Question

Laplace Transformation

Answer 1

\[L ^{-1} [\frac{ s }{ 2 +2s−3} ] \]

Answer 2

I think you last posted \( s^2 + 2s - 3 \) and that seems like a better question than \(2 + 2s - 3\).

But, does partial fraction decomposition sound familiar? We might want to factor the denominator and break it into smaller functions whose Laplace transforms we can find on a table.

Answer 3

yeah ma bad it is s^2 +2s-3

Answer 4

yeah that sounds right

Answer 5

it will give me \[L ^{-1} [ \frac{ s }{ (s+1)^{2}-4 } ]\]

Answer 6

is that right?

Answer 7

That would be completing the square. There is a way to make that work too, if you know hyperbolic trig functions though. That depends on what you know, though or which method you prefer. :)
I meant to take \( s^2 + 2s - 3 \) and convert it into a form: \( (s - a)(s - b) \).

Answer 8

A reminder, the method of partial fractions looks like this:
\( \displaystyle \begin{align}
\dfrac{s}{(s-a)(s-b)} &= \dfrac{A}{s- a} + \dfrac{B}{s - b} \\
s &= A(s - b) + B(s - a) \\
s &= As - Ab + Bs - Ba \\
s &= (A + B) s - Ab - Ba \\
\text{Equate coefficients} \ \ \ \ 1\color{Gray}s &= (A+B)\color{gray}s \qquad -Ab - Ba = 0
\end{align} \)
We would solve the resulting system as if it were a two variable linear system. Then by rewriting our argument as the sum of two terms, we take the inverse Laplace transform of each individually, which should give us two exponential transforms.

Answer 9

\[\frac{ 3 }{ 4(s+3) } +\frac{ 1 }{ s-1}\]

Answer 10

Looks very close, I got a factor of 1/4 on the latter term.
\( \displaystyle \frac{3}{4(s + 3)} + \frac{1}{4(s - 1)} \)

Answer 11

yes u r right, typo

Answer 12

Oh, no worries then.
But yea, we can then use the expanded form and the linearity of inverse Laplace to take:
\( \displaystyle{ \mathcal{L}^ {-1} \left[ \frac{s}{s^2 + 2s - 3} \right] = \mathcal{L}^{-1} \left[ \frac{3}{4(s+3)} + \frac{1}{4(s - 1)} \right] \\
= \frac{3}{4} \mathcal{L}^{-1} \left[ \frac{1}{s+3} \right] +\frac{1}{4} \mathcal{L}^{-1} \left[ \frac{1}{s-1} \right] } \)

Answer 13

that would boil down to \[\frac{ 3 }{ 4 } e ^{-3}+\frac{ 1 }{ 4 }e ^{t}\]

Answer 14

Yup, that looks good to me. :)

Answer 15

thx a lot

Answer 16

Glad to help!