Laplace transform of modulus derivative I'm looking at taking a Laplace transform of a control signal to get the system transfer function. The equation of motion of this system is:
$u(t) = m\ddot{y}(t)+c|\dot{y}(t)|\dot{y}(t)+ky(t) $
where $u$ is the input signal and $y$ is the state variable.
To find $u(s)$ I am unsure how to deal with the $|\dot{y}|\dot{y}$ term. My notes state that 'if the velocity is approximately constant' then:
$u(s) = ms^2y(s)+c|\dot{y}|sy(s) +ky(s) $
Which looks like $|\dot{y}(t)|$ has been treated as a constant, but $\dot{y}(t)$ hasn't. Could anyone explain how this transform has been done?
Many thanks!
 A: This looks like only a hand-wavy approximation. Namely, I would use a first order Taylor series approximation of the nonlinearity, which in this case with $\dot{y}(t)$ near $\bar{\dot{y}}$ would yield
\begin{align}
\left|\dot{y}(t)\right|\,\dot{y}(t) &\approx \left|\bar{\dot{y}}\right|\,\bar{\dot{y}} + \left.\frac{d\,|x|\,x}{dx}\right|_{x=\bar{\dot{y}}}\left(\dot{y}(t) - \bar{\dot{y}}\right), \\
&\approx 2 \left|\bar{\dot{y}}\right| \dot{y}(t) - \left|\bar{\dot{y}}\right|\,\bar{\dot{y}}.
\end{align}
Taking the Laplace transform of this Taylor series approximation would therefore yield
$$
\mathcal{L}\left\{2 \left|\bar{\dot{y}}\right| \dot{y}(t) - \left|\bar{\dot{y}}\right|\,\bar{\dot{y}}\right\}(s) = 2 \left|\bar{\dot{y}}\right| \mathcal{L}\left\{\dot{y}(t)\right\}(s) - \frac{\left|\bar{\dot{y}}\right|\,\bar{\dot{y}}}{s},
$$
with
$$
\mathcal{L}\left\{\dot{y}(t)\right\}(s) = s\,\mathcal{L}\left\{y(t)\right\}(s) - y(0),
$$
but the initial condition is often ignored or assumed to be zero.
If one also ignores the term $\left|\bar{\dot{y}}\right|\,\bar{\dot{y}}/s$ then we do have nearly the same expression as in your question besides the additional factor 2.
