]

Answer 5

am I doing this right?

Answer 6

if so I just need to learn binomial expansion cause im pretty sure that's all this step is

Answer 7

@oldrin.bataku any advice?

Answer 8

just a simple yes im on the right track will suffice then ill watch youtube videos on binomial expansion to try and continue

Answer 9

you multiplied to get rid of the denominators incorrectly :-/

Answer 10

also why the \(Bs\) term?

Answer 11

You would only need that Bx + C numerator if the factor itself is quadratic like (s^2 +3). The squared term on the outside doesn't make it Bs + C, it'd still just be B for that factor.

Answer 12

oh ok thank you @Psymon I will try it again

Answer 13

Are you using partial fraction decomposition?

Answer 14

@abb0t ya

Answer 15

I don't think you need \(\frac{Bs+C}{(s-3)^2}\) I think. Can't you just use \(\frac{B}{(s-3)^2}\)?

Answer 16

Yeah, it should just be B in the numerator.

Answer 17

Oh nvm. I just read that. My bad. Too dam early for ODE.

Answer 18

Agreed, haha.

Answer 19

the reason that works is because we just split into \((s+3),(s+3)^2\) terms individually :-p observe:$$\frac{A}{s+3}+\frac{B}{(s+3)^2}=\frac{A(s+3)}{(s+3)^2}+\frac{B}{(s+3)^2}=\frac{As+(3A+B)}{(s+3)^2}$$

Answer 20

ok I got A=19/6 B=34/3 and C=-19/6 how do I determine what is what from
\[L^{-1}[(19/6)/(S+3)+(34/3)/(S+3)^2-(19/6)/(S+1)]\]
the solution is
\[-e^{-3t}+2te^{-3t}+6e^{-t}\]
so I think my a,b,and c are right but not sure....
@oldrin.bataku

Answer 21

$$F(s)=\frac{5s^2+34s+53}{(s+3)^2(s+1)}=\frac{A}{s+3}+\frac{B}{(s+3)^2}+\frac{C}{s+1}\\A(s+3)(s+1)+B(s+1)+C(s+3)^2=5s^2+34s+53$$Let \(s=-1\) and our equation reduces to:$$C(-1+3)^2=5(-1)^2+34(-1)+53\\4C=5-34+53\\4C=24\\C=6$$Let \(s=-3\) and we get:$$B(-3+1)=5(-3)^2+34(-3)+53\\-2B=45-102+53\\2B=4\\B=2$$To find our last coefficient consider:$$B(s+1)+C(s+3)^2=2(s+1)+6(s+3)^2=2s+2+6(s^2+6s+9)=6s^2+38s+56$$and also:$$(s+3)(s+1)=s^2+4s+3$$hence it makes intuitive sense that \(A=-1\) yielding in conclusion:$$F(s)=-\frac1{s+3}+\frac2{(s+3)^2}+\frac6{s+1}$$

Answer 22

now, we take the inverse transform to return to time domain:$$f(t)=\mathcal{L}^{-1}\left\{-\frac1{s+3}+\frac2{(s+3)^2}+\frac6{s+1}\right\}=-e^{-3t}+2te^{-3t}+6e^{-t}$$

Answer 23

@oldrin.bataku thank you so much im sure that this will help me finish this sections homework. just wondering what your major is and where you go to school???

Answer 24

in high school but I hope to study engineering and math

Answer 25

@SithsAndGiggles can you explain to me how he got A cause apparently it isn't coming to me intuitively :(

Answer 26

I get that (s+3)(s+1) comes from A and the other comes from b and c with the numbers we found....but that's it

Answer 27

i tried adding them together and set equal to the numerator but that didn't help me

Answer 28

\[A(s+3)(s+1)+B(s+1)+C(s+3)^2=5s^2+34s+53\]
I'll solve for the constants using a different method (I usually make mistakes when I use oldrin's method)

First thing to do would be to expand all the binomial expressions:
\[As^2+4As+3A+Bs+B+Cs^2+6Cs+9C=5s^2+34s+53\]
Group the terms with common powers of \(s\):
\[(A+C)s^2+(4A+B+6C)s+(3A+B+9C)=5s^2+34s+53\]
Match up terms containing same powers of \(s\) on the other side of the equation. Doing so yields the system
\[\begin{cases}A+C=5\\4A+B+6C=34\\3A+B+9C=53\end{cases}\]
Solving yields \(A=-1,~B=2,~C=6\). I'll leave the details of solving the system to you... Anyway, I get the same results as oldrin above.

Answer 29

thanks for helping!

Answer 30

I just considered that since I had \(6s^2\) without the \(A\) term and wanted \(5s^2\) that we'd need \(A=-1\) to since \(6+-1=5\)