Maximum overshoot and settling time for a 2nd order system
In a nutshell, the error is the difference between desired output and the one you got:
$$e(t) = y^*(t) - y(t)$$
Then you resolve $$e(t) &lt; ErrorMax$$.
To simplify the comparison note that
$$sin(a + b) = sin(a)cos(b) + sin(b)cos(a)$$
so
$$ a cos(\omega t) + b sin(\omega t) = M sin(\omega t + \phi)$$ where $$ M = \sqrt{a^2 + b^2}$$ and $$\phi = atan(a/b)$$
In consequence, you get something in the form
$$M sin(\omega t + \phi) e^{-t} &lt; ErrorMax$$