Drain Brain said:
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!
Drain Brain said:
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!
evinda said:$$\frac{4s-2}{s^2-6s+18}=\frac{4s-2}{s^2-6s+9+9}=\frac{4s-2}{(s-3)^2+9}=4 \frac{s}{(s-3)^2+9}-2 \frac{1}{(s-3)^2+9} \\ =4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}$$
$$\mathscr{L}^{-1} \left ( 4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}
\right )=4 \mathscr{L}^{-1} \left( \frac{s-3}{(s-3)^2+9} \right )+\frac{10}{3} \mathscr{L}^{-1}\left (\frac{3}{(s-3)^2+9}
\right ) \\ =4 e^{3t} \cos(3t)+\frac{10}{3} e^{3t} \sin(3t)$$
evinda said:$$\frac{4s-2}{s^2-6s+18}=\frac{4s-2}{s^2-6s+9+9}=\frac{4s-2}{(s-3)^2+9}=4 \frac{s}{(s-3)^2+9}-2 \frac{1}{(s-3)^2+9} \\ =4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}$$
$$\mathscr{L}^{-1} \left ( 4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}
\right )=4 \mathscr{L}^{-1} \left( \frac{s-3}{(s-3)^2+9} \right )+\frac{10}{3} \mathscr{L}^{-1}\left (\frac{3}{(s-3)^2+9}
\right ) \\ =4 e^{3t} \cos(3t)+\frac{10}{3} e^{3t} \sin(3t)$$
evinda said:$$\frac{4s-2}{s^2-6s+18}=\frac{4s-2}{s^2-6s+9+9}=\frac{4s-2}{(s-3)^2+9}=4 \frac{s}{(s-3)^2+9}-2 \frac{1}{(s-3)^2+9} \\ =4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}$$
$$\mathscr{L}^{-1} \left ( 4 \frac{s-3}{(s-3)^2+9}+10 \frac{1}{(s-3)^2+9}
\right )=4 \mathscr{L}^{-1} \left( \frac{s-3}{(s-3)^2+9} \right )+\frac{10}{3} \mathscr{L}^{-1}\left (\frac{3}{(s-3)^2+9}
\right ) \\ =4 e^{3t} \cos(3t)+\frac{10}{3} e^{3t} \sin(3t)$$
Drain Brain said:
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 said: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 said:
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) $
The inverse Laplace transform is a mathematical operation that converts a function in the Laplace domain back to its original form in the time domain. It is the reverse process of the Laplace transform and is used to solve differential equations and analyze dynamic systems.
The inverse Laplace transform is calculated using a table of Laplace transforms or through the use of partial fraction decomposition and the residue theorem. The specific method used depends on the complexity of the function in the Laplace domain.
The Laplace transform and the inverse Laplace transform are inverse operations of each other. The Laplace transform converts a function from the time domain to the Laplace domain, while the inverse Laplace transform converts it back to the time domain.
The inverse Laplace transform is used in various fields such as engineering, physics, and mathematics. It is particularly useful in solving differential equations and analyzing dynamic systems, making it a valuable tool in modeling and control systems.
Yes, there are limitations to using the inverse Laplace transform. It may not be possible to calculate the inverse Laplace transform for some functions, and the calculation can be complex for highly intricate functions. Additionally, the inverse Laplace transform assumes that the function being transformed is of exponential order, which may not always be the case.