MHB Simplifying the inverse Laplace Transform using the inverse shift formula

[Drain Brain]

before I go to bed(it's 11:30pm in my place), here is the last problem that I need help with

find the inverse Laplace Transform

$\frac{4s-2}{s^2-6s+18}$

the denominator is a non-factorable quadratic. I don't know what to do.

thanks!

 

[MarkFL]

Hint: Try completing the square on the denominator...:D

 

[Prove It]

before I go to bed(it's 11:30pm in my place), here is the last problem that I need help with

find the inverse Laplace Transform

$\frac{4s-2}{s^2-6s+18}$

the denominator is a non-factorable quadratic. I don't know what to do.

thanks!

Hmmm, 11:30pm... You wouldn't happen to live in Perth would you? :P

Mark is right, completing the square is the way to go. You will then need to apply a shift...

 

[evinda]

before I go to bed(it's 11:30pm in my place), here is the last problem that I need help with

find the inverse Laplace Transform

$\frac{4s-2}{s^2-6s+18}$

the denominator is a non-factorable quadratic. I don't know what to do.

thanks!

$$\frac{4s-2}{s^2-6s+18}=\frac{4s-2}{s^2-6s+9+9}=\frac{4s-2}{(s-3)^2+9}=4 \frac{s}{(s-3)^2+9}-2 \frac{1}{(s-3)^2+9} \\ =4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}$$

$$\mathscr{L}^{-1} \left ( 4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}
\right )=4 \mathscr{L}^{-1} \left( \frac{s-3}{(s-3)^2+9} \right )+\frac{10}{3} \mathscr{L}^{-1}\left (\frac{3}{(s-3)^2+9}
\right ) \\ =4 e^{3t} \cos(3t)+\frac{10}{3} e^{3t} \sin(3t)$$

 

[Prove It]

$$\frac{4s-2}{s^2-6s+18}=\frac{4s-2}{s^2-6s+9+9}=\frac{4s-2}{(s-3)^2+9}=4 \frac{s}{(s-3)^2+9}-2 \frac{1}{(s-3)^2+9} \\ =4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}$$

$$\mathscr{L}^{-1} \left ( 4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}
\right )=4 \mathscr{L}^{-1} \left( \frac{s-3}{(s-3)^2+9} \right )+\frac{10}{3} \mathscr{L}^{-1}\left (\frac{3}{(s-3)^2+9}
\right ) \\ =4 e^{3t} \cos(3t)+\frac{10}{3} e^{3t} \sin(3t)$$

Looks good :)

 

[MarkFL]

$$\frac{4s-2}{s^2-6s+18}=\frac{4s-2}{s^2-6s+9+9}=\frac{4s-2}{(s-3)^2+9}=4 \frac{s}{(s-3)^2+9}-2 \frac{1}{(s-3)^2+9} \\ =4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}$$

$$\mathscr{L}^{-1} \left ( 4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}
\right )=4 \mathscr{L}^{-1} \left( \frac{s-3}{(s-3)^2+9} \right )+\frac{10}{3} \mathscr{L}^{-1}\left (\frac{3}{(s-3)^2+9}
\right ) \\ =4 e^{3t} \cos(3t)+\frac{10}{3} e^{3t} \sin(3t)$$

Impeccable...but now the OP doesn't get to "play." :D

 

[Drain Brain]

$$\frac{4s-2}{s^2-6s+18}=\frac{4s-2}{s^2-6s+9+9}=\frac{4s-2}{(s-3)^2+9}=4 \frac{s}{(s-3)^2+9}-2 \frac{1}{(s-3)^2+9} \\ =4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}$$

$$\mathscr{L}^{-1} \left ( 4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}
\right )=4 \mathscr{L}^{-1} \left( \frac{s-3}{(s-3)^2+9} \right )+\frac{10}{3} \mathscr{L}^{-1}\left (\frac{3}{(s-3)^2+9}
\right ) \\ =4 e^{3t} \cos(3t)+\frac{10}{3} e^{3t} \sin(3t)$$

I was kind of happy when I saw the solution to this last problem thinking that I'll just copy it and wulah! my assignment is complete but then I noticed something and tried to solve the problem myself.

Instead of replacing $s$ with $s-3$ I replaced it with $s+3$ to have $s^2+3^2$

$e^{3t}\mathscr{L}^{-1}\left[4\frac{s+3}{s^2+3^2}-2\frac{1}{s^2+3^2}\right]$

$e^{3t} \mathscr{L}^{-1}\left[4\left(\frac{s}{s^2+3^2}+\frac{3}{s^2+3^2}\right)-2\frac{1}{s^2+3^2}\right]$

$e^{3t} \left[4\left(\cos(3t)+\frac{3}{3}\sin(3t)\right)-\frac{2}{3}\sin(3t)\right]$$e^{3t}\left[4\cos(3t)+4\sin(3t)-\frac{2}{3}\sin(3t)\right]$

$4e^{3t}\cos(3t)+4e^{3t}\sin(3t)-\frac{2}{3}e^{3t}\sin(3t)$ ---->>> this is my final answer please check :)

 

[evinda]

I was kind of happy when I saw the solution to this last problem thinking that I'll just copy it and wulah! my assignment is complete but then I noticed something and tried to solve the problem myself.

Instead of replacing $s$ with $s-3$ I replaced it with $s+3$ to have $s^2+3^2$

$e^{3t}\mathscr{L}^{-1}\left[4\frac{s+3}{s^2+3^2}-2\frac{1}{s^2+3^2}\right]$

$e^{3t} \mathscr{L}^{-1}\left[4\left(\frac{s}{s^2+3^2}+\frac{3}{s^2+3^2}\right)-2\frac{1}{s^2+3^2}\right]$

$e^{3t} \left[4\left(\cos(3t)+\frac{3}{3}\sin(3t)\right)-\frac{2}{3}\sin(3t)\right]$$e^{3t}\left[4\cos(3t)+4\sin(3t)-\frac{2}{3}\sin(3t)\right]$

$4e^{3t}\cos(3t)+4e^{3t}\sin(3t)-\frac{2}{3}e^{3t}\sin(3t)$ ---->>> this is my final answer please check :)

How did you get to this relation: $e^{3t}\mathscr{L}^{-1}\left[4\frac{s+3}{s^2+3^2}-2\frac{1}{s^2+3^2}\right]$ ?

 

[Drain Brain]

How did you get to this relation: $e^{3t}\mathscr{L}^{-1}\left[4\frac{s+3}{s^2+3^2}-2\frac{1}{s^2+3^2}\right]$ ?

using the Inverse shift formula $\mathscr{L}^{-1}[F(s)]=e^{\alpha t}\mathscr{L}^{-1}[F(s+\alpha)]$. I want to get rid of -3 in my denominator to simplify it to $\frac{4(s+3)}{(s+3-3)^2+3^2}=\frac{4(s+3)}{(s)^2+3^2}$ you can see that this matches with
$\frac{s}{s^2+\beta^{2}}=\cos(\beta t)$ and $\frac{\beta}{s^2+\beta^{2}}=\sin(\beta t) $

 

[evinda]

using the Inverse shift formula $\mathscr{L}^{-1}[F(s)]=e^{\alpha t}\mathscr{L}^{-1}[F(s+\alpha)]$. I want to get rid of -3 in my denominator to simplify it to $\frac{4(s+3)}{(s+3-3)^2+3^2}=\frac{4(s+3)}{(s)^2+3^2}$ you can see that this matches with
$\frac{s}{s^2+\beta^{2}}=\cos(\beta t)$ and $\frac{\beta}{s^2+\beta^{2}}=\sin(\beta t) $

I checked it...It is correct! (Yes)