I'm using Nise for my control systems class. Finding a linearised system is all gravy baby, but when it comes to finding the transfer function Nise does some stuff which confounds me:
See page 6/7, exercise 2.13 here: http://higheredbcs.wiley.com/legacy/college/nise/0470547561/skills/sol_skills_assessment.pdf
$\delta_v$ is his notation for a small movement about an equilibrium point, ie. $v = v_0 + \delta_v$.
So everything's fine until the last line, where he just swaps out $\delta_v(s)$ for $V(s)$ (ie swapping $\mathcal{L}(v - v_0)$ for $\mathcal{L}(v)$). Why can we do that? Shouldn't there also be a nonzero $-\mathcal{L}(v_0)$ constant in there? He does the same thing in other places too. I could understand if $v_0 \approx 0$, but $v_0 = \ln(2)$, which isn't really negligible.
Hint: $e^{ln(2)}=2$
So, $e^{ln(2)}=2$ cancels out the -2 in the equation.
Update: $\frac{\partial ({{v}_{o}}+\delta (t))}{\partial t}=\frac{\partial \delta (t)}{\partial t}$ Since $v_o$ is constant, always.