Question

F(s)=7s^2+23s+30/(s-2)(s^2+2s+5)
Inverse laplace transform
just wondering if i should start it like....

Answer 1

complete the square for the
\[s^2+2s+5\]
in the denominator giving
\[(s+1)^2+4\]

Answer 2

but partial fraction decomposition would give me C/4 and don't know if that's acceptable...

Answer 3

@abb0t assistance please???

Answer 4

Wait, what?

Answer 5

just wondering if C/4 would be acceptable for partial fraction decomp...

Answer 6

\(\frac{c}{4}\) ?

Answer 7

Oh, yeah. It is. It's like \(\frac{1}{4}C\) like multiplied by constant.

Answer 8

oh ok cool.. thanks!

Answer 9

@oldrin.bataku if you get a chance and could help me that would be awesome. Im not even sure if im starting this one right

Answer 10

@abb0t could you help with this one? i don't think how i tried to start it is right...

Answer 11

and if it is then i just messed up the partial fractions

Answer 12

Hmm..ok. Well, you could use partial fraction decomp, and substitute (s+1)\(\rightarrow\)s correct?

Answer 13

s+1?

Answer 14

from completing the squre?

Answer 15

oh do i have to take derivative of something?

Answer 16

is latex showing up for you guys @abb0t @rperez36 ?

Answer 17

I think so. Yeah.

Answer 18

word ty

Answer 19

You actually could separate each function. Yes.

Answer 20

\[\frac{ A }{(s-2) }+\frac{ Bs+C }{ (s^2+2s+5) }\]
so u find the constants A, B and C then take inverse laplace of each fraction

Answer 21

You could use the derivative on the last one I believe. I forgot the theorems and corollaries..

Answer 22

ah ok @litchlani that should be it. i'll try it and see if i can derive the answer.

Answer 23

@litchlani i got close to the answer but not quite. Was definitely the way to approach it though!

Answer 24

sorry I wasn't home!
$$F(s)=\frac{7s^2+23s+30}{(s-2)(s^2+2s+5)}=\frac{A}{s-2}+\frac{Bs+C}{s^2+2s+5}\\A(s^2+2s+5)+(Bs+C)(s-2)=7s^2+23s+30\\(A+B)s^2+(2A-2B+C)s+(5A-2C)=7s^2+23s+30\\A+B=7\\2A-2B+C=23\\5A-2C=30$$
solving we use:$$A+B=7\\2A+2B=14\\2A-2B+C=23\\2A-2B+C+2A+2B=23+14\\4A+C=37\\8A+2C=74\\5A-2C=30\\13A=104\\A=8\\B=7-8=-1\\C=37-4(8)=5$$i.e. our decomposition yields:$$F(s)=\frac8{s-2}-\frac{s-5}{s^2+2s+5}$$note however that \(s^2+2s+5\) can be rewritten by completing the square \(s^2+2s+5=s^2+2s+1+4=(s+1)^2+2^2\). We can also rewrite \(s-5=s+1-6\):$$F(s)=\frac8{s-2}-\frac{s+1}{(s+1)^2+2^2}+3\frac2{(s+1)^2+2^2}$$

Answer 25

now, the inverse Laplace is straightforward in this form:$$f(t)=\mathcal{L}^{-1}\left\{\frac8{s-2}-\frac{s+1}{(s+1)^2+2^2}+3\cdot\frac2{(s+1)^2+2^2}\right\}=8e^{2t}-e^{-t}\cos2t+3e^{-t}\sin2t$$

Answer 26

\(f(t)=8e^{2t}-e^{-t}\cos2t+3e^{-t}\sin2t\)