Editing Laplace transform (section)

=== Phase delay ===
{| class="wikitable"
|-
 ! scope="col" | Time function
 ! scope="col" | Laplace transform
|-
 | <math>\sin{(\omega t + \varphi)}</math>
 | <math>\frac{s\sin(\varphi) + \omega \cos(\varphi)}{s^2 + \omega^2}</math>
|-
 | <math>\cos{(\omega t + \varphi)}</math>
 | <math>\frac{s\cos(\varphi) - \omega \sin(\varphi)}{s^2 + \omega^2}.</math>
|}

Starting with the Laplace transform,
<math display=block>X(s) = \frac{s\sin(\varphi) + \omega \cos(\varphi)}{s^2 + \omega^2}</math>
we find the inverse by first rearranging terms in the fraction:
<math display=block>\begin{align}
  X(s) &= \frac{s \sin(\varphi)}{s^2 + \omega^2} + \frac{\omega \cos(\varphi)}{s^2 + \omega^2} \\
       &= \sin(\varphi) \left(\frac{s}{s^2 + \omega^2} \right) + \cos(\varphi) \left(\frac{\omega}{s^2 + \omega^2} \right).
\end{align}</math>

We are now able to take the inverse Laplace transform of our terms:
<math display=block>\begin{align}
  x(t) &= \sin(\varphi) \mathcal{L}^{-1}\left\{\frac{s}{s^2 + \omega^2} \right\} + \cos(\varphi) \mathcal{L}^{-1}\left\{\frac{\omega}{s^2 + \omega^2} \right\} \\
       &= \sin(\varphi)\cos(\omega t) + \cos(\varphi)\sin(\omega t).
\end{align}</math>

This is just the [[Trigonometric identity#Angle sum and difference identities|sine of the sum]] of the arguments, yielding:
<math display=block>x(t) = \sin (\omega t + \varphi).</math>

We can apply similar logic to find that
<math display=block>\mathcal{L}^{-1} \left\{ \frac{s\cos\varphi - \omega \sin\varphi}{s^2 + \omega^2} \right\} = \cos{(\omega t + \varphi)}.</math>