MHB Solving Laplace Transform: \[(s+1)^2/(s^2-s+1)\]

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The discussion focuses on solving the Laplace transform of the function \((s+1)^2/(s^2-s+1)\). The transformation simplifies to a combination of exponential and trigonometric functions, specifically involving terms like \(e^{1/2t}\cos(t\frac{\sqrt{3}}{2})\) and \(e^{1/2t}\sin(t\frac{\sqrt{3}}{2})\). The final expression for the inverse transform includes a delta function and additional terms derived from the simplification process. Participants highlight the steps needed to reach the final form of the inverse transform, emphasizing the importance of breaking down the function into manageable parts. The thread concludes with a clear derivation of the time-domain function \(h(t)\).
Dustinsfl
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\[
\frac{(s+1)^2}{s^2 - s + 1}
\]
I have simplified it down to
\[
\frac{s - \frac{1}{2} + s^2 + s + \frac{3}{2}}{(s - 1/2)^2 + \frac{3}{4}} =
e^{1/2t}\cos\Big(t\frac{\sqrt{3}}{2}\Big) + \sqrt{3}e^{1/2t}\sin\Big(t\frac{\sqrt{3}}{2}\Big) + \frac{s^2 + s}{(s - 1/2)^2 + \frac{3}{4}}
\]
but I can't figure out the last transform.
 
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dwsmith said:
\[
\frac{(s+1)^2}{s^2 - s + 1}
\]
I have simplified it down to
\[
\frac{s - \frac{1}{2} + s^2 + s + \frac{3}{2}}{(s - 1/2)^2 + \frac{3}{4}} =
e^{1/2t}\cos\Big(t\frac{\sqrt{3}}{2}\Big) + \sqrt{3}e^{1/2t}\sin\Big(t\frac{\sqrt{3}}{2}\Big) + \frac{s^2 + s}{(s - 1/2)^2 + \frac{3}{4}}
\]
but I can't figure out the last transform.

With symple steps You obtain...

$\displaystyle H(s) = \frac{(s+1)^{2}}{s^{2} - s + 1} = 1 + \frac{3\ s}{s^{2} - s + 1} = 1 + 3\ \frac{s - \frac{1}{2}}{(s - \frac{1}{2})^{2} + \frac{3}{4}} + 3\ \frac{\frac{1}{2}}{(s - \frac{1}{2})^{2} + \frac{3}{4}}\ (1) $

... and from (1) You derive...

$\displaystyle h(t) = \delta (t) + 3\ e^{\frac{t}{2}}\ \cos \frac{\sqrt{3}}{2}\ t + \sqrt{3}\ e^{\frac{t}{2}}\ \cos \frac{\sqrt{3}}{2}\ t\ (2)$

Kind regards

$\chi$ $\sigma$
 

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